Time optimal control of evolution equations 9783319953625, 9783319953632

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Progress in Nonlinear Differential Equations and Their Applications Subseries in Control 92

Gengsheng Wang Lijuan Wang Yashan Xu Yubiao Zhang 

Time Optimal Control of Evolution Equations

Progress in Nonlinear Differential Equations and Their Applications: Subseries in Control Volume 92

Editor Jean-Michel Coron, Université Pierre et Marie Curie, Paris, France Editorial Board Viorel Barbu, Facultatea de Matematic˘a, Universitatea “Alexandru Ioan Cuza” din Ia¸si, Romania Piermarco Cannarsa, Department of Mathematics, University of Rome “Tor Vergata”, Italy Karl Kunisch, Institute of Mathematics and Scientific Computing, University of Graz, Austria Gilles Lebeau, Laboratoire J.A. Dieudonné, Université de Nice Sophia-Antipolis, France Tatsien Li, School of Mathematical Sciences, Fudan University, China Shige Peng, Institute of Mathematics, Shandong University, China Eduardo Sontag, Department of Mathematics, Rutgers University, USA Enrique Zuazua, Departamento de Matemáticas, Universidad Autónoma de Madrid, Spain

More information about this series at http://www.springer.com/series/15137

Gengsheng Wang • Lijuan Wang • Yashan Xu Yubiao Zhang

Time Optimal Control of Evolution Equations

Gengsheng Wang Center for Applied Mathematics Tianjin University Tianjin, China

Lijuan Wang School of Mathematics and Statistics Wuhan University Wuhan, China

Yashan Xu School of Mathematical Sciences Fudan University Shanghai, China

Yubiao Zhang Center for Applied Mathematics Tianjin University Tianjin, China

ISSN 1421-1750 ISSN 2374-0280 (electronic) Progress in Nonlinear Differential Equations and Their Applications Subseries in Control ISBN 978-3-319-95362-5 ISBN 978-3-319-95363-2 (eBook) https://doi.org/10.1007/978-3-319-95363-2 Library of Congress Control Number: 2018950836 Mathematics Subject Classification (2010): 37L05, 49-02, 49J15, 49J20, 49J30, 93B05, 93C15, 93C20 © Springer International Publishing AG, part of Springer Nature 2018 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 book is published under the imprint Birkhäuser, www.birkhauser-science.com by the registered company Springer Nature Switzerland AG part of Springer Nature. The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Preface

The studies on time optimal controls started with finite dimensional systems in the 1950s, and then extended to infinite dimensional systems in the 1960s. Time optimal control is a very important field in optimal control theory, from perspectives of both application and mathematics. Optimal control theory can be viewed as a branch of optimization theory or variational theory. In general, an optimal control problem is to ask for a control from an available set so that a certain optimality criterion, related to a given dynamic system, is achieved. When the above-mentioned criterion is elapsed time, such an optimal control problem is called a time optimal control problem. From the perspective of application, there are many practical examples which can be described as time optimal control problems. For instance, reach the top fastest by an elevator, warm the room by the stove quickly, and stop the harmonic oscillator in the shortest time. Systematic studies on time optimal control problems may help us to design the best strategy to solve these practical problems. Theoretically, time optimal control problems connect with several other fields of control theory, such as controllability, dynamic programming, and so on. The studies on time optimal controls may develop studies on these related fields. These can be viewed as the motivations to study time optimal controls. The time optimal control problems studied in this monograph can be stated roughly as follows: Given a controlled system over the right half time interval, we ask for such a control (from a constraint set) that satisfies the following three conditions: (i) It is only active from a starting time point to an ending time, point; (ii) It drives the solution of the controlled system from an initial state set (at the starting time) to a target set (at the ending time); (iii) The length of the interval, where the control is active, is the shortest among all such candidates. Thus, each time optimal control problem has four undetermined variables: starting time, initial states, ending time and controls. They form a family of tetrads. When the starting time point is fixed, such a time optimal control problem is called a minimal time control problem. It is to initiate control (in a constraint set) from the beginning so that the corresponding solution (of a controlled system) reaches a given target set in the shortest time. When the ending time point is fixed, such a time optimal

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control problem is called a maximal time control problem. It is to delay the initiation of active control (in a constraint set) as late as possible, so that the corresponding solution (of a controlled system) reaches a given target set in a fixed ending time. Most parts of this monograph focus on the above-mentioned two kinds of time optimal control problems for linear controlled evolution equations. In this monograph, we set up a general mathematical framework on time optimal control problems governed by some evolution equations. This framework has four ingredients: a controlled system, a control constraint set, a starting set, and an ending set. We introduce five themes on time optimal control problems under this framework: the existence of admissible controls and optimal controls, Pontryagin’s maximum principle for optimal controls, the equivalence of several different kinds of optimal control problems, and bang-bang properties. The aim of this monograph is to summarize our, our seniors’ and our collaborators’, ideas, methods, and results in the aforementioned themes. Many of these ideas, methods, and results are the recent advances in the field of time optimal control. These may lead us to comprehensive understanding of the field on time optimal control. We tried our best to make the monograph self-contained and wish that it could interest specialists and graduate students in this field, as well as in other related fields. Indeed, the prerequisite is to know a little about the functional analysis and differential equations. There are a lot of literatures on time optimal control problems for evolution systems. We have not attempted to give a complete list of references. All the references in this monograph are closely related to the materials introduced here. It may possibly happen that some important works in this field have been overlooked. More than thirteen years ago, H. O. Fattorini in his well-known book: H. O. Fattorini, Infinite dimensional linear control systems, the time optimal and norm optimal problems, North-Holland Mathematics Studies, 201, Elsevier Science B. V., Amsterdam, 2005. introduced minimal time control problems and minimal norm control problems. The controlled system there reads: y  = Ay + u, where A generates a C0 -semigroup in a Banach space. (Hence, the state space and the control space are the same, and the control operator is the identity.) The book is mainly concerned with three topics: First, the Pontryagin Maximum Principle with multipliers in different spaces which are called the regular space and the singular space; second, the bang-bang property of minimal time and the minimal norm controls; and third, connections between minimal time and minimal norm control problems. Also, the existence of optimal controls, based on the assumption that there is an admissible control, is presented. In the period after the book, the studies on time optimal control problems were developed greatly. We believe that it is the right time to summarize some of these developments. Differing from the book, the objects in the current monograph are the minimal time and the maximal time control problems; the controlled system in the current monograph reads: y  = Ay + Bu, where A generates a C0 -semigroup in a Hilbert space (which is the state space), B is a linear and bounded operator from the state space to another Hilbert space (which is the control space). Though many topics in the current monograph, such as the Pontryagin Maximum Principle, the bang-bang property, and the equivalence of different optimal control problems, are

Preface

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quite similar to those in the book, we study these issues from different perspectives and by using different ways. Moreover, studies on some topics are developed a lot, for instance, the equivalence of the minimal time control problems and the minimal norm control problems, and the bang-bang property for minimal time controls. The topic on the existence of admissible controls was not touched upon in the book, while in the current monograph we introduce several ways to study it. Let us now be more precise on the contents of the different chapters of this monograph. Chapter 1. This chapter introduces some very basic elements of preliminaries such as functional analysis, evolution equations, controllability, and observability estimates. Chapter 2. In this chapter, we first set up a general mathematical framework of time optimal control problems for evolution equations and present four ingredients of this framework; we next introduce connections between minimal and maximal time control problems; we then show main subjects (on time optimal control problems) which will be studied in this monograph. Several examples are given to illustrate the framework, its ingredients, and main subjects. What deserves to be mentioned is as follows: At the end of Section 2.1, we present a minimal blowup time control problem for some nonlinear ODEs with the blowup behavior. It is a special time optimal control problem (where the target is outside of the state space) and is not under our framework. Chapter 3. The subject of this chapter is the existence of admissible controls and optimal controls for time optimal control problems. The key of this subject is the existence of admissible controls. We study it from three perspectives: the controllability, minimal norm problems, and reachable sets. Consequently, we provide three ways to obtain the existence of admissible controls. We also show how to derive the existence of optimal controls from that of admissible controls. We end this chapter with the existence of optimal controls for a minimal blowup time control problem governed by nonlinear ODEs with the blowup behavior. This problem is very interesting from the viewpoints of application and mathematical theory. However, the studies of such a problem are more difficult than those of problems under our framework. Chapter 4. This chapter is devoted to the Pontryagin Maximum Principle, which is indeed a kind of first-order necessary conditions on time optimal controls. In general, there are two methods to derive the Pontryagin Maximum Principle. They are the analytical method and the geometric method. We focus on the second one and show how to use it to derive the Pontryagin Maximum Principle for both the minimal time control problems and the maximal time control problems. Three different kinds of maximum principles are introduced. They are the classical Pontryagin Maximum Principle, the local Pontryagin Maximum Principle, and the weak Pontryagin Maximum Principle. We find a way consisting of two steps to derive these maximum principles. In step 1, using the Hahn-Banach Theorem for different cases, we separate different objects in different spaces. In step 2, we build up different representation formulas for different cases. Besides, we give connections among the above-mentioned maximum principles.

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Chapter 5. This chapter develops two equivalence theorems for several different kinds of optimal control problems. The first one is concerned with the equivalence of the minimal time control problem and the minimal norm control problem. The second one deals with the equivalence of the maximal time control problem, the minimal norm control problem, and the optimal target control problem. Some applications of these equivalence theorems are referred to in some examples and the miscellaneous note. The above equivalence theorems allow us to obtain desired properties for one optimal control problem through studying another simpler one. Two facts deserve to be mentioned: First, between the minimal norm control problem and the minimal time control problem, the first one is simpler. Second, among the maximal time control problem, the minimal norm control problem, and the optimal target control problem, the first one is the most complicated one and the last one is the simplest one. Chapter 6. In this chapter, we present the bang-bang property for time optimal control problems. In plain language, this property means that each optimal control reaches the boundary of a control constraint set at almost every time. It can be compared with such a property of a function that its extreme points stay only on the boundary of its domain. We begin with finite dimensional cases and then turn to infinite dimensional cases. We introduce two ways to approach the bang-bang property. The first way is the use of the Pontryagin Maximum Principle, together with a certain unique continuation. The second way is the use of a certain null controllability from measurable sets in time. In this monograph, we give some miscellaneous notes at the end of each chapter. In these notes, we review the history and the related works of the involved study or point out some open problems in the related fields. Besides, after most theorems, we provide examples to help readers to understand the theorems better. Tianjin, China Wuhan, China Shanghai, China Tianjin, China July 2018

Gengsheng Wang Lijuan Wang Yashan Xu Yubiao Zhang

Acknowledgments

The authors wish to thank Professor Jean-Michel Coron for the kind invitation to write this book for the Control/PNLDE series. The following colleagues deserve special acknowledgments: Ping Lin (Northeast Normal University, China), Kim Dang Phung (Université d’Orléans, France), Shulin Qin (Wuhan University, China), Can Zhang (Wuhan University, China), and Enrique Zuazua (Universidad Autónoma de Madrid, Spain). Some materials in this monograph are based on our joint works with them. Finally, the authors would like to acknowledge support from the Natural Science Foundation of China under grants 11371285, 11471080, 11571264, 11631004, and 11771344.

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Contents

1

Mathematical Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.1 Elements in Functional Analysis . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.1.1 Spaces and Operators .. . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.1.2 Some Geometric Aspects of Banach Spaces . . . . . . . . . . . . . . . . . . 1.1.3 Semigroups and Evolution Equations.. . . . .. . . . . . . . . . . . . . . . . . . . 1.1.4 Minimization of Functionals . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2 Set-Valued Map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.3 Controllability and Observability Estimate . . . . . . . .. . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

1 1 1 4 6 12 13 23 34

2 Time Optimal Control Problems .. . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.1 Overview on Time Optimal Control Problems . . . .. . . . . . . . . . . . . . . . . . . . 2.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.1.2 Different Ingredients in Time Optimal Control Problems . . . . 2.2 Connections of Minimal and Maximal Time Controls . . . . . . . . . . . . . . . . 2.3 Several Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.4 Main Subjects on Time Optimal Control Problems . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

37 37 37 41 43 52 55 62

3 Existence of Admissible Controls and Optimal Controls .. . . . . . . . . . . . . . . 65 3.1 Admissible Controls and Controllability.. . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 66 3.1.1 General Results in Abstract Setting . . . . . . . .. . . . . . . . . . . . . . . . . . . . 66 3.1.2 Linear ODEs with Ball-Type Control Constraints . . . . . . . . . . . . 67 3.1.3 Heat Equations with Ball-Type Control Constraints . . . . . . . . . . 80 3.2 Admissible Controls and Minimal Norm Problems . . . . . . . . . . . . . . . . . . . 85 3.2.1 General Results in Abstract Setting . . . . . . . .. . . . . . . . . . . . . . . . . . . . 86 3.2.2 Heat Equations with Ball-Type Control Constraints . . . . . . . . . . 93 3.3 Admissible Controls and Reachable Sets . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 99 3.3.1 General Results in Abstract Setting . . . . . . . .. . . . . . . . . . . . . . . . . . . . 99 3.3.2 Reachable Sets of Linear ODEs. . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 104

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3.4 Existence of Optimal Controls . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.4.1 Existence of Optimal Controls for Regular Cases. . . . . . . . . . . . . 3.4.2 Existence of Optimal Controls for Blowup Case . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

112 112 117 122

4 Maximum Principle of Optimal Controls . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.1 Classical Maximum Principle of Minimal Time Controls . . . . . . . . . . . . 4.1.1 Geometric Intuition .. . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.1.2 Classical Maximum Principle .. . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.1.3 Conditions on Separation in Classical Maximum Principle . . 4.2 Local Maximum Principle and Minimal Time Controls . . . . . . . . . . . . . . 4.2.1 Some Properties on Controllable Sets and Reachable Sets . . . 4.2.2 Separability and Local Maximum Principle .. . . . . . . . . . . . . . . . . . 4.2.3 Conditions on Separation in Local Maximum Principle .. . . . . 4.3 Weak Maximum Principle and Minimal Time Controls . . . . . . . . . . . . . . 4.3.1 Separability and Weak Maximum Principle .. . . . . . . . . . . . . . . . . . 4.3.2 Two Representation Theorems Related to the Separability .. . 4.3.3 Conditions on Separation in Weak Maximum Principle .. . . . . 4.4 Maximum Principle for Maximal Time Controls... . . . . . . . . . . . . . . . . . . . 4.4.1 Classical Maximum Principle for Maximal Time Controls.. . 4.4.2 Local Maximum Principle for Maximal Time Controls . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

125 125 126 127 136 148 150 154 157 170 170 175 188 194 195 199 202

5 Equivalence of Several Kinds of Optimal Controls . .. . . . . . . . . . . . . . . . . . . . 5.1 Minimal Time Controls, Minimal Norm Controls, and Time Invariant Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.2 Minimal Time Controls, Minimal Norm Controls, and Time-Varying Systems: Part I. . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.2.1 Properties of Minimal Time and Minimal Norm Functions . . 5.2.2 Proof of the Main Result . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.3 Minimal Time Controls, Minimal Norm Controls, and Time-Varying Systems: Part II . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.4 Maximal Time, Minimal Norm, and Optimal Target Controls.. . . . . . . 5.4.1 Existence of Optimal Controls for These Problems .. . . . . . . . . . 5.4.2 Connections Between Problems of Optimal Target Control and Minimal Norm Control . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.4.3 Connections Between Problems of Minimal Norm Control and Maximal Time Control . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.4.4 Proof of the Main Result . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

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6 Bang-Bang Property of Optimal Controls . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.1 Bang-Bang Property in ODE Cases . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 6.1.1 Bang-Bang Property for Minimal Time Controls . . . . . . . . . . . . . 6.1.2 Bang-Bang Property for Maximal Time Controls .. . . . . . . . . . . .

206 218 220 235 240 253 256 259 266 274 278 281 281 283 293

Contents

6.2 Bang-Bang Property and Null Controllability .. . . .. . . . . . . . . . . . . . . . . . . . 6.2.1 Bang-Bang Property of Minimal Time Control . . . . . . . . . . . . . . . 6.2.2 Bang-Bang Property of Maximal Time Control .. . . . . . . . . . . . . . 6.3 Bang-Bang Property and Maximum Principle .. . . .. . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

xiii

303 304 315 318 328

Index . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 331

Acronyms

R C y Y u U U coU Aad QS QE YE Re z Rn Rn×m ·, · ·, ·X∗ ,X · A−1 2Ω |E| V⊥ L(a, b; U ) L(a, b; U) Lp (a, b; U ) L∞ (a, b; U )

the set of all real numbers the set of all complex numbers a state variable a state space a control variable a control space a control constraint set the convex hull of the set U a set of admissible tetrads a starting set an ending set a target the real part of the complex number z the n-dimensional real Euclidean space the set of all n × m real matrices an inner product in some Hilbert space the duality pairing between X∗ and X a norm in some normed space the inverse of the reversible matrix A the set of all subsets of Ω the Lebesgue measure of the set E the orthogonal complement space of the subspace V in a Hilbert space the set of Lebesgue measurable functions ϕ : (a, b) → U the set of Lebesgue measurable functions ϕ : (a, b) → U with values on U the set of Lebesgue measurable functions ϕ : (a, b) → U so that  b p ϕ(t)U dt < +∞ (p ∈ [1, +∞)) a

the set of essential bounded Lebesgue measurable functions ϕ : (a, b) → U xv

xvi

L (U, Y ) L (Y ) D(A) {et A}t ≥0 A∗ D(A∗ ) C([a, b]; Y ) C(Ω) C(· · · ) σ (A) N (A) R(A) η X∗ N N+ I ntS S ∂S Br (x0 ) Or (x0 ) χω

Acronyms

the space of linear continuous operators from U to Y the space of linear continuous operators from Y to Y the domain of the operator A the C0 semigroup generated by A the adjoint of the operator A the dual space of D(A∗ ) the space of continuous functions in the interval [a, b] with the value in Y the space of continuous functions in the closure of Ω a positive constant dependent on those enclosed in the bracket the spectrum of the operator A the kernel of the operator A the range of the operator A the transpose of the Euclidean vector η the dual space of the normed space X the set of all natural numbers the set of all positive integers the interior of the set S the closure of the set S the boundary of the set S the closed ball with center at x0 and of radius r the open ball with center at x0 and of radius r the characteristic function of a nonempty set ω

Chapter 1

Mathematical Preliminaries

In this chapter, we recall some basic concepts and results which are necessary for the presentation of the theories in the later chapters.

1.1 Elements in Functional Analysis In this section, some basic results of functional analysis are collected. Most proofs for the standard results will be omitted.

1.1.1 Spaces and Operators We begin with the following definition. Definition 1.1 Let X be a linear space over F (F = R or C). (i) A map ϕ : X → R is called a norm on X if it satisfies the following: ⎧ ⎪ ⎪ ⎨ϕ(x) ≥ 0 for all x ∈ X; ϕ(x) = 0 ⇐⇒ x = 0; ϕ(αx) = |α|ϕ(x) for all α ∈ F and x ∈ X; ⎪ ⎪ ⎩ϕ(x + y) ≤ ϕ(x) + ϕ(y) for all x, y ∈ X.

(1.1)

(ii) A map ψ : X × X → F is called an inner product on X if it satisfies the following:

© Springer International Publishing AG, part of Springer Nature 2018 G. Wang et al., Time Optimal Control of Evolution Equations, Progress in Nonlinear Differential Equations and Their Applications 92, https://doi.org/10.1007/978-3-319-95363-2_1

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2

1 Mathematical Preliminaries

⎧ ⎪ ψ(x, x) ≥ 0 for all x ∈ X; ψ(x, x) = 0 ⇐⇒ x = 0; ⎪ ⎪ ⎪ ⎨ψ(x, y) = ψ(y, x) for all x, y ∈ X; ⎪ ψ(αx + βy, z) = αψ(x, z) + βψ(y, z) for all α, β ∈ F ⎪ ⎪ ⎪ ⎩ and x, y, z ∈ X,

(1.2)

where ψ(y, x) is the complex conjugate of ψ(y, x). Hereafter, we denote a norm on X (if it exists) by  · . Sometimes,  · X is used to indicate a norm defined on X. Similarly, we denote an inner product on X (if it exists) by ·, · or ·, ·X if the underlying space X needs to be emphasized. If X has a norm  · , (X,  · ) is called a normed linear space. In most time, we simply write X for (X,  · ). Definition 1.2 A normed linear space (X,  · ) is called a Banach space if it is complete, i.e., for any Cauchy sequence {xn }n≥1 ⊆ X, there exists an x ∈ X, so that lim xn − x = 0.

n→+∞

Definition 1.3 Let X be a linear space√over F with an inner product ·, · and let  ·  be the induced norm (i.e., x = x, x for any x ∈ X). When (X,  · ) is complete, X is called a Hilbert space. Let us now recall some standard terminologies in Banach spaces. Definition 1.4 Let X be a Banach space and let G be a subset of X.  (i) G is open if for any x ∈ G, Or (x)  {y ∈ X  y − x < r} ⊆ G for some r > 0.  (ii) G is closed if X \ G  {x ∈ X  x ∈ / G} is open.  (iii) The set IntG  {x ∈ G  ∃ r > 0, Or (x) ⊆ G} is called the interior of G; the smallest closed set containing G is called the closure of G, denoted by G; and ∂G  G \ IntG is called the boundary of G.   (iv) G is compact if for any family of open sets {Gα  α ∈ Λ} with G ⊆ Gα , α∈Λ  there is a subset Λ1 (of Λ) with finite number of elements so that G ⊆ Gα . α∈Λ1

(v) G is relatively compact if the closure of G is compact. (vi) G is separable if it admits a countable dense subset. When X is separable, it is called a separable Banach space. Proposition 1.1 Let X be a Banach space over F and G be a subset of X. Suppose that G is separable. Then any nonempty subset of G is also separable. Let X and Y be two Banach spaces and let D be a subspace of X (not necessarily closed). A map A : D ⊆ X → Y is called a linear operator if the following holds: A(αx + βy) = αAx + βAy for all x, y ∈ D and α, β ∈ F. The subspace D is called the domain of A. Thereafter, we denote it by D(A). If D(A) = X and A maps any bounded subset (of X) into a bounded subset (of Y ), we say that A is bounded.

1.1 Elements in Functional Analysis

3

For any linear operator A : D(A) ⊆ X → Y , we define ⎧ ⎪ ⎪ ⎨N (A)  {x ∈ D(A) : Ax = 0}, R(A)  {Ax : x ∈ D(A)}, ⎪ ⎪ ⎩Graph(A)  {(x, Ax) : x ∈ D(A)}. They are called the kernel, the range, and the graph of the operator A, respectively. Let us now state some important results concerning linear bounded operators. Theorem 1.1 (Closed Graph Theorem) Let X and Y be Banach spaces. Let A : X → Y be a linear operator with D(A) = X. Suppose that Graph(A) is closed in X × Y . Then A ∈ L (X, Y ). Theorem 1.2 (Principle of Uniform Boundedness) Let X and Y be Banach spaces and let A ⊆ L (X, Y ) be a nonempty family. Then, sup Ax < +∞ for each x ∈ X ⇒ sup A < +∞. A∈A

A∈A

Let us introduce linear functionals and dual spaces. Let X∗ = L (X, F ). It is called the dual space of X. Each f ∈ X∗ is called a linear bounded (or continuous) functional on X. Sometimes, we denote f ∈ X∗ by f (x) = f, xX∗ ,X for all x ∈ X. The symbol ·, ·X∗ ,X is referred to as the duality pairing between X∗ and X. The following results are very important in later applications. Theorem 1.3 (Riesz Representation Theorem) Let X be a Hilbert space. Then X∗ = X. More precisely, for any f ∈ X∗ , there is y ∈ X so that f (x) = x, y for all x ∈ X;

(1.3)

on the other hand, for any y ∈ X, by defining f as in (1.3), one has f ∈ X∗ . Theorem 1.4 (Riesz Representation Theorem in Lp ) (i) Let Ω ⊆ Rn be a domain. Then, for any 1 ≤ p < +∞ and 1/p + 1/q = 1, (Lp (Ω))∗ = Lq (Ω). (ii) Let −∞ ≤ a < b ≤ +∞ and X be a real separable Hilbert space. Then, for any 1 ≤ p < +∞ and 1/p + 1/q = 1, (Lp (a, b; X))∗ = Lq (a, b; X). Theorem 1.5 (Hahn-Banach Theorem) Let X be a Banach space and X0 be a subspace of X inherited the norm from X. Let f0 ∈ X0∗ . Then there exists an extension f ∈ X∗ of f0 so that f X∗ = f0 X0∗ . Let X be a Banach space and X∗ be its dual space. We denote the dual of X∗ by

X∗∗ .

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1 Mathematical Preliminaries

Definition 1.5 A Banach space X is said to be reflexive if X = X∗∗ . Let Ω ⊆ Rn be a domain. For each p ∈ (1, +∞) and m ∈ N+ , we denote  α1 +···+αn f p W m,p (Ω) = f : Ω → R is measurable  ∂ α1 αn ∈ L (Ω), ∂x1 ···∂xn

α1 , . . . , αn ∈ N and

n

 αi ≤ m .

i=1

Lp (Ω)

W m,p (Ω)

Then, the spaces and are reflexive. Moreover, any Hilbert space is reflexive. But C(Ω), L1 (Ω), and L∞ (Ω) are not reflexive. Theorem 1.6 (James Theorem) Let X be a real Banach space. Then X is reflexive if and only if for any f ∈ X∗ , sup f (x) = max f (x). x≤1

X∗

x≤1

Let X and Y be Banach spaces and A ∈ L (X, Y ). We define a map A∗ : Y ∗ → by the following manner: A∗ y ∗ , xX∗ ,X  y ∗ , AxY ∗ ,Y for all y ∗ ∈ Y ∗ and x ∈ X.

Clearly, A∗ is linear and bounded. We call A∗ the adjoint operator of A. When X is a Hilbert space and A ∈ L (X) satisfies that A∗ = A, A is called self-adjoint. We end this subsection with the following definition on compact operators. Definition 1.6 Let X and Y be Banach spaces and A ∈ L (X, Y ). We say that A is compact if A maps any bounded set of X into a relatively compact set in Y , i.e., if G is bounded in X, then the closure of A(G) is compact in Y .

1.1.2 Some Geometric Aspects of Banach Spaces In this subsection, we present some results concerning certain geometric properties of Banach spaces and their subsets. Let X be a Banach space. A subset G of X is said to be convex if for any x, y ∈ G and λ ∈ [0, 1], one has λx + (1 − λ)y ∈ G. Denote coS to be the smallest convex set containing S. We call coS the convex hull of the set S. Important examples of convex sets are balls and subspaces. It should be pointed out that the intersection of convex sets is convex; but the union of two convex sets is not necessarily convex. Also, if G1 and G2 are convex, then for any λ1 , λ2 ∈ R, the set λ1 G1 + λ2 G2  {λ1 x1 + λ2 x2 : x1 ∈ G1 , x2 ∈ G2 } is convex. Theorem 1.7 (Carathéodory Theorem) Let S be a subset of Rn . Let x ∈ coS. Then there exist xi ∈ S and αi ≥ 0 (0 ≤ i ≤ n) so that x=

n

i=0

αi xi and

n

i=0

αi = 1.

1.1 Elements in Functional Analysis

5

Before presenting the next result, we introduce the concept of weak closedness. Let G be a subset of Banach space X. We say G is weakly closed if for any weakly convergent sequence {xn }n≥1 in G, its weakly convergent limit belongs to G. Theorem 1.8 (Mazur Theorem) Let X be a Banach space and G be a convex and closed set in X. Then G is weakly closed in X. Consequently, if {xn }n≥1 ⊆ X satisfies that xn → x weakly in X, then there exist αn,k ≥ 0 (k = 1, . . . , Kn ) with Kn αn,k = 1 so that k=1

K

n



αn,k xk+n − x = 0. lim n→+∞

k=1

We now give the next definition on the separability of two sets in a Banach space. Definition 1.7 Let S1 and S2 be two nonempty subsets in a Banach space X. They are said to be separable in X if there is ψ ∈ X∗ \ {0} and c ∈ R so that inf Reψ, xX∗ ,X ≥ c ≥ sup Reψ, xX∗ ,X .

x∈S1

(1.4)

x∈S2

   When (1.4) holds, we say that the hyperplane x ∈ X  Reψ, xX∗ ,X = c (or the vector ψ) separates S1 from S2 in X; we call this hyperplane as a separating hyperplane; and we call ψ as a vector separating S1 from S2 in X, or simply, a separating vector. (A separating vector is a normal vector of a separating hyperplane.) Sometimes, we say that the vector ψ separates S1 from S2 in X. Let us introduce some basic results on the separability. First we introduce the following result for finite dimensional spaces: Theorem 1.9 ([29]) Let S be a nonempty convex subset in Rn . Assume z0 ∈ ∂S, where ∂S is the boundary of S in Rn . Then, there exists f ∈ Rn with f Rn = 1 so that inf f (s) ≥ f (z0 ).

s∈S

We next introduce the following result for infinite dimensional spaces: Theorem 1.10 ([19]) Let X be a Banach space and S be a nonempty convex subset of X. Assume I ntS = ∅ and z0 ∈ / I ntS. Then there exists an f ∈ X∗ with f X∗ = 1 so that inf Ref (z) ≥ Ref (z0 ).

z∈S

6

1 Mathematical Preliminaries

Theorem 1.11 (Hahn-Banach Theorem (Geometry Form) [24]) Suppose that S1 and S2 are disjoint, nonempty, convex sets in a normed linear space X. The following statements hold: (i) If S1 is closed and S2 is compact, then there exists f ∈ X∗ (with f X∗ = 1), c1 ∈ R and c2 ∈ R so that Ref (s1 ) > c2 > c1 > Ref (s2 ) for all s1 ∈ S1 and s2 ∈ S2 . (ii) If S1 is open, then there exists f ∈ X∗ (with f X∗ = 1) and c ∈ R so that Ref (s2 ) ≥ c > Ref (s1 ) for all s1 ∈ S1 and s2 ∈ S2 . Remark 1.1 Theorem 1.11 is still true when X is a locally convex topological vector space (see Theorem 3.4 in [24]).

1.1.3 Semigroups and Evolution Equations Through this subsection, X and Y are Banach spaces. We will present some basic properties on C0 semigroups. We begin with the next definition. Definition 1.8 Let A : D(A) → Y be a linear operator with D(A) a linear subspace of X (not necessarily closed). (i) We say that A is densely defined if D(A) is dense in X. (ii) We say that A is closed if its graph    Graph(A) = (x, y) ∈ X × Y  x ∈ D(A), y = Ax

(1.5)

is closed in X × Y . We know that any linear bounded operator A ∈ L (X) is densely defined and closed. However, there are many linear, densely defined, closed operators that map some bounded sets to unbounded sets. These operators are called unbounded linear operators. We cannot define any norm for an unbounded operator. A typical linear, densely defined, closed, and unbounded operator is as: on X = L2 (0, 1)  A  d/dx 2 with D(A) = {y is absolutely continuous on [0, 1]  y˙ ∈ L (0, 1), y(0) = 0}. Let us first introduce the following definition. Definition 1.9 (i) Let {T (t)}t ≥0 ⊆ L (X). We call it a C0 semigroup on X if ⎧ ⎪ T (0) = I, ⎪ ⎨ T (t + s) = T (t)T (s) for all t, s ≥ 0, ⎪ ⎪ ⎩ lim T (s)x − x = 0 for each x ∈ X. s→0+

1.1 Elements in Functional Analysis

7

(ii) Let {T (t)}t ∈R ⊆ L (X). We call it a C0 group on X if ⎧ ⎪ T (0) = I, ⎪ ⎨ T (t + s) = T (t)T (s) for all t, s ∈ R, ⎪ ⎪ ⎩ lim T (s)x − x = 0 for each x ∈ X. s→0

Proposition 1.2 Let {T (t)}t ≥0 be a C0 semigroup on X. Then there exist two constants M ≥ 1 and ω ∈ R, so that T (t)L (X) ≤ Meωt for all t ≥ 0. Definition 1.10 Let {T (t)}t ≥0 be a C0 semigroup on X. Let  ⎧  ⎪ ⎨D(A)  x ∈ X  s− lim ⎪ ⎩Ax  s− lim

t →0+

T (t )−I x t t →0+

T (t )−I x t

 exists

,

for all x ∈ D(A),

where s−lim stands for the strong limit in X. The operator A : D(A) ⊆ X → X is called the generator of the semigroup {T (t)}t ≥0 . In general, the operator A may not be bounded. We now present some results for evolution equations. Let T > 0, f (·) ∈ L1 (0, T ; X), and y0 ∈ X. Consider  y(t) ˙ = Ay(t) + f (t), t ∈ (0, T ), (1.6) y(0) = y0 . Here, A : D(A) ⊆ X → X generates a C0 semigroup {eAt }t ≥0 on X. We first introduce some notions of solutions. Definition 1.11 For the Equation (1.6), (i) y ∈ C([0, T ]; X) is called a strong solution of (1.6) if y(0) = y0 ; y is strongly differentiable almost everywhere on (0, T ); y(t) ∈ D(A) for a.e. t ∈ (0, T ); and the equation in (1.6) is satisfied for almost every t ∈ (0, T ). (ii) y ∈ C([0, T ]; X) is called a weak solution of (1.6) if for any x ∗ ∈ D(A∗ ), x ∗ , y(·) is absolutely continuous on [0, T ] and x ∗ , y(t) = x ∗ , y0  +



t

 ∗ ∗  A x , y(s) + x ∗ , f (s) ds for all t ∈ [0, T ].

0

(iii) y ∈ C([0, T ]; X) is called a mild solution of (1.6) if  t y(t) = eAt y0 + eA(t −s)f (s)ds for all t ∈ [0, T ]. 0

8

1 Mathematical Preliminaries

We next consider the following semilinear equation:  y(t) ˙ = Ay(t) + f (t, y(t)),

t ∈ (0, T ),

y(0) = y0 .

(1.7)

where A : D(A) ⊆ X → X generates a C0 semigroup {eAt }t ≥0 on X and f : [0, T ] × X → X satisfies the following conditions: (i) For each x ∈ X, f (·, x) is strongly measurable; (ii) There exists a function (·) ∈ L1 (0, T ) so that  f (t, x1 ) − f (t, x2 ) ≤ (t)x1 − x2  for all x1 , x2 ∈ X and a.e. t ∈ (0, T ), f (t, 0) ≤ (t) for a.e. t ∈ (0, T ). We call y ∈ C([0, T ]; X) a mild solution of (1.7) if  t y(t) = eAt y0 + eA(t −s)f (s, y(s)) for all t ∈ [0, T ].

(1.8)

0

The following result is concerned with the existence and uniqueness of the solution to (1.8). Proposition 1.3 Let the above assumptions concerning A and f hold. Then, for any y0 ∈ X, (1.8) admits a unique solution y. Moreover, if we let y(·; y0 ) be the solution corresponding to y0 , and let the C0 semigroup {eAt }t ≥0 satisfy eAt L (X) ≤ Meωt for all t ≥ 0, where M ≥ 1 and ω ∈ R, then y(t; y0 ) ≤ Meωt +M

t 0

(s)ds

(1 + y0 ) for all 0 ≤ t ≤ T and y0 ∈ X,

and y(t; y0,1) − y(t; y0,2) ≤ Meωt +M

t 0

(s)ds

y0,1 − y0,2 

for all 0 ≤ t ≤ T and y0,1 , y0,2 ∈ X. When we discuss perturbed linear evolution equations, i.e., the Equation (1.7) with f (t, y) = D(t)y, we have the following variation of constants formula: Lemma 1.1 Let {eAt }t ≥0 be a C0 semigroup on X and D(·) ∈ L1 (0, T ; L (X)). Then there exists a unique strongly continuous function G :  → L (X) with  = {(t, s) ∈ [0, T ] × [0, T ] : 0 ≤ s < t ≤ T }, so that  G(t, t) = I for all t ∈ [0, T ], (1.9) G(t, r)G(r, s) = G(t, s) for all 0 ≤ s ≤ r ≤ t ≤ T ,

1.1 Elements in Functional Analysis

9

and so that for any 0 ≤ s ≤ t ≤ T and x ∈ X, G(t, s)x =eA(t −s)x + =eA(t −s)x +

 

t

eA(t −r)D(r)G(r, s)xdr

s t

G(t, r)D(r)eA(r−s)xdr.

s

Because of the property (1.9), we refer to the operator valued function G(·, ·) as the evolution operator generated by A + D(·). Next, we consider the following nonhomogeneous linear equations: 

t

ξ(t) = e ξ0 + At

e

A(t −s)



t

D(s)ξ(s)ds +

0

eA(t −s)g(s)ds, t ∈ [0, T ],

(1.10)

0

and  T ∗ ∗ eA (s−t )D(s)∗ η(s)ds η(t) = eA (T −t ) ηT + t  T A∗ (s−t ) − e h(s)ds, t ∈ [0, T ],

(1.11)

t

where ξ0 ∈ X and ηT ∈ X∗ . It is not hard to see that both (1.10) and (1.11) admit unique solutions. The next result gives the variation of constants formula. Proposition 1.4 Let G(·, ·) be the evolution operator generated by A + D(·). Then the solutions ξ of (1.10) and η of (1.11) can be represented respectively by 

t

ξ(t) = G(t, 0)ξ0 +

G(t, s)g(s) ds for all t ∈ [0, T ],

0

and ∗



T

η(t) = G(T , t) ηT −

G(s, t)∗ h(s) ds for all t ∈ [0, T ].

t

Moreover, for any 0 ≤ s ≤ t ≤ T , it holds that  η(t), ξ(t) − η(s), ξ(s) =

t

  η(r), g(r) + h(r), ξ(r) dr.

s

Finally, we give the well posedness for the following two heat equations: ⎧ ⎨ ∂t y − Δy + a(x, t)y = g in Ω × (0, T ), y=0 on ∂Ω × (0, T ), ⎩ in Ω, y(0) = y0

(1.12)

10

1 Mathematical Preliminaries

and ⎧ ⎨ ∂t y − Δy + f (y) = g in Ω × (0, T ), y=0 on ∂Ω × (0, T ), ⎩ in Ω. y(0) = y0

(1.13)

Here, Ω ⊆ Rn (n ≥ 1) is a bounded domain with a C 2 boundary ∂Ω, T > 0, a ∈ L∞ (Ω × (0, T )), g ∈ L2 (0, T ; L2 (Ω)), y0 ∈ L2 (Ω), and f : R → R is a globally Lipschitz function satisfying that f (r)r ≥ 0 for all r ∈ R. Theorem 1.12 The following statements are true: (i) The Equation (1.12) has a unique solution y W 1,2 (0, T ; H −1 (Ω)) and

∈

L2 (0, T ; H01 (Ω)) ∩

yL2 (0,T ;H 1 (Ω)) + yW 1,2 (0,T ;H −1 (Ω)) ≤ C(y0 L2 (Ω) + gL2 (0,T ;L2 (Ω)) ). 0 (1.14) Here C is a constant independent of y0 and g, but depending on aL∞ (Ω×(0,T )) . (ii) The Equation (1.13) has a unique solution y ∈ L2 (0, T ; H01(Ω)) ∩ W 1,2 (0, T ; H −1 (Ω)) and yL2 (0,T ;H 1 (Ω)) + yW 1,2 (0,T ;H −1 (Ω)) ≤ C(y0 L2 (Ω) + gL2 (0,T ;L2 (Ω)) ). 0 (1.15) Here C is a constant independent of y0 and g, but depending on the Lipschitz constant of f . Moreover, if g = 0, then y(T )L2 (Ω) ≤ e−λ1 T y0 L2 (Ω) ,

(1.16)

where λ1 is the first eigenvalue of −Δ with the homogeneous boundary condition. Proof We prove the conclusions one by one. (i) The existence and the uniqueness of the solution to (1.12) follow from Theorem 3 and Theorem 4 of Section 7.1 in Chapter 7, [10]. We omit the detailed proofs. The remainder is to show (1.14). For this purpose, we multiply the first equation of (1.12) by y and integrate it over Ω × (0, t) (t ∈ (0, T )). Then we obtain that  t y(t)2L2 (Ω) + 2 ∇y(s)2L2 (Ω) ds 0  t ≤ y0 2L2 (Ω) + g2L2 (0,T ;L2 (Ω)) + 2(aL∞ (Ω×(0,T )) + 1) y(s)2L2 (Ω) ds. 0

1.1 Elements in Functional Analysis

11

This, along with Gronwall’s inequality, implies that y2C([0,T ];L2 (Ω)) + y2L2 (0,T ;H 1 (Ω)) ≤ C(y0 2L2 (Ω) + g2L2 (0,T ;L2 (Ω)) ). 0

The above inequality and (1.12) lead to (1.14) immediately. (ii) The existence and the uniqueness of the solution to (1.13) follow from Theorem 2.6 of Chapter 7, [4]. The remainder is to show (1.15) and (1.16). We start with the proof of (1.15). Since f (r)r ≥ 0 for each r ∈ R, by multiplying the first equation of (1.13) by y and then by integrating it over Ω × (0, t) (t ∈ (0, T )), we obtain that  y(t)2L2 (Ω) + 2

≤

t

∇y(s)2L2 (Ω) ds 0  t 2 2 y(s)2L2 (Ω) ds. y0 L2 (Ω) + gL2 (0,T ;L2 (Ω)) + 0

This, along with Gronwall’s inequality, implies that y2C([0,T ];L2 (Ω)) + ∇y2L2 (0,T ;L2 (Ω))

≤ 2eT (T + 1)(y0 2L2 (Ω) + g2L2 (0,T ;L2 (Ω)) ).

(1.17)

Meanwhile, since f (r)r ≥ 0 for each r ∈ R and f : R → R is globally Lipschitz, we have that f (0) = 0 and |f (r)| ≤ L(f )|r| for each r ∈ R. Here, L(f ) denotes the Lipschitz constant of the function f . Hence, f (y)L2 (0,T ;L2 (Ω)) ≤ L(f )yL2 (0,T ;L2 (Ω)) .

(1.18)

By (1.17), (1.18), and (1.13), we get (1.15). We next prove (1.16). Since g = 0 and f (r)r ≥ 0 for each r ∈ R, by multiplying the first equation of (1.13) by y and then by integrating it over Ω, we see that d y(t)2L2 (Ω) + 2∇y(t)2L2 (Ω) ≤ 0 for a.e. t ∈ (0, T ). dt

(1.19)

Noting that ∇w2L2 (Ω) ≥ λ1 w2L2 (Ω) for each w ∈ H01 (Ω), by (1.19), we get that d y(t)2L2 (Ω) + 2λ1 y(t)2L2 (Ω) ≤ 0 for a.e. t ∈ (0, T ). dt

(1.20)

Multiplying (1.20) by e2λ1t and integrating it over (0, T ), we obtain (1.16). Thus, we finish the proof of Theorem 1.12.  

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1 Mathematical Preliminaries

1.1.4 Minimization of Functionals In this subsection, we introduce a theorem concerning with the existence of solutions for the following optimization problem: inf F (u), with F a functional over X.

u∈X

(1.21)

Theorem 1.13 Assume that the following conditions hold: (i) X is a real reflexive Banach space; (ii) F : X → (−∞, +∞) is convex, lower semicontinuous; (iii) F (u) → +∞ when uX → +∞. Then the problem (1.21) has a solution. Proof Let d  inf F (u). We claim that u∈X

d > −∞.

(1.22)

Indeed, if (1.22) was not true, then there would be a sequence {v }≥1 ⊆ X so that F (v ) → −∞ when  → +∞.

(1.23)

From (1.23) and the condition (iii), we find that {v }≥1 is bounded. This, together with the condition (i), implies that there exists a subsequence of {v }≥1 , still denoted by itself, so that v weakly in X for some  v ∈ X. v →  The above, along with the condition (ii) and (1.23), yields that F ( v ) ≤ lim inf F (v ) = −∞, →+∞

which leads to a contradiction, since we assumed that F takes values over (−∞, +∞). Hence, (1.22) is true. According to (1.22), there exists a sequence {u }≥1 ⊆ X so that d = lim F (u ). →+∞

(1.24)

By the condition (iii) and (1.24), we find that {u }l≥1 is bounded. This, together with the condition (i), implies that there is a subsequence of {u }l≥1 , denoted in the same manner, so that u →  u weakly in X for some  u ∈ X.

(1.25)

1.2 Set-Valued Map

13

It follows from (1.24), the condition (ii), and (1.25) that  d = lim inf F (u ) ≥ F ( u). →+∞

(1.26)

Since  u ∈ X and  d = inf F (u), it follows by (1.26) that  u is a solution to the u∈X

problem (1.21). In summary, we finish the proof of Theorem 1.13.

 

1.2 Set-Valued Map Let X and Y be two Banach spaces. Let S be a nonempty subset of X. Denote by 2Y the set consisting of all subsets of Y . We call any map Γ : S → 2Y \{∅} a set-valued map (see [3]). We now present the concept of upper semicontinuous set-valued map, which was introduced by Bouligand and Kuratowski (see [6] and [18]). Definition 1.12 Let X and Y be Banach spaces. Let S ⊆ X. Let Γ : S → 2Y \ {∅} be a set-valued map. We say that Γ is upper semicontinuous at x0 ∈ S, if for any  neighborhood G of Γ (x0 ), there is η > 0 so that Γ (x) ⊆ G for all x ∈ Oη (x0 ) S. When Γ is upper semicontinuous at each point in S, it is said to be upper semicontinuous on S. Let

  Γ −1 (M)  {x ∈ S  Γ (x) M = ∅}

and  Γ +1 (M)  {x ∈ S  Γ (x) ⊆ M}. The subset Γ −1 (M) is called the inverse image of M by Γ , and Γ +1 (M) is called the core of M by Γ . They naturally coincide when Γ is single-valued. We observe that Γ +1 (Y \ M) = S \ Γ −1 (M) and Γ −1 (Y \ M) = S \ Γ +1 (M). We can use the concepts of inverse images and cores to characterize upper semicontinuous maps (see Proposition 1.4.4 in [3]). Proposition 1.5 A set-valued map Γ : S → 2Y \ {∅} is upper semicontinuous at x ∈ S if the core of any neighborhood of Γ (x) by Γ is a neighborhood of x. Hence, Γ is upper semicontinuous if and only if the core of any open subset by Γ is open. In another word, Γ is upper semicontinuous if and only if the inverse image of any closed subset by Γ is closed.

14

1 Mathematical Preliminaries

Let Γ : S → 2Y \ {∅} be a set-valued map. Its graph is defined by    Graph(Γ )  (x, y) ∈ S × Y  y ∈ Γ (x) . Then we have the following result (see Propositions 1.4.8 and 1.4.9 in [3]): Proposition 1.6 Let S be a nonempty closed subset of X and Γ : S → 2Y \ {∅} be a set-valued map. (i) If Γ is upper semicontinuous and of closed values (i.e., for each x ∈ S, Γ (x) is a closed subset of Y ), then Graph(Γ ) is closed. (ii) Let K be a compact subset of Y . If Γ (x) ⊆ K for each x ∈ S and satisfies that Graph(Γ ) is closed, then Γ is upper semicontinuous. Theorem 1.14 (Kakutani-Fan-Glicksberg Theorem) Let K be a nonempty, compact, and convex subset of a Banach space X. Let Γ : K → 2K \ {∅} be a mapping with convex values (i.e., for each x ∈ K, Γ (x) is a nonempty and convex subset of K). Then there is at least one x ∈ K so that x ∈ Γ (x) when either of the following two conditions holds: (i) Γ is upper semicontinuous and of closed values; (ii) Graph(Γ ) is closed. Proof By Proposition 1.6, we observe that if (i) holds, then (ii) is true. The converse is also true. Thus, we can assume, without loss of generality, that (i) is true. Arbitrarily fix ε > 0 and x ∈ K. Set    Uε (x)  y ∈ K  Γ (y) ⊆ Γ (x) + Oε (0) . Since Γ is upper semicontinuous,  by Proposition 1.5, we have that Uε (x) is open. Consider the open cover {Uε (x) Oε (x)}x∈K of K. Since K is paracompact, there  is a neighborhood-finite barycentric refinement {V } , i.e., if V =  ∅, then λ λ∈Λ λ i   there is x0 ∈ K so that Vλi ⊆ Uε (x0 ) Oε (x0 ) (see Chapter 7 in [13]). We now claim that for the above ε, there is a continuous function f ε : K → K so that   Graph(Γ ) = ∅ for all x ∈ K, (1.27) Oε (x) × Oε (f ε (x)) i.e., for each x ∈ K, there is (xε , yε ) ∈ K × K so that yε ∈ Γ (xε ),

xε − xX < ε and yε − f ε (x)X < ε.

(1.28)

To this end, let {βλ }λ∈Λ be a partition of unity subordinate to the cover {Vλ }λ∈Λ . For each λ ∈ Λ, we choose a zλ ∈ Γ (Vλ ) ⊆ K and define a function f ε (x) 

λ∈Λ

βλ (x)zλ for all x ∈ K.

(1.29)

1.2 Set-Valued Map

15

It is obvious that f ε : K → K is continuous, where the convexity of K is used. Then the function f ε (·) defined in (1.29) satisfies (1.27) (or (1.28)). Indeed, we arbitrarily fix x ∈ K and let Vλ1 , . . . , Vλm be all the sets in {Vλ }λ∈Λ containing x. Since {Vλ }λ∈Λ is a barycentric refinement, there is xε ∈ K so that x∈

m 

Vλi ⊆ Uε (xε )



Oε (xε ).

i=1

In particular, x ∈ Oε (xε ) and m 

   Vλi ⊆ Uε (xε ) = y ∈ K  Γ (y) ⊆ Γ (xε ) + Oε (0) .

(1.30)

i=1

From (1.30) it follows that zλi ∈ Γ (Vλi ) ⊆ Γ (xε ) + Oε (0).

(1.31)

Since Γ (xε ) + Oε (0) is convex, by (1.29) and (1.31), we see that f ε (x) ∈ Γ (xε ) + Oε (0). Thus, there is yε ∈ Γ (xε ) ⊆ K so that yε − f ε (x)X < ε. From the above arguments, we see that (1.27) (or (1.28)) holds. Since f ε : K → K is continuous and K is a compact subset of X, according to the Schauder Fixed Point Theorem, there is  xε ∈ K so that f ε ( xε ) =  xε . This, together with (1.28), implies that there is ( xε ,  yε ) ∈ K × K so that  xε −  xε X < ε,

 yε −  xε X < ε and  yε ∈ Γ ( xε ).

(1.32)

Since K is compact, without loss of generality, we can assume that x ∈ K,  xε → 

 xε →  x and  yε →  x when ε → 0.

(1.33)

Since Graph(Γ ) is closed, by (1.33) and the third conclusion in (1.32), we obtain that  x ∈ Γ ( x ). Hence, we finish the proof of Theorem 1.14.   Remark 1.2 Theorem 1.14 is a weakened version of Corollary 17.55 in [1]. Theorem 1.15 (Lyapunov Theorem) Let y ∈ L1 (0, T ; Rn ). If λ : [0, T ] → [0, 1] is Lebesgue measurable, then there is a Lebesgue measurable set E ⊆ [0, T ] so that 

T 0

 λ(t)y(t) dt =

y(t) dt. E

Theorem 1.15 is a weakened version of Lyapunov’s theorem for vector measures (see [21] and [14]), which states that the set of values of an n-dimensional,

16

1 Mathematical Preliminaries

nonatomic, and bounded vector-measure is a convex compact set. The proof of this theorem is adapted from [2] (see Pages 580 and 581 in [2]). Proof We denote by F the σ -field, which consists of all Lebesgue measurable sets in [0, T ]. Let 

 y(t) dt  E ∈ F

μ(F ) 

 (1.34)

E

and    T  ϕ ∈ L∞ (0, T ; R)  0 ≤ ϕ(t) ≤ 1 a.e. t ∈ (0, T ) . Furthermore, we define a continuous linear functional L : L∞ (0, T ; R) → Rn by 

T

L(ϕ) 

ϕ(t)y(t)dt for all ϕ ∈ L∞ (0, T ; R),

(1.35)

0

and denote 

T

L(T ) 

 ϕ(t)y(t)dt  ϕ ∈ T

 .

0

It is clear that L(T ) is a nonempty, convex, and compact subset of Rn . We shall conclude by proving that μ(F ) = L(T ). Since μ(F ) is obviously contained in L(T ), it remains to prove that L(T ) ⊆ μ(F ). To this end, let z ∈ L(T ) and define    Tz  ϕ ∈ T  L(ϕ) = z . One can directly check that Tz is a nonempty, convex, and weakly star compact subset of L∞ (0, T ; R). Then, by the Krein-Milman Theorem, Tz has an extremal point. If we could prove that any extremal point of Tz is a characteristic function,

(1.36)

then there exists E ∈ F so that z = L(χE ). This, together with (1.35) and (1.34), implies that z ∈ μ(F ) and leads to the conclusion. We next show (1.36). By contradiction, suppose that it was not true. Then there would exist an extremal point of Tz , denoted by ϕ0 , which is not a characteristic function. Then, there is a constant c0 ∈ (0, 1/2) and a set E0 ∈ F with |E0 | > 0 so that c0 ≤ ϕ0 (t) ≤ 1 − c0 for all t ∈ E0 .

(1.37)

1.2 Set-Valued Map

17

 Consider the subspace V  {χE0 ϕ  ϕ ∈ L∞ (0, T ; R)}. Since the dimension of V is bigger than n, the kernel of the operator L|V : V → Rn defined by  L|V (χE0 ϕ) 

E0

ϕ(t)y(t)dt for all χE0 ϕ ∈ V ,

is nonzero. Therefore, there exists a function χE0 ϕ1 = 0 with ϕ1 ∈ L∞ (0, T ; R) so that  ϕ1 (t)y(t)dt = 0. (1.38) E0

Denote ϕ2 (t)  c0 χE0 ϕ1 (t)/χE0 ϕ1 L∞ (0,T ;R) for a.e. t ∈ (0, T ). Since ϕ0 ∈ Tz , it follows from (1.38) and (1.37) that ϕ0 ± ϕ2 ∈ T and L(ϕ0 ± ϕ2 ) = z. Hence, ϕ0 + ϕ2 and ϕ0 − ϕ2 belong to Tz . These lead to a contradiction since ϕ0 is an extremal point of Tz . Thus, we finish the proof of Theorem 1.15.   Let us introduce the measurability of a set-valued map. Definition 1.13 Let D ⊆ Rn be a nonempty, Lebesgue measurable set. Let X be a separable Banach space. A set-valued map Γ : D → 2X \ {∅} is Lebesgue measurable if for any nonempty closed set F ⊆ X, the set Γ −1 (F ) is Lebesgue measurable in D. Theorem 1.16 (Kuratowski and Ryll-Nardzewski Theorem) Let D ⊆ Rn be a nonempty, Lebesgue measurable set and X be a separable Banach space. If Γ : D → 2X \ {∅} is Lebesgue measurable and takes closed set values (i.e., Γ (r) is closed for each r ∈ D), then it admits a measurable selection. That is, there is a Lebesgue measurable map f : D → X so that f (r) ∈ Γ (r) for a.e. r ∈ D. Proof Without loss of generality, we assume that Γ (r)



O1 (0) = ∅ for any r ∈ D.

(1.39)

Indeed, by setting    Ok (0) = ∅ , D1  S1 Sk  r ∈ D  Γ (r) and Dk+1 = Sk+1 \ Sk for all k ∈ N+ , we observe that Γ admits a measurable selection if and only if Γ |Dk admits a measurable selection for each k ∈ N+ .

18

1 Mathematical Preliminaries

We will use the recursive method to define a sequence of Lebesgue measurable functions {fn (·)}n≥1 with fn (·) : D → X so that d(fn (r), Γ (r)) < 2−n for all r ∈ D,

(1.40)

d(fn (r), fn−1 (r)) < 2−n+1 for all r ∈ D.

(1.41)

and so that

(Here and throughout the proof of this theorem, d(x1 , x2 ) denotes the distance between x1 and x2 in X.) When this is done, it follows from (1.41) that {fn (r)}n≥1 is a Cauchy sequence in the Banach space X. Thus, there is f (r) ∈ X so that fn (r) → f (r). From this, as well as the facts that Γ (r) is closed and each fn is Lebesgue measurable, we can pass to the limit for n → +∞ in (1.40) to see that f (·) is a measurable selection of Γ . This ends the proof of Theorem 1.16. The remainder is to construct the desired {fn }n≥1 . Let f0 (·), f1 (·) : D → X be given by f0 (·) = f1 (·) ≡ 0. It is clear that f0 (·) and f1 (·) are Lebesgue measurable. Further, it follows from (1.39) that (1.40) and (1.41) hold for n = 1. Given n ≥ 2, suppose that for each m < n, fm with (1.40) and (1.41) has been defined. We now define the desired fn (·). Since X is a separable Banach space, there is a dense subset {xk }k≥1 ⊆ X. For each k ≥ 1, we set  (1.42) Ek  r ∈ D : d(xk , fn−1 (r)) < 2−n+1 and d(xk , Γ (r)) < 2−n . Since fn−1 (·) and Γ (·) are Lebesgue measurable, we obtain that  −1 (O2−n+1 (xk )) Γ −1 (O2−n (xk )) Ek = fn−1 is Lebesgue measurable. We further claim that  Ek = D.

(1.43)

k≥1

Indeed, given r ∈ D, d(fn−1 (r), Γ (r)) < 2−n+1 and Γ (r) is closed. Thus there is a y ∈ Γ (r) so that d(fn−1 (r), y) < 2−n+1 .

(1.44)

Since {xk }k≥1 is dense, it follows from (1.44) that there exists an  ∈ N+ so that d(x , y) < min{2−n+1 − d(fn−1 (r), y), 2−n }. −n+1 and d(x , Γ (r)) ≤ d(x , y) < 2−n . This implies that d(x , fn−1 (r))    < 2 Thus, we have that r ∈ E ⊆ k≥1 Ek , which leads to (1.43).

1.2 Set-Valued Map

19

Next, by (1.43), we define fn (r)  xk when r ∈ Ek \

k−1 

Ei .

(1.45)

i=1

It is clear that fn (·) is measurable. Moreover, (1.40) and (1.41) follow from (1.45) and (1.42). Hence, we finish the proof of Theorem 1.16.   Remark 1.3 The proof of Theorem 1.16 is adapted from Theorem 5.2.1 in [26] and Theorem 2.23 in Chapter 3 of [19]. Let D ⊆ Rn be a nonempty and Lebesgue measurable set, and X and Y be two separable Banach spaces. Let the map ϕ : D × X → Y satisfy the following conditions: ⎫ (i) For any x ∈ X, the map ϕ(·, x) is measurable; ⎪ ⎪ ⎪ ⎪ ⎪ (ii) ϕ(·, ·) is uniformly continuous in u locally at any (r0 , x0 ) ∈ D × X,⎪ ⎪ ⎪ ⎪ ⎪ ⎪ i.e., for any (r0 , x0 ) ∈ D × X, there exists a constant δ > 0 and a ⎪ ⎬ modulus of continuity ω(·)  ω(·, r0 , x0 , δ) so that ϕ(r, x) − ϕ(r,  x )Y ≤ ω(x −  x X )    D × Oδ (x0 ). for all (r, x), (r,  x ) ∈ Oδ (r0 )

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ (1.46)

Theorem 1.17 (Measurable Selection Theorem) Let D ⊆ Rn be a nonempty, Lebesgue measurable set with |D| < +∞. Let X and Y be two separable Banach spaces. Let Γ : D → 2X \ {∅} be a measurable set-valued map taking closed set values (i.e., for each r ∈ D, Γ (r) is closed). Let ϕ : D × X → Y satisfy the above condition (1.46). Then for every Lebesgue measurable function h : D → Y satisfying h(r) ∈ ϕ(r, Γ (r))

for almost all r ∈ D,

(1.47)

there exists a Lebesgue measurable map f : D → X so that f (r) ∈ Γ (r) for a.e. r ∈ D and so that h(r) = ϕ(r, f (r))

for almost all r ∈ D.

Proof Let ϕ (r, x)  ϕ(r, x) − h(r) 

for any (r, x) ∈ D × X.

(1.48)

20

1 Mathematical Preliminaries

It is clear that  ϕ satisfies the condition (1.46). Define a set-valued map Λ : D → 2X by    ϕ (r, x) = 0 for all r ∈ D. Λ(r)  x ∈ X  

(1.49)

For each r ∈ D, it follows from the continuity of  ϕ (r, ·), (1.47), and (1.49) that Λ(r) and Λ(r)



Γ (r) are nonempty and closed.

(1.50)

Here, we used the fact that Γ (r) is closed.  By (1.50), we can define a set-valued map Λ Γ : D → 2X \ {∅} in the following manner: 

Γ )(r)  Λ(r)



Γ (r) for all r ∈ D.

(1.51)

Λ : D → 2X \ {∅} is Lebesgue measurable.

(1.52)

(Λ We now claim that

When (1.52) is proved, (1.48) follows from the Lebesgue measurability of Γ (·), (1.51), (1.50), Theorem 1.16, and (1.49). To show (1.52), we arbitrarily fix a constant ρ > 0. Since X is a separable Banach space, there is a sequence {xk }k≥1, which is dense in X. For each k, according to the Lusin Theorem, there exists a compact set Dρ,k ⊆ D with |D \ Dρ,k | < ρ/2k , so that  ϕ (·, xk ) is continuous on Dρ,k . Set Dρ 

+∞ 

Dρ,k .

k=1

We see that Dρ is closed, |D \ Dρ | < ρ,

(1.53)

ϕ (·, xk ) is continuous on Dρ for all k ∈ N+ . 

(1.54)

and

The rest of the proof will be carried out by the following two steps: Step 1. We prove that  ϕ (·, ·) is continuous on Dρ × X. To this end, we arbitrarily fix (r0 , x0 ) ∈ Dρ × X. Since  ϕ satisfies the condition (1.46), there is a constant δ > 0 and a modulus of continuity ω(·) so that  ϕ (r, x) −  ϕ (r,  x )Y ≤ ω(x −  x X )    for all (r, x), (r,  x ) ∈ Oδ (r0 ) D × Oδ (x0 ). (1.55)

1.2 Set-Valued Map

21

Moreover, for arbitrarily fixed ε > 0, there is  δ  δ (ε) ∈ (0, δ/2) so that ω(2 δ) ≤ ε/3.

(1.56)

Meanwhile, since {xk }k≥1 is dense in X, there is kε ∈ N+ so that xkε − x0 X ≤  δ.

(1.57)

Moreover, according to (1.54), there is  δ  δ (ε) ∈ (0,  δ ) so that  ϕ (r, xkε ) −  ϕ (r0 , xkε )Y ≤ ε/3 for all r ∈ Oδ (r0 )



Dρ .

(1.58)

Now  it follows from (1.57), (1.55), (1.56), and (1.58) that for any (r, x) ∈ (Oδ (r0 ) Dρ ) × Oδ (x0 ), x − xkε X ≤ x − x0 X + x0 − xkε X ≤  δ + δ ≤ 2 δ ≤ δ, and ϕ (r, x) −  ϕ (r, xkε )Y  ϕ (r, x) −  ϕ (r0 , x0 )Y ≤  + ϕ (r, xkε ) −  ϕ (r0 , xkε )Y +  ϕ (r0 , xkε ) −  ϕ (r0 , x0 )Y ≤ ε/3 + ε/3 + ε/3 = ε. Since (r0 , x0 ) is taken arbitrarily from Dρ × X, we complete the proof of Step 1. Step 2. We finish the proof of (1.52). By Step 1, (1.49) and the first conclusion in (1.53), we find that  Graph(Λ|Dρ )  {(r, x)  r ∈ Dρ and x ∈ Λ(r)} is closed.

(1.59)

Set  D

+∞ 

D1/m .

(1.60)

m=1

It follows from (1.53) that  is a Borel set and |D \ D|  = 0. D

(1.61)

Meanwhile, by (1.59) and (1.60), we see that    and x ∈ Λ(r)} Graph(Λ|D  )  {(r, x) r ∈ D +∞  Graph(Λ|D1/m ) is a Borel set. = m=1

(1.62)

22

1 Mathematical Preliminaries

Let PD : D × X → D be the projection operator defined by PD (r, x)  r for all (r, x) ∈ D × X. Then, for any nonempty and closed set S ⊆ X, we have that    Λ−1 (S) = PD Graph(Λ) (D × S)        × S)  × S) , = PD Graph(Λ) (D PD Graph(Λ) ((D \ D) which indicates that        × S) .  × S) (D Λ−1 (S) = PD Graph(Λ|D PD Graph(Λ) ((D \ D) ) (1.63) By the first conclusion of (1.61) and (1.62), we see that the first term in the righthand side of (1.63) is Souslinian. Further, it is Lebesgue measurable. For the second term in the right-hand side of (1.63), we observe that       × S) ⊆ PD (D \ D)  × S = D \ D.  PD Graph(Λ) ((D \ D) This, together with the second conclusion of (1.61), implies that     × S) is a Lebesgue measurable set. PD Graph(Λ) ((D \ D) Hence, by (1.63), Λ−1 (S) is Lebesgue measurable. This implies (1.52). Thus, we finish the proof of Theorem 1.17.

 

Remark 1.4 The proof of Theorem 1.17 is based on Theorem 2.2 in [15] and Theorem 2.24 in Chapter 3 of [19]. Corollary 1.1 (Filippov Lemma) Let Y and U be separable Banach spaces. Let D ⊆ Rn be a Lebesgue measurable set with |D| < +∞. Let Γ : D → 2U \ {∅} be a Lebesgue measurable set-valued map taking closed set values. Let f : D×Y ×U → Y satisfy the conditions: ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ (ii)f is uniformly continuous in u locally at (r0 , y0 , u0 ) for a.e. r0 ∈ D, ⎬ (i)f is measurable in (r, y, u);

⎪ ⎪ ⎪ ⎪ ⎪ ⎭ (iii) For each u ∈ U and for a.e. r ∈ D, f (r, ·, u) : Y →

Y is continuous. (1.64) each y0 ∈ Y and each u0 ∈ U ;

Then for every Lebesgue measurable function h : D → Y and z ∈ C(D; Y ) satisfying

1.3 Controllability and Observability Estimate

h(r) ∈ f (r, z(r), Γ (r))

23

for a.e. r ∈ D,

there exists a measurable selection u(r) ∈ Γ (r), a.e. r ∈ D, so that h(r) = f (r, z(r), u(r))

for a.e. r ∈ D.

Proof We define ϕ : D × U → Y by ϕ(r, u)  f (r, z(r), u) for all (r, u) ∈ D × U. By Theorem 1.17, we only need to check that ϕ satisfies the conditions in (1.46). To this end, we use the measurability of z(·) to find a sequence of simple functions zm (·) : D → Y converging pointwisely to z(·). Then, according to (i) and (iii) in (1.64), for each u ∈ U , the function r → f (r, zm (r), u) is measurable and lim f (r, zm (r), u) = f (r, z(r), u) = ϕ(r, u) for a.e. r ∈ D.

m→+∞

These imply that ϕ satisfies the first condition in (1.46). Next, according to the second condition in (1.64), for a.e. r0 ∈ D and for each u0 ∈ U , there is a constant δ > 0 and a modulus of continuity ω(·)  ω(·; r0 , z(r0 ), u0 , δ) so that uU ), f (r, y, u) − f (r, y, u)Y ≤ ω(u − 

(1.65)

 for all r ∈ Oδ (r0 ) D, y ∈ Oδ (z(r0 )), and u, u ∈ Oδ (u0 ). Since z ∈ C(D; Y ), there is  δ ∈ (0, δ) so that z(r) − z(r0 )Y < δ for all r ∈ Oδ (r0 )



D.

This, together with (1.65), implies that uU ) f (r, z(r), u) − f (r, z(r), u)Y ≤ ω(u −   for all (r, u), (r, u) ∈ (Oδ (r0 ) D) × Oδ (u0 ), i.e., ϕ satisfies the second condition in (1.46). Hence, we finish the proof of Corollary 1.1.  

1.3 Controllability and Observability Estimate In this section, some results on controllability and observability estimates for linear differential equations are presented. We start with some basic concepts and results on controllability and observability.

24

1 Mathematical Preliminaries

Let X and U be two reflexive Banach spaces. Let A : D(A) ⊆ X → X be a linear operator, which generates a C0 semigroup {eAt }t ≥0 on X. Consider the following controlled system: 

t

y(t) = eAt x +

eA(t −s)f (s, y(s), u(s))ds,

t ≥ 0,

(1.66)

0

where f satisfies proper conditions so that for any x ∈ X and u ∈ Lp (0, +∞; U ) with p ∈ [1, +∞], (1.66) admits a unique solution y(·)  y(·; 0, x, u) ∈ C([0, +∞); X). This is the case if, for instance, f (t, x, u) is Lipschitz continuous and grows at most linearly in x, and uniformly in (t, u). Let T > 0, x ∈ X, and p ∈ [1, +∞]. We define the reachable set of the system (1.66) with the initial state x at T and with the control set Lp (0, T ; U ) by    YR (T , x; p)  y(T ; 0, x, u)  u ∈ Lp (0, T ; U ) . We now introduce definitions of various controllabilities. Definition 1.14 The system (1.66) is said to be (i) exactly controllable over (0, T ) with the control set Lp (0, T ; U ) if YR (T , x; p) = X for each x ∈ X; (ii) approximately controllable over (0, T ) with the control set Lp (0, T ; U ) if YR (T , x; p) = X for each x ∈ X; (iii) null controllable over (0, T ) with the control set Lp (0, T ; U ) if 0 ∈ YR (T , x; p) for each x ∈ X. The following theorem plays an important role in the study of the controllability. Theorem 1.18 (Range Comparison Theorem) Let W , V , and Z be Banach spaces where W is reflexive. Let F ∈ L (V , Z) and G ∈ L (W, Z). Then R(F ) ⊆ R(G)

(1.67)

if and only if there is δ > 0 so that G∗ z∗ W ∗ ≥ δF ∗ z∗ V ∗ for all z∗ ∈ Z ∗ .

(1.68)

Furthermore, if there is a constant C > 0 so that (1.68) holds with δ = 1/C, then for any v ∈ V , there is w ∈ W with wW ≤ CvV

(1.69)

1.3 Controllability and Observability Estimate

25

so that F (v) = G(w).

(1.70)

Proof We first show that (1.67)⇒(1.68). By contradiction, we suppose that it was not true. Then, for any n ∈ N+ , there would exist zn∗ ∈ Z ∗ so that G∗ zn∗ W ∗ < Denoting  zn∗ 

1 ∗ ∗ F zn V ∗ . n

(1.71)

√ ∗ nzn /F ∗ zn∗ V ∗ , by (1.71), we have that

√ 1 G∗ zn∗ W ∗ ≤ √ → 0 and F ∗ zn∗ V ∗ = n → +∞. n

(1.72)

According to (1.67), for any v ∈ V , there exists a w ∈ W so that F v = Gw. Then by the first conclusion of (1.72), we obtain that F ∗ zn∗ , vV ∗ ,V =  zn∗ , F vZ ∗ ,Z =  zn∗ , GwZ ∗ ,Z = G∗ zn∗ , wW ∗ ,W → 0. This, together with Principle of Uniform Boundedness (see Theorem 1.2), implies zn∗ }n≥1 must be bounded, which leads to a contradiction with that the sequence {F ∗ the second conclusion of (1.72). We next show that (1.68)⇒(1.67). For this purpose, we arbitrarily fix v ∈ V and define fv : R(G∗ ) → R by fv (G∗ z∗ )  F ∗ z∗ , vV ∗ ,V , z∗ ∈ Z ∗ .

(1.73)

It is clear that fv : R(G∗ ) → R is well defined. Indeed, for any z1∗ , z2∗ ∈ Z ∗ with G∗ z1∗ = G∗ z2∗ , by (1.68), we observe that F ∗ (z1∗ − z2∗ )V ∗ ≤

1 ∗ ∗ G (z1 − z2∗ )W ∗ = 0, δ

which indicates that F ∗ (z1∗ ) = F ∗ (z2∗ ). Moreover, it follows from (1.73) and (1.68) that fv : R(G∗ ) → R is linear and |fv (G∗ z∗ )| = |F ∗ z∗ , vV ∗ ,V | ≤

1 vV G∗ z∗ W ∗ . δ

Then, by the Hahn Banach Theorem (see Theorem 1.5) and the reflexivity of W , we can find w ∈ W so that wW ≤ vV /δ

(1.74)

26

1 Mathematical Preliminaries

and so that fv (G∗ z∗ ) = G∗ z∗ , wW ∗ ,W for all z∗ ∈ Z ∗ .

(1.75)

From (1.73) and (1.75) it follows that z∗ , F vZ ∗ ,Z = F ∗ z∗ , vV ∗ ,V = G∗ z∗ , wW ∗ ,W = z∗ , GwZ ∗ ,Z for each z∗ ∈ Z ∗ . This implies that F v = Gw.

(1.76)

Hence, R(F ) ⊆ R(G). Finally, (1.69) and (1.70) follow from (1.74) and (1.76). Thus, we finish the proof of Theorem 1.18.   Let T > 0, D(·) ∈ L1 (0, T ; L (X)), and B(·) ∈ L∞ (0, T ; L (U, X)). Consider the following controlled linear equation:  y(t) ˙ = Ay(t) + D(t)y(t) + B(t)u(t), t ∈ (0, T ), y(0) ∈ X.

(1.77)

 Write {Φ(t, s)  t ≥ s ≥ 0} for the evolution operator generated by A+D(·) over X. We define two operators FT ∈ L (X) and GT ∈ L (Lp (0, T ; U ), X) (p ≥ 1) by: FT (y0 )  Φ(T , 0)y0 for all y0 ∈ X,

(1.78)

Φ(T , s)B(s)u(s) ds for all u ∈ Lp (0, T ; U ),

(1.79)

and 

T

GT (u(·))  0

respectively. Then by Definition 1.14, one can easily check the following fact: • The system (1.77) is exactly (or null, or approximately) controllable over (0, T ) with the control set Lp (0, T ; U ) if and only if X ⊆ R(GT ) (or R(FT ) ⊆ R(GT ), or R(GT ) = X). Next, based on Theorem 1.18 and the above fact, we give another equivalent condition on the exact controllability (or the null controllability) of the system (1.77). To this end, we introduce the adjoint equation of (1.77): 

ϕ(t) ˙ = −A∗ ϕ(t) − D(t)∗ ϕ(t), t ∈ (0, T ), ϕ(T ) ∈ X∗ .

(1.80)

1.3 Controllability and Observability Estimate

27

We write ϕ(·; T , ϕT ) for the solution of (1.80) over [0, T ] with the terminal condition that ϕ(T ) = ϕT ∈ X∗ . It is clear that G∗T (ϕT ) = B(·)∗ ϕ(·; T , ϕT ) for all ϕT ∈ X∗ .

(1.81)

Let p > 1, W  Lp (0, T ; U ), V = Z  X, G  GT , and F  I (or F  FT ). Then, by Theorem 1.18 and (1.81), we have the following result: Corollary 1.2 Let p, q ∈ (1, +∞) and 1/p + 1/q = 1. (i) The system (1.77) is exactly controllable over (0, T ) with the control set Lp (0, T ; U ) if and only if ϕT X∗ ≤ CB(·)∗ ϕ(·; T , ϕT )Lq (0,T ;U ∗ ) for all ϕT ∈ X∗ ,

(1.82)

where C > 0 is a constant. Furthermore, if (1.82) holds, then for any z ∈ X and y0 ∈ X, there is a control u ∈ Lp (0, T ; U ) so that y(T ; 0, y0 , u) = z and uLp (0,T ;U ) ≤ Cz − Φ(T , 0)y0 X . (ii) The system (1.77) is null controllable over (0, T ) with the control set Lp (0, T ; U ) if ϕ(0; T , ϕT )X∗ ≤ CB(·)∗ ϕ(·; T , ϕT )Lq (0,T ;U ∗ ) for all ϕT ∈ X∗ , (1.83) where C > 0 is a constant. Furthermore, if (1.83) holds, then for any y0 ∈ X, there is a control u ∈ Lp (0, T ; U ) so that y(T ; 0, y0, u) = 0

and uLp (0,T ;U ) ≤ Cy0 X .

We call (1.82) and (1.83) as observability estimates. About the approximate controllability, we have the next theorem. Theorem 1.19 Let p ∈ [1, +∞]. Then the following conclusions are equivalent: (i) The system (1.77) is approximately controllable over (0, T ) with the control set Lp (0, T ; U ); (ii) R(GT ) = X; (iii) The operator G∗T is injective, i.e., N (G∗T ) = {0}; (iv) ϕT = 0 if and only if B(·)∗ ϕ(·; T , ϕT ) = 0. Proof It is clear that (i), (ii), and (iii) are equivalent. We now show that (iii)⇒(iv)⇒(ii). (iii)⇒(iv). On one hand, if ϕT = 0, then it is obvious that B(·)∗ ϕ (·; T , ϕT ) = 0. On the other hand, let ϕT ∈ X∗ satisfy B(·)∗ ϕ(·; T , ϕT ) = 0. This, together with (1.81), implies that G∗T (ϕT ) = 0. Then by (iii), we have that ϕT = 0.

28

1 Mathematical Preliminaries

(iv)⇒(ii). By contradiction, we suppose that R(GT ) = X. Then, there would exist z ∈ X so that z ∈ R(GT ). Since R(GT ) is a nonempty, convex, and closed subset of X, by Theorem 1.11, we can find y ∗ ∈ X∗ with y ∗ X∗ = 1 so that y ∗ , zX∗ ,X ≤ y ∗ ,



!

T

Φ(T , s)B(s)u(s)ds 0

X ∗ ,X

for all u ∈ Lp (0, T ; U ).

Replacing u in the above inequality by ku (k > 0), we find that 

T

B(s)∗ Φ(T , s)∗ y ∗ , u(s)ds = 0 for all u ∈ Lp (0, T ; U ).

0

Consequently, B(·)∗ Φ(T , ·)∗ y ∗ = 0. Then, by (iv), we have that y ∗ = 0, which leads to a contradiction. Hence, we finish the proof of Theorem 1.19.   We next introduce connections between L∞ -null controllability and L1 observability in Hilbert spaces. Let Y and U be two real separable Hilbert spaces. The controlled system is as: y(t) ˙ = Ay(t) + D(t)y(t) + B(t)u(t),

t ∈ (0, +∞),

(1.84)

where D(·) ∈ L1loc (0, +∞; L (Y )), A : D(A) ⊆ Y → Y generates a C0 semigroup {eAt }t ≥0 on Y and B(·) ∈ L∞ (0, +∞; L (U, Y )). Given τ ∈ [0, +∞), y0 ∈ Y and u ∈ L∞ (τ, +∞; U ), write y(·; τ, y0 , u) for the solution of (1.84) over [τ, +∞), with the initial condition that y(τ ) = y0 . Moreover, write {Φ(t, s) : t ≥ s ≥ 0} for the evolution operator generated by A + D(·) over Y . Then we have the following equivalence between null controllability and observability estimate. Theorem 1.20 The following statements are equivalent: (i) For any T2 > T1 ≥ 0, there exists a constant C1 (T1 , T2 ) ∈ (0, +∞) so that Φ(T2 , T1 )∗ zY ≤ C1 (T1 , T2 )B(·)∗ Φ(T2 , ·)∗ zL1 (T1 ,T2 ;U ) for all z ∈ Y. (1.85) (ii) For any T2 > T1 ≥ 0, there exists a constant C2 (T1 , T2 ) ∈ (0, +∞) so that for each y0 ∈ Y , there is a control u ∈ L∞ (T1 , T2 ; U ) satisfying that y(T2 ; T1 , y0 , u) = 0 and uL∞ (T1 ,T2 ;U ) ≤ C2 (T1 , T2 )y0 Y .

(1.86)

(iii) For any T2 > T1 ≥ 0 and y0 ∈ Y , there exists a control u ∈ L∞ (T1 , T2 ; U ) satisfying that y(T2 ; T1 , y0 , u) = 0. Furthermore, when one of the above three conclusions is valid, the constants C1 (T1 , T2 ) in (1.85) and C2 (T1 , T2 ) in (1.86) can be taken as the same number.

1.3 Controllability and Observability Estimate

29

Proof We divide the proof into the following four steps. Step 1. We show that (i)⇒(ii). Suppose that (i) holds. Let T2 > T1 ≥ 0 and let C1 (T1 , T2 ) be given by (1.85). Arbitrarily fix y0 ∈ Y and set  XT1 ,T2  {B(·)∗ Φ(T2 , ·)∗ z|(T1 ,T2 )  z ∈ Y } ⊆ L1 (T1 , T2 ; U ). Define a map FT1 ,T2 ,y0 : XT1 ,T2 → R in the following manner:   FT1 ,T2 ,y0 B(·)∗ Φ(T2 , ·)∗ z|(T1 ,T2 )  y0 , Φ(T2 , T1 )∗ zY for each z ∈ Y.

(1.87)

We first claim that FT1 ,T2 ,y0 is well defined. In fact, if z1 , z2 ∈ Y so that B(·)∗ Φ(T2 , ·)∗ z1 = B(·)∗ Φ(T2 , ·)∗ z2 over (T1 , T2 ), then by (1.85), it follows that Φ(T2 , T1 )∗ z1 = Φ(T2 , T1 )∗ z2 in Y . Hence, FT1 ,T2 ,y0 is well defined. Besides, one can easily check that FT1 ,T2 ,y0 is linear. By making use of (1.85) again, we can find that    FT ,T ,y B(·)∗ Φ(T2 , ·)∗ z|(T ,T )  1

2

0

1

∗

2

∗

≤ C1 (T1 , T2 )y0 Y B(·) Φ(T2 , ·) zL1 (T1 ,T2 ;U ) for all z ∈ Y. From this, we see that FT1 ,T2 ,y0 L (XT1 ,T2 ,R) ≤ C1 (T1 , T2 )y0 Y .

(1.88)

Since XT1 ,T2 is a subspace of L1 (T1 , T2 ; U ), we can apply the Hahn-Banach T1 ,T2 ,y0 ∈ (L1 (T1 , T2 ; U ))∗ so Theorem (see Theorem 1.5) to find a functional F that T ,T ,y (L1 (T ,T ;U ))∗ FT1 ,T2 ,y0 L (XT1 ,T2 ,R) = F 1 2 0 1 2 and T1 ,T2 ,y0 (g) for all g ∈ XT1 ,T2 . FT1 ,T2 ,y0 (g) = F From these, we can apply the Riesz Representation Theorem (see Theorem 1.4) to find a function u ∈ L∞ (T1 , T2 ; U ) so that FT1 ,T2 ,y0 L (XT1 ,T2 ,R) = uL∞ (T1 ,T2 ;U )

(1.89)

and so that  FT1 ,T2 ,y0 (g) =

T2 T1

g(t), −u(t)U dt for all g ∈ XT1 ,T2 .

(1.90)

30

1 Mathematical Preliminaries

From (1.87) and (1.90), we see that for each z ∈ Y ,  T2 Φ(T2 , T1 )y0 , zY = −u(t), B(t)∗ Φ(T2 , t)∗ zU dt T1



=−

T2

! Φ(T2 , t)B(t)u(t)dt, z , Y

T1

which indicates that  Φ(T2 , T1 )y0 +

T2

Φ(T2 , t)B(t)u(t)dt = 0.

T1

This, along with (1.89) and (1.88), yields (1.86) with C2 (T1 , T2 ) = C1 (T1 , T2 ). Step 2. We prove that (ii)⇒(i). Suppose that (ii) holds. Let T2 > T1 ≥ 0 and let C2 (T1 , T2 ) be given by (ii). Arbitrarily fix y0 ∈ Y . By (ii), there is u ∈ L∞ (T1 , T2 ; U ) so that y(T2 ; T1 , y0 , u) = 0 and uL∞ (T1 ,T2 ;U ) ≤ C2 (T1 , T2 )y0 Y .

(1.91)

By the equality in (1.91), we find that y0 , Φ(T2 , T1 )∗ zY = −



T2

u(t), B(t)∗ Φ(T2 , t)∗ zU dt for all z ∈ Y.

T1

This, along with the inequality in (1.91), yields that |y0 , Φ(T2 , T1 )∗ zY | ≤ C2 (T1 , T2 )y0 Y B(·)∗ Φ(T2 , ·)∗ zL1 (T1 ,T2 ;U ) for all z ∈ Y.

Since y0 is arbitrarily taken from Y , the above implies that for all z ∈ Y , Φ(T2 , T1 )∗ zY ≤ C2 (T1 , T2 )B(·)∗ Φ(T2 , ·)∗ zL1 (T1 ,T2 ;U ) , which leads to (1.85) with C1 (T1 , T2 ) = C2 (T1 , T2 ). Step 3. We show that (ii)⇔(iii). It is clear that (ii)⇒(iii). We now show the reverse. Suppose that (iii) holds. Let T2 > T1 ≥ 0. Define a linear operator GT1 ,T2 : L∞ (T1 , T2 ; U ) → Y by setting  GT1 ,T2 (u) 

T2

Φ(T2 , t)B(t)u(t)dt for each u ∈ L∞ (T1 , T2 ; U ). (1.92)

T1

Then it is clear that GT1 ,T2 is bounded. By (iii), we know that for each y0 ∈ Y , there is u ∈ L∞ (T1 , T2 ; U ) so that y(T2 ; T1 , y0 , u) = 0, i.e.,  0 = Φ(T2 , T1 )y0 +

T2

Φ(T2 , t)B(t)u(t)dt. T1

(1.93)

1.3 Controllability and Observability Estimate

31

From (1.92) and (1.93), we see that Range Φ(T2 , T1 ) ⊆ Range GT1 ,T2 .

(1.94)

Write QT1 ,T2 for the quotient space of L∞ (T1 , T2 ; U ) with respect to Ker GT1 ,T2 , i.e., QT1 ,T2  L∞ (T1 , T2 ; U )/Ker GT1 ,T2 . Let πT1 ,T2 : L∞ (T1 , T2 ; U ) → QT1 ,T2 be the quotient map. Then πT1 ,T2 is surjective and it holds that πT1 ,T2 (u)QT1 ,T2   = inf wL∞ (T1 ,T2 ;U ) : w ∈ u + Ker GT1 ,T2 for each u ∈ L∞ (T1 , T2 ; U ). (1.95) Define a map GT1 ,T2 : QT1 ,T2 → Y in the following manner: GT1 ,T2 (πT1 ,T2 (u))  GT1 ,T2 (u) for each πT1 ,T2 (u) ∈ QT1 ,T2 .

(1.96)

One can easily check that GT1 ,T2 is linear and bounded. By (1.96) and (1.94), we see that GT1 ,T2 is injective and that Range Φ(T2 , T1 ) ⊆ Range GT1 ,T2 . From these, we find that for each y0 ∈ Y , there is a unique πT1 ,T2 (uy0 ) ∈ QT1 ,T2 so that   Φ(T2 , T1 )y0 = GT1 ,T2 πT1 ,T2 (uy0 ) .

(1.97)

We next define another map TT1 ,T2 : Y → QT1 ,T2 by TT1 ,T2 (y0 )  πT1 ,T2 (uy0 ) for each y0 ∈ Y.

(1.98)

One can easily check that TT1 ,T2 is well defined and linear. We will use the Closed Graph Theorem (see Theorem 1.1) to show that TT1 ,T2 is bounded. For this purpose, we let {yk }k≥1 ⊆ Y satisfy that yk →  y in Y and TT1 ,T2 (yk ) →  h in QT1 ,T2 as k → +∞.

(1.99)

Because GT1 ,T2 and Φ(T1 , T2 ) are linear and bounded, it follows from (1.99), (1.98), and (1.97) that     h) = lim GT1 ,T2 TT1 ,T2 (yk ) = lim GT1 ,T2 πT1 ,T2 (uyk ) GT1 ,T2 ( k→+∞

k→+∞

= lim Φ(T2 , T1 )yk = Φ(T2 , T1 ) y. k→+∞

(1.100)

32

1 Mathematical Preliminaries

Meanwhile, by (1.97) and (1.98), we find that      Φ(T2 , T1 ) y = GT1 ,T2 πT1 ,T2 (u y) . y ) = GT1 ,T2 TT1 ,T2 (   h) = GT1 ,T2 TT1 ,T2 ( y ) , which, This, together with (1.100), yields that GT1 ,T2 ( combined with the injectivity of GT1 ,T2 , indicates that  h = TT1 ,T2 ( y ). So the graph of TT1 ,T2 is closed. Now we can apply the Closed Graph Theorem (see Theorem 1.1) to see that TT1 ,T2 is bounded. Hence, there is a constant C(T1 , T2 ) ∈ (0, +∞) so that TT1 ,T2 (y0 )QT1 ,T2 ≤ C(T1 , T2 )y0 Y for all y0 ∈ Y . This, along with (1.98), implies that πT1 ,T2 (uy0 )QT1 ,T2 ≤ C(T1 , T2 )y0 Y for each y0 ∈ Y.

(1.101)

Meanwhile, by (1.95), we see that for each y0 ∈ Y , there is  uy0 so that  uy0 ∈ uy0 + Ker GT1 ,T2 and  uy0 L∞ (T1 ,T2 ;U ) ≤ 2πT1 ,T2 (uy0 )QT1 ,T2 .

(1.102)

From (1.97), (1.96), (1.102), and (1.101), we find that for each y0 ∈ Y , there is a control  uy0 ∈ L∞ (T1 , T2 ; U ) so that Φ(T2 , T1 )y0 = GT1 ,T2 ( uy0 ) and  uy0 L∞ (T1 ,T2 ;U ) ≤ 2C(T1 , T2 )y0 Y .

(1.103)

Then by (1.92) and (1.103), we see that for each y0 ∈ Y , there is a control  uy 0 ∈ L∞ (T1 , T2 ; U ) so that y(T2 ; T1 , y0 , − uy0 ) = 0 and  uy0 L∞ (T1 ,T2 ;U ) ≤ 2C(T1 , T2 )y0 Y . These lead to (1.86) with C2 (T1 , T2 ) = 2C(T1 , T2 ). Step 4. About the constants C1 (T1 , T2 ) and C2 (T1 , T2 ). From the proofs in Step 1 to Step 3, we find that the constants C1 (T1 , T2 ) in (1.85) and C2 (T1 , T2 ) in (1.86) can be taken as the same number, provided that one of the conclusions (i)–(iii) holds. In summary, we end the proof of Theorem 1.20.   We next restrict our discussions on linear heat equations. Let Ω ⊆ Rn (n ≥ 1) be a bounded domain with a C 2 boundary ∂Ω. Let ω ⊆ Ω be an open and nonempty subset with its characteristic function χω . Let E ⊆ (0, T ) be a subset of positive measure. Let χE be its characteristic function. Consider the following heat equation: ⎧ ⎨ ∂t y − Δy + ay = χω χE u in Ω × (0, T ), (1.104) y=0 on ∂Ω × (0, T ), ⎩ in Ω. y(0) = y0

1.3 Controllability and Observability Estimate

33

Here, y0 ∈ L2 (Ω) and a ∈ L∞ (Ω × (0, T )), with the norm: a∞  aL∞ (Ω×(0,T )) . The adjoint equation of (1.104) reads ⎧ ⎨ ∂t ϕ + Δϕ − aϕ = 0 in Ω × (0, T ), ϕ=0 on ∂Ω × (0, T ), ⎩ ϕ(T ) ∈ L2 (Ω).

(1.105)

Using the similar arguments as those used in the proof of Theorem 1.20, we can obtain the following theorem: Theorem 1.21 The following statements are equivalent: (i) For any T > 0, there exists a constant C1 (T ) ∈ (0, +∞) so that any solution ϕ to (1.105) satisfies that ϕ(0)L2 (Ω) ≤ C1 (T )χω χE ϕL1 (0,T ;L2 (Ω)) .

(1.106)

(ii) For any T > 0, there exists a constant C2 (T ) ∈ (0, +∞) so that for each y0 ∈ L2 (Ω), there is a control u ∈ L∞ (0, T ; L2 (Ω)) satisfying that y(T ; 0, y0 , u) = 0 and uL∞ (0,T ;L2 (Ω)) ≤ C2 (T )y0 L2 (Ω) .

(1.107)

Furthermore, when one of the above two conclusions is valid, the constants C1 (T ) in (1.106) and C2 (T ) in (1.107) can be taken as the same number. The estimate in (i) of Theorem 1.21 can be derived from the interpolation inequality in the next lemma. Lemma 1.2 ([23]) There are two positive constants C  C (Ω, ω) and β  β (Ω, ω) ∈ (0, 1) so that for any T > 0 and ϕT ∈ L2 (Ω), 1−β

β

ϕ(0)L2 (Ω) ≤ NϕT L2 (Ω) ϕ(0)L2 (ω) ,

(1.108)

where ϕ is the solution to Equation (1.105), with ϕ(T ) = ϕT , and where " " ## 2/3 N  exp C 1 + T −1 + T a∞ + a∞ . Based on Lemma 1.2, we obtain the following estimate from measurable set in time: Theorem 1.22 ([23]) Any solution ϕ to Equation (1.105) satisfies the following observability estimate:  ϕ(0)L2 (Ω) ≤ C (Ω, ω, E, T , a∞ )

|ϕ(x, t)|dxdt, ω×E

(1.109)

34

1 Mathematical Preliminaries

where C (Ω, ω, E, T , a∞ )  exp (C(Ω, ω, E)) " " ## 2/3 exp C (Ω, ω) 1 + T a∞ + a∞ . Remark 1.5 The inequality (1.108) is a unique continuation estimate at one time. It is a strong estimate from two perspectives as follows: (i) It gives the unique continuation for Equation (1.105). Indeed, if ϕ(0) = 0 in a nonempty open set of Ω, then from (1.108), we have that ϕ(0) = 0 in Ω. This, along with the backward uniqueness of parabolic equations (see [5]), implies that ϕ = 0 in Ω × (0, T ). (ii) It implies the observability estimate (1.109). Finally, we consider the following controlled ODE:  y˙ = Ay(t) + Bu(t), t ∈ (0, T ), y(0) ∈ Rn ,

(1.110)

where T > 0, A ∈ Rn×n and B ∈ Rn×m . Then we have the following result. Theorem 1.23 ([8]) For each y0 ∈ Rn and y1 ∈ Rn , there exists a control u ∈ L∞ (0, T ; Rm ) so that y(T ; 0, y0 , u) = y1 if and only if rank(B, AB, . . . , An−1 B) = n (Kalman controllability rank condition). Here, y(·; 0, y0, u) is the solution to (1.110) with the initial condition y(0) = y0 .

Miscellaneous Notes The material of Section 1.1 is almost standard. Most of them are taken from [19]. We recommend the following references on functional analysis: [7, 24] and [28]. Some of the materials in Section 1.2 are adapted from [1–3, 6, 18, 19, 26], and [15]. The notion of controllability was introduced by R. E. Kalman for finite dimensional systems in [16] (see also [17]). Then it was extended to infinite dimensional systems. We would like to mention the following references on controllability for infinite dimensional systems: [9, 11, 12, 20, 25, 27], and [22].

References 1. C.D. Aliprantis, K.C. Border, Infinite Dimensional Analysis, A Hitchhiker’s Guide, 3rd edn. (Springer, Berlin, 2006) 2. J. Aubin, Mathematical Methods of Game and Economic Theory, Reprint of the 1982 revised edition, with a new preface by the author (Dover Publications, Mineola, NY, 2007)

References

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3. J. Aubin, H. Frankowska, Set-valued Analysis, Reprint of the 1990 edition. Modern Birkhäuser Classics (Birkhäuser Boston, Inc, Boston, MA, 2009) 4. V. Barbu, Nonlinear Semigroup and Differential Equations in Banach Spaces (Noordhoff International Publishing, Leyden, 1976) 5. C. Bardos, L. Tartar, Sur l’unicité rétrograde des équations parabliques et quelques questions voisines (French). Arch. Ration. Mech. Anal. 50, 10–25 (1973) 6. G. Bouligand, Sur la semi-continuité d’inclusions et quelques sujets conn exes. Enseignement Mathématique 31, 14–22 (1932) 7. J.B. Conway, A Course in Functional Analysis. Graduate Texts in Mathematics, vol. 96 (Springer, New York, 1985) 8. J.M. Coron, Control and Nonlinearity. Mathematical Surveys and Monographs, vol. 136 (American Mathematical Society, Providence, RI, 2007) 9. T. Duyckaerts, X. Zhang, E. Zuazua, On the optimality of the observability inequalities for parabolic and hyperbolic systems with potentials. Ann. Inst. H. Poincaré Anal. Non Linéaire 25, 1–41 (2008) 10. L.C. Evans, Partial Differential Equations. Graduate Studies in Mathematics, vol. 19 (American Mathematical Society, Providence, RI, 1998) 11. H.O. Fattorini, Some remarks on complete controllability. SIAM J. Control 4, 686–694 (1966) 12. A.V. Fursikov, O. Yu. Imanuvilov, Controllability of Evolution Equations. Lecture Notes Series, vol. 34 (Seoul National University, Seoul, 1996) 13. A. Granas, J. Dugundji, Fixed Point Theory. Springer Monographs in Mathematics (Springer, New York, 2003) 14. P.R. Halmos, The range of a vector measure. Bull. Am. Math. Soc. 54, 416–421 (1948) 15. M.Q. Jacobs, Remarks on some recent extensions of Filippov’s implicit functions lemma. SIAM J. Control 5, 622–627 (1967) 16. R.E. Kalman, Contributions to the theory of optimal control. Bol. Soc. Mat. Mexicana 5, 102–119 (1960) 17. R.E. Kalman, Mathematical description of linear dynamical systems. J. SIAM Control A 1, 152–192 (1963) 18. K. Kuratowski, Les fonctions semi-continues dans l’espace des ensembles fermés. Fundam. Math. 18, 148–159 (1932) 19. X. Li, J. Yong, Optimal Control Theory for Infinite Dimensional Systems (Birkhäuser, Boston, 1995) 20. J.-L. Lions, Exact controllability, stabilization and perturbations for distributed systems. SIAM Rev. 30, 1–68 (1988) 21. A. Lyapunov, Sur les fonctions-vecteurs complètement additives (Russian). Bull. Acad. Sci. URSS Sér. Math. 4, 465–478 (1940) 22. K.D. Phung, G. Wang, An observability estimate for parabolic equations from a measurable set in time and its application. J. Eur. Math. Soc. 15, 681–703 (2013) 23. K.D. Phung, L. Wang, C. Zhang, Bang-bang property for time optimal control of semilinear heat equation. Ann. Inst. H. Poincaré Anal. Non Linéaire 31, 477–499 (2014) 24. W. Rudin, Functional Analysis. International Series in Pure and Applied Mathematics, 2nd edn. (McGraw-Hill Inc., New York, 1991) 25. D.L. Russell, Controllability and stabilizability theory for linear partial differential equations: recent progress and open questions. SIAM Rev. 20, 639–739 (1978) 26. S.M. Srivastava, A Course on Borel Sets. Graduate Texts in Mathematics, vol. 180 (Springer, New York, 1998) 27. M. Tucsnak, G. Weiss, Observation and Control for Operator Semigroups (Birkhäuser Verlag, Basel, 2009) 28. K. Yosida, Functional Analysis. Grundlehren der Mathematischen Wissenschaften, 6th edn., vol. 123 (Springer, Berlin, 1980) 29. X. Zhang, X. Li, Z. Chen, Differential Equation Theory for Optimal Control Systems (Chinese) (Higher Education Publishing House, Shanghai, 1989)

Chapter 2

Time Optimal Control Problems

2.1 Overview on Time Optimal Control Problems 2.1.1 Introduction Optimization is a selection of the best “candidate” from a set of available alternatives, in virtue of some criterion. As a branch of optimization, optimal control problem is to ask for a control from an available set so that a certain optimality criterion, related to a given dynamic system, is achieved. When the criterion is the elapsed time, we call such problem a time optimal control problem. This book is concerned with time optimal control theory. Let us give a practical example of time optimal controls. Consider a car traveling through a hilly road. How should the driver press the accelerator pedal in order to minimize the total traveling time? In this example, the control law refers to the way via which the driver presses the accelerator pedal and shifts the gears. The dynamic system consists of both the car and the road. The optimality criterion is the minimization of the total traveling time. In general, control problems include some ancillary constraints. For instance, in the above example, the amount of available fuel and speed of the car should be limited, and the accelerator pedal cannot be pushed through the floor of the car, and so on. We next introduce a mathematical framework of time optimal control problems. Our framework contains the following four parts (or four ingredients): The first part is a controlled system (or a state equation) which reads: y(t) ˙ = Ay(t) + f (t, y(t), u(t)),

t ∈ (0, +∞),

(2.1)

where and throughout this monograph, the following Basic assumptions are effective:

© Springer International Publishing AG, part of Springer Nature 2018 G. Wang et al., Time Optimal Control of Evolution Equations, Progress in Nonlinear Differential Equations and Their Applications 92, https://doi.org/10.1007/978-3-319-95363-2_2

37

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2 Time Optimal Control Problems

(H1 ) State space Y is a real separable Hilbert space and control space U is another real separable Hilbert space. (H2 ) The operator A : D(A) ⊆ Y → Y generates a C0 semigroup {eAt }t ≥0 on Y and f : (0, +∞) × Y × U → Y is a given map. About the Equation (2.1), we would like to mention what follows: (i) Recall that L(a, b; U ), with 0 ≤ a < b < +∞, is the space of all Lebesgue measurable functions defined on (a, b) ⊆ (0, +∞) and taking values in U . If for each y0 ∈ Y , each pair (a, b) (with 0 ≤ a < b < +∞) and each u ∈ L(a, b; U ), the Equation (2.1), with the initial condition that y(a) = y0 (or with the initial data (a, y0 )), has a unique mild solution y(·; a, y0, u, b) over [a, b], then we call this solution the state trajectory over [a, b], corresponding to the control u and the initial data (a, y0). (ii) Recall that y(·)  y(·; a, y0, u, b) is a mild solution to (2.1), if y ∈ C([a, b]; Y ) and y(t) = eA(t −a)y0 +



t

eA(t −s)f (s, y(s), u(s)) ds,

t ∈ [a, b].

(2.2)

a

(iii) Given y0 ∈ Y and a ≥ 0, we simply write y(·; a, y0, u) for y(·; a, y0, u, b), provided that y(·; a, y0, u, b) exists for some b > a and u ∈ L(a, b; U ). (iv) When we write y(T ; a, y0, u) (with T > a ≥ 0, y0 ∈ Y and u ∈ L(a, T ; U )), we agree that the Equation (2.1), with the initial condition that y(a) = y0 , has a unique mild solution y(·; a, y0, u) on [a, T ]. Hence, y(T ; a, y0, u) denotes the value of this solution at time T . The second part is a control constraint set (or control constraint). Generally, in time optimal control problems, controls are always constrained. Most kinds of control constraints can be put into the framework that u(t) ∈ U a.e. t ∈ (0, +∞), where U is a nonempty subset of U . The third part is a starting set QS , which is a nonempty subset of [0, +∞) × Y . In many literatures, it is taken as {t0 } × {y0 }, with t0 ∈ [0, +∞) and y0 ∈ Y . The last part is an ending set QE , which is also a nonempty subset of (0, +∞) × Y . In many literatures, it is taken as (0, +∞) × S, with S a subset of Y . With the aid of the above four parts, we define the following set:  Aad  (tS , y0 , tE , u) ∈ QS × (tS , +∞) × L(tS , tE ; U)   (tE , y(tE ; tS , y0 , u)) ∈ QE ,

(2.3)

where L(tS , tE ; U) is the set of all functions v in L(tS , tE ; U ) so that v(t) ∈ U for a.e. t ∈ (tS , tE ). Then the time optimal control problem studied in most parts of this monograph can be stated as follows: Problem (TP):

Find a tetrad (tS∗ , y0∗ , tE∗ , u∗ ) in Aad so that

2.1 Overview on Time Optimal Control Problems

t ∗  tE∗ − tS∗ =

39

inf

(tS ,y0 ,tE ,u)∈Aad

(tE − tS ).

(2.4)

In this problem, (a) Each (tS , y0 , tE , u) in Aad is called an admissible tetrad, while Aad is called the set of admissible tetrads (or simply the admissible set). When (tS , y0 , tE , u) ∈ Aad , u is called an admissible control. (b) When the infimum in (2.4) is reached at (tS∗ , y0∗ , tE∗ , u∗ ), the tetrad (tS∗ , y0∗ , tE∗ , u∗ ) is called an optimal tetrad; the number tE∗ − tS∗ is called the optimal time; u∗ is called an optimal control, which is a function, defined on (tS∗ , tE∗ ) and taking values on U; the solution y(·; tS∗ , y0∗ , u∗ ) (to the Equation (2.1) over [tS∗ , tE∗ ]) is called an optimal trajectory (or an optimal state). Remark 2.1 (i) According to definitions of Aad and the optimal tetrad, respectively, each admissible tetrad (tS , y0 , tE , u) satisfies that tE > tS , and that each optimal tetrad (tS∗ , y0∗ , tE∗ , u∗ ) satisfies that tE∗ > tS∗ . (ii) When Aad = ∅, we agree that t ∗ = +∞. Two special cases will be introduced as follows: (i) The first case is that tS is fixed. We call the corresponding Problem (TP) a QS ,QE to denote Problem (TP). It minimal time control problem. We use (T P )min covers the following important minimal time control problem: (P )min

 t ∗  inf{t ∈ (0, +∞)  y(t; 0, y0, u) ∈ YE and u ∈ L(0, t; U)}, (2.5)

where y0 ∈ Y and YE ⊆ Y are fixed. (The set YE is called the target of this problem.) The above problem can be treated as the following time optimal control problem: (It will be explained later.) (P)min

  t ∗  inf{t ∈ (0, +∞)  y(t; 0, y0, u) ∈ YE and u ∈ L(0, +∞; U)}. (2.6)

In this problem, • a tetrad (0, y0 , t, u) is called admissible, if t ∈ (0, +∞), u ∈ L(0, +∞; U) and y(t; 0, y0 , u) ∈ YE ; • a tetrad (0, y0 ,  t ∗ , u∗ ) is called optimal if  t ∗ ∈ (0, +∞),  u∗ ∈ L(0, +∞; U) and y( t ∗ ; 0, y0, u∗ ) ∈ YE ; t ∗ , u∗ ) is an optimal tetrad,  t ∗ and  u∗ are called the optimal • when (0, y0 ,  time and an optimal control, respectively.

40

2 Time Optimal Control Problems

The following three facts can be checked easily: t ∗; • t∗ =  • The infimum in (2.5) can be reached if and only if the infimum in (2.6) can be reached; • The zero extension of each optimal control to (P )min over (0, +∞) is an optimal control to (P)min ; the restriction of each optimal control to (P)min over (0, t ∗ ) is an optimal control to (P )min . Because of the above three facts, we can treat (P )min and (P)min as the same problem. (ii) The second case is that tE is fixed. We call the corresponding Problem (TP) a QS ,QE to denote Problem (TP). It maximal time control problem. We use (T P )max covers the following standard maximal time control problem (see, for instance, [33] and [38]):  (P )max t ∗  sup{t ∈ [0, tE )  y(tE ; t, y(t; 0, y0, 0), u) ∈ O (2.7) and u ∈ L(t, tE ; U)}, where tE > 0, y0 ∈ Y , and O ⊆ Y are fixed. (The set O is called the target of this problem.) The above (P )max is indeed equivalent to the following problem: (We will explain the reasons later.)   (P)max t ∗  sup{t ∈ [0, tE )  y(tE ; 0, y0, χ(t,tE ) u) ∈ O (2.8) and u ∈ L(0, tE ; U)}. In this problem, • we call (0, y0 , tE , χ(t,tE ) u) an admissible tetrad, if t ∈ [0, tE ), u ∈ L(0, tE ; U) and y(tE ; 0, y0, χ(t,tE ) u) ∈ O; • we call (0, y0 , tE , χ(t ∗ ,tE ) u∗ ) an optimal tetrad, if  u∗ ∈ L(0, tE ; U) and y(tE , 0, y0 , χ(t ∗ ,tE ) u∗ ) ∈ O; t ∗ ∈ [0, tE ),  • when (0, y0 , tE , χ(t ∗ ,tE ) u∗ ) is an optimal tetrad, tE − t ∗ and  u∗ are called the optimal time and an optimal control, respectively. We can easily check the following three facts: • t∗ =  t ∗; • The supreme in (2.7) can be reached if and only if the supreme in (2.8) can be reached; • The zero extension of each optimal control to (P )max over (0, tE ) is an optimal control to (P)max , while the restriction of each optimal control to (P)max over (t ∗ , tE ) is an optimal control to (P )max . Based on the above-mentioned facts, we can treat (P )max and (P)max as the same problem.

2.1 Overview on Time Optimal Control Problems

41

2.1.2 Different Ingredients in Time Optimal Control Problems As mentioned in the above subsection, Problem (TP) consists of four ingredients. Different time optimal control problems have different ingredients. We now are going to introduce some special but important ingredients. • For state equations, we will focus on what follows: y(t) ˙ = Ay(t) + f (t, y(t)) + B(t)u(t),

t ∈ (0, +∞),

(2.9)

where A : D(A) ⊆ Y → Y generates a C0 semigroup {eAt }t ≥0 on Y , f : (0, +∞) × Y → Y is a given map, and B(·) ∈ L∞ (0, +∞; L (U, Y )). Equation (2.9) is a semilinear time-varying controlled equation where controls enter the equation linearly. It contains the following important cases: – Semilinear time-invariant evolution equation: y(t) ˙ = Ay(t) + f (y(t)) + Bu(t), t ∈ (0, +∞).

(2.10)

– Linear time-varying evolution equation: y(t) ˙ = Ay(t) + D(t)y(t) + B(t)u(t), t ∈ (0, +∞),

(2.11)

where D(·) ∈ L1loc (0, +∞; L (Y )). – Linear time-invariant evolution equation: y(t) ˙ = Ay(t) + Bu(t), t ∈ (0, +∞).

(2.12)

• For the control constraint sets, we will focus on the following two important types: – Ball type: U is the closed ball Br (0) in U , centered at 0 and of radius r > 0. – Rectangle type: When U = Rm , U is the following rectangle:    U  u ∈ Rm  −λj ≤ u, ej Rm ≤ λj for any j = 1, 2, . . . , m , (2.13) where λj , 1 ≤ j ≤ m, are positive constants and {ej }m j =1 forms the standard basis of Rm ; When dimU = +∞, U is the following rectangle:    U  u ∈ U  −λj ≤ u, ej U ≤ λj for any j ≥ 1 ,

(2.14)

where λj , j = 1, 2 . . . , are positive numbers and {ej }j ≥1 is an orthonormal basis of U .

42

2 Time Optimal Control Problems

• For starting sets and ending sets, we mainly concern the following cases: – We take QS  {0} × YS

and QE ⊆ (0, +∞) × Y,

(2.15)

where YS and QE are respectively nonempty subsets of Y and (0, +∞) × Y . In this case, tS = 0. – We take QS  {(tS , y(tS ; 0, y0, 0)) : tS ∈ [0, tE )}

and QE  {tE } × Br (zd ), (2.16)

where tE > 0, r > 0, y0 ∈ Y and zd ∈ Y are fixed so that y(tE ; 0, y0 , 0) ∈ Br (zd ). In this case, Br (zd ) is called the target set (or the target). Corresponding to (2.15), Problem (TP) is a minimal time control problem S ,QE (T P )Q . Corresponding to (2.16), Problem (TP) is a maximal time control min S ,QE problem (T P )Q max . In general, a minimal time control problem is to ask for a control (from a constraint set) driving the corresponding solution of a controlled equation from the initial state at the beginning to a target in the shortest time, while a maximal time control problem is to ask for a control (from a constraint set) driving the corresponding solution of a controlled equation from the initial state at the beginning to a given target at the fixed ending time so that the initiation of the control is delayed as late as possible. We end this subsection with introducing a special time optimal control problem which is not under the above framework but is important. It is called a minimal blowup time control problem and can be stated as follows:  inf{ t ∈ (0, +∞)  lim y(t; 0, y0, u|(0,t ))Y = +∞ and u ∈ L(0,  t; U)}, t → t−

(2.17)

where y0 ∈ Y . The problem (2.17) can be formally put into our framework by setting   QS  (0, y0 ) and QE  (0, +∞) × {∞}.

(2.18)

By a convention, “(tE , y(tE ; 0, y0 , u)) ∈ QE ” means that the following conclusions are true: (i) For each t ∈ (0, tE ) and each u ∈ L(0, tE ; U), the Equation (2.1), with the initial condition that y(0) = y0 , has a unique mild solution y(·; 0, y0, u|(0,t )) on [0, t]; (ii) It holds that lim y(t; 0, y0 , u|(0,t ))Y = +∞.

t →tE −

(2.19)

2.2 Connections of Minimal and Maximal Time Controls

43

The minimal blowup time control problem (2.17) differs essentially from the S ,QE defined by (2.4) with tS fixed. The problem (2.17) cannot be problem (T P )Q min really put into the framework of (2.4). When QE = (0, +∞)×YE , with YE a subset S ,QE (defined by (2.4) with tS fixed) is the subset YE in of Y , the target of (T P )Q min the state space Y , while the target of the problem (2.17) is outside the state space Y . The problem (2.17) is independently important. We explain it as follows: The differential systems whose solutions have the behavior of blowup replicate a large class of phenomena in applied science. In certain cases, the blowup of a solution is desirable. Thus, it is obviously appealing to minimize the blowup time by making use of controls in certain circumstances. For instance, the optimal process of starting a car engine can be thought of as minimizing the blowup time.

2.2 Connections of Minimal and Maximal Time Controls First of all, we give an example. When a lady plans to drive her car from the city Shanghai to the city Wuhan along a fixed road, how can she arrive at Wuhan in the shortest time? This problem can be formulated as a time optimal control problem. When the departure time is fixed, it is a minimal time control problem. When the arrival time is fixed, it is a maximal time control problem. Under some conditions, these two problems are exchangeable. This hints us that some minimal and maximal time control problems are equivalent in a certain sense. The main purpose of this section is to present several cases, where a minimal time control problem is equivalent to a maximal time control problem in a certain sense, or a minimal time control problem is equivalent to another minimal time control problem in a certain sense. We begin with studying the connection of time optimal control problems S ,Q E Q S ,QE (T P )Q and (T P )max in the following CASE ONE: min • The controlled system is (2.1), with f (t, y, u) = f (y, u) (i.e., (2.1) is timeinvariant); • The control constraint set U ⊆ U is nonempty; • The starting sets and the ending sets are as: QS = {0}×Y1 , QE = (0, +∞)×Y2 ; S = [0,  E = { Q tE ) × Y1 , and Q tE } × Y2 , where Y1 and Y2 are nonempty subsets  of Y and tE > 0. Q ,Q

 ,Q  Q

S E S E In CASE ONE, the admissible sets for (T P )min and (T P )max are respectively as follows:  Aad = {(0, y0 , tE , u)  y0 ∈ Y1 , tE > 0, u ∈ L(0, tE ; U), y(tE ; 0, y0, u) ∈ Y2 }

and  Aad = {( tS ,  y0 ,  tE , u)  0 ≤  tS <  tE ,  y0 ∈ Y1 ,  u ∈ L( tS ,  tE ; U), y( tE ;  tS ,  y0 , u) ∈ Y2 }.

44

2 Time Optimal Control Problems  

Q S ,Q E S ,QE Definition 2.1 In CASE ONE, the problems (T P )Q and (T P )max are said min to be equivalent if the following conditions are valid: S ,QE (i) If (0, y0∗ , tE∗ , u∗ ) is an optimal tetrad to (T P )Q and  tE ≥ tE∗ , then ( tE − min

 

Q S ,Q E tE , v ∗ ), with v ∗ (·) = u∗ (·− tE +tE∗ ), is an optimal tetrad to (T P )max . tE∗ , y0∗ ,  S ,Q E Q ∗ ∗ ∗ ∗ ∗ ∗     (ii) If (t S ,  y0 , tE , u ) is an optimal tetrad to (T P )max , then (0,  y0 , tE − t S , v ), S ,QE with  v ∗ (·) =  u∗ (· +  t ∗S ), is an optimal tetrad to (T P )Q . min Q ,QE

S Theorem 2.1 In CASE ONE, the problems (T P )min alent.

 ,Q E Q

S and (T P )max

are equiv-

Proof We first prove (i) in Definition 2.1. Assume that (0, y0∗ , tE∗ , u∗ ) is an optimal S ,QE tetrad to (T P )Q and  tE ≥ tE∗ . In order to prove that ( tE − tE∗ , y0∗ ,  tE , v ∗ ), with min  ,Q E Q

S tE + tE∗ ), is an optimal tetrad to (T P )max v ∗ (·)  u∗ (· − 

, it suffices to verify that

tE , v ∗ ) ∈ Aad ( tE − tE∗ , y0∗ , 

(2.20)

tE −  tS for all ( tS ,  y0 ,  tE , u) ∈ Aad . tE∗ ≤ 

(2.21)

and

To prove (2.20), we notice that (0, y0∗ , tE∗ , u∗ ) ∈ Aad , which implies that y0∗ ∈ Y1 , tE∗ > 0, u∗ ∈ L(0, tE∗ ; U) and y(tE∗ ; 0, y0∗ , u∗ ) ∈ Y2 .

(2.22)

tE + tE∗ ), we find that Meanwhile, because (2.1) is time-invariant and v ∗ (·) = u∗ (· − tE + tE∗ ; 0, y0∗ , u∗ ) for all t ∈ [ tE − tE∗ ,  tE ], y(t;  tE − tE∗ , y0∗ , v ∗ ) = y(t −  from which, it follows that y( tE ;  tE − tE∗ , y0∗ , v ∗ ) = y(tE∗ ; 0, y0∗ , u∗ ).

(2.23)

This, combined with (2.22), indicates (2.20). To show (2.21), we arbitrarily fix ( tS ,  y0 ,  tE , u) ∈ Aad . Then we have that tE ,  y0 ∈ Y1 ,  u ∈ L( tS ,  tE ; U) and y( tE ;  tS ,  y0 , u) ∈ Y2 . 0 ≤ tS < 

(2.24)

Let  v (·) =  u(· +  tS ). Then by the time-invariance of (2.1), we obtain that v ) = y(t +  tS ;  tS ,  y0 , u) for all t ∈ [0,  tE −  tS ], y(t; 0,  y0, from which, it follows that y( tE −  tS ; 0,  y0, v ) = y( tE ;  tS ,  y0 , u).

(2.25)

2.2 Connections of Minimal and Maximal Time Controls

45

This, combined with (2.24), implies that (0,  y0 ,  tE −  tS , v ) ∈ Aad , S ,QE , leads to (2.21). which, along with the optimality of tE∗ to (T P )Q min We next show (ii) in Definition 2.1. Assume that ( t ∗S ,  y0∗ ,  tE , u∗ ) is an optimal

 

Q S ,Q E . In order to prove that (0,  y0∗ ,  tE −  t ∗S , v ∗ ), with  v ∗ (·) =  u∗ (·+ tetrad to (T P )max QS ,QE ∗  t S ), is an optimal tetrad to (T P )min , it suffices to verify that

tE −  t ∗S , v ∗ ) ∈ Aad (0,  y0∗ , 

(2.26)

 tE −  t ∗S ≤ tE for all (0, y0 , tE , u) ∈ Aad .

(2.27)

and

y0∗ ,  tE , u∗ ) ∈ Aad , which implies that To prove (2.26), we note that ( t ∗S ,  0≤ t ∗S <  tE ,  y0∗ ∈ Y1 ,  u∗ ∈ L( t ∗S ,  tE ; U) and y( tE ;  t ∗S ,  y0∗ , u∗ ) ∈ Y2 . (2.28) By a very similar way to that used to prove (2.25), we can verify that t ∗S ; 0,  y0∗ , v ∗ ) = y( tE ;  t ∗S ,  y0∗ , u∗ ). y( tE −  This, combined with (2.28), implies (2.26). To show (2.27), we arbitrarily fix (0, y0 , tE , u) ∈ Aad . Then it follows that y0 ∈ Y1 , tE > 0, u ∈ L(0, tE ; U) and y(tE ; 0, y0 , u) ∈ Y2 .

(2.29)

tE , (2.27) is trivial. In the case that tE ∈ (0,  tE ), by a similar In the case that tE ≥  way to that used to prove (2.23), we can verify that y( tE ;  tE − tE , y0 , v) = y(tE ; 0, y0 , u), with v(·) = u(· −  tE + tE ). This, combined with (2.29), implies that tE , v) ∈ Aad , ( tE − tE , y0 ,   

Q S ,Q E which, together with the optimality of ( tE −  t ∗S ) to (T P )max , leads to that

 tE −  t ∗S ≤  tE − ( tE − tE ). Hence, (2.27) is also true in the second case. Thus, we complete the proof of Theorem 2.1.

 

46

2 Time Optimal Control Problems Q ,QE

S We next study the connection of time optimal control problems (T P )min

S ,Q E Q (T P )max

and

in the following CASE TWO:

• The operator A in (2.1) generates a C0 group on Y . Arbitrarily fix T > 0 and a nonempty subset Y1 ⊆ Y ; • The control constraint set U ⊆ U is nonempty; • The starting sets and the ending sets are as: QS = {0} × Y1 , QE ⊆ (0, T ] × Y, and    S = ( E = {T } × Y1 ; Q tS , z0 )  (T −  tS , z0 ) ∈ QE , Q Q ,QE

S • The controlled system for (T P )min

is (2.1) (which is time-varying, in  

Q S ,Q E general), while the controlled system for (T P )max is as follows:

z˙ (t) = −Az(t) − f (T − t, z(t), u(t)),

t ∈ (0, T ).

(2.30)

Write z(·; t0 , z0 , u) for the solution of (2.30) with the initial data (t0 , z0 ) ∈ [0, T ) × Y .  

Q S ,Q E S ,QE and (T P )max are In CASE TWO, the sets of admissible tetrads for (T P )Q min respectively as follows:

 Aad = {(0, y0 , tE , u)  y0 ∈ Y1 , u ∈ L(0, tE ; U), (tE , y(tE ; 0, y0 , u)) ∈ QE } (2.31) and  S , tS , z0 , T , u)  ( tS , z0 ) ∈ Q u ∈ L( tS , T ; U), z(T ;  tS , z0 , u) ∈ Y1 }. Aad = {( (2.32) One can easily check that  tS , tS , z0 , T , u)  (T −  z0 ) ∈ QE , u ∈ L( tS , T ; U), z(T ;  tS , z0 , u) ∈ Y1 }. Aad = {( (2.33) Q ,QE

S Definition 2.2 In CASE TWO, the problems (T P )min to be equivalent if the following conditions are true: S (i) If (0, y0∗ , tE∗ , u∗ ) is an optimal tetrad to (T P )min

 ,Q E Q

S and (T P )max

are said

, then (T −tE∗ , y(tE∗ ; 0, y0∗ , u∗ ),

Q ,QE

 

Q S ,Q E T , v ∗ ), with v ∗ (·) = u∗ (T − ·), is an optimal tetrad to (T P )max .   Q S ,Q E ∗ ∗ ∗ (ii) If ( t S , z0 , T , u ) is an optimal tetrad to (T P )max , then (0, z(T ;  t ∗S , z0∗ , u∗ ), Q ,Q ∗ S E T − t S , v ∗ ), with  v ∗ (·) =  u∗ (T − ·) is an optimal tetrad to (T P )min .

2.2 Connections of Minimal and Maximal Time Controls

47  

Q S ,Q E S ,QE Theorem 2.2 In CASE TWO, problems (T P )Q and (T P )max are equivamin lent.

Proof We begin with proving (i) in Definition 2.2. Assume that (0, y0∗ , tE∗ , u∗ ) is an S ,QE . In order to prove that (T −tE∗ , y(tE∗ ; 0, y0∗ , u∗ ), T , v ∗ ), optimal tetrad to (T P )Q min  ,Q E Q

S with v ∗ (·) = u∗ (T − ·), is an optimal tetrad to (T P )max

, it suffices to verify that

(T − tE∗ , y(tE∗ ; 0, y0∗ , u∗ ), T , v ∗ ) ∈ Aad

(2.34)

tS for all ( tS , z0 , T , u) ∈ Aad . tE∗ ≤ T − 

(2.35)

and

To show (2.34), we note that (0, y0∗ , tE∗ , u∗ ) ∈ Aad , which implies that y0∗ ∈ Y1 , u∗ ∈ L(0, tE∗ ; U) and (tE∗ , y(tE∗ ; 0, y0∗ , u∗ )) ∈ QE .

(2.36)

Meanwhile, since the operator A (in (2.1)) generates a C0 group, we find that y(T − t; 0, y0∗ , u∗ ) = z(t; T − tE∗ , y(tE∗ ; 0, y0∗, u∗ ), v ∗ ) for all t ∈ [T − tE∗ , T ]. Especially, we have that z(T ; T − tE∗ , y(tE∗ ; 0, y0∗ , u∗ ), v ∗ ) = y0∗ . This, combined with (2.36), implies (2.34). z0 , T , u) ∈ Aad . Then we have that To prove (2.35), we arbitrarily fix ( tS , z0 ) ∈ QE ,  u ∈ L( tS , T ; U) and z(T ;  tS , z0 , u) ∈ Y1 . (T −  tS ,

(2.37)

Let  v (·) =  u(T − ·). Since the operator A (in (2.1)) generates a C0 group, we see that z0 , u), v ) = z(T − t;  tS , z0 , u) for all t ∈ [0, T −  tS ]. y(t; 0, z(T ;  tS , This, along with (2.37), indicates that (0, z(T ;  tS , z0 , u), T −  tS , v ) ∈ Aad , S ,QE , leads to (2.35). which, together with the optimality of tE∗ to (T P )Q min We next show (ii) in Definition 2.2. Assume that ( t ∗S , z0∗ , T , u∗ ) is an optimal

 

Q S ,Q E . In order to prove that (0, z(T ;  t ∗S , z0∗ , u∗ ), T −  t ∗S , v ∗ ), with tetrad of (T P )max Q ,Q S E  v ∗ (·) =  u∗ (T − ·), is an optimal tetrad to (T P )min , it suffices to show that

(0, z(T ;  t S∗ , z0∗ , u∗ ), T −  t ∗S , v ∗ ) ∈ Aad

(2.38)

48

2 Time Optimal Control Problems

and T − t ∗S ≤ tE for all (0, y0 , tE , u) ∈ Aad .

(2.39)

z0∗ , T , u∗ ) ∈ Aad , which implies that To prove (2.38), we notice that ( t ∗S , z0∗ ) ∈ QE , u∗ ∈ L( t ∗S , T ; U), z(T ;  t ∗S , z0∗ , u∗ ) ∈ Y1 . (T −  t ∗S ,

(2.40)

At the same time, since A in (2.1) generates a C0 group, we obtain that y(t; 0, z(T ;  t ∗S , z0∗ , u∗ ), v ∗ ) = z(T − t;  t ∗S , z0∗ , u∗ ) for all t ∈ [0, T −  t ∗S ], which, combined with (2.40), leads to (2.38). To show (2.39), we let (0, y0 , tE , u) ∈ Aad . Then we have that y0 ∈ Y1 , u ∈ L(0, tE ; U), (tE , y(tE ; 0, y0 , u)) ∈ QE .

(2.41)

Write v(·) = u(T − ·). Since A in (2.1) generates a C0 group, we see that y(T − t; 0, y0 , u) = z(t; T − tE , y(tE ; 0, y0 , u), v) for all t ∈ [T − tE , T ]. This, together with (2.41), implies that (T − tE , y(tE ; 0, y0 , u), T , v) ∈ Aad ,  

Q S ,Q E which, along with the optimality of (T −  t ∗S ) to (T P )max , leads to (2.39). Thus, we complete the proof of Theorem 2.2.

 

We finally study the connection of two different minimal time control problems S ,Q E Q QS ,QE (T P )min and (T P )min in the following CASE THREE: • The controlled system is (2.1), with f (t, y, u) = D(t)y + f1 (t, u), t ∈ (0, +∞),

(2.42)

where D(·) ∈ L∞ (0, +∞; L (Y )). Write Φ(·, ·) for the evolution operator generated by A + D(·); • The control constraint set U ⊆ U is nonempty; • The starting set and the ending set are as: QS = {0}×Y1 and QE ⊆ (0, +∞)×Y , where Y1 is a nonempty subset of Y . Define    TE  tE > 0  there exists y1 ∈ Y so that (tE , y1 ) ∈ QE ,

(2.43)

and define a map YE : TE → 2Y by  YE (tE )  {y1 ∈ Y  (tE , y1 ) ∈ QE } for each tE ∈ TE ;

(2.44)

2.2 Connections of Minimal and Maximal Time Controls

49

• Arbitrarily fix  y0 ∈ Y . Take S  Q S ( Q y0 ) = {(0,  y0 )}; E ( E  Q y0 ) Q    Φ(tE , 0)(h −  y0 ) . = (tE , z1 − z2 )  tE ∈ TE , z1 ∈ YE (tE ), z2 ∈ h∈Y1

Two notes are given in order: S ,QE is not a point in the state space, in general. Can we • The set Y1 of (T P )Q min transfer this problem into a minimal time control problem, where the starting set is {(0, y0 )} for some point y0 in Y ? The answer is positive. Indeed, we will see S ,Q E Q QS ,QE that the above (T P )min and (T P )min are equivalent. • From (2.43) and (2.44), we see that TE is the projection of QE into (0, +∞) and YE (tE ) is the y-slice of QE at tE .

 

Q S ,Q E One can easily check that the admissible set of (T P )min is as:

y0 ) Aad  Aad (

 E }. tE , u)   tE > 0, u ∈ L(0,  tE ; U), ( tE , y( tE ; 0,  y0 , u)) ∈ Q = {(0,  y0 ,   (  ( Q y ),Q y )

S 0 E 0 S ,QE and (T P )min , Definition 2.3 In CASE THREE, the problems (T P )Q min with  y0 ∈ Y arbitrarily fixed, are said to be equivalent if the following conditions are true: S (i) If (0, y0∗ , tE∗ , u∗ ) is an optimal tetrad to (T P )min

Q ,QE

optimal tetrad to

S ( E ( Q y0 ),Q y0 ) (T P )min .



, then (0,  y0 , tE∗ , u∗ ) is an



QS ( y0 ),QE ( y0 ) (ii) If (0,  y0 ,  tE∗ , u∗ ) is an optimal tetrad to (T P )min , then (0, h,  tE∗ , u∗ ) is QS ,QE an optimal tetrad to (T P )min . Here, h ∈ Y1 satisfies that there is z1 ∈ YE ( tE∗ ) and z2 ∈ Y so that

y0, u∗ ) = z1 − z2 and z2 = Φ( tE∗ , 0)(h −  y0 ). y( tE∗ ; 0,   ( E ( Q y0 ),Q y0 )

S S ,QE Theorem 2.3 In CASE THREE, the problems (T P )Q and (T P )min min with  y0 ∈ Y arbitrarily fixed, are equivalent.

,

Proof Arbitrarily fix  y0 ∈ Y . We first show (i) in Definition 2.3. Assume S ,QE ∗ ∗ ∗ . In order to prove that that (0, y0 , tE , u ) is an optimal tetrad to (T P )Q min  ( E ( Q y0 ),Q y0 )

S (0,  y0 , tE∗ , u∗ ) is an optimal tetrad to (T P )min

(0,  y0 , tE∗ , u∗ ) ∈ Aad ( y0 )

, it suffices to show that (2.45)

50

2 Time Optimal Control Problems

and that tE for each (0,  y0 ,  tE , u) ∈ Aad ( y0 ). tE∗ ≤ 

(2.46)

To prove (2.45), we use the assumption that (0, y0∗ , tE∗ , u∗ ) ∈ Aad to get that y0∗ ∈ Y1 , tE∗ > 0, u∗ ∈ L(0, tE∗ ; U), tE∗ ∈ TE and y(tE∗ ; 0, y0∗ , u∗ ) ∈ YE (tE∗ ). (2.47) Meanwhile, because of (2.42), we find that y0 , u∗ ) = y(tE∗ ; 0, y0∗, u∗ ) − y(tE∗ ; 0, y0∗ −  y0 , 0) y(tE∗ ; 0,  y0 ). = y(tE∗ ; 0, y0∗, u∗ ) − Φ(tE∗ , 0)(y0∗ − 

(2.48)

This, together with (2.47), implies (2.45). To prove (2.46), we arbitrarily fix (0,  y0 ,  tE , u) ∈ Aad ( y0 ). Then we have that  u ∈ L(0,  tE ; U),  tE ∈ TE , tE > 0, 

(2.49)

and that y0 , u) = z1 − z2 , z1 ∈ YE ( tE ), z2 = Φ( tE , 0)(h −  y0 ) and h ∈ Y1 . y( tE ; 0,  (2.50) By the same way as that used to prove (2.48), we can verify that y0 , u) = y( tE ; 0, h, u) − Φ( tE , 0)(h −  y0 ). y( tE ; 0,  This, along with (2.49) and (2.50), yields that y( tE ; 0, h, u) = z1 ∈ YE ( tE ) and u) ∈ Aad . (0, h,  tE ,

(2.51)

S E , (2.46) follows at once. By (2.51) and the optimality of tE∗ to (T P )min We next show (ii) in Definition 2.3. Assume that (0,  y0 ,  tE∗ , u∗ ) is an optimal

Q ,Q

 ( E ( Q y0 ),Q y0 )

S tetrad to (T P )min we have that

. Let h, z1 and z2 be given in (ii) of Definition 2.3. Then

 tE∗ > 0 and  u∗ ∈ L(0,  tE∗ ; U)

(2.52)

2.2 Connections of Minimal and Maximal Time Controls

51

and that y( tE∗ ; 0,  y0, u∗ ) = z1 − z2 , z1 ∈ YE ( tE∗ ), z2 = Φ( tE∗ , 0)(h −  y0 ) and h ∈ Y1 . (2.53) QS ,QE In order to show that (0, h,  tE∗ , u∗ ) is an optimal tetrad to the problem (T P )min , it suffices to verify that u∗ ) ∈ Aad (0, h,  tE∗ ,

(2.54)

 tE∗ ≤ tE for each tetrad (0, y0 , tE , u) ∈ Aad .

(2.55)

and

We first show (2.54). By (2.53) and by the same way as that used to prove (2.48), we can easily verify that tE∗ ; 0,  y0, u∗ ) + Φ( tE∗ , 0)(h −  y0 ) = y( tE∗ ; 0, h, u∗ ) z1 = y( and tE∗ ; 0, h, u∗ )) ∈ QE . ( tE∗ , y(

(2.56)

Since h ∈ Y1 , (2.54) follows from (2.52) and (2.56) at once. We next show (2.55). Arbitrarily fix (0, y0 , tE , u) ∈ Aad . Then we have that y0 ∈ Y1 , tE > 0, u ∈ L(0, tE ; U) and (tE , y(tE ; 0, y0 , u)) ∈ QE .

(2.57)

From (2.57), we see that tE ∈ TE and y(tE ; 0, y0 , u) ∈ YE (tE ).

(2.58)

Then, by a very similar way to that used to show (2.48), we can verify that y0, u) = y(tE ; 0, y0 , u) − Φ(tE , 0)(y0 −  y0 ). y(tE ; 0, 

(2.59)

It follows from (2.57), (2.58), and (2.59) that y0 ). (0,  y0 , tE , u) ∈ Aad (  ( E ( Q y0 ),Q y0 )

S This, along with the optimality of  tE∗ to (T P )min Hence, we end the proof of Theorem 2.3.

, leads to (2.55).  

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2 Time Optimal Control Problems

2.3 Several Examples In this section, we will present several examples of time optimal control problems. Example 2.1 ([23]) How can we reach the top fastest by an elevator? In this problem, the controlled system is as: d2 y(t) = u(t) − g, t ∈ (0, +∞), dt 2

(2.60)

where y(t) stands for the height of the mass at time t, u(t) is the force of a unit mass given by an elevator at time t, and g is the acceleration of gravity. We can rewrite (2.60) as the linear equation: d dt

$

y1 (t) y2 (t)

%

$ =

01 00

%$

% $ % $ % 0 y1 (t) 0 + u(t) + , t ∈ (0, +∞). y2 (t) 1 −g

(2.61)

Take the control constraint set U as [−20, 20]; the initial time as tS = 0; the initial state as: (y1 (0), y2 (0)) = (0, 0) ; the terminal condition as (y1 (tE ), y2 (tE )) = (h, 0) , with a fixed h > 0. Then we can put this problem into our framework S ,QE (T P )Q in the following manner: The controlled system is (2.61); min $ % $ % 0 h U = [−20, 20]; QS = {0} × ; QE = (0, +∞) × . 0 0 We now end Example 2.1. Example 2.2 ([37]) Consider the undamped harmonic oscillator d dt

$

y1 (t) y2 (t)

%

$ =

0 1 −1 0

%$

y1 (t) y2 (t)

% +

$ % 0 u(t), t ∈ (0, +∞), 1

(2.62)

where y1 is the displacement from its equilibrium position and y2 is its velocity. Let U = [−1, 1]. We are asked for a control (with the constraint U) driving an initial state (y0 , 0) at time t = 0 to a target (yE , 0) in the shortest time. This problem S ,QE can be put into our framework (T P )Q in the following manner: The controlled min system is (2.62); $ U = [−1, 1]; QS = {0} ×

y0 0

$

% ;

QE = (0, +∞) ×

yE 0

% .

We end Example 2.2. Example 2.3 ([19]) Observations of children playing on swings show that children who are good at this task all follow a similar strategy. Making the reasonable assumption that the objective is to get as high as possible and as quickly as possible,

2.3 Several Examples

53

the strategy used by children is to crouch when the swing is at its highest point and to stand-up when the swing passes its lowest point. This is done both on the forward and return cycle. To understand the physics behind this pumping, we shall make some simplified assumptions. The swing is ideally modeled as a pendulum. The rider is modeled as a point mass, and the variable distance of the center of mass of the rider from the fixed support is denoted by l(t). The angle that the swing makes with the vertical is denoted by θ (t). Conservation of angular momentum for a point mass undergoing planar motion gives: d (l(t)2 θ˙ (t)) = −gl(t) sin(θ (t)), t ∈ (0, +∞). dt We rewrite the above equation as the following system of ordinary differential equations: ⎧ ⎨ z˙ 1 = z2 , t ∈ (0, +∞), ˙ g sin(z1 ) 2lz ⎩ z˙ 2 = − 2 − , t ∈ (0, +∞). l l Now by making the substitution x2 = z2 l 2 and rewriting z1 by x1 , we have that  x2 x˙1 = 2 , t ∈ (0, +∞), l x˙2 = −gl sin(x1 ), t ∈ (0, +∞). Let us now make the following approximation: l(t) = L(1 + εu(t)), with L > 0, 0 < ε 0, u(t) ∈ [0, M] for all t ∈ (0, +∞). In order to pay off the loan on the schedule, when will he have to begin working? This problem can be put into the QS ,QE in the following manner: The controlled system is (2.63); framework (T P )max U = [0, M]; QS = [0, 1) × {y0 }; QE = {1} × [10000, +∞). Now we end Example 2.4. Example 2.5 ([7, 44]) We consider a solid fuel model. The system is governed by canonical equations from combustion theory such as the nonstationary semilinear one-dimensional Frank-Kamenetskii equation: ⎧ (x, t) ∈ (0, 1) × (0, +∞), ⎨ yt = yxx + ey(x,t ), y(0, t) = y(1, t) = 0, t ∈ (0, +∞), ⎩ x ∈ (0, 1), y(x, 0) = y0 (x),

(2.64)

where y stands for the temperature. A force enters the first equation of (2.64) in the following way: yt = yxx + ey(x,t ) + χ(1/3,1/2)(x)u(x, t), (x, t) ∈ (0, 1) × (0, +∞), where u satisfies that u(t)L2 (0,1) ≤ 1 for a.e. t > 0. The problem now is to find a control, with the aforementioned constraint, so that the corresponding solution explodes as soon as possible. This problem can be put into our framework (2.17). We end Example 2.5.

2.4 Main Subjects on Time Optimal Control Problems

55

2.4 Main Subjects on Time Optimal Control Problems This section presents main subjects (on time optimal control problems) which will be studied in this monograph. They are the existence of admissible controls and optimal controls; the Pontryagin Maximum Principle for optimal controls; the bang-bang property of optimal controls; and connections of time optimal control problems with other kinds of optimal control problems, such as minimal norm control problems and optimal target problems. We begin with introducing the existence of admissible controls and optimal controls. Problem (TP) can be treated as an optimization problem, where the admissible set Aad is the set of available alternatives. When Aad = ∅, this problem does not make sense. Hence, one of the main purposes in the studies of Problem (TP) is to prove that Aad = ∅, which is the existence of admissible controls. It connects with some controllability of controlled equations. Both are to ask for a control driving the corresponding solution of a controlled equation from an initial set to a target set. Their main differences are as follows: First, in general, the controllability is to ask for a control (without any constraint) driving the corresponding solution of a controlled system from an initial set to a target in a fixed time interval, while the existence of admissible controls is to ask for a control (in a constraint set), which drives the corresponding solution of a controlled system from an initial set to a target in an unfixed time interval. Second, the aim of most controllability problems is to drive the solution of a controlled system from any initial state y0 ∈ Y to a given target, with the aid of controls, while in most time optimal control problems, the starting set is fixed. The next example shows that for some time optimal control problems, admissible sets can be empty. Q ,QE

S Example 2.6 Consider the time optimal control problem (T P )min controlled system is as:

, where the

y(t) ˙ = y(t) + u(t), t ∈ (0, +∞), and where U = [−1, 1]; QS = {(0, 1)}; QE = (0, +∞) × {0}. One can easily check that for any u ∈ L(0, +∞; U), y(t; 0, 1, u|(0,t )) ≥ 1 for all t ∈ (0, +∞). (Here, u|(0,t ) denotes the restriction of u over (0, t).) Hence, the corresponding admissible set Aad = ∅. We now end Example 2.6. In many cases, the existence of optimal controls can be implied by the existence of admissible controls, as well as some properties on the controlled equations. This

56

2 Time Optimal Control Problems

will be explained in the next chapter. A natural question is as follows: When Aad = ∅, is it necessary that the corresponding time optimal control problem has an optimal control? The answer is negative. This can be seen from the following example: Q ,QE

S Example 2.7 Consider the time optimal control problem (T P )min controlled system is as:

, where the

y(t) ˙ = u(t), t ∈ (0, +∞), and where U = R;

QS = {(0, 1)}; QE = (0, +∞) × {0}.

For any ε > 0, we define a control uε over (0, ε) by 1 uε (t) = − , t ∈ (0, ε). ε Then one can directly check that for each ε > 0, (0, 1, ε, uε ) is an admissible tetrad S ,QE , i.e., for the above problem (T P )Q min (0, 1, ε, uε ) ∈ Aad ,

(2.65)

which leads to that Aad = ∅. On the other hand, from (2.65), we see that the infimum in the corresponding time optimal control problem cannot be reached. QS ,QE Hence, the above problem (T P )min has no optimal control. We end Example 2.7. We now turn to the Pontryagin Maximum Principle (or Pontryagin’s maximum principle). In general, the standard Pontryagin Maximum Principle for Problem (TP) is stated as follows: If u∗ is an optimal control (associated with an optimal tetrade (tS∗ , y0∗ , tE∗ , u∗ )) to Problem (TP), then there is a vector h ∈ Y with h = 0 so that ϕ(t), f (t, y ∗ (t), u∗ (t)) = maxϕ(t), f (t, y ∗ (t), u) for a.e. t ∈ (tS∗ , tE∗ ), u∈U

(2.66) where

y ∗ (·)

is the optimal state given by y ∗ (t) = y(t; tS∗ , y0∗ , u∗ ), t ∈ (tS∗ , tE∗ ),

and where ϕ(·) is the solution to the following adjoint equation (or co-state equation): 

ϕ(t) ˙ = −A∗ ϕ(t) − fy (t, y ∗ (t), u∗ (t))∗ ϕ(t), ϕ(tE∗ ) = h.

t ∈ (tS∗ , tE∗ ),

(2.67)

2.4 Main Subjects on Time Optimal Control Problems

57

Several notes on the standard Pontryagin Maximum Principle are given in order: • This principle may not hold for some time optimal control problems, in particular, in infinitely dimensional settings. We will provide such a counterexample in Chapter 4. We also refer readers to [24] and [11] for such counterexamples. • This principle has a weaker version, where h may not be a vector in the state space Y , but the Equation (2.67) is still well posed in some sense and (2.66) still holds (see, for instance, [11, 41] and [42]). • In general, there are two ways to derive Pontryagin’s maximum principle for time optimal control problems. One is a geometric way which will be introduced in Chapter 4 of this monograph. (We also refer the readers to [35] and [11] for this way.) Another is an analytical way. We will not introduce it in our monograph. We refer readers to books [24] and [3] for this method. We next explain the bang-bang property. An optimal control u∗ (associated with an optimal tetrad (tS∗ , y0∗ , tE∗ , u∗ )) of Problem (TP) is said to have the bang-bang property if u∗ (t) ∈ ∂U for a.e. t ∈ (tS∗ , tE∗ ).

(2.68)

From geometric point of view, it says that the optimal control u∗ takes its value on the boundary of the constraint set U at almost every time in the domain of u∗ . Problem (TP) is said to have the bang-bang property if any optimal control has the bang-bang property. So the bang-bang property is indeed a property of optimal controls. This property is not only important from perspective of applications, but also very interesting from perspective of mathematics. In some cases, from the bangbang property of Problem (TP), one can easily get the uniqueness of the optimal control to this problem. The bang-bang property may help us to do better numerical analyses and algorithm on optimal controls in some cases (see, for instance, [20, 21] and [31]). We now are going to explain how the bang-bang property implies the uniqueness of the optimal control. Definition 2.4 A convex subset Z of a Banach space X is said to be strictly convex, provided that for any two points z1 and z2 on ∂Z (the boundary of Z in X), the whole line segment connecting z1 and z2 meets ∂Z only at z1 and z2 . Theorem 2.4 Let U be strictly convex and closed. Let QS and QE be convex and closed subsets of (0, +∞) × Y . Suppose that the minimal time control problem QS ,QE QS ,QE (T P )min (or the maximal time control problem (T P )max ), associated with the controlled equation (2.11), has the bang-bang property. Then the optimal control to QS ,QE QS ,QE (T P )min (or (T P )max ) is unique. Q ,Q

S E . The Proof We only prove the uniqueness of the optimal control to (T P )min QS ,QE uniqueness of the optimal control to (T P )max can be verified by the same way. QS ,QE y0∗ ,  tE∗ , u∗ ) be two optimal tetrads to (T P )min . It is Let (tS , y0∗ , tE∗ , u∗ ) and (tS ,  obvious that

58

2 Time Optimal Control Problems

tS < tE∗ =  tE∗ , u∗ ∈ L(tS , tE∗ ; U),  u∗ ∈ L(tS , tE∗ ; U), y0∗ ) ∈ QS , (tS , y0∗ ), (tS , 

(2.69) (2.70)

and that (tE∗ , y(tE∗ ; tS , y0∗ , u∗ )) ∈ QE , (tE∗ , y(tE∗ ; tS ,  y0∗ , u∗ )) ∈ QE .

(2.71)

We claim that $ % y0∗ +  y0∗ ∗ u∗ +  u∗ , tE , tS , ∈ Aad . 2 2

(2.72)

Indeed, it follows from the convexity of QS and U, (2.69) and (2.70) that $

y∗ +  y0∗ tS , 0 2

% ∈ QS and

u∗ u∗ +  ∈ L(tS , tE∗ ; U). 2

(2.73)

Meanwhile, the linearity of the Equation (2.11) implies that $ % y∗ +  y0∗ u∗ +  u∗ 1 1 , y tE∗ ; tS , 0 y0∗ , u∗ ), = y(tE∗ ; tS , y0∗ , u∗ ) + y(tE∗ ; tS ,  2 2 2 2 which, combined with the convexity of QE and (2.71), indicates that $ $ %% y∗ +  y0∗ u∗ +  u∗ , tE∗ , y tE∗ ; tS , 0 ∈ QE . 2 2

(2.74)

Since tS < tE∗ , (2.72) follows from (2.73) and (2.74) at once. QS ,QE Because (tE∗ − tS ) is the optimal time of (T P )min , we obtain from (2.72) that the following tetrad: y0∗ )/2, tE∗ , (u∗ +  u∗ )/2) (tS , (y0∗ +  S ,QE is also an optimal tetrad of the problem (T P )Q . Then, by the bang-bang min property of this problem, we have that

u∗ (t) ∈ ∂U,  u∗ (t) ∈ ∂U and (u∗ +  u∗ )/2 ∈ ∂U for a.e. t ∈ (tS , tE∗ ). From this and the strict convexity of U, we can easily verify that u∗ (t) for a.e. t ∈ (tS , tE∗ ). u∗ (t) =  This completes the proof of Theorem 2.4.

 

2.4 Main Subjects on Time Optimal Control Problems

59

The next example presents a time optimal control problem, which holds the Pontryagin Maximum Principle, but does not hold the bang-bang property. Q ,QE

S Example 2.8 Consider the time optimal control problem (T P )min controlled equation is as:

d dt

$

y1 (t) y2 (t)

and where U = [−1, 1], QS = {0} ×

% =

, where the

⎧$ % ⎪ 1 ⎪ ⎪ u(t), t ∈ [0, 1), ⎪ ⎪ ⎨ 0 $ % ⎪ ⎪ ⎪ 0 ⎪ ⎪ u(t), t ∈ [1, +∞), ⎩ 1

$ % $ % 0 0 and QE = (0, +∞) × . 1 0

(2.75)

(2.76)

By the second equality in (2.76) and by (2.75), after some computations, we get that ⎧ t ⎪ ⎪ ⎨ u(s)ds for all t ∈ [0, 1), (2.77) y1 (t) = 0 1 ⎪ ⎪ ⎩ u(s)ds for all t ≥ 1 0

and y2 (t) =

⎧ ⎨1



t

⎩1 +

for all t ∈ [0, 1), u(s)ds for all t ≥ 1.

(2.78)

1

We first show that this problem holds the Pontryagin Maximum Principle. To this end, we let (tS∗ , y0∗ , tE∗ , u∗ ) be an optimal tetrad to this problem. Then, by (2.76), (2.77), and (2.78), we obtain that $ %  1  t∗ E 0 tS∗ = 0; y0∗ = u∗ (t)dt = 0 and 1 + u∗ (t)dt = 0. ; tE∗ > 1, 1 0 1 (2.79) From the first equality in (2.76) and by (2.79), we see that  1 tE∗ = 2, u∗ (t)dt = 0, and u∗ (t) = −1 for a.e. t ∈ (1, 2). 0

Hence, we have that % $ $ % 0 (tS∗ , y0∗ , tE∗ , u∗ ) = 0, , 2, u∗ . 1

(2.80)

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2 Time Optimal Control Problems

Consider the equation: % $ % ⎧ $ d ϕ1 (t) 0 ⎪ ⎪ = , ⎨ ϕ2 (t) 0 dt $ % $ % ⎪ ϕ1 (2) 0 ⎪ ⎩ = . ϕ2 (2) −1

t ∈ (0, 2) (2.81)

From$(2.75), % one can easily check that (2.81) is the dual equation (2.67), with 0 h= , corresponding to the above optimal tetrad. By solving (2.81), we obtain −1 that ϕ1 (t) = 0, ϕ2 (t) = −1 for all t ∈ [0, 2].

(2.82)

Then, by the third equality in (2.80) and by (2.82), we can easily check that &$ &$ % $ ∗ %' % $ %' ϕ1 (t) ϕ1 (t) u (t) u = max , , for a.e. t ∈ (0, 1), ϕ2 (t) 0 ϕ2 (t) u∈[−1,1] 0 and

&$

ϕ1 (t) ϕ2 (t)

&$ % $ %' % $ %' ϕ1 (t) 0 0 , = max , for a.e. t ∈ (1, 2). ϕ2 (t) u∗ (t) u u∈[−1,1]

From (2.81), (2.75), and (2.76), we see that the above two equalities are exactly the equality (2.66) corresponding to the above optimal tetrad (tS∗ , y0∗ , tE∗ , u∗ ) and the dual equation (2.81). Since (tS∗ , y0∗ , tE∗ , u∗ ) was assumed to be an arbitrary S ,QE optimal tetrad, we see that this time optimal control problem (T P )Q holds the min Pontryagin Maximum Principle. We next show that this problem has no bang-bang property. To this end, we take  0, t ∈ [0, 1), v ∗ (t) = −1, t ∈ [1, 2]. % $ $ % 0 One can directly check that 0, , 2, v ∗ is an optimal tetrad to the problem 1 QS ,QE ∗ (T P )min . It is clear that v does not hold the bang-bang property. Hence, the QS ,QE problem (T P )min does not have the bang-bang property. Now we end Example 2.8. We end this section with introducing briefly connections between time optimal control problems with other kind of optimal control problems. Among all kinds of optimal control problems, the time optimal control problem is one of the most QS ,QE QS ,QE difficult problems. In some cases, Problem (T P )min (or Problem (T P )max ) is equivalent to a norm optimal control problem (or a target optimal problem, as

2.4 Main Subjects on Time Optimal Control Problems

61

well as a norm optimal control problem). The aforementioned norm optimal control problems and target optimal control problems will be introduced in Chapter 5. In many cases, the studies of the above-mentioned norm optimal control problems (or target optimal control problems) are easier than those of time optimal control problems. Thus, if we can build up connections between time optimal control problems with norm optimal control problems (or target optimal control problems), then we could get desired information on time optimal control problems through studies of norm optimal control problems (or target optimal control problems).

Miscellaneous Notes The studies on time optimal control problem for ordinary differential equations can be dated back to 1950s (see, for instance, [5, 6] and [35]). Then such studies extended to infinitely dimensional cases in 1960s (see, for instance, [9] and [10]). In [35], the problem is stated in the following manner: Given two points x0 and x1 in the state space, a control constraint set U, a controlled ordinary differential equation (which is called a controlled system) and a functional f 0 of states and controls, we ask for an admissible control u (i.e., it drives the corresponding solution of the system from x(t0 ) = x0 to x(t1 ) = x1 for some t0 < t1 ) so that it minimizes the functional:  t1 J = f 0 (x(t), u(t))dt. t0

Notice that the above t0 and t1 are not fixed and depend on admissible controls. When f 0 ≡ 1, the aforementioned problem can be put into our framework. From this point of view, our monograph focuses on a special kind of time optimal control problems. But this kind of problems is extremely important. It contains the minimal time control problems and the maximal time control problems. There have been many literatures studying the minimal time control problems (see, for instance, [2, 3, 8, 12–15, 17, 20, 27, 32, 36] and [39]). About studies on the maximal time control problems, we refer the readers to [33, 34] and [38]. Though many years have passed, the study on the time optimal controls, in particular, for infinite dimensional cases, is not very mature. There are still many interesting open problems. The differential equations whose solutions have the behavior of blowup replicate a large class of phenomena in applied science (see [16]). In [28], an optimal control problem governed by the following differential equation was studied: ∂t y − Δy = y λ + u in Ω × (0, T ),

(2.83)

where Ω is a bounded domain in Rd , T > 0 is a fixed constant, and λ = 2, 3. For a given control u, solutions to the above equation may blow up. To deal with such kind of optimal control problem, the author introduced the concept of admissible pair: the pair (u, y) is called as an admissible one if y exists globally.

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Time Optimal Control Problems

The restriction of global existence of y precludes the behavior of blowup. The same method was used in [1]. For the controlled equation (2.83), an interesting problem is as: how to find a control, which is supported on a small ball of Ω, to eliminate the blowup? On the other hand, the blowup of a solution is desirable in certain cases. For example, one may desire to end a chemical reaction as fast as possible with the aid of some catalyst. The blowup of the solution might be thought of the end of a chemical reaction, while the control might play the role of a catalyst. The minimal blowup time control problem introduced at the end of Section 2.1 was first studied for controlled ODEs in [25], where the existence of optimal controls and Pontryagin’s maximum principle were obtained. Then it was studied for PDEs in [26], where the existence of optimal controls was obtained. However, the corresponding Pontryagin’s maximum principle was unsolved. This might be an interesting problem. Before [25], a maximal blowup time control problem was studied in [4], where a necessary condition for an optimal control was derived by the dynamic programming method. To our surprise, the published papers on minimal (or maximal) blowup time control problems are very limited (see, for instance, [29] and [30]). We think that this direction has bright development prospect. The numerical analysis on time optimal controls and optimal time is an interesting topic in the field of time optimal control problems. However, we do not touch it in this monograph since it is little bit beyond the scope of this book. We would like to refer readers to [13, 18, 20–22, 31, 40], and [43] for this subject. An interesting problem is what happens for bang-bang control from the numerical perspective. Several issues on time optimal control problems deserve to be studied: • • • •

Numerical analysis on time optimal control problems. Time optimal control problems with noise in some data. Stability of time optimal controls with respect to perturbations of systems. Properties of the Bellman function, i.e., regarding the optimal time as a function of the initial data (in the case that the starting set is a singleton set). • Time optimal control problems for some concrete controlled systems, such as affine controlled ODEs, differential equations with delay, impulsive differential equations, differential equations with sampled-data controls or impulse controls, wave equation, and wave-like equations.

References 1. H. Amann, P. Quittner, Optimal control problems with final observation governed by explosive parabolic equations. SIAM J. Control Optim. 44, 1215–1238 (2005) 2. V. Barbu, The dynamic programming equation for the time-optimal control problem in infinite dimensions. SIAM J. Control Optim. 29, 445–456 (1991) 3. V. Barbu, Analysis and Control of Nonlinear Infinite Dimensional Systems. Mathematics in Science and Engineering, vol. 190 (Academic, Boston, MA, 1993) 4. E.N. Barron, W.X. Liu, Optimal control of the blowup time. SIAM J. Control Optim. 34, 102–123 (1996)

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5. R. Bellman, I. Glicksberg, O. Gross, On the “bang-bang” control problem. Q. Appl. Math. 14, 11–18 (1956) 6. V.G. Boltyanskiˇi, R.V. Gamkrelidz, L.S. Pontryagin, On the theory of optimal processes (Russian). Dokl. Akad. Nauk. SSSR 110, 7–10 (1956) 7. C.J. Budd, V.A. Galaktionov, J.F. Williams, Self-similar blow-up in higher-order semilinear parabolic equations. SIAM J. Appl. Math. 64, 1775–1809 (2004) 8. O. Cârjˇa, The minimal time function in infinite dimensions. SIAM J. Control Optim. 31, 1103–1114 (1993) 9. J.V. Egorov, Optimal control in Banach spaces (Russian). Dokl. Akad. Nauk SSSR 150, 241–244 (1963) 10. H.O. Fattorini, Time-optimal control of solution of operational differential equations. J. SIAM Control 2, 54–59 (1964) 11. H.O. Fattorini, Vanishing of the costate in Pontryagin’s maximum principle and singular time optimal controls. J. Evol. Equ. 4, 99–123 (2004) 12. H.O. Fattorini, Infinite Dimensional Linear Control Systems, the Time Optimal and Norm Optimal Problems. North-Holland Mathematics Studies, vol. 201 (Elsevier Science B.V., Amsterdam, 2005) 13. W. Gong, N. Yan, Finite element method and its error estimates for the time optimal controls of heat equation. Int. J. Numer. Anal. Model. 13, 265–279 (2016) 14. F. Gozzi, P. Loreti, Regularity of the minimum time function and minimum energy problems: the linear case. SIAM J. Control Optim. 37, 1195–1221 (1999) 15. B. Guo, D. Yang, Optimal actuator location for time and norm optimal control of null controllable heat equation. Math. Control Signals Syst. 27, 23–48 (2015) 16. B. Hu, Blow-Up Theories for Semilinear Parabolic Equations. Lecture Notes in Mathematics (Springer, Heidelberg, 2011) 17. K. Ito, K. Kunisch, Semismooth newton methods for time-optimal control for a class of ODEs. SIAM J. Control Optim. 48, 3997–4013 (2010) 18. G. Knowles, Finite element approximation of parabolic time optimal control problem. SIAM J. Control Optim. 20, 414–427 (1982) 19. J.E. Kulkarni, Time-optimal control of a swing. https://courses.cit.cornell.edu/ee476/ideas/ swing.pdf 20. K. Kunisch, D. Wachsmuth, On time optimal control of the wave equation, its regularization and optimality system. ESAIM Control Optim. Calc. Var. 19, 317–336 (2013) 21. K. Kunisch, D. Wachsmuth, On time optimal control of the wave equation and its numerical realization as parametric optimization problem. SIAM J. Control Optim. 51, 1232–1262 (2013) 22. I. Lasiecka, Ritz-Galerkin approximation of the time optimal boundary control problem for parabolic systems with Dirichlet boundary conditions. SIAM J. Control Optim. 22, 477–500 (1984) 23. D. Levy, M. Yadin, A. Alexandrovitz, Optimal control of elevators. Int. J. Syst. Sci. 8, 301–320 (1977) 24. X. Li, J. Yong, Optimal Control Theory for Infinite-Dimensional Systems. Systems & Control: Foundations & Applications (Birkhäuser Boston, Boston, MA, 1995) 25. P. Lin, G. Wang, Blowup time optimal control for ordinary differential equations. SIAM J. Control Optim. 49, 73–105 (2011) 26. P. Lin, G. Wang, Some properties for blowup parabolic equations and their application. J. Math. Pures Appl. 101, 223–255 (2014) 27. J.-L. Lions, Optimisation pour certaines classes d’équations d’évolution non linéaires (French). Ann. Mat. Pura Appl. 72, 275–293 (1966) 28. J.-L. Lions, Contrôle des systèmes distribués singuliers. Méthodes Mathématiques de l’Informatique, vol. 13 (Gauthier-Villars, Montrouge, 1983) 29. H. Lou, W. Wang, Optimal blowup/quenching time for controlled autonomous ordinary differential equation. Math. Control Relat. Fields 5, 517–527 (2015) 30. H. Lou, W. Wang, Optimal blowup time for controlled ordinary differential equations. ESAIM Control Optim. Calc. Var. 21, 815–834 (2015)

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31. X. Lü, L. Wang, Q. Yan, Computation of time optimal control problems governed by linear ordinary differential equations. J. Sci. Comput. 73, 1–25 (2017) 32. S. Micu, I. Roventa, M. Tucsnak, Time optimal boundary controls for the heat equation. J. Funct. Anal. 263, 25–49 (2012) 33. V.J. Mizel, T.I. Seidman, An abstract bang-bang principle and time optimal boundary control of the heat equation. SIAM J. Control Optim. 35, 1204–1216 (1997) 34. K.D. Phung, G. Wang, An observability estimate for parabolic equations from a measurable set in time and its application. J. Eur. Math. Soc. 15, 681–703 (2013) 35. L.S. Pontryagin, V.G. Boltyanskii, R.V. Gamkrelidze, E.F. Mischenko, The Mathematical Theory of Optimal Processes, Translated from the Russian by K. N. Trirogoff, ed. by L.W. Neustadt (Interscience Publishers Wiley, New York, London, 1962) 36. J.P. Raymond, H. Zidani, Time optimal problems with boundary controls. Differ. Integral Equ. 13, 1039–1072 (2000) 37. E.D. Sontag, Mathematical Control Theory. Deterministic Finite-Dimensional Systems, 2nd edn. Texts in Applied Mathematics, vol. 6 (Springer, New York, 1998) 38. G. Wang, Y. Xu, Equivalence of three different kinds of optimal control problems for heat equations and its applications. SIAM J. Control Optim. 51, 848–880 (2013) 39. G. Wang, Y. Zhang, Decompositions and bang-bang problems. Math. Control Relat. Fields 7, 73–170 (2017) 40. G. Wang, G. Zheng, An approach to the optimal time for a time optimal control problem of an internally controlled heat equation. SIAM J. Control Optim. 50, 601–628 (2012) 41. G. Wang, E. Zuazua, On the equivalence of minimal time and minimal norm controls for the internally controlled heat equations. SIAM J. Control Optim. 50, 2938–2958 (2012) 42. G. Wang, Y. Xu, Y. Zhang, Attainable subspaces and the bang-bang property of time optimal controls for heat equations. SIAM J. Control Optim. 53, 592–621 (2015) 43. G. Wang, D. Yang, Y. Zhang, Time optimal sampled-data controls for the heat equation. C. R. Math. Acad. Sci. Paris 355, 1252–1290 (2017) 44. Y.B. Zel’dovich, G.I. Barenblatt, V.B. Librovich, G.M. Makhviladze, The mathematical theory of combustion and explosions, Translated from the Russian by Donald H. McNeill (Consultants Bureau [Plenum], New York, 1985)

Chapter 3

Existence of Admissible Controls and Optimal Controls

In this chapter, we will study the existence of admissible controls and optimal controls for Problem (TP) (given by (2.4) in Chapter 1) for some special cases. Since minimal time control problems and maximal time control problems are two kinds of most important time optimal control problems and because these two kinds of problems can be mutually transformed in many cases (see Theorem 2.1, Theorem 2.2, and Theorem 2.3), we will only study minimal time control problems S ,QE throughout this chapter. More precisely, we will focus on the problem (T P )Q min under the following framework (AT P ): (i) The state space Y and the control space U are real separable Hilbert spaces. (ii) The controlled system (or the state system) is the Equation (2.1), i.e., y(t) ˙ = Ay(t) + f (t, y(t), u(t)),

t ∈ (0, +∞),

where A : D(A) ⊆ Y → Y generates a C0 semigroup {eAt }t ≥0 on Y and f : (0, +∞) × Y × U → Y is a given map. Moreover, for any 0 ≤ a < b < +∞, y0 ∈ Y and u(·) ∈ L(a, b; U), this equation, with the initial condition y(a) = y0 , has a unique mild solution y(·; a, y0, u) on [a, b]. (iii) The starting set is as: QS = {0} × YS with YS a nonempty subset of Y . (iv) The ending set QE is a nonempty subset of (0, +∞) × Y . (v) The control constraint set U is a nonempty subset of U . The main part of this chapter is devoted to the existence of admissible controls S ,QE for the problem (T P )Q . We will study the above existence from three different min viewpoints: the controllability, the minimal norm, and reachable sets. At the last section of this chapter, we introduce ways deriving the existence of optimal controls from the existence of admissible controls. In each section of this chapter (except for the last one), we will first present our main theorem in an abstract setting which can be embedded into the framework of this book (see the beginning of Chapter 1), then apply the main theorem to some examples. © Springer International Publishing AG, part of Springer Nature 2018 G. Wang et al., Time Optimal Control of Evolution Equations, Progress in Nonlinear Differential Equations and Their Applications 92, https://doi.org/10.1007/978-3-319-95363-2_3

65

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3 Existence of Admissible Controls and Optimal Controls

3.1 Admissible Controls and Controllability Q ,QE

S In this section, we will discuss the existence of admissible controls of (T P )min from the perspective of controllability.

3.1.1 General Results in Abstract Setting With the aid of (AT P ) (given at the beginning of Chapter 3), we can define the following two sets:    Q ,U YC E (t)  y0 ∈ Y  ∃ u ∈ L(0, t; U) s.t. (t, y(t; 0, y0, u)) ∈ QE , t ∈ TE ; (3.1) 

YCQE ,U 

YCQE ,U (t).

(3.2)

t ∈TE

Here, TE was defined by (2.43). We call respectively YCQE ,U (t) and YCQE ,U the controllable set at time t and the controllable set for the Equation (2.1), associated with QE and U. The main result of this subsection is as: Q ,QE

S Theorem 3.1 The problem (T P )min  QE ,U YS YC = ∅.

has an admissible control if and only if

Q ,Q

S E has an admissible control u (associated with an Proof Suppose that (T P )min admissible tetrad (tS , y0 , tE , u)). Then we have that

(tS , y0 , tE , u) ∈ Aad , where Aad is defined by (2.3). From (2.3), (iii) of (AT P ) and (2.43), we have that tS = 0, y0 ∈ YS , tE ∈ TE , u ∈ L(0, tE ; U) and (tE , y(tE ; 0, y0 , u)) ∈ QE . These, along with (3.1) and (3.2), yield that y0 ∈ YS



Q ,U

YC E (tE ) ⊆ YS

Conversely, we assume that y0 ∈ YS so that Q ,U





Q ,U

YC E .

YCQE ,U . Then by (3.2), there is tE ∈ TE

y0 ∈ YC E (tE ) and y0 ∈ YS .

3.1 Admissible Controls and Controllability

67

These, along with (iii) of (AT P ) and (3.1), yield that (0, y0 ) ∈ QS ; and (tE , y(tE ; 0, y0, u)) ∈ QE for some u ∈ L(0, tE ; U). From these and (2.3), we find that (0, y0 , tE , u) ∈ Aad . Hence, u is an admissible S ,QE control of (T P )Q . This ends the proof of Theorem 3.1.   min Two facts deserve to be mentioned: First, both YCQE ,U (t) (with t ∈ TE ) and YCQE ,U are independent of QS . Second, it is hard to characterize controllable Q ,U Q ,U sets YC E (t) and YC E , in general. In the next subsection, we will discuss the controllable sets for some linear systems.

3.1.2 Linear ODEs with Ball-Type Control Constraints In this subsection, we will use Theorem 3.1 to study the existence of some minimal time control problems, where the controlled system is as: y(t) ˙ = A(t)y(t) + B(t)u(t), t ∈ (0, +∞),

(3.3)

n×n ) and B(·) ∈ L∞ (0, +∞; Rn×m ). Equation (3.3) with A(·) ∈ L∞ loc (0, +∞; R loc can be put into the framework of the Equation (2.1), in the following manner:

• Set Y  Rn ; U  Rm ; U  Br (0) (the closed ball in Rm , centered at the origin and of radius r > 0); • Take A = 0 and f (t, y, u) = A(t)y + B(t)u. Q ,Q

S E , where The minimal time control problem studied in this subsection is (T P )min the controlled system is (3.3); QS  {0} × YS with YS  { y0 } ⊆ Rn \ {0}; S ,QE QE  (0, +∞) × {0}; and U  Br (0). It is clear that (T P )Q can be put min into the framework (AT P ) in this case. Throughout this subsection, we simply write  y QS ,QE (T P )r 0 for (T P )min ; write y(·; 0, y0, u) for the solution of (3.3), with the initial condition that y(0) = y0 . We now define a special kind of null controllability for (3.3) (see [24]).

Definition 3.1 Several definitions are given in order. (i) The Equation (3.3) is r-ball-null controllable at y0 ∈ Rn if there exists u ∈ L(0, +∞; Br (0)) so that y(·)  y(·; 0, y0, u) satisfies that y(T ) = 0 for some T ∈ (0, +∞). (ii) The Equation (3.3) is globally r-ball-null controllable if it is r-ball-null controllable at each y0 ∈ Rn . By Definition 3.1, we can easily get the following result.

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3 Existence of Admissible Controls and Optimal Controls

Theorem 3.2 Let ( y0 , r) ∈ Rn × (0, +∞). Then the Equation (3.3) is r-ball-null  y controllable at  y0 if and only if the problem (T P )r 0 has an admissible control. Several notes are given in order: Q ,U

Q ,U

• The controllable sets YC E and YC E (t) for the Equation (3.3), associated with QE  (0, +∞) × {0} and U  Br (0), depend only on the Equation (3.3) and r (see (3.1) and (3.2)). To stress such dependence, we denote them by YCr and YCr (t) respectively, and call them respectively the controllable set and the controllable set at t for (3.3), associated with r.  y • It follows by the definition of TE (see (2.43)) that for the problem (T P )r 0 , TE = (0, +∞). Thus, we have that 

YCr =

YCr (t) for each r > 0.

(3.4)

t ∈(0,+∞)

• Let Φ(·, ·) be the evolution operator generated by A(·). Then, by (3.1), we can easily check that for each r > 0 and t ∈ (0, +∞),  YCr (t)

t

=

Φ(s, 0)

−1

   B(s)u(s)ds u(·) ∈ L(0, t; Br (0)) .

(3.5)

0  y

Notice that YS = { y0 } in the problem (T P )r 0 . Thus, we can easily obtain the following consequence of Theorem 3.1:  y

Corollary 3.1 Let ( y0 , r) ∈ Rn × (0, +∞). Then the problem (T P )r 0 has an admissible control if and only if  y0 ∈ YCr . We now give an example where the controllable set can be precisely expressed.  y

Example 3.1 Consider the time optimal control problem (T P )10 , where the controlled system is as: y(t) ˙ = y(t) + u(t), t ∈ (0, +∞), with y(·) and u(·) taking values on R and [−1, 1], respectively. Then we see that YC1 = (−1, 1).

(3.6)

Indeed, given α ∈ YC1 , it follows by (3.5) and (3.4) that there is t0 ∈ (0, +∞) and u ∈ L(0, t0 ; B1 (0)) so that  t0 α= e−s u(s)ds, 0

which indicates that

 |α| ≤ 0

t0

e−s ds = 1 − e−t0 < 1.

3.1 Admissible Controls and Controllability

69

Hence YC1 ⊆ (−1, 1).

(3.7)

To show the reverse of (3.7), we arbitrarily fix β ∈ (−1, 1). Then there exists u by t0 ∈ (0, +∞) so that |β| < 1 − e−t0 . We define a control   u(s) 

β , s ∈ (0, t0 ). 1 − e−t0

Then we have that 

t0

 u ∈ L(0, t0 ; B1 (0)) and β =

e−s u(s)ds.

0

These, together with (3.5) and (3.4), yield that β ∈ YC1 (t0 ) ⊆ YC1 . Hence, (−1, 1) ⊆ YC1 .

(3.8)

Finally, (3.6) follows from (3.7) and (3.8) at once. we now end Example 3.1. The next lemma is concerned with some properties on controllable sets of (3.3). Lemma 3.1 Let r > 0. Then the following conclusions are true: (i) For each t > 0, YCr (t) is convex, closed and symmetric with respect to 0 (i.e., y0 ∈ YCr (t) if and only if −y0 ∈ YCr (t)). (ii) When 0 < t1 ≤ t2 , YCr (t1 ) ⊆ YCr (t2 ). (iii) The set YCr is convex and symmetric with respect to 0. Proof The conclusions (i)–(iii) can be directly verified from (3.5) and (3.4). This ends the proof of Lemma 3.1.   Theorem 3.3 The following statements are equivalent: (i) For any r > 0, the Equation (3.3) is globally r-ball-null controllable. (ii) The pair [A(·), B(·)] satisfies the following Conti’s condition (see [24] and [5]):  0

+∞

B(s)∗ ϕ(s)Rm ds = +∞

(3.9)

for all nonzero solutions ϕ to the adjoint equation: ϕ(t) ˙ = −A(t)∗ ϕ(t), t ∈ (0, +∞).

(3.10)

(iii) For any r > 0, the controllable set YCr of (3.3), associated with r, is Rn .  y (iv) For any r > 0 and  y0 = 0, (T P )r 0 has an admissible control.

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3 Existence of Admissible Controls and Optimal Controls

Proof (i)⇒(ii). Suppose that the conclusion (i) is true. Arbitrarily fix ϕ0 ∈ Rn \ {0}. Let ϕ be the solution of (3.10) with the initial condition ϕ(0) = ϕ0 . Arbitrarily fix r > 0. Since (3.3) is globally r-ball-null controllable, we see that (3.3) is r-ball-null controllable at each −kϕ0 , with k ∈ N+ . Arbitrarily fix k ∈ N+ . Then, there is tk ∈ (0, +∞) and uk ∈ L(0, tk ; Br (0)) so that 

tk

Φ(tk , 0)(−kϕ0 ) +

Φ(tk , s)B(s)uk (s)ds = 0.

0

This implies that 

tk

kϕ0 =

Φ(s, 0)−1 B(s)uk (s)ds.

(3.11)

0

Since uk (t)Rm ≤ r for a.e. t ∈ (0, tk ), we see that 

+∞

∗



B(s) ϕ(s)

0

Rm

tk

ds ≥ 0

≥

1 r

B(s)∗ ϕ(s)Rm ds



tk 0

B(s)∗ ϕ(s), uk (s)Rm ds.

Since k ∈ N+ was arbitrarily fixed, the above, along with (3.11), yields that for all k ∈ N+ , 

+∞ 0

B(s)∗ ϕ(s)Rm ds ≥

1 r



tk 0

ϕ0 , Φ(s, 0)−1 B(s)uk (s)Rn ds =

k ϕ0 2Rn . r

Sending k → +∞ in the above leads to (3.9). Hence, the conclusion (ii) is true. (ii)⇒(iii). By contradiction, suppose that the conclusion (ii) was true, but the r conclusion (iii) did not hold. Then there would be r0 > 0 so that YC0 = Rn . Thus, r0 r0 we have that ∂YC = ∅. This, together with the convexity of YC (see Lemma 3.1), r yields that for an arbitrarily fixed z ∈ ∂YC0 , there is λ ∈ Rn \ {0} so that λ, yRn ≤ λ, zRn for all y ∈ YCr0 .

(3.12)

Meanwhile, we arbitrarily fix t ∈ (0, +∞), and then define a control u by ⎧ ⎨

B(s)∗ (Φ(s, 0)−1 )∗ λ , if B(s)∗ (Φ(s, 0)−1 )∗ λ = 0, u(s)  for a.e. s ∈ (0, t). B(s)∗ (Φ(s, 0)−1 )∗ λRm ⎩ 0, if B(s)∗ (Φ(s, 0)−1 )∗ λ = 0, r0

Then one can easily check that &  t ' λ, Φ(s, 0)−1 B(s)u(s)ds 0

 Rn

t

= r0 0



B(s)∗ ϕ(s) m ds, R

(3.13)

3.1 Admissible Controls and Controllability

71

where ϕ(·) is the solution of (3.10) with the initial condition that ϕ(0) = λ. Now it follows from (3.12) and (3.13) that 

t 0



B(s)∗ ϕ(s) m ds ≤ 1 λ, zRn , R r0

which indicates that 

B(s)∗ ϕ(s) m ds ≤ 1 λ, zRn . R r0

+∞

0

This contradicts (ii). (iii)⇒(iv). It follows from Corollary 3.1. (iv)⇒(i). It follows from Definition 3.1. Hence, we end the proof of Theorem 3.3.

 

Corollary 3.2 For the time-invariant system: y(t) ˙ = Ay(t) + Bu(t), t ∈ (0, +∞)

(3.14)

(where A and B are n × n and n × m constant matrices), the condition (3.9) is equivalent to the combination of the following two conditions: rank(B, AB, A2 B, . . . , An−1 B) = n (Kalman controllability rank condition, see [12]);

(3.15)

   σ (A) ⊆ C−  c ∈ C  Re(c) ≤ 0 (The spectral condition, see [25]). (3.16) Proof We show it by two steps as follows: Step 1. We prove that (3.9)⇒ (3.15) and (3.16). By contradiction, suppose that (3.9) was true, but either (3.15) or (3.16) did not hold. In the case that (3.15) is not true, there is η = 0 so that η Ak B = 0, k = 0, 1, . . . , n − 1.

(3.17)

This, along with the Hamilton-Cayley Theorem, yields that η Ak B = 0 for all k ≥ 0. From this, one can easily check that η et A B = 0 for all t ∈ R. Let  ϕ be the solution to 

ϕ(t) ˙ = −A∗ ϕ(t), t ∈ (0, +∞), ϕ(0) = η.

(3.18)

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3 Existence of Admissible Controls and Optimal Controls

Then it follows by (3.18) that B ∗ ϕ (t) = 0 for all t ∈ (0, +∞), which contradicts (3.9). We now turn to the case when (3.16) is not true. Since σ (A) = σ (A∗ ), there is α+iβ ∈ C (where (α, β) ∈ (0, +∞)×R) and ξ1 +iξ2 (where ξ1 2Rn +ξ2 2Rn = 0) so that A∗ (ξ1 + iξ2 ) = (α + iβ)(ξ1 + iξ2 ).

(3.19)

If ξ1 = 0, then by (3.19), we have that A∗ ξ2 = αξ2 and β = 0.

(3.20)

Since

∗ ∗

B ∗ e−A t ξ2 Rm ≤ BL (Rm ,Rn ) e−A t ξ2

Rn

,

it follows by (3.20) that ∗

B ∗ e−A t ξ2 Rm ≤ e−αt BL (Rm ,Rn ) ξ2 Rn for all t ∈ (0, +∞). This contradicts (3.9). If ξ1 = 0, then by (3.19), we can use a very similar way to the above to obtain that ∗

∗

B ∗ e−A t ξ1 Rm ≤ B ∗ e−A t (ξ1 + iξ2 )Rm ≤ BL (Rm ,Rn ) e−(α+iβ)t (ξ1 + iξ2 )Rn ( ≤ e−αt BL (Rm ,Rn ) ξ1 2Rn + ξ2 2Rn for all t ∈ (0, +∞),

which contradicts (3.9). Hence, we have finished Step 1. Step 2. We show that (3.15) and (3.16) imply (3.9). Assume that (3.15) and (3.16) hold. According to Theorem 3.3, it suffices to show that for any r > 0, the controllable set YCr of (3.14), associated with r, is Rn . By contradiction, suppose that it was not true. Then there would be r0 > 0 so that YCr0 = Rn . Thus, by the convexity of YCr0 (see Lemma 3.1), there exists y0 ∈ Rn \YCr0 and λ0 ∈ Rn \ {0} so that r

λ0 , yRn ≤ λ0 , y0 Rn for all y ∈ YC0 .

(3.21)

Define ∗

v(t)  B ∗ e−t A λ0 , t ≥ 0. By (3.15), we have that v(·) = 0 (i.e., v is not a zero function).

(3.22)

3.1 Admissible Controls and Controllability

We next claim that



+∞ 0

73

v(t)Rm dt = +∞.

(3.23)

For this purpose, we set 

+∞

g(t) 

v(s)ds, t > 0.

(3.24)

t

By contradiction, suppose that (3.23) was not true. Then we would have that  +∞ v(t)Rm dt < +∞. 0

This, along with (3.24), yields that  +∞ |g(t)| ≤ v(s)Rm ds → 0 as t → +∞.

(3.25)

t

Let P (·) be the characteristic polynomial of A. According to the Hamilton-Cayley Theorem, P (A) = 0. Then by (3.22), we get that " " " d # −t A∗ # d# λ0 v(t) = B ∗ P − e P − dt dt " # ∗ = B ∗ P (A∗ )e−t A λ0 = 0 for each t > 0. This, together with (3.24), yields that −

" " d " d# d #" dg(t) # d# P − g(t) = P − − =P − v(t) = 0, dt dt dt dt dt

i.e., g(·) is a solution to the following (n + 1)−order ordinary differential equation: d " d# P − z(t) = 0. dt dt

(3.26)

In the case that g(·) ≡ 0, it follows by (3.24) that v(·) ≡ 0, which leads to a contradiction (since v(·) = 0). In the case when g(·) = 0, let μ1 , μ2 , . . . , μn+1 be solutions to the equation: μP (−μ) = 0, which is the characteristic equation of (3.26). Without loss of generality, we can assume that μn+1 = 0 and μk = −λk (1 ≤ k ≤ n), where λ1 , λ2 , . . . , λn are eigenvalues of the matrix A. On one hand, it follows by (3.16) that Reμk ≥ 0, for each 1 ≤ k ≤ n.

(3.27)

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3 Existence of Admissible Controls and Optimal Controls

On the other hand, since g(t) =

n+1

Pk (t)eμk t ,

k=1

where Pk (t)(1 ≤ k ≤ n) is a certain polynomial, we can easily get a contradiction from (3.27) and (3.25). Finally, by (3.23), we can choose t0 > 0 so that 

t0

r0 0

v(t)Rm dt > λ0 , y0 Rn .

(3.28)

We set  u(t) 

r0 v(t)/v(t)Rm , if v(t) = 0, t ∈ (0, t0 ). 0, if v(t) = 0,

(3.29)

It is obvious that u(·) ∈ L(0, t0 ; Br0 (0)). This, together with (3.5) and (3.4), yields that  t0 r e−t A Bu(t)dt ∈ YC0 , 0

which, combined with (3.21), (3.22), and (3.29), indicates that 

t0

r0 0

v(t)Rm dt ≤ λ0 , y0 Rn .

This leads to a contradiction with (3.28). Hence, we end the proof of Corollary 3.2.

 

 y From Corollary 3.1, we see that  y0 ∈ YCr if and only if (T P )r 0 has an admissible  y control. Thus, for the problem (T P )r 0 , it is important and interesting to give some necessary and sufficient conditions on ( y0 , r) so that  y0 ∈ YCr . We next

provide several such criteria for the case when the controlled system (3.3) is timeinvariant, i.e., A(t) ≡ A and B(t) ≡ B. (In other words, we now turn to consider the Equation (3.14).) For this purpose, we define, for each A ∈ Rn×n and each B ∈ Rn×m , the following three subsets of Rn :   VA,B  y0 ∈ Rn  ∃ T ∈ (0, +∞) and v ∈ L2 (0, T ; Rm )  T  so that y0 = − e−sA Bv(s)ds ,

(3.30)

0

  ∗ NA,B  ξ ∈ Rn  B ∗ e−A s ξ = 0 for all s > 0

(3.31)

3.1 Admissible Controls and Controllability

75

and ⊥ DA,B  NA,B



∂B1 (0).

(3.32)

Theorem 3.4 Let A ∈ Rn×n and B ∈ Rn×m . Let ( y0 , r) ∈ Rn × (0, +∞). Then the following conclusions are equivalent: (i) The pair ( y0 , r) satisfies that  y0 ∈ YCr . Here, YCr is the controllable set of (3.14), associated with r. (ii) The pair ( y0 , r) satisfies that  y0 ∈ VA,B and that   y0 , ξ 

Rn

+∞

ζ, wRn for all w ∈ VA,B .

(3.35)

By (3.30), one can easily check that 0 ∈ VA,B . This, along with (3.35), yields that ∗ ⊥ , the above implies that ζ ∈ ζ, ηRn > 0. Since η ∈ NA,B / NA,B , i.e., B ∗ e−A s0 ζ = 0 for some s0 > 0. Hence, there is δ0 ∈ (0, s0 ) so that ∗

B ∗ e−A s ζ = 0 for each s ∈ (s0 − δ0 , s0 + δ0 ).

(3.36)

76

3 Existence of Admissible Controls and Optimal Controls

Meanwhile, for each k > 0, we set ∗

uk (s)  kχ(0,s0+δ0 ) (s)B ∗ e−A s ζ, s > 0; wk 



s0 +δ0

e−sA Buk (s)ds.

0

It is obvious that wk ∈ VA,B for all k ≥ 1. Thus, by (3.35), we have that  ζ, ηRn > ζ, wk Rn = k

s0 +δ0 0

∗

B ∗ e−sA ζ 2Rm ds for all k ≥ 1.

(3.37)

Now, it follows from (3.36) and (3.37) that ζ, ηRn = +∞, which leads to a contradiction. This ends the proof of Lemma 3.2.   The next lemma is quoted from [11]. Lemma 3.3 ([11]) Let Z be a closed and convex set in Rn . Then Z=



(CH S)Z,ξ ,

ξ ∈∂B1 (0)

where   (CH S)Z,ξ  η ∈ Rn  η, ξ Rn ≤ supz, ξ Rn for each ξ ∈ ∂B1 (0). z∈Z

We now are on the position to show Theorem 3.4. Proof (Proof of Theorem 3.4) To show that (i)⇒(ii), we arbitrarily fix a pair ( y0 , r) with  y0 ∈ YCr . Then, by (3.4) and (3.5), we see that there is T > 0 so that   y0 = −

T

e−sA Bu(s)ds for some u ∈ L(0, T ; Br (0)) ⊆ L2 (0, T ; Rm ),

0

which, along with (3.30), implies that  y0 ∈ VA,B . Moreover, for any ξ ∈ Rn , we have that   y0 , ξ Rn = −

0

T

∗

B ∗ e−sA ξ, u(s)Rm ds ≤ r

 0

T

∗

B ∗ e−sA ξ Rm ds.

(3.38)

∗

Since s → B ∗ e−A s ξ , s ≥ 0, is an analytic function, we see that 

T 0

∗

B ∗ e−A s ξ Rm ds
0 is a given constant; f : R → R satisfies the following hypothesis: (A1 ) (A2 )

f is globally Lipschitz and continuously differentiable; f (r)r ≥ 0 for all r ∈ R.

3.1 Admissible Controls and Controllability

81

The Equation (3.59) can be put into our framework (2.1) (with the initial condition that y(0) =  y0 ) in the following manner: • Let Y  L2 (Ω); U  L2 (Ω); • Let A  Δ, with its domain D(A)  H 2 (Ω) ∩ H01 (Ω). It generates an analytic semigroup on Y ; • The function χω is treated as a linear and bounded operator on U ; • The function f provides a nonlinear operator f : Y → Y via f(y)(x)  f (y(x)), x ∈ Ω. Thus, the Equation (3.59) can be rewritten as 

y˙ − Ay + f(y) = χω u, t ∈ (0, +∞), y(0) =  y0 .

(3.60)

S ,QE , where the controlled In this subsection, we shall study the problem (T P )Q min system is (3.60), and where

 QS  {0} × { y0 }, QE  (0, +∞) × {0} and U  {u ∈ U  uL2 (Ω) ≤ ρ}. Q ,Q

S E can be put into the framework One can easily check that the problem (T P )min  y (AT P ) (given at the beginning of Chapter 2). We will simply write (T P )ρ0 ρ QS ,QE Q ,U for the above (T P )min . We denote by YC the controllable set YC E of the ρ Equation (3.60), associated with QE and U. The same can be said about YC (t). The main result of this subsection is as follows:

ρ

Theorem 3.5 For each ρ > 0, it holds that YC = Y . Remark 3.1 By Theorem 3.5 and Theorem 3.1, we see that for any  y0 ∈ L2 (Ω)\{0}  y0 and ρ > 0, the problem (T P )ρ has an admissible control. To prove Theorem 3.5, we need the next Proposition 3.1. Proposition 3.1 There exists a positive constant κ so that for any z0 ∈ L2 (Ω), there is a control u ∈ L∞ (0, 1; L2 (Ω)), with uL∞ (0,1;L2(Ω)) ≤ κz0 L2 (Ω) , so that the solution z ∈ C([0, 1]; L2(Ω)) to the equation: ⎧ ⎨ ∂t z − Δz + f (z) = χω u in Ω × (0, 1), z=0 on ∂Ω × (0, 1), ⎩ in Ω z(0) = z0 satisfies that z(1) = 0 over Ω.

82

3 Existence of Admissible Controls and Optimal Controls

Proof We will use the Kakutani-Fan-Glicksberg Theorem (see Theorem 1.14) to prove this proposition. To this end, we define ⎧ ⎨ f (r) if r =  0, a(r)  r ⎩ f  (0) if r = 0. Set  K  {ξ ∈ L2 (0, 1; L2(Ω))  ξ L2 (0,1;H 1 (Ω)) + ξ W 1,2 (0,1;H −1(Ω)) ≤ c0 }, 0

where c0 > 0 will be determined later. By assumptions (A1 ) and (A2 ) (given at the beginning of Section 3.1.3), we have that |a(r)| ≤ L, for all r ∈ R,

(3.61)

where L > 0 is the Lipschitz constant of the function f . For each ξ ∈ K , we consider the following linear equation: ⎧ ⎨ zt − Δz + a(ξ(x, t))z = χω u in Ω × (0, 1), z=0 on ∂Ω × (0, 1), ⎩ in Ω. z(0) = z0

(3.62)

Simply write z(·) for the solution of (3.62). According to Theorems 1.22 and 1.21, there is a positive constant κ (independent of z0 and ξ ) and a control u so that uL∞ (0,1;L2(Ω)) ≤ κz0 L2 (Ω) and z(1) = 0.

(3.63)

Define a set-valued mapping Φ : K → 2L

2 (0,1;L2 (Ω))

by setting  Φ(ξ )  {z  there exists a control u ∈ L∞ (0, 1; L2 (Ω)) so that (3.62) and (3.63) hold}, ξ ∈ K . One can easily check that Φ(ξ ) = ∅ for each ξ ∈ K . The rest of the proof will be organized by several steps as follows: Step 1. We show that K is compact and convex in L2 (0, 1; L2 (Ω)), and that each Φ(ξ ) is convex in L2 (0, 1; L2 (Ω)). These can be directly checked. Step 2. We show that Φ(K ) ⊆ K .

3.1 Admissible Controls and Controllability

83

Given ξ ∈ K , there is a control u satisfying (3.62) and (3.63). By (i) of Theorem 1.12 and (3.61)–(3.63), we can easily check that zL2 (0,1;H 1 (Ω)) + zW 1,2 (0,1;H −1(Ω)) ≤ c1 z0 L2 (Ω) , 0

for a positive constant c1 (independent of z0 and ξ ). Hence, z ∈ K if c0 = c1 z0 L2 (Ω) . Step 3. We show that Graph(Φ) is closed. It suffices to show that z ∈ Φ(ξ ), provided that ξ ∈ K → ξ strongly in L2 (0, 1; L2 (Ω)) and z ∈ Φ(ξ ) → z strongly in L2 (0, 1; L2 (Ω)). To this end, we first observe that ξ ∈ K , since K is convex and closed. Next we claim that there exists a subsequence of {}≥1, denoted in the same manner, so that a(ξ )z → a(ξ )z strongly in L2 (0, 1; L2(Ω)).

(3.64)

Indeed, since ξ → ξ strongly in L2 (0, 1; L2 (Ω)), we have a subsequence of {}≥1, denoted in the same manner, so that ξ (x, t) → ξ(x, t) for a.e. (x, t) ∈ Ω × (0, 1). Then, by the definition of the function a, we conclude that a(ξ (x, t)) → a(ξ(x, t)) for a.e. (x, t) ∈ Ω × (0, 1). By this and (3.61), we can apply the Lebesgue Dominated Convergence Theorem to obtain that a(ξ )z − a(ξ )z2L2 (0,1;L2(Ω)) ≤ 2a(ξ )(z − z)2L2 (0,1;L2(Ω)) + 2(a(ξ ) − a(ξ ))z2L2 (0,1;L2(Ω)) ≤ 2L2 z − z2L2 (0,1;L2(Ω)) + 2(a(ξ ) − a(ξ ))z2L2 (0,1;L2(Ω)) → 0. This leads to (3.64).

84

3 Existence of Admissible Controls and Optimal Controls

Finally, for each  ≥ 1, since z ∈ Φ(ξ ) ⊆ K , there is u ∈ L∞ (0, 1; L2(Ω)) with u L∞ (0,1;L2(Ω)) ≤ κz0 L2 (Ω) ,

(3.65)

so that ⎧ ∂t z − Δz + a(ξ (x, t))z = χω u ⎪ ⎪ ⎨ z = 0 ⎪ z (0) = z0 ⎪ ⎩  z (1) = 0

in Ω × (0, 1), on ∂Ω × (0, 1), in Ω, in Ω

(3.66)

z L2 (0,1;H 1(Ω)) + z W 1,2 (0,1;H −1(Ω)) ≤ c0 .

(3.67)

and 0

By (3.65) and (3.67), there is a control u and a subsequence of {}≥1, denoted in the same manner, so that u → u weakly star in L∞ (0, 1; L2 (Ω)), z → z weakly in L2 (0, 1; H01(Ω)) ∩ W 1,2 (0, 1; H −1(Ω)),

(3.68) (3.69)

and z (1) → z(1) strongly in L2 (Ω).

(3.70)

Passing to the limit for  → +∞ in (3.65) and (3.66), making use of (3.64) and (3.68)–(3.70), we obtain that z ∈ Φ(ξ ). Step 4. We apply the Kakutani-Fan-Glicksberg Theorem. From the conclusions in the above three steps, we find that the map Φ satisfies conditions of the Kakutani-Fan-Glicksberg Theorem. Thus we can apply this theorem to obtain such z that z ∈ Φ(z). Then, by the definition of Φ and using the fact that a(z(x, t))z(x, t) = f (z(x, t)), one can easily get the desired results of this proposition. Hence, we end the proof of Proposition 3.1.

 

We now give the proof of Theorem 3.5. Proof (Proof of Theorem 3.5) First, for any y0 ∈ L2 (Ω) \ {0}, we consider the following equation: ⎧ ⎨ ∂t y − Δy + f (y) = 0 in Ω × (0, t0 ), (3.71) y=0 on ∂Ω × (0, t0 ), ⎩ in Ω, y(0) = y0

3.2 Admissible Controls and Minimal Norm Problems

85

where t0 > 0 will be determined later. Write y(·) for the solution of (3.71). By (ii) of Theorem 1.12 and the assumption (A2 ), we have that y(t0 )L2 (Ω) ≤ e−λ1 t0 y0 L2 (Ω) ,

(3.72)

where λ1 > 0 is the first eigenvalue of −Δ with the zero Dirichlet boundary condition. Next, by Proposition 3.1, there is u ∈ L∞ (0, 1; L2 (Ω)) and z ∈ C([0, 1]; L2(Ω)) so that uL∞ (0,1;L2(Ω)) ≤ κy(t0 )L2 (Ω)

(3.73)

⎧ ∂t z − Δz + f (z) = χω u in Ω × (0, 1), ⎪ ⎪ ⎨ z=0 on ∂Ω × (0, 1), ⎪ ) in Ω, z(0) = y(t 0 ⎪ ⎩ z(1) = 0 in Ω.

(3.74)

and

Finally, we set t0 

1 λ1

 $ %   κy0 L2 (Ω)  0, t ∈ (0, t0 ),  + 1 and  ln u (t)    u(t − t0 ), t ∈ (t0 , t0 + 1). ρ

Then, by (3.71)–(3.74), we can easily check that ρ

ρ

y0 ∈ YC (t0 + 1) ⊆ YC .  

This completes the proof of Theorem 3.5.

3.2 Admissible Controls and Minimal Norm Problems S ,QE In this section, we will study the existence of admissible controls of (T P )Q min from the perspective of minimal norm problems. Throughout this section, we focus QS ,QE on the problem (T P )min under the following framework (HT P ):

(i) The state space Y and the control space U are real separable Hilbert spaces. (ii) The controlled system (or the state system) is the Equation (2.1), i.e., y(t) ˙ = Ay(t) + f (t, y(t), u(t)),

t ∈ (0, +∞),

where A : D(A) ⊆ Y → Y generates a C0 semigroup {eAt }t ≥0 on Y and f : (0, +∞) × Y × U → Y is some given map. Moreover, for any 0 ≤ a < b < +∞, y0 ∈ Y and u(·) ∈ L(a, b; U), this equation, with the initial condition y(a) = y0 , has a unique mild solution y(·; a, y0, u) on [a, b].

86

3 Existence of Admissible Controls and Optimal Controls

(iii) The starting set is as: QS = {0} × YS , with YS a nonempty subset of Y . (iv) The ending set is as: QE = (0, +∞) × YE , with YE a nonempty subset of Y . (v) The control constraint set U ⊆ U is a nonempty, bounded, and convex closed subset, which is symmetric with respect to 0 (i.e., u ∈ U ⇔ −u ∈ U) and has a nonzero in-radius. The in-radius of U is defined to be inf

sup

u∈U\{0} λ>0,λu∈U

λuU .

(3.75)

Two facts deserve to be mentioned: First, Br (0) ⊆ U (with r > 0) is an example of such U that has properties in the (v) of (HT P ). Second, (HT P ) implies (AT P ) (given at the beginning of Chapter 2). Indeed, (i)–(iii) in (HT P ) are exactly (i)–(iii) in (AT P ), while (iv) and (v) in (HT P ) are special cases of (iv) and (v) in (AT P ), respectively.

3.2.1 General Results in Abstract Setting S ,QE To study the problem (T P )Q , we need to introduce a norm optimal control min problem. For this purpose, we will present several lemmas, as well as one corollary. The proofs of these results rely on (v) of (HT P ) (given at the beginning of Section 3.2). Let λ ∈ [0, +∞) and let    λU  λu  u ∈ U ⊆ U.

Since U is convex and symmetric with respect to 0, we have that λU ⊆ U for each λ ∈ (0, 1). Set  V  λU ⊆ U (3.76) λ≥0

and define the Minkowski functional p : V → [0, +∞) by    p(v)  min λ ∈ [0, +∞)  v ∈ λU , v ∈ V .

(3.77)

Lemma 3.4 The following conclusions are true: (i) V = spanU. (ii) (V ,  · p ) is a normed linear space, where vp  p(v) for each v ∈ V . Proof We prove the conclusions one by one. (i) It is obvious that V ⊆ spanU. To show the reverse, we arbitrarily fix u in spanU. Then, there are two sequences {uj }kj =1 ⊆ U and {j }kj =1 ⊆ R \ {0} so k k that u = j uj . By setting   |j |, we see that j =1

j =1

3.2 Admissible Controls and Minimal Norm Problems

87

$ % k

|j | j uj . u=  |j | j =1

This, along with the symmetry and the convexity of U, yields that u ∈ V . Thus, we obtain that spanU ⊆ V . Hence, the conclusion (i) is true. (ii) It suffices to show that p(·) defines a norm on V . To this end, we first observe from (3.77) that p(v) = 0 if and only if v = 0,

(3.78)

and that p(μv) = |μ|p(v) for all v ∈ V and μ ∈ R.

(3.79)

We next claim that v1 + v2 p ≤ v1 p + v2 p for all v1 , v2 ∈ V \ {0}.

(3.80)

Indeed, given v1 , v2 ∈ V \ {0}, we let c1  v1 p > 0 and c2  v2 p > 0.

(3.81)

By (3.81) and (3.77), we find that v1 /c1 , v2 /c2 ∈ U.

(3.82)

Since U is a convex subset of U , it follows by (3.82) that c1 v1 c2 v2 v1 + v2 = · + · ∈ U. c1 + c2 c1 + c2 c1 c1 + c2 c2 Hence, we have that v1 + v2 p = p(v1 + v2 ) ≤ c1 + c2 , which, together with (3.81), leads to (3.80). Finally, we see from (3.78), (3.79), and (3.80) that p(·) defines a norm on V . This completes the proof of Lemma 3.4.   The in-radius of U (see (3.75)) has more intuitive expression. Indeed, one can easily check that inf

sup

u∈U\{0} λ>0,λu∈U

λuU =

inf

sup r,

w∈∂B1 (0)∩V rw∈U

(3.83)

88

3 Existence of Admissible Controls and Optimal Controls

where B1 (0) is the closed unit ball in U . We now define C1  sup uU and C2  u∈U

inf

sup r.

w∈∂B1 (0)∩V rw∈U

(3.84)

The above C1 is the radius of U, while C2 is the in-radius of U (see (3.75) and (3.83)). Several lemmas on properties of V and p(·) are given in order. Lemma 3.5 Let C1 and C2 be given by (3.84). Then C2 vp ≤ vU ≤ C1 vp for each v ∈ V . Proof It suffices to show that C2 vp ≤ vU ≤ C1 vp for all v ∈ V \ {0}.

(3.85)

To this end, we arbitrarily fix v ∈ V \ {0}. On one hand, by (3.77), there is u ∈ U so that v = vp u. From which it follows that vU = vp · uU ≤ C1 vp .

(3.86)

On the other hand, it is obvious that v ∈ ∂B1 (0) ∩ V . vU Since U is closed and convex, the above, along with (3.84) and the fact that 0 ∈ U, yields that (C2 v)/vU ∈ U, which implies that (C2 vp )/vU ≤ 1. This, along with (3.86), leads to (3.85). Thus, we end the proof of Lemma 3.5.   Define two normed spaces as follows: VU  (V ,  · U ) and Vp  (V ,  · p ). It deserves mentioning that when U = Br (0) (with r > 0), Vp = (U, r −1  · U ). Lemma 3.6 The following conclusions are true: (i) VU is a Hilbert space with the inner product inherited from U . (ii) Vp is a reflexive Banach space. (iii) The identity operator on V , denoted by Ip , is an isomorphism from VU to Vp . Proof We prove the conclusions one by one. (i) It suffices to show that V is a closed subset of U . To this end, we let {v }≥1 ⊆ V satisfy that v → v¯ strongly in U.

(3.87)

By (3.77) and (3.76), there is a sequence {u }≥1 ⊆ U so that v = v p u for all  ≥ 1.

(3.88)

3.2 Admissible Controls and Minimal Norm Problems

89

Since U is a bounded, convex, and closed subset of U , there is a subsequence of {u }≥1 , denoted in the same way, so that u → u¯ weakly in U for some u¯ ∈ U.

(3.89)

Meanwhile, by (3.87) and Lemma 3.5, there is a subsequence of {}≥1, denoted in the same manner, so that v p → a0 for some a0 ≥ 0.

(3.90)

Now, by (3.87), (3.89), and (3.90), we can pass to the limit for  → +∞ ¯ This, along with (3.76), yields that v¯ ∈ V . in (3.88) to obtain that v¯ = a0 u. Hence, V is a closed subspace of U . So VU is a Hilbert space. (ii) By Lemma 3.4, Vp is a normed space. We now show that Vp is a Banach space. To this end, we let {v }≥1 be a Cauchy sequence in Vp . Then by Lemma 3.5, we find that it is a Cauchy sequence in VU . Since VU is a Hilbert space, we have that v → v¯ strongly in VU for some v¯ ∈ V . By this and by making use of Lemma 3.5 again, we find that Vp is a Banach space. We next show that Vp is reflexive. To this end, we let f be a linear and continuous functional on Vp . It follows from Lemma 3.5 that f is also a linear and continuous functional on VU . Since VU is a Hilbert space, it follows by the Riesz Representation Theorem (see Theorem 1.3) that there is uf ∈ V so that f (v) = uf , vU for all v ∈ V .

(3.91)

Arbitrarily fix f ∈ Vp∗ and consider the following problem: sup f(v). Then vp ≤1

by (3.91), there exists a sequence {v }≥1 ⊆ Vp with v p ≤ 1 so that sup f(v) = lim f(v ) = lim uf, v U . →+∞

vp ≤1

(3.92)

→+∞

Since v p ≤ 1, it follows from Lemma 3.5 that {v }≥1 is bounded in VU . Hence, there is a subsequence of {v }≥1 , denoted in the same way, so that v weakly in VU for some  v ∈ V. v → 

(3.93)

Since v p ≤ 1, it follows by (3.91) and (3.93) that  v p =

sup

f V ∗ ≤1

f ( v) =

p

=

sup

lim f (v ) ≤

f V ∗ ≤1 →+∞ p

sup uf , v U =

f V ∗ ≤1 p

sup

sup

lim uf , v U

f V ∗ ≤1 →+∞ p

lim f Vp∗ v p

f V ∗ ≤1 →+∞ p

≤ 1. (3.94)

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3 Existence of Admissible Controls and Optimal Controls

Meanwhile, it follows from (3.92), (3.93), and (3.91) that sup f(v) = uf, v U = f( v ).

vp ≤1

This, together with (3.94), implies that sup f(v) = max f(v).

vp ≤1

vp ≤1

(3.95)

Since we have proved that Vp is a Banach space, it follows from (3.95) and the James Theorem (see Theorem 1.6) that Vp is a reflexive Banach space. (iii) It follows from Lemma 3.5 that Ip is an isomorphism from VU to Vp . Thus, we end the proof of Lemma 3.6.   Based on Lemma 3.5 and Lemma 3.6, we can easily verify the following result: Corollary 3.4 It holds that L∞ (0, T ; Vp ) = L∞ (0, T ; VU ). To introduce the desired norm optimal control, we also need to make the  following additional assumption (H T P ): • For any y0 ∈ YS , t > 0 and v ∈ L∞ (0, t; Vp ), the controlled system (2.1) has a unique mild solution on [0, t], denoted by y(·; 0, y0, v). Now, given y0 ∈ YS , t ∈ (0, +∞) and YE ⊆ Y , we define the following norm E optimal control problem (NP )t,Y y0 :   N(t, y0 ; YE )  inf vL∞ (0,t ;Vp )  y(t; 0, y0 , v) ∈ YE .

(3.96)

E Several notes related to the problem (NP )t,Y y0 are given in order:

(a) When the set in the right-hand side of (3.96) is empty, we agree that N(t, y0 ; YE ) = +∞. QS ,QE (b) It is a norm optimal control problem corresponding to the problem (T P )min . (c) In the studies of this problem, Corollary 3.4 will be used. E The next theorem gives connections between the problem (NP )t,Y y0 and admissible QS ,QE controls of (T P )min .

QS ,QE  Theorem 3.6 Suppose that (H has an T P ) holds. Then the problem (T P )min admissible control if and only if either of the following two conditions is true:

(i) 1 > infy0 ∈YS ,t >0 N(t, y0 ; YE ). (ii) 1 = infy0 ∈YS ,t >0 N(t, y0 ; YE ); there exists  y0 ∈ YS and  t ∈ (0, +∞) satisfying  t ,YE  that N(t ,  y0 ; YE ) = infy0 ∈YS ,t >0 N(t, y0 ; YE ); and the problem (NP ) y0 has a solution.

3.2 Admissible Controls and Minimal Norm Problems

91

t,YE  Proof Let (H T P ) hold. Then the problem (NP )y0 is well defined. We first show S ,QE the necessity. Assume that the problem (T P )Q has an admissible control. Then min by (HT P ) (given at the beginning of Section 3.2), we see that there exists  y0 ∈ YS ,  t ∈ (0, +∞), and  u ∈ L(0,  t; U) so that y( t; 0,  y0 , u) ∈ YE . This implies that

inf

y0 ∈YS ,t >0

N(t, y0 ; YE ) ≤ N( t,  y0 ; YE ) ≤  uL∞ (0,t;Vp ) .

(3.97)

Meanwhile, since  u(t) ∈ U for a.e. t ∈ (0,  t ), it follows by (3.77) that p( u(t)) ≤ 1 for a.e. t ∈ (0,  t ). This, along with Lemma 3.4, yields that  uL∞ (0,t;Vp ) ≤ 1, from which and (3.97), it follows that inf

y0 ∈YS ,t >0

N(t, y0 ; YE ) ≤ N( t,  y0 ; YE ) ≤  uL∞ (0,t;Vp ) ≤ 1.

(3.98)

When infy0 ∈YS ,t >0 N(t, y0 ; YE ) = 1, we see from (3.98) that inf

y0 ∈YS ,t >0

N(t, y0 ; YE ) = N( t,  y0 ; YE ) =  uL∞ (0,t;Vp ) .

From these, one can easily check that (ii) is true. From this and (3.98), we find that either (i) or (ii) is true. We now show the sufficiency. We first prove that (i) implies the existence of QS ,QE admissible controls for the problem (T P )min . To this end, we suppose that (i) holds. Then we can find ε > 0 so that 2ε < 1 −

inf

y0 ∈YS ,t >0

N(t, y0 ; YE ).

(3.99)

Thus, there exists y0ε ∈ YS and tε ∈ (0, +∞) so that inf

y0 ∈YS ,t >0

N(t, y0 ; YE ) ≤ N(tε , y0ε ; YE )
0

N(t, y0 ; YE ) + ε.

(3.100)

By the definition of N(tε , y0ε ; YE ), there exists a control vε , with vε ∈ L∞ (0, tε ; Vp ) and y(tε ; 0, y0ε , vε ) ∈ YE ,

(3.101)

N(tε , y0ε ; YE ) ≤ vε L∞ (0,tε ;Vp ) < N(tε , y0ε ; YE ) + ε.

(3.102)

so that

It follows from (3.100) and (3.102) that inf

y0 ∈YS ,t >0

N(t, y0 ; YE ) ≤ vε L∞ (0,tε ;Vp )
0

N(t, y0 ; YE ) + 2ε.

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3 Existence of Admissible Controls and Optimal Controls

This implies that vε (t)Vp
0

N(t, y0 ; YE ) + 2ε for a.e. t ∈ (0, tε ),

which, together with the definition of Vp and (3.99), indicates that p(vε (t))
0

N(t, y0 ; YE ) + 2ε < 1 for a.e. t ∈ (0, tε ).

(3.103)

Meanwhile, since vε (t) ∈ V for a.e. t ∈ (0, tε ), it follows from (3.77) that vε (t) ∈ p(vε (t))U for a.e. t ∈ (0, tε ). Since U is convex and 0 ∈ U, it follows by the above and (3.103) that vε (t) ∈ U for a.e. t ∈ (0, tε ).

(3.104)

By the first conclusion in (3.101) and by Corollary 3.4, we see that vε ∈ L∞ (0, tε ; U ), which, along with (3.104), shows that vε ∈ L(0, tε ; U). From this and the second conclusion in (3.101), we can easily verify that (0, y0ε , tε , vε ) is an S ,QE admissible tetrad of the problem (T P )Q . Hence, vε is an admissible control to min this problem. We next show that (ii) implies the existence of admissible controls for the S ,QE . To this end, we suppose that (ii) holds. Then there is  y0 ∈ YS , problem (T P )Q min  t > 0 and a control  v , with t; Vp ) and y( t; 0,  y0, v ) ∈ YE ,  v ∈ L∞ (0, 

(3.105)

so that 1 =  v L∞ (0,t;Vp ) = N( t,  y0 ; YE ). From the latter, it follows that t).  v (t)Vp ≤ 1 for a.e. t ∈ (0,  This implies that 0 ≤ p( v (t)) ≤ 1 for a.e. t ∈ (0,  t).

(3.106)

Since U is convex, 0 ∈ U and  v (t) ∈ p( v (t))U for a.e. t ∈ (0,  t), by (3.106), we obtain that  v (t) ∈ U for a.e. t ∈ (0,  t ), which, combined with (3.105) and Corollary 3.4, indicates that (0,  y0 ,  t, v ) is an QS ,QE admissible tetrad of the problem (T P )min . Hence,  v is an admissible control to this problem. In summary, we end the proof of Theorem 3.6.  

3.2 Admissible Controls and Minimal Norm Problems

93

3.2.2 Heat Equations with Ball-Type Control Constraints In this subsection, we will apply Theorem 3.6 to a minimal time control problem for heat equations. Throughout this subsection, Ω ⊆ Rd , d ≥ 1, is a bounded domain with a C 2 boundary ∂Ω, and ω ⊆ Ω is a nonempty open subset with its characteristic function χω . Consider the following controlled heat equation:  ∂t y − Δy + a(x, t)y = χω u in Ω × (0, +∞), (3.107) y=0 on ∂Ω × (0, +∞), where a ∈ L∞ (Ω × (0, +∞)) and u ∈ L∞ (0, +∞; L2 (Ω)). The Equation (3.107) can be put into our framework (2.1) in the same manner as that used to deal with the Equation (3.59) in Section 3.1.3, except for the term a(x, t)y. To deal with it, we notice that since a ∈ L∞ (Ω × (0, +∞)), there is a measurable set E ⊆ Ω × (0, +∞) with |E| = 0 and a positive constant c0 so that |a(x, t)| ≤ c0 for each (x, t) ∈ (Ω × (0, +∞)) \ E. Define   a (x, t) 

a(x, t), for all (x, t) ∈ (Ω × (0, +∞)) \ E, 0, for all (x, t) ∈ E.

It is obvious that  a ∈ L∞ (Ω × (0, +∞));  a (x, t) = a(x, t) for a.e. (x, t) ∈ Ω × (0, +∞), (3.108) and | a (x, t)| ≤ c0 for all (x, t) ∈ Ω × (0, +∞).

(3.109)

By (3.109), we can define, for each t ∈ (0, +∞), a linear and bounded operator D(t) : Y → Y in the following manner: (D(t)y)(x) =  a (x, t)y(x) for each x ∈ Ω.

(3.110)

With the aid of (3.108) and (3.110), the Equation (3.107) can be put into our framework (2.1) in the following manner: y˙ − Δy + D(t)y = χω u,

t ∈ (0, +∞).

(3.111)

S ,QE In this subsection, we will use Theorem 3.6 to study (T P )Q , where the min controlled system is (3.111) (or equivalently (3.107)); and where  QS  {0} × YS , QE  (0, +∞) × {0}, U  {u ∈ U  uL2 (Ω) ≤ ρ},

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3 Existence of Admissible Controls and Optimal Controls

with ρ > 0, YS a nonempty subset of Y \ {0}. For simplicity, we will denote S ,QE S by (T P )Q (T P )Q ρ . We aim to give a sufficient and necessary condition for min S the existence of admissible controls of (T P )Q ρ . We would like to mention what follows:  • Both (HT P ) (given at the beginning of Section 3.2) and (H T P ) (given after Corollary 3.4) hold for this case.  The condition (HT P ) can be easily verified in this case. Let us explain why (H TP) holds in the current case. By (3.76), (3.77), and Lemma 3.4, we can easily check that V = L2 (Ω); p(v) =

vL2 (Ω) ρ

for each v ∈ L2 (Ω),

and vL∞ (0,t ;Vp) =

1 vL∞ (0,t ;L2(Ω)) for each v ∈ L∞ (0, t; Vp ). ρ

From these, we see that for any y0 ∈ YS , t ∈ (0, +∞), and v ∈ L∞ (0, t; Vp ), the Equation (3.111), with the initial condition that y(0) = y0 , has a unique mild solution on [0, t]. E The corresponding norm optimal control problem (NP )t,Y (see (3.96)), with y0 y0 ∈ YS and t ∈ (0, +∞), is as follows:   N(t, y0 ; {0})  inf vL∞ (0,t ;Vp )  y(t; 0, y0, v) = 0 . (3.112)

(NP )yt,{0} 0

Notice that (3.112) can be rewritten as: (NP )yt,{0} 0

N(t, y0 ; {0}) =

   1 inf vL∞ (0,t ;L2(Ω))  y(t; 0, y0, v) = 0 . ρ (3.113)

In order to use Theorem 3.6 to study the existence of admissible controls for the minimal time control problem of this subsection, we introduce another norm optimal control problem as follows:    (NP )ty0 N(t, y0 )  inf vL∞ (0,t ;L2(Ω))  y (t; 0, y0 , v) = 0 . (3.114) From (3.113) and (3.114), we see that N(t, y0 ; {0}) =

1 N(t, y0 ) for all y0 ∈ YS and t ∈ (0, +∞). ρ

(3.115)

In the problem (NP )ty0 , N (t, y0 ) is called the optimal norm; a function v ∗ is called an optimal control if

  y t; 0, y0 , v ∗ = 0 and v ∗ L∞ (0,t ;L2(Ω)) = N (t, y0 ) ;

3.2 Admissible Controls and Minimal Norm Problems

95

a function  v is called an admissible control if y (t; 0, y0 , v ) = 0 and  v ∈ L∞ (0, t; L2 (Ω)). We say that the problem (NP )ty0 has the bang-bang property if any optimal control v ∗ to this problem verifies that

∗

v (s)

L2 (Ω)

= N (t, y0 ) for a.e. s ∈ (0, t).

Lemma 3.7 Let y0 ∈ L2 (Ω)\{0} and t ∈ (0, +∞). Then the following conclusions are true: (i) The problem (NP )ty0 has the bang-bang property. (ii) The problem (NP )ty0 has a unique optimal control. Proof We prove the conclusions one by one. (i) By contradiction, suppose that (NP )ty0 did not hold the bang-bang property. Since N(t, y0 ) = 0 (see (i) of Remark 1.5 after Theorem 1.22), there would be an optimal control v ∗ (to (NP )ty0 ), ε ∈ (0, N(t, y0 )) and a subset E ⊆ (0, t) of positive measure so that

∗

v (s) 2 ≤ N(t, y0 ) − ε for all s ∈ E. L (Ω)

(3.116)

By Theorems 1.22 and 1.21, we can find a δ > 0 so that for each initial state z, with zL2 (Ω) ≤ δ, there is a control v z verifying that

  y t; 0, z, χE v z = 0 and v z L∞ (0,t ;L2(Ω)) ≤ ε. Here χE is the characteristic function of E. By taking z = δ1 y0 , where δ1 > 0 verifies δ1 y0 L2 (Ω) ≤ δ, in the above, we find that   y t; 0, δ1 y0 , χE v δ1 y0 = 0.

(3.117)

Since v ∗ is an optimal control to (NP )ty0 , we see that

  y t; 0, y0 , v ∗ = 0 and v ∗ L∞ (0,t ;L2(Ω)) = N (t, y0 ) .

(3.118)

From (3.117) and the first equality in (3.118), it follows that   y t; 0, δ1 y0 + y0 , v ∗ + χE v δ1 y0 = 0, which leads to "  # y t; 0, y0 , (1 + δ1 )−1 v ∗ + χE v δ1 y0 = 0.

(3.119)

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3 Existence of Admissible Controls and Optimal Controls

Meanwhile, since

δy

v 1 0

L∞ (0,t ;L2(Ω))

≤ ε,

it follows from the second equality in (3.118) and (3.116) that

 

(1 + δ1 )−1 v ∗ + χE v δ1 y0

L∞ (0,t ;L2(Ω))

≤

1 N (t, y0 ) . 1 + δ1

This, along with (3.119), contradicts the optimality of N(t, y0 ). Hence, the problem (NP )ty0 has the bang-bang property. (ii) The existence of admissible controls follows from Theorems 1.22 and 1.21. To prove the existence of optimal controls, we let {v }≥1 ⊆ L∞ (0, t; 2 L (Ω)) be a minimization sequence of (NP )ty0 . Then y (t; 0, y0 , v ) = 0 and

lim v L∞ (0,t ;L2(Ω)) = N (t, y0 ) .

→+∞

(3.120)

From the second equality of (3.120), there is a subsequence of {v }≥1 , still denoted in the same way, so that " # v → v ∗ weakly star in L∞ 0, t; L2 (Ω)

(3.121)

and   y (t; 0, y0 , v ) → y t; 0, y0 , v ∗ strongly in L2 (Ω). This, together with (3.120) and (3.121), implies that   y t; 0, y0 , v ∗ = 0 and v ∗ L∞ (0,t ;L2(Ω)) ≤ N(t, y0 ). These show that v ∗ is an optimal control of the problem (NP )ty0 . To show the uniqueness, we let v1 and v2 be two optimal controls of the problem (NP )ty0 . Then one can easily check that (v1 + v2 )/2 is also an optimal control to the problem (NP )ty0 . By (i), we find that v1 (s)L2 (Ω) = v2 (s)L2 (Ω) = (v1 + v2 )(s)/2L2 (Ω) = N (t, y0 ) for a.e. s ∈ (0, t). From the above and from the parallelogram law in L2 (Ω), we can easily verify that v1 (s) = v2 (s) for a.e. s ∈ (0, t). Thus, we complete the proof of Lemma 3.7.

 

3.2 Admissible Controls and Minimal Norm Problems

97

Lemma 3.8 For each y0 ∈ L2 (Ω) \ {0}, the function N (·, y0 ) is strictly monotonically decreasing over (0, +∞). Furthermore, lim N (t, y0 ) = +∞

(3.122)

(y0 ) ∈ [0, +∞) . lim N (t, y0 )  N

(3.123)

t →0+

and t →+∞

Proof Let 0 < t1 < t2 . By (ii) of Lemma 3.7, we can let v1 be the optimal control to the problem (NP )ty10 . Then we have that v1 L∞ (0,t1 ;L2 (Ω)) = N (t1 , y0 ) and y (t1 ; 0, y0 , v1 ) = 0. Construct a new control v2 by setting  v1 (s) for a.e. s ∈ (0, t1 ) , v2 (s)  0 for a.e. s ∈ (t1 , t2 ) .

(3.124)

(3.125)

By the first equality of (3.124) and (3.125), we see that v2 L∞ (0,t2;L2 (Ω)) = N (t1 , y0 ) .

(3.126)

From the second equality of (3.124) and (3.125), we find that y (t2 ; 0, y0 , v2 ) = 0.

(3.127)

This yields that v2 is an admissible control to (NP )ty20 . Then by (3.126) and the optimality of N (t2 , y0 ), it follows that N (t1 , y0 ) = v2 L∞ (0,t2 ;L2 (Ω)) ≥ N (t2 , y0 ) . Namely, N (·, y0 ) is monotonically decreasing. We next show that N (t1 , y0 ) > N (t2 , y0 ) . By contradiction, suppose that it was not true. Then by the above monotonicity of N (·, y0 ), we would have that N (t1 , y0 ) = N (t2 , y0 ) , which, along with (3.126), shows that v2 L∞ (0,t2 ;L2 (Ω)) = v1 L∞ (0,t1 ;L2 (Ω)) = N (t1 , y0 ) = N (t2 , y0 ) ,

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3 Existence of Admissible Controls and Optimal Controls

where v1 is the optimal control to the problem (NP )ty10 and v2 is given by (3.125). This, together with (3.127), shows that v2 is the optimal control to the problem (NP )ty20 . From (3.125), we see that v2 is not a bang-bang control. This contradicts the bang-bang property of the problem (NP )ty20 (see Lemma 3.7). The conclusion (3.123) follows from the monotonicity of N (·, y0 ) at once. Finally, we will show (3.122). By contradiction, we suppose that (3.122) was not true. Then there would be a sequence {t }≥1 ⊆ (0, 1) so that t  0 and N (t , y0 )

 ∈ (0, +∞). N

t

Let u be the optimal control to (NP )y0 . Then we have that  for all  ≥ 1, u L∞ (0,t ;L2 (Ω)) = N (t , y0 ) ≤ N

(3.128)

0 = y (t ; 0, y0 , u ) = y(t ; 0, y0, 0) + y(t ; 0, 0, u ).

(3.129)

and

Since t  0 and because of (3.128), we can easily verify that y(t ; 0, 0, u ) → 0, through multiplying (3.107) (with y(0) = 0 and u = u ) by y, and then integrating over Ω × (0, t), t ∈ (0, t ), and finally using Gronwall’s inequality. Meanwhile, since t  0, we can use the continuity of the solution y(·; 0, y0, 0) at time t = 0 to get that y(t ; 0, y0, 0) → y0 . These, along with (3.129), yield that 0 = y0 , which leads to a contradiction (since we assumed that y0 = 0). Hence (3.122) holds. This completes the proof of Lemma 3.8.   Q . Theorem 3.7 The problem (T P )ρ S has an admissible control if and only if ρ > N Here,

(y0 ) given by (3.123).   inf N(y  0 ), with N N y0 ∈YS

(3.130)

Proof We first claim that inf

y0 ∈YS ,t >0

N(t, y0 ; {0}) =

1 N. ρ

(3.131)

Indeed, by (3.115), we get that inf

y0 ∈YS ,t >0

N(t, y0 ; {0}) =

1 ρ

inf

y0 ∈YS ,t >0

N(t, y0 ) =

1 inf inf N(t, y0 ). ρ y0 ∈YS t >0

(3.132)

Meanwhile, by the strict monotonicity of the function N(·, y0 ) (see Lemma 3.8) and by (3.123), we find that  0 ); inf N(t, y0 ) = N(y

t >0

 0 ) for all t > 0 and y0 ∈ L2 (Ω) \ {0}. N(t, y0 ) > N(y

(3.133) (3.134)

3.3 Admissible Controls and Reachable Sets

99

Hence, (3.131) follows from (3.132), (3.133), and (3.130). We next claim that for any  y0 ∈ YS and  t ∈ (0, +∞), inf

y0 ∈YS ,t >0

N(t, y0 ; {0}) < N( t,  y0 ; {0}).

(3.135)

By contradiction, suppose that it was not true. Then there would exist  y0 ∈ YS and  t ∈ (0, +∞) so that inf

y0 ∈YS ,t >0

N(t, y0 ; {0}) = N( t,  y0 ; {0}).

This, together with (3.115), (3.134), and (3.130), implies that inf

y0 ∈YS ,t >0

N(t, y0 ; {0}) =

1 1 1 N( t,  y0 ) > N( y0 ) ≥ N , ρ ρ ρ

which contradicts (3.131). Hence, (3.135) is true. Finally, from (3.131), (3.135), and Theorem 3.6, we can easily get the desired result of this theorem. This completes the proof of Theorem 3.7.  

3.3 Admissible Controls and Reachable Sets Q ,QE

S In this section, we will study the existence of admissible controls of (T P )min from perspective of reachable sets.

3.3.1 General Results in Abstract Setting With the aid of (AT P ) (given at the beginning of Chapter 3), we can define the following three sets:   QR  (t, y(t; 0, y0, u)) ∈ (0, +∞) × Y  y0 ∈ YS and u ∈ L(0, t; U)} ;   YR (t)  y ∈ Y  (t, y) ∈ QR , for each t ∈ (0, +∞);  YR  YR (t). 

(3.136) (3.137) (3.138)

t >0

These sets depend on YS and U. We call respectively QR , YR (t), and YR the reachable horizon, the reachable set at the time t and the reachable set of the Equation (2.1), associated with QS and U. In most time, we simply call them

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3 Existence of Admissible Controls and Optimal Controls

the reachable horizon, the reachable set at the time t and the reachable set of the Equation (2.1), if there is no risk causing any confusion. For any t ∈ TE (Recall (2.43) for the definition of TE ), we define the distance between YR (t) and YE (t) (Recall (2.44) for the definition of YE (t)) by d(t)  dist{YR (t), YE (t)} =

inf

y1 ∈YR (t ),y2∈YE (t )

y1 − y2 Y for all t ∈ TE . (3.139)

Now we introduce an optimization problem: d ∗  inf d(t). t ∈TE

(3.140)

The connection between admissible controls and the reachable sets is presented in the following theorem. Q ,QE

S Theorem 3.8 The problem (T P )min following conditions hold:

has an admissible control if and only if the

(i) d ∗ = 0; (ii) The optimization problem (3.140) admits a solution; (iii) There exists tE ∈ TE , which is a solution of the problem (3.140), so that inf

(y1 ,y2 )∈YR (tE )×YE (tE )

y1 − y2 Y

(3.141)

admits a solution. Proof Necessity. Assume that u is an admissible control (associated with an QS ,QE admissible tetrad (0, y0 , tE , u)) to (T P )min . Then y(tE ; 0, y0, u) ∈ YR (tE ) ∩ YE (tE ).

(3.142)

By (3.142), (3.139), and (3.140), we see that 0 = d ∗ = dist{YR (tE ), YE (tE )}. From the above and (3.142), we find that the conditions (i), (ii), and (iii) hold. Sufficiency. Assume that the conditions (i), (ii), and (iii) hold. Then there is tE ∈ TE , y1∗ ∈ YR (tE ) and y2∗ ∈ YE (tE ) so that d(tE ) = 0 and y1∗ − y2∗ Y = dist{YR (tE ), YE (tE )}.

(3.143)

By (3.139) and (3.143), we find that y1∗ = y2∗ . Hence y1∗ ∈ YE (tE ). Since y1∗ ∈ YR (tE ), there exists y0 ∈ YS and u ∈ L(0, tE ; U) so that y1∗ = y(tE ; 0, y0 , u), which, combined with the fact that y1∗ ∈ YE (tE ), implies that (0, y0 , tE , u) is an S ,QE . Hence, u is an admissible control to admissible tetrad of the problem (T P )Q min this problem. Thus, we end the proof of Theorem 3.8.  

3.3 Admissible Controls and Reachable Sets

101

We now give some applications of Theorem 3.8 by the following three examples. Q ,QE

S Example 3.2 Consider the time optimal control problem (T P )min controlled system is as:

, where the

y(t) ˙ = y(t) + u(t), t ∈ (0, +∞), and where U  [−1, 1], QS  {(0, 2)} and QE  (0, +∞) × {0}. It is clear that (AT P ) holds for this case. Moreover, by (2.43) and (2.44), we have that TE = (0, +∞) and YE (t) = {0} for all t ∈ (0, +∞).

(3.144)

Meanwhile, it follows from (3.137) and (3.136) that  $ %  t  −s t  2+ e u(s)ds e u ∈ L(0, t; U) for all t ∈ (0, +∞). YR (t) = 0

This, together with (3.139) and (3.144), implies that d(t) =

inf

u∈L(0,t ;U)

$ %  t −s 2+ e u(s)ds et for all t ∈ (0, +∞), 0

from which, one can easily check that $ %  t −s d(t) = 2 + e (−1)ds et = 1 + et for all t ∈ (0, +∞). 0

This, combined with (3.140), indicates that d ∗ = 2. So the condition (i) in Theorem 3.8 does not hold. Thus, by Theorem 3.8, we find that the problem S ,QE has no admissible control. (T P )Q min S ,QE Besides, for this example, we can directly check that (T P )Q has no min admissible control. Indeed, since % $  t e−s u(s)ds et ≥ 2 > 0 y(t; 0, 2, u) = 2 + 0

for all t ∈ (0, +∞) and u ∈ L(0, t; U), S ,QE (T P )Q has no admissible control. Now we end Example 3.2. min

Q ,QE

S Example 3.3 Consider the time optimal control problem (T P )min controlled system is as:

y˙ = u(t)y(t), t ∈ (0, +∞),

, where the

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3 Existence of Admissible Controls and Optimal Controls

and where U  [−1, 1], QS  {(0, 2)} and QE  (0, +∞) × {0}. It is clear that (AT P ) (given at the beginning of Chapter 2) holds. Moreover, by (2.43) and (2.44), we have that TE = (0, +∞) and YE (t) = {0} for all t ∈ (0, +∞).

(3.145)

It follows from (3.137) and (3.136) that  t  YR (t) = 2e 0 u(s)ds  u ∈ L(0, t; U) for all t ∈ (0, +∞). This, together with (3.139) and (3.145), implies that d(t) =

inf

u∈L(0,t ;U)

2e

t 0

u(s)ds

= 2e−t for all t ∈ (0, +∞),

(3.146)

which, combined with (3.140), indicates that d ∗ = 0, i.e., the condition (i) in Theorem 3.8 holds. However, (ii) in Theorem 3.8 does not hold for this case. Indeed, by (3.146), one can easily check that for any t ∈ (0, +∞), d(t) > d ∗ . So the optimization QS ,QE problem (3.140) has no solution. Thus, by Theorem 3.8, this problem (T P )min has no admissible control. QS ,QE Besides, we can directly check that this problem (T P )min has no admissible control. This can be seen from what follows: y(t; 0, 2, u) = 2e

t 0

u(s)ds

> 0 for all t ∈ (0, +∞) and u ∈ L(0, t; U).

We now end Example 3.3. Example 3.4 Let Ω ⊆ Rd , d ≥ 1 be a bounded domain with a C 2 boundary ∂Ω, and ω ⊆ Ω is a nonempty open subset with its characteristic function χω . Consider the following controlled heat equation:  ∂t y − y = χω u, in Ω × (0, +∞), (3.147) y = 0, on ∂Ω × (0, +∞). We can put the Equation (3.147) into our framework (2.1) in the same manner as that used to deal with the Equation (3.107) (with a ≡ 0). In this way, the Equation (3.147) can be rewritten as y˙ − Δy = χω u, t ∈ (0, +∞). Q ,QE

S Consider the time optimal control problem (T P )min is (3.148), and where

(3.148)

, where the controlled system

 QS  {(0, 0)}; QE  (0, +∞)×{z ∈ L2 (Ω)\H01(Ω)  zL2 (Ω) ≤ r}, with r > 0;  U  {u ∈ L2 (Ω)  uL2 (Ω) ≤ ρ}, with ρ > 0.

3.3 Admissible Controls and Reachable Sets

103

It is clear that (AT P ) (given at the beginning of Chapter 2) holds. Moreover, by (2.43) and (2.44), we have that, for all t ∈ (0, +∞),  (3.149) TE = (0, +∞) and YE (t) = {z ∈ L2 (Ω) \ H01 (Ω)  zL2 (Ω) ≤ r}. Meanwhile, it follows from (3.137) and (3.136) that   t  (t −s)Δ  YR (t) = e χω u(s)ds u ∈ L(0, t; U) for all t ∈ (0, +∞). 0

This, together with (3.139) and (3.149), implies that

 t



e(t −s)Δχω u(s)ds − y2 d(t) = inf

u∈L(0,t ;U),y2∈YE (t )

≤

inf

y2 ∈YE (t )

0

y2 L2 (Ω) = 0 for all t ∈ (0, +∞).

L2 (Ω)

(3.150)

By (3.150) and (3.140), one can easily check that d ∗ = 0 and that every t ∈ (0, +∞) is a solution to the problem (3.140). So the conditions (i) and (ii) in Theorem 3.8 hold. However, the condition (iii) in Theorem 3.8 does not hold for this case. Indeed, by contradiction, suppose that (iii) was true. Then there would exist tE ∈ (0, +∞), y1 ∈ YR (tE ) and y2 ∈ YE (tE ) so that y1 − y2 L2 (Ω) = d(tE ). This, together with (3.150), implies that y1 = y2 . However, y1 ∈ YR (tE ) ⊆ H01 (Ω) and y2 ∈ H01 (Ω). S ,QE Thus, we obtain a contradiction. Hence, by Theorem 3.8, this problem (T P )Q min has no admissible control. S ,QE Besides, for this example, we can directly check that (T P )Q has no min admissible control. Indeed, since  t e(t −s)Δχω u(s)ds ∈ H01 (Ω) y(t; 0, 0, u) =

0

for any t ∈ (0, +∞) and u ∈ L(0, t; U), we see that (t, y(t; 0, 0, u)) ∈ QE for all t ∈ (0, +∞) and u ∈ L(0, t; U). S ,QE This indicates that the problem (T P )Q has no admissible control. We end min Example 3.4.

Remark 3.2 We would like to mention that conditions (i), (ii), and (iii) in Theorem 3.8 are not verifiable in many cases. In the next subsection, we will present some properties on reachable sets and the function d(·) for some special cases.

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3 Existence of Admissible Controls and Optimal Controls

3.3.2 Reachable Sets of Linear ODEs Reachable sets are very important in studies of the existence of admissible control. Unfortunately, our knowledge on reachable sets of the Equation (2.1) is quite limited. The main purpose of this subsection is to present some properties on reachable sets of the Equation (3.3), under the following assumptions: (L1) U is a nonempty compact subset of Rm ; (L2) YS is a nonempty, convex, and closed subset of Rn . Besides, we will give some properties on the function d(·) (defined by (3.139)). We start with studying reachable sets of the Equation (3.3). Write Φ(·, ·) for the evolution operator generated by A(·). Then the reachable set at the time t of the Equation (3.3) is as:    t   Φ(t, s)B(s)u(s)ds y0 ∈ YS , u ∈ L(0, t; U) . YR (t) = Φ(t, 0)y0 + 0

(3.151)

Write co(U) for the convex hull of U. Let    QR (co(U))  (t, y(t; 0, y0 , u)) ∈ (0, +∞)×Y  y0 ∈ YS and u ∈ L(0, t; co(U)) and    YR (t; co(U))  y ∈ Y  (t, y) ∈ QR (co(U)) for all t ∈ (0, +∞). Some properties on reachable sets of (3.3) are given in the next theorem. Theorem 3.9 The following statements are true: (i) Suppose that (L1) holds. Then YR (t) = YR (t, co(U)) for all t ∈ (0, +∞).

(3.152)

(ii) Suppose that (L1) and (L2) are true. Then for each t ∈ (0, +∞), YR (t) is a convex and closed subset in Rn . Proof We prove the conclusions one by one. (i) Suppose that (L1) holds. Arbitrarily fix t ∈ (0, +∞) and y0 ∈ YS . We define two sets    YR (t, y0 )  y(t; 0, y0 , u) ∈ Y  u ∈ L(0, t; co(U)) and    R (t, y0 )  y(t; 0, y0 , u) ∈ Y  u ∈ L(0, t; U) . Y

3.3 Admissible Controls and Reachable Sets

105

We organize the proof by the following two steps: Step 1. We show that R (t, y0 ). YR (t, y0 ) = Y R (t, y0 ). We now show that It is obvious that YR (t, y0 ) ⊇ Y R (t, y0 ). YR (t, y0 ) ⊆ Y Let y ∈ YR (t, y0 ). Then there is u ∈ L(0, t; co(U)) so that  t y = Φ(t, 0)y0 + Φ(t, s)B(s)u(s)ds.

(3.153)

(3.154)

0

Since u(s) ∈ co(U) for a.e. s ∈ (0, t), according to the Carathéodory Theorem and the Measurable Selection Theorem (see Theorems 1.7 and 1.17), there are measurable functions uj (·) and λj (·), j = 0, 1, . . . , m, so that m

λj (s) = 1 and u(s) =

j =0

m

λj (s)uj (s) for a.e. s ∈ (0, t)

(3.155)

j =0

and so that uj (s) ∈ U and λj (s) ≥ 0 for a.e. s ∈ (0, t) and for all j = 0, 1, . . . , m. (3.156) It follows from (3.154) and (3.155) that y = Φ(t, 0)y0 +

m 

t

λj (s)Φ(t, s)B(s)uj (s)ds.

(3.157)

j =0 0

Furthermore, by the Lyapunov Theorem (see Theorem 1.15), there are measurable sets Ej ⊆ (0, t), j = 0, 1, . . . , m, so that m 

t

λj (s)Φ(t, s)B(s)uj (s)ds =

j =0 0

m 

Φ(t, s)B(s)uj (s)ds,

(3.158)

j =0 Ej

and so that m 

Ej = (0, t) and Ei



Ej = ∅ when i = j.

j =0

Write  u(s)  uj (s) for a.e. s ∈ Ej , j = 0, 1, . . . , m.

(3.159)

106

3 Existence of Admissible Controls and Optimal Controls

Then, by (3.156), (3.157), (3.158), and (3.159), we see that  t R (t, y0 ), y = Φ(t, 0)y0 + Φ(t, s)B(s) u(s)ds ∈ Y 0

R (t, y0 ). which leads to (3.153). Hence, we have proved that YR (t, y0 ) = Y Step 2. We show (3.152). Note that  R (t, y0 ) and YR (t) = Y y0 ∈YS

YR (t, co(U)) =



YR (t, y0 ) for all t ∈ (0, +∞).

y0 ∈YS

From these and the result in Step 1, (3.152) follows at once. (ii) Suppose that (L1) and (L2) hold. Arbitrarily fix t ∈ (0, +∞). We first show the convexity of YR (t). To this end, we let y1 , y2 ∈ YR (t) and λ ∈ (0, 1). Then by (3.152), there exist two vectors y0,1 , y0,2 ∈ YS and two functions u1 , u2 ∈ L(0, t; co(U)), so that y1 = y(t; 0, y0,1, u1 ) and y2 = y(t; 0, y0,2, u2 ).

(3.160)

Set y0,λ  λy0,1 + (1 − λ)y0,2 and uλ  λu1 + (1 − λ)u2 ∈ L(0, t; co(U)). By (L2), we have that y0,λ ∈ YS . Then by (3.152), we deduce that yλ  λy1 + (1 − λ)y2 = y(t; 0, y0,λ , uλ ) ∈ YR (t; co(U)) = YR (t), which implies that YR (t) is convex. We next show that YR (t) is closed. For this purpose, we let {yk }k≥1 ⊆ YR (t) satisfy that yk →  y . By (3.152), there are two sequences {y0,k }k≥1 ⊆ YS and {uk }k≥1 ⊆ L(0, t; co(U)) so that 

t

yk = Φ(t, 0)y0,k +

Φ(t, s)B(s)uk (s)ds for all k ≥ 1.

(3.161)

0

Since {uk }k≥1 ⊆ L(0, t; co(U)), it follows by (L1) that {uk }k≥1 is bounded in L2 (0, t; Rm ). Meanwhile, since yk →  y , it follows from (L1), the boundedness of {uk }k≥1 and (3.161) that {y0,k } is bounded in Rn . Then by (L1) and (L2), there exists a subsequence of {k}k≥1, denoted in the same manner, so that for some  y0 ∈ YS and  u ∈ L2 (0, t; Rm ), y0,k →  y0 and uk →  u weakly in L2 (0, t; Rm ).

(3.162)

3.3 Admissible Controls and Reachable Sets

107

From the second conclusion in (3.162), we can use the Mazur Theorem (see Theorem 1.8) to find, for each k ≥ 1, a finite sequence of nonnegative numbers Jk J {αkj }j k=1 , with αkj = 1, so that j =1

Jk

αkj uk+j →  u strongly in L2 (0, t; Rm ).

(3.163)

j =1 J

Since {uk+j }j k=1 ⊆ L(0, t; co(U)), it follows by (3.163) that  u ∈ L(0, t; coU).

(3.164)

y and because of (3.162), (3.152), and (3.164), we can pass to the Since yk →  limit for k → +∞ in (3.161) to get that 

t

 y = Φ(t, 0) y0 +

Φ(t, s)B(s) u(s)ds = y(t; 0,  y0 , u) ∈ YR (t, co(U)) = YR (t),

0

which implies that YR (t) is closed. This completes the proof of Theorem 3.9.

 

Remark 3.3 In infinitely dimensional cases, YR (t) may be not convex (see Example 5.2 on page 306 of [15]). At the end of this subsection, we will study the function d(·) (see (3.139)) in the framework that the controlled system is (3.3), the assumptions (L1) and (L2) (given at the beginning of Section 3.3.2) hold, and QE = (0, +∞) × YE , with YE ⊆ Rn a nonempty, convex, and compact subset. We begin with introducing the next J. von Neumann Theorem. Theorem 3.10 ([1]) Suppose that the following two conditions hold: (i) E and F are nonempty, convex, and compact subsets of Rn ; (ii) The function g(·, ·) : Rn × Rn → R satisfies that for each y ∈ F , the function g(·, y) : Rn → R is convex and lower semi-continuous; and that for each x ∈ E, the function g(x, ·) : Rn → R is concave and upper semi-continuous. Then there exists a saddle point (x ∗ , y ∗ ) ∈ E × F so that min max g(x, y) = g(x ∗ , y ∗ ) = max min g(x, y). x∈E y∈F

y∈F x∈E

(3.165)

Proposition 3.2 Suppose that the following assumptions hold: (i) U satisfies (L1). (ii) QE = (0, +∞) × YE and YE ⊆ Rn is nonempty, convex, and compact. (iii) YS satisfies (L2) and is bounded. Then for each t > 0,

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3 Existence of Admissible Controls and Optimal Controls

d(t) = max



* + min Φ(t, 0)∗ ψ, y0 Rn − max ψ, y1 Rn

ψ∈B1 (0) y0 ∈YS

 +

y1 ∈YE

t

* +  min B(s)∗ Φ(t, s)∗ ψ, u Rm ds .

0 u∈U

Here, B1 (0) is the closed unit ball in Rn . Proof Arbitrarily fix t > 0. Since YS is a bounded subset of Rn , by (L1) and (L2), we can apply Theorem 3.9 to see that YR (t) is convex and compact. This, together with the convexity and compactness of YE , implies that the following set: YRE (t)  YR (t) − YE (t) is also convex and compact. Hence, it follows from (3.139) that d(t) =

inf

z∈YRE (t )

zRn =

min zRn =

z∈YRE (t )

min

max ψ, zRn .

z∈YRE (t ) ψ∈B1 (0)

Then, we apply Theorem 3.10, with E = YRE (t), F = B1 (0) and g(x, y) = x, yRn , x, y ∈ Rn , to get that d(t) = max

min ψ, zRn

ψ∈B1 (0) z∈YRE & (t )

= max



ψ, Φ(t, 0)y0 +

min

ψ∈B1 (0) y0 ∈YS ,y1 ∈YE u∈L(0,t ;U)

t

' Φ(t, s)B(s)u(s)ds − y1

0

. Rn

(3.166) Meanwhile, since B(·) ∈ C([0, t]; Rn×m ) so that

n×m ), L∞ loc (0, +∞, R



t 0

for any ε > 0, there exists Bε (·) ∈

Bε (s) − B(s)Rn×m ds ≤ ε.

This, together with the Measurable Selection Theorem (see Theorem 1.17), implies that  t * + min B(s)∗ Φ(t, s)∗ ψ, u(s) Rm ds u∈L(0,t  t ;U) 0 (3.167) * + min B(s)∗ Φ(t, s)∗ ψ, u Rm ds. = 0 u∈U

From (3.166) and (3.167), the desired result follows at once. This ends the proof of Proposition 3.2.   Corollary 3.5 Suppose that U = Br (0), with r ∈ (0, +∞); YS = {y0 } ⊆ Rn \ {0} and YE = {0}.

3.3 Admissible Controls and Reachable Sets

has an admissible control if and only if there is t ∗ > 0 so that

Q ,QE

S Then (T P )min

∗

109



∗

Φ(t , 0) ψ, y0 Rn ≤ r

t∗

0

B(s)∗ Φ(t ∗ , s)∗ ψRm ds for all ψ ∈ B1 (0). (3.168)

Proof Since YR (t) is compact and YE (t) = {0} for any t ∈ (0, +∞), by QS ,QE has an admissible control if Theorem 3.8, we can easily check that (T P )min and only if there is t ∗ > 0 so that d(t ∗ ) = 0. This, together with Proposition 3.2, QS ,QE has an admissible control if and only if implies that (T P )min Φ(t ∗ , 0)∗ ψ, y0 Rn ≤ r



t∗ 0

B(s)∗ Φ(t ∗ , s)∗ ψRm ds for all ψ ∈ B1 (0).  

This ends the proof of Corollary 3.5.

Remark 3.4 When A(·) ≡ A and B(·) ≡ B, we can prove that (3.168), with t ∗ > 0, is equivalent to r > N(y0 ).

(3.169)

Here, N(y0 ) is given by (3.48). Hence, in this special case, Corollary 3.5 coincides with Corollary 3.3. We first show that (3.168), with t ∗ > 0, implies (3.169). Observe that in this case, (3.168), with t ∗ > 0, is equivalent to what follows: e

t ∗ A∗

 ψ, y0 

Rn

t∗

≤r

B ∗ e(t

∗ −s)A∗

0

ψRm ds for all ψ ∈ Rn .

(3.170)

Since (3.170) is equivalent to the inequality:  ψ, y0 

Rn

t∗

≤r 0

∗

B ∗ e−sA ψRm ds for all ψ ∈ Rn ,

(3.171)

we see that when t ∗ > 0, (3.168) is equivalent to (3.171). Now, we suppose that (3.168) holds for some t ∗ > 0. Then (3.171) is true for this t ∗ . Define  ∗ A1  {B ∗ e−sA ψ, s ∈ (0, t ∗ )  ψ ∈ Rn } ⊆ L1 (0, t ∗ ; Rm ) and  ∗ A2  {B ∗ e−sA ψ, s ∈ (t ∗ , 2t ∗ )  ψ ∈ Rn } ⊆ L1 (t ∗ , 2t ∗ ; Rm ).

110

3 Existence of Admissible Controls and Optimal Controls ∗

It is obvious that A1 and A2 are finite dimensional. Since s → B ∗ e−sA , s ∈ (0, +∞), is an analytic function, we can easily check that the following operator is well defined and linear: F : A2 → A1 ∗ −sA∗

B e

∗

ψ, s ∈ (t ∗ , 2t ∗ ) → B ∗ e−sA ψ, s ∈ (0, t ∗ ).

Hence, there exists a positive constant c0 so that  2t ∗  t∗ ∗ ∗ B ∗ e−sA ψRm ds ≥ c0 B ∗ e−sA ψRm ds for all ψ ∈ Rn . t∗

0

This, along with (3.171), yields that  +∞ ∗ B ∗ e−sA ψRm ds r 0 t ∗  ∗ ≥r B ∗ e−sA ψRm ds + r 0

2t ∗ t∗

∗

B ∗ e−sA ψRm ds

(3.172)

≥ (1 + c0 )ψ, y0 Rn for all ψ ∈ Rn . By (3.48) and (3.172), we get that r ≥ (1 + c0 )N(y0 ), from which, (3.169) follows at once. We next show that (3.169) implies (3.168) for some t ∗ > 0. Suppose that (3.169) holds. Then, by (3.47), we have that )  +∞ ∗ r > y0 , ψRn B ∗ e−sA ψRm ds for all ψ ∈ Rn . (3.173) 0

Then, it follows from (3.31), (3.47), and (3.173) that y0 , ψRn ≤ 0 for all ψ ∈ NA,B , which indicates that ⊥ y0 ∈ NA,B .

(3.174)

By making use of (3.31) and (3.173) again, we get that 

+∞

r 0

∗

⊥ B ∗ e−sA ψRm ds > y0 , ψRn for all ψ ∈ NA,B ∩ ∂B1 (0).

(3.175)

 ∈ N ⊥ ∩ ∂B1 (0), we see from (3.175) that there is Now for an arbitrarily fixed ψ A,B tψ ∈ (0, +∞) so that  r 0

tψ 

∗  Rn .  Rm ds > y0 , ψ B ∗ e−sA ψ

(3.176)

3.3 Admissible Controls and Reachable Sets

111

Since the following two functions are continuous on Rn : 

tψ 

ψ→ 0

∗

B ∗ e−sA ψRm ds, ψ ∈ Rn ;

z → y0 , zRn , z ∈ Rn ,

 so that there exists a positive constant δ(ψ) 

tψ 

r 0

∗

B ∗ e−sA ϕRm ds > y0 , ϕRn for all ϕ ∈ Uδ(ψ) ,

(3.177)

where  ⊥ Rn < δ(ψ )}. ∩ ∂B1 (0)  ϕ − ψ Uδ(ψ)  {ϕ ∈ NA,B ⊥ ∩ Meanwhile, by the finite covering theorem, there are ψ1 , ψ2 , . . . , ψ0 ∈ NA,B ∂B1 (0) so that

⊥ NA,B

∩ ∂B1 (0) ⊆

0 

Uδ(ψi ) .

(3.178)

 t ∗  max{tψi  1 ≤ i ≤ 0 }.

(3.179)

i=1

Let

⊥ ∩∂B (0), there exists 1 ≤ i ≤  Then it follows by (3.178) that for each ψ ∈ NA,B 1 0 0 so that ψ ∈ Uδ(ψi0 ) . Hence, by (3.177), we have that

 y0 , ψRn < r ≤r

tψi

0

0 t ∗

∗

B ∗ e−sA ψRm ds ∗ −sA∗

B e

0

ψ

Rm

ds for all ψ ∈

(3.180)

⊥ NA,B

∩ ∂B1 (0).

⊥ so Moreover, for each ψ ∈ Rn , there are two vectors ψ1 ∈ NA,B and ψ2 ∈ NA,B that

ψ = ψ1 + ψ2 . This, together with (3.174), (3.180), and (3.31), implies that  t∗  ∗ ∗ −sA y0 , ψRn = y0 , ψ2 Rn ≤ r B e ψ2 Rm ds = r 0

From the latter, we see that (3.171), where Remark 3.4.

0

t∗

t∗

∗

B ∗ e−sA ψRm ds.

is given by (3.179), is true. We end

112

3 Existence of Admissible Controls and Optimal Controls

3.4 Existence of Optimal Controls In this section, we will present ways deriving the existence of optimal controls QS ,QE QS ,QE of (T P )min from the existence of admissible controls for (T P )min . From Theorem 1.13 and its proof, we can conclude the following procedure to show the existence of solutions for the optimization problem (1.21): The first step is to find a minimizing sequence. (In this step, we need to show the existence of minimizing sequences.) The second step is to find a convergent subsequence (from the minimizing sequence) in some suitable topology. The last step is to show that the limit of the above subsequence is a solution of the optimization problem. We can borrow the above-mentioned procedure to show the existence of optimal controls for some time optimal control problems, provided that these problems have admissible controls. We will explain this in detail in the forthcoming subsection.

3.4.1 Existence of Optimal Controls for Regular Cases In this subsection, we study the existence of optimal controls for the problem S ,QE . This problem is as follows: (T P )Q min tE∗ 

inf

(0,y0 ,tE ,u)∈Aad

tE ,

(3.181)

where the admissible tetrad set is defined by  Aad  (0, y0 , tE , u) ∈ QS × (0, +∞) × L(0, tE ; U) :

 (tE , y(tE ; 0, y0 , u)) ∈ QE ,

(3.182)

with QS = {0} × YS . We will present a procedure to show the existence of optimal controls for the problem (3.181), under the assumption that Aad = ∅. The first step is to study the admissible tetrad set. Since Aad = ∅, there are only two possibilities on the set Aad : Either it has finitely many elements or it has infinite elements. In the case that Aad has finitely many elements, it is obvious that the problem (3.181) has an optimal control. And the procedure is finished. In the case when Aad has infinite elements, there is a sequence of admissible tetrads {(0, y0,, tE, , u )}≥1 so that {tE, }≥1 is monotone decreasing and converges to t ∗ . We call such a sequence of admissible tetrads a minimizing sequence of the problem (3.181). The second step is to find a subsequence of the minimizing sequence so that it converges in some topology. More precisely, we need to show that there exists a subsequence of {}≥1, still denoted in the same way, so that for some (0,  y0 ) ∈ QS and  u ∈ L(0, tE∗ ; U), y0, u) in some suitable topology. y(·; 0, y0,, u ) → y(·; 0,  y0, u)) ∈ QE . The final step is to show that (tE∗ , y(tE∗ ; 0, 

3.4 Existence of Optimal Controls

113

The following lemma is quoted from [15] (see Lemma 3.2 on page 104 of [15]). It plays an important role in the study of the existence of optimal controls for the problem (3.181). 

Lemma 3.9 Let {eAt }t ≥0 be a compact C0 semigroup on a Banach space Z. Let p > 1 and t0 ∈ (0, +∞). Then the operator L : Lp (0, t0 ; Z) → C([0, t0 ]; Z) defined by 

t

L (u)(t) 



eA(t −s)u(s) ds for all t ∈ [0, t0 ]

(3.183)

0

is compact. The following assumptions will be effective in the rest of this subsection. (H 1) A : D(A) ⊆ Y → Y generates a compact semigroup {eAt }t ≥0 on Y . (H 2) f : (0, +∞) × Y × U → Y is measurable in (t, y, u); uniformly continuous in u locally at (t0 , y0 , u0 ) for almost every t0 ∈ (0, +∞), for each y0 ∈ Y and each u0 ∈ U ; and continuous in y uniformly with respect to u ∈ U for almost every t ∈ (0, +∞). (H 3) The set f (t, y, U)  {f (t, y, u)|u ∈ U} is convex and closed. (H 4) f (t, y, u)Y ≤ l(t)yY + ϕ(t) for all t ∈ (0, +∞) and u ∈ U, where l(·) ∈ L∞ (0, +∞) and ϕ(·) ∈ Lp (0, +∞) are nonnegative and p > 1. (H 5) YS is a nonempty convex, bounded, and closed subset of Y . (H 6) QE is a nonempty closed subset of (0, +∞) × Y . The main result of this subsection is as follows. Theorem 3.11 Suppose that (H 1) − (H 6) hold. Assume that Aad (given by (3.182)) is not empty. Then the problem (3.181) has an optimal control. Proof Since Aad = ∅, we have that tE∗ < +∞. Thus, there is a sequence {(0, y0,k , tE,k , uk )}k≥1 in Aad so that tE,k  tE∗

(3.184)

and  tE,k , y(tE,k ; 0, y0,k , uk ) ∈ QE for all k ≥ 1. (3.185) By the first conclusion in (3.185) and by (H 5), we can find a subsequence of {k}k≥1, denoted in the same manner, so that y0,k ∈ YS , uk ∈ L(0, tE,k ; U),

y0,k →  y0



weakly in Y for some  y0 ∈ Y.

(3.186)

Claim One: We have that  y0 ∈ YS .

(3.187)

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3 Existence of Admissible Controls and Optimal Controls

Indeed, since YS is convex and closed in Y , it is weakly closed in Y , which, along with (3.186), leads to (3.187). This ends the proof of Claim One. We now let, for each k ∈ N+ ,   yk (t)  y t; 0, y0,k , uk for each t ∈ [0, tE,k ]

(3.188)

and fk (t)  f (t, yk (t), uk (t)) for a.e. t ∈ (0, tE,k ).

(3.189)

Then we have that for each k ∈ N+ , 

t

yk (t) = eAt y0,k +

eA(t −τ ) fk (τ ) dτ for each t ∈ [0, tE,k ].

(3.190)

0

Claim Two: There is f ∈ Lp (0, tE∗ ; Y ) and a subsequence of {k}k≥1 , denoted in the same way, so that fk → f weakly in Lp (0, tE∗ ; Y ),

(3.191)

and so that y (tE∗ ) yk (tE∗ ) → 

strongly in Y,

(3.192)

where  y (·) is the function defined by 

t

 y (t)  eAt  y0 + 0

eA(t −τ )f(τ ) dτ for all t ∈ [0, tE∗ ].

(3.193)

To this end, we first use (3.190), (3.189), (H 1), and (H 4) to get that for all k ∈ N+ and t ∈ [0, tE,k ],  t

A(t −τ )

fk (τ ) dτ

e Y 0 t   C(t −τ ) Ct l(τ ) yk (τ )Y + ϕ(τ ) dτ. Ce ≤ Ce y0,k Y +

yk (t)Y ≤ eAt y0,k Y +

(3.194)

0

(Here and throughout the proof of this theorem, C denotes a generic positive constant independent of k.) Since y0,k ∈ YS for all k ≥ 1 (see (3.185)) and YS is bounded (see (H 5)), we can apply Gronwall’s inequality to (3.194) to get that yk (t)Y ≤ C for all t ∈ [0, tE,k ] and all k ≥ 1.

(3.195)

3.4 Existence of Optimal Controls

115

By (3.189), (3.195), and (H 4), we find that {fk }k≥1 is bounded in Lp (0, tE∗ ; Y ). Thus, there is f ∈ Lp (0, tE∗ ; Y ) and a subsequence of {k}k≥1 , denoted in the same way, satisfying (3.191). Next, by (3.191), (H 1), and Lemma 3.9, there is a subsequence of {k}k≥1 , denoted in the same manner, so that 

t

eA(t −τ ) fk (τ ) dτ →

0



t

eA(t −τ )f(τ ) dτ

0

strongly in C([0, tE∗ ]; Y ).

(3.196) Meanwhile, by (H 1) and (3.186), we can find a subsequence of {k}k≥1, denoted in the same way, so that ∗

∗

eAtE y0,k → eAtE  y0

strongly in Y.

(3.197)

Then, by (3.190), (3.196), (3.193), and (3.197), we can easily obtain (3.192). This ends the proof of Claim Two. Claim Three: we have that y (tE∗ )) ∈ QE , where  y (·) is given by (3.193). (tE∗ , 

(3.198)

In fact, by (3.188), (H 1), (H 4), and (3.195), we have that y (tE∗ )Y yk (tE,k ) −   tE,k

∗

A(tE,k −τ )

≤ eA(tE,k −tE ) yk (tE∗ ) −  y (tE∗ ) + f (τ, yk (τ ), uk (τ )) dτ

e ∗ Y Y t



E

A(tE,k −tE∗ )

A(tE,k −tE∗ ) ∗

∗ ∗ ∗ ≤ e (yk (tE ) −  y (tE )) + e  y (tE ) −  y (tE ) Y Y  tE,k + CeC(tE,k −τ ) [l(τ ) + ϕ(τ )] dτ tE∗

∗ ∗

C(t ≤ Ce E,k −tE ) yk (tE∗ ) −  y (tE∗ )Y + eA(tE,k −tE ) y (tE∗ ) −  y (tE∗ ) Y  tE,k +C [l(τ ) + ϕ(τ )] dτ. tE∗

This, together with (3.192) and (3.184), yields that y (tE∗ ) yk (tE,k ) → 

strongly in Y.

(3.199)

Meanwhile, by (3.185) and (3.188), we have that (tE,k , yk (tE,k )) ∈ QE . From this, (H 6), (3.184), and (3.199), one can easily obtain (3.198). This ends the proof of Claim Three.

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3 Existence of Admissible Controls and Optimal Controls

Claim Four: There is  u ∈ L(0, tE∗ ; U) so that u) =  y (t) for each t ∈ [0, tE∗ ]. y (t; 0,  y0 ,

(3.200)

In fact, by (3.191), we can use the Mazur Theorem (see Theorem 1.8) to find, for Jk each k ≥ 1, a finite sequence of nonnegative numbers {αkj }Jj k=1 , with αkj = 1, j =1

so that gk 

Jk

αkj fk+j → f strongly in Lp (0, tE∗ ; Y ).

(3.201)

j =1

Meanwhile, by the same way as that used to prove (3.192), we can verify that for any t ∈ (0, tE∗ ], there exists a subsequence of {k}k≥1 , still denoted in the same way, so that yk (t) →  y (t).

(3.202)

By (H 2) and (3.189), we see that there is a subset E ⊆ (0, tE∗ ), with |E| = 0 (i.e., the measure of E is zero), so that for each t ∈ (0, tE∗ ) \ E, f (t, ·, u) is continuous uniformly with respect to u, and so that fk (t) = f (t, yk (t), uk (t)) for all t ∈ (0, tE∗ ) \ E.

(3.203)

Arbitrarily fix t ∈ (0, tE∗ ) \ E. Then by the aforementioned continuity of f (t, ·, u) and by (3.202), we find that for an arbitrarily fixed ε ∈ (0, 1), there exists a positive integer k0 (t, ε) so that y (t), u)Y ≤ ε for all k ≥ k0 (t, ε) and u ∈ U. f (t, yk (t), u) − f (t,  This, along with (3.203) and (3.185), yields that for all k ≥ k0 (t, ε), y (t), uk+j (t)) + εB1 (0) fk+j (t) = f (t, yk+j (t), uk+j (t)) ∈ f (t,  ⊆ f (t,  y (t), U) + εB1 (0).

(3.204)

Here B1 (0) denotes the closed ball in Y centered at 0 and of radius 1. Since the set f (t,  y (t), U) is convex and closed (see (H 3)), it follows from (3.204) that gk (t) 

Jk

αkj fk+j (t) ∈ f (t,  y (t), U) + εB1 (0) for all k ≥ k0 (t, ε).

(3.205)

j =1

Since t ∈ (0, tE∗ ) \ E and ε ∈ (0, 1) was arbitrarily fixed, it follows by (3.201), (3.205), and the closedness of f (t,  y (t), U) that f(t) ∈ f (t,  y (t), U) for a.e. t ∈ (0, tE∗ ).

3.4 Existence of Optimal Controls

117

From this and the condition (H 2), we can apply the Filippov Lemma (see Corollary 1.1) to find  u ∈ L(0, t ∗ ; U) so that f(t) = f (t,  y (t), u(t)) for a.e. t ∈ (0, tE∗ ). This, together with (3.193), leads to (3.200). This ends the proof of Claim Four. Finally, by (3.187), (3.198), and (3.200), we see that (0,  y0 , tE∗ , u) ∈ Aad . This completes the proof of Theorem 3.11.  

3.4.2 Existence of Optimal Controls for Blowup Case The time optimal control problem studied in this subsection is a special case of the problem (2.18). Namely, it is a minimal blowup time control problem. We will derive the existence of optimal controls for this problem, under the assumption that the set of admissible controls is not empty. We begin with introducing this problem. Let y0 ∈ Rn be arbitrarily fixed. Consider the following controlled system: ⎧ ⎨ dy = f (t, y(t)) + B(t)u(t), t ∈ (0, +∞), (3.206) dt ⎩ y(0) = y . 0

Here, u ∈ L(0, +∞; U), B(·), f (·, ·), and U satisfy the following assumptions: (S1) B(·) ∈ L1loc (0, +∞; Rn×m ); (S2) f : [0, +∞) × Rn → Rn satisfies that for each y ∈ Rn , f (·, y) is measurable on [0, +∞), and that for any r > 0, there exists a positive constant Mr so that f (t, y1 ) − f (t, y2 )Rn ≤ Mr y1 − y2 Rn , for all y1 , y2 ∈ Br (0) and t ≥ 0. Here Br (0) denotes a closed ball in Rn , centered at 0 and of radius r; (S3) U is a nonempty bounded, convex, and closed set in Rm . We write y(·; 0, y0, u) for the solution of (3.206). The minimal blowup time control problem under study is as: tE∗ 

inf

(0,y0 ,tE ,u)∈Aad

tE ,

(3.207)

where the set of admissible tetrads Aad is given by   Aad  (0, y0 , tE , u) ∈ {(0, y0 )} × (0, +∞) × L(0, tE ; U) 

 lim y(t; 0, y0 , u|(0,t ))Rn = +∞ .

t →tE −

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3 Existence of Admissible Controls and Optimal Controls

One can easily check that the problem (3.207) can be put into the framework of the problem (2.18). Lemma 3.10 Suppose that (S1)–(S3) hold. Let {u }≥1 ⊆ L(0, +∞; U) satisfy that u → u weakly star in L∞ (0, +∞; Rm ) for some u ∈ L(0, +∞; U). (3.208) Further assume that y(·; 0, y0, u) exists on [0, t0∗ ] for some t0∗ ∈ (0, +∞). Then there is δ ∈ (0, +∞) and 0 ∈ N+ so that {y(·; 0, y0, u )}≥0 is bounded in C([0, t0∗ + δ]; Rn ). Proof Let y(·; 0, y0 , u) exist on [0, t0∗ ] for some t0∗ ∈ (0, +∞). Then there is δ ∈ (0, +∞) so that y(·; 0, y0, u) ∈ C([0, t0∗ + δ]; Rn ). Let y(t)  y(t; 0, y0, u), t ∈ [0, t0∗ + δ], and let r

∗

max∗ y(t)Rn + 1 and ε0  re−M2r (t0 +δ) .

0≤t ≤t0 +δ

By (S1), one can easily check that the following function b(·) is absolutely continuous on [0, t0∗ + δ]:  t B(s)L (Rm ,Rn ) ds, t ≥ 0. t → b(t)  0

Thus, there are ti , i = 0, . . . , k0 (for some k0 ∈ N+ ), with 0  t0 < t1 < t2 < · · · < tk0 = t0∗ + δ, so that

 max

tj+1

0≤j ≤k0 −1 tj

B(s)L (Rm ,Rn ) ds ≤

ε0 , 4(c0 + 1)

(3.209)

 with c0  sup{uRm  u ∈ U}. Meanwhile, by (S1) and (3.208), there is a positive integer 0 so that when  ≥ 0 ,





tj 0

B(s)(u (s) − u(s))ds

Rn

≤

ε0 for all 1 ≤ j ≤ k0 . 2

This, along with (3.209), yields that

 t



max B(s)(u (s) − u(s))ds

t ∈[0,t ∗ +δ] 0

0

Rn

≤ ε0 for all  ≥ 0 .

(3.210)

3.4 Existence of Optimal Controls

119

Now we claim that max

t ∈[0,t0∗ +δ]

y(t; 0, y0 , u )Rn < 2r for all  ≥ 0 .

(3.211)

By contradiction, we suppose that (3.211) was not true. Then there would be a t ∈ (0, t0∗ + δ] so that positive integer   ≥ 0 and    y(t; 0, y0 , u  )Rn < 2r for all t ∈ [0, t ); and y(t; 0, y0 , u  )Rn = 2r. (3.212) Let    y  (t)  y(t; 0, y0 , u  ), t ∈ 0, t . Then by (3.206), (3.212), (S2), and (3.210), we get that y  (t) − y(t)Rn



t

= [f (s, y  (s)) − f (s, y(s)) + B(s)(u  (s) − u(s))]ds 0  Rn t

  ∗

 y ≤ M2r  (s) − y (s) Rn ds + ε0 for all t ∈ 0, t , 0

which, combined with Gronwall’s inequality, indicates that M2r t   y ≤ r.  (t ) − y(t )Rn ≤ ε0 e

(3.213)

Since y( t )Rn ≤ r − 1, it follows by (3.213) that     y  (t )Rn ≤ y  (t ) − y(t )Rn + y(t )Rn ≤ 2r − 1. This contradicts (3.212). Hence, (3.211) holds. Thus we end the proof of Lemma 3.10.   Theorem 3.12 Suppose that (S1)–(S3) hold. Assume that Aad = ∅. Then the problem (3.207) has an optimal control. Proof Since Aad = ∅, we have that tE∗ < +∞. Thus, there exist two sequences {tE, }≥1 ⊆ (0, +∞) and {u }≥1 ⊆ L(0, tE, ; U) so that lim y(t; 0, y0 , u |(0,t ))Rn = +∞

t →tE, −

(3.214)

and tE,  tE∗ . Arbitrarily fix u0 ∈ U. Define   u (t) 

u (t), t ∈ (0, tE, ), u0 , t ∈ [tE, , +∞).

(3.215)

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3

Existence of Admissible Controls and Optimal Controls

Then we have that y(t; 0, y0 , u ) = y(t; 0, y0 , u ) for all t ∈ [0, tE, ).

(3.216)

Meanwhile, by (S3), there exists a subsequence of {}≥1, denoted in the same manner, so that  u → u∗ weakly star in L∞ (0, +∞; Rm ) for some u∗ ∈ L∞ (0, +∞; U). (3.217) We now claim that lim y(t; 0, y0 , u∗ |(0,t ))Rn = +∞.

t →tE∗ −

(3.218)

By contradiction, suppose that (3.218) was not true. Then y(·; 0, y0, u∗ ) would exist on [0, tE∗ ]. Thus, by Lemma 3.10 and (3.217), there are two positive constants δ and 0 so that {y(·; 0, y0, u )}≥0 is bounded in C([0, tE∗ + δ]; Rn ). Hence, by (3.215) and (3.216), there are two positive constants 1 ≥ 0 and C so that max y(t; 0, y0 , u )Rn ≤ C for all  ≥ 1 ,

t ∈[0,tE, )

which contradicts (3.214). Hence, (3.218) is true. From (3.218), we find that u∗ is an optimal control for the problem (3.207). This completes the proof of Theorem 3.12.  

Miscellaneous Notes The existence of admissible controls and optimal controls is an important subject in the studies of time optimal control problems. It is independent of other issues, such as Pontryagin’s maximum principle, the bang-bang property, and so on. In many cases, the existence of optimal controls can be derived from the existence of admissible controls, through using some standard methods which are introduced in Section 3.4.1. These methods are summarized from the previous works in this direction (see, for instance, [2, 3, 20, 22, 28], and [23]). The existence of admissible controls can be treated as a kind of constraint controllability. Indeed, an admissible control is a control (in a constraint set) driving the corresponding solution (of a system) to a given target in some time, while the general controllability is to ask for a control, without any constraint, so that the corresponding solution (of a system) reaches a given target set at a fixed ending time. Hence, we study the existence of admissible controls of time optimal control problems from three different viewpoints: the controllability, the minimal norm control problems, and reachable sets.

3.4 Existence of Optimal Controls

121

Several notes on Section 3.1 are given in order: • Many materials in Section 3.1 are based on some parts of [5, 12, 24], and [25] (which are about finite dimensional cases) and some parts of [23] and [30] (which are about infinite dimensional cases). Some materials in Section 3.1 (see, for instance, Theorem 3.4) are developed in this monograph, based on known results and methods in some existing literatures. • Materials in Section 3.1.3 are based on [23]. There, the existence of admissible controls is global, i.e., the initial state can be arbitrarily given. (Notice that the controlled system is a semi-linear heat equation.) In general, it is not easy to get the global null or approximate controllability for nonlinear equations in a fixed time interval. For such works, we refer readers to [6, 7] and [4]. It should be more difficult to obtain the global controllability for nonlinear equations with constraint controls even in a mobile time interval, in other words, it should be harder to get the existence of admissible controls for nonlinear equations. To the best of our knowledge, such studies are very limited (see, for instance, [30]). Our global result (obtained in Section 3.1.3) is due to the particularity of the semi-linear term f in the Equation (3.59). An interesting question is as follows: Can we have the existence of admissible controls when f (r)  r| ln r|3/2 ? This growth rate appeared in both [4] and [7]. For Section 3.2, we would like to mention what follows: • The usual minimal norm control problem is to ask for a control having the minimal norm among all such controls that drive the corresponding solutions of a controlled system from a given initial state to a given target in a fixed time interval. Such problems have been widely studied (see, for instance, [8– 10], and [26]). The norm optimal control problem studied in Section 3.2.1 is a generalization of the above-mentioned minimal norm control problem. We devote this generalization to the current monograph. • The materials in Section 3.2.2 are essentially taken from [32] (see also [26] and [31]). The materials in Section 3.3 are based on some development of the related materials in Section 5 of Chapter 7 in [15]. About Section 3.4, we would like to mention the following facts: • The materials in Section 3.4.2 are simplified from the materials in [18]. More general results on minimal blowup time control problems are presented in [19]. • It is worth mentioning the paper [16], where the author studied some minimal quenching time control problems of some ODEs. It is the first paper studying such interesting problems, which have some connections with, but differ from the minimal time control problems. For more results about the existence of admissible controls and optimal controls, we would like to mention [13, 14, 21, 27, 29], and [17].

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At the end of this miscellaneous note, we would like to present several open problems: • Given infinite dimensional system (A, B) with the ball-type control constraint, can we find a criterion on the existence of admissible controls (such as some counterpart of Corollary 3.2)? For a finite or infinite dimensional system (A, B) with other kinds of control constraints, what is a criterion on the existence of admissible controls? • For controlled systems with some state constraints (for instance, heat equations with nonnegative states), can we find some reasonable condition to ensure the existence of admissible controls?

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Chapter 4

Maximum Principle of Optimal Controls

In this chapter, we will present some first-order necessary conditions on time optimal controls for some evolution systems. Such necessary conditions are referred to as the Pontryagin Maximum Principle (or Pontryagin’s maximum principle), which provides some information on optimal controls. Differing from analysis methods (see [3, 12, 14] and [4]), there are other geometric ways to approach the Pontryagin Maximum Principle (see, for instance, [1] and [16]). Usually, these geometric methods are based on the use of separation theorems and representation theorems. In the first three sections of this chapter, we introduce geometric methods S ,QE in several different cases. Three different through studying the problem (T P )Q min S ,QE kinds of Pontryagin’s maximum principles for (T P )Q are given in order. min They are respectively called as: the classical Pontryagin Maximum Principle, the local Pontryagin Maximum Principle, and the weak Pontryagin Maximum Principle. These Pontryagin’s maximum principles are obtained by separating different objects: separating the target from the reachable set in the state space at the optimal time; separating a reachable set from a controllable set in the state space before the optimal time; separating the target from the reachable set in the reachable space at the optimal time. We then discuss the classical Pontryagin Maximum Principle and S ,QE in the final section, where the local Pontryagin Maximum Principle for (T P )Q max QS ,QE the methods are similar to those used for (T P )min .

4.1 Classical Maximum Principle of Minimal Time Controls S ,QE This section studies the problem (T P )Q , under the following framework min (AT P ):

(i) The state space Y and the control space U are real separable Hilbert spaces.

© Springer International Publishing AG, part of Springer Nature 2018 G. Wang et al., Time Optimal Control of Evolution Equations, Progress in Nonlinear Differential Equations and Their Applications 92, https://doi.org/10.1007/978-3-319-95363-2_4

125

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4 Maximum Principle of Optimal Controls

(ii) The controlled system is as: y(t) ˙ = Ay(t) + D(t)y(t) + B(t)u(t), t ∈ (0, +∞),

(4.1)

where A : D(A) ⊆ Y → Y generates a C0 semigroup {eAt }t ≥0 on Y ; D(·) ∈ L∞ (0, +∞; L (Y )) and B(·) ∈ L∞ (0, +∞; L (U, Y )). Let {Φ(t, s) : t ≥ s ≥ 0} be the evolution system generated by A + D(·) over Y . Given τ ≥ 0, y0 ∈ Y and u ∈ L∞ (τ, +∞; U ), denote by y(·; τ, y0 , u) the unique mild solution of (4.1) over [τ, +∞), with the initial condition y(τ ) = y0 , i.e., for each T > τ , y(·; τ, y0 , u)|[τ,T ] is the mild solution of (4.1) over [τ, T ], with the initial condition y(τ ) = y0 . (The existence and uniqueness of such solutions are ensured by Proposition 1.3.) (iii) The control constraint set U is nonempty, bounded, convex, and closed in U . (iv) Let QS = {0} × YS and QE = (0, +∞) × YE , where YS and YE are two nonempty, bounded, convex, and closed in Y so that YS ∩ YE = ∅. S ,QE We always assume that the problem (T P )Q has an optimal control. The aim of min S ,QE this section is to derive the classical Pontryagin Maximum Principle of (T P )Q min by separating the target from the reachable set (at the optimal time) in the state space.

Remark 4.1 Several notes are given in order. S ,QE is a kind of necessary (i) The Pontryagin Maximum Principle of (T P )Q min condition for an optimal control. It is indeed an Euler equation associated with a minimizer of a variational problem. Thus, in studies of this subject, it is not necessary to assume the existence of optimal control in general. In this chapter, for the sake of convenience, we make this assumption. S ,QE , (ii) Recall what follows: When (0, y0∗ , tE∗ , u∗ ) is an optimal tetrad for (T P )Q min ∗ ∗ u is called an optimal control and tE is called the optimal time (or the minimal time). We write y ∗ (·) for y(·; 0, y0∗ , u∗ ), and call it as the corresponding optimal trajectory (or the optimal state). (iii) Throughout this chapter, we will use tE∗ to denote the optimal time of S ,QE . (T P )Q min

4.1.1 Geometric Intuition Let us recall the reachable set at the time t (with 0 < t < +∞) for the system (4.1) (see Section 3.3.1):    (4.2) YR (t)  y(t; 0, y0, u)  y0 ∈ YS , u ∈ L(0, t; U) . We define a distance function d(·) between YR (·) and the target YE as follows: d(t)  dist{YR (t), YE } =

inf

y1 ∈YR (t ),y2 ∈YE

y1 − y2 Y ,

t ∈ [0, tE∗ ].

(4.3)

4.1 Classical Maximum Principle of Minimal Time Controls

127

YE YR(t1)

YR(t2)

YR(t*E)

t1 < t2 < … < t*E

Fig. 4.1 Evolution of reachable sets

Here, we denote YR (0)  YS . From geometric intuition, we observe what follows: At the initial time t = 0, the set YS is away from the target YE (since we assumed that YE ∩YS = ∅). Thus, we have that d(0) > 0. As time t goes on, the reachable set YR (t) is getting closer and closer to the target YE . The first time t at which YR (t) hits QS ,QE YE is exactly the optimal time tE∗ . (Here, we used the assumption that (T P )min ∗ has an optimal control.) Hence, tE is the first zero point of the function d(·). Since YR (tE∗ ) is convex (see Lemma 4.3), there is a hyperplane so that YR (tE∗ ) and YE stay on two sides of this hyperplane respectively. Mathematically, we say that this hyperplane separates YR (tE∗ ) and YE . In this book, we prefer to say that a normal vector of this hyperplane separates YR (tE∗ ) and YE . In finitely dimensional cases, such hyperplane exists, provided that YR (tE∗ ) is convex. For infinitely dimensional cases, the convexity of YR (tE∗ ) cannot ensure the existence of such hyperplanes. The above-mentioned separation will play an important role in deriving the Pontryagin QS ,QE Maximum Principle for (T P )min (Figure 4.1).

4.1.2 Classical Maximum Principle First of all, we give the definition of the classical Pontryagin Maximum Principle S ,QE for the problem (T P )Q . min Definition 4.1 Several definitions are given in order. (i) An optimal control u∗ (associated with an optimal tetrad (0, y0∗ , tE∗ , u∗ )) to QS ,QE (T P )min is said to satisfy the classical Pontryagin Maximum Principle, if there exists ϕ(·) ∈ C([0, tE∗ ]; Y ), with ϕ(·) = 0 (i.e., ϕ is not a zero function), so that ϕ(t) ˙ = −A∗ ϕ(t) − D(t)∗ ϕ(t) for a.e. t ∈ (0, tE∗ );

(4.4)

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4 Maximum Principle of Optimal Controls

H (t, y ∗(t), u∗ (t), ϕ(t)) = max H (t, y ∗ (t), u, ϕ(t)) for a.e. t ∈ (0, tE∗ ), u∈U

(4.5) where y ∗ ∈ C([0, tE∗ ]; Y ) is the corresponding optimal trajectory, and where H (t, y, u, ϕ)  ϕ, D(t)y + B(t)uY

(4.6)

for a.e. t ∈ (0, tE∗ ) and all (y, u, ϕ) ∈ Y × U × Y ; ϕ(0), y0 − y0∗ Y ≤ 0 for each y0 ∈ YS ;

(4.7)

+ ϕ(tE∗ ), z − y ∗ (tE∗ ) Y ≥ 0 for each z ∈ YE .

(4.8)

and *

Here, (4.4) is called the dual (or adjoint, or co-state) equation; (4.5) is called the maximum condition; the function H (·) defined by (4.6) is called QS ,QE ); (4.7) and (4.8) are called the the Hamiltonian (associated with (T P )min transversality conditions. QS ,QE (ii) The problem (T P )min is said to satisfy the classical Pontryagin Maximum Principle, if any optimal control satisfies the classical Pontryagin Maximum Principle. Recall Definition 1.7 for the separability. We have the following result: Q ,Q

S E satisfies the classical Pontryagin Maximum Theorem 4.1 The problem (T P )min ∗ Principle if and only if YR (tE ) and YE are separable in Y .

Remark 4.2 Several notes are given in order. (i) From Theorem 4.1, we see that the key to get the classical Pontryagin Maximum Principle is to find conditions so that YR (tE∗ ) and YE are separable in Y . (ii) The proof of Theorem 4.1 is based on the classical separation (of reachable set at the optimal time and the target) and the representation formula in the next lemma, which is indeed Proposition 5.7 in Chapter 2 of [12]. We omit its proof. ˙ = Lemma 4.1 Let T > 0 and ψT ∈ Y . Let ψ be the solution of the equation: ψ(t) −A∗ ψ(t) − D(t)∗ ψ(t), t ∈ (0, T ), with ψ(T ) = ψT . Then for each y0 ∈ Y and each u ∈ L(0, T ; U), the following representation formula holds:  y(T ; 0, y0, u), ψT Y = y0 , ψ(0)Y +

T

B(s)∗ ψ(s), u(s)U ds.

(4.9)

0

We now are in a position to prove Theorem 4.1. Proof (Proof of Theorem 4.1) We first show the sufficiency. Assume that YR (tE∗ ) and QS ,QE . Let YE are separable in Y . Let (0, y0∗ , tE∗ , u∗ ) be an optimal tetrad to (T P )min

4.1 Classical Maximum Principle of Minimal Time Controls

129

y ∗ be the corresponding optimal trajectory. It follows by Definition 1.7 that there is ϕ0 ∈ Y with ϕ0 Y = 1 and C ∈ R so that z1 , ϕ0 Y ≥ C ≥ z2 , ϕ0 Y for all z1 ∈ YR (tE∗ ) and z2 ∈ YE .

(4.10)

Since y ∗ (tE∗ ) ∈ YR (tE∗ ) ∩ YE , it follows from (4.10) that min z1 , ϕ0 Y = y ∗ (tE∗ ), ϕ0 Y = max z2 , ϕ0 Y .

z1 ∈YR (tE∗ )

z2 ∈YE

(4.11)

Let ϕ be the solution to the Equation (4.4) with ϕ(tE∗ ) = −ϕ0 . Then by Lemma 4.1, we see that for each y0 ∈ YS and u ∈ L(0, tE∗ ; U), y(tE∗ ; 0, y0, u), −ϕ0 Y

 = y0 , ϕ(0)Y +

tE∗

B(t)∗ ϕ(t), u(t)U dt.

(4.12)

0

By the first equality of (4.11) and (4.12), we get that y0∗ , ϕ(0)Y = max y0 , ϕ(0)Y y0 ∈YS

(4.13)

and  max ∗

u(·)∈L(0,tE ;U) 0

tE∗

B(t)∗ ϕ(t), u(t)U dt =



tE∗

B(t)∗ ϕ(t), u∗ (t)U dt.

(4.14)

0

We now claim that B(t)∗ ϕ(t), u∗ (t)U = maxB(t)∗ ϕ(t), uU a.e. t ∈ (0, tE∗ ). u∈U

(4.15)

Indeed, since U is separable, by Proposition 1.1, there exists a countable subset U0 = {u }≥1 so that U0 is dense in U. From (4.14) it follows that for each u ∈ L(0, tE∗ ; U), 

tE∗

[B(t)∗ ϕ(t), u∗ (t)U − B(t)∗ ϕ(t), u(t)U ]dt ≥ 0.

0

For each u ∈ U0 , we define the following function: g (t)  B(t)∗ ϕ(t), u∗ (t)U − B(t)∗ ϕ(t), u U , t ∈ (0, tE∗ ).

(4.16)

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4 Maximum Principle of Optimal Controls

Then g (·) ∈ L1 (0, tE∗ ). Thus, there exists a measurable set E ⊆ (0, tE∗ ) with |E | = tE∗ , so that any point in E is a Lebesgue point of g (·). Namely, 1 δ→0+ δ lim



t +δ t −δ

|g (s) − g (t)|ds = 0 for each t ∈ E .

Now, for any t ∈ E and δ > 0, we define  uδ (s) 

u∗ (s), if s ∈ (0, tE∗ ) \ Bδ (t), if s ∈ (0, tE∗ ) ∩ Bδ (t). u ,

Then, by (4.16), we get that  (0,tE∗ )∩Bδ (t )

[B(s)∗ ϕ(s), u∗ (s)U − B(s)∗ ϕ(s), u U ]ds ≥ 0.

Dividing the above by δ > 0 and then sending δ → 0, we obtain that g (t) ≥ 0. From this, we see that B(t)∗ ϕ(t), u∗ (t)U − B(t)∗ ϕ(t), u U ≥ 0  E and u ∈ U0 . for all t ∈ E 

(4.17)

≥1

Since U0 is countable and dense in U, by (4.17), we have that |E| = tE∗ and that B(t)∗ ϕ(t), u∗ (t)U − B(t)∗ ϕ(t), uU ≥ 0 for all t ∈ E and u ∈ U. From these, we obtain (4.15). Finally, (4.5), (4.7), and (4.8) follow from (4.15), (4.13), and the second equality of (4.11), respectively. Then by Definition 4.1, u∗ satisfies the classical Pontryagin S ,QE Maximum Principle, consequently, so does (T P )Q . Hence, we have proved the min sufficiency. S ,QE holds the classical PonWe next show the necessity. Suppose that (T P )Q min S ,QE ∗ ∗ ∗ tryagin Maximum Principle. Let (0, y0 , tE , u ) be an optimal tetrad to (T P )Q . min ∗ Then by Definition 4.1, there is ϕ(·) ∈ C([0, tE ]; Y ) with ϕ(·) = 0 so that (4.4)– (4.8) hold. Then by (4.7) and (4.5), we obtain that *

Φ(tE∗ , 0)∗ ϕ(tE∗ ), y0∗

+ Y

* + = max Φ(tE∗ , 0)∗ ϕ(tE∗ ), y0 Y y0 ∈YS

and 

tE∗ 0

*

+

Φ(tE∗ , s)∗ ϕ(tE∗ ), B(s)u∗ (s) Y ds



tE∗

= 0

* + max Φ(tE∗ , s)∗ ϕ(tE∗ ), B(s)u Y ds. u∈U

4.1 Classical Maximum Principle of Minimal Time Controls

131

From the above two equalities it follows that , ϕ(tE∗ ), Φ(tE∗ , 0)y0∗



tE∗

+

, ≥ ϕ(tE∗ ), Φ(tE∗ , 0)y0 +

0



tE∗ 0

Φ(tE∗ , s)B(s)u∗ (s)ds Y

Φ(tE∗ , s)B(s)u(s)ds Y

for all y0 ∈ YS and u ∈ L(0, tE∗ ; U). This, along with the definition of YR (tE∗ ) (see (4.2)), implies that + ϕ(tE∗ ), z Y .

(4.18)

* + + ϕ(tE∗ ), y ∗ (tE∗ ) Y = max∗ ϕ(tE∗ ), z Y .

(4.19)

*

+ ϕ(tE∗ ), y ∗ (tE∗ ) Y ≥

sup

z∈YR (tE∗ )

*

Since y ∗ (tE∗ ) ∈ YR (tE∗ ), it follows by (4.18) that *

z∈YR (tE )

Since y ∗ (tE∗ ) ∈ YE , it follows from (4.8) and (4.19) that * * + * + + min ϕ(tE∗ ), z Y = ϕ(tE∗ ), y ∗ (tE∗ ) Y = max∗ ϕ(tE∗ ), z Y .

z∈YE

z∈YR (tE )

This shows that YR (tE∗ ) and YE are separable in Y . Thus, we have proved the necessity. Hence, we complete the proof of Theorem 4.1.   Remark 4.3 Several notes are given in order. (i) The definition of the classical Pontryagin Maximum Principle (see Definition 4.1) can be extended easily to the case where the controlled system is the nonlinear system (2.1). In that case, we only need to define the Hamiltonian H in the following way: H (t, y, u, ϕ) = ϕ, f (t, y, u)Y for each (t, y, u, ϕ) ∈ (0, tE∗ ) × Y × U × Y. (ii) It follows from (4.6) that the maximum condition (4.5) is indeed as: u∗ (t), B(t)∗ ϕ(t)U = maxu, B(t)∗ ϕ(t)U for a.e. t ∈ (0, tE∗ ). u∈U

(4.20)

The corresponding separating vector (from YR (tE∗ ) and YE ) is −ϕ(tE∗ )  ϕ0 . It is also called a separating vector w.r.t. u∗ . In general, ϕ(·) = 0 cannot ensure that B(·)∗ ϕ(·) = 0. When B(t)∗ ϕ(t) = 0 for a.e. t ∈ (0, tE∗ ), (4.20) provides nothing about the optimal control u∗ . In this case, ϕ0 is called a non-qualified

132

4 Maximum Principle of Optimal Controls

separating vector w.r.t. u∗ . When B(t)∗ ϕ(t) = 0 for a.e. t ∈ (0, tE∗ ), (4.20) may provide information for u∗ over the whole interval (0, tE∗ ). In this case, ϕ0 is called a qualified separating vector w.r.t. u∗ . When B(t)∗ ϕ(t) = 0 for a.e. t ∈ G, with G ⊆ (0, tE∗ ) a measurable set with 0 < |G| < tE∗ , and B(t)∗ ϕ(t) = 0 for a.e. t ∈ (0, tE∗ ) \ G, (4.20) may provide information for u∗ over G. In this case, ϕ0 is called a semi-qualified separating vector w.r.t. u∗ . (iii) It may happen that for an optimal control, there are two separating vectors w.r.t. the optimal control. One is qualified, while another is non-qualified. These can be seen from the next Example 4.1. (iv) When an optimal control holds the maximum condition, any separating vector is qualified, provided that the following condition (Hq ) is true: If a solution ϕ(·) ∈ C([0, tE∗ ]; Y ) to (4.4) satisfies that B(·)∗ ϕ(·) = 0 over a subset G of positive measure, then ϕ(t) = 0 for a.e. t ∈ (0, tE∗ ). (v) When D(·) = 0, the operator A generates an analytic semigroup, B(·) ≡ B and the system (4.1) is L∞ -null controllable for each interval (i.e., for any T2 > T1 ≥ 0 and y0 ∈ Y , there exists u ∈ L∞ (T1 , T2 ; U ) so that y(T2 ; T1 , y0 , u) = 0), the condition (Hq ) (see (iv) of Remark 4.3) holds. To prove it, two facts are given in order: First, for any solution ϕ to the adjoint equation with ϕ(tE∗ ) ∈ Y , the function t → B ∗ ϕ(t) is real analytic over [0, tE∗ ]. This follows from the analyticity of the semigroup. Second, by the null controllability, we observe from Theorem 1.20 that for any t ∈ [0, tE∗ ), there is a constant C(t) > 0 so that 

tE∗

ϕ(t)Y ≤ C(t)

B ∗ ϕ(s)U ds

t

for any solution ϕ to the adjoint equation with ϕ(tE∗ ) ∈ Y . These two facts clearly imply the condition (Hq ) given in (iv) of this remark. Example 4.1 Let B=

$ % 1 , 0

Q ,QE

S Consider the problem (T P )min

y0 =

$ % 1 . 0

, where the controlled system is as:

y(t) ˙ = Bu(t), t ∈ (0, +∞), with y(t) ∈ R2 , u(t) ∈ R, and where QS = {0} × {y0 }, QE = (0, +∞) × {0}, U = {u ∈ R : |u| ≤ 1}. S ,QE can be put into the framework (AT P ) (given One can easily check that (T P )Q min at the beginning of Section 4.1). By direct calculations, we can find that

tE∗ = 1,

u∗ (·) ≡ −1,

y ∗ (t) = (1 − t, 0) , t ∈ [0, 1],

4.1 Classical Maximum Principle of Minimal Time Controls

133

and that both ϕ0 = (0, 1) and ϕ0 = (1, 0) are separating vectors w.r.t. u∗ , the first one is non-qualified and the second one is qualified. We end Example 4.1. The next example shows that in certain cases, any separating vector w.r.t. any optimal control is qualified. Q ,QE

S Example 4.2 Consider the problem (T P )min is (3.59) (with f = 0) and where

, where the controlled system

QS = {0}×{y0} (with y0 ∈ L2 (Ω)\Br (0)); QE = (0, +∞)×Br (0); U = Bρ (0). Q ,Q

S E can be put into the framework (AT P ) One can easily check that (T P )min (given at the beginning of Section 4.1). Moreover, by the result in Remark 3.1 QS ,QE has an admissible control. Then using the similar (after Theorem 3.5), (T P )min arguments as those in the proof of Theorem 3.11, we can easily show that QS ,QE has an optimal control. (T P )min Since the target is a closed ball, we can use the Hahn-Banach Theorem (see Theorem 1.11) to find that the target and the reachable set at tE∗ are separable in L2 (Ω). Then by Theorem 4.1, this problem holds the classical Pontryagin Maximum Principle. Besides, since this controlled equation is null controllable (see Theorems 1.22 and 1.21), and because {eΔt }t ≥0 is an analytic semigroup, we find that any separating vector w.r.t. any optimal control is qualified. We now end Example 4.2.

The next example gives a non-qualified separating vector w.r.t. an optimal control. Example 4.3 Let $ A=

% $ % $ % $ % −1 0 1 0 0 ; B= ; E = span . ; y0 = 0 −1 0 1 1

Let B1/10 (0) be the closed ball in R2 , centered at the origin, and of radius 1/10. QS ,QE , where the controlled system is as: Consider the problem (T P )min y(t) ˙ = Ay(t) + Bu(t), t ∈ (0, +∞), with y(t) ∈ R2 , u(t) ∈ R, and where QS = {0} × {y0 }, QE = (0, +∞) × (E ∩ B1/10 (0)), U = {u ∈ R : |u| ≤ 1}. Q ,Q

S E can be put into the framework (AT P ) (given One can easily check that (T P )min at the beginning of Section 4.1). After some direct computation, we find the following facts on this problem: First, u∗ ≡ 0 is an optimal control and ln 10 is the optimal time; Second, the vector

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4 Maximum Principle of Optimal Controls

(0, 1) separates the target E ∩B1/10 (0) from the reachable set at time ln 10. Hence, by Theorem 4.1, this problem holds the classical Pontryagin Maximum Principle; Third, the solution ϕ to the equation: ϕ(t) ˙ = −A∗ ϕ(t), t ∈ [0, ln 10]; ϕ(ln 10) = (0, −1) , satisfies that B ∗ ϕ ≡ 0 over [0, ln 10]. Hence, the separating vector (0, 1) is nonqualified w.r.t. the optimal control u∗ . We end Example 4.3. The following example is taken from [9] (see Sections 2.6, 2.7 in Chapter 2 of [9]). We will provide some conclusions and omit their proofs. Example 4.4 Consider the controlled equation:  ∂t y(x, t) = −∂x y(x, t) + u(x, t), 0 ≤ x, t < +∞, y(x, 0) = y0 (x), y(0, t) = 0,

(4.21)

where u ∈ L∞ (0, +∞; L2 (0, +∞)) and u(·, t) ∈ B1 (0) for a.e. t ∈ (0, +∞). (Here, B1 (0) is the closed unit ball in L2 (0, +∞).) This is a controlled transport equation which can be put into the framework (4.1) in the following manner: Let Y = U = L2 (0, +∞); let B = I (the identity operator from L2 (0, +∞) to L2 (0, +∞)); D(t) ≡ 0; let A be the operator on Y defined by Ay(x) = −y  (x), x ∈ (0, +∞) with the domain D(A) = {all absolutely continuous y(·), with y  (·) ∈ L2 (0, +∞) and y(0) = 0}. The operator A generates a C0 semigroup {S(t)}t ≥0 which is expressed by  y(x − t) as x ≥ t, S(t)y(x) = 0 as x < t. Q ,QE

S Consider the problem (T P )min where

, where the controlled system is (4.21) and

QS = {0} × {0}; QE = (0, +∞) × { y } with  y ∈ L2 (0, +∞) and  y = 0; U = B1 (0). S ,QE One can easily check that (T P )Q can be put into the framework (AT P ) (given min at the beginning of Section 4.1). We are still in Example 4.4. The next proposition is exactly Theorem 2.7.1 in [9] (see Page 87 in [9]).

Proposition 4.1 Given T > 0 and 0 < δ < T , there is  y ∈ L2 (0, +∞) so that the control  u ∈ L∞ (0, T ; L2 (0, +∞)) driving 0 to  y in the optimal time T satisfies that

4.1 Classical Maximum Principle of Minimal Time Controls

 u(t) =

135

S(T − t)∗ ϕ0 , T − δ < t ≤ T, S(T − t)∗ ϕ0 L2 (0,+∞)

(4.22)

with ϕ0 ∈ L2 (0, +∞) \ {0} so that  ϕ(t) 

 0 as T − δ < t ≤ T , S(T − t)∗ ϕ0 = S(T − t)∗ ϕ0 = 0 as 0 ≤ t ≤ T − δ.

(4.23)

From Proposition 4.1, we see that for an arbitrarily fixed T > 0, we can choose a QS ,QE has the optimal time T and target  y so that the corresponding problem (T P )min an optimal control  u which satisfies (4.22) and (4.23). So (0, 0, T , u) is an optimal tetrad to this problem. Meanwhile, by (4.22) and (4.23), we can easily see that  u(t), ϕ(t)L2 (0,+∞) = maxu, ϕ(t)L2 (0,+∞) for a.e. t ∈ (0, T ),

(4.24)

u∈U

where ϕ(t)  S(T −t)∗ ϕ0 , t ∈ [0, T ], is the solution of the adjoint equation with the terminal condition that ϕ(T ) = ϕ0 . Since ϕ(·) = 0 (see (4.23)), the equality (4.24) is exactly the classical maximum condition. Furthermore, from (4.23), we see that the separating vector ϕ0 is not qualified but is semi-qualified w.r.t.  u. We end Example 4.4. At the end of this subsection, we will present an example to correct a possible error. This error is as: the derivative of any optimal trajectory y ∗ (·) at tE∗ should be a normal vector of some hyperplane separating YR (tE∗ ) and YE (see Figure 4.2). t1 < t2 < ... < t*E

Ys

y0*

YR(t1)

Fig. 4.2 Wrong geometric intuition

YR(t2)

YR(t*E)

y*(t*E)

YE

136

4 Maximum Principle of Optimal Controls Q ,Q

S E Example 4.5 Consider the problem (T P )min , where the controlled system is as: % $ % ⎧ $ d y1 (t) 1 ⎪ ⎪ = u(t), t ∈ (0, +∞), ⎨ (t) dt y2 % $ %0 $ (4.25) ⎪ 0 y (0) ⎪ ⎩ 1 = , y2 (0) 0    U = [−1, 1], YS = (0, 0) and YE = (y1 , y2 )  (y1 − 1)2 + (y2 − 1)2 ≤ 1 . QS ,QE One can easily check that (T P )min can be put into the framework (AT P ) (given at the beginning of Section 4.1) in the current case. Furthermore, we can check that

YR (t) = [−t, t] × {0}, t ∈ (0, +∞); tE∗ = 1; u∗ (t) = 1, y ∗ (t) = (t, 0) , t ∈ (0, tE∗ ). Any hyperplane, which separates YR (tE∗ ) and YE , can be written as   y ∈ R2  ψ, yR2 = 0 , with ψ = c(0, 1) and c ∈ R \ {0}. It is obvious that y˙ ∗ (tE∗ ) = (1, 0) , which is not parallel to the above ψ. Now we end Example 4.5.

4.1.3 Conditions on Separation in Classical Maximum Principle The main result of this subsection is as follows: S ,QE Theorem 4.2 For the problem (T P )Q , the set YR (tE∗ ) and the target YE are min separable in Y , if one of the following conditions holds:

(i) The space Y is of finite dimension. (ii) The space Y is of infinite dimension and I nt (YR (tE∗ ) − YE ) = ∅. To prove Theorem 4.2, some preliminaries are needed. First, we introduce the following Hausdorff metric in 2Y \ {∅}: 1 ρH (Y1 , Y2 )  2



. sup dist(y1 , Y2 ) + sup dist(y2 , Y1 ) ,

y1 ∈Y1

y2 ∈Y2

Y1 , Y2 ⊆ Y, Y1 = ∅, Y2 = ∅. Lemma 4.2 It holds that lim ρH (YR (s), YR (t)) = 0 for all t ∈ (0, +∞).

s→t

4.1 Classical Maximum Principle of Minimal Time Controls

137

Proof We only prove that lim ρH (YR (s), YR (t)) = 0.

s→t +

The similar arguments can be applied to the case when s → t−. First, we claim that YR (·) is locally bounded, i.e., for any given T ∈ (0, +∞), there is C(T ) > 0 so that y(t; 0, y0 , u)Y ≤ C(T ) for all t ∈ (0, T ], y0 ∈ YS and u ∈ L(0, t; U). (4.26) (Here and throughout the proof of this lemma, C(T ) denotes a generic positive constant dependent on T .) Indeed, by (iv) in (AT P ) (given at the beginning of Section 4.1), there is a positive constant C so that sup y0 Y ≤ C. (Here and y0 ∈YS

throughout the proof of this lemma, C is a generic positive constant.) This, together with (ii) and (iii) in (AT P ) (given at the beginning of Section 4.1), implies that for any t ∈ (0, T ], y0 ∈ YS and u ∈ L(0, t; U),

 t / 0

tA

(t −τ )A

D(τ )y(τ ; 0, y0 , u) + B(τ )u(τ ) dτ e y(t; 0, y0 , u)Y = e y0 +

0  Y t ≤ C(T ) + C(T ) y(τ ; 0, y0, u)Y dτ, 0

which, along with the Gronwall’s inequality, yields (4.26). Next, we arbitrarily fix t ∈ (0, +∞). Take t < s < 2t and y1 ∈ YR (s). Then there is  y0 ∈ YS and u ∈ L(0, s; U) so that y1 = y(s; 0,  y0, u).

(4.27)

Denote  y (·)  y(·; 0,  y0, u). Then we have that 

s

 y (s) = eA(s−t ) y (t) +

/ 0 y (τ ) + B(τ )u(τ ) dτ. eA(s−τ ) D(τ )

(4.28)

t

Note that dist(y1 , YR (t)) ≤ y1 − y (t)Y . From this, (4.27), (4.28), and (ii) in (AT P ) (given at the beginning of Section 4.1), it follows that dist(y1 , YR (t))

 s / 0

A(s−t ) A(s−τ ) e  y (t) −  y (t) + e D(τ ) y (τ ) + B(τ )u(τ ) dτ ≤



≤ eA(s−t ) − I

L (Y )

t

(4.29)

Y



s

 y (t)Y + C t

eA(s−τ )L (Y ) (y(τ ; 0,  y0, u)Y + 1)dτ.

138

4 Maximum Principle of Optimal Controls

By (4.29) and (4.26), we obtain that dist(y1 , YR (t)) ≤ C(t)(eA(s−t ) − I L (Y ) + s − t)  γ (s, t).

(4.30)

Similarly, for any y2 ∈ YR (t), we can show that dist(y2 , YR (s)) ≤ γ (s, t), which, combined with (4.30), indicates that ρH (YR (s), YR (t)) ≤ γ (s, t) → 0 as s → t + .  

Hence, we end the proof of Lemma 4.2. The next lemma presents some properties on the set YR (t) with t ∈ (0, +∞).

Lemma 4.3 For each t ∈ (0, +∞), the set YR (t) is bounded, convex, and closed. Proof Arbitrarily fix t ∈ (0, +∞). The boundedness of YR (t) follows from (4.26). The convexity of YR (t) follows from (4.1), (4.2), (iii), and (iv) in (AT P ) (given at the beginning of Section 4.1). We now prove YR (t) is closed. Let {yk }k≥1 ⊆ YR (t) satisfy that yk →  y strongly in Y . By (4.2), there are two sequences {y0,k }k≥1 ⊆ YS and {uk }k≥1 ⊆ L(0, t; U) so that  t yk = Φ(t, 0)y0,k + Φ(t, s)B(s)uk (s)ds for all k ≥ 1. (4.31) 0

Here Φ(·, ·) denotes the evolution operator generated by A + D(·). Since {y0,k }k≥1 ⊆ YS and {uk }k≥1 ⊆ L(0, t; U), by (i), (iii), and (iv) in (AT P ) (given at the beginning of Section 4.1), there exists a subsequence of {k}k≥1 , denoted in the same manner, so that for some  y0 ∈ YS and  u ∈ L2 (0, t; U ), y0,k →  y0 weakly in Y and uk →  u weakly in L2 (0, t; U ).

(4.32)

From the second conclusion in (4.32), we can use the Mazur Theorem (see Theorem 1.8) to find, for each k ≥ 1, a finite sequence of nonnegative numbers Jk {αkj }Jj k=1 with αkj = 1, so that j =1

Jk

αkj uk+j →  u strongly in L2 (0, t; U ).

(4.33)

j =1

Since {uk+j }Jj k=1 ⊆ L(0, t; U), it follows by (iii) in (AT P ) and (4.33) that  u ∈ L(0, t; U).

(4.34)

4.1 Classical Maximum Principle of Minimal Time Controls

139

Since yk →  y strongly in Y , and because of (4.31), (4.32), (4.34), and (ii) in (AT P ) (given at the beginning of Section 4.1), we can pass to the limit for k → +∞ in (4.31) to get that 

t

 y = Φ(t, 0) y0 +

Φ(t, s)B(s) u(s)ds = y(t; 0,  y0 , u) ∈ YR (t),

0

which implies that YR (t) is closed. Hence, we complete the proof of Lemma 4.3.

 

Proposition 4.2 It holds that   0 ∈ ∂ YR (tE∗ ) − YE .

(4.35)

Z(t)  YR (t) − YE , t ∈ [0, tE∗ ].

(4.36)

Proof Let

Two facts are given in order. Fact one: Z(t) is convex and closed for each t ∈ [0, tE∗ ]. Fact two: Z(·) is Hausdorff continuous. Fact one follows from Lemma 4.3 and (iv) in (AT P ) (given at the beginning of Section 4.1). Fact two follows by Lemma 4.2 at once. From the fact two, we find that for any δ ∈ (0, 1), there exists tδ ∈ [0, tE∗ ) so that ρH (Z(tδ ), Z(tE∗ )) ≤ δ/6.

(4.37)

/ Z(tδ ). This, together with Meanwhile, by the optimality of tE∗ , we see that 0 ∈ Theorem 1.11, and the convexity and the closedness of Z(tδ ) (see the fact one above), implies that there exists an f ∈ Y with f Y = 1 so that inf f, zY > 0.

z∈Z(tδ )

(4.38)

Because f Y = sup f, zY and f Y = 1, zY =1

there exists  z ∈ Y so that  zY = 1 and f, zY ≤ −1 + δ/2.

(4.39)

By setting zδ = δ z, (4.38), and (4.39), we obtain that zδ ∈ Bδ (0)

(4.40)

140

4 Maximum Principle of Optimal Controls

and f, z − zδ Y ≥ (1 − δ/2)δ ≥ δ/2 for each z ∈ Z(tδ ). From the latter inequality it follows that z − zδ Y ≥ δ/2 for each z ∈ Z(tδ ), which, combined with (4.37) and (4.36), indicates that / Z(tE∗ ) = YR (tE∗ ) − YE . zδ ∈

(4.41)

Since 0 ∈ YR (tE∗ ) − YE and because δ was arbitrarily taken from (0, 1), (4.35) follows from (4.40) and (4.41) at once. Hence, we end the proof of Proposition 4.2.   We now turn to prove Theorem 4.2. Proof (Proof of Theorem 4.2) Let S  YR (tE∗ ) − YE and z0  0. By the definition of YR (tE∗ ) (see (4.2)) and (AT P ) (given at the beginning of Section 4.1), one can easily check that S is a (nonempty) convex subset in Y . Meanwhile, it follows by Proposition 4.2 that 0 ∈ ∂S. Thus, if (i) is true, then the desired separability follows from Theorem 1.9, while if (ii) is true, then the desired separability follows from Theorem 1.10. This ends the proof of Theorem 4.2.   We now explain the condition that I nt (YR (tE∗ ) − YE ) = ∅. (See (ii) of Theorem 4.2.) This condition holds, if either YR (tE∗ ) or YE has a nonempty interior. S ,QE , YE is already assumed to (One can verify it directly.) For some problem (T P )Q min have a nonempty interior. In the case when I ntYE = ∅, we need to consider the set YR (tE∗ ). The condition that I ntYR (tE∗ ) = ∅ is connected with some controllability for the controlled system. For instance, when YS is a singleton, this condition is equivalent to the exact controllability of the controlled system. The latter is too restrictive and is not satisfied by many controlled equations, such as the internally controlled heat equations. To relax the condition that I nt (YR (tE∗ ) − YE ) = ∅, the concept of the finite codimensionality is introduced in [12]. For more details on this concept, we refer readers to Chapter 4 in [12]. We next give some examples, which may help us to understand Theorem 4.2 and Theorem 4.1 better. Example 4.6 All minimal time control problems given in Examples 4.1, 4.2, 4.3, and 4.5 can be put into the framework (AT P ) (given at the beginning of Section 4.1). And they have optimal controls. (These can be checked directly.) Thus we can apply Theorem 4.2 to find that the sets YR (tE∗ ) and YE in each of these problems are separable in Y . Then, according to Theorem 4.1, each of them satisfies the classical Pontryagin Maximum Principle. We end Example 4.6.

4.1 Classical Maximum Principle of Minimal Time Controls

141

Example 4.7 Let A(·) = (aij (·))n×n and B(·) = (bij (·))n×m (where aij (·) ∈ L∞ (0, +∞) and bij (·) ∈ L∞ (0, +∞) for all 1 ≤ i ≤ n, 1 ≤ j ≤ m. Let y0 ∈ Rn and y1 ∈ Rn satisfy that y0 = y1 . Let Y = Rn and U = Rm . Consider the problem S ,QE , where the controlled system is as: (T P )Q min y(t) ˙ = A(t)y(t) + B(t)u(t),

t ∈ (0, +∞), with y(t) ∈ Rn , u(t) ∈ Rm ,

and where QS = {0} × {y0 }; QE = (0, +∞) × {y1 }; U = Bρ (0) with ρ > 0. One can easily check that the corresponding minimal time control problem QS ,QE (T P )min can be put into the framework (AT P ) (given at the beginning of Section 4.1). QS ,QE We further assume that (T P )min has optimal controls. Then by Theorem 4.2, ∗ we obtain that the sets YR (tE ) and YE in this problem are separable in Y . Hence, QS ,QE satisfies the classical Pontryagin Maximum according to Theorem 4.1, (T P )min Principle. We now end Example 4.7. Example 4.8 Let the controlled system be the heat equation (3.107) with a(·, ·) ∈ C(Ω × [0, +∞)) ∩ L∞ (0, +∞; L∞ (Ω)). Let Y = L2 (Ω) and U = L2 (Ω). Let QS = {0} × {y0 }; QE = (0, +∞) × Br (0) and U = Bρ (0), where y0 ∈ L2 (Ω) \ Br (0), r > 0, and ρ > 0. One can easily check that QS ,QE can be put into the the corresponding minimal time control problem (T P )min framework (AT P ) (given at the beginning of Section 4.1). QS ,QE has optimal controls. Then by Theorem 4.2, We further assume that (T P )min ∗ we see that the sets YR (tE ) and YE in this problem are separable in Y . Hence, S ,QE satisfies the classical Pontryagin Maximum according to Theorem 4.1, (T P )Q min Principle. We end Example 4.8. The final example gives a minimal time control problem satisfying what follows: (i) (ii) (iii) (iv)

It is under the framework (AT P ) (given at the beginning of Section 4.1); It has an optimal control; It does not satisfy the classical Pontryagin Maximum Principle; It satisfies the local Pontryagin Maximum Principle (which will be introduced in Definition 4.2 in the next section).

The proof of the above conclusion (iv) will be given in Example 4.10 of the next section. Because of the above (iv), we introduce the local Pontryagin Maximum Principle in the next subsection.

142

4 Maximum Principle of Optimal Controls

Example 4.9 Consider the following controlled heat equation: 

∂t y − Δy = u in Ω × (0, +∞), y=0 on ∂Ω × (0, +∞),

(4.42)

where Ω = (0, π). As what we did in Section 3.2.2, the Equation (4.42) can be put into framework (4.1). Thus, (4.42) can be rewritten as y˙ − Δy = u,

t ∈ (0, +∞).

(4.43)

Let ek (x) 

1

2/πsinkx, x ∈ [0, π].

(It constitutes an orthonormal basis in L2 (Ω).) Let z

+∞



1

bk ek , with bk  0

k=1

1 −2k 2 (1−t ) e dt for all k ≥ 1, d(t)

(4.44)

where d(t) 

+∞ "

e−2k

2 (1−t )

#1 2

, t ∈ [0, 1).

(4.45)

k=1

Let  Y = U = L2 (Ω); QS  {(0, 0)}; QE = (0, +∞) × {z}; U = {u ∈ U  uU ≤ 1}.

One can easily check that the corresponding minimal time control problem S ,QE can be put into the framework (AT P ) (given at the beginning of (T P )Q min Section 4.1). We will see later that this problem has an optimal control. First we claim that there exists a control  u ∈ L(0, 1; U) so that y(1; 0, 0, u) = z.

(4.46)

To this end, we define ϕ (t)  

+∞

e−k

2 (1−t )

ek ,

t ∈ (−∞, 1).

(4.47)

k=1

It is obvious that  ϕ (·) satisfies 

∂t  ϕ + Δ ϕ = 0 in Ω × (−∞, 1), ϕ=0  on ∂Ω × (−∞, 1).

(4.48)

4.1 Classical Maximum Principle of Minimal Time Controls

143

Since 

1 0

 ϕ (t)L2 (Ω) dt =

 1"

+∞ 

0 1

=

e

−2k 2 (1−t )

√

2

dt ≤

 1"

+∞ 0

k=1 e−(1−t )

0

#1

1 − e−2(1−t )

e−2k(1−t )

#1 2

dt

k=1

dt < +∞, (4.49)

we see that  ϕ ∈ L1 (0, 1; L2(Ω)). Now, we define the following function:  u(t)   ϕ (t)/ ϕ (t)L2 (Ω) for a.e. t ∈ (0, 1).

(4.50)

Then  u ∈ L∞ (0, 1; L2 (Ω)) and 

1

y(1; 0, 0, u) = 0

e(1−t )Δ ϕ (t)/ ϕ (t)L2 (Ω) dt.

(4.51)

By (4.51), (4.44), and (4.45), we get that for each k ≥ 1,  y(1; 0, 0, u) − z, ek L2 (Ω) = 

1 0 1

=

e(1−t )Δek ,  ϕ (t)/d(t)L2 (Ω) dt − bk e−k

2 (1−t )

0

ek ,  ϕ (t)/d(t)L2 (Ω) dt − bk .

This, together with (4.47) and (4.45), implies that y(1; 0, 0, u) − z, ek L2 (Ω) = 0 for each k ≥ 1, which leads to (4.46). Next, by (4.46) and using the similar arguments as those used in the proof of QS ,QE has an optimal control. Theorem 3.11, we can find that (T P )min Finally, we prove the following CLAIM A: For any optimal tetrad (0, 0, tE∗ , u∗ ), there is no ϕt∗∗ ∈ L2 (Ω) \ {0} satisfying that for a.e. t ∈ (0, tE∗ ), E

ϕ ∗ (t), u∗ (t)L2 (Ω) = maxϕ ∗ (t), uL2 (Ω) = ϕ ∗ (t)L2 (Ω) , u∈U

(4.52)

where ϕ ∗ (·)  ϕ(·; tE∗ , ϕt∗∗ ). Here and throughout this example, we will use E

ϕ(·; T , ξ ), with T > 0 and ξ ∈ L2 (Ω), to denote the solution of the following equation: ⎧ ⎨ ∂t ϕ + Δϕ = 0 in Ω × (0, T ), ϕ=0 on ∂Ω × (0, T ), ⎩ ϕ(T ) = ξ in Ω.

144

4 Maximum Principle of Optimal Controls

By contradiction, suppose that CLAIM A did not hold. Then there would be an optimal tetrad (0, 0, tE∗ , u∗ ) so that (4.52) is true for some ϕt∗∗ ∈ L2 (Ω) \ {0}. We E will get a contradiction by the following five steps. Step 1. We show that |z, ξ L2 (Ω) | ≤ ϕ(·; tE∗ , ξ )L1 (0,t ∗ ;L2 (Ω)) for each ξ ∈ L2 (Ω).

(4.53)

E

Indeed, since z, ξ L2 (Ω) = y(tE∗ ; 0, 0, u∗ ), ϕ(tE∗ ; tE∗ , ξ )L2 (Ω)  t∗ E = u∗ (t), ϕ(t; tE∗ , ξ )L2 (Ω) dt for each ξ ∈ L2 (Ω),

(4.54)

0

and because u∗ (t)L2 (Ω) ≤ 1 for a.e. t ∈ (0, tE∗ ), (4.53) follows immediately. Step 2. We claim that z, ξ L2 (Ω) ∗ ξ ∈L2 (Ω)\{0} ϕ(·; tE , ξ )L1 (0,tE∗ ;L2 (Ω)) sup

= 1.

(4.55)

≤ 1.

(4.56)

Indeed, on one hand, it follows by (4.53) that z, ξ L2 (Ω)

sup

∗ ξ ∈L2 (Ω)\{0} ϕ(·; tE , ξ )L1 (0,t ∗ ;L2 (Ω)) E

On the other hand, by (4.54) and (4.52), we obtain that  t∗ E ∗ z, ϕt ∗ L2 (Ω) = u∗ (t), ϕ ∗ (t)L2 (Ω) dt = ϕ ∗ L1 (0,t ∗ ;L2 (Ω)) , E

E

0

(4.57)

which indicates that sup ξ ∈L2 (Ω)\{0}

z, ξ L2 (Ω) ϕ(·; tE∗ , ξ )L1 (0,t ∗ ;L2 (Ω)) E

≥

z, ϕt∗∗ L2 (Ω) E

ϕ ∗ L1 (0,t ∗ ;L2 (Ω))

= 1.

E

This, along with (4.56), leads to (4.55). Step 3. We prove that tE∗ = 1.

(4.58)

By contradiction, we suppose that (4.58) was not true. Then, since tE∗ ≤ 1, we would have that tE∗ < 1. For each ε > 0, we define the following function: ξε 

+∞

k=1

e−k ε ek . 2

(4.59)

4.1 Classical Maximum Principle of Minimal Time Controls

145

First, we show that  lim z, ξε L2 (Ω) =

ε→0+

0

1

 u(t),  ϕ (t)L2 (Ω) dt =  ϕ L1 (0,1;L2(Ω)) .

(4.60)

Indeed, by (4.46) and using the similar arguments as those used to show (4.54), we can verify that z, ξε L2 (Ω) = y(1; 0, 0, u), ϕ(1; 1, ξε )L2 (Ω)  1 =  u(t), ϕ(t; 1, ξε )L2 (Ω) dt.

(4.61)

0

Meanwhile, it follows from (4.47) and (4.59) that ϕ(t; 1, ξε ) =  ϕ (t − ε) for each t ∈ [0, 1],

(4.62)

which indicates that for a.e. t ∈ (0, 1), ϕ (t − ε)L2 (Ω) ≤  ϕ (t)L2 (Ω) . | u(t), ϕ(t; 1, ξε )L2 (Ω) | ≤ 

(4.63)

Since  ϕ ∈ L1 (0, 1; L2(Ω)) ∩ C([−1, 1); L2(Ω)), by (4.61)–(4.63), we can apply the Lebesgue Dominated Convergence Theorem, as well as (4.50), to obtain (4.60). Second, we show that  1−ε ∗  ϕ (t)L2 (Ω) dt. (4.64) ϕ(·; tE , ξε )L1 (0,t ∗ ;L2 (Ω)) = 1−ε−tE∗

E

Indeed, by (4.47) and (4.59), we can check that ϕ(t; tE∗ , ξε ) =  ϕ (1 − ε − tE∗ + t) for a.e. t ∈ (0, tE∗ ), which leads to (4.64). Next, since  ϕ ∈ L1 (0, 1; L2(Ω)), it follows by (4.64) that ϕ L1 (1−t ∗ ,1;L2(Ω)) . lim ϕ(·; tE∗ , ξε )L1 (0,t ∗ ;L2 (Ω)) =  E

ε→0+

E

(4.65)

Finally, by (4.60) and (4.65), we see that lim

ε→0+

z, ξε L2 (Ω) ϕ(·; tE∗ , ξε )L1 (0,t ∗ ;L2 (Ω)) E

=

 ϕ L1 (0,1;L2(Ω))  ϕ L1 (1−t ∗ ,1;L2 (Ω))

> 1,

(4.66)

E

which contradicts (4.55). Hence, (4.58) is true. Step 4. We show that  ϕ (1) ∈ L2 (Ω).

(4.67)

146

4 Maximum Principle of Optimal Controls

To this end, we first introduce the following problem:  (P1 )

V1 

inf

ξ ∈L2 (Ω)

 1 ϕ(·; 1, ξ )2L1 (0,1;L2(Ω)) − z, ξ L2 (Ω) . 2

(4.68)

By (4.53) and (4.58), (P1 ) is well defined. It follows from (4.68), (4.58), and (4.53) that 1 V1 ≥ − . 2

(4.69)

Meanwhile, by (4.68), (4.58), and (4.57), we obtain that V1 ≤

1 ϕ(·; tE∗ , ϕt∗∗ /ϕ ∗ L1 (0,t ∗ ;L2 (Ω)) )2L1 (0,t ∗ ;L2 (Ω)) E E E 2 1 1 ∗ ∗ ∗ −z, ϕt ∗ /ϕ L1 (0,t ;L2 (Ω)) L2 (Ω) = − 1 = − . E E 2 2

(4.70)

From (4.69) and (4.70), we find that 1 V1 = − . 2

(4.71)

Let X  {ϕ(·; 1, ξ ) | ξ ∈ L2 (Ω)} and Z  X

·L1 (0,1;L2 (Ω))

.

Define a functional on Z by F (ψ)   u, ψL∞ (0,1;L2(Ω)),L1(0,1;L2(Ω)) , ψ ∈ Z.

(4.72)

By (4.72) and (4.54), we can easily obtain that F (ϕ(·; 1, ξ )) = z, ξ L2 (Ω) for each ξ ∈ L2 (Ω).

(4.73)

We next introduce the following problem: V2  inf J (ψ),

(P2 )

(4.74)

ψ∈Z

where J (·) : Z → R is defined by J (ψ) 

1 ψ2L1 (0,1;L2(Ω)) − F (ψ), 2

ψ ∈ Z.

(4.75)

It follows from (4.68), (4.71), (4.73)–(4.75), and the density of X in Z that 1 V2 = V1 = − . 2

(4.76)

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147

By (4.62) and (4.59), we have that ϕ (· − ε) ∈ X. 

(4.77)

Since  ϕ ∈ L1 (0, 1; L2 (Ω))∩C([−1, 1); L2 (Ω)), by the second inequality in (4.63) and (4.77), we can apply the Lebesgue Dominated Convergence Theorem to obtain that 

1 0

 ϕ (t − ε) −  ϕ (t)L2 (Ω) dt → 0.

This implies that ϕ ∈ Z. 

(4.78)

Meanwhile, by (4.45), (4.47), and (4.49), we have that 

1

0< 0

d(t)dt =  ϕ L1 (0,1;L2(Ω)) < + ∞.

(4.79)

This, together with (4.78), (4.75), and (4.72), implies that 1 J ( ϕ /dL1 (0,1)) = − 2



1 0

 u,  ϕ L2 (Ω) dt/dL1 (0,1),

which, combined with (4.50) and (4.79), indicates that 1 J ( ϕ /dL1 (0,1)) = − . 2

(4.80)

Besides, it follows from (4.75), (4.58), (4.73), and (4.70) that 1 J (ϕ(·; tE∗ , ϕt∗∗ /ϕ ∗ L1 (0,t ∗ ;L2 (Ω)) )) = − . E E 2

(4.81)

By (4.74), (4.76), (4.80), and (4.81), we see that the following two functions are solutions to the problem (P2 ):  ϕ (·)/dL1 (0,1) and ϕ(·; tE∗ , ϕt∗∗ /ϕ ∗ L1 (0,t ∗ ;L2 (Ω)) ). E

E

Since J (·) is strictly convex (see [18]), there is at most one solution to (P2 ). Hence, ϕ (·)/dL1 (0,1) = ϕ(·; tE∗ , ϕt∗∗ /ϕ ∗ L1 (0,t ∗ ;L2 (Ω)) ).  E

This, along with (4.58), leads to (4.67).

E

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4 Maximum Principle of Optimal Controls

Step 5. We find a contradiction. Obviously, (4.67) contradicts (4.47). By Step 5, we see that CLAIM A is true. From it, we find that any optimal S ,QE does not satisfy the classical Pontryagin Maximum Principle. control to (T P )Q min Furthermore, from (4.50), (4.46), and Step 3, we see that (1, 0, 0, u) is an optimal S ,QE . We now end Example 4.9. tetrad to (T P )Q min

4.2 Local Maximum Principle and Minimal Time Controls S ,QE In this section, we will study the problem (T P )Q under the framework (AT P ) min (given at the beginning of Section 4.1). As what we did in Section 4.1, we always S ,QE has an optimal control. The aim of this section is to assume that (T P )Q min S ,QE obtain the local Pontryagin Maximum Principle of (T P )Q , through separating min a reachable set from a controllable set in the state space before the optimal time. Before giving the definition of the above-mentioned local Pontryagin Maximum S ,QE , we define, for any 0 ≤ t1 < t2 < +∞, the Principle for the problem (T P )Q min following two sets:

   YC (t1 , t2 )  y0 ∈ Y  ∃ u ∈ L(t1 , t2 ; U) so that y(t2 ; t1 , y0 , u) ∈ YE ; C (t1 , t2 )  Y



YC (t1 , τ ).

(4.82) (4.83)

τ ∈(t1 ,t2 )

We call YC (t1 , t2 ) the controllable set in the interval (t1 , t2 ) for the Equation (4.1). Definition 4.2 Several definitions are given in order. (i) An optimal control u∗ (associated with an optimal tetrad (0, y0∗ , tE∗ , u∗ )) to the QS ,QE problem (T P )min is said to satisfy the local Pontryagin Maximum Principle, if for any T ∈ (0, tE∗ ), there exists ϕT (·) ∈ C([0, T ]; Y ), with ϕT (·) = 0 (i.e., ϕT is not a zero function), so that ϕ˙T (t) = −A∗ ϕT (t) − D(t)∗ ϕT (t) for a.e. t ∈ (0, T );

(4.84)

H (t, y ∗(t), u∗ (t), ϕT (t)) = max H (t, y ∗ (t), u, ϕT (t)) for a.e. t ∈ (0, T ), u∈U

(4.85) where y ∗ ∈ C([0, tE∗ ]; Y ) is the corresponding optimal trajectory and H (t, y, u, ϕ)  ϕ, D(t)y + B(t)uY

(4.86)

for a.e. t ∈ (0, tE∗ ) and all (y, u, ϕ) ∈ Y × U × Y . Besides, ϕT (0), y0 − y0∗ Y ≤ 0 for each y0 ∈ YS ,

(4.87)

4.2 Local Maximum Principle and Minimal Time Controls

149

and ϕT (T ), z − y ∗ (T )Y ≥ 0 for each z ∈ YC (T , tE∗ ).

(4.88)

Here (4.84) is called the dual (or adjoint, or co-state) equation; (4.85) is called the maximum condition; the function H (·) defined by (4.86) is called the S ,QE ); (4.87) and (4.88) are called the Hamiltonian (associated with (T P )Q min transversality condition. S ,QE (ii) The problem (T P )Q is said to satisfy the local Pontryagin Maximum min Principle, if any optimal control satisfies the local Pontryagin Maximum Principle. S ,QE Remark 4.4 For the problem (T P )Q , the classical Pontryagin Maximum min Principle cannot imply the local Pontryagin Maximum Principle. The reason is that a nontrivial solution to the adjoint equation on [0, tE∗ ] may be identically zero over [0, T ] for some T < tE∗ . This can be seen from Proposition 4.1 and (4.24). We will see from Corollary 4.2 that under some condition, the classical Pontryagin Maximum Principle implies the local Pontryagin Maximum Principle. S ,QE Example 4.10 The minimal time control problem (T P )Q in Example 4.9 min satisfies the local Pontryagin Maximum Principle. This can be verified by two steps as follows: Step 1. This problem has the unique optimal tetrad (0, 0, 1, u). Let  u and  ϕ be given by (4.50) and (4.47), respectively. In Example 4.9, we already see that (0, 0, 1, u) is an optimal tetrad of the minimal time control problem. To show the desired uniqueness, we arbitrarily fix an optimal control u∗ . Then

y(1; 0, 0, u∗) = z and u∗ L∞ (0,1;L2(Ω)) ≤ 1.

(4.89)

Because of (4.46), it follows by the equality in (4.89) that 

1 0

 u(t) − u∗ (t),  ϕ (t)L2 (Ω) dt = 0.

(4.90)

From (4.50), (4.90), and the inequality in (4.89), we obtain that ϕ (t)/ ϕ (t)L2 (Ω) for a.e. t ∈ (0, 1), u∗ (t) =  which indicates that u∗ (t) =  u(t) for a.e. t ∈ (0, 1). S ,QE Step 2. (T P )Q satisfies the local Pontryagin Maximum Principle. min By Definition 4.2 and Step 1, it suffices to show that (0, 0, 1, u) satisfies the local Pontryagin Maximum Principle. For this purpose, we arbitrarily fix T ∈ (0, 1). Define ϕT (t)   ϕ (t),

t ∈ [0, T ].

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4 Maximum Principle of Optimal Controls

It is obvious that ϕT (·) ∈ C([0, T ]; L2 (Ω)) satisfies ϕT (·) = 0 and (4.84)–(4.87) (see (4.47)–(4.50)). The reminder is to show (4.88). Arbitrarily take  z ∈ YC (T , 1). According to (4.82), there exists  u ∈ L∞ (T , 1; L2 (Ω)) so that y(1; T , z, u) = z and  uL∞ (T ,1;L2(Ω)) ≤ 1.

(4.91)

Since y(1; T , y(T ; 0, 0, u), u) = z, it follows from the equality in (4.91) that y(1; T , z − y(T ; 0, 0, u), u − u) = 0, which, combined with (4.50) and the inequality in (4.91), indicates that  1  u(t) −  u(t),  ϕ (t)L2 (Ω) dt ≤ 0. − ϕ (T ), z − y(T ; 0, 0, u)L2 (Ω) = T

Then z − y(T ; 0, 0, u)L2 (Ω) ≥ 0 for each  z ∈ YC (T , 1). ϕT (T ), This implies that (4.88) holds for this case. QS ,QE satisfies the local Pontryagin Maximum From the above, we see that (T P )min Principle. We now end Example 4.10.

4.2.1 Some Properties on Controllable Sets and Reachable Sets Recall (4.2), (4.82), and (4.83) for the definitions of sets YR (t), YC (t1 , t2 ) and C (t1 , t2 ). In this subsection, we will present some basic properties related to these Y QS ,QE , subsets. First, by (4.82), (4.83), and the definition of the optimal time of (T P )min one can easily prove the next Proposition 4.3. S ,QE if and only if one of Proposition 4.3 Time t ∗ is the optimal time tE∗ to (T P )Q min the following statements holds:

(i) YR (t ∗ ) ∩ YE = ∅ and YR (t1 ) ∩ YC (t1 , t2 ) = ∅ for all 0 ≤ t1 < t2 < t ∗ . (ii) YS ∩ YC (0, t ∗ ) = ∅ and YS ∩ YC (0, t) = ∅ for all t ∈ (0, t ∗ ). C (0, t ∗ ) = ∅. (iii) YS ∩ YC (0, t ∗ ) = ∅ and YS ∩ Y Proposition 4.4 Let t ∈ [0, tE∗ ). Then YR (t) ∩ YC (t, tE∗ )   QS ,QE = y(t; 0, y0∗ , u∗ )  (0, y0∗ , tE∗ , u∗ ) is an optimal tetrad for (T P )min (4.92) and C (t, tE∗ ) = ∅. YR (t) ∩ Y

(4.93)

4.2 Local Maximum Principle and Minimal Time Controls

151

Proof We only prove this proposition for the case when t ∈ (0, tE∗ ). For the case that t = 0, one can use the similar way, with a little modification, to show it. First, we prove (4.92). The proof of “⊇” is obvious. We now turn to the proof of “⊆.” To this end, we arbitrarily fix z ∈ YR (t) ∩ YC (t, tE∗ ). Since z ∈ YR (t), there is y0 ∈ YS and u1 ∈ L(0, t; U) so that z = y(t; 0, y0, u1 ).

(4.94)

Meanwhile, since z ∈ YC (t, tE∗ ), there is u2 ∈ L(t, tE∗ ; U) so that y(tE∗ ; t, z, u2 ) ∈ YE .

(4.95)

Define the following control:  u(s) 

u1 (s), s ∈ (0, t), u2 (s), s ∈ (t, tE∗ ).

(4.96)

Then by (4.94)–(4.96), we can directly check that (0, y0 , tE∗ , u) is an optimal tetrad QS ,QE and z = y(t; 0, y0 , u). Hence, (4.92) is proved. for (T P )min We now show (4.93). By contradiction, suppose that it was not true. Then, there would exist  z ∈ Y so that C (t, tE∗ ).  z ∈ YR (t) ∩ Y y0 ∈ YS and  u1 ∈ L(0, t; U) so that Since  z ∈ YR (t), there is   z = y(t; 0,  y0, u1 ).

(4.97)

C (t, t ∗ ), there is t0 ∈ (t, t ∗ ) and  u2 ∈ L(t, t0 ; U) so that Meanwhile, since  z∈Y E E y(t0 ; t, z, u2 ) ∈ YE .

(4.98)

Define the following control:   u(s) 

 u1 (s), s ∈ (0, t),  u2 (s), s ∈ (t, t0 ).

(4.99)

By (4.97)–(4.99), we can directly check that (0,  y0 , t0 , u) is an admissible tetrad for S ,QE ∗ . Thus, (4.93) follows. , which contradicts that t < t (T P )Q 0 min E Hence, we end the proof of Proposition 4.4.   We next introduce the concept of viability domain and the related results.

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4 Maximum Principle of Optimal Controls

Definition 4.3 Let S be a nonempty subset of Y . We call S a viability domain associated with (4.1) if for any y0 ∈ S and τ ∈ (0, +∞), there is u ∈ L(0, +∞; U) so that y(t; τ, y0 , u) ∈ S for each t ≥ τ. Example 4.11 When 0 ∈ U, one can easily check that {0} is a viability domain associated with (4.1). Proposition 4.5 The following statements are true: (i) Let 0 ≤ t < τ < +∞. Then YC (t, τ ) is a convex and closed subset in Y , and C (t, t1 ) ⊆ Y C (t, t2 ) for all t < t1 < t2 < +∞. Y

(4.100)

(ii) Suppose that YE is a viability domain associated with (4.1). Then for any t ≥ 0, YC (t, t1 ) ⊆ YC (t, t2 ) for all t < t1 < t2 < +∞.

(4.101)

C (t, τ ) is convex in Y and Moreover, for any τ ∈ (t, +∞) (with t ≥ 0), Y C (t, τ ) ⊆ YC (t, τ ). Y

(4.102)

Proof Without loss of generality, we assume that the sets involved in the proof of this proposition are not empty. (i) First, we prove the convexity of YC (t, τ ). To this end, let z1 , z2 ∈ YC (t, τ ). Then there are u1 , u2 ∈ L(t, τ ; U) so that y(τ ; t, z1 , u1 ) ∈ YE and y(τ ; t, z2 , u2 ) ∈ YE .

(4.103)

For each λ ∈ (0, 1), we write zλ  (1 − λ)z1 + λz2 and uλ  (1 − λ)u1 + λu2 .

(4.104)

From the convexity of U, we see that uλ ∈ L(t, T ; U).

(4.105)

By (4.1), the convexity of YE , (4.104), and (4.103), we find that y(τ ; t, zλ, uλ ) = (1 − λ)y(τ ; t, z1 , u1 ) + λy(τ ; t, z2 , u2 ) ∈ YE . This, together with (4.105), implies that zλ ∈ YC (t, τ ). Hence, the set YC (t, τ ) is convex.

4.2 Local Maximum Principle and Minimal Time Controls

153

Second, we show that YC (t, τ ) is closed. For this purpose, we assume that {z }≥1 ⊆ YC (t, τ ) satisfies that z → z

strongly in Y.

(4.106)

Since {z }≥1 ⊆ YC (t, τ ), there exists a sequence {u }≥1 ⊆ L(t, τ ; U) so that 

τ

Φ(τ, t)z +

Φ(τ, s)B(s)u (s)ds ∈ YE .

(4.107)

t

Here Φ(·, ·) denotes the evolution operator generated by A + D(·). Using the similar arguments to those used in the proofs of the second result in (4.32) and (4.34), we can prove that there exists a subsequence of {}≥1, still denoted by itself, and a function  u ∈ L(t, τ ; U), so that u →  u weakly in L2 (t, τ ; U ).

(4.108)

Since YE is convex and closed, by (4.106) and (4.108), we can pass to the limit for  → +∞ in (4.107) to get that 

τ

y(τ ; t, z, u) = Φ(τ, t)z +

Φ(τ, s)B(s) u(s)ds ∈ YE .

t

This indicates that the set YC (t, τ ) is closed. Finally, (4.100) follows from (4.83) directly. (ii) First, we prove (4.101). For this purpose, we arbitrarily fix z1 ∈ YC (t, t1 ). Then there is u1 ∈ L(t, t1 ; U) so that z2  y(t1 ; t, z1 , u1 ) ∈ YE .

(4.109)

Since YE is a viability domain associated with (4.1), it follows by Definition 4.3 and (4.109) that there exists u2 ∈ L(t1 , t2 ; U) so that z3  y(t2 ; t1 , z2 , u2 ) ∈ YE .

(4.110)

Denote  u(s) 

u1 (s), s ∈ (t, t1 ), u2 (s), s ∈ (t1 , t2 ).

It follows from (4.109)–(4.111) and (4.1) that u ∈ L(t, t2 ; U) and y(t2 ; t, z1 , u) = z3 ∈ YE . These imply that z1 ∈ YC (t, t2 ). Hence, (4.101) is true.

(4.111)

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4 Maximum Principle of Optimal Controls

Second, since YC (t, s) is convex for any s > t, it follows by (4.83) C (t, τ ) is also convex for any τ > t. and (4.101) that Y Finally, by (4.83) and (4.101), we find that C (t, τ ) ⊆ YC (t, τ ) for any τ > t. Y Since YC (t, τ ) is closed, (4.102) follows from (4.112) immediately. Hence, we complete the proof of Proposition 4.5.

(4.112)  

4.2.2 Separability and Local Maximum Principle In this subsection, we show some connection between the local Pontryagin Maximum Principle and the separation of YR (t) (see (4.2)) and YC (t, tE∗ ) (see (4.82)), with t ∈ (0, tE∗ ), in the state space Y (see Theorem 4.3). We also present some relationship between the classical Pontryagin Maximum Principle and the local Pontryagin Maximum Principle (see Corollary 4.2). The main result of this subsection is as follows: Q ,Q

S E satisfies the local Pontryagin Maximum Theorem 4.3 The problem (T P )min Principle if and only if YR (t) and YC (t, tE∗ ) are separable in Y for any t ∈ (0, tE∗ ).

Proof We first show the sufficiency. Assume that YR (t) and YC (t, tE∗ ) are separable QS ,QE in Y for any t ∈ (0, tE∗ ). Let (0, y0∗ , tE∗ , u∗ ) be an optimal tetrad to (T P )min . ∗ Arbitrarily fix T ∈ (0, tE ). Then by the similar arguments to those used in the proof of Theorem 4.1 (where YE and YR (tE∗ ) are replaced by YC (T , tE∗ ) and YR (T ), respectively), we can prove that there exists ϕT (·) ∈ C([0, T ]; Y ) with ϕT (·) = 0 so that (4.84)–(4.88) hold. From this and Definition 4.2, we see that (0, y0∗ , tE∗ , u∗ ) satisfies the local Pontryagin Maximum Principle, consequently, so QS ,QE does (T P )min . Hence, we have proved the sufficiency. QS ,QE We next show the necessity. Suppose that (T P )min satisfies the local PontryaQS ,QE gin Maximum Principle. Let (0, y0∗ , tE∗ , u∗ ) be an optimal tetrad to (T P )min . ∗ Then by Definition 4.2, we see that for any T ∈ (0, tE ), there exists ϕT (·) ∈ C([0, T ]; Y ) with ϕT (·) = 0 so that (4.84)–(4.88) hold. Then by the similar arguments as those used in the proof of Theorem 4.1, we can prove that for each T ∈ (0, tE∗ ), YR (T ) and YC (T , tE∗ ) are separable in Y . Hence, we have proved the necessity. Thus, we complete the proof of Theorem 4.3.   Remark 4.5 From Theorem 4.3, we see that the key to derive the local Pontryagin QS ,QE is to find conditions ensuring the separation of Maximum Principle for (T P )min ∗ YR (t) and YC (t, tE ) in Y at each time t ∈ (0, tE∗ ).

4.2 Local Maximum Principle and Minimal Time Controls

155

We next discuss the relationship between the classical Pontryagin Maximum Principle and the local Pontryagin Maximum Principle. For this purpose, we need some preparations. First of all, we define YC (tE∗ , tE∗ )  YE . Lemma 4.4 Let T ∈ (0, tE∗ ]. Let φ ∈ Y be a nonzero vector separating YR (T ) from YC (T , tE∗ ) in Y . Let ϕ(·) be the solution to the equation: 

  ϕ(t) ˙ = − A∗ + D(t)∗ ϕ(t), t ∈ (0, T ), ϕ(T ) = φ.

Then for any t ∈ [0, T ) (with ϕ(t) = 0), ϕ(t) separates YR (t) from YC (t, tE∗ ) in Y . Proof Since φ separates YR (T ) from YC (T , tE∗ ) in Y , we have that z1 , φY ≥ z2 , φY for all z1 ∈ YR (T ) and z2 ∈ YC (T , tE∗ ).

(4.113)

Arbitrarily fix t ∈ [0, T ) with ϕ(t) = 0. By (4.113), we see that &  Φ(T , t)Φ(t, 0)y0 +  +

T

Φ(T , s)B(s)u1 (s)ds t

t

'

Φ(T , s)B(s)u2 (s)ds − z2 , φ

0

≥0 Y

for all y0 ∈ YS , u1 ∈ L(t, T ; U), u2 ∈ L(0, t; U) and z2 ∈ YC (T , tE∗ ). This yields that ' &  t Φ(t, s)B(s)u2 (s)ds, Φ(T , t)∗ φ Φ(t, 0)y0 + 0 Y ' & T (4.114) + Φ(T , s)B(s)u1 (s)ds − z2 , φ ≥ 0 t

Y

for all y0 ∈ YS , u1 ∈ L(t, T ; U), u2 ∈ L(0, t; U) and z2 ∈ YC (T , tE∗ ). Then, by (4.114), we find that inf ξ, ϕ(t)Y

ξ ∈YR (t )

≥

&

sup

z2 ∈YC (T ,tE∗ ),u1 ∈L(t,T ;U)



'

T

z2 −

Φ(T , s)B(s)u1 (s)ds, φ t

.

(4.115)

Y

Next, for any η ∈ YC (t, tE∗ ), there is  u ∈ L(t, tE∗ ; U) so that Φ(tE∗ , t)η



tE∗

+ t

Φ(tE∗ , s)B(s) u(s)ds ∈ YE .

(4.116)

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4 Maximum Principle of Optimal Controls

Denote   z2  Φ(T , t)η +

T

Φ(T , s)B(s) u1 (s)ds t and  u1 (·) =  u(·)(t,T ) ∈ L(t, T ; U).

(4.117)

It follows from (4.116) that  z2 ∈ YC (T , tE∗ ).

(4.118)

By (4.118), the second conclusion in (4.117), and by (4.115) with (z2 , u1 ) = ( z2 , u1 ), we have that for any η ∈ YC (t, tE∗ ), inf ξ, ϕ(t)Y ≥ Φ(T , t)η, φY = η, ϕ(t)Y .

ξ ∈YR (t )

This yields that ϕ(t) separates YR (t) from YC (t, tE∗ ) in Y . Hence, we end the proof of Lemma 4.4. For each t ∈

[0, tE∗ ],

(4.119)

 

we define the following subset:

 Ψt  {ψ ∈ Y \ {0}  ψ separates YR (t) from YC (t, tE∗ ) in Y }.

(4.120)

When YR (t) and YC (t, tE∗ ) are not separable in Y , we set Ψt  ∅. Based on Lemma 4.4, we have the following result. Corollary 4.1 Let Ψt , t ∈ [0, tE∗ ], be defined by (4.120). Then 

 Φ(t, s)∗ Ψt \ {0} ⊆ Ψs , when 0 ≤ s ≤ t ≤ tE∗ .

Here, we agree that Φ(t, s)∗ Ψt  ∅, when Ψt = ∅. Now we give the following assumption: (HB ) ϕ(·) = 0 in (0, tE∗ ), provided that ϕ(t0 ) = 0 for some t0 ∈ [0, tE∗ ) and ϕ(·) solves the following equation: 

  ϕ(t) ˙ = − A∗ + D(t)∗ ϕ(t), t ∈ (0, tE∗ ), ϕ(tE∗ ) ∈ Y.

(4.121)

Q ,Q

S E Corollary 4.2 Assume that the problem (T P )min satisfies the classical PontryaQS ,QE gin Maximum Principle. Suppose that (HB ) is true. Then (T P )min also satisfies the local Pontryagin Maximum Principle.

Q ,Q

S E satisfies the classical Pontryagin Maximum Proof Since the problem (T P )min Principle, by Theorem 4.1, Corollary 4.1, and (HB ), we observe that YR (t) and YC (t, tE∗ ) are separable in Y for any t ∈ (0, tE∗ ). This, together with Theorem 4.3, implies the desired result. Thus, we end the proof.  

4.2 Local Maximum Principle and Minimal Time Controls

157

S E Remark 4.6 When an optimal tetrad (0, y0∗ , tE∗ , u∗ ) to (T P )min satisfies the classical Pontryagin Maximum Principle, the assumption (HB ) ensures the following geometric property for the corresponding ϕ(·): For each t ∈ [0, tE∗ ], ϕ(t) is a vector separating YR (t) from YC (t, tE∗ ) in Y . (This follows from Theorem 4.1, (HB ), and Corollary 4.1.)

Q ,Q

Example 4.12 Consider all minimal time control problems in Examples 4.6, 4.7, and 4.8. One can directly check that they satisfy the classical Pontryagin Maximum Principle. Besides, the assumption (HB ) holds for all of the above problems. For the ODE cases in the above, (HB ) follows from the theory of existence and uniqueness for ODEs; For the heat equation cases, (HB ) follows from (i) of Remark 1.5 (after Theorem 1.22). Then by Corollary 4.2, all of the above problems satisfy the local Pontryagin Maximum Principle. We end Example 4.12.

4.2.3 Conditions on Separation in Local Maximum Principle In this subsection, we will give two conditions to guarantee the separation of YR (t) (see (4.2)) and YC (t, tE∗ ) (see (4.82)) for any t ∈ (0, tE∗ ). We begin with the next definition. Definition 4.4 Equation (4.1) is said to be L∞ -null controllable at any interval if for any 0 ≤ t1 < t2 < +∞ and y0 ∈ Y , there is u ∈ L∞ (t1 , t2 ; U ) so that y(t2 ; t1 , y0 , u) = 0. The first main result of this subsection is the next theorem. Theorem 4.4 Assume that Equation (4.1) is time-invariant and L∞ -null controllable at any interval. Suppose that YE = {0}, and 0 ∈ I ntU. Then for any t ∈ (0, tE∗ ), YR (t), and YC (t, tE∗ ) are separable in Y . Proof Arbitrarily fix t ∈ (0, tE∗ ). Since (4.1) is L∞ -null controllable at any interval, by Theorem 1.20, there exists a positive constant C(t, tE∗ ) (dependent on t and tE∗ ) so that ϕ(t)Y ≤ C(t, tE∗ )



tE∗

B ∗ ϕ(s)U ds,

(4.122)

t

for any solution ϕ(·) to the following equation: 

  ϕ(s) ˙ = − A∗ + D ∗ ϕ(s), ϕ(tE∗ ) ∈ Y.

s ∈ (t, tE∗ ),

(4.123)

Meanwhile, since 0 ∈ I ntU, there is a positive constant δ so that Bδ (0) ⊆ U.

(4.124)

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4 Maximum Principle of Optimal Controls

Denote  δ  δ/C(t, tE∗ ).

(4.125)

For each y0 ∈ Bδ (0), we define N(y0 ) 

min∗

u∈L∞ (t,tE ;U )

  uL∞ (t,tE∗ ;U )  y(tE∗ ; t, y0, u) = 0 .

The rest of the proof will be organized by four steps. Step 1. We claim that N(y0 ) ≤ C(t, tE∗ )y0 Y for all y0 ∈ Bδ (0).

(4.126)

It suffices to show (4.126) for y0 ∈ Bδ (0) \ {0}. To this end, we set    X  B ∗ ϕ(·)  ϕ(·) solves (4.123) ⊆ L1 (t, tE∗ ; U ). It is clear that X is a linear subspace of L1 (t, tE∗ ; U ). We define a map F : X → R by F (B ∗ ϕ(·))  −ϕ(t), y0 /y0 Y Y .

(4.127)

ϕ (·) over (t, tE∗ ), then We first claim that F is well defined. In fact, if B ∗ ϕ(·) = B ∗  by (4.122), we obtain that ϕ(t) =  ϕ (t) in Y . Hence, F is well defined. Besides, one can easily check that F is linear. It follows from (4.127) and (4.122) that |F (B ∗ ϕ(·))| ≤ C(t, tE∗ )B ∗ ϕ(·)L1 (t,t ∗ ;U ) , E

which yields that F L (X,R) ≤ C(t, tE∗ ). Thus, F is a bounded linear functional on X. By the Hahn-Banach Theorem (see Theorem 1.5), there exists a bounded linear functional G : L1 (t, tE∗ ; U ) → R so that G = F on X

(4.128)

and so that GL (L1 (t,t ∗ ;U ),R) = F L (X,R) ≤ C(t, tE∗ ). E

Then by making use of the Riesz Representation Theorem (see Theorem 1.4), there is a function  u ∈ L∞ (t, tE∗ ; U ) so that G(f ) =  u, f L∞ (t,t ∗ ;U ),L1(t,t ∗ ;U ) E

E

(4.129)

4.2 Local Maximum Principle and Minimal Time Controls

159

and so that  uL∞ (t,tE∗ ;U ) = GL (L1 (t,t ∗ ;U ),R) ≤ C(t, tE∗ ). E

(4.130)

Now, by (4.129), (4.128), and (4.127), we see that  u, B ∗ ϕ(·)L∞ (t,t ∗ ;U ),L1(t,t ∗ ;U ) = −ϕ(t), y0 /y0 Y Y . E

E

This implies that u) = 0, y(tE∗ ; t, y0 /y0 Y , which, combined with (4.130), indicates (4.126). Step 2. We show that 0 ∈ I nt (YC (t, tE∗ )).

(4.131)

By (4.126) and (4.125), we obtain that N(y0 ) ≤ δ for all y0 ∈ Bδ (0). This, together with (4.124), yields that Bδ (0) ⊆ YC (t, tE∗ ), which leads to (4.131). Step 3. We show that 0 ∈ ∂(YR (t) − YC (t, tE∗ )).

(4.132)

For this purpose, we first prove that YR (τ ) ∩ YC (t, tE∗ ) = ∅ for all τ ∈ [0, t).

(4.133)

By contradiction, suppose that it was not true. Then there would exist  τ ∈ [0, t) so that YR ( τ ) ∩ YC (t, tE∗ ) = ∅.

(4.134)

Since Equation (4.1) is time-invariant, we can directly check that τ , tE∗ − t +  τ ). YC (t, tE∗ ) = YC (

(4.135)

160

4 Maximum Principle of Optimal Controls

Now it follows from (4.134) and (4.135) that YR ( τ ) ∩ YC ( τ , tE∗ − t +  τ ) = ∅. The latter implies that τ < tE∗ , tE∗ ≤ tE∗ − t +  which leads to a contradiction. Hence, (4.133) is true. Furthermore, it follows from (4.92) that YR (t) ∩ YC (t, tE∗ ) = ∅.

(4.136)

Finally, by (4.136) and (i) in Proposition 4.5, we can use the similar arguments as those used in the proof of Proposition 4.2 (In (4.36), we denote Z(s)  YR (s) − YC (t, tE∗ ), s ∈ [0, t].) to obtain (4.132). Step 4. We show the conclusion in this theorem. By (4.131), we have that I nt (YR (t) − YC (t, tE∗ )) = ∅.

(4.137)

Since YR (t) − YC (t, tE∗ ) is a nonempty convex subset of Y (see Lemma 4.3 and (i) in Proposition 4.5), the desired result follows by (4.137), (4.132), and Theorem 1.10. Hence, we complete the proof of Theorem 4.4.   We next give some examples, which will be helpful for us to understand Theorem 4.4 and Theorem 4.3. Example 4.13 Let A ∈ Rn×n and B ∈ Rn×m with n, m ∈ N+ . Assume that (A, B) satisfies Kalmann controllability rank condition: rank(B, AB, . . . , An−1 B) = n. QS ,QE Let Y  Rn , U = Rm , and y0 ∈ Rn with y0 = 0. Consider the problem (T P )min , where the controlled system is as: y(t) ˙ = Ay(t) + Bu(t),

t ∈ (0, +∞),

and where QS = {0} × {y0 }; QE = (0, +∞) × {0}; U = Bρ (0) with ρ > 0. Q ,Q

S E One can easily check that (T P )min can be put into the framework (AT P ) (given at the beginning of Section 4.1). QS ,QE We further assume that (T P )min has optimal controls. Then by Theorem 1.23 and Theorem 4.4, we see that for each t ∈ (0, tE∗ ), YR (t) and YC (t, tE∗ ) are separable QS ,QE satisfies the local Pontryagin in Y . Hence, according to Theorem 4.3, (T P )min Maximum Principle. We end Example 4.13.

4.2 Local Maximum Principle and Minimal Time Controls

161

Q ,Q

S E Example 4.14 Consider the problem (T P )min , where the controlled system is the heat equation (3.59) (with f = 0); and where

QS = {0} × {y0 }; QE = (0, +∞) × {0} and U = Bρ (0), with y0 ∈ L2 (Ω) \ {0}, ρ > 0. S ,QE can be put into the framework (AT P ) One can easily check that (T P )Q min (given at the beginning of Section 4.1) in the current case. Moreover, according S ,QE has admissible controls, which, to Remark 3.1 (after Theorem 3.5), (T P )Q min along with the similar arguments as those in the proof of Theorem 3.11, leads to the existence of optimal controls. Then by Theorems 1.22 and 1.21, we can apply Theorem 4.4 to see that for each t ∈ (0, tE∗ ), YR (t) and YC (t, tE∗ ) are separable S ,QE satisfies the local Pontryagin in Y . Hence, according to Theorem 4.3, (T P )Q min Maximum Principle. We end Example 4.14.

The second main result of this subsection is as follows. Theorem 4.5 Suppose that YE is a viability domain associated with (4.1) (see Definition 4.3). Assume that   (4.138) I nt YC (t1 , t2 ) = ∅ for all 0 < t1 < t2 < tE∗ , and C (t1 , t2 ) for all 0 < t1 < t2 ≤ tE∗ . YC (t1 , t2 ) = Y

(4.139)

C (t, t ∗ ) are separable in Y ; (ii) YR (t) and Then for any t ∈ (0, tE∗ ), (i) YR (t) and Y E YC (t, tE∗ ) are separable in Y . C (t, t ∗ ) are Proof When (i) is proved, we have that for any t ∈ (0, tE∗ ), YR (t) and Y E separable in Y . This, along with (4.139), yields (ii). The remainder is to show (i). Arbitrarily fix t ∈ (0, tE∗ ). First, since YE is a viability domain associated with (4.1), it follows by (4.83), (4.138), and (ii) in C (t, t ∗ ) is a nonempty convex subset in Y . Second, it follows Proposition 4.5 that Y E by Lemma 4.3 that YR (t) is a nonempty convex subset in Y . Hence, we have that C (t, tE∗ ) − YR (t) is a nonempty convex subset in Y. Y Third, it follows by (4.93), (4.83), and (4.138) that   C (t, tE∗ ) − YR (t) and I nt Y C (t, tE∗ ) − YR (t) = ∅. 0 ∈ Y

(4.140)

(4.141)

From (4.141), we find that   C (t, tE∗ ) − YR (t) either 0 ∈ ∂ Y

(4.142)

C (t, t ∗ ) − YR (t). or 0 ∈ Y E

(4.143)

162

4 Maximum Principle of Optimal Controls

In the case when (4.142) is true, it follows by Theorem 1.10, (4.140), the second C (t, t ∗ ) are separable in Y . In the case result in (4.141) and (4.142) that YR (t) and Y E that (4.143) is true, it follows by Theorem 1.11, (4.140), and (4.143), that YR (t) and C (t, t ∗ ) are separable in Y . Y E Hence, we end the proof of Theorem 4.5.   The next result is comparable with Corollary 4.1. To state it, we define, for each t ∈ [0, tE∗ ), the following set:    t  ψ ∈ Y \ {0}  ψ is a vector separating YR (t) from Y C (t, tE∗ ) in Y . Ψ (4.144) C (t, t ∗ ) t  ∅ and Φ(t, s)∗ Ψ t  ∅, provided that YR (t) and Y Here, we agree that Ψ E are not separable in Y . Proposition 4.6 Assume that YE is a viability domain associated with (4.1). Then 

 t \ {0} ⊆ Ψ s for all 0 ≤ s ≤ t < tE∗ . Φ(t, s)∗ Ψ

(4.145)

Proof Since YE is a viability domain associated with (4.1), by Definition 4.3, (4.83), and (4.82), we have that C (t, tE∗ ) = ∅ for all t ∈ [0, tE∗ ). Y Arbitrarily fix s and t so that 0 ≤ s < t < tE∗ . Without loss of generality, we can t = ∅. Given ψ ∈ Ψ t , it follows by (4.144) that assume that Ψ &



'

t

Φ(t, 0)y0 +

≥ η, ψY ,

Φ(t, r)B(r)u(r)dr, ψ 0

(4.146)

Y

C (t, t ∗ ). Arbitrarily fix  C (s, t ∗ ). for all y0 ∈ YS , u ∈ L(0, t; U) and η ∈ Y z ∈ Y E E ∗ By (4.83) and (4.82), there is τ ∈ (s, tE ) and  v ∈ L(s, τ ; U) so that 

τ

Φ(τ, s) z+

Φ(τ, r)B(r) v (r)dr ∈ YE .

(4.147)

s

We next claim that there exists a function  u ∈ L(s, t; U) so that  Φ(t, s) z+ s

t

C (t, tE∗ ). Φ(t, r)B(r) u(r)dr ∈ Y

(4.148)

It will be proved by the following three cases related to τ : Case 1. τ > t. In this case, we set 

t

z  Φ(t, s) z+

Φ(t, r)B(r) v (r)dr, s

(4.149)

4.2 Local Maximum Principle and Minimal Time Controls

163

which, combined with (4.147), indicates that  τ Φ(τ, t)z + Φ(τ, r)B(r) v (r)dr ∈ YE . t

This, together with (4.82) and (4.83), implies that C (t, tE∗ ). z ∈ YC (t, τ ) ⊆ Y

(4.150)

By setting  u(·)   v (·)|(s,t ), we obtain (4.148) from (4.150) and (4.149) for Case 1. Case 2. τ = t. In this case, we set  τ Φ(τ, r)B(r) v (r)dr, z  Φ(τ, s) z+ s

which, combined with (4.147), indicates that  t z = Φ(t, s) z+ Φ(t, r)B(r) v (r)dr ∈ YE .

(4.151)

s

Since YE is a viability domain associated with (4.1), by Definition 4.3 and (4.151), there exists a function w ∈ L(t, (t + tE∗ )/2; U) so that y((t + tE∗ )/2; t, z, w) ∈ YE . This, along with (4.82) and (4.83), yields that C (t, tE∗ ). z ∈ YC (t, (t + tE∗ )/2) ⊆ Y

(4.152)

By setting  u(·)   v (·), we obtain (4.148) from (4.152) and (4.151) for Case 2. Case 3. τ < t. Since YE is a viability domain associated with (4.1), by Definition 4.3 and (4.147), there exists a function w ∈ L(τ, t; U) so that $  y t; τ, Φ(τ, s) z+

τ

% Φ(τ, r)B(r) v (r)dr, w ∈ YE .

s

Set   u(r) 

 v (r), r ∈ (s, τ ), w(r), r ∈ (τ, t).

(4.153)

164

4 Maximum Principle of Optimal Controls

Then  u ∈ L(s, t; U), and by (4.153), we get that 

t

Φ(t, s) z+

Φ(t, r)B(r) u(r)dr ∈ YE .

(4.154)

s

By (4.154) and using the similar arguments as those used in Case 2, we obtain (4.148) for Case 3. In summary, we conclude that (4.148) is true. Finally, we show (4.145). It will be proved by the following two possibilities related to s: s > 0 or s = 0. When s > 0, for each v ∈ L(0, s; U), we define  u(r) 

v(r), r ∈ (0, s),  u(r), r ∈ (s, t).

It is obvious that u ∈ L(0, t; U). Then, by (4.146) and (4.148), we obtain that &



s

Φ(s, 0)y0 +

∗

'

≥  z, Φ(t, s)∗ ψY

Φ(s, r)B(r)v(r)dr, Φ(t, s) ψ 0

Y

C (s, t ∗ ). Hence, (4.145) follows at once. for all y0 ∈ YS , v ∈ L(0, s; U) and  z∈Y E When s = 0, replacing u in (4.146) by  u, using (4.146) and (4.148), we find that C (0, tE∗ ). y0 , Φ(t, 0)∗ ψY ≥  z, Φ(t, 0)∗ ψY for all y0 ∈ YS and  z∈Y Thus, (4.145) is true. Hence, we complete the proof of Proposition 4.6.

 

The next Proposition 4.7 will give a condition to ensure the condition (4.139) when YE  {0} and U  Br (0). Here, Br (0) is the closed ball in U with center at the origin and of radius r ∈ (0, +∞). For this purpose, we define a minimum norm function N(·, ·; ·) from {(t, T ) ∈ R2 : 0 ≤ t < T < +∞} × Y to [0, +∞] by    N(t, T ; y0 )  inf uL∞ (t,T ;U )  y(T ; t, y0 , u) = 0 , for all 0 ≤ t < T < +∞, y0 ∈ Y.

(4.155)

When the set in the right-hand side of (4.155) is empty, we agree that N(t, T ; y0 ) = +∞. Proposition 4.7 Let YE  {0} and U  Br (0). Suppose that for any t ∈ [0, +∞) and y0 ∈ Y , the function

4.2 Local Maximum Principle and Minimal Time Controls

165

N(t, ·; y0 ) : (t, +∞) → [0, +∞] is left continuous. Then C (t, T ) for all 0 ≤ t < T < +∞. YC (t, T ) = Y Proof Arbitrarily fix t and T with 0 ≤ t < T < +∞. Since YE = {0} and 0 ∈ U, it follows by Definition 4.3 (or Example 4.11) that YE is a viability domain associated with (4.1). Hence, by (ii) of Proposition 4.5, we see that in order to prove this proposition, it suffices to show that C (t, T ). YC (t, T ) ⊆ Y

(4.156)

To show (4.156), we first observe from (4.82) that 0 ∈ YC (t, T ). So we have that YC (t, T ) = ∅. Thus, we can arbitrarily fix z ∈ YC (t, T ). Then, by (4.82) (where U = Br (0)), there exists u ∈ L(t, T ; Br (0)) so that y(T ; t, z, u) = 0. This, along with (4.155), indicates that N(t, T ; z) ≤ r.

(4.157)

Meanwhile, since N(t, ·; z) : (t, +∞) → [0, +∞] is left continuous, we find that for any n ∈ N+ , there is δn ∈ (0, T − t) so that N(t, T; z) ≤ N(t, T ; z) + 1/n for all T ∈ (T − δn , T ). From the above and (4.157) it follows that N(t, T; z) ≤ r + 1/n for all T ∈ (T − δn , T ), which, along with (4.155), implies that "

N t, T;

$

nr nr + 1

%2 # z < r for all T ∈ (T − δn , T ).

(4.158)

C (t, T ) (see (4.83)), we have that By (4.158) and the definition of Y $

nr nr + 1

%2

C (t, T ). z∈Y

C (t, T ). Hence, (4.156) is true. This completes the proof of This indicates that z ∈ Y Proposition 4.7.   The next two examples may help us to understand Proposition 4.7, Theorem 4.5, and Theorem 4.3. Example 4.15 Let Y = U = R and let  y0 ∈ R \ {0}. Consider the problem S ,QE , where the controlled system is as: (T P )Q min

166

4 Maximum Principle of Optimal Controls

y(t) ˙ = −ty(t) + u(t),

t ∈ (0, +∞),

and where QS = {0} × { y0 }; QE = (0, +∞) × {0}; U = Bρ (0) with ρ > 0. Q ,Q

S E One can easily check that the above problem (T P )min can be put into the framework (AT P ) (given at the beginning of Section 4.1). Moreover, by Theorem 3.3, QS ,QE has admissible controls. Then, by the similar one can easily check that (T P )min arguments as those in the proof of Theorem 3.11, one can show the existence of optimal controls. After some calculations, we see that for any t ∈ [0, +∞) and y0 ∈ R, " T 2 #−1 N(t, T ; y0 ) = |y0 | eτ /2 dτ for each T > t,

t

which indicates that the function N(t, ·; y0 ) : (t, +∞) → [0, +∞] is continuous. Then, by Proposition 4.7, we have that C (t, T ) for all t < T < +∞. YC (t, T ) = Y

(4.159)

Meanwhile, one can easily check that 0 ∈ I nt (YC (t, T )) for all t < T < +∞.

(4.160)

It follows from (4.159), (4.160), Example 4.11, and Theorem 4.5 that YR (t) and YC (t, tE∗ ) are separable in Y for each t ∈ (0, tE∗ ). Hence, according to S ,QE satisfies the local Pontryagin Maximum Principle. We Theorem 4.3, (T P )Q min end Example 4.15. S ,QE Example 4.16 Consider the problem (T P )Q , where the controlled system is min the heat equation (3.107) (with a(x, t) = a1 (x)+a2 (t), a1 (·) ∈ L∞ (Ω) and a2 (·) ∈ L∞ (0, +∞)), and where

QS = {0}×{y0}, QE = (0, +∞)×{0}, U = Bρ (0), with y0 ∈ L2 (Ω)\{0}, ρ > 0. Q ,Q

S E One can easily check that the above (T P )min can be put into the framework QS ,QE (AT P ) (given at the beginning of Section 4.1). We further assume that (T P )min has an optimal control. Notice that when ρ > 0 is large enough, this problem has an optimal control (see Theorem 3.7). We now prove that YR (t) and YC (t, tE∗ ) are separable in Y for each t ∈ (0, tE∗ ) in the current case. For this purpose, we will use a fact whose proof can be found from Example 5.4 and Proposition 5.3 in the next chapter of this book. This fact is as: for any t ∈ [0, +∞) and y0 ∈ L2 (Ω), the function N(t, ·; y0 ) : (t, +∞) → [0, +∞] is continuous. From this and Proposition 4.7, we have that

4.2 Local Maximum Principle and Minimal Time Controls

C (t, T ) for all t < T < +∞. YC (t, T ) = Y

167

(4.161)

Meanwhile, by Theorems 1.22 and 1.21, one can easily check that 0 ∈ I nt (YC (t, T )) for all t < T < +∞.

(4.162)

Now, it follows from (4.161), (4.162), Example 4.11, and Theorem 4.5 that YR (t) and YC (t, tE∗ ) are separable in Y for each t ∈ (0, tE∗ ). Hence, according to QS ,QE satisfies the local Pontryagin Maximum Principle. We Theorem 4.3, (T P )min end Example 4.16. The final example of this subsection gives a minimal time control problem satisfying what follows: (i) (ii) (iii) (iv)

It is under the framework (AT P ) (given at the beginning of Section 4.1); It has an optimal control; It satisfies neither the classical nor the local Pontryagin Maximum Principle; It satisfies the weak Pontryagin Maximum Principle (which will be given in Definition 4.7).

The proof of the above conclusion (iv) will be given in Example 4.18 of the next section. Because of the above (iv), we introduce the weak Pontryagin Maximum Principle in the next section. Example 4.17 Consider the following controlled equation: 

∂t y = x 2 u in Ω × (0, +∞), y=0 on ∂Ω × (0, +∞),

(4.163)

where Ω = (0, 1). We can put the Equation (4.163) into our framework (4.1) in the following manner: Let Y = U = L2 (0, 1); let A be the operator on Y defined by A = 0 with D(A) = L2 (0, 1); let B be the operator on Y defined by Bu = x 2 u for each u ∈ L2 (0, 1). QS ,QE Consider the problem (T P )min , where the controlled equation is (4.163) and where √ QS = {0} × {y0 } with y0 (x) = − 3x 3 , x ∈ (0, 1); QE = (0, +∞) × {0}; U = B1 (0). Q ,Q

S E One can easily check that (T P )min can be put into the framework (AT P ) (given at the beginning of Section 4.1) in the current case. Moreover, we will see that QS ,QE has an optimal control later. (T P )min QS ,QE The aim now is to show that (T P )min satisfies neither the classical nor the local Pontryagin Maximum Principle. The proofs will be carried out by three steps. Step 1. We show that (0, y0 , 1, u) with

168

4 Maximum Principle of Optimal Controls

 u(x, t) 

√

3x, (x, t) ∈ (0, 1) × (0, 1),

(4.164)

S ,QE . is the unique optimal tetrad of (T P )Q min Indeed, after some simple calculations, we can directly check that (0, y0 , 1, u) is QS ,QE an admissible tetrad of (T P )min . Then by the similar arguments as those in the proof of Theorem 3.11, one can easily show the existence of optimal controls. We arbitrarily fix an optimal tetrad (0, y0 , tE∗ , u∗ ). It follows that

y(tE∗ ; 0, y0, u∗ ) = 0 and u∗ L∞ (0,t ∗ ;L2 (0,1)) ≤ 1. E

(4.165)

For each k ∈ N+ , we define zk (x) 

√ 3χ(1/2k,1)(x)/x, x ∈ (0, 1).

(4.166)

From (4.166) and the equality in (4.165), we get that for each k ∈ N+ , 0 = y(tE∗ ; 0, y0 , u∗ ), zk L2 (0,1)  t∗  1 √ E = [(8k 3)−1 − 1] + 3x · u∗ (x, t)dxdt, 1/2k

0

which, combined with the inequality in (4.165), indicates that  t∗  1 √ E 3x · u∗ (x, t)dxdt = 1. 0

0

This, together with the inequality in (4.165), yields that  1=

tE∗

  1√ 3x · u∗ (x, t)dxdt ≤

0 0 t ∗ √ E ≤  3xL2 (0,1)dt = tE∗ .

tE∗ 0

√ u∗ (x, t)L2 (0,1) 3xL2 (0,1)dt

0

(4.167)

Since (0, y0 , tE∗ , u∗ ) is an optimal tetrad and (0, y0 , 1, u) is an admissible tetrad, (4.167) indicates that tE∗ = 1. Furthermore, by (4.167) and (4.168), we obtain that  1  1 1√ √ 3x · u∗ (x, t)dxdt = u∗ (x, t)L2 (0,1) 3xL2 (0,1)dt 0

0



0

1

= 0

√  3xL2 (0,1)dt.

(4.168)

4.2 Local Maximum Principle and Minimal Time Controls

169

Thus, we get that there exists a function λ(·) : (0, 1) → [0, +∞) so that for a.e. t ∈ (0, 1), √ u∗ (x, t) = λ(t) · 3x and u∗ (x, t)L2 (0,1) = 1. From these, we conclude that for a.e. (x, t) ∈ (0, 1) × (0, 1), √ u∗ (x, t) = 3x.

(4.169)

Hence, by (4.164), (4.168), (4.169), and the optimality of (0, y0 , tE∗ , u∗ ), we complete the proof of Step 1. S ,QE does not satisfy the local Pontryagin Maximum Step 2. We prove that (T P )Q min Principle. S ,QE did satisfy the local Pontryagin By contradiction, we suppose that (T P )Q min Maximum Principle. Arbitrarily fix T ∈ (0, 1). According to Definition 4.2 and Step 1, there exists ϕT ∈ L2 (0, 1) \ {0} so that  1  1 2 x  u(x, t)ϕT (x)dx = max x 2 v(x)ϕT (x)dx for a.e. t ∈ (0, T ). vL2 (0,1) ≤1 0

0

This, along with (4.164), yields that √  3x, x 2 ϕT L2 (0,1) = x 2 ϕT L2 (0,1).

(4.170)

Since √ √  3x, x 2 ϕT L2 (0,1) ≤  3xL2 (0,1)x 2 ϕT L2 (0,1) = x 2 ϕT L2 (0,1), according to (4.170), there exists a constant λT ∈ [0, +∞) so that √ x 2 ϕT (x) = λT · 3x for a.e. x ∈ (0, 1). This yields that ϕT (x) =

√ 3λT /x for a.e. x ∈ (0, 1).

Since ϕT ∈ L2 (0, 1), the above indicates that λT = 0 and ϕT (·) = 0, which leads to a contradiction with ϕT (·) = 0. Hence, we finish the proof of Step 2. QS ,QE Step 3. We show that (T P )min does not satisfy the classical Pontryagin Maximum Principle. Indeed, Step 3 follows from Step 2 and Corollary 4.2 at once. We now end Example 4.17.

170

4 Maximum Principle of Optimal Controls

4.3 Weak Maximum Principle and Minimal Time Controls Q ,QE

S This section studies the problem (T P )min (HT P ):

under the following framework

(i) The state space Y and the control space U are real separable Hilbert spaces. (ii) The controlled system is as: y(t) ˙ = Ay(t) + Bu(t), t ∈ (0, +∞),

(4.171)

where A : D(A) ⊆ Y → Y generates a C0 semigroup {eAt }t ≥0 on Y ; B ∈ L (U, Y ). Given τ ≥ 0, y0 ∈ Y and u ∈ L∞ (τ, +∞; U ), denote by y(·; τ, y0 , u) the unique mild solution of (4.1) over [τ, +∞), with the initial condition y(τ ) = y0 , i.e., for each T > τ , y(·; τ, y0 , u)|[0,T ] is the mild solution of (4.1) over [τ, T ], with the initial condition y(τ ) = y0 . (The existence and uniqueness of such solutions are ensured by Proposition 1.3.) (iii) The control constraint set U = BM (0) (i.e., the closed ball in U with centered at 0 and of radius M > 0). (iv) Let QS = {0} × { y0 } and QE = (0, +∞) × {0}, where  y0 ∈ Y \ {0} is fixed. Notice that (HT P ) is covered by (AT P ) (given at the beginning of Section 4.1). As what we did in the previous two sections, we always assume that the problem QS ,QE (T P )min has an optimal control. The aim of this section is to define a weak Pontryagin Maximum Principle for QS ,QE (T P )min and to derive the weak Pontryagin Maximum Principle by separating the target from the reachable set at the optimal time tE∗ in the reachable subspace RtE∗ (see (4.172) below).

4.3.1 Separability and Weak Maximum Principle First, for each t ∈ (0, +∞), we define   Rt  y(t; 0, 0, u) ∈ Y | u ∈ L∞ (0, t; U ) ,

(4.172)

which is endowed with the following norm   zRt  Min uL∞ (0,t ;U ) | y(t; 0, 0, u) = z , z ∈ Rt .

(4.173)

We call (4.172) the reachable subspace of the system (4.171) at t. Let us recall the reachable set of the system (4.171) at time t ∈ (0, +∞) (see (4.2)): y0, u) | u ∈ L(0, t; U)}. YR (t)  {y(t; 0, 

(4.174)

4.3 Weak Maximum Principle and Minimal Time Controls

171

It is obvious that YR (tE∗ ) ⊆ RtE∗ . We endow YR (tE∗ ) with the topology inherited from RtE∗ . In this subsection, we will establish the weak Pontryagin Maximum Principle

S E of (T P )min by separating the target {0} from YR (tE∗ ) in RtE∗ . For this purpose, we need to introduce the following two definitions.

Q ,Q

Definition 4.5 The sets YR (tE∗ ) and {0} are separable in RtE∗ if there exists ψ ∈ (RtE∗ )∗ \ {0} so that sup ψ, z(Rt ∗ )∗ ,Rt ∗ ≤ 0.

z∈YR (tE∗ )

E

E

Definition 4.6 An element ψ ∈ (RtE∗ )∗ is regular if there exists fψ ∈ L1 (0, tE∗ ; U ) so that the following representation formula holds: ψ, y(tE∗ ; 0, 0, u)(Rt ∗ )∗ ,Rt ∗ E

E



tE∗

=

u(t), fψ (t)U dt

(4.175)

0

for all u ∈ L∞ (0, tE∗ ; U ). Remark 4.7 For each T > 0, the space RT may be neither separable nor reflexive (even the system (4.171) has the L∞ -null controllability). Such examples can be found in [9] (see, for instance, [9, Theorem 2.8.8]). When RT is not reflexive for some T > 0, the dual space of RT is very large and some elements in RT may be very “irregular.” Some elements in this dual space may not have the representation formula (4.175). It should be pointed out that fψ in Definition 4.6 has a good regularity, since it belongs to a special subspace (see Corollary 4.3). This fact can be deduced from the next lemma. Lemma 4.5 Let T > 0 and f ∈ L1 (0, T ; U ). The following two statements are equivalent: (i) For any u1 , u2 ∈ L∞ (0, T ; U ), with y(T ; 0, 0, u1 ) = y(T ; 0, 0, u2 ), it holds that 

T



T

f (t), u1 (t)U dt =

0

f (t), u2 (t)U dt.

(4.176)

0

(ii) The function f belongs to the following space: OT  XT

·L1 (0,T ;U)

∗ (T −·)

with XT  {B ∗ eA

z|(0,T ) : z ∈ D(A∗ )}. (4.177)

Proof (i)⇒(ii). By contradiction, suppose that f satisfied (i) but f ∈ / OT . Then, since OT is a closed subspace of L1 (0, T ; U ) (see (4.177)), it would follow by the Hahn-Banach Theorem and the Riesz Representation Theorem (see Theorems 1.11 and 1.4) that there is  u ∈ L∞ (0, T ; U ) so that

172

4 Maximum Principle of Optimal Controls



T



T

g(t), u(t)U dt = 0
0), the function f in (4.180) belongs to OtE∗ which consists of all functions fsatisfying that

4.3 Weak Maximum Principle and Minimal Time Controls

175

⎧ ϕ (t) a.e. t ∈ (0, tE∗ ), ⎨ f(t) = B ∗  ϕ ∈ L1 (0, tE∗ ; U ),  ϕ ∈ C([0, tE∗ ); Y ), B ∗  ⎩  ϕ satisfies the adjoint equation over (0, tE∗ ). The internally controlled time-varying heat equation is such an example (see [18, Lemma 2.1]).

4.3.2 Two Representation Theorems Related to the Separability The first representation theorem is as follows: Theorem 4.7 For each T > 0, there is a linear isometry ΦT from RT to OT∗ (the dual space of OT ) so that for all yT ∈ RT (given by (4.172)) and f ∈ OT (given by (4.177)),  yT , f RT ,OT  ΦT (yT ), f OT∗ ,OT =

T

u(t), f (t)U dt,

(4.188)

0

where u is any control in L∞ (0, T ; U ) satisfying yT = y(T ; 0, 0, u). Here, OT is equipped with the norm  · L1 (0,T ;U ) . Proof Arbitrarily fix a T > 0. For each yT ∈ RT , we define the following set: UyT  {v ∈ L∞ (0, T ; U ) | y(T ; 0, 0, v) = yT }.

(4.189)

From (4.172) and (4.189) it follows that UyT = ∅ for any yT ∈ RT ,

(4.190)

and yT = y(T ; 0, 0, v) for all yT ∈ RT and v ∈ UyT .

(4.191)

By (4.190) and (4.191), we see that for each yT ∈ RT , z ∈ D(A∗ ) and v ∈ UyT , 

T

yT , zY = 0

∗ (T −t )

v(t), B ∗ eA

zU dt

∗ (T −·)

≤ vL∞ (0,T ;U ) B ∗ eA

(4.192)

zL1 (0,T ;U ) .

Given yT ∈ RT and v ∈ UyT , define a map FyT : XT → R in the following manner:

176

4 Maximum Principle of Optimal Controls

  ∗ FyT B ∗ eA (T −·) z|(0,T ) 



T

∗ (T −t )

v(t), B ∗ eA

zU dt, z ∈ D(A∗ ),

(4.193)

0

where XT is given by (4.177). By the equality in (4.192), we observe that the definition of FyT (·) is independent of the choice of v ∈ UyT . Thus it is well defined. Moreover, by (4.193), the inequality in (4.192) and (4.177), we find that FyT (·) can yT ∈ O ∗ satisfying be uniquely extended to be an element F T yT O ∗ ≤ vL∞ (0,T ;U ) for all v ∈ UyT . F T This, together with (4.173), implies that y O ∗ ≤ inf{uL∞ (0,T ;U ) | u ∈ Uy } = yT R for all yT ∈ RT . F T T T T (4.194) Meanwhile, we define a map ΦT : RT → OT∗ in the following manner: yT , yT ∈ RT . ΦT (yT )  F

(4.195)

It is obvious that ΦT is well defined and linear. Next, we claim that ΦT is an isometry from RT to OT∗ .

(4.196)

The proof of (4.196) will be carried out by three steps. Step 1. We show that ΦT : RT → OT∗ is surjective.

(4.197)

Arbitrarily fix g ∈ OT∗ . Since OT ⊆ L1 (0, T ; U ) (see (4.177)), by the HahnBanach Theorem (see Theorem 1.5), there exists  g ∈ (L1 (0, T ; U ))∗ so that  g (ψ) = g(ψ) for all ψ ∈ OT ; and  g (L1 (0,T ;U ))∗ = gOT∗ .

(4.198)

Then, by the Riesz Representation Theorem (see Theorem 1.4), there is  v ∈ L∞ (0, T ; U ) so that  g (w) =  v , wL∞ (0,T ;U ),L1(0,T ;U ) for all w ∈ L1 (0, T ; U ), and so that  g (L1 (0,T ;U ))∗ =  v L∞ (0,T ;U ) . These, combined with (4.198), indicate that

4.3 Weak Maximum Principle and Minimal Time Controls



T

∗ (T −t )

 v (t), B ∗ eA

∗ (T −·)

zU dt = g(B ∗ eA

177

z|(0,T ) ) for all z ∈ D(A∗ ),

0

(4.199) and  v L∞ (0,T ;U ) = gOT∗ . Write  yT  y(T ; 0, 0, v ). Then  v ∈ with (4.199), (4.193), and (4.177), yields that

(4.200)

U yT (see (4.189)). This, along

∗  g=F yT in OT ,

which, together with (4.195), leads to (4.197). Step 2. We show that ΦT : RT → OT∗ is injective.

(4.201)

Let yT ∈ RT satisfy that ΦT (yT ) = 0. By (4.195), we get that yT = 0 in O ∗ , F T which, combined with (4.193) and (4.192), indicates that yT , zY = 0 for all z ∈ D(A∗ ). Since D(A∗ ) is dense in Y , the above equality yields that yT = 0. So (4.201) is true. Step 3. We show that ΦT : RT → OT∗ preserves norms. ∗  yT =  y (T ; 0, 0, v ) with Let g ∈ OT∗ . Then we have that g = F yT in OT , where  ∞  v ∈ L (0, T ; U ) satisfying (4.199) and (4.200). This, along with (4.173), yields  that  yT RT ≤ F yT OT∗ . From this and (4.194), we see that ΦT preserves norms. Finally, the conclusion of the theorem follows from (4.196), (4.195), (4.193), and (4.177), immediately. Hence, we complete the proof of Theorem 4.7.   Remark 4.9 Several notes are given in order. (i) For a time-varying heat equation, a similar result to Theorem 4.7 has been built up in [18, (i) of Theorem 1.4]. (ii) In Theorem 4.6, we assume that the target {0} and YR (tE∗ ) are separable in RtE∗ by a regular vector ψ ∈ (RtE∗ )∗ \ {0}. This regular vector corresponds to a function f ∈ OtE∗ (see Definition 4.6 and Corollary 4.3). By Theorem 4.7, one can regard RtE∗ as the dual space of OtE∗ . Then (RtE∗ )∗ = (OtE∗ )∗∗ ⊇ OtE∗ . In general, one cannot obtain that (RtE∗ )∗ = OtE∗ . How do we check the assumption in Theorem 4.6? This assumption says that

178

4 Maximum Principle of Optimal Controls

the target {0} and YR (tE∗ ) are separable in RtE∗ under the weak star topology σ (RtE∗ , OtE∗ ).

(4.202)

It will be seen later that we cannot obtain (4.202) by using the result in Remark 1.1 (after Theorem 1.11). This will lead to the other representation theorem (see Theorem 4.8). As we point out in (ii) of Remark 4.9, to get the weak Pontryagin Maximum Principle via Theorem 4.6, we must face the separation of the target {0} and YR (tE∗ ) in RtE∗ under the weak star topology σ (RtE∗ , OtE∗ ). By (4.172), (4.173) and (4.174), one can directly check that  ∗ y0 } + {z ∈ RtE∗  zRt ∗ ≤ M}. YR (tE∗ ) = {eAtE  E

If we want to use the result in Remark 1.1 (after Theorem 1.11) to obtain the aforementioned separation, we need to answer the following problem: Is the unit closed ball in RtE∗ has a nonempty interior in the weak star topology? The next lemma is concerned with this problem. Lemma 4.6 Let T > 0 and RT be given by (4.172). Let BRT be the unit closed ball in RT , i.e., BRT  {yT ∈ RT | yT RT ≤ 1}. Assume that BRT has a nonempty interior in the weak star topology of RT (i.e., σ (RT , OT )). Then RT is of finite dimension. Proof Assume that BRT has a nonempty interior with respect to the weak star ◦∗

◦∗

topology σ (RT , OT ). Write B RT for this interior. Take yT ∈ B RT . By the definition ◦∗

of B RT , there exists {fj }kj =1 ⊆ OT , with a positive integer k, and r0 > 0 so that   yT ∈ {yT } + z ∈ RT | max |fj (z)| < r0 ⊆ BRT . 1≤j ≤k

(4.203)

From the latter it follows that   − yT ∈ {−yT } + z ∈ RT | max |fj (z)| < r0 ⊆ BRT . 1≤j ≤k

By (4.203) and (4.204), we have that   0 ∈ {yT /2 + (−yT )/2} + z ∈ RT | max |fj (z)| < r0 ⊆ BRT . 1≤j ≤k

Then

(4.204)

4.3 Weak Maximum Principle and Minimal Time Controls



 z ∈ RT | max |fj (z)| < r0 ⊆ BRT . 1≤j ≤k

179

(4.205)

We now claim that zRT ≤

1 max |fj (z)| for all z ∈ RT . r0 1≤j ≤k

(4.206)

By contradiction, we suppose that (4.206) was not true. Then there would exist z0 ∈ RT so that z0 RT >

1 max |fj (z0 )|. r0 1≤j ≤k

(4.207)

It is obvious that z0 = 0 in RT and 2 zz0R ∈ BRT . Hence, by (4.205), we get that T

2

  z0 ∈ z ∈ RT | max |fj (z)| < r0 , 1≤j ≤k z0 RT

which implies that 1 z0 RT max |fj (2z0 /z0 RT )| 2 1≤j ≤k 1 ≥ r0 z0 RT > 0. 2

c0  max |fj (z0 )| = 1≤j ≤k

(4.208)

Arbitrarily fix ε ∈ (0, r0 ). From (4.208) it follows that max |fj ((r0 − ε)z0 /c0 )| =

1≤j ≤k

r0 − ε max |fj (z0 )| = r0 − ε < r0 . c0 1≤j ≤k

This, together with (4.205), implies that r0 − ε z0 ∈ BRT . c0

(4.209)

By (4.209) and (4.208), we see that z0 RT

r0 − ε 1 c0 c0

max |fj (z0 )|. = z0

≤ r − ε = r − ε 1≤j ≤k r0 − ε c0 0 0 RT

(4.210)

Passing ε → 0+ in (4.210), we obtain a contradiction with (4.207). Hence, (4.206) is true. We next show that (RT )∗ = span {f1 , . . . , fk }.

(4.211)

180

4 Maximum Principle of Optimal Controls

To this end, we arbitrarily fix f ∈ (RT )∗ . By (4.206), there exists a positive constant Cf so that |f (z)| ≤ Cf max |fj (z)| for all z ∈ RT . 1≤j ≤k

(4.212)

Define a map as follows:   Gf f1 (z), . . . , fk (z)  f (z), z ∈ RT .

(4.213)

By (4.212), one can check that Gf is well defined. Moreover, Gf can be extended to be a linear map from Rk to R. Then there exists {cf,j }kj =1 ⊆ R so that k  

Gf f1 (z), . . . , fk (z) = cf,j fj (z) for all z ∈ RT . j =1

This, together with (4.213), implies that f (z) =

k

cf,j fj (z) for all z ∈ RT ,

j =1

which implies that f =

k

cf,j fj .

j =1

Thus, (4.211) follows immediately. Finally, by (4.211), we see that (RT )∗ is of finite dimension. Then we conclude that RT is also of finite dimension. Hence, we finish the proof of Lemma 4.6.   From Lemma 4.6, we see that in general, one cannot directly get the desired separation in Theorem 4.6 by the result in Remark 1.1 (after Theorem 1.11). To overcome this difficulty, we introduce the following set: For each T > 0,  RT0  y(T ; 0, 0, u) ∈ Y | u ∈ L∞ (0, T ; U ) with

 lim uL∞ (s,T ;U ) = 0 .

s→T −

(4.214)

is a subspace of RT . This subspace By (4.214) and (4.172), we can check that is related to the second representation theorem in this subsection (see Theorem 4.8 below) and will be used to obtain the separation in Theorem 4.6. Before stating the second representation theorem, we make the following hypothesis: RT0

4.3 Weak Maximum Principle and Minimal Time Controls

181

(H1) The system (4.171) is L∞ -null controllable over (0, T ) for each T > 0, i.e., for each T > 0 and y0 ∈ Y , there exists u ∈ L∞ (0, T ; U ) so that y(T ; 0, y0 , u) = 0. Based on the hypothesis (H1), we have the second representation theorem as follows: Theorem 4.8 Assume that (H1) holds. Then for each T ∈ (0, +∞), there is an isometry ΨT : OT → (RT0 )∗ so that for all yT ∈ RT0 and f ∈ OT ,  ΨT (f ), yT (R 0 )∗ ,R 0  f, yT OT ,R 0 = T

T

T

T

u(t), f (t)U dt,

(4.215)

0

where u is any element in L∞ (0, T ; U ) satisfying yT = y(T ; 0, 0, u). Before proving Theorem 4.8, we introduce the next lemma. Lemma 4.7 Suppose that (H1) holds. Then the following conclusions are true: (i) For any T > t > 0, there exists a constant C(T , t) ∈ (0, +∞) so that for each u ∈ L∞ (0, T ; U ), there is vu ∈ L∞ (0, T ; U ) so that  y(T ; 0, 0, χ(0,t )u) = y(T ; 0, 0, χ(t,T ) vu ); vu L∞ (0,T ;U ) ≤ C(T , t)uL2 (0,T ;U ) .

(4.216)

(ii) Let T ∈ (0, +∞), f ∈ L1 (0, T ; U ), and f |(0,t ) ∈ Ot for each t ∈ (0, T ). Then f ∈ OT . Here and throughout this subsection, C(·) denotes a positive constant depending on what are enclosed in the bracket. Proof We prove the conclusions one by one. (i) Since (H1) holds, we can apply Theorem 1.20 to find: for each s ∈ (0, +∞), there exists a constant C(s) ∈ (0, +∞) so that for each y0 ∈ Y , there is a control uy0 ∈ L∞ (0, s; U ) so that y(s; 0, y0, uy0 ) = 0 and uy0 L∞ (0,s;U ) ≤ C(s)y0 Y .

(4.217)

Now, for any T > t > 0 and u ∈ L∞ (0, T ; U ), we set y0,u  y(t; 0, 0, u).

(4.218)

By (4.217) (where s = T − t and y0 = y0,u ), we can find vy0,u ∈ L∞ (0, T − t; U ) so that y(T − t; 0, y0,u, vy0,u ) = 0 and vy0,u L∞ (0,T −t ;U ) ≤ C(T − t)y0,u Y . These, together with (4.218), imply that

182

4 Maximum Principle of Optimal Controls

vy0,u L∞ (0,T −t ;U )

 t

A(t −τ )

≤ C(T − t) e Bu(τ )dτ

0

Y

≤ C(T , t) max eAτ L (Y ;Y ) BL (U ;Y )uL2 (0,t ;U ) 0≤τ ≤t

(4.219) and that y(T ; 0, 0, χ(0,t )u) = eA(T −t ) y(t; 0, 0, u) = eA(T −t ) y0,u = y(T − t; 0, 0, −vy0,u ). Define

 vu (τ ) 

(4.220)

0, τ ∈ (0, t], −vy0,u (τ − t), τ ∈ (t, T ).

By (4.219) and (4.220), one can directly check that (4.216) holds for the above u and vu . (ii) Arbitrarily fix u ∈ L∞ (0, T ; U ) satisfying that y(T ; 0, 0, u) = 0.

(4.221)

By Lemma 4.5, we see that it suffices to show 

T

f (t), u(t)U dt = 0.

(4.222)

0

To prove (4.222), we arbitrarily fix ε ∈ (0, +∞). Then choose δε ∈ (0, min{ε, T /2}) so that y(δε ; 0, 0, u)Y < ε.

(4.223)

zε  y(δε ; 0, 0, u).

(4.224)

Set

By (4.217), there exists vuε ∈ L∞ (0, T /2; U ) so that y(T /2; 0, zε , vuε ) = 0 and vuε L∞ (0,T /2;U ) ≤ C(T /2)zε Y ,

(4.225)

where C(T /2) is given in (4.217). Define a new control  vε (t) 

u(t + δε ) − vuε (t), t ∈ (0, T /2), u(t + δε ), t ∈ (T /2, T − δε ).

From (4.225) and (4.226), we find that

(4.226)

4.3 Weak Maximum Principle and Minimal Time Controls



T −δε

y(T − δε ; 0, 0, vε ) = 0



 =

0 T

eA(T −δε −t ) Bu(t + δε )dt

T /2

−

183

eA(T −δε −t ) Bvuε (t)dt

eA(T −t ) Bu(t)dt + eA(T /2−δε ) eAT /2 zε ,

δε

which, combined with (4.224) and (4.221), indicates that  y(T − δε ; 0, 0, vε ) =

T

eA(T −t ) Bu(t)dt = 0.

(4.227)

0

Since f |(0,T −δε ) ∈ OT −δε , by (4.227) and Lemma 4.5, we get that 

T −δε

f (t), vε (t)U dt = 0.

0

This, along with (4.226), (4.225), (4.224), and (4.223), yields that   

T −δε

0

  =

T /2

0

  f (t), u(t + δε )U dt 

  f (t), vuε (t)U dt  ≤ C(T /2)f L1 (0,T /2;U )ε.

Since ε > 0 was arbitrarily taken and δε < ε, the above yields that 

T −δε

f (t), u(t + δε )U dt → 0, as ε → 0.

(4.228)

0

Now, by the density of C([0, T ]; U ) in L1 (0, T ; U ), we obtain (4.222) from (4.228). Hence, we complete the proof of Lemma 4.7.   We now are on the position to prove Theorem 4.8. Proof Let T > 0. According to Theorem 4.7, each g ∈ OT induces a linear and g over R 0 , via bounded functional F T g (yT )  yT , gR ,O , yT ∈ R 0 , F T T T

(4.229)

where ·, ·RT ,OT is given by (4.188). Thus, we obtain a map ΨT from OT to (RT0 )∗ defined by g for all g ∈ OT . ΨT (g)  F

(4.230)

184

4 Maximum Principle of Optimal Controls

One can easily check that ΨT : OT → (RT0 )∗ is linear. The rest of the proof is organized by the following three steps. Step 1. We show that gOT = ΨT (g)(R 0 )∗ for all g ∈ OT .

(4.231)

T

Arbitrarily fix g ∈ OT . On one hand, it follows from (4.229) that g  0 ∗ = F (R ) T

yT , gRT ,OT ≤ gOT ,

sup

(4.232)

yT ∈BR 0 (0,1) T

where BR 0 (0, 1) is the closed unit ball in RT0 . T On the other hand, we arbitrarily fix t0 ∈ (0, T ). Then according to the Hahn-Banach Theorem and the Riesz Representation Theorem (see Theorems 1.5 and 1.4), there is a control ut0 ∈ L∞ (0, t0 ; U ) so that gL1 (0,t0 ;U ) = ut0 , gL∞ (0,t0;U ),L1(0,t0 ;U ) and ut0 L∞ (0,t0 ;U ) = 1.

(4.233)

Write  ut0 for the zero extension of ut0 over (0, T ). Then it follows from (4.214) that y(T ; 0, 0, ut0 ) ∈ RT0 . Now, by (4.233), (4.188), (4.229), and (4.173), one can directly check that ut0 , gL∞ (0,T ;U ),L1(0,T ;U ) = y(T ; 0, 0, ut0 ), gRT ,OT gL1 (0,t0 ;U ) =    g y(T ; 0, 0, g  0 ∗ y(T ; 0, 0, ut ) ≤ F =F ut )R (RT )

0

0

T

g  0 ∗   ∞ ≤ F (R ) ut0 L (0,T ;U ) = Fg (R 0 )∗ . T

T

Since t0 was arbitrarily taken from (0, T ), the above yields that g  0 ∗ . gOT = gL1 (0,T ;U ) ≤ F (R ) T

This, together with (4.232), implies (4.231). Step 2. We claim that ΨT : OT → (RT0 )∗ is surjective. g ∈ OT so that Let f ∈ (RT0 )∗ . We aim to find  f = ΨT ( g ) in (RT0 )∗ .

(4.234)

u the In what follows, for each u ∈ L∞ (0, t0 ; U ), with t0 ∈ (0, T ), we denote by  zero extension of u over (0, T ). Then it follows from (4.214) that y(T ; 0, 0, u) ∈ RT0 . We define, for each t0 ∈ (0, T ), a map Gf,t0 from L∞ (0, t0 ; U ) to R by setting Gf,t0 (u)  f, y(T ; 0, 0, u)(R 0 )∗ ,R 0 , u ∈ L∞ (0, t0 ; U ). T

From (4.235), we see that for each t0 ∈ (0, T ),

T

(4.235)

4.3 Weak Maximum Principle and Minimal Time Controls

|Gf,t0 (u)| ≤ f(R 0 )∗ y(T ; 0, 0, u)RT for all u ∈ L∞ (0, t0 ; U ). T

185

(4.236)

Arbitrarily fix t0 ∈ (0, T ). By (i) of Lemma 4.7, there exists a constant C(T , t0 ) > 0 so that for each u ∈ L∞ (0, t0 ; U ), there is a control vu ∈ L∞ (0, T ; U ) satisfying that y(T ; 0, 0, u) = y(T ; 0, 0, χ(t0,T ) vu ); vu L∞ (0,T ;U ) ≤ C(T , t0 ) uL2 (0,T ;U ) . (4.237) From the equality in (4.237) and (4.173), we find that y(T ; 0, 0, u)RT ≤ vu L∞ (0,T ;U ) , which, combined with the inequality in (4.237), indicates that y(T ; 0, 0, u)RT ≤ C(T , t0 )uL2 (0,t0;U ) for all u ∈ L∞ (0, t0 ; U ). This, together with (4.236), implies that for each t0 ∈ (0, T ), |Gf,t0 (u)| ≤ C(T , t0 )f(R 0 )∗ uL2 (0,t0 ;U ) for all u ∈ L∞ (0, t0 ; U ). T

(4.238)

By (4.238) and the Hahn-Banach Theorem (see Theorem 1.5), we can uniquely extend Gf,t0 to be an element in (L2 (0, t0 ; U ))∗ , denoted in the same manner, so that |Gf,t0 (u)| ≤ C(T , t0 )f(R 0 )∗ uL2 (0,t0;U ) for all u ∈ L2 (0, t0 ; U ). T

(4.239)

By (4.239), using the Riesz Representation Theorem (see Theorem 1.4), we find that for each t0 ∈ (0, T ), there exists a gt0 ∈ L2 (0, t0 ; U ) so that  Gf,t0 (u) =

t0 0

gt0 (t), u(t)U dt for all u ∈ L2 (0, t0 ; U ).

(4.240)

We now take v ∈ L∞ (0, t0 ; U ) so that y(T ; 0, 0, v ) = 0. (Here,  v is the zero extension of v over (0, T ).) By (4.240) and (4.235), we see that 

t0 0

gt0 (t), v(t)U dt = Gf,t0 (v) = 0.

This, together with Lemma 4.5, implies that gt0 ∈ Ot0 for each t0 ∈ (0, T ).

(4.241)

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4 Maximum Principle of Optimal Controls

Meanwhile, by (4.240), (4.236), and (4.173), one can easily check that for each u ∈ L∞ (0, t0 ; U ), 

t0 0

gt0 (t), u(t)U dt ≤ f(R 0 )∗ y(T ; 0, 0, u)RT ≤ f(R 0 )∗ uL∞ (0,t0;U ) . T

T

This, together with (4.241), implies that gt0 Ot0 = gt0 L1 (0,t0;U ) ≤ f(R 0 )∗ for all t0 ∈ (0, T ). T

(4.242)

We next define a function  g : (0, T ) → U in the following manner: For each t0 ∈ (0, T ),  g (t)  gt0 (t) for all t ∈ (0, t0 ).

(4.243)

Then, the map  g is well defined on (0, T ). In fact, when 0 < t1 < t2 < T , it follows from (4.240) and (4.235) that for each u ∈ L∞ (0, t1 ; U ), 

t1 0

gt1 (t), u(t)U dt = Gf,t1 (u) = f, y(T ; 0, 0, u)(R 0 )∗ ,R 0 = Gf,t2 ( u|(0,t2) ) T



t2

= 0

 gt2 (t), u(t)U dt =

t1 0

T

gt2 (t), u(t)U dt,

which indicates that gt1 (·) = gt2 (·) over (0, t1 ). So we can check from (4.243) that the function  g is well defined. By (4.243) and (4.242), we see that  g L1 (0,T ;U ) ≤ f(R 0 )∗ .

(4.244)

T

From (H1), (4.244), (4.243), (4.241), and (ii) of Lemma 4.7, it follows that g OT ≤ f(R 0 )∗ .  g ∈ OT and  T

(4.245)

Moreover, by (4.235), (4.240), and (4.243), we deduce that for each t0 ∈ (0, T ), f, y(T ; 0, 0, u)(R 0 )∗ ,R 0 = T

T



T

 g (t), u(t)U dt for all u ∈ L∞ (0, t0 ; U ).

0

(4.246)

Finally, for each yT ∈ RT0 , it follows from (4.214) that there is uyT ∈ so that

L∞ (0, T ; U )

yT = y(T ; 0, 0, uyT ) and lim uyT L∞ (s,T ;U ) = 0. s→T

4.3 Weak Maximum Principle and Minimal Time Controls

187

By these and (4.173), we can check that when s → T , y(T ; 0, 0, χ(0,s)uyT ) − yT RT = y(T ; 0, 0, χ(s,T ) uyT )RT ≤ uyT L∞ (s,T ;U ) → 0, which indicates that y(T ; 0, 0, χ(0,s)uyT ) → yT in RT .

(4.247)

g ∈ OT (see (4.245)). Thus, Notice that y(T ; 0, 0, χ(0,s)uyT ) ∈ RT0 and  by (4.247), (4.246), (4.188), and the Lebesgue Dominated Convergence Theorem, we find that for each yT ∈ RT0 , f, yT (R 0 )∗ ,R 0 = lim f, y(T ; 0, 0, χ(0,s)uyT )(R 0 )∗ ,R 0 T

s→T

T

T



T

= lim

s→T



T

= 0

0

T

 g (t), χ(0,s) (t)uyT (t)U dt

 g (t), uyT (t)U dt = yT ,  g RT ,OT .

This, together with (4.229), implies that 0 0 ∗   f, yT (R 0 )∗ ,R 0 = F g (yT ) for all yT ∈ RT , i.e., f = F g in (RT ) . T

T

Hence, (4.234) follows immediately from the latter and (4.230). Step 3. We show the second equality in (4.215). The second equality in (4.215) follows from (4.188). Hence, we finish the proof of Theorem 4.8.

 

Remark 4.10 The assumption (H1) in Theorem 4.8 can be replaced by the following weaker condition (see [19, Theorem 2.6 and Proposition 9]). (H2) There is p0 ∈ [2, +∞) so that Ap0 (T ,  t) ⊆ A∞ (T ,  t) for all T ,  t, with 0 <  t < T < +∞, where t)  {y(T ; 0, 0, u) | u ∈ Lp0 (0, T ; U ) with u|(t,T ) = 0}; Ap0 (T ,  A∞ (T ,  t)  {y(T ; 0, 0, v) | v ∈ L∞ (0, T ; U ) with v|(0,t) = 0}. It deserves to mention that (H2) is essentially weaker than (H1), since the next two facts are proved in [19]. Fact one: (H1) implies (H2), while (H1) does not hold when (H2) is true for many cases. Fact two: In finite dimensional cases, any pair of matrices (A, B) in Rn×n × Rn×m satisfies (H2).

188

4 Maximum Principle of Optimal Controls

4.3.3 Conditions on Separation in Weak Maximum Principle The aim of this subsection is to give a condition that ensures the separation in Theorem 4.6. Theorem 4.9 Suppose that (H1) is true. Then the target {0} and YR (tE∗ ) are separable in RtE∗ by a regular vector ψ ∈ (RtE∗ )∗ \ {0}. The proof of Theorem 4.9 needs the next lemma. σ (RT ,OT )

Lemma 4.8 For each T > 0, it holds that BRT = B R 0

. Here, BRT and BR 0 T

T

σ (R ,O )

are the closed unit balls in RT and RT0 , respectively, and the set B R 0 T T is the T closure of BR 0 in the space RT , under the weak star topology σ (RT , OT ). T

Proof Let T > 0. We first prove that σ (RT ,OT )

BRT ⊆ B R 0

(4.248)

.

T

To this end, let yT ∈ BRT . By (4.172) and (4.173), there exists a sequence {vk }k≥1 ⊆ L∞ (0, T ; U ) so that for all k ≥ 1, yT = y(T ; 0, 0, vk ) and yT RT ≤ vk L∞ (0,T ;U ) ≤ yT RT + T /2k. (4.249) For each k ≥ 1, we set λk 

yT RT and uk  χ(0,T −T /2k) λk vk . yT RT + T /2k

(4.250)

On one hand, it follows from (4.250) and the second inequality in (4.249) that uk L∞ (0,T ;U ) ≤ yT RT ≤ 1 for all k ≥ 1.

(4.251)

This, together with (4.214), (4.250), and (4.173), implies that y(T ; 0, 0, uk ) ∈ BR 0 for all k ≥ 1. T

(4.252)

On the other hand, by the equality in (4.249), (4.188), and (4.250), we find that for each f ∈ OT , + + y(T ; 0, 0, uk ) − yT , f R ,O = y(T ; 0, 0, uk − vk ), f R ,O T T T T  T * + uk (t) − vk (t), f (t) U dt → 0 as k → +∞. = 0

σ (RT ,OT )

This, together with (4.252), implies that yT ∈ B R 0 T

. Hence, (4.248) follows.

4.3 Weak Maximum Principle and Minimal Time Controls

189

We next show that σ (RT ,OT )

BRT ⊇ B R 0

(4.253)

.

T

For this purpose, we let yT ∈ RT and {yk }k≥1 ⊆ BR 0 so that T

yk → yT in the topology σ (RT , OT ) as k → +∞. Since RT = OT∗ (see Theorem 4.7), we have that yk → yT in the weak star topology as k → +∞. Hence, yT RT ≤ lim inf yk RT ≤ 1, k→+∞

which indicates that yT ∈ BRT . Then (4.253) holds.

σ (RT ,OT )

Finally, it follows from (4.248) and (4.253) that BRT = B R 0 T completes the proof of Lemma 4.8.

. This  

We now are in a position to prove Theorem 4.9. Proof (Proof of Theorem 4.9) The proof will be carried out by the following three stages. Stage 1. We show that ∗

∗

y0 ∈ Rt0∗ and eAtE  y0 Rt ∗ = M. eAtE  E

(4.254)

E

Its proof will be divided into the following two steps: Step 1. We claim that ∗

y0 ∈ Rt0∗ . eAtE  E

(4.255)

In fact, it follows by (H1) that there exists u1 ∈ L∞ (0, tE∗ /2; U ) so that y0, u1 ) = 0. y(tE∗ /2; 0, 

(4.256)

Write  u1 for the zero extension of u1 over (0, +∞). Then it follows from (4.256) that ∗

y(tE∗ ; 0,  y0 , u1 ) = eAtE /2 y(tE∗ /2; 0,  y0, u1 ) = 0,

(4.257)

which indicates that ∗

y0 = −y(tE∗ ; 0, 0, u1 ). eAtE  Since  u1 = 0 over (tE∗ /2, tE∗ ), by the above equality and the definition of Rt0∗ E (see (4.214)), we obtain (4.255).

190

4 Maximum Principle of Optimal Controls

Step 2. We show that ∗

eAtE  y0 R 0∗ = M. t

(4.258)

E

S ,QE To this end, we let (0,  y0 , tE∗ , u∗ ) be an optimal tetrad to the problem (T P )Q . min Then we have that

y(tE∗ ; 0,  y0, u∗ ) = 0 and u∗ L∞ (0,tE∗ ;U ) ≤ M. These yield that ∗

y0 = −y(tE∗ ; 0, 0, u∗ ) ∈ RtE∗ eAtE  and ∗

y0 Rt ∗ ≤ M. eAtE 

(4.259)

E

Now, by contradiction, we suppose that (4.258) was not true. Then by (4.259), we would find that ∗

eAtE  y0 Rt ∗ < M. E

This, together with (4.172) and (4.173), implies that there is a control u2 ∈ L∞ (0, tE∗ ; U ) so that ∗

eAtE  y0 = y(tE∗ ; 0, 0, u2) and u2 L∞ (0,tE∗ ;U ) < M.

(4.260)

Choose ε > 0 small enough so that 0 < ε ≤ M − u2 L∞ (0,tE∗ ;U ) .

(4.261)

Then, by (H1) we can apply Theorem 1.20 to find δ ∈ (0, tE∗ /2) and uδ ∈ L∞ (0, +∞; U ) so that y(tE∗ /2; 0, zδ , uδ ) = 0 and uδ L∞ (0,+∞;U ) ≤ ε,

(4.262)

where zδ   y0 − y(δ; 0,  y0, −u2 ).

(4.263)

We define a new control in the following manner: u3 (t)  χ(0,tE∗ /2) uδ (t) − u2 (t + δ) for a.e. t ∈ (0, tE∗ − δ).

(4.264)

4.3 Weak Maximum Principle and Minimal Time Controls

191

By (4.263) and (4.264), we have that y(tE∗ − δ; 0,  y 0 , u3 )  t ∗ −δ E ∗ ∗ = eA(tE −δ) y0 + eA(tE −δ−t ) Bu3 (t)dt 0

=e

A(tE∗ −δ)



[zδ + y(δ; 0,  y0, −u2 )]

tE∗ −δ

+ 0

∗

eA(tE −δ−t ) B[χ(0,tE∗ /2) uδ (t) − u2 (t + δ)]dt,

which, combined with the equalities in (4.260) and (4.262), indicates that y(tE∗ − δ; 0,  y 0 , u3 )  t ∗ −δ E ∗ ∗ = eA(tE −δ) zδ + eA(tE −δ−t ) Bχ(0,tE∗ /2) uδ (t)dt 0  δ " #  tE∗ ∗ ∗ −δ) A(δ−t ) A(t Aδ E +e e  y0 − e Bu2 (t)dt − eA(tE −t ) Bu2 (t)dt 0

(4.265)

δ

= y(tE∗ − δ; 0, zδ , χ(0,tE∗ /2) uδ ) + y(tE∗ ; 0,  y0 , −u2 ) = 0. Moreover, it follows from (4.264), (4.262), and (4.261) that u3 L∞ (0,tE∗ −δ;U ) ≤ uδ L∞ (0,+∞;U ) + u2 L∞ (0,tE∗ ;U ) ≤ M. This, together with (4.265), contradicts the optimality of tE∗ . Hence, (4.258) holds. Finally, (4.254) follows from (4.255) and (4.258) immediately. Stage 2. We claim that there exists a regular vector ϕ ∗ ∈ (RtE∗ )∗ \ {0} so that ∗

y0 (Rt ∗ )∗ ,Rt ∗ . sup ϕ ∗ , z(Rt ∗ )∗ ,Rt ∗ ≤ ϕ ∗ , −eAtE  E

z∈AM

E

E

(4.266)

E

Here AM  {z ∈ Rt0∗ | zRt ∗ ≤ M}. E

(4.267)

E

By (4.267), (4.254), and Theorem 1.11, there exists  ϕ ∈ (Rt0∗ )∗ \ {0} so that E

sup  ϕ , z(R 0∗ )∗ ,R 0∗ ≤  ϕ , −e tE

z∈AM

AtE∗

tE

 y0 (R 0∗ )∗ ,R 0∗ . tE

(4.268)

tE

Meanwhile, by Theorem 4.8, there exists f ϕ ∈ OtE∗ \ {0} so that f ϕ (R 0∗ )∗ and f ϕ , z(R 0∗ )∗ ,R 0∗ for all z ∈ Rt0∗ . ϕ Ot ∗ =  ϕ , zO ∗ ,R 0∗ =  E

t

E

tE

t

E

t

E

t

E

E

192

4 Maximum Principle of Optimal Controls

These, together with (4.268), imply that ∗

At sup f y0 O ∗ ,R 0∗ . ϕ , zO ∗ ,R 0∗ ≤ f ϕ , −e E  tE

z∈AM

t

tE

E

t

(4.269)

E

Two observations are given in order: First, it follows from Theorem 4.8 and Theorem 4.7, that 0 f ϕ , zO ∗ ,R 0∗ = z, f ϕ Rt ∗ ,Ot ∗ for all z ∈ RtE ∗ . t

E

tE

E

E

Second, it follows from Lemma 4.8 and (4.267) that sup z, f ϕ Rt ∗ ,Ot ∗ = E

z∈AM

E

sup

z∈Rt ∗ ,zR ∗ ≤M t

E

z, f ϕ Rt ∗ ,Ot ∗ . E

E

E

These two observations, together with (4.269), imply that sup

z∈Rt ∗ ,zR ∗ ≤M E

≤ f ϕ , −e

z, f ϕ Rt ∗ ,Ot ∗ = sup f ϕ , zO ∗ ,R 0∗ E

tE AtE∗

E

 y0 O ∗ ,R 0∗ = −e tE

t

z∈AM

AtE∗

tE

(4.270)

 y0 , f ϕ Rt ∗ ,Ot ∗ . E

tE

E

E

Meanwhile, according to Theorem 4.7, there exists  ϕ ∈ (RtE∗ )∗ so that ϕ , z(Rt ∗ )∗ ,Rt ∗ for all z ∈ RtE∗ ; and f ϕ (Rt ∗ )∗ . z, f ϕ Rt ∗ ,Ot ∗ =  ϕ Ot ∗ =  E

E

E

E

E

E

This, along with (4.270), Definition 4.6, and Theorem 4.7, leads to (4.266), with ϕ. ϕ∗ =  Stage 3. We show the conclusions in Theorem 4.9. First, we claim that ∗

y0 } + {z ∈ RtE∗ | zRt ∗ ≤ M}. YR (tE∗ ) = {eAtE 

(4.271)

E

To this end, we arbitrarily fix y1 ∈ YR (tE∗ ). By the definition of YR (tE∗ ) (see (4.174)), there exists u ∈ L∞ (0, tE∗ ; U ) so that y1 = y(tE∗ ; 0,  y0, u) and uL∞ (0,tE∗ ;U ) ≤ M. It follows from the latter, and definitions of RtE∗ and ·Rt ∗ (see (4.172) and (4.173)) E that ∗

∗

y0 + y(tE∗ ; 0, 0, u) ∈ {eAtE  y0 } + {z ∈ RtE∗ | zRt ∗ ≤ M}, y1 = eAtE  E

which indicates that ∗

y0 } + {z ∈ RtE∗ | zRt ∗ ≤ M}. YR (tE∗ ) ⊆ {eAtE  E

(4.272)

4.3 Weak Maximum Principle and Minimal Time Controls

193

On the other hand, for any z ∈ RtE∗ with zRt ∗ ≤ M, by definitions of RtE∗ and E  · Rt ∗ (see (4.172) and (4.173)), there exists v ∈ L∞ (0, tE∗ ; U ) so that E

z = y(tE∗ ; 0, 0, v) and vL∞ (0,tE∗ ;U ) ≤ M. These, together with (4.174), imply that ∗

y0 + z = y(tE∗ ; 0,  y0 , v) ∈ YR (tE∗ ), eAtE  which indicates that ∗

{eAtE  y0 } + {z ∈ RtE∗ | zRt ∗ ≤ M} ⊆ YR (tE∗ ).

(4.273)

E

Hence, (4.271) follows from (4.272) and (4.273) immediately. Next, by (4.266) and (4.267), we obtain that ∗

ϕ ∗ , eAtE  y0 + z(Rt ∗ )∗ ,Rt ∗ ≤ 0 for all z ∈ Rt0∗ with zRt ∗ ≤ M. E

E

E

(4.274)

E

We claim that ∗

ϕ ∗ , eAtE  y0 + z(Rt ∗ )∗ ,Rt ∗ ≤ 0 for all z ∈ RtE∗ with zRt ∗ ≤ M. E

E

(4.275)

E

Indeed, on one hand, by Lemma 4.8, for any fixed z ∈ RtE∗ with zRt ∗ ≤ M, there E

exists a sequence {zk }k≥1 ⊆ Rt0∗ with zk Rt ∗ ≤ M so that E

E

zk , ψRt ∗ ,Ot ∗ → z, ψRt ∗ ,Ot ∗ E

E

E

for all ψ ∈ OtE∗ .

E

(4.276)

On the other hand, since ϕ ∗ ∈ (RtE∗ )∗ \ {0} is regular, by Definition 4.6, Corollary 4.3, and Theorem 4.7, we obtain that there exists fϕ ∗ ∈ OtE∗ so that for all u ∈ L∞ (0, tE∗ ; U ), ϕ ∗ , y(tE∗ ; 0, 0, u)(Rt ∗ )∗ ,Rt ∗ = y(tE∗ ; 0, 0, u), fϕ ∗ R ∗∗ ,Ot ∗ . E

tE

E

(4.277)

E

It follows from (4.277) and (4.276) that ϕ ∗ , z(Rt ∗ )∗ ,Rt ∗ = z, fϕ ∗ Rt ∗ ,Ot ∗ E

E

E

E

= lim zk , fϕ ∗ Rt ∗ ,Ot ∗ = lim ϕ ∗ , zk (Rt ∗ )∗ ,Rt ∗ , k→+∞

E

E

k→+∞

which, combined with (4.274), indicates (4.275).

E

E

194

4 Maximum Principle of Optimal Controls

By (4.275) and (4.271), we see that sup ϕ ∗ , z(Rt ∗ )∗ ,Rt ∗ ≤ 0.

z∈YR (tE∗ )

E

E

From the above inequality and Definitions 4.5 and 4.6, it follows that YR (tE∗ ) and the target {0} are separable in RtE∗ by a regular vector ϕ ∗ . Hence, we finish the proof of Theorem 4.9.   At the end of this subsection, we give two examples, which may help us to understand Theorem 4.9 and Theorem 4.6 better. S ,QE Example 4.19 Consider the problem (T P )Q in Example 4.13. One can easily min check two facts as follows: First, this problem can be put into the framework (HT P ) (given at the beginning of Section 4.3); Second, (H1) (given before Theorem 4.8) is S ,QE has optimal true. (Here we used Theorem 1.23.) We further assume that (T P )Q min controls. According to Theorem 4.9, the target {0} and the set YR (tE∗ ) are separable in RtE∗ S ,QE by a regular vector in (RtE∗ )∗ \ {0}. Hence, by Theorem 4.6, (T P )Q satisfies the min weak Pontryagin Maximum Principle. We end Example 4.19.

S ,QE Example 4.20 Consider the problem (T P )Q in Example 4.14. One can easily min check that it can be put into the framework (HT P ) (given at the beginning of Section 4.3) and that (H1) (given before Theorem 4.8) holds. (Here we used S ,QE Theorems 1.22 and 1.21.) Moreover, (T P )Q has optimal controls. Then min according to Theorem 4.9, the target {0} and the set YR (tE∗ ) are separable in RtE∗

S ,QE by a regular vector in (RtE∗ )∗ \ {0}. Hence, by Theorem 4.6, (T P )Q satisfies the min weak Pontryagin Maximum Principle. We end Example 4.20.

4.4 Maximum Principle for Maximal Time Controls Q ,QE

S This section studies the maximal time control problem (T P )max max following framework (AT P ):

under the

(i) The state space Y and the control space U are real separable Hilbert spaces. (ii) Let T > 0. Let A : D(A) ⊆ Y → Y generate a C0 semigroup {eAt }t ≥0 on Y . Let D(·) ∈ L∞ (0, T ; L (Y )) and B(·) ∈ L∞ (0, T ; L (U, Y )). The controlled system is as: y(t) ˙ = Ay(t) + D(t)y(t) + B(t)u(t), t ∈ (0, T ).

(4.278)

Let {Φ(t, s) : t ≥ s ≥ 0} be the evolution system generated by A + D(·) over Y . Given τ ∈ [0, T ), y0 ∈ Y and u ∈ L∞ (τ, T ; U ), write y(·; τ, y0 , u) for the solution of (4.278) with the initial condition that y(τ ) = y0 .

4.4 Maximum Principle for Maximal Time Controls

195

(iii) The control constraint set U is a nonempty, bounded, convex, and closed subset in U . (iv) Let QS = [0, T )×YS and QE = {T }×YE , where YS and YE are two nonempty, bounded, convex, and closed subsets in Y so that YS ∩ YE = ∅. As what we did in the previous sections, we always assume that the problem QS ,QE (T P )max has an optimal control. The aim of this section is to derive the classical QS ,QE and the local Pontryagin Maximum Principles for (T P )max , through using the similar methods to those used in Sections 4.1 and 4.2. QS ,QE Recall that (T P )max is defined as follows:  sup{tS  tS ∈ [0, T ), y0 ∈ YS , u ∈ L(tS , T ; U) and y(T ; tS , y0 , u) ∈ YE }. S ,QE ∗ When (tS∗ , y0∗ , T , u∗ ) is an optimal tetrad for (T P )Q max , u is called an optimal control and tS∗ is called the maximal time. (Throughout this section, we use tS∗ to S ,QE ∗ ∗ ∗ ∗ denote the maximal time of (T P )Q max .) We write y (·) for y(·; tS , y0 , u ), and call it as the corresponding optimal trajectory (or the optimal state).

4.4.1 Classical Maximum Principle for Maximal Time Controls We first give the definition of classical Pontryagin Maximum Principle for the QS ,QE . problem (T P )max Definition 4.8 Several definitions are given in order. (i) An optimal control u∗ (associated with an optimal tetrad (tS∗ , y0∗ , T , u∗ )) to S ,QE (T P )Q is said to satisfy the classical Pontryagin Maximum Principle, if max there exists ϕ(·) ∈ C([tS∗ , T ]; Y ), with ϕ(·) = 0 (i.e., ϕ is not a zero function), so that ϕ(t) ˙ = −A∗ ϕ(t) − D(t)∗ ϕ(t) for a.e. t ∈ (tS∗ , T );

(4.279)

H (t, y ∗ (t), u∗ (t), ϕ(t)) = max H (t, y ∗ (t), u, ϕ(t)) for a.e. t ∈ (tS∗ , T ), u∈U

(4.280) where y ∗ ∈ C([tS∗ , T ]; Y ) is the corresponding optimal trajectory, and where H (t, y, u, ϕ)  ϕ, D(t)y + B(t)uY

(4.281)

for a.e. t ∈ (tS∗ , T ) and all (y, u, ϕ) ∈ Y × U × Y ; ϕ(tS∗ ), y0 − y0∗ Y ≤ 0 for each y0 ∈ YS ;

(4.282)

196

4 Maximum Principle of Optimal Controls

and * + ϕ(T ), z − y ∗ (T ) Y ≥ 0 for each z ∈ YE .

(4.283)

Here, (4.279) is called the dual (or adjoint) equation; (4.280) is called the maximum condition; the function H (·) defined by (4.281) is called the Hamiltonian S ,QE (associated with (T P )Q max ); (4.282) and (4.283) are called the transversality conditions. S ,QE (ii) The problem (T P )Q is said to satisfy the classical Pontryagin Maximum max Principle, if any optimal control satisfies the classical Pontryagin Maximum Principle. Before presenting main results of this subsection, we define, for any 0 ≤ t1 < t2 ≤ T , the following two sets:    YRmax (t2 , t1 )  y(t2 ; t1 , y0 , u)  y0 ∈ YS , u ∈ L(t1 , t2 ; U)

(4.284)

and    YCmax (t1 )  y0 ∈ Y  ∃ u ∈ L(t1 , T ; U) so that y(T ; t1 , y0 , u) ∈ YE . (4.285) We call YRmax (t2 , t1 ) the reachable set at the time t2 from the time t1 of the Equation (4.278). We call YCmax (t1 ) the controllable set at the time t1 for the Equation (4.278). We now present the first main result of this subsection. Q ,Q

S E Theorem 4.10 The problem (T P )max satisfies the classical Pontryagin Maximax ∗ mum Principle if and only if YR (T , tS ) and YE are separable in Y .

Proof We first show the sufficiency. Assume that YRmax (T , tS∗ ) and YE are separable QS ,QE in Y . Let (tS∗ , y0∗ , T , u∗ ) be an optimal tetrad to (T P )max . By the similar argu∗ ments to those used to show Theorem 4.1 (where YR (tE ) is replaced by YR (T , tS∗ )), we can prove that there exists ϕ(·) ∈ C([tS∗ , T ]; Y ) with ϕ(·) = 0 so that (4.279)– (4.283) hold. Then by Definition 4.8, we find that (tS∗ , y0∗ , T , u∗ ) satisfies the classiQS ,QE cal Pontryagin Maximum Principle, consequently, so does (T P )max . Hence, we have proved the sufficiency. QS ,QE satisfies the classical PonWe next show the necessity. Suppose that (T P )max QS ,QE ∗ ∗ ∗ . tryagin Maximum Principle. Let (tS , y0 , T , u ) be an optimal tetrad to (T P )max ∗ By Definition 4.8, there exists ϕ(·) ∈ C([tS , T ]; Y ) with ϕ(·) = 0 so that (4.279)– (4.283) hold. Then using the similar arguments as those used in the proof of Theorem 4.1, we can prove that YRmax (T , tS∗ ) and YE are separable in Y . Hence, the necessity has been shown. Thus, we complete the proof of Theorem 4.10.  

4.4 Maximum Principle for Maximal Time Controls

197

We next give the second main result of this subsection. Theorem 4.11 Assume that A : D(A) ⊆ Y → Y generates a C0 group. Then the QS ,QE problem (T P )max satisfies the classical Pontryagin Maximum Principle if and only if YCmax (tS∗ ) and YS are separable in Y .  

Q S ,Q E , where the Proof We define a minimal time optimal control problem (T P )min controlled system is

 z˙ (t) =

−Az(t) − D(T − t)z(t) − B(T − t)v(t), t ∈ (0, T ), 0, t ∈ [T , +∞),

(4.286)

S  {0} × YE and Q E  (0, T ] × YS . Q

(4.287)

and where

Write z(·; τ, z0 , v) for the solution of (4.286) with the initial condition that z(τ ) = z0 , where τ ∈ [0, +∞) and z0 ∈ Y are arbitrarily fixed. We now show the sufficiency. Assume that YCmax (tS∗ ) and YS are separable in S ,QE Y . Let (tS∗ , y0∗ , T , u∗ ) be an optimal tetrad to (T P )Q max . Then it follows by Theorem 2.2 and Definition 2.2 that  ,Q  Q

S E (0, y(T ; tS∗ , y0∗ , u∗ ), T − tS∗ , u∗ (T − ·)) is an optimal tetrad to (T P )min . (4.288)

Meanwhile, by (4.285), we can easily check that    YCmax (tS∗ ) = z(T − tS∗ ; 0, z0 , v)  z0 ∈ YE , v ∈ L(0, T − tS∗ ; U) .

(4.289)

Since YCmax (tS∗ ) is the reachable set at T − tS∗ of the system (4.286) (see (4.289))  

Q S ,Q E and YS is the target set of (T P )min (see (4.287)), we can apply Theorem 4.1 ∗ to find ϕ(·) ∈ C([tS , T ]; Y ) with ϕ(·) = 0 so that (4.279)–(4.283) hold. From this and Definition 4.8, we see that (tS∗ , y0∗ , T , u∗ ) satisfies the classical Pontryagin S ,QE Maximum Principle, consequently, so does (T P )Q max . Hence, the sufficiency has been proved. S ,QE satisfies the classical PonWe next show the necessity. Suppose that (T P )Q max S ,QE tryagin Maximum Principle. Let (tS∗ , y0∗ , T , u∗ ) be an optimal tetrad to (T P )Q max . Then by Definition 4.8, there exists ϕ(·) ∈ C([tS∗ , T ]; Y ) with ϕ(·) = 0 so that (4.279)–(4.283) hold. Define the following function:

ψ(t)  −ϕ(T − t),

t ∈ [0, T − tS∗ ].

Then ψ(·) = 0, and it follows by (4.279)–(4.283) that ˙ ψ(t) = A∗ ψ(t) + D(T − t)∗ ϕ(t) a.e. t ∈ (0, T − tS∗ ),

(4.290)

198

4 Maximum Principle of Optimal Controls

−B(T − t)∗ ψ(t), u∗ (T − t)U = max−B(T − t)∗ ψ(t), uU a.e. t ∈ (0, T − tS∗ ),

(4.291)

u∈U

ψ(0), z − y(T ; tS∗ , y0∗ , u∗ )Y ≤ 0 for all z ∈ YE ,

(4.292)

and for all y0 ∈ YS , ψ(T − tS∗ ), y0 − z(T − tS∗ ; 0, y(T ; tS∗ , y0∗ , u∗ ), u∗ (T − ·))Y ≥ 0.

(4.293)

Meanwhile, by Theorem 2.2 and Definition 2.2, we obtain that  ,Q  Q

S E (0, y(T ; tS∗ , y0∗ , u∗ ), T − tS∗ , u∗ (T − ·)) is an optimal tetrad to (T P )min . (4.294)

By (4.290)–(4.293), using the similar arguments to those used in the proof of Theorem 4.1 (replacing respectively YR (tE∗ ) and YE by YCmax (tS∗ ) and YS ), we can prove that YCmax (tS∗ ) and YS are separable in Y . Thus, the necessity is true. Hence, we finish the proof of Theorem 4.11.   The next two results of this subsection are comparable with Theorem 4.2. S ,QE max ∗ Theorem 4.12 For the problem (T P )Q max , the set YR (T , tS ) and the target YE are separable in Y , provided that one of the following conditions holds:

(i) The space Y is of finite dimension. (ii) The space Y is of infinite dimension and I nt (YRmax (T , tS∗ ) − YE ) = ∅.  

Proof Its proof is the same as that of Theorem 4.2. We omit it here.

Theorem 4.13 Assume that A : D(A) ⊆ Y → Y generates a C0 group. Then for QS ,QE the problem (T P )max , the set YCmax (tS∗ ) and YS are separable in Y , if one of the following conditions holds: (i) The space Y is of finite dimension. (ii) The space Y is of infinite dimension and I nt (YCmax (tS∗ ) − YS ) = ∅. 



Q S ,Q E be defined in the proof of Theorem 4.11. As we pointed Proof Let (T P )min  ,Q E Q

S out there, T − tS∗ is the optimal time of (T P )min

(see (4.288)), YCmax (tS∗ ) is the  

Q S ,Q E reachable set at T − tS∗ of the system (4.286) and YS is the target set of (T P )min . max ∗ By Theorem 4.2, we see that if either (i) or (ii) holds, then YC (tS ) and YS are separable in Y . This completes the proof.  

At the end of this subsection, we give some examples, which may help us to understand the main results of this subsection. Example 4.21 Let A(·) = (aij (·))n×n and B(·) = (bij (·))n×m , with aij (·) ∈ L∞ (0, T ) (1 ≤ i, j ≤ n) and bij (·) ∈ L∞ (0, T ) (1 ≤ i ≤ n, 1 ≤ j ≤ m). Let

4.4 Maximum Principle for Maximal Time Controls

199 Q ,QE

S y0 , y1 ∈ Rn , with y0 = y1 . Consider the maximal time control problem (T P )max where the controlled system is as:

y(t) ˙ = A(t)y(t) + B(t)u(t),

,

t ∈ (0, T ), with y(t) ∈ Rn , u(t) ∈ Rm , (4.295)

and where QS = [0, T ) × {y0 }; QE = {T } × {y1 }; U = Bρ (0) with ρ > 0. S ,QE can be put into the framework One can easily check that the above (T P )Q max max (AT P ) (given at the beginning of Section 4.4). S ,QE has an optimal control. Then We further assume that the above (T P )Q max according to Theorem 4.12 and Theorem 4.10 (or Theorem 4.13 and Theorem 4.11), S ,QE the problem (T P )Q satisfies the classical Pontryagin Maximum Principle. We max end Example 4.21.

Example 4.22 Let Ω ⊆ Rd (with d ≥ 1) be a bounded domain with a C 2 -boundary ∂Ω. Let ω ⊆ Ω be a nonempty and open subset with its characteristic function χω . Consider the following controlled heat equation:  ∂t y − Δy + a(x, t)y = χω u in Ω × (0, T ), y=0 on ∂Ω × (0, T ), where a(·, ·) ∈ C(Ω × [0, T ]). By setting Y = U = L2 (Ω), we can easily put the above equation into the framework (4.278). QS ,QE We now consider a time optimal control problem (T P )max , where the controlled system is the above heat equation, and where QS = [0, T ) × {y0 } with y0 ∈ L2 (Ω) \ Br (0) and r > 0; QE = {T } × Br (0); U = Bρ (0) with ρ > 0. Q ,Q

S E One can easily check that (T P )max can be put into the framework (ATmax P ) (given QS ,QE at the beginning of Section 4.4). We further assume that (T P )max has an optimal control. Notice that when ρ > 0 is large enough, this problem has an optimal control (see Theorem 3.7). S ,QE According to Theorem 4.12 and Theorem 4.10, the problem (T P )Q satisfies max the classical Pontryagin Maximum Principle. We end Example 4.22.

4.4.2 Local Maximum Principle for Maximal Time Controls We first give the definition of the local Pontryagin Maximum Principle for the S ,QE problem (T P )Q max .

200

4 Maximum Principle of Optimal Controls

Definition 4.9 Several definitions are given in order. (i) An optimal control u∗ (associated with an optimal tetrad (tS∗ , y0∗ , T , u∗ )) to QS ,QE (T P )max is said to satisfy the local  Pontryagin   Maximum Principle, if for  any T ∈ (tS∗ , T ), there exists ϕT(·) ∈ C tS∗ , T ; Y , with ϕT(·) = 0 (i.e., ϕT is not a zero function), so that   ϕ˙T(t) = −A∗ ϕT(t) − D(t)∗ ϕT(t) for a.e. t ∈ tS∗ , T ;

(4.296)

H (t, y ∗ (t), u∗ (t), ϕT(t))   = max H (t, y ∗ (t), u, ϕT(t)) for a.e. t ∈ tS∗ , T ,

(4.297)

u∈U

where y ∗ ∈ C([tS∗ , T ]; Y ) is the corresponding optimal trajectory and H (t, y, u, ϕ)  ϕ, D(t)y + B(t)uY

(4.298)

for a.e. t ∈ (tS∗ , T ) and all (y, u, ϕ) ∈ Y × U × Y . Besides, ϕT(tS∗ ), y0 − y0∗ Y ≤ 0 for each y0 ∈ YS ,

(4.299)

 +   ϕT(T), z − y ∗ T Y ≥ 0 for each z ∈ YCmax T .

(4.300)

and *

Here, (4.296) is called the dual (or adjoint) equation; (4.297) is called the maximum condition; the function H (·) defined by (4.298) is called the Hamiltonian S ,QE (associated with (T P )Q max ); (4.299) and (4.300) are called the transversality condition. S ,QE (ii) The problem (T P )Q is said to satisfy the local Pontryagin Maximum max Principle, if any optimal control satisfies the local Pontryagin Maximum Principle. The main result of this subsection is as follows. Q ,Q

S E Theorem 4.14 The problem (T P )max satisfies the local Pontryagin Maximum max ∗ Principle if and only if YR (t, tS ) and YCmax (t) are separable in Y for any t ∈ (tS∗ , T ).

Proof We first show the sufficiency. Assume that YRmax (t, tS∗ ) and YCmax (t) are separable in Y for any t ∈ (tS∗ , T ). Let (tS∗ , y0∗ , T , u∗ ) be an optimal tetrad to QS ,QE (T P )max . For any T ∈ (tS∗ , T ), by the similar arguments to those used in  the proof of Theorem 4.1 (replacing respectively YE and YR (tE∗ ) by YCmax T  ∗     max ∗   and YR T , tS ), we can show that there exists ϕT(·) ∈ C tS , T ; Y with ϕT(·) = 0 so that (4.296)–(4.300) hold. From this and Definition 4.9, we see that (tS∗ , y0∗ , T , u∗ ) satisfies the local Pontryagin Maximum Principle, consequently, so QS ,QE does (T P )max . Hence, we have proved the sufficiency.

4.4 Maximum Principle for Maximal Time Controls

201 Q ,Q

S E We next show the necessity. Suppose that (T P )max satisfies the local PonQS ,QE ∗ ∗ ∗ . tryagin Maximum Principle. Let (tS , y0 , T , u ) be an optimal tetrad to∗ (T P )max  ∗ Then by Definition 4.9, for any T ∈ (tS , T ), there exists ϕT(·) ∈ C tS , T ; Y with ϕT(·) = 0 so that (4.296)–(4.300) hold. Next, by the similar arguments as    those  used in the proof of Theorem 4.1, we can show that YRmax T, tS∗ and YCmax T are separable in Y . Hence, we have proved the necessity. Thus, we complete the proof.  

The next result is concerned with relations between the classical Pontryagin Maximum Principle and the local Pontryagin Maximum Principle for the problem QS ,QE (T P )max . We start with the following assumption: (HC ) ϕ(·) = 0 in (tS∗ , T ), provided that ϕ(t0 ) = 0 for some t0 ∈ (tS∗ , T ] and ϕ(·) solves the equation: 

ϕ(t) ˙ = −[A∗ + D(t)∗ ]ϕ(t), ϕ(T ) ∈ Y.

t ∈ (tS∗ , T ),

S ,QE satisfies the classical PontryaCorollary 4.4 Assume that the problem (T P )Q max S ,QE gin Maximum Principle. Suppose that (HC ) is true. Then (T P )Q also satisfies max the local Pontryagin Maximum Principle.

Proof The proof is the same as that of Corollary 4.2. We omit it here.

 

Example 4.23 Consider the maximal time control problems in Examples 4.21 and 4.22. In those examples, we have seen that these problems satisfy the classical Pontryagin Maximum Principle. Moreover, one can easily check that the assumption (HC ) holds for these two cases. (For the case of ODEs, we use the theory of existence and uniqueness for ODEs; for the case of the heat equation, we use (i) of Remark 1.5 (after Theorem 1.22).) Then by Corollary 4.4, these problems satisfy the local Pontryagin Maximum Principle.

Miscellaneous Notes The studies on the Pontryagin Maximum Principle can be dated back to 1950s (see, for instance, [5, 6] and [14]). Due to the importance of the Pontryagin Maximum Principle, many mathematicians have worked on it. Here, we mention literatures [2–4, 7, 8, 10, 12, 13, 15], and [11]. In particular, we would like to recommend the book [9]. In this book, the author studied the Pontryagin Maximum Principle for infinite dimensional evolution equations in an abstract setting; introduced two different maximum principles: the strong maximum principle and the weak maximum principle. (Notice that the weak maximum principle in [9] differs from that introduced in our monograph.)

202

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Maximum Principle of Optimal Controls

About three kinds of maximum principles introduced in this chapter, several notes are given in order: • Definitions on these maximum principles are summarized from [9, 19] and [18]. • Connections among these maximum principles are as follows: (i) Some problems do not hold the classical maximum principle, but have the local one (see Example 4.9); (ii) The classical one cannot imply the local one (see Remark 4.4); (iii) If the adjoint equation has some unique continuation property, then the classical one implies the local one (see Corollary 4.2); (iv) Some problems hold neither the classical one nor the local one, but have the weak one (see Example 4.17). • Our methods to derive these maximum principles are based on the separability results and the representation formulas introduced in Sections 1.1.2 and 4.3.2. The two representation formulas in Section 4.3.2 are taken from [19], where the first representation formula is originally from [18]. Materials of this chapter come from two parts: First, some of them are taken from papers [19] and [18]; Second, some of them, including several examples, are given specially to this monograph by us. Next, we would like to stress the following fact about our control operator: In this chapter, though it is assumed that B(·) ∈ L∞ (0, +∞; L (U, Y )) or B(·) ∈ L∞ (0, T ; L (U, Y )), many results are still true for the case that B(·) ∈ L∞ (0, +∞; L (U, Y−1 )) or B(·) ∈ L∞ (0, T ; L (U, Y−1 )) is admissible for the semigroup generated by A. (Here, Y−1  D(A∗ ) is the dual space of D(A∗ ).) When B(·) ≡ B, its admissibility has been defined in [17]. This definition can be easily extended to the case that B(·) is time varying. The case that B(·) is admissible covers some time optimal control problems governed by boundary controlled PDEs. With regard to the maximum principles for the case when B(·) ≡ B is admissible, we refer the readers to [19]. Finally, we give several open problems that might interest readers: • Can we have weak maximum principles for time-invariant systems with general control constraints? • Can we have weak maximum principles for general time-varying systems? • For heat equations, what is the best regularity for the multiplier in the local/weak maximum principle?

References 1. A.A. Agrachev, Y.L. Sachkov, Control Theory from the Geometric Viewpoint. Encyclopaedia of Mathematical Sciences, Control Theory and Optimization, II, vol. 87 (Springer, Berlin, 2004) 2. N. Arada, J.P. Raymond, Time optimal problems with Dirichlet boundary controls Discrete Contin. Dyn. Syst. 9, 1549–1570 (2003) 3. V. Barbu, Optimal Control of Variational Inequalities. Research Notes in Mathematics, vol. 100 (Pitman (Advanced Publishing Program), Boston, MA, 1984)

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4. V. Barbu, Analysis and control of nonlinear infinite-dimensional systems, in Mathematics in Science and Engineering, vol. 190 (Academic, Boston, MA, 1993) 5. R. Bellman, I. Glicksberg, O. Gross, On the “bang-bang” control problem. Q. Appl. Math. 14, 11–18 (1956) 6. V.G. Boltyanskiˇi, R.V. Gamkrelidz, L.S. Pontryagin, On the theory of optimal processes (Russian), Dokl. Akad. Nauk. SSSR 110, 7–10 (1956) 7. R. Conti, Time-optimal solution of a linear evolution equation in Banach spaces. J. Optim. Theory Appl. 2, 277–284 (1968) 8. H.O. Fattorini, The time optimal control problems in Banach spaces. Appl. Math. Optim. 1, 163–188 (1974) 9. H.O. Fattorini, Infinite Dimensional Linear Control Systems, the Time Optimal and Norm Optimal Problems. North-Holland Mathematics Studies, vol. 201 (Elsevier Science B.V., Amsterdam, 2005) 10. K. Kunisch, L. Wang, Time optimal controls of the linear Fitzhugh-Nagumo equation with pointwise control constraints. J. Math. Anal. Appl. 395, 114–130 (2012) 11. K. Kunisch, L. Wang, Time optimal control of the heat equation with pointwise control constraints. ESAIM Control Optim. Calc. Var. 19, 460–485 (2013) 12. X. Li, J. Yong, Optimal Control Theory for Infinite-Dimensional Systems. Systems & Control: Foundations & Applications (Birkhäuser Boston, Boston, MA, 1995) 13. J. Lohéac, M. Tucsnak, Maximum principle and bang-bang property of time optimal controls for Schrodinger-type systems. SIAM J. Control Optim. 51, 4016–4038 (2013) 14. L.S. Pontryagin, V.G. Boltyanskii, R.V. Gamkrelidze, E.F. Mischenko, The Mathematical Theory of Optimal Processes, ed. by L.W. Neustadt, (Interscience Publishers Wiley, New York, London, 1962) [Translated from the Russian by K. N. Trirogoff] 15. J.P. Raymond, H. Zidani, Pontryagin’s principle for time-optimal problems. J. Optim. Theory Appl. 101, 375–402 (1999) 16. E. Roxin, A geometric interpretation of Pontryagin’s maximum principle, in International Symposium on Nonlinear Differential Equations and Nonlinear Mechanics (Academic, New York, 1963), pp. 303–324 17. M. Tucsnak, G. Weiss, Observation and control for operator semigroups, (Birkhäuser Verlag, Basel, 2009) 18. G. Wang, Y. Xu, Y. Zhang, Attainable subspaces and the bang-bang property of time optimal controls for heat equations. SIAM J. Control Optim. 53, 592–621 (2015) 19. G. Wang, Y. Zhang, Decompositions and bang-bang problems. Math. Control Relat. Fields 7, 73–170 (2017)

Chapter 5

Equivalence of Several Kinds of Optimal Controls

Connections of time optimal control problems with other kinds of optimal control problems can allow us to get properties for one kind of optimal control problems through studying another kind of optimal control problems. In [4] (see also [5]), the author derived the Pontryagin Maximum Principle for minimal time controls (or minimal norm controls) via the Pontryagin Maximum Principle for minimal norm controls (or minimal time controls). Such connections can also help us in the following subjects: (i) To study the regularity of the Bellman functions associated with minimal time control problems (see for instance, [1, 6] and [3]); (ii) To study the behaviors of optimal time and an optimal control, when a controlled system has a small perturbation (see [19] and [13]); (iii) To get necessary and sufficient conditions on both optimal time and optimal controls for some time optimal control problems (see [17] and [15]); (iv) To find some iterative algorithms for solutions of time optimal control problems (see [15]). To our best knowledge, the terminology “equivalence” on optimal controls arose first in [17]. It means that several optimal control problems share the same optimal control in certain senses. In this chapter, we first present three equivalence theorems for minimal time optimal control problems and minimal norm control problems. Among these equivalent theorems, the first one is concerned with time invariant systems with terminal singleton constraints; the second one is related to time-varying systems with terminal non-singleton constraints; the third one is about time-varying systems with terminal singleton constraints. We then present the equivalence among maximal time optimal control problems, minimal norm control problems, and optimal target control problems.

© Springer International Publishing AG, part of Springer Nature 2018 G. Wang et al., Time Optimal Control of Evolution Equations, Progress in Nonlinear Differential Equations and Their Applications 92, https://doi.org/10.1007/978-3-319-95363-2_5

205

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5 Equivalence of Several Kinds of Optimal Controls

5.1 Minimal Time Controls, Minimal Norm Controls, and Time Invariant Systems S ,QE In this section, we will study the equivalence of the problem (T P )Q and min the corresponding minimal norm control problem under the following framework (A1 ):

(i) The state space Y and the control space U are real separable Hilbert spaces. (ii) The controlled system is as: y(t) ˙ = Ay(t) + Bu(t),

(iii) (iv) (v)

(vi)

t ∈ (0, +∞),

(5.1)

where A : D(A) ⊆ Y → Y generates a C0 semigroup {eAt }t ≥0 on Y and B ∈ L (U, Y ). Given τ ∈ [0, +∞), y0 ∈ Y and u ∈ L∞ (τ, +∞; U ), we write y(·; τ, y0 , u) for the solution of (5.1) over [τ, +∞), with the initial condition that y(τ ) = y0 . Let U = BM (0), where BM (0) is the closed ball in U , centered at 0 and of radius M ≥ 0. (When M = 0, we agree that BM (0)  {0}.) Let QS = {0}×{ y0} and QE = (0, +∞)×{0}, where  y0 ∈ Y \{0} is arbitrarily fixed. The system (5.1) is L∞ -null controllable over [0, T ] for each T ∈ (0, +∞) (i.e., for each T ∈ (0, +∞) and y0 ∈ Y , there exists u ∈ L∞ (0, T ; U ) so that y(T ; 0, y0 , u) = 0). The semigroup {eAt }t ≥0 holds the backward uniqueness property (i.e., if eAt y0 = 0 for some y0 ∈ Y and t ≥ 0, then y0 = 0).  y

S ,QE for (T P )M0 which reads as: In this section, we simply write (T P )Q min

 y

(T P )M0

 y0, u) = 0, u ∈ L(0, T ; U)}. T(M,  y0 )  inf{T ∈ (0, +∞)  y(T ; 0,  (5.2)

We consider the following problem:  y

(T2 P )M0

  y0 , u) = 0 and u ∈ UM }, T(M,  y0 )  inf{T ∈ (0, +∞)  y(T ; 0,  (5.3)

where  UM  {u ∈ L∞ (0, +∞; U )  u(t)U ≤ M a.e. t ∈ (0, +∞)}. In this problem, • a tetrad (0,  y0 , T , u) is called admissible, if T ∈ (0, +∞), u ∈ UM and y(T ; 0,  y0 , u) = 0; • a tetrad (0,  y0 ,  T(M,  y0 ), u∗ ) is called optimal if  T(M,  y0 ) ∈ (0, +∞), u∗ ∈ UM and y( T(M,  y0 ); 0,  y0 , u∗ ) = 0;

5.1 Minimal Time Controls, Minimal Norm Controls, and Time Invariant. . .

207

• when (0,  y0 ,  T(M,  y0 ), u∗ ) is an optimal tetrad,  T(M,  y0 ) and  u∗ are called the optimal time and an optimal control, respectively. The following three facts can be checked easily: T(M,  y0 ); • T(M,  y0 ) =  • The infimum in (5.2) can be reached if and only if the infimum in (5.3) can be reached;  y • The zero extension of each optimal control to (T P )M0 over (0, +∞) is an  y  y optimal control to (T2 P )M0 ; the restriction of each optimal control to (T2 P )M0 over  y0 (0, T(M,  y0 )) is an optimal control to (T P )M .  y  y Because of the above three facts, (T P )M0 and (T2 P )M0 can be treated as the same problem. y The corresponding minimal norm problem (NP )T0 (with T > 0 and y0 ∈ Y arbitrarily fixed) reads:

 N(T , y0 )  inf{vL∞ (0,T ;U )  y(T ; 0, y0 , v) = 0}.

y

(NP )T0

(5.4)

In this problem, • N(T , y0 ) is called the minimal norm; • a control v ∈ L∞ (0, T ; U ) is called admissible, if y(T ; 0, y0, v) = 0; • a control v ∗ ∈ L∞ (0, T ; U ) is called optimal, if it is admissible and satisfies that v ∗ L∞ (0,T ;U ) = N(T , y0 ). Moreover, for each y0 ∈ Y , the map T → N(T , y0 ) is called a minimal norm function, which is denoted by N(·, y0 ).  y  y In this section, we aim to introduce connections between (T P )M0 and (NP )T0 . From (5.4), we see that for each y0 ∈ Y , the minimal norm function N(·, y0 ) is decreasing over (0, +∞). Hence, for each y0 ∈ Y , it is well defined that (y0 )  lim N(T , y0 ). N

(5.5)

T →+∞

 y

 y

We now give the definition about the equivalence between (T P )M0 and (NP )T0 .  y

 y

Definition 5.1 Let M ≥ 0 and T ∈ (0, +∞). Problems (T P )M0 and (NP )T0 are said to be equivalent if the following three conditions hold:  y

 y

(i) Both (T2 P )M0 and (NP )T0 have optimal controls;  y (ii) The restriction of each optimal control to (T2 P )M0 over (0, T ) is an optimal  y control to (NP )T0 ;  y (iii) The zero extension of each optimal control to (NP )T0 over (0, +∞) is an  y optimal control to (T2 P )M0 .

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5 Equivalence of Several Kinds of Optimal Controls

The main result of this section is as follows. Theorem 5.1 The function T → N(T ,  y0 ) is strictly decreasing and continuous ( from (0, +∞) onto (N y0 ), +∞). Moreover, if write  (5.6) y0 ), T ) ∈ R2  T ∈ (0, +∞)}, E y0  {(N(T ,  then the following conclusions are true:  y

 y

0 0 (i) When (M, T ) ∈ E y0 , problems (T P )M and (NP )T are equivalent;  y0  y0 (ii) When (M, T ) ∈ [0, +∞) × (0, +∞) \ E y0 , problems (T P )M and (NP )T are not equivalent.

Before proving Theorem 5.1, some preliminaries are given in order. y

Lemma 5.1 Let y0 ∈ Y . Then for each T ∈ (0, +∞), the problem (NP )T0 has at least one optimal control. Proof Arbitrarily fix T ∈ (0, +∞) and y0 ∈ Y . Without loss of generality, we can assume that y0 = 0, for otherwise the null control is an optimal control. By (v) of (A1 ) (given at the beginning of Section 5.1), there exists a control  v ∈ L∞ (0, T ; U ) so that v ) = 0, y(T ; 0, y0 , i.e., (NP )T0 has admissible controls. So we can let {vk }k≥1 ⊆ L∞ (0, T ; U ) satisfy that y

y(T ; 0, y0, vk ) = 0 and vk L∞ (0,T ;U ) ≤ N(T , y0 ) + k −1 for all k ∈ N+ . (5.7) By the inequality in (5.7), there exists a subsequence of {vk }k≥1 , denoted in the same way, and  v ∈ L∞ (0, T ; U ) so that v weakly star in L∞ (0, T ; U ) as k → +∞. vk → 

(5.8)

From this and the inequality in (5.7), it follows that  v L∞ (0,T ;U ) ≤ lim inf vk L∞ (0,T ;U ) ≤ N(T , y0 ). k→+∞

(5.9)

Meanwhile, by (5.8), we see that y(T ; 0, y0, vk ) → y(T ; 0, y0 , v ) weakly in Y. v ) = 0, i.e.,  v is an This, together with the equality in (5.7), implies that y(T ; 0, y0 , y admissible control to (NP )T0 . Then by the optimality of N(T , y0 ) (see (5.4)) and y (5.9), we find that  v is an optimal control to (NP )T0 . Hence, we end the proof of Lemma 5.1.  

5.1 Minimal Time Controls, Minimal Norm Controls, and Time Invariant. . .

209

The following lemma is concerned with the continuity and monotonicity of minimal norm function N(·, y0 ). Lemma 5.2 The function N(·,  y0 ) is continuous and strictly decreasing from ( (0, +∞) onto (N y0 ), +∞). Proof The proof is organized by the following four steps: Step 1. We show that N(T ,  y0 ) ∈ (0, +∞) for each T ∈ (0, +∞).

(5.10)

By contradiction, suppose that (5.10) was not true. Then we would have that N(T,  y0 ) = 0 for some T ∈ (0, +∞). This, together with Lemma 5.1, implies that the null control is the unique optimal  y control to (NP )T0 . So we have that 

eAT  y0 = y(T; 0,  y0, 0) = 0. This, along with (vi) of (A1 ) (given at the beginning of Section 5.1), yields that  y0 = 0, which leads to a contradiction, since we assumed that  y0 = 0. Thus, (5.10) is true. Step 2. We prove that when 0 < T1 < T2 < +∞, y0 ) > N(T2 ,  y0 ). N(T1 , 

(5.11)

To this end, we arbitrarily fix T1 , T2 ∈ (0, +∞) so that T2 > T1 . By Lemma 5.1,  y the problem (NP )T01 has an optimal control v1∗ . Then we have that y(T1 ; 0,  y0 , v1∗ ) = 0 and v1∗ L∞ (0,T1 ;U ) = N(T1 ,  y0 ).

(5.12)

Meanwhile, by (v) of (A1 ) (given at the beginning of Section 5.1), there is v2 ∈ L∞ (0, T2 − T1 ; U ) so that y(T2 − T1 ; 0, eAT1  y0 , v2 ) = 0.

(5.13)

Since N(T1 ,  y0 ) ∈ (0, +∞) (see (5.10)), we can choose a constant λ ∈ (0, 1) so that y0 )/2. λv2 L∞ (0,T2 −T1 ;U ) < N(T1 , 

(5.14)

We now define a new control in the following manner:  vλ (t) 

(1 − λ)v1∗ (t), t ∈ (0, T1 ), λv2 (t − T1 ), t ∈ (T1 , T2 ).

(5.15)

210

5 Equivalence of Several Kinds of Optimal Controls

By (5.15), we find that  y(T2 ; 0,  y0, vλ ) = e

AT2

 +λ

T1

 y0 + (1 − λ) 0 T2

eA(T2 −t ) Bv1∗ (t)dt

eA(T2 −t ) Bv2 (t − T1 )dt

T1

 " = (1 − λ)eA(T2 −T1 ) eAT1  y0 +

T1 0

 / A(T2 −T1 ) AT1 +λ e (e  y0 ) +

eA(T1 −t ) Bv1∗ (t)dt

T2 −T1

#

0 eA(T2 −T1 −t ) Bv2 (t)dt .

0

This, together with the first equality in (5.12) and (5.13), implies that y(T2 ; 0,  y0, vλ ) = (1−λ)eA(T2 −T1 ) y(T1 ; 0,  y0, v1∗ )+λy(T2 −T1 ; 0, eAT1  y0 , v2 ) = 0,  y

which indicates that vλ is an admissible control to (NP )T02 . Then by the optimality of N(T2 ,  y0 ) (see (5.4)) and (5.15), we get that   y0 ) ≤ vλ L∞ (0,T2 ;U ) ≤ max (1 − λ)v1∗ L∞ (0,T1 ;U ) , λv2 L∞ (0,T2 −T1 ;U ) . N(T2 ,  From this, the second equality in (5.12) and (5.14), we obtain (5.11) at once. Step 3. We show that given η ∈ (0, +∞) and T1 , T2 ∈ (0, +∞) with η < T1 < T2 < η−1 , there exists a constant C(η) ∈ (0, +∞) so that   y0 ) ≤ N(T2 ,  y0 ) + C(η) eA(T2 −T1 ) y0 −  y0 Y + N(T2 ,  y0 )(T2 − T1 ) . N(T1 ,  (5.16) For this purpose, we arbitrarily fix η ∈ (0, +∞) and T1 , T2 ∈ (0, +∞) with  y η < T1 < T2 < η−1 . By Lemma 5.1, the problem (NP )T02 has an optimal control v2∗ . Then we have that y0 , v2∗ ) = 0 and v2∗ L∞ (0,T2 ;U ) = N(T2 ,  y0 ). y(T2 ; 0, 

(5.17)

Write  y  y(T2 − T1 ; 0,  y0, v2∗ ).

(5.18)

By (v) of (A1 ) (given at the beginning of Section 5.1), we can apply Theorem 1.20 to find C(η) ∈ (0, +∞) and v3 ∈ L∞ (0, T1 ; U ) so that y0 −  y , v3 ) = 0 and v3 L∞ (0,T1 ;U ) ≤ C(η) y0 −  y Y . y(T1 ; 0, 

(5.19)

We now define a new control in the following manner: v4 (t)  v2∗ (t + T2 − T1 ) + v3 (t),

t ∈ (0, T1 ).

(5.20)

5.1 Minimal Time Controls, Minimal Norm Controls, and Time Invariant. . .

211

It can be easily verified that  y0, v4 ) = e y(T1 ; 0, 

AT1

0



T1

+ 0



y+ = eAT1  + e

eA(T1 −t ) Bv2∗ (t + T2 − T1 )dt

eA(T1 −t ) Bv3 (t)dt

/

/

T1

 y0 +

AT1

T2 T2 −T1

eA(T2 −t ) Bv2∗ (t)dt 

T1

( y0 −  y) +

0

0 eA(T1 −t ) Bv3 (t)dt .

0

This, together with (5.18) and the first equalities in both (5.17) and (5.19), implies that y0 , v4 ) = y(T2 ; 0,  y0 , v2∗ ) + y(T1 ; 0,  y0 −  y , v3 ) = 0, y(T1 ; 0,   y

which indicates that v4 is an admissible control to (NP )T01 . Then by the optimality of N(T1 ,  y0 ), we get that y0 ) ≤ v4 L∞ (0,T1 ;U ) . N(T1 ,  This, together with (5.20), the second conclusions in both (5.17) and (5.19), yields that N(T1 ,  y0 ) ≤ v2∗ L∞ (0,T2 ;U ) + v3 L∞ (0,T1 ;U ) ≤ N(T2 ,  y0 ) + C(η) y0 −  y Y . From the above inequality, (5.18) and the second equality in (5.17), it follows that " N(T1 ,  y0 ) ≤ N(T2 ,  y0 ) + C(η)  y0 − eA(T2 −T1 ) y0 Y



+ 

T2 −T1 0

#

eA(T2 −T1 −t ) Bv2∗ (t)dt Y

 ≤ N(T2 ,  y0 ) + C(η)  y0 − eA(T2 −T1 ) y0 Y + N(T2 ,  y0 )(T2 − T1 ) , which leads to (5.16). Step 4. We prove that lim N(T ,  y0 ) = +∞.

T →0+

(5.21)

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5 Equivalence of Several Kinds of Optimal Controls

By contradiction, suppose that (5.21) was not true. Then there would exist a sequence {Tk }k≥1 ⊆ (0, +∞) with Tk → 0 and a positive constant C so that N(Tk ,  y0 ) ≤ C for all k ∈ N+ .  y

Thus, according to Lemma 5.1, each (NP )T0k , with k ∈ N+ , has an optimal control vk∗ . Then we have that vk∗ L∞ (0,Tk ;U ) ≤ C and y(Tk ; 0,  y0 , vk∗ ) = 0. These yield that y0 , 0) = lim  y0 = lim y(Tk ; 0,  k→+∞

= lim

k→+∞

k→+∞

y(Tk ; 0,  y0 , vk∗ )



y(Tk ; 0,  y0, vk∗ ) − y(Tk ; 0, 0, vk∗ )



= 0,

which contradicts the assumption that  y0 = 0. So (5.21) is true. Finally, the desired results of this lemma follow from results in the above ( four steps, as well as the definition of N y0 ) (see (5.5)). This ends the proof of Lemma 5.2.    y

( Lemma 5.3 The problem (T2 P )M0 has optimal controls if and only if M > N y0 ).  y0 ) Proof Arbitrarily fix  y0 ∈ Y \ {0}. Then by Lemma 5.2 and the definition of N( (see (5.5)), we have that ( N y0 ) < +∞. The proof will be finished by the following two steps.  y  y0 ), then (T2 Step 1. We show that if M > N( P )M0 has an optimal control.  y0 ). Then by the definition of N ( To this end, we arbitrarily fix M > N( y0 ), we  ∈ (0, +∞) so that can find T ( ( N(T,  y0 ) < N y0 ) + (M − N y0 )) = M.

(5.22)

Meanwhile, according to Lemma 5.1, there exists v ∗ ∈ L∞ (0, T; U ) so that y(T; 0,  y0, v ∗ ) = 0 and v ∗ L∞ (0,T;U ) = N(T,  y0 ).

(5.23)

Write  v ∗ for the zero extension of v ∗ over (0, +∞). From (5.23) and (5.22), we  y see that (0,  y0 , T, v ∗ ) is an admissible tetrad to (T2 P )M0 . So we can let {Tk }k≥1 ⊆ ∞ (0, +∞) and {vk }k≥1 ⊆ L (0, +∞; U ) satisfy that T(M,  y0 ), Tk → 

(5.24)

5.1 Minimal Time Controls, Minimal Norm Controls, and Time Invariant. . .

y(Tk ; 0,  y0 , vk ) = 0 and vk L∞ (0,+∞;U ) ≤ M for all k ∈ N+ .

213

(5.25)

By the inequality in (5.25), there exists a subsequence of {vk }k≥1 , denoted in the same manner, and  v ∈ L∞ (0, +∞; U ) so that v weakly star in L∞ (0, +∞; U ) as k → +∞, vk → 

(5.26)

 v L∞ (0,+∞;U ) ≤ lim inf vk L∞ (0,+∞;U ) ≤ M.

(5.27)

and k→+∞

It follows from (5.24) and (5.26) that y0 , vk ) → y( T(M,  y0 ); 0,  y0 , v ) weakly in Y, y(Tk ; 0,  which, combined with the equality in (5.25), indicates that y( T(M,  y0 ); 0,  y0, v) =  y0 2 0. This, together with (5.27), yields that  v is an optimal control to (T P )M .  y  y0 ), then (T2 Step 2. We show that if 0 ≤ M ≤ N( P )M0 has no optimal control. By contradiction, suppose that it was not true. Then there would be M, with  y0 ), and a control u∗ in L∞ (0, +∞; U ) so that 0 ≤ M ≤ N( y0 , u∗ ) = 0 and u∗ L∞ (0,+∞;U ) ≤ M. y( T(M,  y0 ); 0, 

(5.28)

Since  y0 = 0, the equality of (5.28) implies that  T(M,  y0 ) ∈ (0, +∞) and that  y0 u∗ |(0, is an admissible control to (NP ) . Then, it follows from the T(M, y0 ))  T(M, y0 ) y0 ) that optimality of N( T(M,  y0 ),  N( T(M,  y0 ),  y0 ) ≤ u∗ L∞ (0, T(M, y0 );U ) .  y0 ), the above inequality, combined with the inequality in (5.28), Since M ≤ N( indicates that  y0 ). N( T(M,  y0 ),  y0 ) ≤ M ≤ N(

(5.29)

Now, it follows from Lemma 5.2 and (5.29) that  y0 ) for each T >  T(M,  y0 ),  y0 ) ≤ N( T(M,  y0 ). N(T ,  y0 ) < N( ( However, by Lemma 5.2 and the definition of N y0 ) (see (5.5)), we find that  N(T ,  y0 ) > N( y0 ) for all T > 0. Thus, we get a contradiction. So the conclusion in Step 2 is true. In summary, we end the proof of Lemma 5.3.  

214

5 Equivalence of Several Kinds of Optimal Controls

Remark 5.1 Lemma 5.3 can also be proved by combining Lemma 5.2 with Theorem 3.6. Corollary 5.1 Let (M, T ) ∈ [0, +∞) × (0, +∞). Then M = N(T ,  y0 ) if and only if T =  T(M,  y0 ). Proof Let  y0 ∈ Y \ {0} and (M, T ) ∈ [0, +∞) × (0, +∞). We first show that T(M,  y0 ). M = N(T ,  y0 ) ⇒ T = 

(5.30)

According to Lemma 5.1, there exists a control u∗1 ∈ L∞ (0, T ; U ) so that y(T ; 0,  y0, u∗1 ) = 0 and u∗1 L∞ (0,T ;U ) = N(T ,  y0 ).

(5.31)

y0 ), it Write  u∗1 for the zero extension of u∗1 over (0, +∞). Since M = N(T ,   y follows from (5.31) that (0,  y0 , T , u∗1 ) is an admissible tetrad to (T2 P )M0 . Then by the optimality of  T(M,  y0 ) (see (5.3)), we find that  T(M,  y0 ) ≤ T < +∞.

(5.32)

 y0 ). This, Meanwhile, since M = N(T ,  y0 ), it follows by Lemma 5.2 that M > N(  together with Lemma 5.3, yields that T(M,  y0 ) ∈ (0, +∞) and there exists a control u∗2 ∈ L∞ (0, +∞; U ) satisfying that y( T(M,  y0 ); 0,  y0 , u∗2 ) = 0 and u∗2 L∞ (0,+∞;U ) ≤ M.

(5.33)

Then by making use of (5.33), we see that u∗2 |(0, T(M, y0 )) is an admissible control to  y

0 (NP ) and T(M, y ) 0

y0 ) ≤ u∗2 L∞ (0, N( T(M,  y0 ),  T(M, y0 );U ) ≤ M.

(5.34)

Because M = N(T ,  y0 ), it follows by (5.34) and Lemma 5.2 that  T(M,  y0 ) ≥ T . This, together with (5.32), yields that T =  T(M,  y0 ). We next show that T = T(M,  y0 ) ⇒ M = N(T ,  y0 ).

(5.35)

 y Since  T(M,  y0 ) = T ∈ (0, +∞), we have that (T2 P )M0 has an admissible tetrad. Then by similar arguments as those used in the proof of Lemma 5.3 (see Step 1 in  y Lemma 5.3), we can verify that (T2 P )M0 has an optimal control. This, together with Lemma 5.3, implies that

5.1 Minimal Time Controls, Minimal Norm Controls, and Time Invariant. . .

215

( M >N y0 ), which, combined with Lemma 5.2, indicates that M = N(T,  y0 ) for some T ∈ (0, +∞).

(5.36)

From this and (5.30), it follows that T =  T(M,  y0 ).

(5.37)

Since we have assumed that T =  T(M,  y0 ), it follows from (5.37) and (5.36) that M = N(T ,  y0 ). Hence, (5.35) holds. Finally, by (5.30) and (5.35), we finish the proof of Corollary 5.1.   We now are on the position to prove Theorem 5.1. Proof (Proof of Theorem 5.1) Let E y0 be given by (5.6). It follows from Lemma 5.2 that the function N(·,  y0 ) is strictly decreasing and continuous from (0, +∞) onto  y0 ), +∞). The conclusions (i) and (ii) of Theorem 5.1 will be proved one by (N( one. (i) Arbitrarily fix (M, T ) ∈ E y0 . Three facts are given in order. First, by (5.6), Lemma 5.2, and Corollary 5.1, we have that ( M = N(T ,  y 0 ) ∈ (N y0 ), +∞) and T =  T(M,  y0 ) > 0.

(5.38)

The above two equalities, together with Lemmas 5.3 and 5.1, imply that both  y  y (T2 P )M0 and (NP )T0 have optimal controls.  y Second, each optimal control u∗ to (T2 P )M0 satisfies that y( T(M,  y0 ); 0,  y0 , u∗ ) = 0 and u∗ L∞ (0,+∞;U ) ≤ M.

(5.39)

From (5.39) and (5.38), one can easily see that u∗ |(0,T ) is an optimal control to  y (NP )T0 .  y Third, each optimal control v ∗ to (NP )T0 satisfies that y0 ). y(T ; 0,  y0, v ∗ ) = 0 and v ∗ L∞ (0,T ;U ) = N(T , 

(5.40)

Write  v ∗ for the zero extension of v ∗ over (0, +∞). From (5.40) and (5.38), we see  y ∗ that  v is an optimal control to (T2 P )M0 .  y0 Finally, from the above three facts and Definition 5.1, we see that (T P )M  and  y

(NP )T0 are equivalent. So the conclusion (i) in Theorem 5.1 is true. (ii) By contradiction, suppose that the conclusion (ii) was not true. Then there would be  T) ∈ [0, +∞) × (0, +∞) \ E (M, y0

(5.41)

216

5 Equivalence of Several Kinds of Optimal Controls  y

 y

0 0 so that (T P )M  and (NP )T are equivalent. We claim that

T =  T(M,  y0 ).

(5.42)

 = N(T,  When (5.42) is proved, it follows by Corollary 5.1 that M y0 ), which contradicts (5.41) and completes the proof of (ii).  y0  y0 We now turn to prove (5.42). Since (T P )M  and (NP )T are equivalent, it follows  y

by (i) of Definition 5.1 that (NP )T0 has an optimal control v ∗ . By the optimality of v ∗ , we find that y0 ). y(T; 0,  y0, v ∗ ) = 0 and v ∗ L∞ (0,T;U ) = N(T, 

(5.43)

Moreover, from (iii) of Definition 5.1, we see that the zero extension of v ∗ over  y0 (0, +∞), denoted by  v ∗ , is an optimal control to (T2 P )M  . This, together with (5.43), yields that T ≥  T(M,  y0 ). From this, we see that to show (5.42), it suffices to prove what follows is not true: >   T T(M, y0 ).

(5.44)  y

0 By contradiction, suppose that (5.44) was true. From (i) of Definition 5.1, (T2 P )M  ∗ ∗ has an optimal control  u . Then, by the optimality of  u , we see that

      T(M, y0 ) ∈ (0, +∞), y( T(M, y0 ); 0,  y0, u∗ ) = 0 and  u∗ L∞ (0,+∞;U ) ≤ M. (5.45) Arbitrarily take u0 ∈ U satisfying that  u0 U = M, and then define a new control  ∗    u (t), if 0 < t ≤  T(M, y0 ),  u(t)    y0 ). u0 , if t > T(M, 

(5.46)

(5.47)

It follows from (5.45)–(5.47) that    y( T(M, y0 ); 0,  y0 , u) = 0 and  uL∞ (0,+∞;U ) = M.  y0 These indicate that  u is an optimal control to (T2 P )M  . This, along with the fact  y

 y

0 0 u|(0,T) is an optimal control to that (T P )M  and (NP )T are equivalent, yields that 

 y

(NP )T0 (see (ii) of Definition 5.1). Thus we have that

5.1 Minimal Time Controls, Minimal Norm Controls, and Time Invariant. . .

217

 uL∞ (0,T;U ) = N(T,  y0 ). From this, (5.44) and (5.45)–(5.47), one can easily check that  = N(T,  M y0 ), which, along with (5.6), indicates that  T) ∈ E (M, y0 . This contradicts (5.41). Hence, (5.44) is not true. Thus we have proved (5.42). Hence, we complete the proof of Theorem 5.1.

 

At the end of this section, we give two examples which can be put into the framework (A1 ) (given at the beginning of Section 5.1). Example 5.1 Let (A, B) ∈ Rn×n × Rn×m with n, m ∈ N+ , Y  Rn and U  Rm . Assume that (A, B) satisfies Kalman controllability rank condition: rank (B, AB, . . . , An−1 B) = n. Consider the following controlled ordinary differential equation: y(t) ˙ = Ay(t) + Bu(t),

t ∈ (0, +∞).

Then we can easily check that the corresponding time optimal control problem  y (T P )M0 can be put into the framework (A1 ) (given at the beginning of Section 5.1). Here, we used Theorem 1.23. We end Example 5.1. Example 5.2 Let Ω be a bounded domain in Rd , d ≥ 1, with a C 2 boundary ∂Ω. Let ω ⊆ Ω be a nonempty and open subset. Consider the following controlled heat equation: ⎧ ⎨ ∂t y − Δy = χω u in Ω × (0, +∞), y=0 on ∂Ω × (0, +∞), ⎩ 2 y(0) ∈ L (Ω), where u ∈ L∞ (0, +∞; L2 (Ω)). Let Y  L2 (Ω) and U  L2 (Ω); Let A = Δ with its domain H 2 (Ω) ∩ H01 (Ω); Let B = χω , where χω (the characteristic function of ω) is treated as a linear and bounded operator on U . Then one can  y easily check that the corresponding time optimal control problem (T P )M0 can be put into the framework (A1 ) (given at the beginning of Section 5.1). Here, we used Theorems 1.22, 1.21, and (i) of Remark 1.5 (after Theorem 1.22). Thus Theorem 5.1 holds for this example. With the aid of Theorem 5.1, one can obtain a necessary and  y sufficient condition on the optimal time and the optimal control for (T P )M0 in the current case (see [17]). We now end Example 5.2.

218

5 Equivalence of Several Kinds of Optimal Controls

5.2 Minimal Time Controls, Minimal Norm Controls, and Time-Varying Systems: Part I S ,QE In this section, we will study the equivalence of the problem (T P )Q and min the corresponding minimal norm control problem under the following framework (A2 ):

(i) The state space Y and the control space U are real separable Hilbert spaces. (ii) The controlled system is as: y(t) ˙ = Ay(t) + D(t)y(t) + Bu(t),

t ∈ (0, +∞),

(5.48)

where A : D(A) ⊆ Y → Y generates a C0 semigroup {eAt }t ≥0 on Y , D(·) ∈ L1loc ([0, +∞); L (Y )) and B ∈ L (U, Y ). Given τ ∈ [0, +∞), y0 ∈ Y and u ∈ L∞ (τ, +∞; U ), write y(·; τ, y0, u) for the solution of (5.48) over [τ, +∞), with the initial condition that y(τ ) = y0 . Moreover, write {Φ(t, s) : t ≥ s ≥ 0} for the evolution operator generated by A + D(·) over Y . (iii) Let U = BM (0), where BM (0) is the closed ball in U , centered at 0 and of radius M ≥ 0. (When M = 0, we agree that BM (0)  {0}.) (iv) Let QS = {0} × { y0 } with  y0 ∈ Y \ Q; QE = (0, +∞) × Q, where Q is arbitrarily taken from the following family of sets:  ⊆Y Q  is bounded, closed, convex F  {Q and has nonempty interior in Y }.

(5.49)

(v) The system (5.48) is L∞ -approximately controllable over each interval, i.e., for any T2 > T1 ≥ 0, y0 , y1 ∈ Y and ε ∈ (0, +∞), there exists u ∈ L∞ (T1 , T2 ; U ) so that y(T2 ; T1 , y0 , u) − y1 Y ≤ ε. M, y0

S ,QE In this section, we simply write (T P )Q for (T P )Q min

M, y0

(T P )Q

which reads:

  T(M,  y0 , Q)  inf T ∈ (0, +∞)  y(T ; 0,  y0, u) ∈ Q,  u ∈ L(0, T ; U) .

Similar to what we explained in the previous section (see (5.2) and (5.3)), we can M, y treat (T P )Q 0 as the following problem: (T2 P )Q

M, y0

   y0, u) ∈ Q T(M,  y0 , Q)  inf T ∈ (0, +∞)  y(T ; 0,   and u ∈ UM ,

where    UM  u ∈ L∞ (0, +∞; U )  u(t)U ≤ M a.e. t ∈ (0, +∞) .

(5.50)

5.2 Minimal Time Controls, Minimal Norm Controls, and Time-Varying. . .

219

In this problem, • a tetrad (0,  y0 , T , u) is called admissible, if T ∈ (0, +∞), u ∈ UM and y(T ; 0,  y0 , u) ∈ Q; • a tetrad (0,  y0 ,  T(M,  y0 , Q), u∗ ) is called optimal, if  T(M,  y0 , Q) ∈ (0, +∞), u∗ ∈ UM and y( T(M,  y0 , Q); 0,  y0 , u∗ ) ∈ Q; • when (0,  y0 ,  T(M,  y0 , Q), u∗ ) is an optimal tetrad,  T(M,  y0 , Q) and u∗ are called the optimal time and an optimal control, respectively. T(M,  y0 , Q) (M ≥ 0) is called a minimal Moreover, given  y0 and Q, the map M →  time function. We next introduce the corresponding minimal norm control problem in the following manner: Given T ∈ (0, +∞), y0 ∈ Y and Q ∈ F , we set T ,y0

(NP )Q

 N(T , y0 , Q)  inf{vL∞ (0,T ;U )  y(T ; 0, y0, v) ∈ Q}.

(5.51)

In this problem, • N(T , y0 , Q) is called the minimal norm; • a control v ∈ L∞ (0, T ; U ) is called admissible, if y(T ; 0, y0, v) ∈ Q; • a control v ∗ ∈ L∞ (0, T ; U ) is called optimal, if it is admissible and satisfies that v ∗ L∞ (0,T ;U ) = N(T , y0 , Q). Moreover, given y0 and Q, the map T → N(T , y0 , Q) is called a minimal norm function, which is denoted by N(·, y0 , Q). M, y We now give the definition about the equivalence between (T P )Q 0 and T , y

(NP )Q 0 . Definition 5.2 Let M ≥ 0, T ∈ (0, +∞), Q ∈ F and  y0 ∈ Y \ Q. Problems M, y T , y (T P )Q 0 and (NP )Q 0 are said to be equivalent if the following three conditions hold: M, y T , y (i) Both (T2 P )Q 0 and (NP )Q 0 have optimal controls; M, y (ii) The restriction of each optimal control to (T2 P ) 0 over (0, T ) is an optimal Q

T , y

control to (NP )Q 0 ; T , y0

(iii) The zero extension of each optimal control to (NP )Q M, y optimal control to (T2 P ) 0.

over (0, +∞) is an

Q

The main result of this section is as follows. y0 ∈ Q. Then the function Theorem 5.2 Let  y0 ∈ Y and Q ∈ F satisfy that  N(·,  y0 , Q) is continuous over (0, +∞). Moreover, if write     (G T ) T(M,  y0 , Q) y0 ,Q  (M, T ) ∈ [0, +∞) × (0, +∞) T = 

(5.52)

220

5 Equivalence of Several Kinds of Optimal Controls

and     (K N) y0 , Q) = 0 , y0 ,Q  (M, T ) ∈ [0, +∞) × (0, +∞) M = 0, N(T ,  (5.53) then the following conclusions are true: M, y

T , y

0 (i) When (M, T ) ∈ (G T ) and (NP )Q 0 y0 ,Q \ (K N) y0 ,Q , problems (T P )Q are equivalent and the null controls (over (0, T(M,  y0 , Q)) and (0, T ), respectively) are not optimal controls to these two problems, respectively. M, y0 T , y (ii) When (M, T ) ∈ (K N) and (NP )Q 0 are equivalent y0 ,Q , problems (T P )Q and the null controls (over (0, T(M,  y0 , Q)) and (0, T ), respectively) are the unique optimal controls to these two problems, respectively.   (iii) When (M, T ) ∈ [0, +∞) × (0, +∞) \ (G T ) y0 ,Q ∪ (K N) y0 ,Q , problems M, y T , y (T P )Q 0 and (NP )Q 0 are not equivalent.

We will give its proof later.

5.2.1 Properties of Minimal Time and Minimal Norm Functions In this subsection, we mainly discuss some properties of minimal time functions and minimal norm functions (see Theorems 5.3 and 5.4). We start with the following three lemmas. Lemma 5.4 The following three conclusions are true: (i) If there exists z ∈ Y and T > τ > 0 so that B ∗ Φ(T , t)∗ z = 0 for all t ∈ (τ, T ), then z = 0. (ii) Let T2 > T1 > 0 and yd , y0 ∈ Y . Then for each ε > 0, there exists a positive constant Cε  C(T1 , T2 , yd , y0 , ε) so that for each T ∈ [T1 , T2 ], |yd − Φ(T , 0)y0 , zY |  T ≤ Cε B ∗ Φ(T , t)∗ zU dt + εzY for all z ∈ Y.

(5.54)

0

(iii) Let T2 > T1 > 0, y0 ∈ Y and Q ∈ F . Then there exists a positive constant C  C(T1 , T2 , y0 , Q) so that for each T ∈ [T1 , T2 ], there is uT ∈ L∞ (0, T ; U ) so that y(T ; 0, y0, uT ) ∈ Q and uT L∞ (0,T ;U ) ≤ C.

(5.55)

5.2 Minimal Time Controls, Minimal Norm Controls, and Time-Varying. . .

221

Proof We prove the conclusions one by one. (i) Suppose that there exists z ∈ Y and T > τ > 0 so that B ∗ Φ(T , t)∗ z = 0 for all t ∈ (τ, T ).

(5.56)

By (v) of (A2 ) (given at the beginning of Section 5.2), for each yd ∈ Y and ε ∈ (0, +∞), there is u ∈ L∞ (0, T ; U ) so that y(T ; 0, 0, χ(τ,T ) u) − yd Y ≤ ε. This, along with (5.56), yields that for each yd ∈ Y and ε ∈ (0, +∞), yd , zY = y(T ; 0, 0, χ(τ,T ) u), zY + yd − y(T ; 0, 0, χ(τ,T ) u), zY  T ≤ u(t), B ∗ Φ(T , t)∗ zU dt + εzY τ

≤ εzY , which leads to z = 0. (ii) Let T2 > T1 > 0 and yd , y0 ∈ Y . By contradiction, suppose that (5.54) was not true. Then there would exist ε0 ∈ (0, +∞), {tk }k≥1 ⊆ [T1 , T2 ] and {zk }k≥1 ⊆ Y so that for each k ∈ N+ , 

tk

|yd − Φ(tk , 0)y0 , zk Y | = 1 > k

B ∗ Φ(tk , t)∗ zk U dt + ε0 zk Y , (5.57)

0

which implies that {zk }k≥1 is bounded in Y . Since {tk }k≥1 ⊆ [T1 , T2 ], there exists a subsequence of {k}k≥1 , still denoted in the same manner,  t ∈ [T1 , T2 ] and  z ∈ Y so that lim tk =  t and zk →  z weakly in Y.

k→+∞

These, together with (5.57), imply that t, 0)y0 , zY | and B ∗ Φ( t, t)∗ z = 0 for all t ∈ (0,  t). 1 = |yd − Φ( The above conclusions, combined with the conclusion (i) of this lemma, lead to a contradiction. So the conclusion (ii) of this lemma is true. (iii) Let T2 > T1 > 0, y0 ∈ Y and Q ∈ F . By (5.49), there exists a closed ball Br (yd ) (centered at yd and of radius r ∈ (0, +∞)) so that Br (yd ) ⊆ Q.

(5.58)

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5 Equivalence of Several Kinds of Optimal Controls

Let Cr be given by (5.54) with ε = r. Arbitrarily fix T ∈ [T1 , T2 ]. Define the following subspace of L1 (0, T ; U ) × Y :   O  (Cr B ∗ Φ(T , ·)∗ z, rz) ∈ L1 (0, T ; U ) × Y  z ∈ Y .

(5.59)

And then define an operator R : O → R in the following manner:   R Cr B ∗ Φ(T , ·)∗ z, rz  Φ(T , 0)y0 − yd , zY for each z ∈ Y.

(5.60)

From (5.60), we see that the map R is well defined and linear. By (5.54), we find that RL (O,R) ≤ 1.

(5.61)

Here, the norm of O is inherited from that of the space L1 (0, T ; U ) × Y . Next, according to the Hahn-Banach Theorem (see Theorem 1.5), there exists a bounded linear functional  : L1 (0, T ; U ) × Y → R R so that  = R on O and R  L (L1 (0,T ;U )×Y,R) = RL (O,R). R

(5.62)

Then according to the Riesz Representation Theorem (see Theorem 1.4), there is a pair ( u, ψ) ∈ L∞ (0, T ; U ) × Y so that  g) =  R(f, u, f L∞ (0,T ;U ),L1(0,T ;U ) + ψ, gY for all (f, g) ∈ L1 (0, T ; U ) × Y.

(5.63)

Now, it follows from (5.60), (5.59), (5.62), and (5.63) that for each z ∈ Y ,    Cr B ∗ Φ(T , ·)∗ z, rz Φ(T , 0)y0 − yd , zY = R =  u, Cr B ∗ Φ(T , ·)∗ zL∞ (0,T ;U ),L1 (0,T ;U ) + ψ, rzY . This yields that for each z ∈ Y , u) − yd , zY y(T ; 0, y0, −Cr   T = Φ(T , 0)y0 − yd , zY −  u(t), Cr B ∗ Φ(T , t)∗ zU dt 0

= rψ, zY ,

5.2 Minimal Time Controls, Minimal Norm Controls, and Time-Varying. . .

223

which leads to u) = yd + rψ. y(T ; 0, y0 , −Cr 

(5.64)

Finally, from (5.64) and (5.58), we see that in order to show (5.55), it suffices to prove the inequality: max{ uL∞ (0,T ;U ) , ψY } ≤ 1.

(5.65)

By (5.63), (5.62), and (5.61), we see that for each f ∈ L1 (0, T ; U ) and g ∈ Y ,  T  g) ≤ (f, g)L1 (0,T ;U )×Y ,  u(t), f (t)U dt + ψ, gY = R(f, 0

which leads to (5.65). Hence, we end the proof of Lemma 5.4.

  T ,y0

Lemma 5.5 For any T ∈ (0, +∞), y0 ∈ Y and Q ∈ F , the problem (NP )Q has at least one optimal control.

Proof Since Q ∈ F , it has a nonempty interior. Then it follows from (v) of (A2 ) T ,y that (NP )Q 0 has at least one admissible control. Next, by the same way as that used in the proof of Lemma 5.1, one can easily prove the existence of optimal controls to this problem. This completes the proof.   Lemma 5.6 Let T ∈ (0, +∞), y0 ∈ Y and Q ∈ F . Then the following two conclusions are true: (i) If Φ(T , 0)y0 ∈ Q, then there exists z ∈ Y \ {0} so that sup

y(T ; 0, y0, u), zY ≤ inf q, zY . q∈Q

uL∞ (0,T ;U) ≤N(T ,y0 ,Q)

(ii) Each optimal control v ∗ to (NP )Q

T ,y0

(5.66)

satisfies that

v ∗ L∞ (τ,T ;U ) = N(T , y0 , Q) for any τ ∈ [0, T ).

(5.67)

Proof We prove the conclusions one by one. (i) Define a subset of Y by  A  {y(T ; 0, y0 , u)  uL∞ (0,T ;U ) ≤ N(T , y0 , Q)}.

(5.68)

We claim that ◦

A ∩ Q = ∅ and A ∩ Q = ∅, ◦

where Q denotes the interior of Q in Y .

(5.69)

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5 Equivalence of Several Kinds of Optimal Controls

To show the first conclusion in (5.69), we use Lemma 5.5 to take an optimal T ,y control u1 for the problem (NP )Q 0 . By the optimality of u1 , we have that y(T ; 0, y0, u1 ) ∈ Q and u1 L∞ (0,T ;U ) = N(T , y0 , Q). These, combined with (5.68), lead to the first conclusion in (5.69). We next prove the second conclusion in (5.69). By contradiction, suppose that it was not true. Then by (5.68), there would exist u2 ∈ L∞ (0, T ; U ) so that ◦

y(T ; 0, y0, u2 ) ∈ Q and u2 L∞ (0,T ;U ) ≤ N(T , y0 , Q).

(5.70)

◦

Since y(T ; 0, y0 , u2 ) ∈ Q, we can find a constant λ ∈ (0, 1) so that y(T ; 0, y0, λu2 ) ∈ Q, T ,y

which implies that λu2 is an admissible control to (NP )Q 0 . This, together with the optimality of N(T , y0 , Q) and the inequality in (5.70), yields that N(T , y0 , Q) ≤ λu2 L∞ (0,T ;U ) ≤ λN(T , y0 , Q). Since λ ∈ (0, 1), the above leads to N(T , y0 , Q) = 0, which indicates that Φ(T , 0)y0 ∈ Q. This contradicts the assumption that Φ(T , 0)y0 ∈ Q. Hence, the second conclusion in (5.69) is true. ◦

Since A and Q are nonempty convex subsets in Y (see (5.68), the first conclusion in (5.69) and (5.49)), by the second conclusion in (5.69), we can apply the Hahn-Banach Theorem (see Theorem 1.11) to find z ∈ Y \ {0} so that sup y, zY ≤ inf q, zY . y∈A

q∈Q

This, along with (5.68), leads to (5.66). T ,y (ii) Let v ∗ be an optimal control to (NP )Q 0 . Then 0 ≤ N(T , y0 , Q) < +∞. When N(T , y0 , Q) = 0, it is clear that v ∗ is the null control and (5.67) is true. Next we show (5.67) when N(T , y0 , Q) ∈ (0, +∞). In this case, we have that Φ(T , 0)y0 ∈ Q.

(5.71)

By contradiction, suppose that (5.67) was not true. Then there would be  τ ∈ T ,y [0, T ) and an optimal control  v ∗ to (NP )Q 0 so that

5.2 Minimal Time Controls, Minimal Norm Controls, and Time-Varying. . .

225

 v ∗ L∞ ( τ ,T ;U ) < N(T , y0 , Q).

(5.72)

Since  v ∗ is an optimal control to (NP )Q 0 , we have that T ,y

y(T ; 0, y0 , v ∗ ) ∈ Q and  v ∗ L∞ (0,T ;U ) = N(T , y0 , Q). These, along with (5.71) and (i) of this lemma, yield that for some z ∈ Y \ {0}, y(T ; 0, y0 , u), zY = y(T ; 0, y0, v ∗ ), zY ,

max

uL∞ (0,T ;U) ≤N(T ,y0 ,Q)

which implies that  max

=

uL∞ (0,T ;U) ≤N(T ,y0 ,Q) 0  T ∗ ∗ ∗

T

u(t), B ∗ Φ(T , t)∗ zU dt (5.73)

 v (t), B Φ(T , t) zU dt.

0

By (5.73), we can use the similar way as that used in the proof of (4.15) to get that  v ∗ (t), B ∗ Φ(T , t)∗ zU =

max

w, B ∗ Φ(T , t)∗ zU for a.e. t ∈ (0, T ).

wU ≤N(T ,y0 ,Q)

The above, along with (5.72), yields that B ∗ Φ(T , t)∗ z = 0 for all t ∈ ( τ , T ). This, together with (i) of Lemma 5.4, implies that z = 0, which leads to a contradiction. Hence, we complete the proof of Lemma 5.6.   Let y0 ∈ Y and Q ∈ F . For each M ≥ 0, we define    y JM0  t ∈ (0, +∞)  N(t, y0 , Q) ≤ M .

(5.74)

We agree that y

y

inf JM0  +∞, when JM0 = ∅.

(5.75)

The following theorem presents some connection between the minimal time and the minimal norm functions. Such connection plays an important role in our studies.  y

Theorem 5.3 Let  y0 ∈ Y and Q ∈ F satisfy that  y0 ∈ Q. Let JM0 , with M ≥ 0, be defined by (5.74). Then

226

5 Equivalence of Several Kinds of Optimal Controls  y  T(M,  y0 , Q) = inf JM0 for all M ≥ 0.

(5.76)

y0 ∈ Q. Let M ≥ 0. Then Proof Arbitrarily fix  y0 ∈ Y and Q ∈ F satisfying   y  y either JM0 = ∅ or JM0 = ∅.  y In the case that JM0 = ∅, we first claim that y(t; 0,  y0, u) ∈ / Q for all t ∈ (0, +∞) and u ∈ UM .

(5.77)

By contradiction, suppose that (5.77) was not true. Then there would exist  t ∈ (0, +∞) and  u ∈ L∞ (0, +∞; U ) so that y( t; 0,  y0 , u) ∈ Q and  uL∞ (0,+∞;U ) ≤ M.

(5.78)

The first conclusion in (5.78) implies that  u|(0,t) is an admissible control to  t , y0 t,  y0 , Q) and the second conclusion (NP )Q . This, along with the optimality of N( in (5.78), yields that uL∞ (0,t;U ) ≤ M, N( t,  y0 , Q) ≤   y

which, combined with (5.74), indicates that  t ∈ JM0 . This leads to a contradiction  y0 since we are in the case that JM = ∅. So (5.77) is true. M, y Now, from (5.77), we see that (T2 P )Q 0 has no admissible tetrad. Thus, we have that  T(M,  y0 , Q) = +∞. This, together with (5.75), leads to (5.76).  y In the case where JM0 = ∅, we first show that  y  T(M,  y0 , Q) ≤ inf JM0 .

(5.79)

 y To this end, we arbitrarily fix  t ∈ JM0 . Then it follows by (5.74) that

N( t,  y0 , Q) ≤ M.  t , y

Thus (NP )Q 0 has an optimal control v. Write  v for the zero extension of v over (0, +∞). One can easily check that v ) ∈ Q and  v L∞ (0,+∞;U ) = N( t,  y0 , Q) ≤ M. y( t; 0,  y0 ,

(5.80)

M, y From (5.80), we see that (0,  y0 ,  t, v ) is an admissible tetrad to (T2 P )Q 0 . This, along with the optimality of  T(M,  y0 , Q), yields that

5.2 Minimal Time Controls, Minimal Norm Controls, and Time-Varying. . .

227

 t. T(M,  y0 , Q) ≤   y

Since  t is arbitrarily taken from the set JM0 , the above inequality leads to (5.79). We next show the reverse of (5.79). For this purpose, we define     y0 , u) ∈ Q and u ∈ UM . T(M, y0 ,Q)  t ∈ (0, +∞) y(t; 0, 

(5.81)

From (5.80), it follows that T(M, y0 ,Q) = ∅. Then by (5.81), we see that given  t ∈ T(M, u ∈ y0 ,Q) , there is a control  L∞ (0, +∞; U ) so that u) ∈ Q and  uL∞ (0,+∞;U ) ≤ M. y( t; 0,  y0 ,

(5.82)

The first conclusion in (5.82) implies that  u|(0,t) is an admissible control to  t , y0 (NP )Q . This, along with the optimality of N( t,  y0 , Q) and the second conclusion in (5.82), yields that N( t,  y0 , Q) ≤  uL∞ (0,t;U ) ≤ M,  y

which, combined with (5.74), indicates that  t ∈ JM0 . Hence,  y

t. inf JM0 ≤  Since  t was arbitrarily taken from T(M, y0 ,Q) , the above implies that  y

inf JM0 ≤ inf T(M, y0 ,Q) . This, along with (5.81) and (5.50), leads to the reverse of (5.79). Finally, (5.76) follows from (5.79) and its reverse at once. This ends the proof of Theorem 5.3.   Remark 5.2 For better understanding of Theorem 5.3, we explain it with the aid of Figure 5.1 (see below), where the curve denotes the graph of the minimal norm function. Suppose that the minimal norm function is continuous over (0, +∞) (which will be proved in Theorem 5.4). A beam (which is parallel to the t-axis and has the distance M with the t-axis) moves from the left to the right. The first M, y time point at which this beam reaches the curve is the optimal time to (T2 P )Q 0 . Thus, we can treat Theorem 5.3 as a “falling sun theorem” (see, for instance, raising sun lemma-Lemma 3.5 and Figure 5 on Pages 121–122 in [12]): If one thinks of the sun falling down the west (at the left) with the rays of light parallel to the t-axis,

228

5 Equivalence of Several Kinds of Optimal Controls

Fig. 5.1 Falling sum theorem

then the points ( T(M,  y0 , Q), M), with M ≥ 0, are precisely the points which are in the bright part on the curve. (These points constitute the part outside bold on the curve in Figure 5.1.) To study some properties about minimal norm function N(·, y0 , Q) (with y0 ∈ Y and Q ∈ F ), we consider the following minimization problem: T ,y0

T ,y0

V (T , y0 , Q)  inf JQ

(J P )Q

z∈Y

T ,y0

where T ∈ (0, +∞) and JQ T ,y0

JQ

(z) 

(z),

(5.83)

: Y → R is defined by

 #2 1" T B ∗ Φ(T , t)∗ zU dt 2 0 +y0 , Φ(T , 0)∗ zY + sup q, −zY , z ∈ Y.

(5.84)

q∈Q T ,y

For the problem (J P )Q 0 , we have the following proposition: Proposition 5.1 Let T2 > T1 > 0, T ∈ [T1 , T2 ], y0 ∈ Y , and Q ∈ F . Let Br (yd ) be a closed ball in Y , centered at yd and of radius r > 0, so that Br (yd ) ⊆ Q. Then the following conclusions are true: (i) There exists a constant C  C(T1 , T2 , y0 , Q) ∈ (0, +∞) so that T ,y0

JQ

(z) ≥

r zY − C for all z ∈ Y. 2

(5.85)

(ii) There exists z∗ ∈ Y so that T ,y0

V (T , y0 , Q) = JQ

(z∗ ).

Besides, the above z∗ can be chosen as 0 if and only if Φ(T , 0)y0 ∈ Q.

(5.86)

5.2 Minimal Time Controls, Minimal Norm Controls, and Time-Varying. . .

229

(iii) It holds that 1 V (T , y0 , Q) = − N(T , y0 , Q)2 . 2

(5.87)

Proof We will prove the conclusions (i)–(iii) one by one. (i) Arbitrarily fix z ∈ Y . From (5.84), we see that  #2 1" T ∗ T ,y JQ 0 (z) ≥ B Φ(T , t)∗ zU dt 2 0 +y0 , Φ(T , 0)∗ zY +

sup q, −zY q∈Br (yd )

 #2 1" T ∗ = B Φ(T , t)∗ zU dt 2 0

+Φ(T , 0)y0 − yd , zY + rzY .

(5.88)

Meanwhile, by (ii) of Lemma 5.4, there is C  C(T1 , T2 , y0 , yd , r) ∈ (0, +∞) so that  T r Φ(T , 0)y0 − yd , zY ≥ −C B ∗ Φ(T , t)∗ zU dt − zY . 2 0 This, together with (5.88), implies that  #2 1" T ∗ T ,y JQ 0 (z) ≥ B Φ(T , t)∗ zU dt 2 0 T r −C B ∗ Φ(T , t)∗ zU dt + zY 2 0 r 1 ≥ zY − C 2 . 2 2

(5.89)

Because Br (yd ) was arbitrarily taken from Y , the above C depends only on T1 , T2 , y0 , and Q, i.e., C = C(T1 , T2 , y0 , Q). This, together with (5.89), gives (5.85) immediately. (ii) By (5.83), (5.84), (5.85), and by (i) and (ii) of (A2 ) (given at the beginning of Section 5.2), we can use Theorem 1.13 to find z∗ ∈ Y holding (5.86). We now prove that z∗ = 0 ⇒ Φ(T , 0)y0 ∈ Q. Indeed, if z∗ = 0, then by (5.83), (5.84), and (5.86), we have that for each z ∈ Y ,  1  T ,y0 T ,y JQ (λz) − JQ 0 (0) = Φ(T , 0)y0 , zY + sup q, −zY . λ→0+ λ q∈Q

0 ≤ lim

This implies that sup q, zY ≥ Φ(T , 0)y0 , zY for all z ∈ Y. q∈Q

From this and Theorem 1.11, we can easily check that Φ(T , 0)y0 ∈ Q.

230

5 Equivalence of Several Kinds of Optimal Controls

We next show that Φ(T , 0)y0 ∈ Q implies that z∗ can be chosen as 0. In fact, if Φ(T , 0)y0 ∈ Q, then y0 , Φ(T , 0)∗ zY + sup q, −zY q∈Q

≥ y0 , Φ(T , 0)∗ zY + Φ(T , 0)y0 , −zY = 0 for all z ∈ Y. From this, (5.83) and (5.84), we see that T ,y0

V (T , y0 , Q) = 0 = JQ

(0).

Hence, we can choose z∗ = 0. In summary, the conclusion (ii) has been proved. (iii) We first show that 1 V (T , y0 , Q) ≥ − N(T , y0 , Q)2 . 2 T ,y0

By Lemma 5.5, (NP )Q we have that

(5.90)

has an optimal control v ∗ . By the optimality of v ∗ ,

y(T ; 0, y0, v ∗ ) ∈ Q and v ∗ L∞ (0,T ;U ) = N(T , y0 , Q).

(5.91)

It follows from the first conclusion in (5.91) that for each z ∈ Y , sup q, −zY ≥ y(T ; 0, y0, v ∗ ), −zY q∈Q

= −y0 , Φ(T , 0)∗ zY −



T

v ∗ (t), B ∗ Φ(T , t)∗ zU dt.

0

This, together with the second conclusion in (5.91), implies that for each z ∈ Y ,  T ∗ sup q, −zY ≥ −y0 , Φ(T , 0) zY − N(T , y0 , Q) B ∗ Φ(T , t)∗ zU dt 0

q∈Q

≥ −y0 , Φ(T , 0)∗ zY −

 1" 2

T

B ∗ Φ(T , t)∗ zU dt

#2

0

1 − N(T , y0 , Q)2 . 2 The above, along with (5.83) and (5.84), yields (5.90). We next show that 1 V (T , y0 , Q) ≤ − N(T , y0 , Q)2 . 2 Notice that either Φ(T , 0)y0 ∈ Q or Φ(T , 0)y0 ∈ Q.

(5.92)

5.2 Minimal Time Controls, Minimal Norm Controls, and Time-Varying. . .

231

In the case that Φ(T , 0)y0 ∈ Q, we have that y(T ; 0, y0 , 0) ∈ Q. This implies T ,y that the null control is an optimal control to (NP )Q 0 . So we have that N(T , y0 , Q) = 0.

(5.93)

Meanwhile, by (5.83) and (5.84), we see that T ,y0

V (T , y0 , Q) ≤ JQ

(0) = 0,

which, combined with (5.93), indicates (5.92) in this case. In the case where Φ(T , 0)y0 ∈ Q, we can use (i) of Lemma 5.6 to find z ∈ Y \ {0} so that y(T ; 0, y0 , u), zY ≤ inf q, zY .

sup

(5.94)

q∈Q

uL∞ (0,T ;U) ≤N(T ,y0 ,Q)

This yields that y0 , Φ(T , 0)∗ zY + N(T , y0 , Q)



T

B ∗ Φ(T , t)∗ zU dt

0

≤ inf q, zY = − sup q, −zY . q∈Q

q∈Q

The above, along with (5.84), shows that for each λ ≥ 0, T ,y JQ 0 (λz)

λ2 " ≤ 2



T

B ∗ Φ(T , t)∗ zU dt

0



T

−λN(T , y0 , Q) =

1/ λ 2

#2

B ∗ Φ(T , t)∗ zU dt

(5.95)

0



T

B ∗ Φ(T , t)∗ zU dt − N(T , y0 , Q)

02

0

1 − N(T , y0 , Q)2 . 2 Meanwhile, since z = 0, we can use (i) of Lemma 5.4 to find that B ∗ Φ(T , ·)∗ z = 0 in L1 (0, T ; U ). This, along with (5.83) and (5.95), yields (5.92). Finally, (5.87) follows from (5.90) and (5.92) immediately. Thus, we complete the proof of Proposition 5.1.

 

The following result is mainly concerned with the continuity of the minimal norm function N(·, y0 , Q).

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5 Equivalence of Several Kinds of Optimal Controls

Theorem 5.4 Let y0 ∈ Y and Q ∈ F . Then the minimal norm function t → N(t, y0 , Q) is continuous over (0, +∞). Furthermore, if y0 ∈ Q, then lim N(t, y0 , Q) = +∞. t →0+

Proof Arbitrarily fix y0 ∈ Y and Q ∈ F . The proof is organized by three steps. Step 1. We show that for each T ∈ (0, +∞), N(T , y0 , Q) ≤ lim inf N(t, y0 , Q). t →T

(5.96)

Arbitrarily fix T ∈ (0, +∞), and then arbitrarily take a sequence {tk }k≥1 ⊆ (T /2, 2T ) so that tk → T . It follows from (iii) of Lemma 5.4 (where T1 = T /2 and T2 = 2T ) that sup N(tk , y0 , Q) < +∞.

k∈N+

(5.97)

t ,y

By Lemma 5.5, each (NP )Qk 0 , with k ≥ 1, has an optimal control uk . By the optimality of uk , we have that y(tk ; 0, y0 , uk ) ∈ Q and uk L∞ (0,tk ;U ) = N(tk , y0 , Q).

(5.98)

For each k ≥ 1, let  uk be the zero extension of uk over (0, +∞). Then the second conclusion in (5.98), together with (5.97), implies that { uk }k≥1 is bounded in L∞ (0, +∞; U ). Thus there exists a subsequence of {k }≥1 of {k}k≥1 and  u ∈ L∞ (0, +∞; U ) so that  uk →  u weakly star in L∞ (0, +∞; U ) as  → +∞. From this, we find that y(tk ; 0, y0, uk ) → y(T ; 0, y0, u) weakly in Y as  → +∞

(5.99)

and uk L∞ (0,+∞;U ).  uL∞ (0,+∞;U ) ≤ lim inf  →+∞

Since Q is convex and closed, it follows by (5.98), (5.99), and (5.100) that y(T ; 0, y0, u) ∈ Q and  uL∞ (0,+∞;U ) ≤ lim inf N(tk , y0 , Q). →+∞

From these and the optimality of N(T , y0 , Q), we see that uL∞ (0,+∞;U ) ≤ lim inf N(tk , y0 , Q), N(T , y0 , Q) ≤  →+∞

which leads to (5.96).

(5.100)

5.2 Minimal Time Controls, Minimal Norm Controls, and Time-Varying. . .

233

Step 2. We show that for each T ∈ (0, +∞), N(T , y0 , Q) ≥ lim sup N(t, y0 , Q).

(5.101)

t →T

According to (iii) of Proposition 5.1, (5.101) is equivalent to the following inequality: V (T , y0 , Q) ≤ lim inf V (t, y0 , Q) for each T ∈ (0, +∞). t →T

(5.102)

To show (5.102), we arbitrarily fix T ∈ (0, +∞), and then arbitrarily take a sequence {tk }k≥1 ⊆ (T /2, 2T ) so that tk → T . It follows from (iii) of Proposition 5.1 that V (tk , y0 , Q) ≤ 0 for all k ≥ 1.

(5.103) t ,y

Meanwhile, according to (ii) of Proposition 5.1, each (J P )Qk 0 (with k ≥ 1) has a minimizer zk∗ . Then, by (5.86), (5.85), and (5.103), we find that sup zk∗ Y < +∞.

k∈N+

z ∈ Y so that Thus, there exists a subsequence {k }≥1 of {k}k≥1 and  zk∗ →  z weakly in Y as  → +∞, which, combined with (5.84), indicates that T ,y0

JQ

tk ,y0

( z) ≤ lim inf JQ →+∞

(zk∗ ).

From this, (5.83), and (5.86), it follows that V (T , y0 , Q) ≤ lim inf V (tk , y0 , Q), →+∞

which leads to (5.102). Step 3. We prove that lim N(T , y0 , Q) = +∞ when y0 ∈ Q. T →0+

By contradiction, suppose that it was not true. Then there would exist {Tk }k≥1 ⊆ (0, +∞) so that lim Tk = 0 and sup N(Tk , y0 , Q) < +∞.

k→+∞

k∈N+

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5 Equivalence of Several Kinds of Optimal Controls

These yield that y0 = lim Φ(Tk , 0)y0 = lim y(Tk ; 0, y0 , uk ) ∈ Q, k→+∞

k→+∞ T ,y

where uk is an optimal control to (NP )Qk 0 . (The existence of uk is ensured by Lemma 5.5.) This contradicts the fact that y0 ∈ Q. Hence, we end the proof of Theorem 5.4.   Based on Theorems 5.4 and 5.3, we can prove the following Proposition 5.2, which will be used in the proof of Theorem 5.2. Proposition 5.2 Let  y0 ∈ Y and Q ∈ F satisfy that  y0 ∈ Q. Then the following two conclusions are true: (i) For each M ∈ [0, +∞),  T(M,  y0 , Q) ∈ (0, +∞]. (ii) For each M ∈ [0, +∞) with  T(M,  y0 , Q) < +∞, N( T(M,  y0 , Q),  y0 , Q) = M.

(5.104)

Proof We prove the conclusions one by one. (i) By contradiction, we suppose that the conclusion (i) was not true. Then there would be M0 ∈ [0, +∞) so that  T(M0 ,  y0 , Q) = 0. This, along with (5.50), yields that there exist two sequences {tk }k≥1 ⊆ (0, +∞) and {uk }k≥1 ⊆ UM0 so that lim tk = 0, y(tk ; 0,  y 0 , uk ) ∈ Q

k→+∞

and uk L∞ (0,+∞;U ) ≤ M0 for all k ∈ N+ .

(5.105)

By (5.105), we obtain that y0, uk ) ∈ Q,  y0 = lim y(tk ; 0,  k→+∞

which contradicts the assumption that  y0 ∈ Q. So the conclusion (i) is true. (ii) Arbitrarily fix M ∈ [0, +∞) so that  T(M,  y0 , Q) < +∞. By the conclusion (i) in this proposition, we have that  T(M,  y0 , Q) ∈ (0, +∞). This, along with  y (5.76) (see Theorem 5.3), yields that JM0 = ∅. Thus, by (5.76) and (5.74), there is a sequence {tk }k≥1 ⊆ (0, +∞) so that T(M,  y0 , Q) and N(tk ,  y0 , Q) ≤ M for all k ∈ N+ . lim tk = 

k→+∞

5.2 Minimal Time Controls, Minimal Norm Controls, and Time-Varying. . .

235

Since  T(M,  y0 , Q) ∈ (0, +∞), the above conclusions, together with the continuity of the minimal norm function N(·,  y0 , Q) at  T(M,  y0 , Q) (see Theorem 5.4), yield that y0 , Q) ≤ M. N( T(M,  y0 , Q), 

(5.106)

We next prove (5.104). By contradiction, suppose that it was not true. Then by (5.106), we would have that y0 , Q) < M. N( T(M,  y0 , Q),  This, along with the continuity of the minimal norm function N(·,  y0 , Q) at  T(M,  y0 , Q), yields that there is δ0 ∈ (0,  T(M,  y0 , Q)) so that N( T(M,  y0 , Q) − δ0 ,  y0 , Q) < M. Then it follows from (5.74) that  y  T(M,  y0 , Q) − δ0 ∈ JM0 ,

which contradicts (5.76). Hence, (5.104) is true. Thus, we complete the proof of Proposition 5.2.

 

5.2.2 Proof of the Main Result We now on the position to prove Theorem 5.2. y0 ∈ Q. First of Proof (Proof of Theorem 5.2) Let  y0 ∈ Y and Q ∈ F satisfy that  all, by Theorem 5.4, the function N(·,  y0 , Q) is continuous over (0, +∞). We now show the conclusions (i)–(iii) one by one. (i) Arbitrarily fix (M, T ) so that (M, T ) ∈ (G T ) y0 ,Q \ (K N) y0 ,Q .

(5.107)

Then it follows from (5.52) that 0 T1 ≥ 0, there exists a constant C(T1 , T2 ) ∈ (0, +∞) so that Φ(T2 , T1 )∗ zY ≤ C(T1 , T2 )B ∗ Φ(T2 , ·)∗ zL1 (T1 ,T2 ;U ) for all z ∈ Y.

(5.127)

5.3 Minimal Time Controls, Minimal Norm Controls, and Time-Varying. . .

243

This implies that B ∗ Φ(T , ·)∗ zL1 (0,T ;U ) = 0 for all z ∈ Y \ {0},

(5.128)

and y0 , Φ(T , 0)∗ zY < +∞. ∗ ∗ z∈Y \{0} B Φ(T , ·) zL1 (0,T ;U )

M  sup

(5.129)

We now claim that M ≤ N(T , y0 ).

(5.130)

By Lemma 5.7, (NP )T0 has an optimal control u∗ . By the optimality of u∗ , we have that y

y(T ; 0, y0, u∗ ) = 0 and u∗ L∞ (0,T ;U ) = N(T , y0 ). These imply that for each z ∈ Y , y0 , Φ(T , 0)∗ zY = y(T ; 0, y0 , u∗ ), zY −



T

u∗ (t), B ∗ Φ(T , t)∗ zU dt

0



T

≤ N(T , y0 )

B ∗ Φ(T , t)∗ zU dt.

0

This, together with (5.128) and (5.129), gives (5.130). We next show that M ≥ N(T , y0 ).

(5.131)

For this purpose, we set  XT  {B ∗ Φ(T , ·)∗ z|(0,T )  z ∈ Y } ⊆ L1 (0, T ; U ). It is clear that XT is a linear subspace of L1 (0, T ; U ). We define a map F : XT → R in the following manner:   F B ∗ Φ(T , ·)∗ z|(0,T )  y0 , Φ(T , 0)∗ zY for all z ∈ Y.

(5.132)

From this and (5.127), one can check that F is well defined and linear. Furthermore, it follows from (5.129) and (5.132) that   ∗  F B Φ(T , ·)∗ z|(0,T )  ≤ M

 0

T

B ∗ Φ(T , t)∗ zU dt for all z ∈ Y.

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5 Equivalence of Several Kinds of Optimal Controls

From this, we can apply the Hahn-Banach Theorem and the Riesz Representation Theorem (see Theorems 1.5 and 1.4) to get  u ∈ L∞ (0, T ; U ), with  uL∞ (0,T ;U ) ≤ M

(5.133)

so that   F B ∗ Φ(T , ·)∗ z|(0,T ) =



T

 u(t), B ∗ Φ(T , t)∗ zU dt for all z ∈ Y.

(5.134)

0

From (5.134) and (5.132), we see that ∗



T

y(T ; 0, y0, − u), zY = y0 , Φ(T , 0) zY −

 u(t), B ∗ Φ(T , t)∗ zU dt

0

= 0 for all z ∈ Y, which indicates that y(T ; 0, y0, − u) = 0. y

This implies that − u is an admissible control to (NP )T0 , which, combined with (5.133), indicates that N(T , y0 ) ≤  uL∞ (0,T ;U ) ≤ M. This leads to (5.131). Finally, the desired equality in Lemma 5.8 follows from (5.131), (5.130), and (5.129) at once. This completes the proof of Lemma 5.8.   The following proposition is concerned with the continuity and monotonicity of the minimal norm function N(·, y0 ). Proposition 5.3 Let y0 ∈ Y \ {0}. Then the function N(·, y0 ) is strictly decreasing  0 ), +∞). and continuous from (0, +∞) onto (N(y Proof Let y0 ∈ Y \ {0}. The proof is organized by several steps. Step 1. We show that N(·, y0 ) is strictly decreasing from (0, +∞) to (0, +∞). First of all, we claim that N(T , y0 ) ∈ (0, +∞) for each T ∈ (0, +∞).

(5.135)

 > 0 so By contradiction, suppose that (5.135) was not true. Then there would be T that N(T, y0 ) = 0.

(5.136)

5.3 Minimal Time Controls, Minimal Norm Controls, and Time-Varying. . .

245

By Lemma 5.7, (NP )T0 has an optimal control  u∗ . From (5.136), we see that  u∗ = ∗ 0. Then by the optimality of  u , we have that y

Φ(T, 0)y0 = y(T; 0, y0, 0) = 0, which, combined with (vii) of (A3 ) (given at the beginning of Section 5.3), indicates that y0 = 0. This contradicts the assumption that y0 = 0. Therefore, (5.135) is true. y Next, we arbitrarily fix T2 > T1 > 0. According to Lemma 5.7, (NP )T01 has an optimal control v1 . By the optimality of v1 , we see that y(T1 ; 0, y0 , v1 ) = 0 and v1 L∞ (0,T1 ;U ) = N(T1 , y0 ).

(5.137)

Meanwhile, by (v) of (A3 ) (given at the beginning of Section 5.3), there exists v2 ∈ L∞ (0, T2 ; U ) so that y(T2 ; 0, y0, χ(T1 ,T2 ) v2 ) = 0.

(5.138)

By (5.135), there exists a constant λ ∈ (0, 1) so that λv2 L∞ (0,T2 ;U ) ≤ (1 − λ)N(T1 , y0 ).

(5.139)

Write  v1 for the zero extension of v1 over (0, T2 ). Define a new control as follows: v3 (t)  (1 − λ) v1 + λχ(T1 ,T2 ) v2 (t),

t ∈ (0, T2 ).

(5.140)

Then, from (5.140), (5.138), and the first equality in (5.137), we see that v1 ) + λy(T2 ; 0, y0 , χ(T1 ,T2 ) v2 ) y(T2 ; 0, y0, v3 ) = (1 − λ)y(T2 ; 0, y0, = 0. y

Thus, v3 is an admissible control to (NP )T02 . This, together with the optimality of N(T2 , y0 ), implies that N(T2 , y0 ) ≤ v3 L∞ (0,T2 ;U ) . The above inequality, along with (5.140), (5.139), and the second equality in (5.137), yields that N(T2 , y0 ) ≤ (1 − λ)N(T1 , y0 ). Since N(T1 , y0 ) ∈ (0, +∞) (see (5.135)), the above inequality implies that N(T2 , y0 ) < N(T1 , y0 ). This completes the proof of Step 1.

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5 Equivalence of Several Kinds of Optimal Controls

Step 2. We prove the right-continuity of N(·, y0 ). Arbitrarily fix T ∈ (0, +∞), and then arbitrarily choose a sequence {Tk }k≥1 ⊆ (T , T + 1) so that lim Tk = T .

k→+∞

By Lemma 5.7, each (NP )T0k (with k ≥ 1) has an optimal control u∗k . By the optimality of u∗k , we see that y

y(Tk ; 0, y0, u∗k ) = 0 and u∗k L∞ (0,Tk ;U ) = N(Tk , y0 ) for all k ≥ 1.

(5.141)

Meanwhile, it follows from the conclusion of Step 1 that sup N(Tk , y0 ) < +∞.

(5.142)

k∈N+

Write  u∗k for the zero extension of u∗k over (0, +∞). By (5.142) and (5.141), the sequence { u∗k }k≥1 is bounded in L∞ (0, +∞; U ). Then there exists a subsequence ∗ { uk }≥1 and  u ∈ L∞ (0, +∞; U ) so that  u∗k →  u weakly star in L∞ (0, +∞; U ) as  → +∞. From this and (5.141), we can directly check that u∗k ) → y(T ; 0, y0, u) weakly in Y, y(T ; 0, y0 , u) = 0 y(Tk ; 0, y0, and  uL∞ (0,+∞;U ) ≤ lim inf  u∗k L∞ (0,+∞;U ) ≤ lim inf N(Tk , y0 ). →+∞

→+∞

By the above two conclusions and the optimality of N(T , y0 ), we get that N(T , y0 ) ≤  uL∞ (0,T ;U ) ≤ lim inf N(Tk , y0 ). →+∞

This, along with the monotonicity of N(·, y0 ) (see (i) of this proposition), leads to the right-continuity of N(·, y0 ). Step 3. We show that lim N(T , y0 ) = +∞. T →0+

By contradiction, suppose that it was not true. Then there would be a sequence {tk }k≥1 ⊆ (0, +∞), with lim tk = 0, so that k→+∞

sup N(tk , y0 ) < +∞.

k∈N+

(5.143)

5.3 Minimal Time Controls, Minimal Norm Controls, and Time-Varying. . .

247

According to Lemma 5.7, each (NP )tk0 (with k ≥ 1) has an optimal control vk∗ . By the optimality of vk∗ , we see that y

y(tk ; 0, y0 , vk∗ ) = 0 and vk∗ L∞ (0,tk ;U ) = N(tk , y0 ) for all k ≥ 1.

(5.144)

Now, it follows from (5.144) and (5.143) that y0 = lim

k→+∞



y(tk ; 0, y0 , vk∗ ) − y(tk ; 0, 0, vk∗ )



= − lim y(tk ; 0, 0, vk∗ ) = 0, k→+∞

which contradicts the assumption that y0 = 0. Step 4. We prove the left-continuity of N(·, y0 ). For this purpose, we arbitrarily fix a T ∈ (0, +∞), and then arbitrarily take {Tn }n≥1 ⊆ (T /2, T ) so that Tn T . By Lemma 5.8, we can choose {zn }n≥1 ⊆ Y \ {0} so that for each n ∈ N+ , 

Tn

B ∗ Φ(Tn , t)∗ zn U dt = 1

(5.145)

0

and N(Tn , y0 ) − 1/n ≤ y0 , Φ(Tn , 0)∗ zn Y .

(5.146)

We now claim that there exists a subsequence {nk }k≥1 ⊆ {n}n≥1 and B ∗ ϕ ∈ ZT (given by (5.122)) so that for each δ ∈ (0, T ], Φ(Tnk , T − δ)∗ znk → ϕ(T − δ) weakly in Y as k → +∞.

(5.147)

Indeed, on one hand, if we write ψn (·) for the zero extension of B ∗ Φ(Tn , ·)∗ zn over (0, T ), then from (5.145), we find that for each  ∈ N+ , {ψn (·)|(T ,T ) }n≥ is bounded inL1 (T , T ; U ).From this, we can use Theorem 1.20 to see that for each  ∈ N+ , Φ(Tn , T )∗ zn n≥+1 is bounded in Y . (More precisely, in (1.85), we choose T2 , T1 and z as T+1 , T and Φ(Tn , T+1 )∗ zn , respectively.) This, together with the diagonal law, implies that there exists a subsequence {mm }m≥1 ⊆ {n}n≥1 , so that for each  ∈ (0, +∞), there is  z ∈ Y so that z weakly in Y as m → +∞. Φ(Tmm , T )∗ zmm → 

(5.148)

On the other hand, we arbitrarily fix 1 , 2 ∈ N+ . Then we have that Φ(Tmm , T1 )∗ zmm = Φ(T1 +2 , T1 )∗ Φ(Tmm , T1 +2 )∗ zmm for all m ≥ 1 + 2 .

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5 Equivalence of Several Kinds of Optimal Controls

This, together with (5.148), implies that  z1 = Φ(T1 +2 , T1 )∗ z1 +2 .

(5.149)

Define a function ϕ : [0, T ) → Y in the following manner: z for all t ∈ [0, T ] and  ∈ N+ . ϕ(t)  Φ(T , t)∗ From the above, (5.149), (5.148), (5.145), and (5.122), we can verify the following conclusions: ϕ is well defined over [0, T ), B ∗ ϕ ∈ ZT and Φ(Tmm , T )∗ zmm → ϕ(T ) weakly in Y as m → +∞. These lead to (5.147). We next show that for the subsequence {nk }k≥1 given in (5.147), 

∗

T

B ϕ ∈ YT , 0

B ∗ ϕ(t)U dt ≤ 1 and y0 , Φ(Tnk , 0)∗ znk Y → y0 , ϕ(0)Y . (5.150)

B ∗ϕ

In fact, since ∈ ZT , the first conclusion in (5.150) follows from (vi) of (A3 ) (given at the beginning of Section 5.3). The second conclusion in (5.150) follows from (5.147) and (5.145). The third conclusion in (5.150) follows from (5.147). Finally, by (5.146) and (5.150), we get that lim sup N(Tnk , y0 ) ≤ lim y0 , Φ(Tnk , 0)∗ znk Y = y0 , ϕ(0)Y . k→+∞

k→+∞

(5.151)

From the above, we can show that  lim sup N(Tnk , y0 ) ≤ N(T , y0 ) k→+∞

T

B ∗ ϕ(t)U dt ≤ N(T , y0 ).

(5.152)

0

Indeed, on one hand, since B ∗ ϕ ∈ YT = ZT (see (5.150) and (5.121)), and because of (5.122), there is  zT /2 ∈ Y and a sequence { zn }n≥1 ⊆ Y so that ϕ(·) = Φ(T /2, ·)∗ zT /2 over [0, T /2]

(5.153)

and zn − B ∗ ϕ(·)L1 (0,T ;U ) → 0 as n → +∞. B ∗ Φ(T , ·)∗

(5.154)

5.3 Minimal Time Controls, Minimal Norm Controls, and Time-Varying. . .

249

On the other hand, according to (i) of Theorem 1.20, there is a positive constant C(T ) (independent of n) so that Φ(T /2, 0)∗ (Φ(T , T /2)∗ zn − zT /2 )Y ≤ C(T )B ∗ Φ(T /2, ·)∗ (Φ(T , T /2)∗ zn − zT /2 )L1 (0,T /2;U ) = C(T )B ∗ Φ(T , ·)∗ zn − B ∗ Φ(T /2, ·)∗ zT /2 L1 (0,T /2;U ), which, combined with (5.153) and (5.154), indicates that Φ(T , 0)∗ zn − ϕ(0)Y → 0. From the above, Lemma 5.8 and (5.154), it follows that zn Y y0 , ϕ(0)Y = lim y0 , Φ(T , 0)∗ n→+∞

≤ N(T , y0 ) lim sup B ∗ Φ(T , ·)∗ zn L1 (0,T ;U ) n→+∞

= N(T , y0 )B ∗ ϕL1 (0,T ;U ) . This, together with (5.151) and (5.150), yields (5.152). From (5.152) and the monotonicity of N(·, y0 ) (see step 1), we conclude that N(·, y0 ) is left-continuous. Hence, we end the proof of Proposition 5.3.   With the help of Proposition 5.3, we can have the following existence result on  y optimal controls of (T2 P )M0 :  y Lemma 5.9 The problem (T2 P )M0 (M ∈ (0, +∞)) has optimal controls if and only ( ( if M > N y0 ). Here N y0 ) is given by (5.125).

Proof The proof is quite similar to that of Lemma 5.3. We omit it here.

 

Based on Lemma 5.9 and Proposition 5.3, we have the following consequence: Corollary 5.2 Let (M, T ) ∈ [0, +∞) × (0, +∞). Then M = N(T ,  y0 ) if and only if T =  T(M,  y0 ). Proof The proof is quite similar to that of Corollary 5.1. We omit it here.

 

We now are in the position to prove Theorem 5.5. Proof (Proof of Theorem 5.5) By the above several lemmas, as well as Proposition 5.3 and Corollary 5.2, we can use the same way as that used in the proof of Theorem 5.1 to prove Theorem 5.5. We omit the details here.   We end this section with the next example, which can be put into the framework of this section.

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5 Equivalence of Several Kinds of Optimal Controls

Example 5.4 Let Ω be a bounded domain in Rd , d ≥ 1, with a C 2 boundary ∂Ω. Let ω ⊆ Ω be a nonempty and open subset with its characteristic function χω . Let a1 ∈ L∞ (Ω) and a2 ∈ L∞ (0, +∞). Consider the controlled heat equation: ⎧ ⎨ ∂t y − Δy + (a1 (x) + a2 (t))y = χω u in Ω × (0, +∞), y=0 on ∂Ω × (0, +∞), ⎩ y(0) = y0 ∈ L2 (Ω),

(5.155)

where u ∈ L∞ (0, +∞; L2 (Ω)). Let Y = U = L2 (Ω); Let A = Δ with its domain H 2 (Ω) ∩ H01 (Ω); Let D(·) = −a1 − a2 (·); Let B = χω , where χω is treated as a linear and bounded operator on U . Thus, the Equation (5.155) can be rewritten as 

y(t) ˙ = Ay(t) + D(t)y(t) + Bu(t), t ∈ (0, +∞), y(0) = y0 .

To verify that the above setting can be put into the framework (A3 ) (given at the beginning of Section 5.3), we only need to show the following two facts: (a) The above D(·) satisfies that D ∈ L1loc ([0, +∞); L (L2 (Ω))).

(5.156)

(b) (vi) of (A3 ) (given at the beginning of Section 5.3) is true. (Here we used Theorems 1.22, 1.21, and Remark 1.5 after Theorem 1.22.) We will prove (a) and (b) one by one. We start with proving (a). Choose a sequence of step functions {a2,k }k≥1 ⊆ L1 (0, +∞) so that lim |a2,k (t) − a2 (t)| = 0 for a.e. t ∈ (0, +∞).

k→+∞

(5.157)

It is clear that for each k ∈ N+ , Dk (·)  −a1 − a2,k (·) ∈ L1loc ([0, +∞); L (L2 (Ω))).

(5.158)

This, along with (5.157), yields that for a.e. t ∈ (0, +∞), lim D(t) − Dk (t)L (L2 (Ω)) = lim |a2,k (t) − a2 (t)| = 0,

k→+∞

k→+∞

which, combined with (5.158), indicates that D(·) is strongly measurable from (0, +∞) to L (L2 (Ω)). Now, (5.156) can be easily checked. Hence, the conclusion (a) is true. To prove the conclusion (b), we arbitrarily fix T > 0. We first show that YT ⊆ ZT . (see (5.122) for the definitions of YT , ZT .)

(5.159)

5.3 Minimal Time Controls, Minimal Norm Controls, and Time-Varying. . .

251

For this purpose, we arbitrarily take ψ ∈ YT . By the definition of YT (see (5.122)), there exists a sequence {zn }n≥1 ⊆ Y so that χω Φ(T , ·)∗ zn → ψ strongly in L1 (0, T ; U ). Let {Tk }k≥1 ⊆ (0, T ) be such that Tk see that there exists  zk ∈ Y so that

(5.160)

T . For each k ∈ N+ , by Theorem 1.20, we

Φ(T , Tk )∗ zn →  zk as n → +∞.

(5.161)

zk+ for each k,  ∈ N+ .  zk = Φ(Tk+ , Tk )∗

(5.162)

This yields that

For each k ≥ 1, we define the following function: zk , t ∈ [0, Tk ]. ϕ(t)  Φ(Tk , t)∗

(5.163)

By (5.162) and (5.163), we see that ϕ ∈ C([0, T ); Y ) is well defined. Moreover, it follows from (5.161) and (5.163) that χω Φ(T , ·)∗ zn → χω ϕ(·) strongly in L1 (0, Tk ; U ). This, together with (5.160), implies that ψ(·) = χω ϕ(·) ∈ L1 (0, T ; U ), which, along with (5.163) and the definition of ZT (see (5.122)), leads to (5.159). We next show that ZT ⊆ YT .

(5.164)

Observe that ψ ∈ C([0, T ); L2 (Ω)) ∩ L1 (0, T ; L2 (ω)) solves the equation:  ∂t ψ + Δψ − (a1 (x) + a2 (t))ψ = 0 in Ω × (0, T ), (5.165) ψ =0 on ∂Ω × (0, T ) if and only if ϕ ∈ C([0, T ); L2 (Ω)) ∩ L1 (0, T ; L2 (ω)) solves 

∂t ϕ + Δϕ − a1 (x)ϕ = 0 in Ω × (0, T ), ϕ=0 on ∂Ω × (0, T ),

where the function ϕ is defined by " T # ϕ(x, t)  exp a2 (s)ds ψ(x, t), (x, t) ∈ Ω × (0, T ). t

(5.166)

(5.167)

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5 Equivalence of Several Kinds of Optimal Controls

 ∈ ZT , let  . Let {Tk }k≥1 ⊆ (0, T ) Given χω ψ ϕ be given by (5.167) where ψ = ψ be such that Tk T . Write ϕk for the solution of the Equation (5.166) with the terminal condition ϕk (T ) =  ϕ (Tk ) (which belongs to L2 (Ω)). Let ψk be given by (5.167) where ϕ = ϕk . Then, ψk ∈ C([0, T ]; L2 (Ω)) solves (5.165). We claim that  strongly in L1 (0, T ; L2 (ω)). χω ψk → χω ψ

(5.168)

 ∈ YT , which leads to (5.164). When (5.168) is proved, we get that χω ψ The remainder is to show (5.168). Clearly, (5.168) is equivalent to χω ϕ k → χω  ϕ strongly in L1 (0, T ; L2 (ω)).

(5.169)

Let  ϕ satisfy that 

∂t  ϕ + Δ ϕ − a1 (x) ϕ = 0 in Ω × (−T , T ), ϕ=0  on ∂Ω × (−T , T )

(5.170)

ϕ (x, t) =   ϕ (x, t), (x, t) ∈ Ω × (0, T ).

(5.171)

     ϕ ∈ C [−T , T ); L2 (Ω) ∩ L1 − T , T ; L2 (ω) .

(5.172)

and

It is clear that

Because the equations satisfied by  ϕ and ϕk are time invariant, one can easily check that ϕ (t − (T − Tk )), when t ∈ (0, T ). ϕk (t) = 

(5.173)

By (5.172), we see that given ε ∈ (0, +∞), there are two positive constants δ(ε) and η(ε) = η(ε, δ(ε)) so that ϕ L1 (a,b;L2(ω)) ≤ ε, when (a, b) ⊆ (−T , T ), |a − b| ≤ δ(ε) χω 

(5.174)

and    ϕ (a) −  ϕ (b)L2 (Ω) ≤ ε, when (a, b) ⊆ − T , T − δ(ε) , |a − b| ≤ η(ε). (5.175) Let k0 = k0 (ε) verify that 0 < T − Tk ≤ η(ε), when k ≥ k0 .

(5.176)

5.4 Maximal Time, Minimal Norm, and Optimal Target Controls

253

From (5.171) and (5.173) it follows that ϕ L1 (0,T ;L2 (ω)) χω ϕk − χω    ϕ (· − (T − Tk )) −  ≤ χω  ϕ (·) L1 (0,T −δ(ε);L2 (ω)) +χω  ϕ L1 (T −δ(ε),T ;L2 (ω)) + χω  ϕ (· − (T − Tk ))L1 (T −δ(ε),T ;L2 (ω)) . This, along with (5.175), (5.176), and (5.174), yields that χω ϕk − χω  ϕ L1 (0,T ;L2 (ω)) ≤ (T − δ(ε))ε + 2ε ≤ (T + 2)ε, when k ≥ k0 , which leads to (5.169), as well as (5.168). Hence, (5.164) is proved. Finally, by (5.159) and (5.164), we obtain the conclusion (b). Hence, this example can be put into the framework (A3 ) (given at the beginning of Section 5.3). We now end Example 5.4.

5.4 Maximal Time, Minimal Norm, and Optimal Target Controls In this section, we will study the equivalence among the maximal time control QS ,QE , an optimal target control problem and a minimal norm control problem (T P )max problem under the following framework (A4 ): (i) The state space Y and the control space U are real separable Hilbert spaces. (ii) Let T ∈ (0, +∞). Let A : D(A) ⊆ Y → Y generate a C0 semigroup {eAt }t ≥0 on Y . Let D(·) ∈ L1 (0, T ; L (Y )) and B ∈ L (U, Y ). The controlled system is as: y(t) ˙ = Ay(t) + D(t)y(t) + Bu(t),

t ∈ (0, T ).

(5.177)

Given τ ∈ [0, T ), y0 ∈ Y and u ∈ L∞ (τ, T ; U ), write y(·; τ, y0 , u) for the solution of (5.177) over [τ, T ], with the initial condition that y(τ ) = y0 . Moreover, write {Φ(t, s) : t ≥ s ≥ 0} for the evolution system generated by A + D(·) over Y . (iii) Let U = BM (0), where BM (0) is the closed ball in U centered at 0 and of radius M ≥ 0. (When M = 0, we agree that BM (0)  {0}.) (iv) Let QS = {(t, y(t; 0, y0, 0)) : t ∈ [0, T )} and QE = {(T , Br (zd ))}, where r ∈ [0, +∞), and where y0 ∈ Y and zd ∈ Y are arbitrarily fixed so that rT  y(T ; 0, y0, 0) − zd Y > 0.

(5.178)

(v) The following unique continuation property holds: If there exist two constants a, b ∈ (0, T ) and z ∈ Y so that B ∗ Φ(T , t)∗ z = 0 for each t ∈ [a, b], then z = 0 in Y .

254

5 Equivalence of Several Kinds of Optimal Controls Q ,QE

S In this section, we simply write (T P )max

for (T P )rM which reads as:

 τ (M, r)  sup{τ ∈ [0, T )  y(T ; τ, y(τ ; 0, y0, 0), u) ∈ Br (zd ) and u ∈ L(τ, T ; U)}. (5.179)

(T P )rM

We next consider the following problem:   τ (M, r)  sup{τ ∈ [0, T )  y(T ; 0, y0, χ(τ,T ) u) ∈ Br (zd )

(T2 P )rM

and u ∈ UM,τ },

(5.180)

where  UM,τ  {u ∈ L∞ (0, T ; U )  u(t)U ≤ M a.e. t ∈ (τ, T )}. In this problem, • we call (0, y0 , T , χ(τ,T ) u) an admissible tetrad, if τ ∈ [0, T ), u ∈ UM,τ and y(T ; 0, y0 , χ(τ,T ) u) ∈ Br (zd ); • we call (0, y0 , T , χ( u∗ ) an optimal tetrad, if τ (M,r),T )  τ (M, r) ∈ [0, T ), u∗ ∈ UM, u∗ ) ∈ Br (zd ); τ (M,r) and y(T , 0, y0 , χ( τ (M,r),T ) u∗ ) is an optimal tetrad,  τ (M, r) and  u∗ are called the • when (0, y0 , T , χ( τ (M,r),T ) optimal time and an optimal control, respectively. We can easily check the following three facts: • τ (M, r) =  τ (M, r); • The supreme in (5.179) can be reached if and only if the supreme in (5.180) can be reached; • The zero extension of each optimal control to (T P )rM over (0, T ) is an optimal control to (T2 P )rM , while the restriction of each optimal control to (T2 P )rM over r (τ (M, r), T ) is an optimal control to (T P )M . Based on the above-mentioned facts, we can treat (T P )rM and (T2 P )rM as the same problem. We now introduce the corresponding optimal target control problems (OP )τM , with M ≥ 0 and τ ∈ [0, T ), in the following manner: (OP )τM

 r(M, τ )  inf{y(T ; 0, y0 , χ(τ,T ) u) − zd Y  u ∈ UM,τ }. (5.181)

In the problem (OP )τM , u∗ is called an optimal control, if u∗ ∈ UM,τ and y(T ; 0, y0, χ(τ,T ) u∗ ) − zd Y = r(M, τ ).

5.4 Maximal Time, Minimal Norm, and Optimal Target Controls

255

We next introduce the corresponding minimal norm control problem (NP )rτ , with τ ∈ [0, T ) and r ∈ [0, +∞), in the following manner: (NP )rτ

 M(τ, r)  inf{vL∞ (τ,T ;U )  y(T ; 0, y0, χ(τ,T ) v) ∈ Br (zd ) and v ∈ L∞ (0, T ; U )}. (5.182)

In the problem (NP )rτ , • we call v ∈ L∞ (0, T ; U ) an admissible control, if y(T ; 0, y0, χ(τ,T ) v) ∈ Br (zd ); • we call v ∗ an optimal control, if y(T ; 0, y0, χ(τ,T ) v ∗ ) ∈ Br (zd ) and v ∗ L∞ (τ,T ;U ) = M(τ, r); • when the set in the right-hand side of (5.182) is empty, we agree that M(τ, r) = +∞. The purpose of this section is to build up connections among (T P )rM , (OP )τM and (NP )rτ . We now give the definition about the equivalence for these problems. Definition 5.4 Let M ≥ 0, τ ∈ [0, T ), and r ∈ (0, +∞). (i) Problems (T P )rM and (OP )τM are said to be equivalent if (T2 P )rM and (OP )τM have the same optimal controls and (τ, r) = ( τ (M, r), r(M, τ )). P )rM and (NP )rτ (ii) Problems (T P )rM and (NP )rτ are said to be equivalent if (T2 have the same optimal controls and (M, τ ) = (M(τ, r), τ (M, r)). (iii) Problems (OP )τM and (NP )rτ are said to be equivalent if they have the same optimal controls and (M, r) = (M(τ, r), r(M, τ )). The main result of this section is presented as follows: Theorem 5.6 The following conclusions are true: ) (i) Given τ ∈ [0, T ) and M ∈ (0, M(τ, 0)), problems (OP )τM , (T P )r(M,τ , and M r(M,τ ) (NP )τ are equivalent; (ii) Given τ ∈ [0, T ) and r ∈ (0, rT ), problems (NP )rτ , (OP )τM(τ,r) , and (T P )rM(τ,r) are equivalent; (iii) Given M ∈ (0, +∞) and r ∈ [r(M, 0), rT ) ∩ (0, rT ), problems (T P )rM ,  τ (M,r) r (NP ) are equivalent. τ (M,r) and (OP )M

256

5 Equivalence of Several Kinds of Optimal Controls

Remark 5.3 The proof will be given in Section 5.4.4. Throughout this section, we agree that the characteristic function of the empty set is identically zero, i.e., χ∅  0.

5.4.1 Existence of Optimal Controls for These Problems In this subsection, we will discuss the existence of optimal controls to problems (OP )τM , (NP )rτ and (T2 P )rM . We start with introducing the following preliminary lemma: Lemma 5.10 The following two conclusions are true: (i) Let τ1 and τ2 be so that T > τ2 > τ1 ≥ 0. Let yd ∈ Y . Then given ε ∈ (0, +∞), there is Cε  C(T , τ1 , τ2 , yd , ε) ∈ (0, +∞) so that for each τ ∈ [τ1 , τ2 ], 

T

|yd , zY | ≤ Cε

B ∗ Φ(T , t)∗ zU dt + εzY for all z ∈ Y. (5.183)

τ

(ii) For each τ ∈ [0, T ), yd ∈ Y and r ∈ (0, +∞), there exists u ∈ L∞ (0, T ; U ) so that y(T ; 0, y0, χ(τ,T ) u) ∈ Br (yd ). Proof The proof of this lemma is similar to that of Lemma 5.4; we omit the details.   The following theorem is concerned with the existence of optimal controls to P )rM . (OP )τM , (NP )rτ and (T2 Theorem 5.7 The following conclusions are true: (i) For any M ≥ 0 and τ ∈ [0, T ), (OP )τM has optimal controls. (ii) (a) For any τ ∈ [0, T ) and r ∈ (0, +∞), (NP )rτ has optimal controls and M(τ, r) ≤ M(τ, 0); (b) For any τ ∈ [0, T ) and r ∈ (0, rT ), it holds that M(τ, r) ∈ (0, +∞). (Here, rT is given by (5.178).) (iii) The problem (T2 P )rM , with r ∈ (0, rT ) and M ≥ 0, has optimal controls if and only if M ≥ M(0, r). Proof We prove the conclusions one by one. (i) Arbitrarily fix M ≥ 0 and τ ∈ [0, T ). By the definition of r(M, τ ) (see (5.181)), there is a sequence {uk }k≥1 ⊆ UM,τ so that y(T ; 0, y0 , χ(τ,T ) uk ) − zd Y → r(M, τ ) as k → +∞.

(5.184)

Meanwhile, since {χ(τ,T ) uk }k≥1 is bounded in L∞ (0, T ; U ), there exists a subsequence of {uk }k≥1 , denoted in the same manner, and  u ∈ L∞ (0, T ; U ) so that

5.4 Maximal Time, Minimal Norm, and Optimal Target Controls

χ(τ,T ) uk → χ(τ,T ) u weakly star in L∞ (0, T ; U )

257

(5.185)

and  uL∞ (τ,T ;U ) ≤ lim inf uk L∞ (τ,T ;U ) ≤ M. k→+∞

From the above, we see that  u ∈ UM,τ . By (5.185), we obtain that y(T ; 0, y0, χ(τ,T ) uk ) → y(T ; 0, y0 , χ(τ,T ) u) weakly in Y. This, together with (5.184) and the definition of r(M, τ ), yields that u) − zd Y r(M, τ ) ≤ y(T ; 0, y0, χ(τ,T ) ≤ lim inf y(T ; 0, y0, χ(τ,T ) uk ) − zd Y = r(M, τ ). k→+∞

u is an optimal control to Since  u ∈ UM,τ , it follows from the above that  (OP )τM . (ii) We first show (a). Arbitrarily fix τ ∈ [0, T ) and r ∈ (0, +∞). By (ii) of Lemma 5.10 and the definition of M(τ, r) (see (5.182)), there is a sequence {uk }k≥1 ⊆ L∞ (0, T ; U ) so that y(T ; 0, y0, χ(τ,T ) uk ) − zd Y ≤ r and

lim uk L∞ (τ,T ;U ) = M(τ, r).

k→+∞

(5.186) Then there exists a subsequence of {uk }k≥1 , denoted in the same manner, and  u ∈ L∞ (0, T ; U ) so that χ(τ,T ) uk → χ(τ,T ) u weakly star in L∞ (0, T ; U )

(5.187)

 uL∞ (τ,T ;U ) ≤ lim inf uk L∞ (τ,T ;U ) = M(τ, r).

(5.188)

and k→+∞

By (5.187), we have that u) weakly in Y. y(T ; 0, y0, χ(τ,T ) uk ) → y(T ; 0, y0 , χ(τ,T ) From this and the inequality in (5.186), we see that u) − zd Y ≤ lim inf y(T ; 0, y0 , χ(τ,T ) uk ) − zd Y ≤ r. y(T ; 0, y0, χ(τ,T ) k→+∞

This, together with (5.188), implies that  u is an optimal control to (NP )rτ . Besides, by (5.182), we can easily check that M(τ, r) ≤ M(τ, 0).

258

5 Equivalence of Several Kinds of Optimal Controls

We next show (b). Arbitrarily fix τ ∈ [0, T ) and r ∈ (0, rT ). By (a) of this theorem, we see that M(τ, r) < +∞. Seeking for a contradiction, we suppose that M(τ, r) = 0 for some r ∈ (0, rT ). Then we would have that y(T ; 0, y0 , 0) − zd Y ≤ r < rT , which, along with (5.178), leads to a contradiction. (iii) Arbitrarily fix r ∈ (0, rT ) and M ≥ 0. We first show that M ≥ M(0, r) ⇒ (T2 P )rM has optimal controls. By (ii) of this theorem, (NP )r0 has an optimal control v ∗ . By the optimality of v ∗ , we see that v ∗ L∞ (0,T ;U ) = M(0, r) ≤ M and y(T ; 0, y0, v ∗ ) ∈ Br (zd ). From these, it follows that (T2 P )rM has admissible controls. Then by the definition of  τ (M, r), there are two sequences {tk }k≥1 ⊆ [0, T ) and {uk }k≥1 ⊆ UM,tk so that lim tk =  τ (M, r) and y(T ; 0, y0, χ(tk ,T ) uk ) ∈ Br (zd ).

k→+∞

(5.189)

Furthermore, there is a subsequence of {uk }k≥1 , denoted in the same manner, and a control  u ∈ L∞ (0, T ; U ) so that χ(tk ,T ) uk → χ( u weakly star in L∞ (0, T ; U ). τ (M,r),T ) Since {uk }k≥1 ⊆ UM,tk , the latter conclusion implies that y(T ; 0, y0, χ(tk ,T ) uk ) → y(T ; 0, y0, χ( u) weakly in Y τ (M,r),T )

(5.190)

and uL∞ (0,T ;U ) ≤ lim inf χ(tk ,T ) uk L∞ (0,T ;U ) ≤ M. χ( τ (M,r),T ) k→+∞

(5.191)

From (5.190), (5.191), and the second conclusion of (5.189), we find that u) ∈ Br (zd ) and  u ∈ UM, y(T ; 0, y0 , χ( τ (M,r),T ) τ (M,r) .

(5.192)

5.4 Maximal Time, Minimal Norm, and Optimal Target Controls

259

Since r ∈ (0, rT ), by the first conclusion of (5.192) and (5.178), we see that  τ (M, r) ∈ [0, T ). This, together with (5.192), yields that  u is an optimal control to (T2 P )rM . We next show that (T2 P )rM has optimal controls ⇒ M ≥ M(0, r). P )rM . Then For this purpose, we assume that u∗ is an optimal control to (T2 ∗ ∗  τ (M, r) ∈ [0, T ), y(T ; 0, y0 , χ( τ (M,r),T ) u ) ∈ Br (zd ) and u ∈ UM, τ (M,r) . (5.193) ∗ By the second conclusion of (5.193), we have that χ( τ (M,r),T ) u is an r admissible control to (NP )0 . Thus, it follows from the optimality of M(0, r) and the third conclusion of (5.193) that

M(0, r) ≤ M.  

Hence, we complete the proof of Theorem 5.7.

5.4.2 Connections Between Problems of Optimal Target Control and Minimal Norm Control This subsection presents connections between (OP )τM and (NP )rτ . These will be used in the proof of Theorem 5.6. We begin with studying (OP )τM . Lemma 5.11 Let (τ, M) ∈ [0, T ) × [0, M(τ, 0)). Then the following conclusions are true: (i) r(M, τ ) ∈ (0, +∞). (ii) u∗ is an optimal control for (OP )τM if and only if u∗ ∈ UM,τ and 

T τ

∗ ∗

∗

B ϕ (t), u (t)U dt =



T

max

v(·)∈UM,τ

B ∗ ϕ ∗ (t), v(t)U dt,

(5.194)

τ

where ϕ ∗ (t)  Φ(T , t)∗ (zd − y(T ; 0, y0, χ(τ,T ) u∗ )),

t ∈ [0, T ].

(iii) Each optimal control u∗ to (OP )τM satisfies that u∗ L∞ (a,b;U ) = M for any interval (a, b) ⊆ (τ, T ).

(5.195)

260

5 Equivalence of Several Kinds of Optimal Controls

Proof Arbitrarily fix (τ, M) ∈ [0, T ) × [0, M(τ, 0)). We will prove the conclusions (i), (ii), and (iii) one by one. (i) According to (i) of Theorem 5.7, there is a control u ∈ L∞ (0, T ; U ) so that y(T ; 0, y0, χ(τ,T ) u) − zd Y = r(M, τ ) and uL∞ (τ,T ;U ) ≤ M. Hence, r(M, τ ) < +∞. Meanwhile, if r(M, τ ) = 0, then by the definition of M(τ, 0), we obtain that M ≥ uL∞ (τ,T ;U ) ≥ M(τ, 0), which contradicts the assumption that M ∈ [0, M(τ, 0)). Thus, r(M, τ ) ∈ (0, +∞). (ii) Arbitrarily fix u1 , u2 ∈ UM,τ and ε ∈ [0, 1]. Set ⎧ ⎨ uε  (1 − ε)u1 + εu2 ; y (t)  y(t; 0, y0 , χ(τ,T ) ui ), t ∈ [0, T ], i = 1, 2; ⎩ i yε (t)  y(t; 0, y0 , χ(τ,T ) uε ), t ∈ [0, T ]. We can easily check that yε (T ) − zd 2Y − y1 (T ) − zd 2Y = ε2 y(T ; 0, 0, χ(τ,T ) (u2 − u1 ))2Y  T +2ε u2 − u1 , B ∗ Φ(T , t)∗ (y1 (T ) − zd )U dt.

(5.196)

τ

We now show the necessity. Let u∗ be an optimal control to (OP )τM . Arbitrarily take u ∈ UM,τ . By choosing u1 = u∗ and u2 = u in (5.196), we find that 

T

u−u∗ , B ∗ Φ(T , t)∗ (zd −y(T ; 0, y0, χ(τ,T ) u∗ ))U dt ≤ 0 for all u ∈ UM,τ .

τ

This implies (5.194). We next show the sufficiency. Suppose that u∗ ∈ UM,τ satisfies (5.194). Arbitrarily fix u ∈ UM,τ . Then by (5.194), and by taking u1 = u∗ and u2 = u in (5.196), we see that yε (T ) − zd 2Y − y(T ; 0, y0 , χ(τ,T ) u∗ ) − zd 2Y = ε2 y(T ; 0, 0, χ(τ,T ) (u − u∗ ))2Y  T +2ε u − u∗ , B ∗ Φ(T , t)∗ (y(T ; 0, y0, χ(τ,T ) u∗ ) − zd )U dt ≥ 0.

τ

(5.197)

5.4 Maximal Time, Minimal Norm, and Optimal Target Controls

261

Choosing ε = 1 in (5.197), we obtain that y(T ; 0, y0 , χ(τ,T ) u)−zd 2Y ≥ y(T ; 0, y0 , χ(τ,T ) u∗ )−zd 2Y for all u ∈ UM,τ . This implies that u∗ is an optimal control to (OP )τM . (iii) By contradiction, suppose that the conclusion (iii) was not true. Then there would be an optimal control  u∗ to (OP )τM and ( a,  b) ⊆ (τ, T ) so that  u∗ L∞ (a, b;U ) < M.

(5.198)

u∗ . Then by the same According to (ii) of this lemma, (5.194) holds for u∗ =  way as that used in the proof of Theorem 4.1, we can verify that B ∗  ϕ ∗ (t), u∗ (t)U = maxB ∗  ϕ ∗ (t), vU for a.e. t ∈ (τ, T ), v∈U

(5.199)

where u∗ )) for each t ∈ [0, T ]. ∗ (t)  Φ(T , t)∗ (zd − y(T ; 0, y0 , χ(τ,T ) ϕ This, along with (5.198), yields that ϕ ∗ (t) = 0 for each t ∈ [ a,  b], B ∗ which, combined with (v) of (A4 ) (given at the beginning of Section 5.4), indicates that u∗ ) = zd . y(T ; 0, y0, χ(τ,T ) This contradicts the conclusion (i) of this lemma. Therefore, (5.195) is true. In summary, we end the proof of Lemma 5.11.   Remark 5.4 Several notes are given in order. (i) We mention that M(τ, 0) ∈ (0, +∞] for each τ ∈ [0, T ) (see (5.182) and (5.178)). (ii) The conclusion (ii) in Lemma 5.11 still stands without the assumption that M < M(τ, 0). That is, a control u ∈ UM,τ satisfying (5.194) must be an optimal control to (OP )τM . This can be observed from the proof of the conclusion (ii) in Lemma 5.11. (iii) To ensure the conclusion (i) in Lemma 5.11, it is necessary to have that 0 ≤ M < M(τ, 0). Lemma 5.12 Let τ ∈ [0, T ). Then the following conclusions are true: (i) The map r(·, τ ) is strictly decreasing and homeomorphic from [0, M(τ, 0)) onto (0, rT ]. (ii) For each r ∈ (0, +∞), it stands that M(τ, r) < M(τ, 0).

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5 Equivalence of Several Kinds of Optimal Controls

(iii) The inverse of r(·, τ ) is the map M(τ, ·), i.e., r = r(M(τ, r), τ ) when r ∈ (0, rT ],

(5.200)

M = M(τ, r(M, τ )) when M ∈ [0, M(τ, 0)).

(5.201)

and

Proof We prove the conclusions one by one. (i) The proof of the conclusion (i) is divided into the following three steps: Step 1. We show that the map r(·, τ ) is strictly decreasing. Let 0 ≤ M1 < M2 < M(τ, 0). We claim that r(M1 , τ ) > r(M2 , τ ). Otherwise, we would have that r(M1 , τ ) ≤ r(M2 , τ ). Then by (i) of Theorem 5.7, there is an optimal control u∗1 ∈ UM1 ,τ to (OP )τM1 so that y(T ; 0, y0, χ(τ,T ) u∗1 ) − zd Y = r(M1 , τ ) ≤ r(M2 , τ ) and u∗1 L∞ (τ,T ;U ) ≤ M1 < M2 .

(5.202)

These imply that u∗1 is also an optimal control to (OP )τM2 . From this and (iii) of Lemma 5.11 it follows that u∗1 (t)L∞ (τ,T ;U ) = M2 . This contradicts (5.202). Step 2. We prove that the map r(·, τ ) is Lipschitz continuous. Let 0 ≤ M1 < M2 < M(τ, 0). According to (i) of Theorem 5.7, (OP )τM2 has an optimal control u∗2 . Then it follows that  T



Φ(T , t)∗ Bu∗2 dt − zd r(M2 , τ ) = Φ(T , 0)y0 + Y τ  T

# "M 1 ∗

∗ ≥ Φ(T , 0)y0 + Φ(T , t) B u2 dt − zd Y M 2  Tτ

M2 − M1

∗ ∗ − Φ(T , t) Bu2 dt .

Y M2 τ

(5.203)

5.4 Maximal Time, Minimal Norm, and Optimal Target Controls

263

Since M1 u∗2 /M2 ∈ UM1 ,τ , by (5.203) and the optimality of r(M1 , τ ), we get that r(M2 , τ ) ≥ r(M1 , τ )−(M2 −M1 )(T −τ ) sup Φ(T , t)L (H,H ) BL (U,H ) . 0≤t ≤T

The above inequality, together with the result of Step 1, yields that |r(M1 , τ )−r(M2 , τ )| ≤ (T −τ )|M1 −M2 | sup Φ(T , t)L (H,H ) BL (U,H ) . 0≤t ≤T

Step 3. We show that the map r(·, τ ) : [0, M(τ, 0)) → (0, rT ] is surjective and homeomorphic. By Steps 1–2, we only need to show that r(0, τ ) = rT and

lim

M→M(τ,0)

r(M, τ ) = 0.

(5.204)

The first equality in (5.204) follows from the definition of rT (see (5.178)). We now show the second equality in (5.204). About M(τ, 0), we have that either M(τ, 0) < +∞ or M(τ, 0) = +∞. In the case that M(τ, 0) < +∞, we arbitrarily fix ε ∈ (0, +∞). By the definition of M(τ, 0), there exists uε ∈ L∞ (0, T ; U ) so that y(T ; 0, y0 , χ(τ,T ) uε ) = zd and uε L∞ (τ,T ;U ) ≤ M(τ, 0) + ε. The inequality in (5.205) implies that the optimality of r(M, τ ), yields that

M M(τ,0)+ε uε

"

r(M, τ ) ≤ y T ; 0, y0, χ(τ,T )

(5.205)

∈ UM,τ . This, together with

# M

uε − zd , Y M(τ, 0) + ε

which, combined with (5.205), indicates that



r(M, τ ) ≤

T 0

|M(τ, 0) + ε − M|

Φ(T , t)Bχ(τ,T ) uε (t)dt Y M(τ, 0) + ε

≤ (T − τ )|M(τ, 0) + ε − M|BL (U,H ) sup Φ(T , t)L (H ) . 0≤t ≤T

Sending ε → 0+ in the above inequality, we obtain that r(M, τ ) ≤ (T − τ )|M(τ, 0) − M|BL (U,H ) sup Φ(T , t)L (H ) . 0≤t ≤T

This implies the second equality in (5.204) in the first case. In the case that M(τ, 0) = +∞, we use (ii) of Lemma 5.10 to find that for each ε ∈ (0, +∞), there exists uε ∈ L∞ (0, T ; U ) so that

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5 Equivalence of Several Kinds of Optimal Controls

y(T ; 0, y0, χ(τ,T ) uε ) ∈ Bε (zd ).

(5.206)

Set Mε  uε L∞ (τ,T ;U ) . It follows from (5.206) and the optimality of r(Mε , τ ) that r(Mε , τ ) ≤ y(T ; 0, y0 , χ(τ,T ) uε ) − zd Y ≤ ε, which, combined with the monotonicity of r(·, τ ) over [0, M(τ, 0)), indicates the second equality in (5.204) for the second case. So we have proved (5.204). Thus, we finish the proof of the conclusion (i). (ii) Let r ∈ (0, +∞) and r0  min(r, rT ). By (i) of this lemma and (i) of Theorem 5.7, we can find M0 ∈ [0, M(τ, 0)) and an optimal control u0 ∈ UM0 ,τ to (OP )τM0 so that y(T ; 0, y0, χ(τ,T ) u0 ) − zd Y = r(M0 , τ ) = r0 ≤ r.

(5.207)

From (5.207) and the optimality of M(τ, r), we see that M(τ, r) ≤ u0 L∞ (τ,T ;U ) ≤ M0 < M(τ, 0) for all r ∈ (0, +∞). Hence, the conclusion (ii) is true. (iii) First of all, it follows by (5.182) and (5.178) that (5.200) holds for r = rT . We next show that (5.200) for an arbitrarily fixed r ∈ (0, rT ). To this end, we let  Ar  {M ∈ [0, M(τ, 0))  r(M, τ ) > r} and  Br  {M ∈ [0, M(τ, 0))  r(M, τ ) ≤ r}. From (i) of this lemma, we see that Ar and Br are nonempty and disjoint subsets of [0, M(τ, 0)), Ar ∪ Br = [0, M(τ, 0)) and sup Ar = inf Br  m.

(5.208)

m = M(τ, r).

(5.209)

We now claim that

Indeed, according to (ii) of Theorem 5.7 and (ii) of this lemma, (NP )rτ has an optimal control u1 . Then by the optimality of u1 , we have that y(T ; 0, y0 , χ(τ,T ) u1 )−zd Y ≤ r and u1 L∞ (τ,T ;U ) = M(τ, r) < M(τ, 0).

5.4 Maximal Time, Minimal Norm, and Optimal Target Controls

265

From these and the definition of r(M(τ, r), τ ), we see that r(M(τ, r), τ ) ≤ r, which indicates that M(τ, r) ∈ Br and m ≤ M(τ, r).

(5.210)

Meanwhile, for each M ∈ (m, M(τ, 0)), we have that M ∈ Br . Then according to (i) of Theorem 5.7, (OP )τM has an optimal control u2 . By the optimality of u2 , we have that y(T ; 0, y0 , χ(τ,T ) u2 ) − zd Y = r(M, τ ) ≤ r and u2 L∞ (τ,T ;U ) ≤ M. These, together with the definition of M(τ, r), imply that M(τ, r) ≤ M, from which, it follows that M(τ, r) ≤ m.

(5.211)

Now, (5.209) follows from (5.210) and (5.211) immediately. Then from (5.208), (5.209), and (i) of this lemma, we see that (5.200) holds for the case when r ∈ (0, rT ). To prove (5.201), we let M ∈ [0, M(τ, 0)). By (i) of this lemma, we see that 0 < r(M, τ ) ≤ rT . Thus, about r(M, τ ), we have that either r(M, τ ) ∈ (0, rT ) or r(M, τ ) = rT . In the case where r(M, τ ) ∈ (0, rT ), it is clear that M ∈ Br(M,τ ) , which, combined with (5.208) and (5.209), indicates that M(τ, r(M, τ )) = inf Br(M,τ ) ≤ M. If M(τ, r(M, τ )) < M, then by the strictly decreasing property of r(·, τ ), we have that  τ ) > r(M, τ ) for each M  ∈ (M(τ, r(M, τ )), M). r(M,  ∈ Ar(M,τ ), which, together with (5.208) and (5.209), This implies that M  ≤ M(τ, r(M, τ )). This leads to a contradiction. Hence, (5.201) yields that M is true for the first case. In the case that r(M, τ ) = rT , we can use (i) of this lemma to find that M = 0.

(5.212)

Meanwhile, by (5.182) and (5.178), we obtain that M(τ, rT ) = 0. This, together with (5.212), yields that M = M(τ, r(M, τ )). Hence, (5.201) is true for the second case. Thus, we have proved (5.201). In summary, we complete the proof of Lemma 5.12.

 

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5 Equivalence of Several Kinds of Optimal Controls

We end this subsection with the following theorem concerning the connections between the optimal target control problem and the minimal norm control problem. Theorem 5.8 The following conclusions are true: (i) When τ ∈ [0, T ) and 0 ≤ M < M(τ, 0), each optimal control to (OP )τM is r(M,τ ) also an optimal control to (NP )τ . (ii) When τ ∈ [0, T ) and r ∈ (0, rT ], each optimal control to (NP )rτ is also an optimal control to (OP )τM(τ,r) . (iii) When τ ∈ [0, T ) and r ∈ (0, rT ], each optimal control u∗ to (NP )rτ satisfies that u∗ L∞ (a,b;U ) = M(τ, r) for any interval (a, b) ⊆ (τ, T ). Proof We prove the conclusions one by one. (i) Arbitrarily fix τ ∈ [0, T ) and 0 ≤ M < M(τ, 0). Then according to (i) of Theorem 5.7, (OP )τM has an optimal control u1 . By the optimality of u1 , it follows that y(T ; 0, y0, χ(τ,T ) u1 ) − zd Y = r(M, τ ) and u1 L∞ (τ,T ;U ) ≤ M< M(τ, 0). ) From these and (5.201), we see that u1 is an optimal control to (NP )r(M,τ . τ (ii) Arbitrarily fix τ ∈ [0, T ) and r ∈ (0, rT ]. Then according to (ii) of Theorem 5.7, (NP )rτ has an optimal control u2 . By the optimality of u2 , we have that

y(T ; 0, y0 , χ(τ,T ) u2 ) − zd Y ≤ r and u2 L∞ (τ,T ;U ) = M(τ, r). These, along with (5.200), yield that u2 is an optimal control to (OP )τM(τ,r) . (iii) The conclusion (iii) follows from (ii) of this theorem and (iii) of Lemma 5.11. Hence, we end the proof of Theorem 5.8.  

5.4.3 Connections Between Problems of Minimal Norm Control and Maximal Time Control This subsection presents connections between (NP )rτ and (T2 P )rM . We start with the following Lemma 5.13. Since its proof is very similar to that of Proposition 5.1, we omit the details here. Lemma 5.13 Let y0 ∈ Y and r ∈ (0, rT ). Let τ1 and τ2 satisfy that 0 ≤ τ1 < τ2 < T . Then the following two conclusions are true: (i) For each τ ∈ [τ1 , τ2 ], there exists zτ∗ ∈ Y \ {0} so that

5.4 Maximal Time, Minimal Norm, and Optimal Target Controls

 #2 1 1" T − M(τ, r)2 = B ∗ Φ(T , t)∗ zτ∗ U dt 2 2 τ +y0 , Φ(T , 0)∗ zτ∗ Y − inf q, zτ∗ Y .

267

(5.213)

q∈Br (zd )

(ii) There exists a constant C(T , y0 , τ1 , τ2 , zd , r) ∈ (0, +∞) so that zτ∗ (given in (5.213)) satisfies zτ∗ Y ≤ C(T , y0 , τ1 , τ2 , zd , r) for all τ ∈ [τ1 , τ2 ]. The next lemma presents some properties of the function M(·, r). Lemma 5.14 Let r ∈ (0, rT ). Then the following conclusions are true: (i) The function M(·, r) is strictly increasing and homeomorphic from [0, T ) onto [M(0, r), +∞). (ii) The inverse of M(·, r) is the function τ (·, r), i.e., M = M( τ (M, r), r) for every M ≥ M(0, r),

(5.214)

τ = τ (M(τ, r), r) for every τ ∈ [0, T ).

(5.215)

and

Proof (i) The proof is organized by four steps as follows: Step 1. We show that the function M(·, r) is strictly increasing. Let 0 ≤ τ1 < τ2 < T . It suffices to show M(τ1 , r) < M(τ2 , r).

(5.216)

Seeking for a contradiction, we suppose that M(τ1 , r) ≥ M(τ2 , r).

(5.217)

According to (ii) of Theorem 5.7, (NP )rτ2 has an optimal control u∗ . Then by the optimality of u∗ and (5.217), we have that y(T ; 0, y0 , χ(τ2 ,T ) u∗ ) ∈ Br (zd ) and u∗ L∞ (τ2 ,T ;U ) = M(τ2 , r) ≤ M(τ1 , r). From the above, we see that χ(τ2 ,T ) u∗ is an optimal control to (NP )rτ1 . Then it follows from (iii) of Theorem 5.8 and (ii) of Theorem 5.7 that χ(τ2 ,T ) u∗ L∞ (τ1 ,τ2 ;U ) = M(τ1 , r) ∈ (0, +∞), which leads to a contradiction, since χ(τ2 ,T ) u∗ (t) = 0 for a.e. t ∈ (τ1 , τ2 ). Hence, (5.216) is true and M(·, r) is strictly increasing.

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5 Equivalence of Several Kinds of Optimal Controls

Step 2. We prove that the function M(·, r) is left-continuous. Let 0 ≤ τ1 < τ2 < · · · < τk < · · · < τ < T and lim τk = τ . It suffices k→+∞

to show that lim M(τk , r) = M(τ, r).

(5.218)

k→+∞

Suppose that (5.218) was not true. Then by Step 1, we would have that lim M(τk , r) = M(τ, r) − δ for some δ ∈ (0, +∞).

k→+∞

(5.219)

Meanwhile, according to (ii) of Theorem 5.7, each (NP )rτk (with k ∈ N+ ) has an optimal control uk . Then we have that uk L∞ (τk ,T ;U ) = M(τk , r) < M(τ, r) and y(T ; 0, y0 , χ(τk ,T ) uk ) ∈ Br (zd ). (5.220) Since lim τk = τ , by the first conclusion in (5.220), there is a subsequence k→+∞

{uk }≥1 ⊆ {uk }k≥1 and a control  u ∈ L∞ (0, T ; U ) so that u weakly star in L∞ (0, T ; U ). χ(τk ,T ) uk → χ(τ,T )

(5.221)

This, along with the first conclusion in (5.220) and (5.219), implies that  uL∞ (τ,T ;U ) ≤ lim inf uk L∞ (τk ,T ;U ) = lim inf M(τk , r) = M(τ, r) − δ. →+∞

→+∞

(5.222) Next, by (5.221), we can easily check that y(T ; 0, y0, χ(τk ,T ) uk ) → y(T ; 0, y0, χ(τ,T ) u) weakly in Y. This, together with the second conclusion in (5.220), yields that u) ∈ Br (zd ). y(T ; 0, y0, χ(τ,T ) From this and the optimality of M(τ, r), we get that  uL∞ (τ,T ;U ) ≥ M(τ, r), which contradicts (5.222). Hence, (5.218) holds. Step 3. We show that the function M(·, r) is right-continuous. Let τ ∈ [0, T ). Take a sequence {τk }k≥1 ⊆ (τ, T ) so that lim τk = τ . By k→+∞

Lemma 5.13, we find that for each k ∈ N+ , there exists zk∗ ∈ Y with zk∗ Y ≤ C (where C is independent of k) so that

5.4 Maximal Time, Minimal Norm, and Optimal Target Controls

 #2 1 1" T ∗ − M(τk , r)2 = B Φ(T , t)∗ zk∗ U dt 2 2 τk +y0 , Φ(T , 0)∗ zk∗ Y − inf q, zk∗ Y .

269

(5.223)

q∈Br (zd )

Since {zk∗ }k≥1 is bounded in Y , there exists a subsequence {zk∗ }≥1 ⊆ {zk∗ }k≥1 and z ∈ Y so that zk∗ → z weakly in Y as  → +∞. This, together with (5.223), yields that  #2 1" T ∗ 1 B Φ(T , t)∗ zU dt − lim supM(τk , r)2 ≥ 2 →+∞ 2 τ +y0 , Φ(T , 0)∗ zY − inf q, zY .

(5.224)

q∈Br (zd )

Meanwhile, by (ii) of Theorem 5.7, (NP )rτ has an optimal control v ∗ . Then by the optimality of v ∗ , we have that y(T ; 0, y0 , χ(τ,T ) v ∗ ) ∈ Br (zd ) and v ∗ L∞ (τ,T ;U ) = M(τ, r). From this and (5.224) it follows that 1 − lim sup M(τk , r)2 ≥ 2 →+∞



T τ

1 v ∗ (t), B ∗ Φ(T , t)∗ zU dt − M(τ, r)2 2

+y0 , Φ(T , 0)∗ zY − y(T ; 0, y0 , χ(τ,T ) v ∗ ), zY 1 = − M(τ, r)2 , 2 which indicates that M(τ, r) ≥ lim sup M(τk , r). →+∞

This, together with the conclusion in Step 1, yields that the function M(·, r) is right-continuous over [0, T ). Step 4. We prove that the function M(·, r) : [0, T ) → [M(0, r), +∞) is surjective and homeomorphic. By the results in Steps 1–3, we see that we only need to show lim M(τ, r) = +∞.

τ →T

(5.225)

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5 Equivalence of Several Kinds of Optimal Controls

Seeking for a contradiction, suppose that (5.225) was not true. Then we would have that lim M(τ, r) ≤ N0 for some N0 ∈ (0, +∞).

τ →T

(5.226)

We take {τk }k≥1 ⊆ [0, T ) with τk → T . According to (ii) of Theorem 5.7, each (NP )rτk (with k ∈ N+ ) has an optimal control uk . By the optimality of uk , we have that y(T ; 0, y0, χ(τk ,T ) uk ) ∈ Br (zd ) and uk L∞ (τk ,T ;U ) = M(τk , r).

(5.227)

Then it follows from (5.226) and (5.227) that {χ(τk ,T ) uk }k≥1 is bounded in L∞ (0, T ; U ). Thus, there is a subsequence of {χ(τk ,T ) uk }k≥1, still denoted in the same way, so that χ(τk ,T ) uk → 0 strongly in L2 (0, T ; U ), from which, it follows that y(T ; 0, y0, χ(τk ,T ) uk ) → y(T ; 0, y0 , 0) strongly in Y. This, together with (5.178) and the first conclusion in (5.227), yields that rT = y(T ; 0, y0 , 0) − zd Y = lim y(T ; 0, y0 , χ(τk ,T ) uk ) − zd Y ≤ r, k→+∞

which leads to a contradiction since r < rT . Hence, (5.225) follows. Therefore, we have proved the conclusion (i) in Lemma 5.14. (ii) Let M ≥ M(0, r). Set   M  {τ ∈ [0, T )  M(τ, r) > M}. AM  {τ ∈ [0, T )  M(τ, r) ≤ M} and B M are nonempty and disjoint From (i) of this lemma, we see that AM and B   subsets of [0, T ), AM ∪ BM = [0, T ) and M   sup AM = inf B t.

(5.228)

 t = τ (M, r).

(5.229)

We now claim that

Indeed, according to (iii) of Theorem 5.7, (T2 P )rM has an optimal control u1 . Thus, we have that y(T ; 0, y0, χ( τ (M,r),T ) u1 ) ∈ Br (zd ) and u1 L∞ ( τ (M,r),T ;U ) ≤ M.

5.4 Maximal Time, Minimal Norm, and Optimal Target Controls

271

From these and the definition of M( τ (M, r), r), we obtain that M( τ (M, r), r) ≤ M. This, together with (5.228), yields that τ (M, r) ≤  t.  τ (M, r) ∈ AM and 

(5.230)

About  t, we have that either  t = 0 or  t > 0. In the case when  t = 0, we have that  t ≤ τ (M, r). This, along with the inequality in (5.230), yields (5.229) for this case. In the case that  t > 0, one can easily check that   τ ∈ AM , when τ ∈ 0,  t . (5.231)   Arbitrarily fix τ ∈ 0,  t . According to (ii) of Theorem 5.7, (NP )rτ has an optimal control u2 . From the optimality of u2 , (5.231) and the definition of AM , we find that y(T ; 0, y0 , χ(τ,T ) u2 ) ∈ Br (zd ) and u2 L∞ (τ,T ;U ) = M(τ, r) ≤ M. The above facts, combined with the definition of  τ (M, r), indicate that τ ≤  τ (M, r), which leads to  t ≤ τ (M, r), (5.232)   since τ was arbitrarily taken from 0,  t . Now, by (5.232) and the inequality in (5.230), we obtain (5.229) for the current case. Thus, we have proved (5.229). We next show that (5.214) and (5.215). It is clear that (5.214) follows from (5.228), (5.229), and (i) of this lemma. To prove (5.215), let τ ∈ [0, T ). It is obvious that τ ∈ AM(τ,r), which, combined with (5.228) and (5.229), indicates that  τ (M(τ, r), r) = sup AM(τ,r) ≥ τ. By contradiction, suppose that  τ (M(τ, r), r) > τ . Then by the strictly increasing property of M(·, r), we would have that M( τ , r) > M(τ, r) for all  τ ∈ (τ, τ (M(τ, r), r)). This implies that M(τ,r) .  τ ∈B From the latter, (5.228) and (5.229) it follows that  τ ≥ τ (M(τ, r), r), which leads to a contradiction. Hence, (5.215) is true. In summary, we end the proof of Lemma 5.14.

 

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5 Equivalence of Several Kinds of Optimal Controls

Lemma 5.15 Let  E1  {(M, r) ∈ (0, +∞) × (0, rT )  r ≥ r(M, 0)};  E2  {(M, r) ∈ (0, +∞) × (0, rT )  M ≥ M(0, r)}. Then the following conclusions are true: (i) It holds that E1 = E2 . (ii) When (M, r) ∈ E1 , M < M( τ (M, r), 0).

(5.233)

Proof We prove the conclusions one by one. (i) We first show that E1 ⊆ E2 .

(5.234)

To this end, we arbitrarily fix (M, r) ∈ E1 . Since r ≥ r(M, 0) and M(0, ·) is decreasing over [0, +∞), we have that M(0, r) ≤ M(0, r(M, 0)).

(5.235)

Meanwhile, according to (i) of Theorem 5.7, (OP )0M has an optimal control u∗ . Then we have that y(T ; 0, y0 , u∗ ) − zd Y = r(M, 0) and u∗ L∞ (0,T ;U ) ≤ M. From these, we find that u∗ is an admissible control to (NP )0 with the optimality of M(0, r(M, 0)), yields that

r(M,0)

. This, along

M(0, r(M, 0)) ≤ M. The above, together with (5.235), indicates that M ≥ M(0, r), which leads to (5.234). We next show that E2 ⊆ E1 .

(5.236)

5.4 Maximal Time, Minimal Norm, and Optimal Target Controls

273

For this purpose, we arbitrarily fix (M, r) ∈ E2 . Since r ∈ (0, rT ), it follows from (ii) of Theorem 5.7 and (5.200) that M(0, r) ∈ (0, +∞) and r = r(M(0, r), 0).

(5.237)

Because M ≥ M(0, r) and r(·, τ ) is decreasing over [0, +∞), it follows from (5.237) that M ≥ M(0, r) > 0 and r = r(M(0, r), 0) ≥ r(M, 0). From these, we see that (M, r) ∈ E1 . Hence, (5.236) is true. Finally, according to (5.234) and (5.236), the conclusion (i) is true. (ii) To show (5.233), we arbitrarily fix (M, r) ∈ E2 (which implies that r ∈ (0, rT ) and M ≥ M(0, r)). Then by (5.214) and (iii) of Theorem 5.7, we have that M = M( τ (M, r), r) and  τ (M, r) ∈ [0, T ).

(5.238)

Meanwhile, by Lemma 5.12, we see that M(τ, ·) is strictly decreasing over (0, rT ) for each τ ∈ [0, T ). This, together with (5.238), yields that M = M( τ (M, r), r) < M( τ (M, r), r/2), which, combined with (ii) of Theorem 5.7, indicates that M < M( τ (M, r), r/2) ≤ M( τ (M, r), 0). This yields (5.233). Hence, we end the proof of Lemma 5.15.

 

The connections between optimal norm and time optimal control problems are presented in the following theorem: Theorem 5.9 The following conclusions are true: (i) When τ ∈ [0, T ) and r ∈ (0, rT ), each optimal control to (NP )rτ is an optimal control to (T2 P )rM(τ,r). (ii) Suppose that either  (M, r) ∈ {(M, r) ∈ (0, +∞) × (0, rT )  r ≥ r(M, 0)} or  (M, r) ∈ {(M, r) ∈ (0, +∞) × (0, rT )  M ≥ M(0, r)}. r Then each optimal control to (T2 P )rM is an optimal control to (NP ) τ (M,r) .

274

5 Equivalence of Several Kinds of Optimal Controls

Proof (i) Arbitrarily fix τ ∈ [0, T ) and r ∈ (0, rT ). Then according to (ii) of Theorem 5.7, (NP )rτ has an optimal control u1 . By the optimality of u1 , we have that y(T ; 0, y0 , χ(τ,T ) u1 ) ∈ Br (zd ) and u1 L∞ (τ,T ;U ) = M(τ, r). From these and (5.215), we see that u1 is an optimal control to (T2 P )rM(τ,r). (ii) Arbitrarily fix (M, r) as required. Then by Lemma 5.15, we have that M ≥ M(0, r). Thus, according to (iii) of Theorem 5.7, (T2 P )rM has an optimal control u2 . By the optimality of u2 , we have that y(T ; 0, y0, χ( τ (M,r),T ) u2 ) ∈ Br (zd ) and u2 L∞ ( τ (M,r),T ;U ) ≤ M. r From these and (5.214), we see that u2 is an optimal control to (NP ) τ (M,r) . Thus we complete the proof of Theorem 5.9.  

5.4.4 Proof of the Main Result This subsection presents the proof of Theorem 5.6. Proof (Proof of Theorem 5.6) Throughout the proof, by (P1 ) = (P2 ), we mean that problems (P1 ) and (P2 ) are equivalent in the sense of Definition 5.4; by (P1 ) ⇒ (P2 ), we mean that each optimal control to (P1 ) is also an optimal control to (P2 ). We organize the proof by three steps as follows: Step 1. We show that r(M,τ )

(OP )τM = (T P )M

) = (NP )r(M,τ , τ

(5.239)

when τ ∈ [0, T ) and 0 < M < M(τ, 0). To this end, we arbitrarily fix (τ, M) as required. On one hand, it follows by (5.181) that r(M, τ ) ≥ r(M, 0). This, together with (i) of Lemma 5.11, Lemma 5.12, and Lemma 5.14, implies that τ ∈ [0, T ), M ∈ (0, M(τ, 0)), r(M, τ ) ∈ [r(M, 0), rT ) ∩ (0, rT )

(5.240)

and M = M(τ, r(M, τ )), τ =  τ (M(τ, r(M, τ )), r(M, τ )) =  τ (M, r(M, τ )). (5.241) Using Theorem 5.8 and (5.240), we obtain that ) ) and (NP )r(M,τ ⇒ (OP )τM(τ,r(M,τ )). (OP )τM ⇒ (NP )r(M,τ τ τ

5.4 Maximal Time, Minimal Norm, and Optimal Target Controls

275

From the above, the first conclusion in (5.241) and Definition 5.4, we see that ) . (OP )τM = (NP )r(M,τ τ

(5.242)

On the other hand, by Theorem 5.9 and (5.240), we obtain that ) ) (NP )r(M,τ ⇒ (T2 P )r(M,τ τ M(τ,r(M,τ ))

and (T2 P )M(τ,r(M,τ )) ⇒ (NP ) τ (M(τ,r(M,τ )),r(M,τ )). r(M,τ )

r(M,τ )

These, together with (5.241) and Definition 5.4, yield that ) ) (NP )r(M,τ = (T P )r(M,τ . τ M

(5.243)

Now, (5.239) follows from (5.242), (5.243), and Definition 5.4 at once. Step 2. We prove that (NP )rτ = (OP )τM(τ,r) = (T P )rM(τ,r) ,

(5.244)

when τ ∈ [0, T ) and r ∈ (0, rT ). For this purpose, we arbitrarily fix (τ, r) as required. Then by (ii) of Theorem 5.7, Lemma 5.12, and Lemma 5.14, one can easily check that τ ∈ [0, T ), r ∈ (0, rT ), M(τ, r) ∈ [M(0, r), M(τ, 0)) ∩ (0, M(τ, 0))

(5.245)

and r = r(M(τ, r), τ ), τ =  τ (M(τ, r), r).

(5.246)

From Theorem 5.8 and (5.245), we have that ) . (NP )rτ ⇒ (OP )τM(τ,r) and (OP )τM(τ,r) ⇒ (NP )r(M(τ,r),τ τ

These, along with (5.246) and Definition 5.4, yield that (NP )rτ = (OP )τM(τ,r) .

(5.247)

On the other hand, it follows by Theorem 5.9 and (5.245) that r (NP )rτ ⇒ (T2 P )rM(τ,r) and (T2 P )rM(τ,r) ⇒ (NP ) τ (M(τ,r),r).

These, along with (5.246) and Definition 5.4, yield that (NP )rτ = (T P )rM(τ,r) . Now, (5.244) follows from (5.247), (5.248), and Definition 5.4 at once.

(5.248)

276

5 Equivalence of Several Kinds of Optimal Controls

Step 3. We show that  τ (M,r)

r (T P )rM = (NP ) τ (M,r) = (OP )M

,

(5.249)

where M ∈ (0, +∞) and r ∈ [r(M, 0), rT ) ∩ (0, rT ). Arbitrarily fix (M, r) as required. Then by Lemma 5.15, (iii) of Theorem 5.7, Lemma 5.12, and Lemma 5.14, one can easily check that τ (M, r), 0)  τ (M, r) ∈ [0, T ), r ∈ [r(M, 0), rT ) ∩ (0, rT ), 0 < M < M( (5.250) and M = M( τ (M, r), r), r = r(M( τ (M, r), r), τ (M, r)) = r(M, τ (M, r)). (5.251) By Theorem 5.9 and (5.250), we obtain that r r 2 r (T2 P )rM ⇒ (NP ) τ (M,r) and (NP ) τ (M,r) ⇒ (T P )M( τ (M,r),r) .

These, along with (5.251) and Definition 5.4, yield that r (T P )rM = (NP ) τ (M,r) .

(5.252)

On the other hand, by Theorem 5.8, (5.250), and (5.251), one can easily verify that  τ (M,r)

r (NP ) τ (M,r) ⇒ (OP )M( τ (M,r),r)

and τ (M,r) r(M( τ (M,r),r), τ (M,r)) . (OP ) M( τ (M,r),r) ⇒ (NP ) τ (M,r)

From these, (5.251) and Definition 5.4, we obtain that  τ (M,r) r . (NP ) τ (M,r) = (OP )M

Now (5.249) follows from (5.252), (5.253), and Definition 5.4 at once. Thus, we end the proof of Theorem 5.6.

(5.253)  

Miscellaneous Notes To our best knowledge, the terminology “equivalence of minimal time and minimal norm control problems” was first formally presented in [17]. (The definition of the equivalence was hidden in Theorem 1.1 of [17].) The equivalence theorems

5.4 Maximal Time, Minimal Norm, and Optimal Target Controls

277

of different optimal control problems play important roles in the studies of these problems. With regard to their applications, we would like to mention the following facts: • With the aid of the equivalence theorem in Section 5.1, one can derive the norm optimality from the time optimality (see [5]); one can build up a necessary and sufficient condition for both the optimal time and the optimal control of a minimal time control problem, through analyzing characteristics for the corresponding minimal norm control problem (see [17]); one can present a numerical algorithm to compute the optimal time of a minimal time control problem (see [9]); one can study a certain approximation property for a minimal time control problem (see [20]); one can obtain some regularity of the Bellman function (see [6]), which is heavily related to the controllability and has been studied in, for instance, [2] and [10]. • By the equivalence theorem in Section 5.2, one can get algorithms for both the optimal time and the optimal controls of some minimal time control problems (see [11]). The same can be said about the equivalence theorem in Section 5.3. By the equivalence theorem in Section 5.4, one can get algorithms for both the optimal time and the optimal controls of some maximal time control problems (see [15]). • With the aid of the second equivalence theorem, one can design a feedback optimal norm control (see [15]). Materials in Section 5.1 are generalized from the related materials in [17]. Materials in Section 5.2 are developed from the materials in [11]. Materials in Section 5.3 are developed from the materials in [18]. Materials in Section 5.4 are summarized from the materials in [15]. More notes on the equivalence theorems are given in order: • The method to derive the equivalence theorem in Section 5.1 does not work for the equivalence theorem in Section 5.2, even when the system in Section 5.2 is time invariant. The main reason is that the minimal norm function in Section 5.2 is no longer decreasing. • The method to derive the equivalence theorem in Section 5.1 does not work for the equivalence theorem in Section 5.3. The main reason is that the system in Section 5.3 is time-varying. • The difference between problems in Section 5.2 and Section 5.3 is that they have different targets. • Though it is assumed that B ∈ L (U, Y ), many results of this chapter can be extended to the case when B ∈ L (U, Y−1 ) is admissible for the semigroup generated by A. (Here, Y−1  D(A∗ ) is the dual space of D(A∗ ).) The admissibility has been defined in [14]. For such extensions, we refer the readers to [16]. • We would like to mention several other related works. (a) We recommend H. O. Fattorini’s book [5], where connections between minimal time and minimal norm optimal controls for abstract equations in Banach spaces were introduced.

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5 Equivalence of Several Kinds of Optimal Controls

(b) Some connections between minimal time and minimal norm controls were also addressed in [6] for abstract equations in Hilbert spaces. (c) Earlier, the relationship between minimal time and minimal norm optimal controls was studied for the heat equation with scalar controls, i.e., controls depending only on the time variable (see [8]). (d) When the target is the ball B(0, ε) in L2 (Ω), the equivalence of minimal time and minimal norm control problems for some semi-linear heat equations was built up in [20]. (e) The equivalence between minimal time and minimal norm optimal controls was obtained for some ODEs, with rectangular control constraint, in [21]. This study improved the work [7] from the perspective of computation of optimal time for some ODEs with cube control constraint. (e) The equivalence theorems were derived for Schrödinger equations in [22]. We end this note with a few open problems which might interest readers: • Can we develop some numerical algorithms for time optimal controls and the optimal time based on the equivalence built up in this chapter? • Can we find more applications for the equivalence obtained in Section 5.2? • How to extend equivalence in Section 5.3 to general time-varying systems?

References 1. O. Cârjˇa, On the minimal time function for distributed control systems in Banach spaces. J. Optim. Theory Appl. 44, 397–406 (1984) 2. O. Cârjˇa, On continuity of the minimal time function for distributed control system. Boll. Un. Mat. Ital. A 4, 293–302 (1985) 3. O. Cârjˇa, The minimal time function in infinite dimensions. SIAM J. Control Optim. 31, 1103–1114 (1993) 4. H.O. Fattorini, Time-optimal control of solution of operational differential equations. J. SIAM Control 2, 54–59 (1964) 5. H.O. Fattorini, Infinite Dimensional Linear Control Systems, the Time Optimal and Norm Optimal Problems. North-Holland Mathematics Studies, vol. 201 (Elsevier Science B.V., Amsterdam, 2005) 6. F. Gozzi, P. Loreti, Regularity of the minimum time function and minimum energy problems: the linear case. SIAM J. Control Optim. 37, 1195–1221 (1999) 7. K. Ito, K. Kunisch, Semismooth newton methods for time-optimal control for a class of ODEs. SIAM J. Control Optim. 48, 3997–4013 (2010) 8. W. Krabs, Optimal control of processes governed by partial differential equations, Part I: heating Processes. Z. Oper. Res. A-B 26, 21–48 (1982) 9. X. Lü, L. Wang, Q. Yan, Computation of time optimal control problems governed by linear ordinary differential equations. J. Sci. Comput. 73, 1–25 (2017) 10. N.N. Petrov, The Bellman problem for a time-optimality problem (Russian), Prikl. Mat. Meh. 34, 820–826 (1970) 11. S. Qin, G. Wang, Equivalence between minimal time and minimal norm control problems for the heat equation. SIAM J. Control Optim. 56, 981–1010 (2018) 12. E.M. Stein, R. Shakarchi, Real Analysis, Measure Theory, Integration, and Hilbert Spaces. Princeton Lectures in Analysis, vol. 3 (Princeton University Press, Princeton, NJ, 2005)

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13. M. Tucsnak, G. Wang, C. Wu, Perturbations of time optimal control problems for a class of abstract parabolic systems. SIAM J. Control Optim. 54, 2965–2991 (2016) 14. M. Tucsnak, G. Weiss, Observation and Control for Operator Semigroups (Birkhäuser Verlag, Basel, 2009) 15. G. Wang, Y. Xu, Equivalence of three different kinds of optimal control problems for heat equations and its applications. SIAM J. Control Optim. 51, 848–880 (2013) 16. G. Wang, Y. Zhang, Decompositions and bang-bang problems. Math. Control Relat. Fields 7, 73–170 (2017) 17. G. Wang, E. Zuazua, On the equivalence of minimal time and minimal norm controls for the internally controlled heat equations. SIAM J. Control Optim. 50, 2938–2958 (2012) 18. G. Wang, Y. Xu, Y. Zhang, Attainable subspaces and the bang-bang property of time optimal controls for heat equations. SIAM J. Control Optim. 53, 592–621 (2015) 19. H. Yu, Approximation of time optimal controls for heat equations with perturbations in the system potential. SIAM J. Control Optim. 52 1663–1692 (2014) 20. H. Yu, Equivalence of minimal time and minimal norm control problems for semilinear heat equations. Syst. Control Lett. 73, 17–24 (2014) 21. C. Zhang, The time optimal control with constraints of the rectangular type for linear timevarying ODEs. SIAM J. Control Optim. 51, 1528–1542 (2013) 22. Y. Zhang, Two equivalence theorems of different kinds of optimal control problems for Schrödinger equations. SIAM J. Control Optim. 53, 926–947 (2015)

Chapter 6

Bang-Bang Property of Optimal Controls

In this chapter, we will study the bang-bang property of some time optimal control problems. The bang-bang property of such a problem says, in plain language, that any optimal control reaches the boundary of the corresponding control constraint set at almost every time. This property not only is mathematically interesting, but also has important applications. For instance, the uniqueness of time optimal control is an immediate consequence of the bang-bang property in certain cases.

6.1 Bang-Bang Property in ODE Cases Q ,QE

S In this section, we will study the bang-bang properties of problems (T P )min QS ,QE (T P )max under the following framework (B1 ):

and

(i) The state space Y and the control space U are Rn and Rm with n, m ∈ N+ , respectively. (ii) The controlled system is as: y(t) ˙ = Ay(t) + Bu(t),

t ∈ (0, +∞),

(6.1)

where A ∈ Rn×n and B ∈ Rn×m . Given τ ∈ [0, +∞), y0 ∈ Rn and u ∈ L∞ (τ, +∞; Rm ), we write y(·; τ, y0 , u) for the solution of (6.1) over [τ, +∞), with the initial condition that y(τ ) = y0 . 3 (iii) The control constraint set U = Bρ (0) with ρ > 0 or U = m j =1 [−aj , aj ] with {aj }m ⊆ (0, +∞). (Roughly speaking, U is either of the ball type or of the j =1 rectangular type.) (iv) Sets QS and QE are disjoint nonempty subsets of [0, +∞)×Rn and (0, +∞)× Rn , respectively.

© Springer International Publishing AG, part of Springer Nature 2018 G. Wang et al., Time Optimal Control of Evolution Equations, Progress in Nonlinear Differential Equations and Their Applications 92, https://doi.org/10.1007/978-3-319-95363-2_6

281

282

6 Bang-Bang Property of Optimal Controls Q ,Q

Q ,Q

S E S E As what we did in Chapter 5, we always assume that (T P )min and (T P )max have optimal controls. QS ,QE We now give the definition of the bang-bang property for problems (T P )min QS ,QE and (T P )max as follows.

Definition 6.1 Several definitions are given in order. (i) An optimal control u∗ (associated with an optimal tetrad (tS∗ , y0∗ , tE∗ , u∗ )) to the QS ,QE QS ,QE problem (T P )min (or (T P )max ) is said to have the bang-bang property if u∗ (t) ∈ exU for a.e. t ∈ (tS∗ , tE∗ ),

(6.2)

where exU denotes the extreme set of U. (In this case, we also say that u∗ is a bang-bang control.) QS ,QE QS ,QE (or (T P )max ) is said to have the bang-bang (ii) The problem (T P )min property, if any optimal control satisfies the bang-bang property. Remark 6.1 In the case when U = Bρ (0) (i.e., the ball-type control constraint), the condition (6.2) is as: u∗ (t)Rm = ρ for a.e. t ∈ (tS∗ , tE∗ ). 3 In the case when U = m j =1 [−aj , aj ] (i.e., the rectangle-type control constraint), the condition (6.2) is as: |u∗j (t)| = aj for each j ∈ {1, . . . , m} and for a.e. t ∈ (tS∗ , tE∗ ). We next present a minimal time control problem, which does not hold the bangbang property but has a bang-bang optimal control. Example 6.1 Let $ A=

$ % $ % % $ % 00 0 0 1 , B= and y1 = . , y0 = 01 1 e 0

S ,QE , where the controlled equation is as: Consider the problem (T P )Q min

y(t) ˙ = Ay(t) + Bu(t), t ∈ (0, +∞) (with y(t) ∈ R2 and u(t) ∈ R) and where QS = {0} × {y0 }, QE = (0, +∞) × {y1 }, U = [−1, 1]. One can easily check the following facts:

6.1 Bang-Bang Property in ODE Cases

283

Q ,Q

S E (i) The problem (T P )min can be put into the framework (B1 ) (given at the beginning of Section 6.1); QS ,QE and the corresponding opti(ii) The tetrad (0, y0 , 1, 0) is optimal for (T P )min QS ,QE mal control does not have the bang-bang property. Consequently, (T P )min does not have the bang-bang property; QS ,QE and u∗ holds the bang-bang (iii) The tetrad (0, y0 , 1, u∗ ) is optimal for (T P )min property, where

u∗ (t) 



1, t ∈ (0, 1/2), −1, t ∈ [1/2, 1).

We end Example 6.1.

6.1.1 Bang-Bang Property for Minimal Time Controls This subsection aims to give some conditions to ensure the bang-bang property for S ,QE (T P )Q . We first study the bang-bang property in the case that U = Bρ (0). To min this end, we introduce the following Kalman controllability rank condition: rank (B, AB, . . . , An−1 B) = n.

(6.3)

The first main theorem (of this subsection) concerns the bang-bang property of minimal time control problems where controls have the ball-type constraint. Theorem 6.1 Let U = Bρ (0) with ρ > 0. Assume that (6.3) holds. Let QS = {0} × YS and QE = (0, +∞) × YE , S ,QE where YS and YE are disjoint nonempty subsets of Rn . Then the problem (T P )Q min has the bang-bang property.

Proof We organize the proof by two steps as follows: S ,QE Step 1. We prove that (T P )Q has the bang-bang property when YS and YE min are different singleton subsets of Rn . Arbitrarily fix y0 , y1 ∈ Rn with y0 = y1 . Let YS = {y0 } and YE = {y1 }. S ,QE . It is clear that tE∗ > 0. Recall Let (0, y0 , tE∗ , u∗ ) be an optimal tetrad of (T P )Q min that (see (3.137) in Chapter 3)

 YR (tE∗ )  {y(tE∗ ; 0, y0, u)  u ∈ L(0, tE∗ ; U)}.

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6 Bang-Bang Property of Optimal Controls

Then by Theorem 4.2, we find that YR (tE∗ ) and {y1 } are separable in Rn . This, QS ,QE satisfies the classical Pontryagin along with Theorem 4.1, yields that (T P )min Maximum Principle. Then according to Definition 4.1, there exists z∗ ∈ Rn \ {0} so that ∗ (t ∗ −t ) E

u∗ (t), B ∗ eA

z∗ Rm =

∗ (t ∗ −t ) E

max v, B ∗ eA

vRm ≤ρ

z∗ Rm for a.e. t ∈ (0, tE∗ ). (6.4)

Meanwhile, it follows from (6.3) and the analyticity of the function t → that ∗ (t ∗ −t ) E

B ∗ eA

∗ B ∗ e A t z∗

z∗ = 0 in Rm for a.e. t ∈ (0, tE∗ ).

This, together with (6.4), yields that u∗ (t)Rm = ρ a.e. t ∈ (0, tE∗ ). Then by Definition 6.1, (0, y0 , tE∗ , u∗ ) satisfies the bang-bang property, conseQS ,QE quently, so does (T P )min . QS ,QE Step 2. We prove that (T P )min has the bang-bang property when YS and YE are disjoint nonempty subsets of Rn . QS ,QE Let (0, y0∗ , tE∗ , u∗ ) be an optimal tetrad to (T P )min . We can directly check that  

Q S ,Q E (0, y0∗ , tE∗ , u∗ ) is also an optimal tetrad to the problem (T P )min , where

E = (0, +∞) × {y(tE∗ ; 0, y0∗ , u∗ )}. S = {0} × {y0∗ } and Q Q Since YS and YE are disjoint, we have that {y0∗ } ∩ {y(tE∗ ; 0, y0∗ , u∗ )} = ∅. Thus, we can use the conclusion of Step 1, as well as Definition 6.1, to find that u∗ has the S ,QE bang-bang property, consequently, so does (T P )Q . min Hence, we end the proof of Theorem 6.1.   The next two examples give auxiliary instructions on Theorem 6.1. Q ,QE

S Example 6.2 Consider the problem (T P )min is as:

, where the controlled equation

y(t) ˙ = −y(t) + u(t), t ∈ (0, +∞) (with y(t) ∈ R and u(t) ∈ R), and where QS = {0} × {2}, QE = (0, +∞) × [−1, 1], U = [−1, 1]. One can easily check the following facts: Q ,Q

S E can be put into the framework (B1 ) (given at the (i) The problem (T P )min beginning of Section 6.1); QS ,QE (ii) The tetrad (0, 2, ln2, 0) is admissible for (T P )min ;

6.1 Bang-Bang Property in ODE Cases

285

Q ,Q

S E (iii) The problem (T P )min has an optimal control; (iv) The assumptions in Theorem 6.1 hold.

Q ,QE

S Then by Theorem 6.1, (T P )min Example 6.2.

has the bang-bang property. We now end

Example 6.3 Let $ A=

$ % $ % % $ % 10 0 0 1 , B= and y1 = . , y0 = 01 1 e 0 Q ,QE

S Consider the problem (T P )min

, where the controlled equation is as:

y(t) ˙ = Ay(t) + Bu(t), t ∈ (0, +∞) (with y(t) ∈ R2 and u(t) ∈ R), and where QS = {0} × {y0 }, QE = (0, +∞) × {y1 }, U = [−1, 1]. One can easily check the following facts: S ,QE can be put into the framework (B1 ) (given at the (i) The problem (T P )Q min beginning of Section 6.1); (ii) It holds that rank (B, AB) = 1 (i.e., (6.3) does not hold); S ,QE (iii) The tetrad (0, y0 , 1, 0) is optimal for (T P )Q and the corresponding min optimal control does not satisfy the bang-bang property.

We now end Example 6.3. The second main theorem (of this subsection) concerns the bang-bang property for minimal time control problems with the rectangle-type control constraint. n Theorem 6.2 Let m ≥ 2 and B  (b1 , . . . , bm ) with {bj }m j =1 ⊆ R . Let 3m m U = j =1 [−aj , aj ] with {aj }j =1 ⊆ (0, +∞). The following two conclusions are true:

(i) If rank(bj , Abj , . . . , An−1 bj ) = n for each j ∈ {1, . . . , m},

(6.5)

and if QS = {0} × YS and QE = (0, +∞) × YE , Q ,QE

S where YS and YE are disjoint nonempty subsets of Rn , then (T P )min the bang-bang property.

holds

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6 Bang-Bang Property of Optimal Controls

(ii) If rank(b1 , Ab1 , . . . , An−1 b1 ) < rank(B, AB, . . . , An−1 B),

(6.6)

S ,QE then there exists  y0 ∈ Rn \ {0} so that the problem (T P )Q , with min

y0 } and QE = (0, +∞) × {0}, QS = {0} × { has at least one optimal control but does not hold the bang-bang property. It deserves to mention what follows: The condition (6.5) means that for each j ∈ {1, . . . , m}, (A, bj ) satisfies the Kalman controllability rank condition (6.3) (where B is replaced by bj ). To prove the conclusion (ii) of Theorem 6.2, we need two lemmas. The first lemma is concerned with some kind of norm optimal control problem. To state it, we define uRmU  max {|uj |/aj } for each u = (u1 , . . . , um ) ∈ Rm ,

(6.7)

1≤j ≤m

3 m m m where U = m j =1 [−aj , aj ] with {aj }j =1 ⊆ (0, +∞). It is obvious that (R , ·RU ) is a Banach space. Then for each y0 ∈ Rn and T > 0, we introduce the following norm optimal control problem:    T ,y (NP )U 0 NU (T , y0 )  inf uL∞ (0,T ;RmU )  y(T ; 0, y0, u) = 0 . (6.8) T ,y0

Several notes on (NP )U

are given in order:

• When the right-hand side of (6.8) is empty, we agree that NU (T , y0 ) = +∞; • A control v ∈ L∞ (0, T ; Rm U ) is called admissible, if y(T ; 0, y0 , u) = 0; • A control v ∗ ∈ L∞ (0, T ; Rm U ) is called optimal, if it is admissible and satisfies that v ∗ L∞ (0,T ;RmU ) = NU (T , y0 ). The first lemma needed is as follows: n Lemma 6.1 Let m ≥ 2 and B = (b1 , . . . , bm ) with {bj }m j =1 ⊆ R . Let U = 3m m j =1 [−aj , aj ] with {aj }j =1 ⊆ (0, +∞). Assume that

rank(b1 , Ab1 , . . . , An−1 b1 ) < rank(B, AB, . . . , An−1 B).

(6.9) T , y

Then for each T > 0, there exists  y0 ∈ Rn \ {0} so that the problem (NP )U 0 has ∗ ∗ an optimal control u (·) = (u1 (·), . . . , u∗m (·)) , with u∗1 (·) = 0 over (0, T ), so that y0 ) = 1. NU (T ,  Proof Arbitrarily fix T > 0. Define two subsets in Rn as follows:    R U,T  y(T ; 0, 0, u)  uL∞ (0,T ;RmU ) ≤ 1

(6.10)

6.1 Bang-Bang Property in ODE Cases

287

and    R0U,T  y(T ; 0, 0, u)  uL∞ (0,T ;RmU ) ≤ 1, u = (u1 , . . . , um ) , u1 = 0 . (6.11) From (6.10) and (6.11), one can easily check the following facts: Fact One: R0U,T ⊆ R U,T ; Fact Two: R0U,T and R U,T are nonempty, bounded, and closed subsets of Rn ; Fact Three: For each z ∈ Rn , m  T

∗ aj |bj eA (T −t ) z|dt (6.12) max y, zRn = y∈R U,T

j =1 0

and max y, zRn =

y∈R0U,T

m 

T

j =2 0

∗ (T −t )

aj |bj eA

z|dt.

(6.13)

We now claim that there exist  z ∈ Rn and  y1 ∈ R0U,T so that max y, z Rn = max y, z Rn =  y1 , z Rn > 0.

y∈R U,T

y∈R0U,T

(6.14)

In fact, by (6.9), we can choose  z ∈ Range (B, AB, . . . , An−1 B) \ {0} so that  z, Ak b1 Rn = 0 for each k ∈ {0, . . . , n − 1}. From this, we can apply the Hamilton-Caley Theorem to verify that ∗

z = 0 for each t ∈ R. b1 eA t This, together with (6.12) and (6.13), leads to the first equality in (6.14). The second equality in (6.14) follows from the fact that R0U,T is closed and bounded. From (6.12), we see that to show the last inequality in (6.14), it suffices to prove that m 

j =1 0

T

∗ (T −t )

aj |bj eA

 z|dt > 0.

By contradiction, we suppose that the above was not true. Then we would have that ∗ (T −t )

bj eA

 z = 0 over [0, T ] for each j ∈ {1, . . . , m},

which indicates that  z, Ak bj Rn = 0 for all k ∈ N+ ∪ {0} and j ∈ {1, . . . , m}.

288

6 Bang-Bang Property of Optimal Controls

This leads to a contradiction since  z ∈ Range(B, AB, · · · , An−1 B) \ {0}. Hence, the last inequality in (6.14) is true. Thus, we have proved (6.14). Let y1 (with  y1 given by (6.14)).  y0  −e−AT 

(6.15)

By the inequality in (6.14), we see that  y0 = 0. The rest of the proof is carried out by the following two steps: Step 1. We show that y0 ) ≤ 1. NU (T , 

(6.16)

Indeed, since  y1 ∈ R0U,T , according to (6.11), there exists a control  u(·) = ( u1 (·), . . . , um (·)) ∈ L∞ (0, T ; Rm u1 (·) = 0 over (0, T ) U ) with  (6.17) so that  y1 = y(T ; 0, 0, u) and  uL∞ (0,T ;RmU ) ≤ 1. These, together with (6.15), imply that u) = 0 and  uL∞ (0,T ;RmU ) ≤ 1. y(T ; 0,  y0 ,

(6.18)

T , y

From (6.18), we see that  u is an admissible control of (NP )U 0 , and consequently (6.16) holds. Step 2. We show that NU (T ,  y0 ) = 1. By contradiction, suppose that the above was not true. Then by (6.16), we would have that NU (T ,  y0 ) < 1. Thus there is a control u∗ ∈ L∞ (0, T ; Rm U ) so that y(T ; 0,  y0, u∗ ) = 0 and u∗ L∞ (0,T ;RmU ) < 1.

(6.19)

By the inequality in (6.19), we can find a constant λ > 1 so that λu∗ L∞ (0,T ;RmU ) ≤ 1. Meanwhile, it follows from (6.15) and the equality in (6.19) that λ y1 = λ(−eAT  y0 ) = y(T ; 0, 0, λu∗ ). This, along with (6.10) and (6.20), implies that λ y1 ∈ R U,T .

(6.20)

6.1 Bang-Bang Property in ODE Cases

289

From the latter and (6.14), we see that z Rn > 0 and  y1 , z Rn ≥ λ y1 , z Rn .  y1 , Since λ > 1, the above leads to a contradiction. Therefore, we have proved the conclusion in Step 2. Finally, by (6.18), (6.17) and the conclusion in Step 2, we see that  u is the desired optimal control and  y0 (given by (6.15)) is the desired initial state. This completes the proof of Lemma 6.1.   The second lemma is concerned with some kind of controllability. Lemma 6.2 Let T > t > 0. Then for each u ∈ L∞ (0, T ; Rm ), there exists vu ∈ L∞ (0, T ; Rm ) so that y(T ; 0, 0, χ(0,t )u) = y(T ; 0, 0, χ(t,T ) vu ).

(6.21)

Conversely, for each v ∈ L∞ (0, T ; Rm ), there exists uv ∈ L∞ (0, T ; Rm ) so that y(T ; 0, 0, χ(t,T ) v) = y(T ; 0, 0, χ(0,t )uv ).

(6.22)

Proof Arbitrarily fix T > t > 0. First of all, we show that there exists a constant C(T , t) > 0 so that 

t

∗ (T −s)

B ∗ eA

0

 zRm ds ≤ C(T , t)

T

∗ (T −s)

B ∗ eA

t

zRm ds for each z ∈ Rn . (6.23)

For this purpose, we define two subspaces of L1 (0, T ; Rm ) as follows: ∗ (T −·)

 z  z ∈ Rn }

∗ (T −·)

 z  z ∈ Rn }.

O0,t  {χ(0,t )(·)B ∗ eA and

Ot,T  {χ(t,T ) (·)B ∗ eA

It is clear that both O0,t and Ot,T are of finite dimension. Meanwhile, for each fixed ∗ z ∈ Rn , since the function s → B ∗ eA s z, s ∈ R, is analytic, we can check that ∗ (T −·)

B ∗ eA

∗ (T −·)

z = 0 over (0, t) if and only if B ∗ eA

z = 0 over (t, T ).

Therefore, (6.23) is true. Next, we arbitrarily fix u ∈ L∞ (0, T ; Rm ) and define ∗ (T −·)

F (χ(t,T ) (·)B ∗ eA



t

z)  0

∗ (T −s)

B ∗ eA

z, u(s)Rm ds for each z ∈ Rn . (6.24)

290

6 Bang-Bang Property of Optimal Controls

From (6.23), we see that F is well defined over Ot,T and is linear bounded from Ot,T to R. Then, according to the Hahn-Banach Theorem and the Riesz Representation Theorem (see Theorems 1.5 and 1.4), there exists vu ∈ L∞ (0, T ; Rm ) so that ∗ (T −·)

F (χ(t,T ) (·)B ∗ eA



T

z) =

∗ (T −s)

B ∗ eA

t

z, vu (s)Rm ds for each z ∈ Rn .

This, together with (6.24), leads to (6.21). Finally, the conclusion (6.22) can be proved by the very similar way to the above. Hence, we finish the proof of Lemma 6.2.   We now are in the position to prove Theorem 6.2. Proof (Proof of Theorem 6.2) We prove the conclusions one by one. (i) It suffices to prove this theorem in the case that both YS and YE are singleton subsets of Rn . (The reason is similar to that explained in the proof of Theorem 6.1.) Arbitrarily fix two different vectors y0 and y1 in Rn . We set YS  {y0 } and YE  {y1 }. S E Let (0, y0 , tE∗ , u∗ ) be an optimal tetrad of (T P )min . It is clear that tE∗ > 0. By the similar arguments to those used in the proof of Theorem 6.1, we can QS ,QE satisfies the classical Pontryagin Maximum Principle. obtain that (T P )min Then according to Definition 4.1, there exists z∗ ∈ Rn \ {0} so that

Q ,Q

∗ (t ∗ −t ) E

u∗ (t), B ∗ eA

∗ (t ∗ −t ) E

z∗ Rm = maxv, B ∗ eA v∈U

z∗ Rm for a.e. t ∈ (0, tE∗ ). (6.25)

A∗ t

Meanwhile, since the function t → bj e z∗ , t ∈ R (with j ∈ {1, . . . , m}) is analytic, it follows by (6.5) that for each j ∈ {1, . . . , m}, ∗

bj eA t z∗ = 0 for a.e. t ∈ (0, tE∗ ). Write u∗ (·) = (u∗1 (·), . . . , u∗m (·)) . From the above and (6.25), we see that |u∗j (t)| = aj for a.e. t ∈ (0, tE∗ ) and each j ∈ {1, . . . , m}. Then by Definition 6.1, (0, y0 , tE∗ , u∗ ) satisfies the bang-bang property, conseS ,QE holds the bang-bang property. quently, (T P )Q min (ii) Assume that (6.6) holds. Arbitrarily fix T > 0. According to Lemma 6.1, T , y there exists  y0 ∈ Rn \ {0} so that (NP )U 0 has an optimal control u∗ (·) = ∗ ∗ (u1 (·), . . . , um (·)) satisfying that

6.1 Bang-Bang Property in ODE Cases

291

y(T ; 0,  y0 , u∗ ) = 0, u∗ L∞ (0,T ;RmU ) = NU (T ,  y0 ) = 1, u∗1 (·) = 0 over (0, T ). (6.26) Denote QS = {0} × { y0 } and QE = (0, +∞) × {0}. We next show that the QS ,QE problem (T P )min has at least one optimal control but does not hold the bangbang property. The proof will be carried out by the following two steps. Step 1. We show that (0,  y0 , T , u∗ ) is an optimal tetrad to the problem QS ,QE (T P )min . From the first two conclusions in (6.26), we see that (0,  y0 , T , u∗ ) is an QS ,QE admissible tetrad to the problem (T P )min . The reminder is to show that T is S ,QE the optimal time of (T P )Q . By contradiction, we suppose that it was not min true. Then there would exist  t ∈ (0, T ) and  v ∈ L(0,  t; U) so that v ) = 0. y( t; 0,  y0,

(6.27)

t, T ; Rm ) so that Meanwhile, according to Lemma 6.2, there exists  v ∈ L∞ (  T  t A(T −t ) e B v (t)dt = eA(T −t ) B v (t)dt. (6.28)  t

0

We choose a small λ ∈ (0, 1) so that 2λ v (t) ∈ U for a.e. t ∈ ( t, T ).

(6.29)

Define a new control as follows:  (1 − λ) v (t), t ∈ (0,  t), vλ (t)   λ v (t), t ∈ (t, T ).

(6.30)

Then, by (6.30) and (6.28), we find that y(T ; 0,  y0, vλ ) = eAT  y0 + (1 − λ)  y0 + = eAT 



 t

eA(T −t ) B v (t)dt + λ

0  t



T

 t

eA(T −t ) B v (t)dt

eA(T −t ) B v (t)dt.

0

This, together with (6.27), implies that t; 0,  y0, v ) = 0. y(T ; 0,  y0, vλ ) = eA(T −t) y( T , y0

Thus, vλ is an admissible control to (NP )U the second conclusion in (6.26) that

(see (6.8)). Then it follows from

vλ L∞ (0,T ;RmU ) ≥ 1.

(6.31)

292

6 Bang-Bang Property of Optimal Controls

Meanwhile, since  v (t) ∈ U for a.e. t ∈ (0,  t), we see from (6.7), (6.30), and (6.29) that vλ (t)RmU ≤ max{1 − λ, 1/2} < 1 for a.e. t ∈ (0, T ), Q ,Q

S E which contradicts (6.31). Therefore, T is the optimal time of (T P )min . Q ,Q S E Hence, (0,  y0 , T , u∗ ) is an optimal tetrad of (T P )min . QS ,QE Step 2. We prove that (T P )min does not hold the bang-bang property. By Step 1, the last conclusion in (6.26) and Definition 6.1, we see that u∗ QS ,QE does not have the bang-bang property, consequently, (T P )min does not hold the bang-bang property. Hence, we finish the proof of Theorem 6.2.  

The next two examples give auxiliary instructions on Theorem 6.2. Example 6.4 Let $ A=

$ % $ % % $ % 0 1 1 0 10 , B= and y1 = . , y0 = −1 0 1 0 01 Q ,QE

S Consider the problem (T P )min

, where the controlled system is as:

y(t) ˙ = Ay(t) + Bu(t), t ∈ (0, +∞) (with y(t) ∈ R2 , u(t) ∈ R2 ) and where QS = {0} × {y0 }, QE = (0, +∞) × {y1 }, U = [−1, 1] × [−1, 1]. One can easily check the following facts: S ,QE can be put into the framework (B1 ) (given at the (i) The problem (T P )Q min beginning of Section 6.1); S ,QE (ii) The problem (T P )Q has at least one admissible control; (This follows min from Theorem 3.3 and Corollary 3.2.) S ,QE (iii) The problem (T P )Q has at least one optimal control; min (iv) The assumptions in (i) of Theorem 6.2 hold. S ,QE Then by (i) of Theorem 6.2, (T P )Q holds the bang-bang property. We end min Example 6.4.

Example 6.5 Let $ A=

% $ % $ % $ % 00 10 1 0 , B= , y0 = and y1 = . 00 01 1 0

6.1 Bang-Bang Property in ODE Cases Q ,QE

S Consider the problem (T P )min

293

, where the controlled equation is as:

y(t) ˙ = Ay(t) + Bu(t), t ∈ (0, +∞) (with y(t) ∈ R2 and u(t) ∈ R2 ) and where QS = {0} × {y0 }, QE = (0, +∞) × {y1 }, U = [−1, 1] × [−2, 2]. One can easily check the following facts: S ,QE can be put into the framework (B1 ) (given at the (i) The problem (T P )Q min beginning of Section 6.1); (ii) The assumption (6.6) in (ii) of Theorem 6.2 holds; S ,QE (iii) The tetrad (0, y0 , 1, u∗ ) is optimal for (T P )Q , where min

∗

u (t) 



(−1, −2) , t ∈ (0, 1/2), (−1, 0) , t ∈ [1/2, 1).

From the above, we see that u∗ is not a bang-bang control. Consequently, S ,QE does not hold the bang-bang property. We now end Example 6.5. (T P )Q min

6.1.2 Bang-Bang Property for Maximal Time Controls This subsection aims to give some conditions to ensure the bang-bang property of S ,QE the maximal time control problem (T P )Q max . We start with the next lemma. Lemma 6.3 Let (A, B) ∈ Rn×n × Rn×m satisfy (6.3). Then for any T > t ≥ 0 and each subset E ⊆ (t, T ) of positive measure, there exists a constant C  C(T , t, E) > 0 so that for each y0 ∈ Rn , there exists u ∈ L∞ (t, T ; Rm ) satisfying that y(T ; t, y0, χE u) = 0 and χE uL∞ (t,T ;Rm ) ≤ Cy0 Rn . Proof Arbitrarily fix T > t > 0. First of all, we show that there exists a constant C  C(T , t, E) > 0 so that  ∗ zRn ≤ C B ∗ eA (T −s) zRm ds for each z ∈ Rn . (6.32) E

(Here and throughout the proof of this lemma, C  C(T , t, E) denotes a generic positive constant depending on T , t, and E.) For this purpose, we define a subspace of L1 (t, T ; Rm ) as follows:  ∗ O  {χE (·)B ∗ eA (T −·) z  z ∈ Rn }.

294

6 Bang-Bang Property of Optimal Controls

It is clear that O is of finite dimension. Meanwhile, for each z ∈ Rn , the function ∗ s → B ∗ eA s z, s ∈ R, is analytic. Thus, by (6.3), we see that ∗ (T −·)

B ∗ eA

z = 0 over E if and only if z = 0.

Therefore, (6.32) is true. Next, we arbitrarily fix y0 ∈ Rn and define ∗ (T −·)

F (χE (·)B ∗ eA

z)  z, −e(T −t )A y0 Rn for each z ∈ Rn .

(6.33)

From (6.32), we see that F is well defined over O. Besides, one can easily check that F is linear. Then by (6.33) and (6.32), we find that ∗

|F (χE (·)B ∗ eA (T −·) z)|  T ∗ ≤ Cy0 Rn χE (s)B ∗ eA (T −s) zRm ds for each z ∈ Rn , t

which indicates that F L(O,R) ≤ Cy0 Rn . Thus, F is a linear bounded functional on O. Then, according to the Hahn-Banach Theorem and the Riesz Representation Theorem (see Theorems 1.5 and 1.4), there exists u ∈ L∞ (t, T ; Rm ) so that  ∗ ∗ F (χE (·)B ∗ eA (T −·) z) = B ∗ eA (T −s) z, u(s)Rm ds for each z ∈ Rn , E

and so that uL∞ (t,T ;Rm ) = F L (O,R) ≤ Cy0 Rn . These, together with (6.33), lead to the desired result. Hence, we finish the proof of Lemma 6.3.

 

The first main theorem of this subsection concerns the bang-bang property for maximal time control problems where controls have the ball-type constraint. Theorem 6.3 Let T > 0 and U = Bρ (0) with ρ > 0. Assume that (6.3) holds. Let  QS = {(t, eAt y0 )  0 ≤ t < T , y0 ∈ YS } and QE = {T } × YE , where YS and YE are two nonempty subsets of Rn so that (eAT YS ) ∩ YE = ∅. Then S ,QE the problem (T P )Q has the bang-bang property. max

6.1 Bang-Bang Property in ODE Cases

295 Q ,Q

S E Proof By contradiction, we suppose that (T P )max did not hold the bangbang property. Then by Definition 6.1, there would exist an optimal tetrad ∗ QS ,QE , a constant ε ∈ (0, ρ) and a subset E ⊆ (tS∗ , T ) (tS∗ , eAtS y0∗ , T , u∗ ) of (T P )max of positive measure so that

u∗ (t)Rm < ρ − ε for a.e. t ∈ E.

(6.34)

∗

Meanwhile, by the optimality of the tetrad (tS∗ , eAtS y0∗ , T , u∗ ), we have that ∗

0 ≤ tS∗ < T , y0∗ ∈ YS , y(T ; tS∗ , eAtS y0∗ , u∗ ) ∈ YE

(6.35)

and that u∗ (t)Rm ≤ ρ for a.e. t ∈ (tS∗ , T ).

(6.36)

Define a subset of E in the following manner:  E1  {t ∈ E  tS∗ + |E|/2 < t < T }.

(6.37)

E1 ⊆ (tS∗ + |E|/2, T ) and |E1 | ≥ |E|/2.

(6.38)

It is obvious that

Let C  C(T , t, E) > 0 be given by Lemma 6.3, where (T , t, E) is replaced by (T , tS∗ + |E|/2, E1 ). We choose a small δ ∈ (0, |E|/2) so that # " y0  −y(tS∗ + δ; tS∗ , 0, u∗ ).  y0 Rn ≤ ε/ C max eAs L (Rn ) with  0≤s≤T

(6.39)

Then by Lemma 6.3, we can find a control  u ∈ L∞ (0, +∞; Rm ) so that y0 , 0), χE1  u) = 0 y(T ; tS∗ + |E|/2, y(tS∗ + |E|/2; tS∗ + δ, 

(6.40)

and so that uL∞ (0,+∞;Rm ) ≤ Cy(tS∗ + |E|/2; tS∗ + δ,  y0 , 0)Rn . χE1 

(6.41)

It follows from (6.38) and (6.40) that y(T ; tS∗ + δ,  y0 , χE1  u) = 0.

(6.42)

Define a new control in the following manner: u(t),  u(t)  u∗ (t) + χE1 (t)

t ∈ (tS∗ + δ, T ).

(6.43)

296

6 Bang-Bang Property of Optimal Controls

On one hand, by (6.36), (6.34), (6.37), (6.41), and (6.39), we can directly check that  u(t)Rm ≤ ρ for a.e. t ∈ (tS∗ + δ, T ).

(6.44)

On the other hand, it follows from (6.43) that ∗

y(T ; tS∗ + δ, eA(tS +δ) y0∗ , u) = ∗



T tS∗ +δ

eA(T −t ) B(χE1  u)(t)dt

∗

+ eA(T −tS −δ) y(tS∗ + δ; tS∗ , eAtS y0∗ , u∗ ) + ∗



T tS∗ +δ

∗

eA(T −t ) Bu∗ (t)dt ∗

+eA(T −tS −δ) (eA(tS +δ) y0∗ − y(tS∗ + δ; tS∗ , eAtS y0∗ , u∗ )), which, along with (6.39), yields that ∗

u) y(T ; tS∗ + δ, eA(tS +δ) y0∗ , ∗

∗

= y(T ; tS∗ , eAtS y0∗ , u∗ ) + eA(T −tS −δ) y0 +



T tS∗ +δ

eA(T −t ) B(χE1  u)(t)dt.

This, combined with (6.42) and the third conclusion of (6.35), indicates that ∗

∗

y(T ; tS∗ + δ, eA(tS +δ) y0∗ , u) = y(T ; tS∗ , eAtS y0∗ , u∗ ) + y(T ; tS∗ + δ,  y0 , χE1  u) AtS∗ ∗ ∗ ∗ = y(T ; tS , e y0 , u ) ∈ YE . (6.45) Since 0 ≤ tS∗ + δ < tS∗ + |E|/2 < T , it follows by the second conclusion of ∗ u) is an admissible tetrad of (6.35), (6.44), and (6.45) that (tS∗ + δ, eA(tS +δ) y0∗ , T , QS ,QE QS ,QE (T P )max , which contradicts the optimality of tS∗ . Thus, (T P )max holds the bang-bang property. This completes the proof of Theorem 6.3.   The next two examples give auxiliary instructions on Theorem 6.3. Example 6.6 Let $ A=

$ % $ % % $ % 01 0 1 0 , B= and y1 = . , y0 = 00 0 0 1 Q ,QE

S Consider the problem (T P )max

, where the controlled equation is as:

y(t) ˙ = Ay(t) + Bu(t), t ∈ (0, +∞) (with y(t) ∈ R2 and u(t) ∈ R) and where  QS = {(t, eAt y0 )  0 ≤ t < 2}, QE = {2} × {y1 }, U = [−1, 1], T = 2.

6.1 Bang-Bang Property in ODE Cases

297

One can easily check the following facts: Q ,Q

S E (i) The problem (T P )max can be put into the framework (B1 ) (given at the beginning of Section 6.1); (ii) It holds that rank (B, AB) = 2; QS ,QE (iii) The tetrad (0, y0 , 2, u) is admissible for (T P )max , where

 u(t) 

1, t ∈ (0, 1), −1, t ∈ [1, 2);

Q ,Q

S E (iv) The problem (T P )max has optimal controls. (This follows from (iii).) QS ,QE holds the bang-bang property. Then by Theorem 6.3, (T P )max

We now end Example 6.6. Example 6.7 Let $ A=

$ % $ % % $ % 00 0 1 1 , B= and y1 = . , y0 = 01 1 e 0 Q ,QE

S Consider the problem (T P )max

, where the controlled system is as:

y(t) ˙ = Ay(t) + Bu(t), t ∈ (0, +∞) (with y(t) ∈ R2 , u(t) ∈ R), and where  QS = {(t, eAt y0 )  0 ≤ t < 1}, QE = {1} × {y1 }, U = [−1, 1], T = 1. One can easily check the following facts: S ,QE can be put into the framework (B1 ) (given at the (i) The problem (T P )Q max beginning of Section 6.1); (ii) It holds that rank (B, AB) = 1 < 2 (i.e., (A, B) does not satisfy (6.3)); S ,QE (iii) The tetrad (0, y0 , 1, 0) is optimal for (T P )Q max . S ,QE Consequently, (T P )Q has no bang-bang property. Now we end Example 6.9. max

The second main result of this subsection concerns the bang-bang property for maximal time control problems where controls have the rectangle-type constraint. n Theorem 6.4 Let m ≥ 2 and B = (b1 , . . . , bm ) with {bj }m j =1 ⊆ R . Let U = 3m m j =1 [−aj , aj ] with {aj }j =1 ⊆ (0, +∞). Then the following two conclusions are true:

(i) Assume that rank(bj , Abj , . . . , An−1 bj ) = n for each j ∈ {1, . . . , m}.

298

6 Bang-Bang Property of Optimal Controls

Let  QS = {(t, eAt y0 )  0 ≤ t < T , y0 ∈ YS } and QE = {T } × YE , where T > 0, YS and YE are two nonempty subsets of Rn with (eAT YS ) ∩ YE = S ,QE ∅. Then the problem (T P )Q holds the bang-bang property. max (ii) Assume that rank(b1 , Ab1 , . . . , An−1 b1 ) < rank(B, AB, . . . , An−1 B).

(6.46)

S ,QE  > 0 so that the problem (T P )Q Then there exists  y0 ∈ Rn \ {0} and T max , with

QS = {(t, eAt  y0 ) : 0 ≤ t < T} and QE = {T} × {0}, has at least one optimal control but does not hold the bang-bang property. Q ,Q

S E did not hold the bangProof (i) By contradiction, we suppose that (T P )max bang property. Then by Definition 6.1, we could assume, without loss of ∗ generality, that there would be an optimal tetrad (tS∗ , eAtS y0∗ , T , u∗ ) (with u∗ = QS ,QE (u∗1 , . . . , u∗m ) ) of (T P )max , an ε ∈ (0, a1 ) and a subset E (in (tS∗ , T )) of positive measure so that

|u∗1 (t)| < a1 − ε for each t ∈ E.

(6.47)

It is clear that ∗

0 ≤ tS∗ < T , y0∗ ∈ YS , y(T ; tS∗ , eAtS y0∗ , u∗ ) ∈ YE

(6.48)

|u∗i (t)| ≤ ai for each i ∈ {1, . . . , m} and a.e. t ∈ (tS∗ , T ).

(6.49)

and

Define a subset of E in the following manner:  E1  {t ∈ E  tS∗ + |E|/2 < t < T }.

(6.50)

E1 ⊆ (tS∗ + |E|/2, T ) and |E1 | ≥ |E|/2.

(6.51)

It is obvious that

Let C  C(T , t, E) > 0 be given by Lemma 6.3, where (A, B) and (B, T , t, E) are replaced by (A, b1 ) and (b1 , T , tS∗ + |E|/2, E1 ) respectively. We choose a small δ ∈ (0, |E|/2) so that

6.1 Bang-Bang Property in ODE Cases

299

# " y0  −y(tS∗ + δ; tS∗ , 0, u∗ ).  y0 Rn ≤ ε/ C max eAs L (Rn ) with  0≤s≤T

(6.52) Then according to Lemma 6.3, there is  u ∈ L∞ (0, +∞; Rm ), with  u = ( u1 , 0, . . . , 0) , so that y(T ; tS∗ + |E|/2, y(tS∗ + |E|/2; tS∗ + δ,  y0 , 0), χE1  u) = 0

(6.53)

and so that u1 L∞ (0,+∞;R) ≤ Cy(tS∗ + |E|/2; tS∗ + δ,  y0 , 0)Rn . χE1 

(6.54)

It follows from (6.51) and (6.53) that y(T ; tS∗ + δ,  y0 , χE1  u) = 0.

(6.55)

Define a new control in the following manner: um (t))   u(t)  u∗ (t)+χE1 (t) u(t), ( u1 (t), . . . ,

t ∈ (tS∗ +δ, T ).

(6.56)

Then it follows by (6.49), (6.47), (6.50), (6.54), and (6.52) that | uj (t)| ≤ aj for a.e. t ∈ (tS∗ + δ, T ) and for each j ∈ {1, . . . , m}.

(6.57)

Meanwhile, from (6.56), we see that y(T ; tS∗

∗ + δ, eA(tS +δ) y0∗ , u)

∗

 =

T tS∗ +δ

eA(T −t ) B(χE1  u)(t)dt

∗

+ eA(T −tS −δ) y(tS∗ + δ; tS∗ , eAtS y0∗ , u∗ ) + ∗



T tS∗ +δ

∗

eA(T −t ) Bu∗ (t)dt ∗

+eA(T −tS −δ) (eA(tS +δ) y0∗ − y(tS∗ + δ; tS∗ , eAtS y0∗ , u∗ )), which, together with (6.52), yields that ∗

u) y(T ; tS∗ + δ, eA(tS +δ) y0∗ , =

∗ y(T ; tS∗ , eAtS y0∗ , u∗ )

+e



A(T −tS∗ −δ)

 y0 +

T tS∗ +δ

eA(T −t ) B(χE1  u)(t)dt.

This, combined with (6.55) and the third conclusion of (6.48), indicates that ∗

∗

y(T ; tS∗ + δ, eA(tS +δ) y0∗ , u) = y(T ; tS∗ , eAtS y0∗ , u∗ ) + y(T ; tS∗ + δ,  y0 , χE1  u) ∗ = y(T ; tS∗ , eAtS y0∗ , u∗ ) ∈ YE . (6.58)

300

6 Bang-Bang Property of Optimal Controls

Finally, since 0 ≤ tS∗ + δ < tS∗ + |E|/2 < T , it follows by the second ∗ u) is an conclusion of (6.48), (6.57), and (6.58) that (tS∗ + δ, eA(tS +δ) y0∗ , T , QS ,QE admissible tetrad of (T P )max , which contradicts the optimality of tS∗ . Thus, QS ,QE holds the bang-bang property. (T P )max (ii) Assume that (6.46) holds. Arbitrarily fix T > 0. According to (6.46) and T , y Lemma 6.1, there exists  y0 ∈ Rn \ {0} so that (NP )U 0 has an optimal control ∗ ∗ ∗ u (·) = (u1 (·), . . . , um (·)) satisfying that y(T ; 0,  y0 , u∗ ) = 0, u∗ L∞ (0,T ;RmU ) = NU (T ,  y0 ) = 1 and u∗1 (·) = 0 over (0, T ).

(6.59)

Let y0 ) : 0 ≤ t < T } and QE = {T } × {0}. QS = {(t, eAt  S ,QE has at least one optimal tetrad but We next show that the problem (T P )Q max does not hold the bang-bang property. The proof will be carried out by the following two steps. Step 1. We show that (0,  y0 , T , u∗ ) is an optimal tetrad to the problem QS ,QE (T P )max . From the first two conclusions in (6.59), we see that (0,  y0 , T , u∗ ) is an QS ,QE admissible tetrad to the problem (T P )max . The reminder is to show that T is QS ,QE the optimal time of (T P )max . By contradiction, we suppose that it was not true. Then there would exist  t ∈ (0, T ) and  v ∈ L( t, T ; U) so that

y0 , v ) = 0. y(T ;  t, eAt 

(6.60)

Meanwhile, according to Lemma 6.2, there exists  v ∈ L∞ (0,  t; Rm ) so that 

T

 t

e

A(T −t )



 t

B v (t)dt =

eA(T −t ) B v (t)dt.

(6.61)

0

We choose a small λ ∈ (0, 1) so that 2λ v (t) ∈ U for a.e. t ∈ (0,  t). Define a new control as follows:  λ v (t), t ∈ (0,  t), vλ (t)   (1 − λ) v (t), t ∈ (t, T ). Then it follows from (6.63) and (6.61) that

(6.62)

(6.63)

6.1 Bang-Bang Property in ODE Cases

y(T ; 0,  y0, vλ ) y0 + (1 − λ) = eAT   = eAT  y0 +

T

 t



T

 t

301

eA(T −t ) B v (t)dt + λ



 t

eA(T −t ) B v (t)dt

0

eA(T −t ) B v (t)dt.

This, together with (6.60), implies that t, eAt  y0 , v ) = 0. y(T ; 0,  y0 , vλ ) = y(T ;  T , y0

Thus, vλ is an admissible control to (NP )U the second conclusion in (6.59) that

(see (6.8)). Then it follows from

vλ L∞ (0,T ;RmU ) ≥ 1.

(6.64)

Meanwhile, since  v (t) ∈ U a.e. t ∈ ( t, T ), from (6.7), (6.63), and (6.62), we see that vλ (t)RmU ≤ max{1 − λ, 1/2} < 1 for a.e. t ∈ (0, T ), S ,QE which contradicts (6.64). Therefore, T is the optimal time of (T P )Q max . QS ,QE ∗ Hence, (0,  y0 , T , u ) is an optimal tetrad of (T P )max . S ,QE does not hold the bang-bang property. Step 2. We prove that (T P )Q max By the conclusion in Step 1, the third conclusion in (6.59), and Definition 6.1, we see that (0,  y0 , T , u∗ ) does not have the bang-bang property, QS ,QE consequently, (T P )max does not hold the bang-bang property. This completes the proof of Theorem 6.4.  

The next two examples give auxiliary instructions on Theorem 6.4. Example 6.8 Let $ A=

$ % $ % % $ % 0 0 0 1 10 and y1 = . , B= , y0 = −1 0 −1 0 01

S ,QE Consider the problem (T P )Q max , where the controlled equation is as:

y(t) ˙ = Ay(t) + Bu(t), t ∈ (0, +∞), with y(t) ∈ R2 and u(t) ∈ R2 , and where  QS = {(t, eAt y0 )  0 ≤ t < 1}, QE = {1}×{y1}, U = [−1, 1]×[−1, 1], T = 1.

302

6 Bang-Bang Property of Optimal Controls

One can easily check the following facts: Q ,Q

S E (i) The problem (T P )max can be put into the framework (B1 ) (given at the beginning of Section 6.1); (ii) The assumptions in (i) of Theorem 6.4 hold; QS ,QE , where (iii) The tetrad (0, y0 , 1, u) is admissible for (T P )max

$ % 1 u(t)  , t ∈ (0, 1); 0 Q ,Q

S E has at least one optimal control. (This follows from (iv) The problem (T P )max the existence of admissible controls, see (iii).)

Q ,QE

S Then by (i) of Theorem 6.4, (T P )max Example 6.8.

holds the bang-bang property. We now end

Example 6.9 Let $ A=

% $ $ % $ % % 00 10 1 0 , B= , y0 = and y1 = . 00 01 1 0 Q ,QE

S Consider the problem (T P )max

, where the controlled system is as:

y(t) ˙ = Ay(t) + Bu(t), t ∈ (0, +∞) (with y(t) ∈ R2 , u(t) ∈ R2 ), and where  QS = {(t, eAt y0 )  0 ≤ t < 1}, QE = {1} × {y1}, U = [−1, 1] × [−2, 2], T = 1. One can easily check the following facts: S ,QE can be put into the framework (B1 ) (given at the (i) The problem (T P )Q max beginning of Section 6.1); (ii) The assumption in (ii) of Theorem 6.4 holds; S ,QE (iii) The tetrad (0, y0 , 1, u∗ ) is optimal for (T P )Q max , where

∗

u (t) 



(−1, −2) , t ∈ (0, 1/2), (−1, 0) , t ∈ [1/2, 1).

S By (iii), we see that u∗ is not a bang-bang control. Thus, (T P )max hold the bang-bang property. We now end Example 6.9.

Q ,QE

does not

6.2 Bang-Bang Property and Null Controllability

303

6.2 Bang-Bang Property and Null Controllability In this section, we will study the bang-bang properties of minimal/maximal time QS ,QE QS ,QE control problems (T P )min /(T P )max from the perspective of the L∞ -null controllability from measurable sets in time. These problems are under the following framework (B2 ): (i) The state space Y and the control space U are two real separable Hilbert spaces. (ii) The controlled system is as: y(t) ˙ = Ay(t) + D(t)y(t) + B(t)u(t), t > 0,

(6.65)

where A : D(A) ⊆ Y → Y generates a C0 semigroup {eAt }t ≥0 on Y ; D(·) ∈ L∞ (0, +∞; L (Y )) and B(·) ∈ L∞ (0, +∞; L (U, Y )). Given τ ∈ [0, +∞), y0 ∈ Y and u ∈ L∞ (τ, +∞; U ), we write y(·; τ, y0 , u) for the solution of (6.65) over [τ, +∞), with the initial condition that y(τ ) = y0 . Moreover, write {Φ(t, s) : t ≥ s ≥ 0} for the evolution system generated by A + D(·) over Y . (iii) The control constraint set U is nonempty and closed in U . (iv) Sets QS and QE are two disjoint nonempty subsets of [0, +∞) × Y and (0, +∞) × Y , respectively. Q ,Q

S E As what we did in the previous section, we always assume that (T P )min and QS ,QE (T P )max have optimal controls. QS ,QE We now give the definition of the bang-bang property for problems (T P )min QS ,QE and (T P )max as follows:

Definition 6.2 (i) An optimal control u∗ (associated with an optimal tetrad (tS∗ , y0∗ , tE∗ , u∗ )) to the QS ,QE QS ,QE problem (T P )min (or (T P )max ) is said to have the bang-bang property if u∗ (t) ∈ ∂U for a.e. t ∈ (tS∗ , tE∗ ),

(6.66)

where ∂U denotes the boundary of U. (We also say that u∗ is a bang-bang control when (6.66) holds.) QS ,QE QS ,QE (ii) The problem (T P )min (or (T P )max ) is said to have the bang-bang property, if any optimal control satisfies the bang-bang property. Remark 6.2 In Section 6.1, we defined the bang-bang property for the finite dimension controlled system (see Definition 6.1). Definition 6.1 differs from Definition 6.2 (see “ex U” in (6.2) and “∂U” in (6.66)). In general, ex U and ∂U are different. For example, when U is a rectangle of Rm (with m ∈ N+ ), they are different. But if U is a nonempty closed ball in a Hilbert space, then ex U and ∂U are the same.

304

6 Bang-Bang Property of Optimal Controls Q ,Q

Q ,Q

S E S E For time optimal control problems (T P )min (or (T P )max ), where the controlled systems are in infinitely dimensional spaces, we will only study the bangbang property given by Definition 6.2. The reason is that in infinitely dimensional cases, the study on the bang-bang property given by Definition 6.1 is very hard. (To our best knowledge, there have not been any result on this issue so far.)

We next give the definition of the L∞ -null controllability from measurable sets in time for the system (6.65). Definition 6.3 The system (6.65) is said to be L∞ -null controllable from measurable sets in time, if for any T > t > 0 and each subset E ⊆ (t, T ) of positive measure, there exists a constant C(T , t, E) > 0 so that for each y0 ∈ Y , there is a control u ∈ L∞ (0, +∞; U ) satisfying that y(T ; t, y0, χE u) = 0 and χE uL∞ (0,+∞;U ) ≤ C(T , t, E)y0 Y .

(6.67)

6.2.1 Bang-Bang Property of Minimal Time Control In this subsection, we aim to use the L∞ -null controllability (from measurable sets in time) to study the bang-bang property for the minimal time control problem QS ,QE 41): . For this purpose, we need the following assumption (H (T P )min (i) D(·) ≡ 0 and B(·) ≡ B with B ∈ L (U, H ). (That is, the system (6.65) is time-invariant.) (ii) The operator A generates an analytic semigroup {eAt }t ≥0 over Y . (iii) There exist two constants d > 0 and σ > 0 so that for each L ∈ (0, 1], ∗

eA L z2Y ≤ ed/L

σ



L

∗ (L−t )

B ∗ eA

0

z2U dt for all z ∈ Y.

41) holds. Then the system (6.65) is L∞ -null Proposition 6.1 Suppose that (H controllable from measurable sets in time. The proof of Proposition 6.1 will be given later. We now use it to prove the following main result of this subsection: 41) holds. Let Theorem 6.5 Suppose that (H QS = {0} × YS and QE = (0, +∞) × YE , where YS and YE are two disjoint nonempty subsets of Y . Then the problem S ,QE (T P )Q has the bang-bang property. min S ,QE did not hold the bang-bang Proof By contradiction, we suppose that (T P )Q min property. Then according to Definition 6.2, there would exist an optimal tetrad

6.2 Bang-Bang Property and Null Controllability S (0, y0∗ , tE∗ , u∗ ) of (T P )min

Q ,QE

305

and a subset E (in (0, tE∗ )) of positive measure so that

/ ∂U for each t ∈ E. u∗ (t) ∈

(6.68)

By the optimality of (0, y0∗ , tE∗ , u∗ ), we find that y0∗ ∈ YS , y(tE∗ ; 0, y0∗ , u∗ ) ∈ YE and u∗ (t) ∈ U for a.e. t ∈ (0, tE∗ ).

(6.69)

For each k ∈ N+ , we define a subset of E in the following manner:    Ek  t ∈ E  dU (u∗ (t), ∂U) > k −1 .

(6.70)

Here and in what follows, dU (·, ∂U) denotes the distance between · and ∂U in U . We first claim that for each k ∈ N+ , Ek is measurable. To this end, We arbitrarily fix k ∈ N+ . Since u∗ (·) : E → U is strongly measurable, there exists {nj }j ≥1 ⊆ + N+ , {a  j  : j ∈ N , 1 ≤  ≤ nj } ⊆ U and a sequence of measurable subsets {Fj   j ∈ N+ , 1 ≤  ≤ nj } ⊆ E, with Fj  ∩ Fj  = ∅, when 1 ≤ ,  ≤ nj ,  =  and j ∈ N+ , so that uj (t) 

nj

aj  χFj (t) → u∗ (t) strongly in U for a.e. t ∈ E.

=1

From this it follows that for a.e. t ∈ E, dU (u∗ (t), ∂U) = lim dU (uj (t), ∂U). j →+∞

(6.71)

Meanwhile, we find that for each j ∈ N+ , dU (uj (t), ∂U) =

nj

χFj (t)dU (aj  , ∂U) for each t ∈ E,

=1

which indicates that for each j ∈ N+ , the function t → dU (uj (t), ∂U) is measurable from E to R. This, together with (6.71), yields that the function t → dU (u∗ (t), ∂U) is also measurable from E to R. From this and (6.70), we see that each Ek is measurable. Next, from (6.68) and (6.70), one can directly check that E=

+∞  k=1

Ek and Ej ⊆ E for all j,  ∈ N+ with j ≤ .

(6.72)

306

6 Bang-Bang Property of Optimal Controls

Two observations are given in order: (i) There exists  k ∈ N+ so that |Ek | > 0. (This follows from (6.72) and the fact that |E| > 0.) (ii) It follows from (6.70) that dU (u∗ (t), ∂U) > 1/ k for each t ∈ Ek .

(6.73)

We define a subset of Ek in the following manner: Ek,1  {t ∈ Ek : |Ek |/3 < t < tE∗ − |Ek |/3}.

(6.74)

Then one can easily check that Ek,1 ⊆ (|Ek |/3, tE∗ − |Ek |/3) and |Ek,1 | ≥ |Ek |/3.

(6.75)

Now we will use the above observations (i)–(ii) and Proposition 6.1 to derive a contradiction. 41), we can apply Proposition 6.1 to find that the system (6.65) is Indeed, by (H L∞ -null controllable from measurable sets in time. Then by Definition 6.3 (where (T , t, E) is replaced by (tE∗ − |Ek |/3, |Ek |/3, Ek,1 )), for each δ ∈ (0, |Ek |/3) satisfying that # " y0  y0∗ − y(δ; 0, y0∗ , u∗ ),  y0 Y ≤ 1/  kC max∗ eAs L (Y ) with  0≤s≤tE

(6.76)

there is  u ∈ L∞ (0, +∞; U ) so that y(tE∗ − |Ek |/3; |Ek |/3, y(|Ek |/3; δ,  y0, 0), χEk,1  u) = 0,

(6.77)

and so that uL∞ (0,+∞;U ) ≤ Cy(|Ek |/3; δ,  y0, 0)Y . χEk,1 

(6.78)

(Here, C is the corresponding constant in Definition 6.3.) It follows from (6.75) and (6.77) that y0, χEk,1  u) = 0. y(tE∗ − |Ek |/3; δ, 

(6.79)

Define a new control in the following manner:  u(t)  u∗ (t + δ) + χEk,1 (t + δ) u(t + δ), 0 < t < tE∗ − δ.

(6.80)

Then we have that  u(t) ∈ U for a.e. t ∈ (0, tE∗ − δ). In fact, by the third conclusion in (6.69) and (6.73), we obtain that

(6.81)

6.2 Bang-Bang Property and Null Controllability

307

u∗ (t) + B1/k (0) ⊆ U for a.e. t ∈ Ek . This, together with (6.80), (6.78), (6.76), and (6.74), yields that u(t + δ)  u(t) = u∗ (t + δ) +  ∈ u∗ (t + δ) + B1/k (0) ⊆ U for a.e. t ∈ Ek,1 − {δ}, which, together with (6.80) and the third conclusion in (6.69), leads to (6.81). Meanwhile, from (6.80), (6.76), and (6.75) it follows that ∗

u) = eA(tE −δ) y0∗ y(tE∗ − δ; 0, y0∗ ,  t ∗ −δ   E ∗ + eA(tE −δ−t ) B u∗ (t + δ) + (χEk,1  u)(t + δ) dt 0

/

∗

= eA(tE −δ) y(δ; 0, y0∗, u∗ ) + /

∗

y0 + + eA(tE −δ)



tE∗ δ

=



tE∗

∗

eA(tE −t ) Bu∗ (t)dt

0

δ

∗

eA(tE −t ) B(χEk,1  u)(t)dt

y(tE∗ ; 0, y0∗ , u∗ ) + eA|Ek |/3 y(tE∗

0

− |Ek |/3; δ,  y0, χEk,1  u),

which, combined with (6.79) and the second conclusion in (6.69), indicates that y(tE∗ − δ; 0, y0∗ , u) = y(tE∗ ; 0, y0∗ , u∗ ) ∈ YE .

(6.82)

u) is an admissible Finally, from (6.82) and (6.81), we see that (0, y0∗ , tE∗ − δ, QS ,QE S ,QE ∗ tetrad of (T P )min , which contradicts the optimality of tE . Hence, (T P )Q min holds the bang-bang property. This completes the proof of Theorem 6.5.   We are now in the position to show Proposition 6.1. For this purpose, some preliminaries are needed. The first preliminary concerns complexifications. Write Y C and U C for the complexifications of Y and U , respectively. Define two operators in the following manner: AC (y1 + iy2)  Ay1 + iAy2 for each y1 , y2 ∈ D(A),

(6.83)

B C (u1 + iu2 )  Bu1 + iBu2 for each u1 , u2 ∈ U.

(6.84)

and

From (6.83) and (6.84), we see that for each z ∈ Y and each u ∈ U , C )∗ t

e(A

∗

z = eA t z for all t ≥ 0 and (B C )∗ u = Bu.

308

6 Bang-Bang Property of Optimal Controls

41), indicates the following two properties: This, along with (ii) and (iii) of (H (ii) The operator AC generates an analytic semigroup {eA t }t ≥0 over Y C . (iii) There exist two constants d > 0 and σ > 0 so that for each L ∈ (0, 1], C

e

(AC )∗ L

z2Y C

≤e

d/Lσ



L

C )∗ (L−t )

(B C )∗ e(A

0

z2U C dt for all z ∈ Y C .

41).) (Here (d, τ ) is given by (iii) of (H Thus we conclude that (ii) and (iii) of (H 1) ⇒ (ii) and (iii) .

(6.85)

The second preliminary is the L1 -observability inequality from measurable sets in time, presented in the next Lemma 6.4, which will be proved after the proof of Proposition 6.1. 41) holds. Let T > t > 0 and let E ⊆ (t, T ) be a subset Lemma 6.4 Assume that (H of positive measure. Then there exists a constant C0  C0 (A, B, T , t, E) > 0 so that  C ∗ C ∗ e(A ) (T −t ) zY C ≤ C0 (B C )∗ e(A ) (T −s) zU C ds for all z ∈ Y C . (6.86) E

Now we give the proof of Proposition 6.1. Proof (Proof of Proposition 6.1) Arbitrarily fix T > t > 0 and a subset E (in (t, T )) of positive measure. According to Lemma 6.4, (6.83), and (6.84), there is C0  C0 (A, B, T , t, E) > 0 so that  ∗ A∗ (T −t ) e zY ≤ C0 B ∗ eA (T −s) zU ds for all z ∈ Y. E

Then by the similar arguments to those used in the proof of Theorem 1.20, we can obtain (6.67). Thus, the system (6.65) is L∞ -null controllable from measurable sets in time (see Definition 6.3). This ends the proof of Proposition 6.1.   Next we go back to the proof of Lemma 6.4. Two other lemmas (Lemma 6.5 and Lemma 6.6) are needed. We will prove them after the proof of Lemma 6.4. 41) holds. Let 0 < t1 < t2 with 0 < t2 − t1 ≤ 1. Let Lemma 6.5 Suppose that (H η ∈ (0, 1) and E be a measurable subset so that |E ∩ (t1 , t2 )| ≥ η(t2 − t1 ).   C(A,  Then there exist two constants C B, η) > 0 and θ  θ (A, B, η) ∈ (0, 1) C so that for each z ∈ Y ,

6.2 Bang-Bang Property and Null Controllability

e

(AC )∗ t2

"

zY C

 C

 (t2 −t1 )σ ≤ Ce



309 C )∗ t

(B C )∗ e(A E∩(t1 ,t2 )

#θ zU C dt

C )∗ t

e(A

1

z1−θ . YC (6.87)

Lemma 6.6 Let E ⊆ (0, +∞) be a subset of positive measure. Let  ∈ (0, +∞) be a Lebesgue density point of E. Then for each λ ∈ (0, 1), there exists an increasing sequence {m }m≥1 ⊆ (0, +∞) ∩ ( − 1, ) so that for each m ∈ N+ ,  − m = λm−1 ( − 1 ) and |E ∩ (m , m+1 )| ≥

1 |m+1 − m |. 3

(The above Lemma 6.6 is quoted from [20, Proposition 2.1].) We now give the proof of Lemma 6.4. Proof (Proof of Lemma 6.4.) First of all, by (6.85), we have (ii) and (iii) . Let  ∈ (t, T ) be a Lebesgue density point of E. Let σ be given by (iii) and −1 −1  θ )  (C(A,  let (C,   B, 3 ), θ (A, B, 3 )) be given by Lemma 6.5. We choose 1/σ λ ∈ (1 − θ ) , 1 . According to Lemma 6.6, there exists an increasing sequence {m }m≥1 ⊆ (t, ) ∩ ( − 1, ) so that for each m ∈ N+ ,  − m = λm−1 ( − 1 ) and |E ∩ (m , m+1 )| ≥

1 |m+1 − m |. 3

(6.88)

Arbitrarily fix z ∈ Y C . Define C )∗ (T −t )

h(t)  e(A

z, t ∈ [0, T ].

(6.89)

For each m ∈ N+ , we apply Lemma 6.5, where (E, t1 , t2 , η) is replaced by ({T } − E ∩ (m , m+1 ), T − m+1 , T − m , 1/3), to find that h(m )Y C

 "  m+1 −m )σ C/(  ≤ Ce E∩(m ,m+1 )

(B C )∗ h(s)U C ds

#θ

h(m+1 )1−θ . YC

By the latter inequality and the Young inequality: ab ≤ εa p + ε

− pr r

b when a > 0, b > 0, ε > 0 with

1 1 + = 1, p > 1, r > 1, p r

we see that for each m ∈ N+ and each ε > 0, h(m )Y C ≤ εh(m+1 )Y C   C σ  + (1−θ)/θ eC/(m+1 −m ) (B C )∗ h(s)U C ds, ε E∩(m ,m+1 )

310

6 Bang-Bang Property of Optimal Controls

which implies that for each m ∈ N+ and each ε > 0, ε(1−θ)/θ e   ≤C

 C σ m+1 −m )

− (

E∩(m ,m+1 )

h(m )Y C − ε1/θ e

 C σ m+1 −m )

− (

h(m+1 )Y C

(B C )∗ h(s)U C ds.

In the above, we choose, for each m ∈ N+ , /

ε  εm  e

0

 C σ m+1−m )

− (

1−λσ λσ +(1−θ )(λσ −1)/θ

.

Then we obtain that for each m ∈ N+ ,  σ

e

Cλ − [λσ +(1−θ )(λσ −1)/θ ](

−e

m+1 −m )

σ

h(m )Y C

 σ Cλ − [λσ +(1−θ )(λσ −1)/θ ](m+2 −m+1 )σ

 ≤C

 E∩(m ,m+1 )

h(m+1 )Y C

(B C )∗ h(s)U C ds.

Summing the above over all m, we see that  σ Cλ

 [λσ +(1−θ )(λσ −1)/θ ](2 −1 )σ h(1 )Y C ≤ Ce

 E∩(1 ,)

(B C )∗ h(s)U C ds.

This, along with (6.88) and (6.89), leads to (6.86). Hence, we finish the proof of Lemma 6.4.

 

The remainder on Proposition 6.1 is to give proofs of Lemma 6.5 and Lemma 6.6. To show Lemma 6.5, we need a propagation of smallness estimate from measurable sets for real analytic functions, built up in [25] (see also [1, Lemma 2] or [2, Lemma 13]). This estimate will be given in Lemma 6.7, without proof. Lemma 6.7 Let f : [a, a + s] → R, where a ∈ R and s > 0, be an analytic function satisfying that |f (β) (x)| ≤ Mβ!(sρ)−β for all x ∈ [a, a + s] and β ∈ N+ ∪ {0}  ⊆ [a, a + s] is a with some constants M > 0 and ρ ∈ (0, 1]. Assume that E  subset of positive measure. Then there are two constants C  C(ρ, |E|/s) ≥ 1 and  ϑ  ϑ(ρ, |E|/s) ∈ (0, 1) so that f L∞ (a,a+s) ≤ CM 1−ϑ We now give the proof of Lemma 6.5.

" 1  #ϑ |f (x)|dx .  E |E|

6.2 Bang-Bang Property and Null Controllability

311

Proof (Proof of Lemma 6.5) First of all, we recall (6.85). Let 0 < t1 < t2 with 0 < t2 − t1 ≤ 1. Let η ∈ (0, 1) and E be a measurable subset so that |E ∩ (t1 , t2 )| ≥ η(t2 − t1 ). Arbitrarily fix z ∈ Y C . The rest of the proof is carried out by the following two steps: Step 1. We show two facts. Fact One: The following function is real analytic: C )∗ t

g(t; z)  (B C )∗ e(A

z2U C ,

t > 0;

Fact Two: There is ρ ∈ (0, 1) (only depending on AC ) so that for all t, s > 0 with 0 < t − s ≤ 1, β! C ∗ e(A ) s z2Y C for all β ∈ N+ . β [ρ(t − s)] (6.90) Recall properties (ii) and (iii) before (6.85). Fact One follows from (ii) . We now show Fact Two. According to (ii) , there exist three positive constants M, ω and δ (only depending on AC ) so that |g (β) (t; z)| ≤ B C 2L (U C ,Y C )

C

eA τ L (Y C ,U C ) ≤ Meωτ for each τ ∈ Δδ  {τ ∈ C : |arg τ | < δ}.

(6.91)

We claim that there exists ρ1 ∈ (0, 1) (only depending on AC ) so that C ∗

[(AC )∗ ]α e(A ) τ yY C α! ≤ yY C for each τ ∈ (0, 1], α ∈ N+ ∪ {0} and y ∈ Y C . (ρ1 τ )α

(6.92)

Indeed, by (ii) and (6.91), we see that AC − ωI generates a bounded analytic semigroup over Y C . (Here and in what follows, I denotes the identity operator from Y C to Y C .) This, together with Theorem 5.2 in Chapter 2 of [19], implies that for each τ > 0, C −ωI )τ

(AC − ωI )e(A

 ≤ C1 /τ (with C1 > 0 depending only on AC ).

From the above inequality, (ii) and (6.91), we see that for each τ > 0, C )∗ τ

(AC )∗ e(A

C

L (Y C ,Y C ) = AC eA τ L (Y C ,Y C )  C ≤ eωτ (AC − ωI )e(A −ωI )τ L (Y C ,Y C )  C +ωe(A −ωI )τ L (Y C ,Y C ) ≤ eωτ (C1 /τ + ωM),

312

6 Bang-Bang Property of Optimal Controls

which indicates that C )∗ τ

(AC )∗ e(A

L (Y C ,Y C ) ≤ eω (C1 + ωM)/τ  C2 /τ for all τ ∈ (0, 1].

This yields that for each α ∈ N+ ∪ {0} and τ ∈ (0, 1], C )∗ τ

[(AC )∗ ]α e(A

C )∗ τ/α

yY C = [(AC )∗ e(A

]α yY C ≤

" C α #α 2 yY C . τ

From this and the Stirling formula: α α  eα α! for each α ∈ N+ ∪ {0}, we obtain (6.92). Next, we arbitrarily fix t, s > 0 with 0 < t − s ≤ 1. For each β ∈ N+ ∪ {0}, since g (β) (t; z) =

β1 +β2 =β

dβ1  C ∗ (AC )∗ t  dβ2  C ∗ (AC )∗ t ! β! (B ) e z , β (B ) e z C, U β1 !β2 ! dt β1 dt 2

it follows that |g (β) (t; z)|

C ∗ C ∗ β! ≤ (B C )∗ [(AC )∗ ]β1 e(A ) t zU C (B C )∗ [(AC )∗ ]β2 e(A ) t zU C β1 !β2 ! β1 +β2 =β

≤ (B C )∗ 2L (Y C ,U C )

β1 +β2 =β

C ∗ C ∗ β! [(AC )∗ ]β1 e(A ) t zY C [(AC )∗ ]β2 e(A ) t zY C . β1 !β2 !

(6.93)

Finally, from (6.93) and (6.92), we see that for each β ∈ N+ ∪ {0}, |g (β) (t; z)| ≤ (B C )∗ 2L (Y C ,U C ) [(AC )∗ ]β2 e

β1 +β2 =β

(AC )∗ (t −s)

≤ (B C )∗ 2L (Y C ,U C )

β! C ∗ C ∗ [(AC )∗ ]β1 e(A ) (t −s)(e(A ) s z)Y C β1 !β2 !

C )∗ s

(e(A

β1 +β2 =β

z)Y C

β1 !β2 ! β! (AC )∗ s 2 zY C β1 +β2 e  β1 !β2 ! ρ1 (t − s)

2 (β + 1)!  C ∗ = (B C )∗ L (Y C ,U C ) (e(A ) s )zY C  β . ρ1 (t − s) Since β + 1 ≤ 2β for each β ∈ N+ ∪ {0}, the above leads to (6.90). So Fact Two is true.

6.2 Bang-Bang Property and Null Controllability

313

Step 2. We prove (6.87). Set τ  t1 +

η   E ∩ [τ, t2 ]. (t2 − t1 ) and E 2

It is clear that  ≥ |E ∩ (t1 , t2 )| − (τ − t1 ) ≥ |E|

η (t2 − t1 ). 2

 be a measurable subset so that 1 ⊆ E Let E 1 | = |E

η (t2 − t1 ). 2

By Step 1, there exist two constants K ≥ 1 and ρ ∈ (0, 1) (only depending on AC and B C ) so that for each t ∈ [τ, t2 ] and β ∈ N+ ∪ {0}, |g (β) (t; z)| ≤ K ≤K

β! C ∗ e(A ) t1 z2Y C [ρ(t − t1 )]β β! C ∗ e(A ) t1 z2Y C . β [ρη(t2 − t1 )/2]

Then according to Lemma 6.7, where  M, ρ, s, a) is replaced by (E,

"

" # # 1 , Ke(AC )∗ t1 z2 C , ρη , 1 − η (t2 −t1 ), τ , E Y 2−η 2

there exists C3 > 0 and ν ∈ (0, 1) (only depending on AC , B C and η) so that for each t ∈ [τ, t2 ], " #1−ν " 1  #ν C ∗ |g(s; z)|ds . |g(t; z)| ≤ C3 Ke(A ) t1 z2Y C 1 | E1 |E Thus, for each t ∈ [τ, t2 ], we have that C )∗ t

(B C )∗ e(A

z2U C ≤

" 2C3 K C ∗ 2(1−ν) e(A ) t1 zY C η(t2 − t1 )



C )∗ s

1 E

(B C )∗ e(A

From this and (iii) , we see that C )∗ t

C ∗

C ∗

z2Y C = e(A ) (t2 −τ ) (e(A ) τ z)2Y C  t2 −τ C ∗ C ∗ d/(t2 −τ )σ ≤e (B C )∗ e(A ) (t2 −τ −s) (e(A ) τ z)2U C ds e(A

2

0

z2U C ds

#ν .

314

6 Bang-Bang Property of Optimal Controls

=e

d/(t2 −τ )σ



t2

C )∗ t

(B C )∗ e(A

τ

≤ ed/(t2−τ ) (t2 − τ ) σ

z2U C dt

" 2C3 K C ∗ e(A ) t1 z2(1−ν) YC η(t2 − t1 )



C )∗ t

1 E

(B C )∗ e(A

z2U C dt

#ν .

This implies that C )∗ t

e(A

2

(2 − η)C3 K (AC )∗ t1 2(1−ν) e zY C × η  " #ν C ∗ C ∗ sup (B C )∗ e(A ) s zU C (B C )∗ e(A ) t zU C dt

z2Y C ≤ ed/(t2−τ )

σ

1 s∈E

E

0 σ (2 − η)C3 K ≤ ed/(t2−τ ) η  (B C )∗ L (U C ,Y C ) /

C )∗ t

×e(A

1

z2−ν YC

"

C )∗ s

sup

s∈[0,t2 −t1 ] C )∗ t

(B C )∗ e(A E

e(A

zU C dt

L (Y C )

ν

#ν ,

which leads to (6.87). This completes the proof of Lemma 6.5.

 

Finally, we prove Lemma 6.6. Proof (Proof of Lemma 6.6.) Arbitrarily fix λ ∈ (0, 1). Write E c for the complement subset of E. Since  is a Lebesgue density point of E, we have that lim

h→0+

1 c |E ∩ ( − h,  + h)| = 0. h

This yields that 1 |E ∩ ( − λm ,  − λm+1 )| − λm+1  1 |E c ∩ ( − λm ,  − λm+1 )| = 1 − lim m m→+∞ λ  − λm+1  2 1 lim |E c ∩ ( − λm ,  + λm )| = 1. = 1− 1 − λ m→+∞ 2λm  lim

m→+∞ λm 

Thus, there exists m0 ∈ N+ so that for each m ≥ m0 , 1 |E ∩ ( − λm ,  − λm+1 )| ≥ 1/3 and λm0  < 1. λm  − λm+1 

(6.94)

6.2 Bang-Bang Property and Null Controllability

315

Let m   − λm+m0  for each m ∈ N+ . We get the desired result from (6.94) immediately. This completes the proof of Lemma 6.6.   We end this subsection with the next example which gives auxiliary instructions on Theorem 6.5. Q ,Q

S E , where the controlled system is Example 6.10 Consider the problem (T P )min the heat equation (3.59) (with f = 0); and where

QS = {0} × {y0 }, QE = (0, +∞) × Br (0) and U = Bρ (0), with y0 ∈ L2 (Ω) \ Br (0), r > 0 and ρ > 0. One can easily check the following facts: Q ,Q

S E (i) The problem (T P )min can be put into the framework (B2 ) (given at the beginning of Section 6.2); QS ,QE (ii) The problem (T P )min has at least one admissible control (see Remark 3.1 after Theorem 3.5), which leads to the existence of optimal controls; 41) is satisfied for this case (see [5, 20] and [21]). (iii) The assumption (H

Q ,QE

S Then according to Theorem 6.5, (T P )min end Example 6.10.

holds the bang-bang property. Now we

6.2.2 Bang-Bang Property of Maximal Time Control In this subsection, we aim to use the L∞ -null controllability (from measurable sets in time) to study the bang-bang property for the maximal control problem S ,QE . For this purpose, we need the following assumption: (T P )Q max 42) The system (6.65) is L∞ -null controllable from measurable sets in time. (H The main result of this subsection is stated as follows: 42) holds. Let T > 0 and let Theorem 6.6 Suppose that the assumption (H  QS = {(t, Φ(t, 0)z)  0 ≤ t < T , z ∈ YS } and QE = {T } × YE , where YS and YE are two nonempty subsets of Y with (Φ(T , 0)YS ) ∩ YE = ∅. Then S ,QE the maximal time control problem (T P )Q holds the bang-bang property. max Q ,Q

S E did not hold the bangProof By contradiction, we suppose that (T P )max bang property. Then by Definition 6.2, there would exist an optimal tetrad S ,QE and a subset E ⊆ (tS∗ , T ) of positive measure (tS∗ , Φ(tS∗ , 0)y0∗ , T , u∗ ) of (T P )Q max so that

u∗ (t) ∈ / ∂U for a.e. t ∈ E.

(6.95)

316

6 Bang-Bang Property of Optimal Controls

It is clear that 0 ≤ tS∗ < T , y0∗ ∈ QS , y(T ; tS∗ , Φ(tS∗ , 0)y0∗ , u∗ ) ∈ YE

(6.96)

and u∗ (t) ∈ U for a.e. t ∈ (tS∗ , T ).

(6.97)

For each k ∈ N+ , we define a subset of E in the following manner:    Ek  t ∈ E  dU (u∗ (t), ∂U) > k −1 ,

(6.98)

where dU (·, ∂U) denotes the distance between · and ∂U in U . By the same way as that used in the proof of Theorem 6.5, we can prove that for each k ∈ N+ , Ek is measurable. Meanwhile, from (6.95) and (6.98), one can directly check that E=

+∞ 

Ek and Ej ⊆ E for all j,  ∈ N+ with j ≤ .

(6.99)

k=1

Two observations are given in order: (i) There exists  k ∈ N+ so that |Ek | > 0; (This follows from (6.99) and the fact that |E| > 0.) (ii) It follows from (6.98) that dU (u∗ (t), ∂U) > 1/ k for each t ∈ Ek .

(6.100)

Define a subset of Ek in the following manner:  Ek,1  {t ∈ Ek  tS∗ + |Ek |/2 < t < T }.

(6.101)

It is obvious that Ek,1 ⊆ (tS∗ + |Ek |/2, T ) and |Ek,1 | ≥ |Ek |/2.

(6.102)

We now use the above observations (i) and (ii) to obtain a contradiction. First, by 42) and Definition 6.3 (where (T , t, E) is replaced by (T , t ∗ + |E|/2, E )), we (H k k,1 S see that for each δ ∈ (0, |Ek |/2) satisfying that " kC  y0 Y ≤ 1/ 

sup

0≤s≤t ≤T

Φ(t, s)L (Y )

#

there is a control  u ∈ L∞ (0, +∞; U ) so that

with  y0  −y(tS∗ + δ; tS∗ , 0, u∗ ), (6.103)

6.2 Bang-Bang Property and Null Controllability

317

y(T ; tS∗ + |Ek |/2, y(tS∗ + |Ek |/2; tS∗ + δ,  y0 , 0), χEk,1  u) = 0

(6.104)

and so that uL∞ (0,+∞;U ) ≤ Cy(tS∗ + |Ek |/2; tS∗ + δ,  y0 , 0)Y . χEk,1 

(6.105)

(Here, C is the corresponding constant given by Definition 6.3.) Then, it follows from (6.102) and (6.104) that y0 , χEk,1  u) = 0. y(T ; tS∗ + δ, 

(6.106)

Define a new control in the following manner: u(t),  u(t)  u∗ (t) + χEk,1 (t)

t ∈ (tS∗ + δ, T ).

(6.107)

We claim that  u(t) ∈ U for a.e. t ∈ (tS∗ + δ, T ).

(6.108)

Indeed, by (6.97) and (6.100), we obtain that u∗ (t) + B1/k (0) ⊆ U for a.e. t ∈ Ek . From the above, (6.107), (6.101), (6.105), and (6.103), we find that  u(t) = u∗ (t) +  u(t) ∈ u∗ (t) + B1/k (0) ⊆ U for a.e. t ∈ Ek,1 . This, along with (6.107) and (6.97), leads to (6.108). Next, by (6.107) and (6.103), one can directly check that y(T ; tS∗ + δ, Φ(tS∗ + δ, 0)y0∗ , u) y0 + = y(T ; tS∗ , Φ(tS∗ , 0)y0∗ , u∗ ) + Φ(T , tS∗ + δ)



T tS∗ +δ

Φ(T , t)B(t)(χEk,1  u)(t)dt,

which, combined with (6.106) and the third conclusion of (6.96), indicates that y(T ; tS∗ + δ, Φ(tS∗ + δ, 0)y0∗ , u) y0 , χEk,1  u) = y(T ; tS∗ , Φ(tS∗ , 0)y0∗ , u∗ ) + y(T ; tS∗ + δ,  = y(T ; tS∗ , Φ(tS∗ , 0)y0∗ , u∗ ) ∈ YE .

(6.109)

Finally, since 0 ≤ tS∗ + δ < T , it follows by the second conclusion of (6.96), (6.109) and (6.108) that (tS∗ + δ, Φ(tS∗ + δ, 0)y0∗ , T , u) is an admissible tetrad of QS ,QE S ,QE ∗ (T P )max , which contradicts the optimality of tS . Thus, (T P )Q holds the max bang-bang property. This completes the proof of Theorem 6.6.  

318

6 Bang-Bang Property of Optimal Controls

We end this subsection with an example which gives auxiliary instructions on Theorem 6.6. Q ,Q

S E Example 6.11 Consider the maximal time control problem (T P )max , where the ∞ ∞ controlled system is the heat equation (3.107) (with a ∈ L (0, +∞; L (Ω))); and where

 QS = {(t, Φ(t, 0)y0 )  0 ≤ t < T }, QE = {T } × Br (0) and U = Bρ (0), with Φ(T , 0)y0 ∈ L2 (Ω) \ Br (0), r > 0 and ρ > 0. One can easily check the following facts: S ,QE (i) (T P )Q can be put into the framework of (B2 ) (given at the beginning of max Section 6.2); 42) (given at the beginning of Section 6.2.2) holds in this (ii) The assumption (H case. (Here, we used Theorems 1.22 and 1.21.)

Q ,Q

S E If we further assume that (T P )max has optimal controls, then according to QS ,QE Theorem 6.6, (T P )max holds the bang-bang property. We end Example 6.11.

6.3 Bang-Bang Property and Maximum Principle In this section, we will study the bang-bang property of the minimal time control QS ,QE problem (T P )min from the perspective of the Pontryagin Maximum Principle. This problem is under the framework (B3 ): (i) The state space Y and the control space U are two real separable Hilbert spaces. (ii) The controlled system is as: y(t) ˙ = Ay(t) + D(t)y(t) + B(t)u(t), t ∈ (0, +∞),

(6.110)

where A : D(A) ⊆ Y → Y generates a C0 semigroup {eAt }t ≥0 over Y ; D(·) ∈ L∞ (0, +∞; L (Y )) and B(·) ∈ L∞ (0, +∞; L (U, Y )). Given τ ∈ [0, +∞), y0 ∈ Y and u ∈ L∞ (τ, +∞; U ), we write y(·; τ, y0 , u) for the solution of (6.110) over [τ, +∞) with the initial condition that y(τ ) = y0 . Moreover, write {Φ(t, s) : t ≥ s ≥ 0} for the evolution system generated by A + D(·) over Y . (iii) The control constraint set U is bounded, convex, and closed in U , and has a nonempty interior in U . (iv) Sets QS and QE are disjoint nonempty subsets of [0, +∞)×Y and (0, +∞)× Y , respectively. Q ,QE

S As what we did in the previous two sections, we always assume that (T P )min optimal controls.

has

6.3 Bang-Bang Property and Maximum Principle

319

We now give the definition of the bang-bang property for the problem S ,QE . (T P )Q min Definition 6.4 (i) An optimal control u∗ (associated with an optimal tetrad (0, y0∗ , tE∗ , u∗ )) to the QS ,QE is said to have the bang-bang property if problem (T P )min u∗ (t) ∈ ∂U for a.e. t ∈ (0, tE∗ ), where ∂U denotes the boundary of U. S ,QE is said to have the bang-bang property, if any optimal (ii) The problem (T P )Q min control satisfies the bang-bang property. 43) plays an In this section, the following unique continuation property (H important role: 43) If there is a measurable subset E ⊆ (0, t0 ) (with t0 > 0 and |E| > 0) and (H z ∈ Y so that B(t)∗ Φ(T , t)∗ z = 0 for each t ∈ E, then z = 0 in Y . The first main result of this section is stated as follows: 43) holds. Let YS ⊆ Y be a nonempty subset and Theorem 6.7 Suppose that (H YE ⊆ Y be a bounded, convex, and closed subset with a nonempty interior. Assume that YS ∩ YE = ∅. Let QS = {0} × YS and QE = (0, +∞) × YE . Q ,QE

S Then the problem (T P )min

holds the bang-bang property.

Proof It suffices to prove this theorem in the case that YS is a singleton subset of Y . The reason is the same as that stated in the proof of Theorem 6.1. Thus, we can let YS = {y0 } with y0 ∈ Y .  QS ,QE Let (0, y0 , tE∗ , u∗ ) be an optimal tetrad to (T P )min . Since YS YE = ∅, we have that tE∗ > 0. According to Theorem 4.1 and Theorem 4.2, the problem QS ,QE satisfies the classical Pontryagin Maximum Principle. Then by Defini(T P )min tion 4.1, there exists z∗ ∈ Y \ {0} so that u∗ (t), B(t)∗ Φ(tE∗ , t)∗ z∗ U = maxv, B(t)∗ Φ(tE∗ , t)∗ z∗ U for a.e. t ∈ (0, tE∗ ). v∈U

(6.111) 43) to derive that We now are going to use (6.111) and (H u∗ (t) ∈ ∂U for a.e. t ∈ (0, tE∗ ).

(6.112)

By contradiction, we suppose that (6.112) was not true. Then there would exist a subset E ⊆ (0, tE∗ ) of positive measure so that u∗ (t) ∈ Int U for a.e. t ∈ E.

320

6 Bang-Bang Property of Optimal Controls

This, together with (6.111), yields that B(t)∗ Φ(tE∗ , t)∗ z∗ = 0 for a.e. t ∈ E, 43), indicates that z∗ = 0 in Y . This leads to a contradiction which, combined with (H ∗ since z = 0. Therefore, (6.112) is true, i.e., u∗ holds the bang-bang property, S ,QE . Thus, we end the proof of Theorem 6.7.   consequently, so does (T P )Q min The next example gives auxiliary instructions on Theorem 6.7. Example 6.12 Consider the controlled system: y(t) ˙ = D(t)y(t) + B(t)u(t), t ∈ (0, +∞), (with y(t) ∈ Rn and u(t) ∈ Rm ) (6.113) where (D(·), B(·)) ∈ L∞ (0, +∞; Rn×n ) × L∞ (0, +∞; Rn×m ) (with n, m ∈ N+ ). S ,QE Consider the problem (T P )Q , where the controlled system is (6.113) and where min QS = {0} × {y0 }, QE = (0, +∞) × Br (0) and U = Bρ (0), S ,QE with y0 ∈ Rn \ Br (0) and r, ρ > 0. Then one can check that (T P )Q can be put min into the framework (B3 ) (given at the beginning of Section 6.3). S ,QE 43) holds has optimal controls and that (H If we further assume that (T P )Q min QS ,QE for this case, then according to Theorem 6.7, (T P )min has the bang-bang property. We now end Example 6.12.

The second main result of this section is as follows: Theorem 6.8 Let D(·) ≡ 0, B(·) ≡ B ∈ L (U, Y ), and let QS = {0} × YS and QE = (0, +∞) × {0}, 43) holds and that / YS ). Suppose that (H where YS is a nonempty subset of Y (with 0 ∈ ∞ the system (6.110) is L -null controllable over any interval. Assume that 0 ∈ Int U. S ,QE Then the corresponding minimal time control problem (T P )Q holds the bangmin bang property. Proof It suffices to prove this theorem in the case that YS is a singleton subset of Y . The reason is the same as that stated in the proof of Theorem 6.1. Thus, we can let YS = {y0 } with y0 = 0. S ,QE Let (0, y0 , tE∗ , u∗ ) be an optimal tetrad to the problem (T P )Q . According to min QS ,QE Theorem 4.4 and Theorem 4.3, (T P )min satisfies the local Pontryagin Maximum

6.3 Bang-Bang Property and Maximum Principle

321

Principle. Then by Definition 4.2, for each T ∈ (0, tE∗ ), there exists zT∗ ∈ Y \ {0} so that ∗ (T −t )

u∗ (t), B ∗ eA

∗ (T −t )

zT∗ U = maxv, B ∗ eA v∈U

zT∗ U for a.e. t ∈ (0, T ).

From this and using the similar arguments as those used in the proof of Theorem 6.7, we obtain that for each T ∈ (0, tE∗ ), u∗ (t) ∈ ∂U for a.e. t ∈ (0, T ), which leads to that u∗ (t) ∈ ∂U for a.e. t ∈ (0, tE∗ ), S ,QE i.e., (0, y0 , tE∗ , u∗ ) holds the bang-bang property, consequently, (T P )Q has the min bang-bang property. This ends the proof of Theorem 6.8.  

Before presenting the last main result of this section, we introduce the following hypothesis: 44) YT = ZT for each T ∈ (0, +∞), where (H ·L1 (0,T ;U)

YT  X T

,

 XT  {B(·)∗ Φ(T , ·)∗ z  z ∈ Y },   ZT  B(·)∗ ϕ ∈ L1 (0, T ; U )  for each s ∈ (0, T ), there exists

 zs ∈ Y so that ϕ(·) = Φ(s, ·)∗ zs over [0, s] .

Theorem 6.9 Let B(·) ≡ B and let QS = {0} × YS , QE = (0, +∞) × {0} and U = Bρ (0), 43) and / YS ) and ρ > 0. Suppose that (H where YS is a nonempty subset of Y (with 0 ∈ 44) hold and that the system (6.110) is L∞ -null controllable over each interval. (H QS ,QE Then the corresponding minimal time control problem (T P )min has the bangbang property. Proof It suffices to prove this theorem in the case that YS is a singleton subset of Y . The reason is the same as that stated in the proof of Theorem 6.1. Thus, we can let YS = { y0 } with  y0 = 0. In order to show this theorem, we only need to prove (4.138) and (4.139) of Theorem 4.5. Indeed, when they are proved, it follows from Theorem 4.5 and S ,QE satisfies the local Pontryagin Maximum Principle. Theorem 4.3 that (T P )Q min Then by the same arguments as those used in the proof of Theorem 6.8, we can QS ,QE verify the bang-bang property for (T P )min .

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6 Bang-Bang Property of Optimal Controls

We now turn to the proof of (4.138) of Theorem 4.5. Since YE = {0} and the system (6.110) is L∞ -null controllable over each interval, it follows from Theorem 1.20 and the definition of YC (t1 , t2 ) (see (4.82)) that 0 ∈ I ntYC (t1 , t2 ). Hence, (4.138) is true for the current case. We next show (4.139) of Theorem 4.5. By Proposition 4.7, we see that in order to show the above-mentioned (4.139), we only need to prove that for each t ∈ [0, +∞) and each y0 ∈ Y , the function T → N(t, T ; y0) (see (4.155)) is left continuous from (t, +∞) to [0, +∞]. To this end, we arbitrarily fix t ≥ 0. Using the similar arguments as those used in the proof of Proposition 5.3, we can see that in order to show the above-mentioned left continuity, it suffices to verify the following condition: Yt,T = Zt,T for each T > t,

(6.114)

where Yt,T and Zt,T are given by ·

1

Yt,T  X t,TL (t,T ;U) ,  Xt,T  {B ∗ Φ(T , ·)∗ z|(t,T )  z ∈ Y },  Zt,T  B ∗ ϕ ∈ L1 (t, T ; U )  for each s ∈ (t, T ), there exists  zs ∈ Y so that ϕ(·) = Φ(s, ·)∗ zs over [t, s] .

(6.115)

44)), Y0,T = YT and Z0,T = ZT , we find that it suffices to Since YT = ZT (see (H show (6.114) for the case that t > 0. The proof will be carried out by the following two steps: Step 1. We show that Yt,T ⊆ Zt,T . To this end, we arbitrarily take f ∈ Yt,T . According to the definition of Yt,T , there is a sequence of {zn }n≥1 ⊆ Y so that B ∗ Φ(T , ·)∗ zn → f (·) strongly in L1 (t, T ; U ).

(6.116)

Since the system (6.110) is L∞ -null controllable over each interval, by Theorem 1.20, there exists a positive constant C  C(T , t) so that ∗



T

Φ(T , t) zY ≤ C

B ∗ Φ(T , s)∗ zU ds for each z ∈ Y.

t

44)), yields that there This, together with (6.116) and the definition of YT (see (H exists g ∈ YT so that B ∗ Φ(T , ·)∗ zn → g(·) strongly in L1 (0, T ; U ) and g|(t,T ) = f. 44) and the definition of Zt,T (see (6.115)), we find that f ∈ Zt,T , From this, (H which leads to Yt,T ⊆ Zt,T .

6.3 Bang-Bang Property and Maximum Principle

323

Step 2. We prove that Zt,T ⊆ Yt,T . Arbitrarily fix f ∈ Zt,T . We first claim that there is f ∈ ZT so that f|(t,T ) = f.

(6.117)

For this purpose, we use the definition of Zt,T (given by (6.115)) to obtain the next two facts. Fact One: f ∈ L1 (t, T ; U );

(6.118)

Fact Two: For each s ∈ (t, T ), there exists zs ∈ Y so that f (τ ) = B ∗ Φ(s, τ )∗ zs for a.e. τ ∈ (t, s).

(6.119)

Meanwhile, from the observability estimate (i) of Theorem 1.20, we see from (6.119) that for each s1 , s2 with t < s1 ≤ s2 < T , B ∗ Φ(s1 , τ )∗ zs1 = B ∗ Φ(s2 , τ )∗ zs2 for each τ ∈ (0, s1 ).

(6.120)

Now, we arbitrarily take  s ∈ (t, T ) and define the following function: f(τ ) 



s, τ )∗ zs , τ ∈ (0, t], B ∗ Φ( f (τ ), τ ∈ (t, T ).

(6.121)

44)), we can easily Then from (6.118)–(6.120) and the definition of ZT (given by (H verify (6.117). 44)) and f ∈ ZT (see (6.117)), we obtain that Next, since YT = ZT (see (H  44)), indicates that there is f ∈ YT . This, along with the definition of YT (see (H {ηn }n≥1 ⊆ Y so that B ∗ Φ(T , ·)∗ ηn → f(·) strongly in L1 (0, T ; U ). This, together with the definition of Yt,T (see (6.115)) and (6.117), yields that f = f|(t,T ) ∈ Yt,T , which leads to that Zt,T ⊆ Yt,T . Finally, (6.114) follows from the conclusions in Step 1 and Step 2. Hence, we finish the proof of Theorem 6.9.   We next give three examples, which may help us to understand Theorems 6.9, 6.7, and 6.8 better. Example 6.13 Let Ω ⊆ Rd (d ≥ 1) be a bounded domain with a C 2 boundary ∂Ω. Let ω ⊆ Ω be a nonempty and open subset with its characteristic function χω . Let a1 ∈ L∞ (Ω) and a2 ∈ L∞ (0, +∞). Consider the controlled heat equation:

324

6 Bang-Bang Property of Optimal Controls

⎧ ⎨ ∂t y − Δy + (a1 (x) + a2 (t))y = χω u in Ω × (0, +∞), y=0 on ∂Ω × (0, +∞), ⎩ y(0) = y0 ∈ L2 (Ω),

(6.122)

where u ∈ L∞ (0, +∞; L2 (Ω)). Let Y = U  L2 (Ω); Let A = Δ with its domain H 2 (Ω) ∩ H01 (Ω); Let D(·) = −a1 − a2 (·); Let B = χω , where χω is treated as a linear and bounded operator on U . Thus, the equation (6.122) can be rewritten as: 

y(t) ˙ = Ay(t) + D(t)y(t) + Bu(t), y(0) = y0 .

t ∈ (0, +∞),

S ,QE Consider the problem (T P )Q , where the controlled system is as: min

y(t) ˙ = Ay(t) + D(t)y(t) + Bu(t), t ∈ (0, +∞)

(6.123)

and where QS = {0} × {y0 }, QE = (0, +∞) × {0} and U = Bρ (0), with y0 ∈ L2 (Ω) \ {0} and ρ > 0. Notice that this problem is exactly the problem given in Example 5.4. One can easily check the following facts: S ,QE can be put into the framework (B3 ), given at the beginning of (i) (T P )Q min Section 6.3 (see Example 5.4); 43) and (H 44) hold in the current case (see Theorem 1.22, Remark 1.5 (ii) Both (H (after Theorem 1.22), and Example 5.4); (iii) The system (6.123) is L∞ -null controllable over each interval (see Theorems 1.22 and 1.21). S ,QE We further assume that (T P )Q has optimal controls. Notice that it has optimal min controls when ρ is large enough (see Theorem 3.7). S ,QE holds the bang-bang property. We Now, according to Theorem 6.9, (T P )Q min end Example 6.13. S ,QE Example 6.14 Consider the problem (T P )Q in Example 6.13, where we min replace respectively QS and QE by {0} × {y0 } and (0, +∞) × Br (0) (with r > 0 and y0 ∈ L2 (Ω) \ Br (0)). One can easily check the following facts: S ,QE (i) (T P )Q can be put into the framework (B3 ), given at the beginning of min Section 6.3 (see Example 5.4); 43) holds (see Theorem 1.22, Remark 1.5 after Theorem 1.22). (ii) (H

Q ,Q

S E has optimal controls. Then, according to Furthermore, we assume that (T P )min QS ,QE Theorem 6.7, (T P )min holds the bang-bang property. We now end Example 6.14.

6.3 Bang-Bang Property and Maximum Principle

325

Q ,Q

S E Example 6.15 For the problem (T P )min in Example 6.13, we choose a1 (·) = a2 (·) = 0 in (6.122). Then D(·) = 0 in (6.123). One can easily check what follows:

Q ,Q

S E (i) (T P )min can be put into the framework of (B3 ) (given at the beginning of Section 6.3); 43) holds (see Theorem 1.22 and Remark 1.5 after Theorem 1.22); (ii) (H (iii) The system (6.123) is L∞ -null controllable over each interval (see Theorems 1.22 and 1.21); QS ,QE (iv) (T P )min has at least one admissible control (see Remark 3.1 after Theorem 3.5), which leads to the existence of optimal controls. S ,QE holds the bang-bang property. We now Then, according to Theorem 6.8, (T P )Q min end Example 6.15.

We have introduced two different methods to derive the bang-bang property in Section 6.2 and Section 6.3 respectively. In the next Remark 6.3, we will compare these two methods. Remark 6.3 Denote by (M1) and (M2) the methods introduced in Section 6.2 and Section 6.3, respectively. Several notes on (M1)and (M2) are given as follows: Q ,Q

S E in finitely dimensional cases, (M1) and (M2) are essentially (i) For (T P )min the same. On one hand, the key in (M1) is the L∞ -null controllability from 43) and measurable sets in time, while the keys in (M2) are the property (H QS ,QE in maximum principles. Maximum principles always hold for (T P )min finitely dimensional cases, provided that U, YS and YE are nonempty convex sets (see Theorem 4.1 and Theorem 4.2). On the other hand, the aforementioned 43) in finitely dimensional cases. We explain controllability is equivalent to (H this by the following controlled system:

y(t) ˙ = D(t)y(t) + Bu(t), t ∈ (0, +∞) (with y(t) ∈ Rn , u(t) ∈ Rm ), (6.124) where D(·) ∈ L∞ (0, +∞; Rn×n ) and B ∈ Rn×m . Indeed, we arbitrarily fix T > t0 ≥ 0 and a measurable subset E ⊆ (t0 , T ) with positive measure. Define the following map: F (B ∗ Φ(T , ·)∗ z|E )  Φ(T , t0 )∗ z for each z ∈ Rn .

(6.125)

43) holds, then by (6.125), we see that the map F is well defined. Since the If (H domain of F is of finite dimension, there exists a constant C  C(T , t0 , E) > 0 so that  Φ(T , t0 )∗ zRn ≤ C B ∗ Φ(T , t)∗ zRm dt for each z ∈ Rn . E

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6 Bang-Bang Property of Optimal Controls

Then using the similar arguments as those used in the proof of Theorem 1.20, we can obtain that the system (6.124) is L∞ -null controllable from measurable subsets in time. Conversely, if the system (6.124) is L∞ -null controllable from measurable subsets in time, then using the similar arguments to those used in 43) holds. the proof of Theorem 1.20, we can see that (H QS ,QE in infinitely dimension cases, (M1) and (M2) are not the same (ii) For (T P )min in general. (M1) works for the case when both U and YE are not convex and the controlled system is time-invariant, while (M2) works for some time-varying controlled systems, provided that both U and YE are convex.

Miscellaneous Notes The studies on the bang-bang property of time optimal control problems started with finite dimensional systems in 1950s (see [4] and [22]), and then extended to infinite dimensional systems in 1960s. To our best knowledge, [9] seems to be the first work dealing with the bang-bang property for abstract evolution equations in infinite dimensional spaces. One of the usual methods to derive the bang-bang property is the use of the controllability from measurable sets in time. (Simply, we call it as the Econtrollability.) The original idea of this method invisibly arose from [9]. This method was first presented formally in [18], which was partially inspired by [24]. More about this method, we would like to mention what follows: • This method works well for the minimal time control problems governed by time invariant systems, i.e., the E-controllability and the time-invariance of the system imply the bang-bang property for the corresponding minimal time control problem. • This method doesn’t seem to work for the minimal time control problems governed by time-varying systems. • This method works for the maximal time control problems governed by timevarying systems (see [18] and [20]). • The E-controllability, combined with a fixed point argument, implies the bangbang property of minimal time control problems governed by some semi-linear heat equations (see [21] and [29]). About studies on the E-controllability, we would like to give the following notes: • The E-controllability was first proved for the boundary controlled 1-dimension heat equation in [18]. • The E-controllability was showed for the internally controlled n-dimension heat equation in [26], with the aid of Lebeau and Robbiano’s spectrum inequality (see [13] and [14]).

6.3 Bang-Bang Property and Maximum Principle

327

• The E-controllability was built up for internally controlled n-dimension heat equations with lower terms (where the physical domain is bounded and convex) in [20], via the frequency function method used in [6] and [23]. Then the convexity condition was dropped in [21]. • The E-controllability was obtained for some evolution equations in [30]. • A much stronger version of the E-controllability for the internally (or boundary) controlled heat equation was built up in [2] (see also [7] and [8]). The corresponding minimal time control problem has the control constraint: |u(x, t)| ≤ M. This strong version implies that any optimal control u∗ satisfies that |u∗ (x, t)| = M for a.e. (x, t) ∈ Ω × (0, T ∗ ) (or |u∗ (x, t)| = M for a.e. (x, t) ∈ ∂Ω × (0, T ∗ )), where Ω is the physical domain and T ∗ is the minimal time. Another usual method to derive the bang-bang property is the use of the Pontryagin Maximum Principle, together with some unique continuation of the adjoint equation. This method was used in [4] for finite dimensional cases. Then it was extended to infinite dimensional cases (see, for instance, [10, 11, 16, 31], and [27]). As what we have seen in Chapter 4, for infinite dimensional cases, it is difficult to derive the Pontryagin Maximum Principle for the cases where controlled systems are infinitely dimensional and the target sets have empty interiors, such as points in the state spaces. Besides, it is not easy to obtain the required unique continuation for infinite dimensional cases, in general. Therefore, the studies on the bang-bang property for infinite dimensional cases via this way are much harder, compared with finite dimensional cases. In [15], the author studied the controlled system: y  = Ay + u, where A is the generator of a group, and the state and the control spaces are the same. For such a system, the bang-bang property of a kind of minimal time control problem is established. More about the bang-bang property, we would like to mention the works [3, 17, 28, 32], and [12]. Materials in Section 6.1 are partially summarized and developed from the related materials in [10, 22] and [33], while most examples in Section 6.1 are given by us. Materials in Section 6.2 are taken from [30] with a bit of modification. Materials in Section 6.3 are developed from [31] and [32]. Finally, we give some open problems which might interest readers: • Can we have the bang-bang property for infinite dimensional controlled systems with general control constraints, especially with the rectangular-type control constraint? • Does the bang-bang property hold for controlled systems with some state constraints? (For instance, heat equations with nonnegative states.) • Does the bang-bang property hold for general time-varying controlled systems in infinite dimensional spaces? • How to estimate switching times of a bang-bang control? • Do controlled wave equations (or controlled wave-like equations) have the bangbang property? (We believe that the key to answer this question is to understand the relation between the bang-bang property and the propagation.) • What happens for bang-bang controls from the numerical perspective?

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Index

A analyticity, 284

B blowup, 42, 117 boundary, 2

C Cauchy sequence, 2, 89 C0 group, 46, 47 characteristic polynomial, 73 closure, 2 compactness, 108 complex conjugate, 2 complexification, 307 condition Conti’s condition, 69 Dirichlet boundary condition, 85 Kalman controllability rank condition, 34, 217, 283, 286 maximum condition, 128, 135, 149, 196 transversality condition, 128, 149, 196 constant Lipschitz constant, 82 constraint ball-type constraint, 67, 282, 294 rectangle-type constraint, 282, 297 singleton constraint, 205 continuity, 235, 244 continuous absolutely continuous, 118 Hausdorff continuous, 139

Lipschitz continuous, 24, 262 upper semi-continuous, 107 control admissible control, 97, 208, 223, 288 bang-bang control, 98, 282 minimal norm control, 55, 259 target control, 254, 259 controllability, 23, 65 null controllability, 67, 303 controllable approximately controllable, 24, 218 exactly controllable, 24 null controllable, 24 convex hull, 4 convexity, 58, 70 core, 13 C0 semigroup, 6, 8, 240

D densely defined, 6 differentiable continuously differentiable, 80 domain, 2 duality pairing, 3

E eigenvalue, 73 equation adjoint equation, 33, 56 controlled equation, 42 dual equation, 60, 128 Euler equation, 126

© Springer International Publishing AG, part of Springer Nature 2018 G. Wang et al., Time Optimal Control of Evolution Equations, Progress in Nonlinear Differential Equations and Their Applications 92, https://doi.org/10.1007/978-3-319-95363-2

331

332

Index

equation (cont.) evolution equation, 6, 8 Frank-Kamenetskii equation, 54 heat equation, 32 linear time-invariant evolution equation, 41 linear time-varying evolution equation, 41 nonhomogeneous linear equation, 9 parabolic equation, 34 semilinear equation, 8 semilinear heat equation, 80 semilinear time-invariant evolution equation, 41 transport equation, 134 existence, 8, 67 extremal point, 16

injective, 31 inner product, 1, 2 in-radius, 86 interior, 2, 178 isomorphism, 88

F Filippov Lemma, 117 finite codimensionality, 140 formula representation formula, 128, 171 Stirling formula, 312 variation of constants formula, 8, 9 function analytic function, 76, 310 characteristic function, 16, 80, 256 distance function, 126 minimal norm function, 227, 232 nonzero function, 80 operator valued function, 9 simple function, 23 functional continuous linear functional, 16 linear bounded functional, 3 functional analysis, 1

M measurability, 17 measurable Lebesgue measurable, 15, 38 method recursive method, 18 modulus of continuity, 20, 23 monotonicity, 97, 209, 244

G generator, 7 globally Lipschitz, 80 graph, 3, 6, 14 H Hamiltonian, 128, 149, 196 Hausdorff metric, 136 homeomorphic, 261 hyperplane, 5, 127 separating hyperplane, 5 I inequality Gronwall’s inequality, 98, 114, 119

K kernel, 3, 17

L Lebesgue point, 130 linear isometry, 175 line segment, 57

N neighborhood, 13 neighbourhood-finite barycentric refinement, 14 norm, 1, 2 induced norm, 2 normal vector, 127 numerical analyses, 57

O observability, 23 observability estimate, 23, 323 operator adjoint operator, 4 closed operator, 6 compact operator, 4 evolution operator, 9, 48, 68, 138 nonlinear operator, 81 projection operator, 22 self-adjoint operator, 4 optimal control, 39, 161, 223 optimality, 264 optimality criterion, 37 optimal norm, 94 optimal state, 39, 56 optimal time, 39, 301

Index optimal trajectory, 39, 128, 135 optimization, 37 orthonormal basis, 41, 142

P paracompact, 14 partition of unity, 14 principle classical Pontryagin Maximum Principle, 127, 290, 319 local Pontryagin Maximum Principle, 148, 321 Principle of Uniform Boundedness, 3, 25 weak Pontryagin Maximum Principle, 170 property backward uniqueness property, 206, 241 bang-bang property, 55, 57, 95, 281, 282, 291, 298 strictly decreasing property, 265

Q quotient map, 31

R radius, 218 range, 3 reachable horizon, 99 reflexive, 4

S saddle point, 107 semigroup, 6, 38 analytic semigroup, 133, 308 compact semigroup, 113 separability, 5 separable, 2 separating vector, 5, 132 set admissible set, 39 admissible tetrad set, 112 Borel set, 21 closed set, 2 compact set, 2, 14 complement set, 314 constraint set, 80 control constraint set, 38, 41, 52, 65, 281 controllable set, 66, 148 convex set, 6 countable set, 129 ending set, 46

333 extreme set, 282 measurable set, 303 open set, 2 reachable set, 24, 65, 99, 148 relatively compact set, 2, 4 starting set, 46 weak closed set, 5 weakly star compact set, 16 set-valued map, 13 upper semicontinuous set-valued map, 13 singleton, 140, 290, 320 solution mild solution, 7, 8, 38 strong solution, 7 weak solution, 7 Souslinian, 22 space Banach space, 2, 286 dual space, 3 finite dimensional space, 5 Hilbert space, 2 infinite dimensional space, 5 linear space, 1, 2 linear subspace, 6 normed linear space, 6 normed space, 88 quotient space, 31 reachable space, 170 reflexive Banach space, 88 separable Hilbert space, 38, 65 topological vector space, 6 state trajectory, 38 strongly differentiable, 7 strongly measurable, 8 surjective, 31

T target, 40, 126 tetrad admissible tetrad, 39, 66, 161, 236 optimal tetrad, 39, 298 theorem Carathéodory Theorem, 105 Closed Graph Theorem, 31, 32 falling sun theorem, 227 finite covering theorem, 111 Hahn-Banach Theorem, 3, 6, 29, 133, 158, 176, 184, 185, 222, 244, 290, 294 Hamilton-Cayley Theorem, 71, 73, 287 J. von Neumann Theorem, 107 James Theorem, 90 Kakutani-Fan-Glicksberg Theorem, 14, 82, 84

334 theorem (cont.) Krein-Milman Theorem, 16 Kuratowski and Ryll-Nardzewski Theorem, 17 Lebesgue Dominated Convergence Theorem, 83, 145, 187 Lusin Theorem, 20 Lyapunov’s theorem for vector measures, 15 Lyapunov Theorem, 15, 105 Mazur Theorem, 107, 116, 138 Measurable Selection Theorem, 19, 105, 108 Range Comparison Theorem, 24 Riesz Representation Theorem, 3, 29, 89, 158, 172, 176, 184, 185, 222, 244, 290, 294 Schauder Fixed Point Theorem, 15

Index U undamped harmonic oscillator, 52 uniformly continuous, 22 unique continuation, 34 uniqueness, 8

V variational problem, 126 viability domain, 151, 161

W weak star topology, 188

Z zero extension, 216, 232