Content: 1 Vector Fields and Control Systems on Smooth Manifolds 1 1.1 Smooth Manifolds 1 1.2 Vector Fields on Smooth Manifolds 4 1.3 Smooth Differential Equations and Flows on Manifolds 8 1.4 Control Systems 12 2 Elements of Chronological Calculus 21 2.1 Points, Diffeomorphisms, and Vector Fields 21 2.2 Seminorms and $C^{\infty }(M)$-Topology 25 2.3 Families of Functionals and Operators 26 2.4 Chronological Exponential 28 2.5 Action of Diffeomorphisms on Vector Fields 37 2.6 Commutation of Flows 40 2.7 Variations Formula 41 2.8 Derivative of Flow with Respect to Parameter 43 3 Linear Systems 47 3.1 Cauchy's Formula for Linear Systems 47 3.2 Controllability of Linear Systems 49 4 State Linearizability of Nonlinear Systems 53 4.1 Local Linearizability 53 4.2 Global Linearizability 57 5 The Orbit Theorem and its Applications 63 5.1 Formulation of the Orbit Theorem 63 5.2 Immersed Submanifolds 64 5.3 Corollaries of the Orbit Theorem 66 5.4 Proof of the Orbit Theorem 67 5.5 Analytic Case 72 5.6 Frobenius Theorem 74 5.7 State Equivalence of Control Systems 76 6 Rotations of the Rigid Body 81 6.1 State Space 81 6.2 Euler Equations 84 6.3 Phase Portrait 88 6.4 Controlled Rigid Body: Orbits 90 7 Control of Configurations 97 7.1 Model 97 7.2 Two Free Points 100 7.3 Three Free Points 101 7.4 Broken Line 104 8 Attainable Sets 109 8.1 Attainable Sets of Full-Rank Systems 109 8.2 Compatible Vector Fields and Relaxations 113 8.3 Poisson Stability 116 8.4 Controlled Rigid Body: Attainable Sets 118 9 Feedback and State Equivalence of Control Systems 121 9.1 Feedback Equivalence 121 9.2 Linear Systems 123 9.3 State-Feedback Linearizability 131 10 Optimal Control Problem 137 10.1 Problem Statement 137 10.2 Reduction to Study of Attainable Sets 138 10.3 Compactness of Attainable Sets 140 10.4 Time-Optimal Problem 143 10.5 Relaxations 143 11 Elements of Exterior Calculus and Symplectic Geometry 145 11.1 Differential 1-Forms 145 11.2 Differential $k$-Forms 147 11.3 Exterior Differential 151 11.4 Lie Derivative of Differential Forms 153 11.5 Elements of Symplectic Geometry 157 12 Pontryagin Maximum Principle 167 12.1 Geometric Statement of PMP and Discussion 167 12.2 Proof of PMP 172 12.3 Geometric Statement of PMP for Free Time 177 12.4 PMP for Optimal Control Problems 179 12.5 PMP with General Boundary Conditions 182 13 Examples of Optimal Control Problems 191 13.1 The Fastest Stop of a Train at a Station 191 13.2 Control of a Linear Oscillator 194 13.3 The Cheapest Stop of a Train 197 13.4 Control of a Linear Oscillator with Cost 199 13.5 Dubins Car 200 14 Hamiltonian Systems with Convex Hamiltonians 207 15 Linear Time-Optimal Problem 211 15.1 Problem Statement 211 15.2 Geometry of Polytopes 212 15.3 Bang-Bang Theorem 213 15.4 Uniqueness of Optimal Controls and Extremals 215 15.5 Switchings of Optimal Control 218 16 Linear-Quadratic Problem 223 16.1 Problem Statement 223 16.2 Existence of Optimal Control 224 16.3 Extremals 227 16.4 Conjugate Points 229 17 Sufficient Optimality Conditions, Hamilton-Jacobi Equation,Dynamic Programming 235 17.1 Sufficient Optimality Conditions 235 17.2 Hamilton-Jacobi Equation 242 17.3 Dynamic Programming 244

##### Citation preview

A. A. Agrachev, Yu. L. Sachkov

Control Theory from the Geometric Viewpoint { Monograph { January 15, 2004

Springer

Berlin Heidelberg NewYork Hong Kong London Milan Paris Tokyo

To our parents Alexander and Elena Agrachev Leonid and Anastasia Sachkov

Preface

This book presents some facts and methods of the Mathematical Control Theory treated from the geometric point of view. The book is mainly based on graduate courses given by the rst coauthor in the years 2000{2001 at the International School for Advanced Studies, Trieste, Italy. Mathematical prerequisites are reduced to standard courses of Analysis and Linear Algebra plus some basic Real and Functional Analysis. No preliminary knowledge of Control Theory or Dierential Geometry is required. What this book is about? The classical deterministic physical world is described by smooth dynamical systems: the future in such a system is completely determined by the initial conditions. Moreover, the near future changes smoothly with the initial data. If we leave room for \free will" in this fatalistic world, then we come to control systems. We do so by allowing certain parameters of the dynamical system to change freely at every instant of time. That is what we routinely do in real life with our body, car, cooker, as well as with aircraft, technological processes etc. We try to control all these dynamical systems! Smooth dynamical systems are governed by dierential equations. In this book we deal only with nite dimensional systems: they are governed by ordinary dierential equations on nite dimensional smooth manifolds. A control system for us is thus a family of ordinary dierential equations. The family is parametrized by control parameters. All dierential equations of the family are dened on one and the same manifold which is called the state space of the control system. We may select any admissible values of the control parameters (i.e. select any dynamical system from the family) and we are free to change these values at every time instant. The way of selection, which is a function of time, is called a control or a control function. As soon as a control is xed, the control system turns into a nonautonomous ordinary dierential equation. A solution of such an equation is uniquely determined by the initial condition and is called an admissible trajectory of the control system (associated with a given control). Thus, an admissible trajectory is a curve in the state space. The initial condition (initial

VIII

Preface

Preface

IX

provides adequate methods for the investigation of variational problems with nonholonomic constraints. The rst chapter of the book is of introductory nature: we recall what smooth manifolds and ordinary dierential equations on manifolds are, and dene control systems. Chapter 2 is devoted to an operator calculus that creates great exibility in handling of nonlinear control systems. In Chapters 3 and 4 we introduce a simple and extremely popular in applications class of linear systems and give an eective characterization of systems that can be made linear by a smooth transformation of the state space. Chapters 5{7 are devoted to the fundamental Orbit Theorem of Nagano and Sussmann and its applications. The Orbit Theorem states that any orbit of the group generated by a family of ows is an immersed submanifold (the group itself may be huge and wild). Chapter 8 contains general results on the structure of attainable sets starting from a simple test to guarantee that these sets are full dimensional. In Chapter 9 we introduce feedback transformations, give a feedback classication of linear systems, and eectively characterize systems that can be made linear by feedback and state transformations. The rest of the book is mainly devoted to the Optimal Control. In Chapter 10 we state the optimal control problem, give its geometric interpretation, and discuss the existence of solutions. Chapter 11 contains basic facts on dierential forms and Hamiltonian systems we need this information to investigate optimal control problems. Chapter 12 is devoted to the intrinsic formulation and detailed proof of the Pontryagin Maximum Principle, a key result in the Optimal Control Theory. Chapters 13{16 contain numerous applications of the Pontryagin Maximum Principle including a curious property of Hamiltonian systems with convex Hamiltonians and more or less complete theories of linear time-optimal problems and linear{quadratic problems with nite horizons. In Chapter 17 we discuss a Hamiltonian version of the theory of elds of extremals, which is suitable for applications in the Optimal Control, and introduce the Hamilton{Jacobi equation. Chapters 18 and 19 are devoted to the moving frames technique for optimal control problems and to problems on Lie groups. The denition and basic facts on Lie groups are given in Chapter 18: they are simple corollaries of the general geometric control techniques developed in previous chapters. Chapters 20 and 21 contain the theory of the Second Variation with second order necessary and sucient optimality conditions for regular and singular extremals. The short Chapter 22 presents an instructive reduction procedure, which establishes a connection between singular and regular extremals. In Chapter 23 we introduce and compute (in simplest low dimensional cases) the curvature, a remarkable feedback invariant of optimal control problems. Finally in Chapter 24 we discuss the control of a classical nonholonomic system: two bodies rolling one on another without slipping or twisting. The Appendix contains proofs of some results formulated in Chapter 2. This is a very brief overview of the contents of the book. In each chapter we try to stay at textbook level, i.e. to present just the rst basic results with

X

Preface

some applications. The topic of practically every chapter has an extensive development, sometimes rather impressive. In order to study these topics deeper the reader is referred to research papers. Geometric Control Theory is a broad subject and many important topics are not even mentioned in the book. In particular, we do not study the feedback stabilization problem and the huge theory of control systems with outputs including fundamental concepts of Observability and Realization. For this and other material see books on Control listed in the Bibliography. Acknowledgments. We wish to thank our Teachers Professor R.V. Gamkrelidze and Professor A.F. Filippov for their Lessons on Mathematics and Control Theory, and for encouragement during our work on this book. We acknowledge support of this project by the International School for Advanced Studies (Trieste, Italy), Steklov Mathematical Institute (Moscow, Russia), and Program Systems Institute (Pereslavl-Zalessky, Russia). We are grateful to the participants of the seminar on geometric control theory at the International School for Advanced Studies, Trieste, especially to Ulysse Serres, Igor Zelenko, and Sergio Rodrigues, for valuable remarks that helped us to improve exposition. We are indebted to Irina and Elena: without patience and constant support of our wives, this book could not be written. Moscow { Pereslavl-Zalessky { Trieste October 2003

A. A. Agrachev Yu. L. Sachkov

Contents

1 Vector Fields and Control Systems on Smooth Manifolds : : 1 1.1 1.2 1.3 1.4

Smooth Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Vector Fields on Smooth Manifolds . . . . . . . . . . . . . . . . . . . . . . . . Smooth Dierential Equations and Flows on Manifolds . . . . . . . Control Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 4 8 12

2 Elements of Chronological Calculus : : : : : : : : : : : : : : : : : : : : : : : : 21 2.1 Points, Dieomorphisms, and Vector Fields . . . . . . . . . . . . . . . . . 21 2.2 Seminorms and C 1 (M)-Topology . . . . . . . . . . . . . . . . . . . . . . . . . 25 2.3 2.4 2.5 2.6 2.7 2.8

Families of Functionals and Operators. . . . . . . . . . . . . . . . . . . . . . Chronological Exponential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Action of Dieomorphisms on Vector Fields . . . . . . . . . . . . . . . . . Commutation of Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Variations Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Derivative of Flow with Respect to Parameter . . . . . . . . . . . . . . .

26 28 37 40 41 43

3 Linear Systems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 47 3.1 Cauchy's Formula for Linear Systems . . . . . . . . . . . . . . . . . . . . . . 47 3.2 Controllability of Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . 49

4 State Linearizability of Nonlinear Systems : : : : : : : : : : : : : : : : : 53

4.1 Local Linearizability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 4.2 Global Linearizability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

5 The Orbit Theorem and its Applications :: : : : : : : : : : : : : : : : : : 63 5.1 5.2 5.3 5.4 5.5 5.6

Formulation of the Orbit Theorem . . . . . . . . . . . . . . . . . . . . . . . . . Immersed Submanifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Corollaries of the Orbit Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . Proof of the Orbit Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Analytic Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Frobenius Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

63 64 66 67 72 74

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5.7 State Equivalence of Control Systems . . . . . . . . . . . . . . . . . . . . . . 76

6 Rotations of the Rigid Body : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 81 6.1 6.2 6.3 6.4

State Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Euler Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Phase Portrait . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Controlled Rigid Body: Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . .

81 84 88 90

7 Control of Congurations :: : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 97 7.1 7.2 7.3 7.4

Model. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 Two Free Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 Three Free Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 Broken Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

8 Attainable Sets : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 109 8.1 8.2 8.3 8.4

Attainable Sets of Full-Rank Systems . . . . . . . . . . . . . . . . . . . . . . 109 Compatible Vector Fields and Relaxations . . . . . . . . . . . . . . . . . . 113 Poisson Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 Controlled Rigid Body: Attainable Sets . . . . . . . . . . . . . . . . . . . . 118

9 Feedback and State Equivalence of Control Systems : : : : : : : 121

9.1 Feedback Equivalence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 9.2 Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 9.3 State-Feedback Linearizability . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

10 Optimal Control Problem :: : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 137 10.1 10.2 10.3 10.4 10.5

Problem Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 Reduction to Study of Attainable Sets . . . . . . . . . . . . . . . . . . . . . 138 Compactness of Attainable Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 Time-Optimal Problem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 Relaxations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

11 Elements of Exterior Calculus and Symplectic Geometry : : 145

11.1 Dierential 1-Forms. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 11.2 Dierential k-Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 11.3 Exterior Dierential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 11.4 Lie Derivative of Dierential Forms . . . . . . . . . . . . . . . . . . . . . . . . 153 11.5 Elements of Symplectic Geometry . . . . . . . . . . . . . . . . . . . . . . . . . 157

12 Pontryagin Maximum Principle :: : : : : : : : : : : : : : : : : : : : : : : : : : : 167 12.1 Geometric Statement of PMP and Discussion . . . . . . . . . . . . . . . 167 12.2 Proof of PMP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 12.3 Geometric Statement of PMP for Free Time . . . . . . . . . . . . . . . . 177 12.4 PMP for Optimal Control Problems . . . . . . . . . . . . . . . . . . . . . . . 179 12.5 PMP with General Boundary Conditions . . . . . . . . . . . . . . . . . . . 182

Contents

XIII

13 Examples of Optimal Control Problems : : : : : : : : : : : : : : : : : : : : 191

13.1 The Fastest Stop of a Train at a Station . . . . . . . . . . . . . . . . . . . . 191 13.2 Control of a Linear Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 13.3 The Cheapest Stop of a Train . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 13.4 Control of a Linear Oscillator with Cost . . . . . . . . . . . . . . . . . . . . 199 13.5 Dubins Car . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200

14 Hamiltonian Systems with Convex Hamiltonians :: : : : : : : : : : 207 15 Linear Time-Optimal Problem : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 211 15.1 15.2 15.3 15.4 15.5

Problem Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 Geometry of Polytopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212 Bang-Bang Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 Uniqueness of Optimal Controls and Extremals. . . . . . . . . . . . . . 215 Switchings of Optimal Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218

16 Linear-Quadratic Problem : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 223 16.1 16.2 16.3 16.4

Problem Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223 Existence of Optimal Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224 Extremals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 Conjugate Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229

17 Su cient Optimality Conditions, Hamilton-Jacobi Equation, and Dynamic Programming : : : : : : : : : : : : : : : : : : : : : 235

17.1 Sucient Optimality Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 17.2 Hamilton-Jacobi Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242 17.3 Dynamic Programming .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244

18 Hamiltonian Systems for Geometric Optimal Control Problems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 247 18.1 Hamiltonian Systems on Trivialized Cotangent Bundle . . . . . . . 247 18.2 Lie Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 18.3 Hamiltonian Systems on Lie Groups . . . . . . . . . . . . . . . . . . . . . . . 260

19 Examples of Optimal Control Problems on Compact Lie Groups : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 265 19.1 Riemannian Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 19.2 A Sub-Riemannian Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267 19.3 Control of Quantum Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271 19.4 A Time-Optimal Problem on SO(3) . . . . . . . . . . . . . . . . . . . . . . . . 284

20 Second Order Optimality Conditions : : : : : : : : : : : : : : : : : : : : : : 293 20.1 20.2 20.3 20.4

Hessian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293 Local Openness of Mappings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297 Dierentiation of the Endpoint Mapping . . . . . . . . . . . . . . . . . . . 304 Necessary Optimality Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . 309

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Contents

20.5 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318 20.6 Single-Input Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321

21 Jacobi Equation : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 333 21.1 21.2 21.3 21.4 21.5

Regular Case: Derivation of Jacobi Equation . . . . . . . . . . . . . . . . 334 Singular Case: Derivation of Jacobi Equation . . . . . . . . . . . . . . . 338 Necessary Optimality Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . 342 Regular Case: Transformation of Jacobi Equation. . . . . . . . . . . . 343 Sucient Optimality Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . 346

22 Reduction : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 355

22.1 Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355 22.2 Rigid Body Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 358 22.3 Angular Velocity Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359

23 Curvature : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 363 23.1 Curvature of 2-Dimensional Systems . . . . . . . . . . . . . . . . . . . . . . . 363 23.2 Curvature of 3-Dimensional Control-Ane Systems . . . . . . . . . . 373

24 Rolling Bodies : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 377 24.1 Geometric Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377 24.2 Two-Dimensional Riemannian Geometry . . . . . . . . . . . . . . . . . . . 379 24.3 Admissible Velocities. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383 24.4 Controllability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384 24.5 Length Minimization Problem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387

A Appendix :: : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 393 A.1 Homomorphisms and Operators in C 1 (M) . . . . . . . . . . . . . . . . . 393 A.2 Remainder Term of the Chronological Exponential . . . . . . . . . . . 395

References : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 399 List of Figures : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 407 Index : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 409

1 Vector Fields and Control Systems on Smooth Manifolds

1.1 Smooth Manifolds We give just a brief outline of basic notions related to the smooth manifolds. For a consistent presentation, see an introductory chapter to any textbook on analysis on manifolds, e. g. 146]. In the sequel, \smooth" (manifold, mapping, vector eld etc.) means C 1 . Denition 1.1. A subset M n is called a smooth k-dimensional submanifold of n, k  n, if any point x 2 M has a neighborhood Ox in n in which M is described in one of the following ways: (1) there exists a smooth vector-function F : Ox ! n;k rank dd Fx = n ; k q

R

R

R

R

such that

Ox \ M = F ;1(0)

(2) there exists a smooth vector-function

R

f : V0 ! n from a neighborhood of the origin 0 2 V0

f(0) = x such that

R

k with

rank dd fx = k

Ox \ M = f(V0 )

and f : V0 ! Ox \ M is a homeomorphism

0

2

1 Vector Fields and Control Systems on Smooth Manifolds

R R

(3) there exists a smooth vector-function

 : Ox ! O0 n n with onto a neighborhood of the origin 0 2 O0 rank dd x = n x

R

such that

(Ox \ M) = k \ O0: Exercise 1.2. Prove that three local descriptions of a smooth submanifold given in (1){(3) are mutually equivalent. Remark 1.3. (1) There are two topologically dierent one-dimensional manifolds: the line 1 and the circle S 1 . The sphere S 2 and the torus 2 = S 1  S 1 are two-dimensional manifolds. The torus can be viewed as a sphere with a handle. Any compact orientable two-dimensional manifold is topologically a sphere with p handles, p = 0 1 2 : : : . (2) Smooth manifolds appear naturally already in the basic analysis. For example, the circle S 1 and the torus 2 are natural domains of periodic and doubly periodic functions respectively. On the sphere S 2 , it is convenient to consider restriction of homogeneous functions of 3 variables. So a smooth submanifold is a subset in n which can locally be dened by a regular system of smooth equations and by a smooth regular parametrization. In spite of the intuitive importance of the image of manifolds as subsets of a Euclidean space, it is often convenient to consider manifolds independently of any embedding in n. An abstract manifold is dened as follows. Denition 1.4. A smooth k-dimensional manifold M is a Hausdor paracompact topological space endowed with a smooth structure: M is covered by

R

T

T

R

a system of open subsets

R

M = O

R

called coordinate neighborhoods, in each of which is de ned a homeomorphism

 : O ! k

R!

called a local coordinate system such that all compositions

  ;1 :  (O \ O )

are di eomorphisms, see Fig. 1.1.

k

 (O \ O )

R

k

As a rule, we denote points of a smooth manifold by q, and its coordinate representation in a local coordinate system by x:

1.1 Smooth Manifolds

3

O 

;1



O M

Rk

Fig. 1.1. Coordinate system in smooth manifold M

R

R

 : O ! k x = (q) 2 k:

q 2 M

R

For a smooth submanifold in n, the abstract Denition 1.4 holds. Conversely, any connected smooth abstract manifold can be considered as a smooth submanifold in n. Before precise formulation of this statement, we give two denitions. Denition 1.5. Let M and N be k- and l-dimensional smooth manifolds re-

R

spectively. A continuous mapping

f : M !N is called smooth if it is smooth in coordinates. That is, let M =  O and N =  V be coverings of M and N by coordinate neighborhoods and

R

 : O ! k

R

 : V ! l

R!

the corresponding coordinate mappings. Then all compositions

  f  ;1 :  (O \ f ;1 (V ))

should be smooth.

k

 (f(O ) \ V )

R

l

Denition 1.6. A smooth manifold M is called dieomorphic to a smooth manifold N if there exists a homeomorphism f : M !N such that both f and its inverse f ;1 are smooth mappings. Such mapping f is called a dieomorphism.

4

1 Vector Fields and Control Systems on Smooth Manifolds

The set of all dieomorphisms f : M ! M of a smooth manifold M is denoted by Di M. A smooth mapping f : M ! N is called an embedding of M into N if f : M ! f(M) is a dieomorphism. A mapping f : M ! N is called proper if f ;1 (K) is compact for any compactum K b N (the notation K b N means that K is a compact subset of N). Theorem 1.7 (Whitney). Any smooth connected k-dimensional manifold can be properly embedded into

R

2k+1.

Summing up, we may say that a smooth manifold is a space which looks locally like a linear space but without xed linear structure, so that all smooth coordinates are equivalent. The manifolds, not linear spaces, form an adequate framework for the modern nonlinear analysis.

1.2 Vector Fields on Smooth Manifolds The tangent space to a smooth manifold at a point is a linear approximation of the manifold in the neighborhood of this point. Denition 1.8. Let M be a smooth k-dimensional submanifold of n and x 2 M its point. Then the tangent space to M at the point x is a k-dimensional linear subspace

Tx M

R

R

n

de ned as follows for cases (1){(3) of De nition 1.1:

(1) Tx M = Ker dd Fx , x d f (2) Tx M = Im d x ,  d  0 ;1 k. (3) Tx M = d x x

R

Remark 1.9. The tangent space is a coordinate-invariant object since smooth

changes of variables lead to linear transformations of the tangent space. In an abstract way, the tangent space to a manifold at a point is the set of velocity vectors to all smooth curves in the manifold that start from this point. Denition 1.10. Let ( ) be a smooth curve in a smooth manifold M starting from a point q 2 M :  : (;" ") ! M a smooth mapping (0) = q: The tangent vector

1.2 Vector Fields on Smooth Manifolds

5

d  = _ (0) d t t=0 to the curve ( ) at the point q is the equivalence class of all smooth curves in M starting from q and having the same 1-st order Taylor polynomial as ( ), for any coordinate system in a neighborhood of q. _ (0) (t) (0)

Fig. 1.2. Tangent vector _ (0)

Denition 1.11. The tangent space to a smooth manifold M at a point q 2 M is the set of all tangent vectors to all smooth curves in M starting at q:   d j  : (;" ") ! M smooth (0) = q : Tq M = d t t=0

_ (0)

Tq M

M

q (t)

Fig. 1.3. Tangent space Tq M

Exercise 1.12. Let M be a smooth k-dimensional manifold and q 2 M. Show that the tangent space Tq M has a natural structure of a linear k-dimensional space.

6

1 Vector Fields and Control Systems on Smooth Manifolds

Denition 1.13. A smooth vector eld on a smooth manifold M is a smooth mapping q 2 M 7! V (q) 2 Tq M that associates to any point q 2 M a tangent vector V (q) at this point. In the sequel we denote by Vec M the set of all smooth vector elds on a smooth manifold M. Denition 1.14. A smooth dynamical system, or an ordinary dierential equation (ODE), on a smooth manifold M is an equation of the form d q = V (q) q 2 M dt or, equivalently,

q_ = V (q)

q 2 M

where V (q) is a smooth vector eld on M . A solution to this system is a smooth mapping where I

R

 : I ! M

is an interval, such that

d  = V ((t)) 8 t 2 I: dt

V ( (t)) (t)

Fig. 1.4. Solution to ODE q_ = V (q)

1.2 Vector Fields on Smooth Manifolds

7

Denition 1.15. Let  : M ! N be a smooth mapping between smooth manifolds M and N . The dierential of  at a point q 2 M is a linear mapping Dq  : Tq M ! T(q) N

de ned as follows:

d



= ddt ((t)) Dq  d t t=0 t=0 where

 : (;" ") ! M q(0) = q is a smooth curve in M starting at the point q, see Fig. 1.5.

q

= (0)

v

= _ (0) (t)

Dq 



Dq  v

(q)

M

( (t)) N

Fig. 1.5. Di erential Dq  Now we apply smooth mappings to vector elds. Let V 2 Vec M be a vector eld on M and q_ = V (q) (1.1) the corresponding ODE. To nd the action of a dieomorphism  : M ! N  : q 7! x = (q) on the vector eld V (q), take a solution q(t) of (1.1) and compute the ODE satised by the image x(t) = (q(t)): _ = (Dq ) V (q(t)) = (D;1 (x) ) V (;1(x(t))): x(t) _ = ddt (q(t)) = (Dq ) q(t) So the required ODE is ;  x_ = D;1 (x)  V (;1(x)): (1.2) The right-hand side here is the transformed vector eld on N induced by the dieomorphism :

8

1 Vector Fields and Control Systems on Smooth Manifolds

;



( V )(x) def = D;1 (x) V (;1(x)): The notation q is used, along with Dq , for dierential of a mapping  at a point q. Remark 1.16. In general, a smooth mapping  induces transformation of tangent vectors, not of vector elds. In order that D transform vector elds to vector elds,  should be a dieomorphism.

1.3 Smooth Dierential Equations and Flows on Manifolds Theorem 1.17. Consider a smooth ODE q_ = V (q) q2M

R

R

n

(1.3)

on a smooth submanifold M of n. For any initial point q0 2 M , there exists a unique solution

q(t q0)

t 2 (a b) a < 0 < b

of equation (1:3) with the initial condition

q(0 q0) = q0 de ned on a suciently short interval (a b). The mapping

(t q0) 7! q(t q0) is smooth. In particular, the domain (a b) of the solution q(  q0) can be chosen smoothly depending on q0.

R

Proof. We prove the theorem by reduction to its classical analog in n.

The statement of the theorem is local. We rectify the submanifold M in the neighborhood of the point q0:

RR R R 2R

n ! O0  : Oq0 (Oq0 \ M) = k:

n

Consider the restriction ' = jM . Then a curve q(t) in M is a solution to (1.3) if and only if its image x(t) = '(q(t)) in k is a solution to the induced system: x_ =  V (x)

x

k:

ut

R

1.3 Smooth Di erential Equations and Flows on Manifolds

Theorem 1.18. Let M

9

n be a smooth submanifold and let

R

R

q 2 n

q_ = V (q)

be a system of ODEs in n such that

(1.4)

q 2 M ) V (q) 2 Tq M:

Then for any initial point q0 2 M , the corresponding solution q(t q0) to (1:4) with q(0 q0) = q0 belongs to M for all suciently small jtj. Proof. Consider the restricted vector eld:

f = V jM : By the existence theorem for M, the system q 2 M

q_ = f(q) has a solution q(t q0), q(0 q0) = q0 , with q(t q0) 2 M

for small jtj:

(1.5)

On the other hand, the curve q(t q0) is a solution of (1.4) with the same initial condition. Then inclusion (1.5) proves the theorem. ut Denition 1.19. A vector eld V 2 Vec M is called complete, if for all q0 2 M the solution q(t q0) of the Cauchy problem

R

is de ned for all t 2 .

RnRf g ;1

q_ = V (q)

q(0 q0) = q0

(1.6)

R

Example 1.20. The vector eld V (x) = x is complete on , as well as on

R

0 , (  0), (0 +1), and f0g, but not complete on other submanifolds of . The vector eld V (x) = x2 is not complete on any submanifolds of except f0g. Proposition 1.21. Suppose that there exists " > 0 such that for any q0 2 M the solution q(t q0) to Cauchy problem (1:6) is de ned for t 2 (;" "). Then the vector eld V (q) is complete. Remark 1.22. In this proposition it is required that there exists " > 0 common for all initial points q0 2 M. In general, " may be not bounded away from zero for all q0 2 M. E.g., for the vector eld V (x) = x2 we have " ! 0 as x0 ! 1.

10

1 Vector Fields and Control Systems on Smooth Manifolds

Proof. Suppose that the hypothesis of the proposition is true. Then we can

introduce the following family of mappings in M: P t : M ! M t 2 (;" ") t P : q0 7! q(t q0): P t(q0) is the shift of a point q0 2 M along the trajectory of the vector eld V (q) for time t. By Theorem 1.17, all mappings P t are smooth. Moreover, the family f P t j t 2 (;" ") g is a smooth family of mappings. A very important property of this family is that it forms a local one-parameter group, i.e., P t(P s(q)) = P s(P t(q)) = P t+s(q) q 2 M t s t + s 2 (;" "): Indeed, the both curves in M: t 7! P t(P s(q)) and t 7! P t+s(q) satisfy the ODE q_ = V (q) with the same initial value P 0(P s(q)) = P 0+s(q) = P s(q). By uniqueness, P t(P s(q)) = P t+s(q). The equality for P s(P t(q)) is obtained by switching t and s. So we have the following local group properties of the mappings P t: P t  P s = P s  P t = P t+s t s t + s 2 (;" ") P 0 = Id P ;t  P t = P t  P ;t = Id t 2 (;" ") ;  ; 1 ; t t P = P  t 2 (;" "): In particular, all P t are dieomorphisms. Now we extend the mappings P t for all t 2 .Any t 2 can be represented as 0  < 2"  K = 0 1 2 : : : : t = 2" K +  We set P t def = P   P| "=2  {z  P "=2}  = sgn t:

R

Then the curve

jK j times

t 7! P t(q0) is a solution to Cauchy problem (1.6).

R

R

t2 

Denition 1.23. For a complete vector eld V 2 Vec M , the mapping t 7! P t t2  is called the ow generated by V .

R

ut

1.3 Smooth Di erential Equations and Flows on Manifolds

11

Remark 1.24. It is useful to imagine a vector eld V 2 Vec M as a eld of

R

velocity vectors of a moving liquid in M. Then the ow P t takes any particle of the liquid from a position q 2 M and transfers it for a time t 2 to the position P t(q) 2 M, see Fig. 1.6.

V q

P q t

( )

M Fig. 1.6. Flow P t of vector eld V Simple sucient conditions for completeness of a vector eld are given in terms of compactness. Proposition 1.25. Let K M be a compact subset, and let V 2 Vec M . Then there exists "K > 0 such that for all q0 2 K the solution q(t q0) to Cauchy problem (1:6) is de ned for all t 2 (;"K  "K ). Proof. By Theorem 1.17, domain of the solution q(t q0) can be chosen continuously depending on q0. The diameter of this domain has a positive inmum 2"K for q0 in the compact set K. ut Corollary 1.26. If a smooth manifold M is compact, then any vector eld V 2 Vec M is complete. Corollary 1.27. Suppose that a vector eld V 2 Vec M has a compact support:

supp V def = f q 2 M j V (q) 6= 0 g is compact.

Then V is complete.

Proof. Indeed, by Proposition 1.25, there exists " > 0 such that all trajectories

of V starting in supp V are dened for t 2 (;" "). But V jM nsupp V = 0, and all trajectories of V starting outside of supp V are constant, thus dened for all t 2 . By Proposition 1.21, the vector eld V is complete. ut

R

12

1 Vector Fields and Control Systems on Smooth Manifolds

Remark 1.28. If we are interested in the behavior of (trajectories of) a vector

eld V 2 Vec M in a compact subset K M, we can suppose that V is complete. Indeed, take an open neighborhood OK of K with the compact closure OK . We can nd a function a 2 C 1 (M) such that

 1

a(q) = 0 q 2 Mqn2OK K: Then the vector eld a(q)V (q) 2 Vec M is complete since it has a compact support. On the other hand, in K the vector elds a(q)V (q) and V (q) coincide, thus have the same trajectories.

1.4 Control Systems For dynamical systems, the future q(t q0), t > 0, is completely determined by the present state q0 = q(0 q0). The law of transformation q0 7! q(t q0) is the ow P t, i.e., dynamics of the system q_ = V (q)

q 2 M

(1.7)

it is determined by one vector eld V (q). In order to be able to aect dynamics, to control it, we consider a family of dynamical systems q_ = Vu(q)

q 2 M u 2 U

(1.8)

with a family of vector elds Vu parametrized by a parameter u 2 U. A system of the form (1.8) is called a control system . The variable u is a control parameter , and the set U is the space of control parameters . A priori we do not impose any restrictions on U, it is an arbitrary set, although, typically U will be a subset of a smooth manifold. The variable q is the state , and the manifold M is the state space of control system (1.8). In control theory we can change dynamics of control system (1.8) at any moment of time by changing values of u 2 U. For any u 2 U, the corresponding vector eld Vu 2 Vec M generates the ow, which is denoted by Put . A typical problem of control theory is to nd the set of points that can be reached from an initial point q0 2 M by choosing various values of u 2 U and switching from one value to another time to time (for dynamical system (1.7), this reachable set is just the semitrajectory q(t q0) = P t(q0 ), t  0). Suppose that we start from a point q0 2 M and use the following control strategy for control system (1.8): rst we choose some control parameter u1 2 U, then we switch to another control parameter u2 2 U. Which points in M can be reached with such control strategy? With the control parameter u1 , we can reach points of the form f Put11 (q0) j t1  0 g

1.4 Control Systems

13

and the whole set of reachable points has the form

f Put22  Put11 (q0) j t1  t2  0 g a piece of a 2-dimensional surface: Put22

Put11

q0

A natural next question is: what points can be reached from q0 by any kind of control strategies? Before studying this question, consider a particular control system that gives a simplied model of a car. Example 1.29. We suppose that the state of a car is determined by the position of its center of mass x = (x1  x2) 2 2 and orientation angle 2 S 1 relative to the positive direction of the axis x1. Thus the state space of our system is a nontrivial 3-dimensional manifold, a solid torus

R j 2R

M = f q = (x ) x

2

R 2R

2 S1 g =

2

S1 :

Suppose that two kinds of motion are possible: we can drive the car forward and backwards with some xed linear velocity u1 , and we can turn the car around its center of mass with some xed angular velocity u2 2 . We can combine these two kinds of motion in an admissible way. The dynamical system that describes the linear motion with a velocity u1 2 has the form 8 x_ 1 = u cos  < 2 1 (1.9) : x __ ==0:u1 sin  Rotation with an angular velocity u2 2 is described as 8 x_ 1 = 0 < 2 (1.10) : x __ ==u02:

R

R

R

14

R

1 Vector Fields and Control Systems on Smooth Manifolds

The control parameter u = (u1 u2) can take any values in the given subset 2. If we write ODEs (1.9) and (1.10) in the vector form: U q_ = u1V1 (q)

where

q_ = u2 V2(q)

0 cos 1 V1 (q) = @ sin A 

001 V2(q) = @ 0 A 

0

(1.11)

1

then our model reads q 2 M u 2 U:

q_ = Vu (q) = u1V1 (q) + u2 V2(q)

C

This model can be rewritten in the complex form: z = x1 + ix2 2  z_ = u1ei 

_ = u2  (u1  u2) 2 U (z ) 2  S 1 :

C

Remark 1.30. Control system (1.8) is often written in another form:

q_ = f(q u)

q 2 M u 2 U:

We prefer the notation Vu (q), which stresses that for a xed u 2 U, Vu is a single object | a vector eld on M. Now we return to the study of the points reachable by trajectories of a control system from an initial point. Denition 1.31. The attainable set (or reachable set) of control system (1.8) with piecewise-constant controls from a point q0 2 M for a time t  0 is de ned as follows:

Aq0 (t) = f Pukk  : : :  Pu11 (q0 ) j i  0

k X i=1

N

i = t ui 2 U k 2 g:

The attainable set from q0 for arbitrary nonnegative time of motion has the form 

Aq0 =

see Fig. 1.7.

t0

Aq0 (t)

1.4 Control Systems

15

Vuk Vu3 Vu1

Vu2

Aq0

q0

Fig. 1.7. Attainable set Aq0 For simplicity, consider rst the smallest nontrivial space of control parameters consisting of two indices: U = f1 2g (even this simple case shows essential features of the reachability problem). Then the attainable set for arbitrary nonnegative times has the form: Aq0 = f P2k  P1k;1  : : :  P22  P11 (q0) j i  0 k 2 g: This expression suggests that the attainable set Aq0 depends heavily upon commutator properties of the ows P1t and P2s. Consider rst the trivial commutative case, i.e., suppose that the ows commute: P1t  P2s = P2s  P1t 8t s 2 : Then the attainable set can be evaluated precisely: since P2k  P1k;1  : : :  P22  P11 = P2k +:::+2  P1k;1 +:::+1  then Aq0 = f P2s  P1t(q0) j t s  0 g: So in the commutative case the attainable set by two control parameters is a piece of a smooth two-dimensional surface, possibly with singularities. It is easy to see that if the number of control parameters is k  2 and the corresponding ows P1t1 , : : : , Pktk commute, then Aq0 is, in general, a piece of a k-dimensional manifold, and, in particular, dim Aq0  k. But this commutative case is exceptional and occurs almost never in real control systems. Example 1.32. In the model of a car considered above the control dynamics is dened by two vector elds (1:11) on the 3-dimensional manifoldM = 2xS1 .

N

R

R

16

1 Vector Fields and Control Systems on Smooth Manifolds 1

q0

x1 0

x0

q1

Fig. 1.8. Initial and nal congurations of the car PtV11 PtV22 PtV31

q0

q1

Fig. 1.9. Steering the car from q0 to q1 It is obvious that from any initial conguration q0 = (x0  0) 2 M we can drive the car to any terminal conguration q1 = (x1  1) 2 M by alternating linear motions and rotations (both with xed velocities), see Fig. 1.9. So any point in the 3-dimensional manifold M can be reached by means of 2 vector elds V1 , V2 . This is due to noncommutativity of these elds (i.e., of their ows). Given an arbitrary pair of vector elds V1  V2 2 Vec M, how can one recognize their commuting properties without nding the ows P1t, P2s explicitly, i.e., without integration of the ODEs q_ = V1 (q), q_ = V2 (q) ? If the ows P1t, P2s commute, then the curve (s t) = P1;t  P2s  P1t(q) = P2s(q) t s 2  (1.12) does not depend on t. It is natural to suggest that a lower-order term in the Taylor expansion of (1:12) at t = s = 0 is responsible for commuting properties of ows of the vector elds V1 , V2 at the point q. The rst-order derivatives @ @ = 0 @ t s=t=0 @ s s=t=0 = V2 (q) are obviously useless, as well as the pure second-order derivatives @2 @2 @ s = 0 @t2 s=t=0 @s2 s=t=0 = @ s s=0 V2 (P2 (q)):

R

1.4 Control Systems

17

The required derivative should be the mixed second-order one @2 @[email protected] s=t=0 : It turns out that this derivative is a tangent vector to M. It is called the Lie bracket of the vector elds V1 , V2 and is denoted by V1 V2](q): @2 P1;t  P2s  P1t(q) 2 Tq M: (1.13) V1 V2](q) def = @[email protected] t=s=0 The vector eld V1 V2] 2 Vec M determines commuting properties of V1 and V2 (it is often called commutator of vector elds V1, V2 ). An eective formula for computing Lie bracket of vector elds in local coordinates is given in the following statement. Proposition 1.33. Let V1, V2 be vector elds on n. Then V1  V2](x) = ddVq2 V1 (x) ; ddVx1 V2 (x): (1.14) The proof is left to the reader as an exercise. Another way to dene Lie bracket of vector elds V1 , V2 is to consider the path (t) = P2;t  P1;t  P2t  P1t(q) see Fig. 1.10.

R

P2t P1t

q

P1;t

(t)

V1  V2 ](q )

P2;t

Fig. 1.10. Lie bracket of vector elds V1 , V2

Exercise 1.34. Show that in local coordinates (t) = x + V1 V2 ](x)t2 + o(t2 )

t ! 0

18

1 Vector Fields and Control Systems on Smooth Manifolds

p

i.e., V1  V2](x) is the velocity vector of the C 1 curve ( t). In particular, this proves that V1 V2](x) is indeed a tangent vector to M: V1  V2](x) 2 Tx M: In the next chapter we will develop an ecient algebraic way to do similar calculations without any coordinates. In the commutative case, the set of reachable points does not depend on the number of switches of a control strategy used. In the general noncommutative case, on the contrary, the greater number of switches, the more points can be reached. Suppose that we can move along vector elds V1 and V2 . Then, innitesimally, we can move in the new direction V1  V2], which is in general linearly independent of the initial ones V1 , V2 . Using the same switching control strategy with the vector elds V1 and V1 V2 ], we add one more innitesimal direction of motion V1 V1 V2]]. Analogously, we can obtain V2 V1 V2]]. Iterating this procedure with the new vector elds obtained at previous steps, we can have a Lie bracket of arbitrarily high order as an innitesimal direction of motion with a suciently large number of switches. Example 1.35. Compute the Lie bracket of the vector elds 0 cos 1 001 0x 1 1 V1 (q) = @ sin A  V2 (q) = @ 0 A  q = @ x2 A 2 2(x1x2 )  S1 0 1

appearing in the model of a car. Recall that the eld V1 generates the forward motion, and V2 the counterclockwise rotation of the car. By (1:14), we have 0 0 0 ; sin 1 0 0 1 V1 V2 ](q) = ddVq2 V1 (q) ; ddVq1 V2 (q) = ; @ 0 0 cos A @ 0 A 00 0 1 0 sin 1 = @ ; cos A : 0 The vector eld V1 V2] generates the motion of the car in the direction perpendicular to orientation of the car. This is a typical maneuver in parking a car: the sequence of 4 motions with the same small amplitude of the form motion forward ! rotation counterclockwise ! motion backward ! ! rotation clockwise results in motion to the right (in the main term), see Fig. 1.11. We show this explicitly by computing the Lie bracket V1  V2] as in Exercise 1.34:

R

1.4 Control Systems

19

P2t P1t P2;t

P1;t

Fig. 1.11. Lie bracket for a moving car

0 x 1 0 x + t(cos ; cos( + t)) 1 1 1 P2;t  P1;t  P2t  P1t @ x2 A = @ x2 + t(sin ; sin( + t)) A

0 x 1 0 sin 1

1 t ! 0 = @ x2 A + t2 @ ; cos A + o(t2 )

and we have once more

0

0 sin 1 V1 V2](q) = @ ; cos A : 0

(1.15)

Of course, we can also compute this Lie bracket by denition as in (1:13):

0 x 1 0 x + t(cos ; cos( + s)) 1 1 1 P1;t  P2s  P1t @ x2 A = @ x2 + t(sin ; sin( + s)) A 0 x 1 0 0 1 0 sin

+1s 1 = @ x2 A + s @ 0 A + ts @ ; cos A + O(t2 + s2 )3=2 t s ! 0

1

and the Lie bracket (1:15) follows.

0

2 Elements of Chronological Calculus

We introduce an operator calculus that will allow us to work with nonlinear systems and ows as with linear ones, at least at the formal level. The idea is to replace a nonlinear object, a smooth manifold M, by a linear, although innite-dimensional one: the commutative algebra of smooth functions on M (for details, see 19], 22]). For basic denitions and facts of functional analysis used in this chapter, one can consult e.g. 144].

2.1 Points, Dieomorphisms, and Vector Fields In this section we identify points, dieomorphisms, and vector elds on the manifold M with functionals and operators on the algebra C 1 (M) of all smooth real-valued functions on M. Addition, multiplication, and product with constants are dened in the algebra C 1 (M), as usual, pointwise: if a b 2 C 1 (M), q 2 M, 2 , then (a + b)(q) = a(q) + b(q) (a b)(q) = a(q) b(q) ( a)(q) = a(q): Any point q 2 M denes a linear functional qb : C 1 (M) !  qba = a(q) a 2 C 1(M): The functionals qb are homomorphisms of the algebras C 1 (M) and : qb(a + b) = qba + qbb a b 2 C 1(M) qb(a b) = (qba) (qbb) a b 2 C 1 (M) qb( a) = qba 2  a 2 C 1(M): So to any point q 2 M, there corresponds a nontrivial homomorphism of algebras qb : C 1(M) ! . It turns out that there exists an inverse correspondence.

R

R

R

R

R

22

R

2 Elements of Chronological Calculus

Proposition 2.1. Let ' : C 1(M) ! be a nontrivial homomorphism of algebras. Then there exists a point q 2 M such that ' = qb. We prove this proposition in the Appendix. Remark 2.2. Not only the manifold M can be reconstructed as a set from the algebra C 1(M). One can recover topology on M from the weak topology in the space of functionals on C 1 (M): 1 lim q = q if and only if nlim n!1 n !1 qbna = qba 8a 2 C (M):

Moreover, the smooth structure on M is also recovered from C 1(M), actually, \by denition": a real function on the set fqb j q 2 M g is smooth if and only if it has a form qb 7! qba for some a 2 C 1 (M). Any dieomorphism P : M ! M denes an automorphism of the algebra C 1(M): Pb : C 1 (M) ! C 1 (M) Pb 2 Aut(C 1 (M)) b = a(P(q)) q 2 M a 2 C 1(M) (Pa)(q)

i.e., Pb acts as a change of variables in a function a. Conversely, any automorphism of C 1 (M) has such a form. Proposition 2.3. Any automorphism A : C 1 (M) ! C 1(M) has a form of Pb for some P 2 Di M . Proof. Let A 2 Aut(C 1 (M)). Take any point q 2 M. Then the composition qb  A : C 1(M) ! is a nonzero homomorphism of algebras, thus it has the form qb1 for some q1 2 M. We denote q1 = P(q) and obtain

R

qb  A = P(q) = qb  Pb

8q 2 M

i.e.,

b A = P and P is the required dieomorphism. ut Now we characterize tangent vectors to M as functionals on C 1 (M). Tangent vectors to M are velocity vectors to curves in M, and points of M are identied with linear functionals on C 1 (M) thus we should obtain linear functionals on C 1 (M), but not homomorphisms into . To understand, which functionals on C 1 (M) correspond to tangent vectors to M, take a smooth curve q(t) of points in M. Then the corresponding curve of functionals qb(t) = qd (t) on C 1 (M) satises the multiplicative rule

R

2.1 Points, Di eomorphisms, and Vector Fields

a b 2 C 1 (M):

qb(t)(a b) = qb(t)a qb(t)b

23

We dierentiate this equality at t = 0 and obtain that the velocity vector to the curve of functionals

def = dd qbt  : C 1 (M) !  t=0 satises the Leibniz rule:

R

(ab) = (a)b(q(0)) + a(q(0)) (b): Consequently, to each tangent vector v 2 Tq M we should put into correspondence a linear functional

: C 1 (M) !

such that

R

a b 2 C 1 (M):

(ab) = ( a)b(q) + a(q)( b)

(2.1)

But there is a linear functional = vb naturally related to any tangent vector v 2 Tq M, the directional derivative along v: bva = ddt a(q(t)) q(0) = q q(0) _ = v t=0 and such functional satises Leibniz rule (2:1). Now we show that this rule characterizes exactly directional derivatives. Proposition 2.4. Let : C 1 (M) ! be a linear functional that satis es Leibniz rule (2:1) for some point q 2 M . Then = bv for some tangent vector v 2 Tq M . Proof. Notice rst of all that any functional that meets Leibniz rule (2:1) is local, i.e., it depends only on values of functions in an arbitrarily small neighborhood Oq of the point q:

R

a ~a 2 C 1 (M):

a~jOq = ajOq ) ~a = a

Indeed, take a cut function b 2 C 1(M) such that bjM nOq  1 and b(q) = 0. Then (~a ; a)b = a~ ; a, thus

(~a ; a) = ((~a ; a)b) = (~a ; a) b(q) + (~a ; a)(q) b = 0: So the statement of the proposition is local, and we prove it in coordinates. Let (x1 : : :  xn) be local coordinates on M centered at the point q. We have to prove that there exist 1 : : :  n 2 such that

R

24

2 Elements of Chronological Calculus

=

n X

i @@x : i 0 i=1

First of all,

(1) = (1 1) = ( 1) 1 + 1 ( 1) = 2 (1) thus (1) = 0. By linearity, (const) = 0. In order to nd the action of on an arbitrary smooth function, we expand it by the Hadamard Lemma: a(x) = a(0) +

n Z 1 @a X i=1 0

where

@ xi (tx)xi dt = a(0) + i=1 bi (x)xi 

bi (x) =

are smooth functions. Now

a =

n X i=1

(bi xi) =

n X i=1

n X

Z 1 @a 0

@ xi (tx) dt

(( bi )xi (0) + bi (0)( xi )) =

n X i @@ xa (0) i i=1

where we denote i = xi and make use of the equality bi (0) = @@xa (0). i

R

ut

So tangent vectors v 2 Tq M can be identied with directional derivatives

bv : C 1(M) ! , i.e., linear functionals that meet Leibniz rule (2:1).

Now we characterize vector elds on M. A smooth vector eld on M is a family of tangent vectors vq 2 Tq M, q 2 M, such that for any a 2 C 1 (M) the mapping q 7! vq a, q 2 M, is a smooth function on M. To a smooth vector eld V 2 Vec M there corresponds a linear operator Vb : C 1(M) ! C 1 (M)

that satises the Leibniz rule

Vb (ab) = (Vb a)b + a(Vb b)

a b 2 C 1 (M)

the directional derivative (Lie derivative) along V . A linear operator on an algebra meeting the Leibniz rule is called a derivation of the algebra, so the Lie derivative Vb is a derivation of the algebra C 1(M). We show that the correspondence between smooth vector elds on M and derivations of the algebra C 1 (M) is invertible. Proposition 2.5. Any derivation of the algebra C 1(M) is the directional derivative along some smooth vector eld on M .

2.2 Seminorms and C 1 (M )-Topology

25

Proof. Let D : C 1 (M) ! C 1(M) be a derivation. Take any point q 2 M.

R

We show that the linear functional dq def = qb  D : C 1 (M) ! is a directional derivative at the point q, i.e., satises Leibniz rule (2:1): dq (ab) = qb(D(ab)) = qb((Da)b + a(Db)) = qb(Da)b(q) + a(q)qb(Db) = (dq a)b(q) + a(q)(dq b) a b 2 C 1 (M):

R

ut

So we can identify points q 2 M, dieomorphisms P 2 Di M, and vector elds V 2 Vec M with nontrivial homomorphisms qb : C 1(M) ! , automorphisms Pb : C 1(M) ! C 1 (M), and derivations Vb : C 1 (M) ! C 1 (M) respectively. b For example, we can write a point P(q) in the operator notation as qb  P. Moreover, in the sequel we omit hats and write q  P. This does not cause ambiguity: if q is to the right of P, then q is a point, P a dieomorphism, and P(q) is the value of the dieomorphism P at the point q. And if q is to the left of P, then q is a homomorphism, P an automorphism, and q  P a homomorphism of C 1 (M). Similarly, V (q) 2 Tq M is the value of the vector eld V at the point q, and q  V : C 1(M) ! is the directional derivative along the vector V (q).

R

2.2 Seminorms and C 1(M )-Topology

We introduce seminorms and topology on the space C 1 (M). By Whitney's Theorem, a smooth manifold M can be properly embedded into a Euclidean space N for suciently large N. Denote by hi, i = 1 : : :  N, the smooth vector eld on M that is the orthogonal projection from N to M of the constant basis vector eld @@xi 2 Vec( N ). So we have N vector elds h1 : : :  hN 2 Vec M that span the tangent space Tq M at each point q 2 M. We dene the family of seminorms k ksK on the space C 1 (M) in the following way:

R

R

R

kaksK = sup fjhil   hi1 a(q)j j q 2 K 1  i1  : : :  il  N 0  l  sg  a 2 C 1 (M) s  0 K b M: This family of seminorms denes a topology on C 1 (M). A local base of this topology is given by the subsets



N



a 2 C 1(M) j kaknKn < n1 

N

n2 

where Kn, n 2 , is a chained system of compacta that cover M:

26

2 Elements of Chronological Calculus

Kn Kn+1 

1 

n=1

Kn = M:

R

This topology on C 1 (M) does not depend on embedding of M into N . It is called the topology of uniform convergence of all derivatives on compacta , or just C 1 (M)-topology . This topology turns C 1 (M) into a Fr"echet space (a complete, metrizable, locally convex topological vector space). A sequence of functions ak 2 C 1 (M) converges to a 2 C 1 (M) as k ! 1 if and only if lim ka ; aksK = 0 8 s  0 K b M: k!1 k For vector elds V 2 Vec M, we dene the seminorms

kV ksK = sup fkV aksK j kaks+1K = 1g  s  0 K b M: (2.2) One can prove that any vector eld V 2 Vec M has nite seminorms kV ksK , and that there holds an estimate of the action of a dieomorphism P 2 Di M on a function a 2 C 1 (M): kPaksK  CsP kaksP (K ) 

s  0 K b M:

(2.3)

Thus vector elds and dieomorphisms are linear continuous operators on the topological vector space C 1 (M).

2.3 Families of Functionals and Operators In the sequel we will often consider one-parameter families of points, dieomorphisms, and vector elds that satisfy various regularity properties (e.g. dierentiability or absolute continuity) with respect to the parameter. Since we treat points as functionals, and dieomorphisms and vector elds as operators on C 1 (M), we can introduce regularity properties for them in the weak sense, via the corresponding properties for one-parameter families of functions t 7! at 

R

at 2 C 1 (M) t 2 :

So we start from denitions for families of functions. Continuity and di erentiability of a family of functions at w.r.t. parameter t are dened in a standard way since C 1(M) is a topological vector space. A family at is called measurable w.r.t. t if the real function t 7! at(q) is measurable for any q 2 M. A measurable family at is called locally integrable if Z t1 katksK dt < 1 8 s  0 K b M t0  t1 2 : t0

A family at is called absolutely continuous w.r.t. t if

R

2.3 Families of Functionals and Operators

at = at0 +

Zt t0

27

b d

for some locally integrable family of functions bt . A family at is called Lipschitzian w.r.t. t if kat ; a ksK  CsK jt ; j 8s  0 K b M t 2  and locally bounded w.r.t. t if katksK  CsKI  8 s  0 K b M I b  t 2 I where CsK and CsKI are some constants depending on s, K, and I. Now we can dene regularity properties of families of functionals and operators on C 1 (M). A family of linear functionals or linear operators on C 1 (M) t 7! At  t2  has some regularity property (i.e., is continuous , di erentiable , measurable , locally integrable , absolutely continuous , Lipschitzian , locally bounded w.r.t. t) if the family t 7! At a t2  1 has the same property for any a 2 C (M). A locally bounded w.r.t. t family of vector elds t 7! Vt  Vt 2 Vec M t 2  is called a nonautonomous vector eld , or simply a vector eld , on M. An absolutely continuous w.r.t. t family of dieomorphisms t 7! P t P t 2 Di M t 2  is called a ow on M. So, for a nonautonomous vector eld Vt, the family of functions t 7! Vt a is locally integrable for any a 2 C 1(M). Similarly, for a ow P t, the family of functions (P ta)(q) = a(P t(q)) is absolutely continuous w.r.t. t for any a 2 C 1 (M). Integrals of measurable locally integrable families, and derivatives of differentiable families are also dened in the weak sense:

R

R

R

R

R R

Zt t0

1

At dt : a 7!

Zt

1

t0

(Ata) dt

a 2 C 1(M)

d d a 2 C 1(M): d t At : a 7! d t (Ata) One can show that if At and Bt are continuous families of operators on C 1(M) which are dierentiable at t0, then the family At  Bt is continuous, moreover, dierentiable at t0 , and satises the Leibniz rule:

28

2 Elements of Chronological Calculus



!



!

d (A  B ) = d A  B + A  d B  t0 t0 d t t0 t t d t t0 t d t t0 t see the proof in the Appendix. If families At and Bt of operators are absolutely continuous, then the composition At  Bt is absolutely continuous as well, the same is true for composition of functionals with operators. For an absolute continuous family of functions at , the family At at is also absolutely continuous, and the Leibniz rule holds for it as well.

2.4 Chronological Exponential In this section we consider a nonautonomous ordinary di erential equation of the form q_ = Vt (q) q(0) = q0 (2.4) where Vt is a nonautonomous vector eld on M, and study the ow determined by this eld. We denote by q_ the derivative dd qt , so equation (2:4) reads in the expanded form as d q(t) = V (q(t)): t dt

2.4.1 ODEs with Discontinuous Right-Hand Side To obtain local solutions to the Cauchy problem (2:4) on a manifold M, we reduce it to a Cauchy problem in a Euclidean space. For details about nonautonomous dierential equations in n with right-hand side discontinuous in t, see e.g. 138]. Choose local coordinates x = (x1 : : :  xn) in a neighborhood Oq0 of the point q0: n  : Oq0 M ! Ox0  : q 7! x (q0) = x0: In these coordinates, the eld Vt reads n X e ( Vt ) (x) = Vt (x) = vi (t x) @@xi  x 2 Ox0  t 2  (2.5) i=1

R

R

R

R

and problem (2:4) takes the form n: x_ = Vet (x) x(0) = x0 x 2 Ox0 (2.6) Since the nonautonomous vector eld Vt 2 Vec M is locally bounded, the components vi (t x), i = 1 : : :  n, of its coordinate representation (2:5) are:

R

2.4 Chronological Exponential

29

(1) measurable and locally bounded w.r.t. t for any xed x 2 Ox0 , (2) smooth w.r.t. x for any xed t 2 , (3) dierentiable in x with locally bounded partial derivatives: @ vi (t x)  C  t 2 I b  x 2 K b Ox0  i = 1 : : :  n: IK @x By the classical Carath"eodory Theorem (see e.g. 8]), the Cauchy problem (2:6) has a unique solution, i.e., a vector-function x(t x0), Lipschitzian w.r.t. t and smooth w.r.t. x0 , and such that: (1) ODE (2:6) is satised for almost all t, (2) initial condition holds: x(0 x0) = x0. Then the pull-back of this solution from n to M

R

R ;

q(t q0) =  1(x(t x0)) is a solution to problem (2:4) in M. The mapping q(t q0) is Lipschitzian w.r.t. t and smooth w.r.t. q0, it satises almost everywhere the ODE and the initial condition in (2:4). For any q0 2 M, the solution q(t q0) to the Cauchy problem (2:4) can be continued to a maximal interval t 2 Jq0 containing the origin and depending on q0. We will assume that the solutions q(t q0) are dened for all q0 2 M and all t 2 , i.e., Jq0 = for any q0 2 M. Then the nonautonomous eld Vt is called complete . This holds, e.g., when all the elds Vt , t 2 , vanish outside of a common compactum in M (in this case we say that the nonautonomous vector eld Vt has a compact support ).

R

R

R

R

2.4.2 Denition of the Right Chronological Exponential Equation (2:4) rewritten as a linear equation for Lipschitzian w.r.t. t families of functionals on C 1 (M): q(t) _ = q(t)  Vt 

R

q(0) = q0

is satised for the family of functionals

q(t q0) : C 1 (M) ! 

q0 2 M t 2

(2.7)

R

constructed in the previous subsection. We prove later that this Cauchy problem has no other solutions (see Proposition 2.9). Thus the ow dened as P t : q0 7! q(t q0) is a unique solution of the operator Cauchy problem

(2.8)

30

2 Elements of Chronological Calculus

P_ t = P t  Vt

P 0 = Id

(2.9)

(where Id is the identity operator) in the class of Lipschitzian ows on M. The ow P t determined in (2:8) is called the right chronological exponential of the eld Vt and is denoted as

;! P t = exp

Zt 0

V d :

Now we develop an asymptotic series for the chronological exponential, which justies such a notation.

2.4.3 Formal Series Expansion We rewrite dierential equation in (2:7) as an integral one: q(t) = q0 +

Zt 0

q( )  V d

(2.10)

then substitute this expression for q(t) into the right-hand side = q0 +

Z t

q0 +

0 Z t

= q0  Id +

0



Z

1

0

q( 2 )  V2 d 2  V1 d 1



ZZ

V dt +

02 1 t

q( 2 )  V2  V1 d 2 d 1

repeat this procedure iteratively, and obtain the decomposition:

0 Zt ZZ B q(t) = q0  @Id + V d + V  V d 2 d 1 + : : :+ 0

(t) 1 Z Z Vn   V d n : : : d 1C A+

n (t) Z Z 2

1

2

1

q( n+1 )  Vn+1   V1 d n+1 : : : d 1: (2.11)

n+1 (t)

Here

R

n(t) = f( 1  : : :  n ) 2 n j 0  n   1  tg is the n-dimensional simplex. Purely formally passing in (2:11) to the limit n ! 1, we obtain a formal series for the solution q(t) to problem (2:7):

2.4 Chronological Exponential

0 1 Z Z 1 X q0  B Vn   V d n : : : d 1C @Id + A

31

1

n=1 n (t)

thus for the solution P t to problem (2:9): Id +

1Z X

Z

Vn   V1 d n : : : d 1:

n=1 n (t)

(2.12)

Exercise 2.6. We obtained the previous series expansion under the condition t > 0, although;! theRchronological exponential is dened for all values of t. Show t V d admits for t < 0 the series expansion that the ow exp 0  1Z Z X Id + (;Vn )   (;V1 ) d n : : : d 1: n=1 n (;t) This series is similar to (2:12), so in the sequel we restrict ourselves by the study of the case t > 0.

2.4.4 Estimates and Convergence of the Series Unfortunately, these series never converge on C 1 (M) in the weak sense (if Vt 6 0): there always exists a smooth function on M, on which they diverge. Although, one can show that series (2:12)R gives an asymptotic expansion for ;! t V d . There holds the following the chronological exponential P t = exp 0

bound of the remainder term: denote the m-th partial sum of series (2:12) as Sm (t) = Id +

mX ;1 Z

Z

Vn   V1 d n : : : d 1 

n=1 n (t)

then for any a 2 C 1 (M), s  0, K b M

 ;! Z t    exp V d ; Sm (t) a 0 sK Z t R C t kV ksK0 d 1  Ce

m

m! 0 kV ks+m;1K 0 d kaks+mK 0 (2.13) = O(tm ) t ! 0 where K 0 b M is some compactum containing K. We prove estimate (2:13) in the Appendix. It follows from estimate (2:13) that 0

 ;! Z t   exp "V d ; Sm" (t) a = O("m ) 0 sK

" ! 0

32

2 Elements of Chronological Calculus

where Sm" (t) is the m-th partial sum of series (2:12) for the eld "Vt . Thus we have an asymptotic series expansion:

;! exp

Zt 0

V d  Id +

1Z X

Z

Vn   V1 d n : : : d 1:

n=1 n (t)

(2.14)

In the sequel we will use terms of the zeroth, rst, and second orders of the series obtained:

;! exp

Zt 0

V d  Id +

Zt

V d +

0

ZZ

02 1 t

V2  V1 d 2 d 1 + :

We prove that the asymptotic series converges to the chronological exponential on any normed subspace L C 1 (M) where Vt is well-dened and bounded: Vt L L kVtk = sup fkVt ak j a 2 L kak  1g < 1: (2.15) We apply operator series (2:14) to any a 2 L and bound terms of the series obtained: a+ We have

1Z X

Z

Vn   V1 a d n : : : d 1:

n=1 n (t)

(2.16)

  Z Z   Vn   V a d n : : : d 1  n(t)  Z Z 1



0n 1 t

kVn k kV1 k d n : : : d 1 kak

by symmetry w.r.t. permutations of indices  : f1 : : :  ng ! f1 : : :  ng

Z

Z

kVn k kV1 k d n : : : d 1 kak 0(n) (1) t

=

passing to the integral over cube

Zt Zt = n!1 : : : kVn k kV1 k d n : : : d 1 kak Z0 t 0 n 1 = n! kV k d kak: 0

2.4 Chronological Exponential

33

So series (2:16) is majorized by the exponential series, thus the operator series (2:14) converges on L. Series (2:16) can be dierentiated termwise, thus it satises the same ODE as the function P ta: a_ t = Vt at a0 = a: Consequently,

1Z

Z

X P ta = a + n=1 n (t)

Vn   V1 a d n : : : d 1:

So in the case (2:15) the asymptotic series converges to the chronological exponential and there holds the bound

R

kP tak  e 0t kV k d kak

a 2 L:

Moreover, one can show that the bound and convergence hold not only for locally bounded, but also for integrable on 0 t] vector elds:

Zt 0

kV k d < 1:

RR

Notice that conditions (2:15) are satised for any nite-dimensional Vtinvariant subspace L C 1 (M). In particular, this is the case when M = n, L is the space of linear vector elds, and Vt is a linear vector eld on n. If M, Vt , and a are real analytic, then series (2:16) converges for suciently small t, see the proof in 19].

2.4.5 Left Chronological Exponential Consider the ;! inverse operator Qt = (P t);1 to the right chronological expoR nential P t = exp t V d . We nd an ODE for the ow Qt by dierentiation of the identity

0

Leibniz rule yields

P t  Qt = Id :

P_ t  Qt + P t  Q_ t = 0 thus, in view of ODE (2:9) for the ow P t, P t  Vt  Qt + P t  Q_ t = 0: We multiply this equality by Qt from the left and obtain Vt  Qt + Q_ t = 0: That is, the ow Qt is a solution of the Cauchy problem

34

2 Elements of Chronological Calculus

d Qt = ;V  Qt  Q0 = Id (2.17) t dt which is dual to the Cauchy problem (2:9) for P t. The ow Qt is called the left chronological exponential and is denoted as

; Qt = exp

Zt 0

(;V ) d :

We nd an asymptotic expansion for the left chronological exponential in the same way as for the right one, by successive substitutions into the right-hand side: Qt = Id + = Id + = Id +

Zt Z0t 0

(;V )  Q d (;V ) d +

mX ;1 Z

Z

ZZ

2 (t)

(;V1 )  (;V2 )  Q2 d 2 d 1 =

(;V1 )   (;Vn ) d n : : : d 1

n=1 n (t)

Z

Z

+ (;V1 )   (;Vm )  Qm d m : : : d 1 :

m (t)

For the left chronological exponential holds an estimate of the remainder term as (2:13) for the right one, and the series obtained is asymptotic:

; exp

Zt 0

(;V ) d  Id +

1Z X

Z

(;V1 )   (;Vn ) d n : : : d 1:

n=1 n (t)

Remark 2.7. (1) Notice that the reverse arrow in the left chronological expo-

; corresponds to the reverse order of the operators (;V )   nential exp 1 (;Vn ), n  : : :  1 . (2) The right and left chronological exponentials satisfy the corresponding dierential equations: d exp ;! Z t V d = exp ;! Z t V d  Vt dt 0 0 d exp ; Z t (;V ) d = ;Vt  exp ; Z t (;V ) d : dt 0

0

The directions of arrows correlate with the direction of appearance of operators Vt , ;Vt in the right-hand side of these ODEs. (3) If the initial value is prescribed at a moment of time t0 6= 0, then the lower limit of integrals in the chronological exponentials is t0 .

2.4 Chronological Exponential

35

(4) There holds the following obvious rule for composition of ows:

;! exp

Zt

1

t0

;! Z t2 V d = exp ;! Z t2 V d : V d  exp t1

t0

Exercise 2.8. Prove that ;! Z t1 V d =  exp ;! Z t0 V d ;1 = exp ; Z t0 (;V ) d : exp t0

t1

t1

(2.18)

2.4.6 Uniqueness for Functional and Operator ODEs We saw that equation (2:7) for Lipschitzian families of functionals has a soR ;! lution q(t) = q0 exp 0t V d . We can prove now that this equation has no other solutions. Proposition 2.9. Let Vt be a complete nonautonomous vector eld on M . Then Cauchy problem (2:7) has a unique solution in the class of Lipschitzian families of functionals on C 1 (M). Proof. Let a Lipschitzian family of functionals qt be a solution to problem (2:7). Then d ;q  (P t);1  = d ;q  Qt = q  V  Qt ; q  V  Qt = 0 t t t t dt t dt t thus qt  Qt  const. But Q0 = Id, consequently, qt  Qt  q0, hence

;! qt = q0  P t = q0  exp

Zt 0

V d

is a unique solution of Cauchy problem (2:7). ut Similarly, the both operator equations P_ t = P t  Vt and Q_ t = ;Vt  Qt have no other solutions in addition to the chronological exponentials.

2.4.7 Autonomous Vector Fields For an autonomous vector eld Vt  V 2 Vec M the ow generated by a complete eld is called the exponential and is denoted as etV . The asymptotic series for the exponential takes the form etV 

1 tn X

2 n = Id +tV + t V  V +  V 2 n=0 n!

36

2 Elements of Chronological Calculus

i.e, it is the standard exponential series. The exponential of an autonomous vector eld satises the ODEs d tV tV tV etV t=0 = Id : d te = e  V = V  e  We apply the asymptotic series for exponential to nd the Lie bracket of autonomous vector elds V W 2 Vec M. We compute the rst nonconstant term in the asymptotic expansion at t = 0 of the curve: q(t) = q  etV  etW  e;tV  e;tW     2 2 t t 2 2 = q  Id +tV + 2 V +  Id +tW + 2 W +     2 2 t t 2 2  Id ;tV + 2 V +  Id ;tW + 2 W +   2 t 2 2 = q  Id +t(V + W) + 2 (V + 2V  W + W ) +   2 t 2 2  Id ;t(V + W) + 2 (V + 2V  W + W ) + = q  (Id +t2 (V  W ; W  V ) + ) :

So the Lie bracket of the vector elds as operators (directional derivatives) in C 1(M) is V W] = V  W ; W  V: This proves the formula in local coordinates: if V= then

n X ai @@x  i i=1

W=

n X bi @@x  i i=1

ai  bi 2 C 1 (M)

 @ dW dV n  @b X @ a i i V W ] = aj @ x ; bj @ x @ x = d x V ; d x W: j j i ij =1

Similarly,

q  etV  esW  e;tV = q  (Id +tV + )  (Id +sW + )  (Id ;tV + ) = q  (Id +sW + tsV W] + ) and

@2 q  etV  esW  e;tV : q  V W] = @[email protected] s=t=0

2.5 Action of Di eomorphisms on Vector Fields

2.5 Action of Dieomorphisms on Vector Fields

37

We have already found counterparts to points, dieomorphisms, and vector elds among functionals and operators on C 1 (M). Now we consider action of dieomorphisms on vector elds. Take a tangent vector v 2 Tq M and a dieomorphism P 2 Di M. The tangent vector P v 2 TP (q) M is the velocity vector of the image of a curve starting from q with the velocity vector v. We claim that P v = v  P v 2 Tq M P 2 Di M (2.19) as functionals on C 1 (M). Take a curve q(t) 2 M q(0) = q ddt q(t) = v t=0 then   Pv a = ddt a(P(q(t))) = ddt q(t)  P a t=0 t=0 = v  Pa a 2 C 1 (M): Now we nd expression for P V , V 2 Vec M, as a derivation of C 1 (M). We have q  P  P V = P(q)  PV = (P V ) (P(q)) = P (V (q)) = V (q)  P = q  V  P q 2 M thus P  P V = V  P i.e., P V = P ;1  V  P P 2 Di M V 2 Vec M: So dieomorphisms act on vector elds as similarities. In particular, dieomorphisms preserve compositions: P(V  W) = P ;1  (V  W)  P = (P ;1  V  P)  (P ;1  W  P ) = P V  P W thus Lie brackets of vector elds: P V W] = P (V  W ; W  V ) = P V  P W ; P W  P V = PV P W ]: If B : C 1 (M) ! C 1(M) is an automorphism, then the standard algebraic notation for the corresponding similarity is Ad B: (Ad B)V def = B  V  B ;1 :

That is,

38

2 Elements of Chronological Calculus

P = Ad P ;1 P 2 Di M: Now we nd an innitesimal version of the operator Ad. Let P t be a ow on M, d t P 0 = Id d t P = V 2 Vec M: t=0

Then so

; t;1 = ;V d d t t=0 P d t)W = d t t ;1 (Ad P d t t=0 d t t=0 (P  W  (P ) ) = V  W ; W  V = V W] W 2 Vec M:

Denote then

d

ad V = ad d t t=0

Pt



= ddt Ad P t t=0

def

(ad V )W = V W] W 2 Vec M: Dierentiation of the equality Ad P t X Y ] = AdP t X Ad P t Y ] X Y 2 Vec M at t = 0 gives Jacobi identity for Lie bracket of vector elds: (ad V )X Y ] = (ad V )X Y ] + X (ad V )Y ] which may also be written as V X Y ]] = V X] Y ] + X V Y ]] V X Y 2 Vec M or, in a symmetric way X Y Z]] + Y Z X]] + Z X Y ]] = 0 X Y Z 2 Vec M: (2.20) The set Vec M is a vector space with an additional operation | Lie bracket, which has the properties: (1) bilinearity:  X + Y Z] = X Z] + Y Z] X Y + Z] = X Y ] + X Z] X Y Z 2 Vec M   2  (2) skew-symmetry:

R

X Y ] = ;Y X]

X Y 2 Vec M

2.5 Action of Di eomorphisms on Vector Fields

39

Zt 0

Z

Z

+ ad Vn   ad V1 d n : : : d 1 + (2.23)

n (t)

then prove uniqueness of the solution, and justify the following notation:





Zt Zt ;! ;! def t exp ad V d = Ad P = Ad exp V d : 0

0

Similar identities for the left chronological exponential are





; Z t ad(;V ) d def ; Z t(;V ) d exp = Ad exp 0 0 1Z Z X  Id+ (; ad V1 )   (; ad Vn ) d n : : : d 1: n=1 n (t)

40

2 Elements of Chronological Calculus

For the asymptotic series (2:23), there holds an estimate of the remainder term similar to estimate (2:13) for the ow P t. Denote the partial sum Tm = Id +

mX ;1 Z

Z

ad Vn   ad V1 d n : : : d 1

n=1 n (t)

then for any X 2 Vec M, s  0, K b M

 ;! Z t    Ad exp V d ; Tm X  0 sK Z t R C t kV ks K0 d 1  C1e

1 0

+1

m!

0

kV ks+mK 0 d

m

kX ks+mK 0 (2.24)

= O(tm )

t ! 0 where K 0 b M is some compactum containing K. For autonomous vector elds, we denote

et ad V def = Ad etV  thus the family of operators et ad V : Vec M ! Vec M is the unique solution to the problem A_ t = At  ad V A0 = Id which admits the asymptotic expansion 2 et ad V  Id +t ad V + t2 ad2 V + : Exercise 2.10. Let P 2 Di M, and let Vt be a nonautonomous vector eld on M. Prove that

;! Z t V d  P ;1 = exp ;! Z t Ad P V d : P  exp   0

2.6 Commutation of Flows

(2.25)

0

R

;! t V d the Let Vt 2 Vec M be a nonautonomous vector eld and P t = exp 0 corresponding ow. We are interested in the question: under what conditions the ow P t preserves a vector eld W 2 Vec M: P t W = W 8t or, which is equivalent, (P t); 1 W = W 8t:

2.7 Variations Formula

41

Proposition 2.11. PtW = W 8t , Vt  W ] = 0 8t: Proof. We have





d (P );1 W = d Ad P tW = d exp ;! Z t ad V d W t dt  dt dt 0

 ;! Z t   ;! Z t  = exp ad V d  ad V W = exp ad V d Vt W] 0 0 t ; 1 = (P ) Vt  W]

thus (P t); 1 W  W if and only if Vt W]  0. ut In general, ows do not commute, neither for nonautonomous vector elds Vt, Wt :

;! exp

Zt 0

1

;! V d  exp

Zt

2

0

;! W d 6= exp

Zt 0

2

;! Z t1 V d  W d  exp 0

nor for autonomous vector elds V , W : et1 V  et2 W 6= et2 W  et1 V : In the autonomous case, commutativity of ows is equivalent to commutativity of vector elds:

R

et1 V  et2 W = et2 W  et1 V  t1  t2 2 

,

V W ] = 0:

We already showed that commutativity of vector elds is necessary for commutativity of ows. Let us prove that it is sucient. Indeed, ;Ad et1V  W = et1 ad V W = W: Taking into account equality (2:25), we obtain tV et1 V  et2 W  e;t1 V = et2 (Ad e 1 )W = et2 W :

2.7 Variations Formula Consider an ODE of the form q_ = Vt (q) + Wt (q): (2.26) We think of Vt as an initial vector eld and Wt as its perturbation. Our aim is to nd a formula for the ow Qt of the new eld Vt + Wt as a perturbation

42

2 Elements of Chronological Calculus

R

;! t V d of the initial eld Vt . In other words, we wish of the ow P t = exp 0 to have a decomposition of the form ;! Qt = exp

Zt 0

(V + W ) d = Ct  P t:

We proceed as in the method of variation of parameters we substitute the previous expression to ODE (2:26): d t t d t Q = Q  (Vt + Wt ) = C_ t  P t + Ct  P t  Vt = C_ t  P t + Qt  Vt  cancel the common term Qt  Vt: Qt  Wt = C_ t  P t and write down the ODE for the unknown ow Ct: ;  C_ t = Qt  Wt  P t ;1 ;  = Ct  P t  Wt  P t ;1 ;  = Ct  Ad P t Wt

 ;! Z t



= Ct  exp ad V d Wt  0 C0 = Id : This operator Cauchy problem is of the form (2:9), thus it has a unique solution: ;! Z t  exp ;! Z  adV d  W d : Ct = exp 0

0

Hence we obtain the required decomposition of the perturbed ow:

 Zt Z t  ;! Z  Zt ;! ;! ;! exp (V + W ) d = exp exp ad V d W d  exp V d : 0

0

0

0

(2.27) This equality is called the variations formula . It can be written as follows:

;! exp

Zt 0

;! (V + W ) d = exp

Zt 0

(Ad P  ) W d  P t:

So the perturbed ow is a composition of the initial ow P t with the ow of the perturbation Wt twisted by P t. Now we obtain another form of the variations formula, with the ow P t to the left of the twisted ow. We have

;! Z t

exp

0

2.8 Derivative of Flow with Respect to Parameter

;! Z t

(V + W ) d = exp

0

43

(Ad P  ) W d  P t

;  = P t  P t ;1 

;! exp

Zt

(Ad P  ) W d  P t

Z t  ; ;1  ;! t = P  exp Ad P t  Ad P  W d 0

Z t  ; ;1  ;! t Ad P t  P  W d : = P  exp 0 0

;P t;1  P  = exp ;! Z  V d 

Since

t

we obtain





;! Z t (V + W ) d = P t exp ;! Z t exp ;! Z  ad V d W d exp 0 0 t  Z Z t ;! t exp ;! Z  ad V d  W d : ;! V d  exp = exp 0

t

0

(2.28) For autonomous vector elds V W 2 Vec M, the variations formulas (2:27), (2:28) take the form:

;! et(V +W ) = exp

Zt 0

;! e ad V W d  etV = etV  exp

In particular, for t = 1 we have eV +W

Zt 0

e( ;t) ad V W d : (2.29)

Z1 ;! = exp e ad V W d  eV : 0

2.8 Derivative of Flow with Respect to Parameter Let Vt (s) be a nonautonomous vector eld depending smoothly on a real parameter s. We study dependence of the ow of Vt (s) on the parameter s. We write

;! exp

Zt 0

;! V (s + ") d = exp

Zt 0

(V (s) + V (s ")) d

(2.30)

with the perturbation V (s ") = V (s + ") ; V (s). By the variations formula (2:27), the previous ow is equal to

 Z t  ;! Z  ;! Z t V (s) d : ;! exp exp adV (s) d V (s ") d  exp  0

0

0

44

2 Elements of Chronological Calculus

Now we expand in ": " ! 0 V (s ") = " @@s V (s) + O("2 )   Z  ;! ad V (s) d V (s ") W (s ") def = exp  ;! 0Z  @ = " exp ad V (s) d @ s V (s) + O("2 ) 0 thus

;! exp

Zt 0

W (s ") d = Id +

Zt

= Id +" Finally,

;! exp

Zt

;! V (s + ") d = exp

Zt ;! = exp V (s) d 0

+"

Z 0t  ;! Z  exp

0

0

W (s ") d + O("2 )

Z0 t  ;! Z  0

Zt 0

" ! 0

exp

0

@ ad V (s) d @ s V (s) d + O("2 ):

;! Ws (") d  exp

Zt 0

V (s) d

@ Zt ;! ad V (s) d @ s V (s) d  exp V (s) d + O("2 ) 0

that is,

@ exp ;! Z t V (s) d  @s 0  Z t  ;! Z  ;! Z t V (s) d : (2.31) = exp ad V (s) d @@s V (s) d  exp 0

0

0

Similarly, we obtain from the variations formula (2:28) the equality

@ exp ;! Z t V (s) d @s 0

 adV (s) d @@s V (s) d : (2.32) 0 0 t For an autonomous vector eld depending on a parameter V (s), formula (2:31) takes the form ;! = exp

Zt

V (s) d 

Z t  ;! Z  exp

@ etV (s) = Z t e ad V (s) @ V d  etV (s) @s @s 0

2.8 Derivative of Flow with Respect to Parameter

and at t = 1:

@ eV (s) = Z 1 e ad V (s) @ V d  eV (s) : @s @s 0

Proposition 2.12. Assume that

Z t

Then

0

;! exp



8t:

V d  Vt = 0

Zt 0

R V d = e 0t V d

45

(2.33) (2.34)

8t:

That is, we state that under assumption (2:34), the chrono;! R t Vthed commutativity t = eR0t V d dened logical exponential exp coincides with the ow Q 0 as follows: Qt = Qt1  @ Qts = Z t V d  Qt  Qt = Id :  s 0 @s 0 Proof. We show that the exponential in the right-hand side satises the same ODE as the chronological exponential in the left-hand side. By (2:33), we have d eR0t V d = Z 1 e ad R0t V d V d  eR0t V d : t dt 0 In view of equality (2:34), R e ad 0t V d Vt = Vt  thus d eR0t V d = V  eR0t V d : t dt By equality (2:34), we can permute operators in the right-hand side: d eR0t V d = eR0t V d  V : t dt Notice the initial condition Rt e 0 V d t=0 = Id : Now the statement follows since the Cauchy problem for ows A_ t = At  Vt  A0 = Id has a unique solution: At = e

Rt V 0

d

Zt ;! = exp V d : 0

ut

3 Linear Systems

In this chapter we consider the simplest class of control systems | linear systems

R

x_ = Ax + c +

m X i=1

R

R

x 2 n u = (u1 : : :  um ) 2 m

ui bi

(3.1)

where A is a constant real n  n matrix and c, b1, : : : , bm are constant vectors in n.

3.1 Cauchy's Formula for Linear Systems Let u(t) = (u1(t) : : :  um (t)) be locally integrable functions. Then the solution of (3:1) corresponding to this control and satisfying the initial condition x(0 x0) = x0 is given by Cauchy's formula :



x(t x0) = etA x0 +

Zt 0

e;A

X m i=1

ui( )bi + c d

!!

R

 t2 :

Here we use the standard notation for the matrix exponential: 2 n etA = Id +tA + t2! A2 + + tn! An + : Cauchy's formula is veried by dierentiation. In view of uniqueness, it gives the solution to the Cauchy problem. Linear system (3:1) is a particular case of a control-ane system :



x_ = x  f0 +

m X i=1

!

uifi 

(3.2)

48

3 Linear Systems

in order to obtain (3:1) from (3:2), one should just take f0 (x) = Ax + c

fi (x) = bi i = 1 : : :  m:

(3.3)

Let us show that Cauchy's formula is actually a special case of the general variations formula.

Proposition 3.1. Cauchy's formula specializes the variations formula for linear systems.

Proof. We restrict ourselves with the case c = 0.

The variations formula for system (3:2) takes the form



!

m ;! Z t f0 + X exp ui( )fi d 0

i=1

 ;! Z t  ;! Z 

= exp

exp

0

m ;! Z t X

= exp

0

i=1

0

 X m

ad f0 d 

!

i=1

!

Zt ;! ui( )fi d  exp f0 d 0

ui ( )e ad f0 fi d  etf0 :

(3.4)

We assume that c = 0, i.e., f0 (x) = Ax. Then x  etf0 = etA x:

(3.5)

Further, since (ad f0 )fi = f0 fi ] = Ax b] = ;Ab then n 2 e ad f0 fi = fi + (ad f0 )fi + 2! (ad f0 )2 fi + + n! (ad f0 )n fi + 2 n = bi ; Abi + 2! (;A)2 bi + + n! (;A)n bi + = e;A bi : In order to compute the left ow in (3:4), recall that the curve

! ! Z t X Z t X m m ;! ;! x0  exp ui( )e ad f0 fi d = x0 exp ui ( )e;A bi d 0

0

i=1

i=1

is the solution to the Cauchy problem x(t) _ = thus (3:6) is equal to

m X i=1

ui (t)e;tA bi

x(0) = x0

(3.6)

3.2 Controllability of Linear Systems

x(t) = x0 +

Z t

e;A

0

m X i=1

!

49

ui( )bi d :

Taking into account (3:5), we obtain Cauchy's formula:



!

m ;! Z t f0 + X x(t) = x0 exp ui( )fi d 0



= x0 + = etA

Z t



e;A

0

x0 +

Z t 0

! !

i=1 m X i=1

ui ( )bi d  etf0

! !

X e;A ui ( )bi d : m

i=1

ut

Notice that in the general case (c 6= 0) Cauchy's formula can be written as follows: x(t x0) = etA x0 + etA = etA x0 + etA where

Zt 0

Zt 0

m

Zt

i=1

0

X e;A ui ( )bi d + etA m ;A X

e

e;A c d

tA ui ( )bi d + e A; Id c i=1

(3.7)

etA ; Id c = tc + t2 Ac + t3 A2c + + tn An;1c + : A 2! 3! n!

3.2 Controllability of Linear Systems Cauchy's formula (3:7) yields that the mapping u 7! x(t u x0) which sends a locally integrable control u = u( ) to the endpoint of the corresponding trajectory, is ane. Thus the attainable set Ax0 (t) of linear system (3:1) for a xed time t > 0 is an ane subspace in n. Denition 3.2. A control system on a state space M is called completely controllable for time t > 0 if

R

Ax0 (t) = M

8x0 2 M:

50

3 Linear Systems

This denition means that for any pair of points x0 x1 2 M exists an admissible control u( ) such that the corresponding solution x(  u x0) of the control system steers x0 to x1 in t units of time: x(0 u x0) = x0 x(t u x0) = x1 : The study of complete controllability of linear systems is facilitated by the following observation. The ane mapping tA ; Id

u 7! etA x0 + e

A

c + etA

is surjective if and only if its linear part u 7! etA

Zt 0

Zt 0

m X

e;A

i=1

e;A

m X i=1

ui ( )bi d

ui ( )bi d

(3.8)

is onto. Moreover, (3:8) is surjective i the following mapping is: u 7!

Zt 0

e;A

m X i=1

ui( )bi d :

(3.9)

Theorem 3.3. The linear system (3:1) is completely controllable for a time

R

t > 0 if and only if (3.10) spanfAj bi j j = 0 : : :  n ; 1 i = 1 : : :  mg = n: Proof. Necessity. Assume, by contradiction, that condition (3:10) is violated. Then there exists a covector p 2 n, p 6= 0, such that pAj bi = 0 j = 0 : : :  n ; 1 i = 1 : : :  m: (3.11) By the Cayley-Hamilton theorem,

R

An =

nX ;1 j =0

j Aj

for some real numbers 0 , : : : , n;1, thus

N

nX ;1 k A = jk Aj j =0

R

for any k 2 and some jk 2 . Now we obtain from (3:11): pAk bi =

nX ;1 j =0

jk pAj bi = 0

k = 0 1 : : :  i = 1 : : :  m:

3.2 Controllability of Linear Systems

That is why and nally p

pe;A bi = 0

Zt 0

i = 1 : : :  m

m

Z tX m

i=1

0 i=1

X e;A ui( )bi d =

51

ui( )pe;A bi d = 0

i.e., mapping (3:9) is not surjective. The contradiction proves necessity. Suciency. By contradiction, suppose that mapping (3:9) is not surjective. Then there exists a covector p 2 n, p 6= 0, such that p

Z tX m 0 i=1

ui( )e;A bi d = 0

R

8u( ) = (u1( ) : : :  um ( )):

(3.12)

Choose a control of the form: u( ) = (0 : : :  0 vs( ) 0 : : :  0) where the only nonzero i-th component is  1 0   s vs ( ) = 0 > s: Then equality (3:12) gives p

Zs 0

e;A bi d = 0

thus

R

s 2  i = 1 : : :  m

R

pe;sA bi = 0 s 2  i = 1 : : :  m: We dierentiate this equality repeatedly at s = 0 and obtain pAk bi = 0

k = 0 1 : : :  i = 1 : : :  m

a contradiction with (3:10). Suciency follows. ut So if a linear system is completely controllable for a time t > 0, then it is completely controllable for any other positive time as well. In this case the linear system is called controllable.

4 State Linearizability of Nonlinear Systems

The aim of this chapter is to characterize nonlinear systems q_ = f0 (q) +

m X i=1

ui fi (q)

R

u = (u1  : : :  um ) 2 m q 2 M

(4.1)

that are equivalent, locally or globally, to controllable linear systems. That is, we seek conditions on vector elds f0 , f1 , : : : , fm that guarantee existence of n) a dieomorphism (global  : M ! n or local  : Oq0 M ! O0 which transforms nonlinear system (4:1) into a controllable linear one (3:1).

R

R

4.1 Local Linearizability We start with the local problem. A natural language for conditions of local linearizability is provided by Lie brackets, which are invariant under dieomorphisms:  V W] =  V W ] V W 2 Vec M: The controllability condition (3:10) can easily be rewritten in terms of Lie brackets: since (;A)j bi = (ad f0)j fi = f|0 :{z: :f0} fi] : : :]] j times

for vector elds (3:3), then the controllability test for linear systems (3:10) reads

R

spanfx0  (ad f0 )j fi j j = 0 : : :  n ; 1 i = 1 : : :  mg = Tx0 n: Further, one can see that the following equality is satised for linear vector elds (3:3):

54

4 State Linearizability of Nonlinear Systems

(adf0 )j1 fi1  (ad f0 )j2 fi2 ] = (;A)j1 bi1  (;A)j2 bi2 ] = 0 0  j1  j2  1  i1  i2  m: It turns out that the two conditions found above give a precise local characterization of controllable linear systems. Theorem 4.1. Let M be a smooth n-dimensional manifold, and let f0 , f1 , : : : , fm 2 Vec M . There exists a di eomorphism

R

 : Oq0 ! O0

of a neighborhood Oq0 M of a point q0 2 M to a neighborhood O0 the origin 0 2 n such that

R

n of

( f0 )(x) = Ax + c x 2 O0 ( fi )(x) = bi x 2 O0 i = 1 : : :  m

R

for some n  n matrix A and c b1 : : :  bm 2 n that satisfy the controllability condition (3:10) if and only if the following conditions hold:

spanfq0  (ad f0 )j fi j j = 0 : : :  n ; 1 i = 1 : : :  mg = Tq0 M (4.2) q  (ad f0 )j1 fi1  (ad f0 )j2 fi2 ] = 0 q 2 Oq0  0  j1  j2  n 1  i1  i2  m: (4.3) Remark 4.2. In other words, the dieomorphism  from the theorem transforms a nonlinear system (4:1) to a linear one (3:1). Before proving the theorem, we consider the following proposition, which we will need later. Lemma 4.3. Let M be a smooth n-dimensional manifold, and let Y1 , : : : , Yk 2 Vec M . There exists a di eomorphism of a neighborhood O0

R

 : O0 ! Oq0

n to a neighborhood Oq 0

 @  @ x = Yi  i

M , q0 2 M , such that

i = 1 : : :  k

if and only if the vector elds Y1  : : :  Yk commute:

Yi  Yj ]  0

i j = 1 : : :  k

and are linearly independent:

dimspan(q0  Y1  : : :  q0  Yk ) = k:

4.1 Local Linearizability

55

Proof. Necessity is obvious since Lie bracket and linear independence are in-

variant with respect to dieomorphisms. Suciency. Choose Yk+1  : : :  Yn 2 Vec M that complete Y1  : : :  Yk to a basis: span(q  Y1 : : :  q  Yn) = Tq M q 2 Oq0 : The mapping (s1  : : :  sn ) = q0  esn Yn   es1 Y1 is dened on a suciently small neighborhood of the origin in n. We have  def @  = @s (s) = @@" q0  e"Yi = q0  Yi :  @@s i s=0 i s=0 "=0

R

R

Hence  js=0 is surjective and  is a dieomorphism of a neighborhood of 0 in n and a neighborhood of q0 in M, according to the implicit function theorem. Now we prove that  recties the vector elds Y1  : : :  Yk . First of all, notice that since these vector elds commute, then their ows also commute, thus P esk Yk   es1 Y1 = e ki=1 si Yi and P (s1  : : :  sn) = q0  esn Yn   esk+1 Yk+1  e ki=1 siYi : Then for i = 1 : : :  k   = @@" (s1  : : :  si + " : : :  sn )  @@s i (s) "=0 Pk @ = @ " q0  esn Yn   esk+1 Yk+1  e j=1 sj Yj  e"Yi "=0 Pk = q0  esn Yn   esk+1 Yk+1  e j=1 sj Yj  @@" e"Yi "=0 = (s)  Yi :

ut Now we can prove Theorem 4.1 on local equivalence of nonlinear systems with linear ones. Proof. Necessity is obvious since Lie brackets are invariant with respect to dieomorphisms, and for controllable linear systems conditions (4:2), (4:3) hold. Suciency. Select a basis of the space Tq0 M among vectors of the form q0  (ad f0 )j fi :

56

4 State Linearizability of Nonlinear Systems

Y = (ad f0 )j fi  = 1 : : :  n 0  j  n ; 1 1  i  m span(q0  Y1  : : :  q0  Yn ) = Tq0 M: By Lemma 4.3, there exists a rectifying dieomorphism:  : Oq0 ! O0 Y = @ @x  = 1 : : :  n: We show that  is the required dieomorphism. (1) First we check that the vector elds  fi , i = 1 : : :  m, are constant. That is, we show that in the decomposition  fi =

n X

=1

 i (x) @ @x 

i = 1 : : :  m

the functions  i (x) are constant. We have

n i X  @@x  fi ] = @@ x @ @x  j =1

(4.4)

on the other hand  @@x  fi ] =  Y   fi ] = Y  fi] =  (ad f0 )j fi  fi ] = 0 (4.5) by hypothesis (4:3). Now we compare (4:4) and (4:5) and obtain @  i @  0 )  i = const i = 1 : : :m = 1 : : :  n @ xj @ x i.e.,  fi , i = 1 : : :  m, are constant vector elds bi , i = 1 : : :  m. (2) Now we show that the vector eld  f0 is linear. We prove that in the decomposition n X  f0 = i (x) @@x i=1

i

all functions i (x), i = 1 : : :  n, are linear. Indeed, n @2 X i @ =  @   @   f ]] @x @x @ x @ x  0  @ xi =1

= Y  Y  f0 ]] = Y  Y  f0 ]] =  (ad f0 )j fi  (ad f0 )j fi  f0]] = ; (ad f0 )j fi  f0 (ad f0 )j fi ]] = ; (ad f0 )j fi  (ad f0 )j +1 fi ] = 0   = 1 : : :  n

4.2 Global Linearizability

57

by hypothesis (4:3). Thus @ 2 i @  0 i   = 1 : : :  n @x @x @ xi i.e.,  f0 is a linear vector eld Ax + c.Pm For the linear system x_ = Ax+ c+ i=1 ui bi, hypothesis (4:2) implies the controllability condition (3:10). ut

4.2 Global Linearizability Now we prove the following statement on global equivalence. Theorem 4.4. Let M be a smooth connected n-dimensional manifold, and let f0 f1  : : :  fm 2 Vec M . There exists a di eomorphism

TR ; T R 2T R ; 2T R ;

 : M ! k  n;k

of M to the product of a k-dimensional torus k with n k for some k  n such that

( f0 )(x) = Ax + c ( fi )(x) = bi x

x

k

k n k n k i = 1 : : :  m

for some n  n matrix A with zero rst k rows:

R

Aei = 0

i = 1 : : :  k

(4.6)

and c b1 : : :  bm 2 n that satisfy the controllability condition (3:10) if and only if the following conditions hold:

(ad f0 )j fi 

j = 0 1 : : :  n ; 1 i = 1 : : :  m are complete vector elds,

spanfq  (adf0 )j fi j j = 0 : : :  n ; 1 q  (adf0 )j1 fi  (ad f0 )j2 fi ] = 0 1

R

i = 1 : : :  mg = Tq M

(4.7) (4.8)

2

q 2 M 0  j1  j2  n 1  i1  i2  m: (4.9)

Remark 4.5. (1) If M is additionally supposed simply connected, then it is

T

dieomorphic to n, i.e., k = 0. (2) If, on the contrary, M is compact, i.e., dieomorphic to n and m < n, then there are no globally linearizable controllable systems on M. Indeed, then A = 0, and the controllability condition (3:10) is violated.

58

4 State Linearizability of Nonlinear Systems

Proof. Suciency. Fix a point q0 2 M and nd a basis in Tq0 M of vectors of

the form

Y = (ad f0 )j fi  = 1 : : :  n span(q0  Y1  : : :  q0  Yn ) = Tq0 M: (1) First we show that the vector elds Y1 : : :  Yn are linearly independent everywhere in M. The set O = fq 2 M j span(q  Y1  : : :  q  Yn ) = Tq M g is obviously open. We show that it is closed. In this set we have a decomposition q  (ad f0 )j fi = q 

n X

=1

aij Y 

q 2 O j = 0 : : :  n ; 1 i = 1 : : :  m

(4.10) for some functions aij 2 C 1 (O). We prove that actually all aij are constant. We have 0 = Y 

n X

=1

aij Y ]

by Leibniz rule X aY ] = (Xa)Y + aX Y ] = =

n X =1

n X

=1

aij Y  Y ] + (Y aij )Y 

n X

(Y aij )Y

=1

 = 1 : : :  n j = 0 : : :  n ; 1 i = 1 : : :  m

thus Y aij = 0 ) aij O = const = 1 : : :  n j = 0 : : :  n ; 1 i = 1 : : :  m: That is why equality (4:10) holds in the closure O. Thus the vector elds Y1 : : :  Yn are linearly independent in O (if this is not the case, then the whole family (ad f0 )j fi , j = 0 : : :  n ; 1, i = 1 : : :  m, is not linearly independent in O). Hence the set O is closed. Since it is simultaneously open and M is connected, O = M i.e., the vector elds Y1  : : :  Yn are linearly independent in M.

4.2 Global Linearizability

59

(2) We dene the \inverse"  of the required dieomorphism as follows: (x1 : : :  xn) = q0  ex1 Y1   exn Yn since the vector elds Y commute

Pn

= q0  e

=1

x Y 

R

x = (x1  : : :  xn) 2 n:

R

(3) We show that the (obviously smooth) mapping  : n ! M is regular, i.e., its dierential is surjective. Indeed, d @ @ x (x) = d " "=0 (x1  : : :  x + " : : :  xn) Pn = dd" q0  e =1 x Y +"Y "=0 Pn = q0  e =1 x Y  Y = (x)  Y  = 1 : : :  n

R

thus

R

x ( n) = T (x) M: The mapping  is regular, thus a local dieomorphism. In particular, ( n) is open. (4) We prove that ( n) is closed. Take any point q 2 ( n). Since the vector elds Y1  : : :  Yn are linearly independent, the image of the mapping

R

Pn

(y1  : : :  yn ) 7! q  e

=1

R

y Y 

R

contains a neighborhood of the point q. Thus there exists y

Pn

qe i.e.,

=1

y Y

2 ( n)

P

R ;P

Pn

q  e n=1 y Y = q0  e for some x = (x1  : : :  xn) 2 n. Then

Pn

q = q0  e =1 xY  e = (x ; y):

RR

R 2R

y = (y1  : : :  yn) 2 n

n y Y =1

=1

n such

that

x Y

Pn

= q0  e

=1

(x ;y )Y

In other words, q 2 ( n). That is why the set ( n) is closed. Since it is open and M is connected,

R

( n) = M:

60

4 State Linearizability of Nonlinear Systems

R

(5) It is easy to see that the preimage  ;1 (q0) = fx 2 n j (x) = q0 g is a subgroup of the Abelian group n. Indeed, let P P (x) = q0  e n=1 x Y = (y) = q0  e n=1 y Y = q0 then P P P (x + y) = q0  e n=1 (x +y )Y = q0  e n=1 x Y  e n=1 y Y = q0: Analogously, if P (x) = q0  e n=1 x Y = q0 then P (;x) = q0  e; n=1 xY = q0 : Finally, (0) = q0:

R

R R

(6) Moreover, G0 =  ;1 (q0) is a discrete subgroup of n, i.e., there are no nonzero elements of  ;1 (q0) in some neighborhood of the origin in n, since  is a local dieomorphism. (7) The mapping  is well-dened on the quotient n=G0 . Indeed, let y 2 G0 . Then P P P (x + y) = q0  e n=1 (x +y )Y = q0  e n=1 y Y  e n=1 x Y Pn = q0  e =1 x Y = (x): So the mapping  : n=G0 ! M (4.11) is well-dened. (8) The mapping (4:11) is one-to-one: if x y 2 n (x) = (y) then P P q0  e n=1 x Y = q0  e n=1 yY  thus Pn q0  e =1 (x ;y )Y = q0 i.e., x ; y 2 G0.

R

R

R

4.2 Global Linearizability

R

61

(9) That is why mapping (4:11) is a dieomorphism. By Lemma 4.6 (see below), the discrete subgroup G0 of n is a lattice: G0 =

(X k i=1

niei j ni 2

Z

)



R TR TR

thus the quotient is a cylinder: n=G0 = k  n;k: Hence we constructed a dieomorphism  =  ;1 : M ! k  n;k: Equalities (4:8) and (4:9) follow exactly as in Theorem 4.1. The vector eld  f0 = Ax + c is well-dened on the quotient k  n;k, that is why equalities (4:6) hold. The suciency follows. Necessity. For a linear system on a cylinder k  n;k, conditions (4:7) and (4:9) obviously hold. If a linear system is controllable on the cylinder, then it is also controllable on n, thus the controllability condition (4:8) is also satised. ut Now we prove the following general statement used in the preceding argument. Lemma 4.6. Let ; be a discrete subgroup in n. Then it is a lattice, i.e., there exist linearly independent vectors e1  : : :  ek 2 n such that

T R

R

;=

(X k i=1

niei j ni

TR

RR ) 2Z :

R

RR

Proof. We prove by induction on dimension n of the ambient group n.

(1) Let n = 1. Since the subgroup ; is discrete, it contains an element e1 6= 0 closest to the origin 0 2 . By the group property, all multiples e1  e1   e1 = ne1 , n = 0 1 2 : ::, are also in ; . We prove that ; contains no other elements. By contradiction, assume that there is an element x 2 ; such that ne1 < x < (n + 1)e1 , n 2 . Then the element y = x ; ne1 2 ; is in the interval (0 e1) . So y 6= 0 is closer to the origin than e1 , a contradiction. Thus ; = e1 = fne1 j n 2 g, q.e.d. (2) We prove the inductive step: let the statement of the lemma be proved for some n ; 1 2 , we prove it for n. Choose an element e1 2 ;, e1 6= 0, closest to the origin 0 2 n. Denote by l the line e1, and by ;1 the lattice e1 ;. We suppose that ; 6= ;1 (otherwise everything is proved).

Z R Z Z N R

Z

R

62

4 State Linearizability of Nonlinear Systems

e2

0

e1

Fig. 4.1. Lattice generated by vectors e1 , e2 Now we show that there is an element e2 2 ; n ;1 closest to l: dist(e2  l) = minfdist(x l) j x 2 ; n lg:

R

R

(4.12)

Take any segment I = ne1  (n + 1)e1] l, and denote by  : n ! l the orthogonal projection from n to l along the orthogonal complement to l in n. Since the segment I is compact and the subgroup ; is discrete, the n-dimensional strip ;1 (I) contains an element e2 2 ; n l closest to I:

R

dist(e2  I) = minfdist(x I) j x 2 (; n l) \ ;1 (I)g:

Then the element e2 is the required one: it satises equality (4:12) since any element that satises (4:12) can be translated to the strip ;1 (I) by elements of the lattice ;1. That is why a suciently small neighborhood of l is free of elements of ; n ;1 . Thus the quotient group ;=;1 is a discrete subgroup in n=l = n;1. By the inductive hypothesis, ;=;1 is a lattice, hence ; is also a lattice. ut

R R

5 The Orbit Theorem and its Applications

5.1 Formulation of the Orbit Theorem

Let F Vec M be any set of smooth vector elds. In order to simplify notation, we assume that all elds from F are complete. Actually, all further denitions and results have clear generalizations to the case of noncomplete elds we leave them to the reader. We return to the study of attainable sets: we study the structure of the attainable sets of F by piecewise constant controls Aq0 = fq0  et1 f1   etk fk j ti  0 fi 2 F  k 2 g q0 2 M: But rst we consider a larger set | the orbit of the family F through a point: Oq0 = fq0  et1 f1   etk fk j ti 2  fi 2 F  k 2 g q0 2 M: In an orbit Oq0 , one is allowed to move along vector elds fi both forward and backwards, while in an attainable set Aq0 only the forward motion is possible, see Figs. 5.1, 5.2. Although, if the family F is symmetric F = ;F (i.e., f 2 F ) ;f 2 F ), then attainable sets coincide with orbits: Oq0 = Aq0 , q0 2 M. In general, orbits have more simple structure that attainable sets. It is described in the following fundamental proposition. Theorem 5.1 (Orbit Theorem, Nagano{Sussmann). Let F Vec M and q0 2 M . Then: (1) Oq0 is a connected immersed submanifold of M , (2) Tq Oq0 = spanfq  (Ad P)f j P 2 P  f 2 Fg, q 2 Oq0 . Here we denote by P the group of dieomorphisms of M generated by ows in F : P = fet1 f1   etk fk j ti 2  fi 2 F  k 2 g Di M: We dene and discuss the notion of immersed manifold in the next section.

R

N N

R

N

64

5 The Orbit Theorem and its Applications

f2 f2

q0

q0

f1

f1

Aq 0

Oq0

Fig. 5.1. Attainable set Aq0

Fig. 5.2. Orbit Oq0

5.2 Immersed Submanifolds Denition 5.2. A subset W of a smooth n-dimensional manifold is called an immersed k-dimensional submanifold of M , k  n, if there exists a one-to-one immersion  : N ! M Ker q = 0 8 q 2 N of a k-dimensional smooth manifold N such that

W = (N): Remark 5.3. An immersed submanifold W of M can also be dened as a manifold contained in M such that the inclusion mapping i : W ! M i : q 7! q is an immersion. Suciently small neighborhoods Ox in an immersed submanifold W of M are submanifolds of M, but the whole W is not necessarily a submanifold of M in the sense of Denition 1.1. In general, the topology of W can be stronger than the topology induced on W by the topology of M. Example 5.4. Let  : ! 2 be a one-to-one immersion of the line into the plane such that limt!+1 (t) = (0). Then W = ( ) is an immersed one-dimensional submanifold of 2, see Fig. 5.3. The topology of W inherited from is stronger than the topology induced by 2. The intervals (;" "), " > 0 small enough, are open in the rst topology, but not open in the second one. The notion of immersed submanifold appears inevitably in the description of orbits of families of vector elds. Already the orbit of one vector eld (i.e., its trajectory) is an immersed submanifold, but may fail to be a submanifold in the sense of Denition 1.1.

R

R R

R

R

R

5.2 Immersed Submanifolds

65

W

(0)

Fig. 5.3. Immersed manifold Example 5.5. Oscillator with 2 degrees of freedom is described by the equa-

tions:

x# + 2x = 0 y# +  2 y = 0

RR

x2  y2 :

In the complex variables z = x ; ix=  _ w = y ; iy= _ these equations read z_ = i z z 2  (5.1) w_ = iw w2  and their solutions have the form z(t) = ei tz(0) w(t) = eit w(0): Any solution (z(t) w(t)) to equations (5:1) belongs to an invariant torus 2 = f(z w) 2 2 j jz j = const jwj = constg: Any such torus is parametrized by arguments of z, w modulo 2, thus it is a group: 2 ' 2=(2 )2. We introduce a new parameter = t, then trajectories (z w) become images of the line f(  (= ) ) j 2 g under the immersion (  (= ) ) 7! ( + 2  (= ) + 2 ) 2 2=(2 )2 thus immersed submanifolds of the torus. If the ratio = is irrational, then trajectories are everywhere dense in the torus: they form the irrational winding of the torus. In this case, trajectories, i.e., orbits of a vector eld, are not submanifolds, but just immersed submanifolds.

CC

T

T R Z

C

Z

R

ZR Z

66

5 The Orbit Theorem and its Applications

Remark 5.6. Immersed submanifolds inherit many local properties of sub-

manifolds. In particular, the tangent space to an immersed submanifold W = Im M,  an immersion, is given by T(q) W = Imq : Further, it is easy to prove the following property of a vector eld V 2 Vec M: V (q) 2 Tq W 8 q 2 W ) q  etV 2 W 8 q 2 W for all t close enough to 0.

5.3 Corollaries of the Orbit Theorem Before proving the Orbit Theorem, we obtain several its corollaries. Let Oq0 be an orbit of a family F Vec M. First of all, if f 2 F , then f(q) 2 Tq Oq0 for all q 2 Oq0 . Indeed, the trajectory q  etf belongs to the orbit Oq0 , thus its velocity vector f(q) is in the tangent space Tq Oq0 . Further, if f1  f2 2 F , then f1  f2](q) 2 Tq Oq0 for all q 2 Oq0 . This follows since the vector f1  f2](q) is tangent to the trajectory t 7! q  etf1  etf2  e;tf1  e;tf2 2 Oq0 : Given three vector elds f1  f2  f3 2 F , we have f1  f2 f3]](q) 2 Tq Oq0 , q 2 Oq0 . Indeed, it follows that f2 f3 ](q) 2 Tq Oq0 , q 2 Oq0 , then all trajectories of the eld f2  f3] starting in the immersed submanifold Oq0 do not leave it. Then we repeat the argument of the previous items. We can go on and consider Lie brackets of arbitrarily high order f1  : : :fk;1 fk ] : : :]](q) as tangent vectors to Oq0 if fi 2 F . These considerations can be summarized in terms of the Lie algebra of vector elds generated by F : Lie F = spanff1  : : :fk;1 fk ] : : :]] j fi 2 F  k 2 g Vec M and its evaluation at a point q 2 M: Lieq F = fq  V j V 2 Lie Fg Tq M: We obtain the following statement.

N

Corollary 5.7. for all q 2 Oq0 .

Lieq F

Tq Oq0

(5.2)

5.4 Proof of the Orbit Theorem

67

Remark 5.8. We show soon that in many important cases inclusion (5:2) turns

into equality. In the general case, we have the following estimate: dimLieq F  dim Oq0  q 2 Oq0 : Another important corollary of the Orbit Theorem is the following proposition often used in control theory. Theorem 5.9 (Rashevsky{Chow). Let M be a connected smooth manifold, and let F Vec M . If the family F is completely nonholonomic: Lieq F = Tq M 8 q 2 M (5.3) then

Oq0 = M 8 q0 2 M: (5.4) Denition 5.10. A family F Vec M that satis es property (5:3) is called

completely nonholonomic or bracket-generating. Now we prove Theorem 5.9. Proof. By Corollary 5.7, equality (5:3) means that any orbit Oq0 is an open set in M. Further, consider the following equivalence relation in M: q1  q2 , q2 2 Oq1  q1 q2 2 M: (5.5) The manifold M is the union of (naturally disjoint) equivalence classes. Each class is an open subset of M and M is connected. Hence there is only one nonempty class. That is, M is a single orbit Oq0 . ut For symmetric families attainable sets coincide with orbits, thus we have the following statement.

Corollary 5.11. A symmetric bracket-generating family on a connected manifold is completely controllable.

5.4 Proof of the Orbit Theorem Introduce the notation: (Ad P )F def = f(Ad P)f j P 2 P  f 2 Fg Vec M: Consider the following subspace of Tq M: q def = spanfq  (Ad P )Fg: This space is a candidate for the tangent space Tq Oq0 .

68

5 The Orbit Theorem and its Applications

Lemma 5.12. dimq = dimq0 for all q 2 Oq0 , q0 2 M . Proof. If q 2 Oq0 , then q = q0  Q for some dieomorphism Q 2 P . Take an arbitrary element q0  (Ad P )f in q0 , P 2 P , f 2 F . Then Q (q0  (Ad P)f) = q0  (Ad P )f  Q = q0  P  f  P ;1  Q = (q0  Q)  (Q;1  P  f  P ;1  Q) = q  Ad(Q;1  P)f 2 q since Q;1  P 2 P . We have Q q0 q , thus dimq0  dimq . But q0 and q can be switched, that is why dimq  dimq0 . Finally, dimq = dim q0 . ut Now we prove the Orbit Theorem. Proof. The manifold M is divided into disjoint equivalence classes of relation (5:5) | orbits Oq . We introduce a new \strong" topology on M in which all orbits are connected components. For any point q 2 M, denote m = dimq and pick elements V1  : : :  Vm 2 (Ad P )F such that span(V1 (q) : : :  Vm (q)) = q : (5.6) Introduce a mapping: Gq : (t1  : : :  tm ) 7! q  et1 V1   etm Vm  We have

R

ti 2 :

@ Gq = V (q) @ ti 0 i thus in a suciently small neighborhood O0 of the origin 0 2 m the vectors @ Gq , : : : , @ Gq are linearly independent, i.e., G j is an immersion. q O0 @ t1 @ tm The sets of the form Gq (O0), q 2 M, are candidates for elements of a topology base on M. We prove several properties of these sets. (1) Since the mappings Gq are regular, the sets Gq (O0 ) are m-dimensional submanifolds of M, may be, for smaller neighborhoods O0. (2) We show that Gq (O0 ) Oq . Any element of the basis (5:6) has the form Vi = (Ad Pi )fi , Pi 2 P , fi 2 F . Then

R

etVi = et(Ad Pi )fi = etPi fi Pi = Pi  etfi  Pi;1 2 P  ;1

thus

Gq (t) = q  etVi 2 Oq 

t 2 O0 :

R

5.4 Proof of the Orbit Theorem

69

(3) We show that Gt(Tt m) = G(t), t 2 O0. Since rank GtjO0 = m and dim G(t) O0 = m, it remains to prove that @@Gtiq t 2 Gq (t) for t 2 O0. We have @ G (t) = @ q  et1 V1   etm Vm @ ti q @ ti = q  et1 V1   eti Vi  Vi  eti+1 Vi+1   etm Vm = q  et1 V1   eti Vi  eti+1 Vi+1   etm Vm  e;tm Vm   e;ti+1 Vi+1  Vi  eti+1 Vi+1   etm Vm (introduce the notation Q = eti+1 Vi+1   etm Vm 2 P ) = Gq (t)  Q;1  Vi  Q = Gq (t)  Ad Q;1 Vi 2 Gq(t) : (4) We prove that sets of the form Gq (O0 ), q 2 M, form a topology base in M. It is enough to prove that any nonempty intersection Gq (O0) \ Gq~(Oe0) contains a subset of the form Gq^(Ob0), i.e., this intersection has the form as at the left gure, not at the right one:

Gq~(Oe0 )

Gq (O0) Let a point q^ belong to Gq (O0). Then dimq^ = dim q = m. Consider the mapping Gq^ : (t1  : : :  tm ) 7! q^  et1 Vb1   etm Vbm  span(^q  Vb1  : : :  q^  Vbm ) = q^:

It is enough to prove that for small enough (t1  : : :  tm ) Gq^(t1 : : :  tm ) 2 Gq (O0)

then we can replace Gq (O0) by Gq~(Oe0 ). We do this step by step. Consider the curve t1 7! qb  et1 Vb1 . By property (3) above, Vb1 (q0) 2 q0 for q0 2 Gq (O0)

70

5 The Orbit Theorem and its Applications

and suciently close to qb. Since Gq (O0) is a submanifold of M and q = Tq Gq (O0 ), the curve qb  et1 Vb1 belongs to Gq (O0 ) for suciently small jt1j. We repeat this argument and show that (qb  et1 Vb1 )  et2 Vb2 2 Gq (O0)

for small jt1j, jt2j. We continue this procedure and obtain the inclusion (qb  et1 Vb1   etm;1 Vbm;1 )  etm Vbm 2 Gq (O0)

R

for (t1  : : :  tm) suciently close to 0 2 m. Property (4) follows, and the sets Gq (O0 ), q 2 M, form a topology base on M. We denote by M F the topological space obtained, i.e., the set M endowed with the \strong" topology just introduced. (5) We show that for any q0 2 M, the orbit Oq0 is connected, open, and closed in the \strong" topology. Connectedness: all mappings t 7! q  etf , f 2 F , are continuous in the \strong" topology, thus any point q 2 Oq0 can be connected with q0 by a path continuous in M F . Openness: for any q 2 Oq0 , a set of the form Gq (O0 ) Oq0 is a neighborhood of the point q in M F . Closedness: any orbit is a complement to a union of open sets (orbits), thus it is closed. So each orbit Oq0 is a connected component of the topological space M F . (6) A smooth structure on each orbit Oq0 is dened by choosing Gq (O0) to be coordinate neighborhoods and G;q 1 coordinate mappings. Since Gq jO0 are immersions, then each orbit Oq0 is an immersed submanifold of M. Notice that dimension of these submanifolds may vary for dierent q0. (7) By property (3) above, Tq Oq0 = q , q 2 Oq0 . The Orbit Theorem is proved. ut The Orbit Theorem provides a description of the tangent space of an orbit: Tq Oq0 = span(q  (Ad P )F ): Such a description is rather implicit since the structure of the group P is quite complex. However, we already obtained the lower estimate Lieq F span(q  (Ad P )F ) (5.7) from the Orbit Theorem. Notice that this inclusion can easily be proved directly. We make use of the asymptotic expansion of the eld Ad etf fb = et ad f f.b Take an arbitrary element ad f1   adfk fb 2 Lie F , fi  fb 2 F . We have Ad(et1 f1   etk fk )fb 2 (Ad P )F , thus

5.4 Proof of the Orbit Theorem

71

k q  @t @ @t Ad(et1 f1   etk fk )fb 1 k 0 k = q  @t @ @t (et1 ad f1   etk ad fk )fb 1 k 0 = q  adf1   ad fk fb 2 span(q  (Ad P )F ): Now we consider a situation where inclusion (5:7) is strict. @  @ 1 2 Example 5.13. Let M = , F = @ x1  a(x ) @ x2 , where the function 1 a 2 C ( ), a 6 0, has a compact support. It is easy to see that the orbit Ox through any point x 2 2 is the whole plane 2. Indeed, the family F  (;F ) is completely controllable in the plane. Given an initial point x0 = (x10 x20) and a terminal point x1 = (x11  x21), we can steer x0 to x1 : rst we go from x0 by a eld  @@x1 to a point (~x1 x20) with a(~x1) 6= 0, then we go by a eld a(~x1 ) @@x2 to a point (~x1  x21), and nally we reach (x11  x21) along  @@x1 , see Fig. 5.4.

R

R

R

R

x2 @ @x1 @ a(~ x1) 2 @x @ @x1 x~1

x1 x0 x1

Fig. 5.4. Complete controllability of the family F On the other hand, we have

 1 x1 2= supp a

dimLie(x1 x2 ) (F ) = 2 a(x1 ) 6= 0: That is, x  (Ad P )F = Tx 2 6= Liex F if x1 2= supp a. Although, such example is essentially non-analytic. In the analytic case, inclusion (5:7) turns into equality. We prove this statement in the next section.

R

72

5 The Orbit Theorem and its Applications

5.5 Analytic Case

The set Vec M is not just a Lie algebra (i.e., a vector space close under the operation of Lie bracket), but also a module over C 1 (M): any vector eld V 2 Vec M can be multiplied by a function a 2 C 1 (M), and the resulting vector eld aV 2 Vec M. If vector elds are considered as derivations of C 1(M), then the product of a function a and a vector eld V is the vector eld (aV )b = a (V b) b 2 C 1 (M): In local coordinates, each component of V at a point q 2 M is multiplied by a(q). Exercise 5.14. Let X Y 2 Vec M, a 2 C 1(M), P 2 Di M. Prove the equalities: (ad X)(aY ) = (Xa)Y + a(ad X)Y (Ad P)(aX) = (Pa) Ad P X:

A submodule V Vec M is called nitely generated over C 1 (M) if it has a nite global basis of vector elds:

9 V1  : : :  Vk 2 Vec M such that V =

(X k i=1

)

ai Vi j ai 2 C 1 (M) :

Lemma 5.15. Let V Vec M be a nitely generated submodule over C 1(M). Assume that (ad X)V = f(adX)V j V 2 Vg V for a vector eld X 2 Vec M . Then ;Ad etX  V = V : Proof. Let V1  : : :  Vk be a basis of V . By the hypothesis of the lemma, X Vi] =

k X j =1

aij Vj

(5.8)

for some functions aij 2 C 1 (M). We have to prove that the vector elds Vi (t) = (Ad etX )Vi = et ad X Vi 

R

t2 

can be expressed as linear combinations of the elds Vi with coecients from C 1(M). We dene an ODE for Vi (t):

5.5 Analytic Case

73

X V_i (t) = et ad X X Vi] = et ad X aij Vj k

k ; X

=

j =1



j =1

etX aij Vj (t):

For a xed q 2 M, dene the k  k matrix: A(t) = (aij (t))

aij (t) = etX aij  i j = 1 : : :  k:

Then we have a linear system of ODEs:

X V_i (t) = aij (t)Vj (t): k

(5.9)

j =1

Find a fundamental matrix ; of this system: ;_ = A(t); ;(0) = Id : Since A(t) smoothly depends on q, then ; depends smoothly on q as well: ; (t) = (ij (t))

R

ij (t) 2 C 1 (M) i j = 1 : : :  k t 2 :

Now solutions of the linear system (5:9) can be written as follows: Vi (t) =

k X j =1

ij (t)Vj (0):

But Vi (0) = Vi are the generators of the module, and the required decomposition of Vi (t) along the generators is obtained. ut A submodule V Vec M is called locally nitely generated over C 1 (M) if any point q 2 M has a neighborhood O M in which the restriction FjO is nitely generated over C 1 (O), i.e., has a basis of vector elds. Theorem 5.16. Let F Vec M . Suppose that the module Lie F is locally

nitely generated over C 1 (M). Then Tq Oq0 = Lieq F 

q 2 Oq0

(5.10)

for any orbit Oq0 , q0 2 M , of the family F .

We prove this theorem later, but now obtain from it the following consequence. Corollary 5.17. If M and F are real analytic, then equality (5:10) holds.

74

5 The Orbit Theorem and its Applications

Proof. In the analytic case, Lie F is locally nitely generated. Indeed, any

module generated by analytic vector elds is locally nitely generated. This is N#otherian property of the ring of germs of analytic functions, see 140]. ut Now we prove Theorem 5.16. Proof. By the Orbit Theorem, n o ;  Tq Oq0 = span q  Ad et1 f1   etk fk fb j fi  fb 2 F  tk 2  k 2 : (5.11) By denition of the Lie algebra Lie F , (adf) Lie F Lie F 8 f 2 F : Apply Lemma 5.15 for the locally nitely generated C 1 (M)-module V = Lie F . We obtain ;Ad etf  Lie F Lie F 8 f 2 F : That is why ;  Ad et1 f1   etk fk fb = Ad et1 f1   Ad etk fk fb 2 Lie F for any fi  fb 2 F , tk 2 . In view of equality (5:11), Tq Oq0 Lieq F : But the reverse inclusion (5:7) was already obtained. Thus Tq Oq0 = Lieq F . Another proof of the theorem can be obtained via local convergence of the exponential series in the analytic case. ut

R N

R

5.6 Frobenius Theorem We apply the Orbit Theorem to obtain the classical Frobenius Theorem as a corollary. Denition 5.18. A distribution  T M on a smooth manifold M is a family of linear subspaces q Tq M smoothly depending on a point q 2 M . Dimension of the subspaces q , q 2 M , is assumed constant. Geometrically, at each point q 2 M there is attached a space q Tq M, i.e., we have a eld of tangent subspaces on M. Denition 5.19. A distribution  on a manifold M is called integrable if for any point q 2 M there exists an immersed submanifold Nq M , q 2 Nq , such that

Tq0 Nq = q0 8 q0 2 Nq  see Fig. 5.5. The submanifold Nq is called an integral manifold of the distribution  through the point q.

5.6 Frobenius Theorem

75

q0

q

q

q

0

Nq Fig. 5.5. Integral manifold Nq of distribution  In other words, integrability of a distribution  T M means that through any point q 2 M we can draw a submanifold Nq whose tangent spaces are elements of the distribution . Remark 5.20. If dimq = 1, then  is integrable by Theorem 1.17 on existence and uniqueness of solutions of ODEs. Indeed, in a neighborhood of any point in M, we can nd a base of the distribution , i.e., a vector eld V 2 Vec M such that q = span(V (q)), q 2 M. Then trajectories of the ODE q_ = V (q) are one-dimensional submanifolds with tangent spaces q . But in the general case (dimq > 1), a distribution  may be nonintegrable. Indeed, consider the family of vector elds tangent to :  = fV 2 Vec M j V (q) 2 q 8 q 2 M g: Assume that the distribution  is integrable. Any vector eld from the family  is tangent to integral manifolds Nq , thus the orbit Oq of the family  restricted to a small enough neighborhood of q is contained in the integral manifold Nq . Moreover, since dim Oq  dimq = dimNq , then locally Oq = Nq : we can go in Nq in any direction along vector elds of the family . By the Orbit Theorem, Tq Oq  Lieq , that is why Lieq  = q : This means that V1 V2 ] 2  8 V1  V2 2 :

(5.12)

Let dimq = k. In a neighborhood Oq0 of a point q0 2 M we can nd a base of the distribution : q = span(f1 (q) : : :  fk (q)) 8 q 2 Oq0 : Then inclusion (5:12) reads as Frobenius condition :

76

5 The Orbit Theorem and its Applications

fi fj ] =

k X l=1

clij fl  clij 2 C 1 (Oq0 ):

(5.13)

We have shown that integrability of a distribution implies Frobenius condition for its base. Conversely, if condition (5:13) holds in a neighborhood of any point q0 2 M, then Lie() = . Thus Lie() is a locally nitely generated module over C 1(M). By Theorem 5.16, Tq Oq0 = Lieq  q 2 Oq0 : So Tq Oq0 = q  q 2 Oq0  i.e., the orbit Oq0 is an integral manifold of  through q0 . We proved the following proposition. Theorem 5.21 (Frobenius). A distribution  T M is integrable if and only if Frobenius condition (5:13) holds for any base of  in a neighborhood of any point q0 2 M . Remark 5.22. (1) In view of the Leibniz rule f ag] = (fa)g + af g] f g 2 Vec M a 2 C 1(M) Frobenius condition is independent on the choice of a base f1  : : :  fk : if it holds in one base, then it also holds in any other base. (2) One can also consider smooth distributions  with non-constant dimq . Such a distribution is dened as a locally nitely generated over C 1 (M) submodule of Vec M. For such distributions Frobenius condition implies integrability but dimension of integrable manifolds becomes, in general, dierent, although it stays constant along orbits of . This is a generalization of phase portraits of vector elds. Although, notice once more that in general distributions with dimq > 1 are nonintegrable.

5.7 State Equivalence of Control Systems In this section we consider one more application of the Orbit Theorem | to the problem of equivalence of control systems (or families of vector elds). Let U be an arbitrary index set. Consider two families of vector elds on smooth manifolds M and N parametrized by the same set U: fU = ffu j u 2 U g Vec M gU = fgu j u 2 U g Vec N: Take any pair of points x0 2 M, y0 2 N, and assume that the families fU , gU are bracket-generating: Liex0 fU = Tx0 M Liey0 gU = Ty0 N:

5.7 State Equivalence of Control Systems

77

Denition 5.23. Families fU and gU are called locally state equivalent if there exists a di eomorphism of neighborhoods  : Ox0 M ! Oy0 N  : x0 7! y0  that transforms one family to another:

fu = gu 8 u 2 U: Notation: (fU  x0) ' (gU  y0). Remark 5.24. Here we consider only smooth transformations of state x 7! y, while the controls u do not change. That is why this kind of equivalence is called state equivalence. We already studied state equivalence of nonlinear and linear systems, both local and global, see Chap. 4. Now, we rst try to nd necessary conditions for local equivalence of systems fU and gU . Assume that (fU  x0) ' (gU  y0): By invariance of Lie bracket, we get  fu1  fu2 ] = fu1  fu2 ] = gu1  gu2 ] u1 u2 2 U i.e., relations between Lie brackets of vector elds of the equivalent families fU and gU must be preserved. We collect all relations between these Lie brackets at one point: dene the systems of tangent vectors

u1 :::uk = fu1  : : :  fuk ] : : :](x0) 2 Tx0 M u1:::uk = gu1  : : :  guk ] : : :](y0 ) 2 Ty0 N: Then we have  jx0 u1 :::uk = u1:::uk  u1 : : :  uk 2 U k 2 : Now we can state a necessary condition for local equivalence of families fU and gU in terms of the linear isomorphism jx0 = A : Tx0 M $Ty0 N: If (fU  x0) ' (gU  y0 ), then there exists a linear isomorphism A : Tx0 M$ Ty0 N that maps the conguration of vectors f u1:::uk g to the conguration fu1:::uk g. It turns out that in the analytic case this condition is sucient. I.e., in the analytic case the combinations of partial derivatives of vector elds fu , u 2 U, that enter f u1:::uk g, form a complete system of state invariants of a family fU .

N

78

5 The Orbit Theorem and its Applications

Theorem 5.25. Let fU and gU be real analytic and bracket-generating families of vector elds on real analytic manifolds M and N respectively. Let x0 2 M , y0 2 N . Then (fU  x0) ' (gU  y0 ) if and only if there exists a linear isomorphism such that

A : Tx0 M $Ty0 N N Af u1 :::uk g = fu1:::uk g 8 u1  : : :  uk 2 U k 2 : (5.14) Remark 5.26. If in addition M, N are simply connected and all the elds fu , gu are complete, then we have the global equivalence. Before proving Theorem 5.25, we reformulate condition (5:14) and provide a method to check it. Let a family fU be bracket-generating: spanf u1 :::uk j u1 : : :  uk 2 U k 2 g = Tx0 M: We can choose a basis: span( 1  : : :  n ) = Tx0 M % i = (u1i  : : :  uki) i = 1 : : :  n (5.15) and express all vectors in the conguration through the base vectors: N u1 :::uk = n X i=1 ciu1 :::uk i : (5.16) If there exists a linear isomorphism A : Tx0 M$ Ty0 N with (5:14), then the vectors  i  i = 1 : : :  n should form a basis of Ty0 N: span( 1  : : :   n ) = Ty0 N (5.17) and all vectors of the conguration  should be expressed through the base vectors with the same coecients as the conguration , see (5:16): u1:::uk =

n X i=1

ciu1 :::uk  i :

(5.18)

It is easy to see the converse implication: if we can choose bases in Tx0 M and Ty0 N from the congurations and  as in (5:15) and (5:17) such that decompositions (5:16) and (5:18) with the same coecients ciu1 :::uk hold, then there exists a linear isomorphism A with (5:14). Indeed, we dene then the isomorphism on the bases: A : i 7!  1  i = 1 : : :  n:

5.7 State Equivalence of Control Systems

79

We can obtain one more reformulation via the following agreement. Congurations f u1 :::uk g and fu1:::uk g are called equivalent if the sets of relations K(fU ) and K(gU ) between elements of these congurations coincide: K(fU ) = K(gU ). We denote here by K(fU ) the set of all systems of coecients such that the corresponding linear combinations vanish:

(

K(fU ) = (bu1 :::uk ) j

X

u1 :::uk

)

bu1 :::uk u1 :::uk = 0 :

Then Theorem 5.25 can be expressed in the following form.

Nagano Principle. All local information about bracket-generating families of analytic vector elds is contained in Lie brackets.

Notice, although, that the conguration u1 :::uk and the system of relations K(fU ) are, in general, immense and cannot be easily characterized. Thus Nagano Principle cannot usually be applied directly to describe properties of control systems, but it is an important guiding principle. Now we prove Theorem 5.25. Proof. Necessity was already shown. We prove suciency by reduction to the Orbit Theorem. For this we construct an auxiliary system on the Cartesian product M  N = f(x y) j x 2 M y 2 N g: For vector elds f 2 Vec M, g 2 Vec N, dene their direct product f  g 2 Vec(M  N) as the derivation (f  g)aj(xy) = (fa1y ) x + (ga2x ) y 

a 2 C 1 (M  N)

(5.19)

where the families of functions a1y 2 C 1 (M), a2x 2 C 1 (N) are dened as follows: a1y : x 7! a(x y) a2x : y 7! a(x y) x 2 M y 2 N: So projection of f  g to M is f, and projection to N is g. Finally, we dene the direct product of systems fU and gU as fU  gU = ffu  gu j u 2 U g Vec(M  N): We suppose that there exists a linear isomorphism A : Tx0 M $Ty0 N that maps the conguration to  as in (5:14), and construct the local equivalence (fU  x0) ' (gU  y0). In view of denition (5:19), Lie bracket in the family fU  gU is computed as fu1  gu1  fu2  gu2 ] = fu1  fu2 ]  gu1  gu2 ] u1 u2 2 U thus 80 5 The Orbit Theorem and its Applications fu1  gu1  : : :  fuk  guk ] : : :](x0 y0 ) = fu1  : : :  fuk ] : : :](x0)  gu1  : : :  guk ] : : :](y0) = u1 :::uk  u1 :::uk = u1 :::uk  A u1 :::uk  u1  : : :  uk 2 U k 2 : N That is why dimLie(x0 y0 ) (fU  gU ) = n where n = dimM. By the analytic version of the Orbit Theorem (Corollary 5.17) for the family fU  gU Vec(M  N), the orbit O of fU  gU through the point (x0 y0 ) is an n-dimensional immersed submanifold (thus, locally a submanifold) of M  N. The tangent space of the orbit is T(x0 y0 ) O = span( u1 :::uk  A u1 :::uk ) = spanfv  Av j v 2 Tx0 g T(x0 y0) M  N = Tx0 M  Ty0 N i.e., the graph of the linear isomorphism A. Consider the canonical projections onto the factors: 1 : M  N ! M 2 : M  N ! N 1(x y) = x 2(x y) = y: The restrictions 1jO , 2 jO are local dieomorphisms since the dierentials 1j(x0 y0 ) : (v Av) 7! v v 2 Tx0 M 2j(x0 y0 ) : (v Av) 7! Av v 2 Tx0 M are one-to-one. Now  = 2  (1jO );1 is a local dieomorphism from M to N with the graph O, and  = 2  (1jO ); 1 : fu 7! gu  Consequently, (fU  x0) ' (gU  y0 ). u 2 U: ut 6 Rotations of the Rigid Body In this chapter we consider rotations of a rigid body around a xed point. That is, we study motions of a body in the three-dimensional space such that:  distances between all points in the body remain xed (rigidity), and  there is a point in the body that stays immovable during motion (xed point). We consider both free motions (in the absence of external forces) and controlled motions (when external forces are applied in order to bring the body to a desired state). Such system is a very simplied model of a satellite in the space rotating around its center of mass. For details about ODEs describing rotations of the rigid body, see 135]. 6.1 State Space The state of the rigid body is determined by its position and velocity. We x an orthonormal frame attached to the body at the xed point (the moving frame), and an orthonormal frame attached to the ambient space at the xed point of the body (the xed frame), see Fig. 6.1. The set of positions of the rigid body is the set of all orthonormal frames in the three-dimensional space with positive orientation. This set can be identied with SO(3), the group of linear orthogonal orientation-preserving transformations of 3, or, equivalently, with the group of 3  3 orthogonal unimodular matrices: SO(3) = fQ : 3 ! 3 j (Qx Qy) = (x y) det Q = 1g = fQ : 3 ! 3 j QQ = Id det Q = 1g: RR RR R !R R 3 transforms the coordinate representation of a The mapping Q : 3 point in the moving frame to the coordinate representation of this point in the xed frame. 82 6 Rotations of the Rigid Body Fig. 6.1. Fixed and moving frames R R (  ). If a pair of vectors x y 2 have coordinates x = (x1  x2 x3), y = (y1  y2  y3) in some orthonormal frame, then (x y) = x1y1 + x2y2 + x3y3 . Notice that the set of positions of the rigid body SO(3) is not a linear space, but a nontrivial smooth manifold. Now we describe velocities of the rigid body. Let Qt 2 SO(3) be position of the body at a moment of time t. Since the operators Qt : 3 ! 3 are orthogonal, then (Qtx Qty) = (x y) x y 2 3 t 2 : We dierentiate this equality w.r.t. t and obtain (Q_ tx Qty) + (Qt x Q_ ty) = 0: (6.1) The matrix t = Q;t 1Q_ t is called the body angular velocity . Since Q_ t = Qt t  then equality (6:1) reads (Qtt x Qty) + (Qt x Qtty) = 0 whence by orthogonality (t x y) + (x ty) = 0 i.e., t = ;t  the matrix t is antisymmetric. So velocities of the rigid body have the form Remark 6.1. We denote above the standard inner product in 3 R R 3 by R R 6.1 State Space 83 Q_ t = Qtt  t = ;t : In other words, we found the tangent space TQ SO(3) = fQ j   = ; g Q 2 SO(3): The space of antisymmetric 3  3 matrices is denoted by so(3), it is the tangent space to SO(3) at the identity: so(3) = f : 3 ! 3 j   = ; g = TId SO(3): The space so(3) is the Lie algebra of the Lie group SO(3). To each antisymmetric matrix  2 so(3), we associate a vector ! 2 3: 0! 1 0 0 ;! ! 1 1 3 2 (6.2)   !  = @ !3 0 ;!1 A  ! = @ !2 A : ;!2 !1 0 !3 Then the action of the operator  on a vector x 2 3 can be represented via the cross product in 3: x = !  x x 2 3: Let x be a point in the rigid body. Then its position in the ambient space 3 is Qt x. Further, velocity of this point is Q_ tx = Qt t x = Qt(!t  x): !t is the vector of angular velocity of the point x in the moving frame: if we x the moving frame Qt at one moment of time t, then the instantaneous velocity of the point x at the moment of time t in the movingframe is Q;t 1 Q_ tx = t x = !t  x, i.e., the point x rotates around the line through !t with the angular velocity k!tk. Introduce the following scalar product of matrices  = (ij ) 2 so(3): R R R R h 1   2i = ; 12 tr( 1  2 ) = 12 R 3 X ij =1 R ij1 ij2 = R X i 0 called inertia tensor of the rigid body. Finally, the functional of action has the form Z t1 Z t1 J(: ) = j(t ) dt = 12 hAt  ti dt 0 0 where 0 and t1 are the initial and terminal moments of motion. Let Q0 and Qt1 be the initial and terminal positions of the moving body. By the least action principle, the motion Qt, t 2 0 t1], of the body should be an extremal of the following problem: J(: ) ! min (6.6) Q_ t = Qtt  Q0  Qt1 xed: We nd these extremals. Let t be angular velocity along the reference trajectory Qt, then Z t1 ;! ; 1 Q0  Qt1 = exp t dt: 0 Consider an arbitrary small perturbation of the angular velocity: t + "Ut + O("2 ) " ! 0: In order that such perturbation was admissible, the starting point and endpoint of the corresponding trajectory should not depend on ": Z t1 ;  ;! ; 1 Q0  Qt1 = exp t + "Ut + O("2 ) dt 0 thus ;! Z t1 ;t + "Ut + O("2 ) dt: (6.7) 0 = @@" Q;0 1  Qt1 = @@" exp 0 "=0 "=0 By formula (2:31) of derivative of a ow w.r.t. parameter, the right- hand side above is equal to Zt 1 0  ;! Z t Ad exp  Z t1 ;! t dt  d Ut dt exp 0 Z t0 ;  = Ad Q;0 1  Qt Ut dt  Q;0 1  Qt 0 Zt 1 1 = Q;0 1 1 0 Ad Qt Ut dt  Qt1 : 86 6 Rotations of the Rigid Body Taking into account (6:7), we obtain Zt 1 0 Denote Vt = Ad Qt Ut dt = 0: Zt 0 Ad Q U d  then admissibility condition of a variation Ut takes the form V0 = Vt1 = 0: (6.8) (6.9) Now we nd extremals of problem (6:6). Z t1 0 = @@" J(:" ) = hAt  Uti dt 0 "=0 by (6:3) = by (6:8) = Zt 1 0 Zt 1 0 h(Ad Qt)At  (Ad Qt)Ut i dt h(Ad Qt)At  V_ti dt integrating by parts with the admissibility condition (6:9) =;  Z t d (Ad Q )A  V t t t dt: dt 1 0 So the previous integral vanishes for any admissible operator Vt, thus d d t (Ad Qt )At = 0 t 2 0 t1]: Hence Ad Qt(t  At] + A_ t ) = 0 t 2 0 t1] that is why A_ t = At  t] t 2 0 t1]: (6.10) Introduce the operator Mt = At  6.2 Euler Equations 87 called kinetic momentum of the body, and denote B = A;1: We combine equations (6:10), (6:5) and come to Euler equations of rotations of a free rigid body:  M_ = M  BM ] M 2 so(3) t t t t Q_ t = QtBMt  Qt 2 SO(3): Remark 6.2. The presented way to derive Euler equations can be applied to the curves on the group SO(n) of orthogonal orientation-preserving n  n matrices with an arbitrary n > 0. Then we come to equations of rotations of a generalized n-dimensional rigid body. Now we rewrite Euler equations via isomorphism (6:2) of so(3) and 3, which is essentially 3-dimensional and does not generalize to higher dimensions. Recall that for an antisymmetric matrix 0 0 ;  1 3 2 M = @ 3 0 ;1 A 2 so(3) ;2 1 0 the corresponding vector  2 3 is 0 1 1  = @ 2 A  M  : 3 Now Euler equations read as follows:  _ =      2 3 t t t t Q_ t = Qtbt  Qt 2 SO(3) R R R R R R$R

where  : 3 ! 3 and b : 3 ! so(3) are the operators corresponding to 3 (6:2). B : so(3) ! so(3) via the isomorphism so(3) Eigenvectors of the symmetric positive denite operator  : 3 ! 3 are called principal axes of inertia of the rigid body. In the sequel we assume that the rigid body is asymmetric, i.e., the operator  has 3 distinct eigenvalues 1 2  3. We order the eigenvalues of : 1 > 2 > 3  and choose an orthonormal frame e1  e2 e3 of the corresponding eigenvectors, i.e., principal axes of inertia. In the basis e1  e2  e3, the operator  is diagonal: 0 1 0  1 1 1 1  @ 2 A = @ 2 2 A  3 3 3

R R

88

6 Rotations of the Rigid Body

and the equation _ t = t  t reads as follows:

8 >

:_ 23 = (12 ; 31)1132:

(6.11)

6.3 Phase Portrait

R

Now we describe the phase portrait of the rst of Euler equations: _ t = t  t 

t 2 3:

This equation has two integrals: energy

(6.12)

(t  t) = const and moment of momentum (t  t) = const : Indeed: d (   ) = 2(     ) = ;2(     ) = 0 t t t t t t dt t t d d t (t  t ) = (t  t  t ) + (t  (t  t )) = 2(t  t  t ) = ;2(t  t  t ) = 0 by the invariance property (6:4) and symmetry of . So all trajectories t of equation (6:12) satisfy the restrictions

(

21 + 22 + 23 = const 121 + 2 22 + 3 23 = const

(6.13)

i.e., belong to intersection of spheres with ellipsoids. Moreover, since the dierential equation (6:12) is homogeneous, we draw its trajectories on one sphere | the unit sphere 21 + 22 + 23 = 1

(6.14)

and all other trajectories are obtained by homotheties. First of all, intersections of the unit sphere with the principal axes of inertia, i.e., the points e1  e2  e3 are equilibria, and there are no other equilibria, see equations (6:11).

6.3 Phase Portrait

89

e3

e2

e1

Fig. 6.2. Phase portrait of system (6:12) Further, the equilibria e1  e3 corresponding to the maximal and minimal eigenvalues 1  3 are stable, more precisely, they are centers, and the equilibria e2 corresponding to 2 are unstable | saddles. This is obvious from the geometry of intersections of the unit sphere with ellipsoids 1 21 + 2 22 + 3 23 = C: Indeed, for C < 3 the ellipsoids are inside the sphere and do not intersect it. For C = 3 , the ellipsoid touches the unit sphere from inside at the points e3 . Further, for C > 3 and close to 3, the ellipsoids intersect the unit sphere by 2 closed curves surrounding e3 and ;e3 respectively. The behavior of intersections is similar in the neighborhood of C = 1 . If C > 1 , then the ellipsoids are big enough and do not intersect the unit sphere for C = 1 , the small semiaxis of the ellipsoid becomes equal to radius of the sphere, so the ellipsoid touches the sphere from outside at e1  and for C < 1 and close to 1 the intersection consists of 2 closed curves surrounding e1 . If C = 2 , then the ellipsoid touches the sphere at the endpoints of the medium semiaxes e2 , and in the neighborhood of each point e2 , ;e2 , the intersection consists of four separatrix branches tending to this point. Equations for the separatrices are derived from the system

(

21 + 22 + 23 = 1 1 21 + 2 22 + 3 23 = 2 :

We multiply the rst equation by 2 and subtract it from the second equation: (1 ; 2 )21 ; (2 ; 3 )23 = 0:

90

6 Rotations of the Rigid Body

Thus the separatrices belong to intersection of the unit sphere with two planes

R

p

p

 def = f(1 2 3) 2 3 j 1 ; 2 1 =  2 ; 3 3 g thus they are arcs of great circles. It turns out that separatrices and equilibria are the only trajectories belonging to a 2-dimensional plane. Moreover, all other trajectories satisfy the following condition:  2=    2= ei )  ^ _ ^ # 6= 0 (6.15) i.e., the vectors , ,_ and # are linearly independent. Indeed, take any trajectory t on the unit sphere. All trajectories homothetic to the chosen one form a cone of the form C(21 + 22 + 23) = 1 21 + 2 22 + 3 23  3  C  1 : (6.16) But a quadratic cone in 3 is either degenerate or elliptic. The conditions  2=  ,  2= ei mean that C 6= i , i = 1 2 3, i.e., cone (6:16) is elliptic. Now inequality (6:15) follows from the next two facts. First,  ^ _ 6= 0, i.e., the trajectory t is not tangent to the generator of the cone. Second, the section of an elliptic cone by a plane not containing the generator of the cone is an ellipse | a strongly convex curve. In view of ODE (6:12), the convexity condition (6:15) for the cone generated by the trajectory is rewritten as follows:  2=    2= ei )  ^ (  ) ^ ((  )   +   (  )) 6= 0: (6.17) The planar separatrix curves in the phase portrait are regular curves on the sphere, hence  2    2= e2 )  ^ _ 6= 0 or, by ODE (6:12),  2    2= e2 )  ^ (  ) 6= 0: (6.18)

R

R

R

R

R

R

6.4 Controlled Rigid Body: Orbits Assume that we can control rotations of the rigid body by applying a torque along a line that is xed in the body. We can change the direction of torque to the opposite one in any moment of time. Then the control system for the angular velocity is written as _ t = t  t  l t 2 3 (6.19) and the whole control system for the controlled rigid body is

R

R

6.4 Controlled Rigid Body: Orbits

 _ t = t  t  l Q_ t = Qt bt

91

t 2 3 (6.20) Qt 2 SO(3) where l 6= 0 is a xed vector along the chosen line. Now we describe orbits and attainable sets of the 6-dimensional control system (6:20). But before that we study orbits of the 3-dimensional system (6:19).

6.4.1 Orbits of the 3-Dimensional System System (6:19) is analytic, thus dimension of the orbit through a point  2 coincides with dimension of the space Lie (    l) = Lie (   l): Denote the vector elds: f() =    g()  l and compute several Lie brackets: g f]() = dd f g() ; dd g f() = l   +   l g g f]]() = l  l + l  l = 2l  l 1 g g f]] g f]]() = l  (l  l) + (l  l)  l: 2

R

3

We apply (6:17) with l =  and obtain that three constant vector elds g, g f], g g f]] g f]] are linearly independent: g() ^ 12 g f]() ^ 21 g g f]] g f]]() = l ^ l  l ^ ((l  l)  l + l  (l  l)) 6= 0 if l 2= , l 2= ei: We obtain the following statement for generic disposition of the vector l. Case 1. l 2= , l 2= ei. Proposition 6.3. Assume that l 2= , l 2= ei. Then Lie (f g) = 3 for any  2 3. System (6:19) has one 3-dimensional orbit, 3. Now consider special dispositions of the vector l. Case 2. Let l 2 +, l 2= e2. Since the plane + is invariant for the free body (6:12) and l 2 + , then the plane + is also invariant for the controlled body (6:19), i.e., the orbit through any point of + is contained in + . On the other hand, implication (6:18) yields l ^ (l  l) 6= 0:

R

R

R

R

R

R

R

92

6 Rotations of the Rigid Body

But the vectors l = g() and l  l = 21 g g f]]() form a basis of the plane + , thus + is in the orbit through any point  2 + . Consequently, the plane + is an orbit of (6:19). If an initial point 0 2= + , then the trajectory t of (6:19) through 0 is not at, thus (t  t ) ^ l ^ (l  l) 6 0:

R

So the orbit through 0 is 3-dimensional. We proved the following statement. Proposition 6.4. Assume that l 2 + n e2. Then system (6:19) has one 2-dimensional orbit, the plane + , and two 3-dimensional orbits, connected components of 3 n + . The case l 2 ; n e2 is completely analogous, and there holds a similar proposition with + replaced by ;. Case 3. Now let l 2 e1 n f0g, i.e., l = ce1 , c 6= 0. First of all, the line e1 is an orbit. Indeed, if  2 e1, then f() = 0, and g() = l is also tangent to the line e1. To nd other orbits, we construct an integral of the control system (6:19) from two integrals (6:13) of the free body. Since g() = l = ce1 , we seek for a linear combination of the integrals in (6:13) that does not depend on 1 . We multiply the rst integral by 1 , subtract from it the second integral and obtain an integral for the controlled rigid body:

R

R

R RR

R

(1 ; 2 )22 + (1 ; 3 )23 = C:

R

(6.21)

Since 1 > 2 > 3 , this is an elliptic cylinder in 3. So each orbit of (6:19) is contained in a cylinder (6:21). On the other hand, the orbit through any point 0 2 3 n e1 must be at least 2-dimensional. Indeed, if 0 2= e2  e3, then the free body has trajectories not tangent to the eld g and if 0 2 e2 or e3, this can be achieved by a small translation of 0 along the eld g. Thus all orbits outside of the line e1 are elliptic cylinders (6:21). Proposition 6.5. Let l 2 e1 nf0g. Then all orbits of system (6:19) have the form (6:21): there is one 1-dimensional orbit | the line e1 (C = 0), and an in nite number of 2-dimensional orbits | elliptic cylinders (6:21) with C > 0,

R RR R R R R

see Fig. 6.3.

R R R

R

R

The case l 2 e3 n f0g is completely analogous to the previous one. Proposition 6.6. Let l 2 e3nf0g. Then system (6:19) has one 1-dimensional orbit | the line e3, and an in nite number of 2-dimensional orbits | elliptic cylinders

(1 ; 3 )21 + (2 ; 3 )22 = C

C > 0:

6.4 Controlled Rigid Body: Orbits

93

e3

e2

e1

Fig. 6.3. Orbits in the case l 2 Re1 n f0g

R

Case 4. Finally, consider the last case: let l 2 e2 n f0g. As above, we obtain an integral of control system (6:19):

(1 ; 2 )21 ; (2 ; 3 )23 = C:

(6.22)

If C 6= 0, this equation determines a hyperbolic cylinder. By an argument similar to that used in Case 3, we obtain the following description of orbits. Proposition 6.7. Let l 2 e2 nf0g. Then there is one 1-dimensional orbit | the line e2, and an in nite number of 2-dimensional orbits of the following

R

R

form: (1) connected components of hyperbolic cylinders (6:22) for C 6= 0 (2) half-planes | connected components of the set (+  ; ) n e2, see Fig. 6.4.

R

e3

e1

e2

Fig. 6.4. Orbits in the case l 2 Re2 n f0g

94

R

6 Rotations of the Rigid Body

So we considered all possible dispositions of the vector l 2 3 nf0g, and in all cases described orbits of the 3-dimensional system (6:19). Now we study orbits of the full 6-dimensional system (6:20).

6.4.2 Orbits of the 6-Dimensional System The vector elds in the right-hand side of the 6-dimensional system (6:20) are

 Qb  f(Q ) =    

R

0

(Q ) 2 SO(3)  3:

g(Q ) = l 

Notice the commutation rule for vector elds of the form that appear in our problem:



R



b i() 2 Vec(SO(3)  3) fi (Q ) = Qv w() i @ w 1 0 @ w 2 1 b b b BB Qw1 w2]so(3) + Q @  v1 ; @  v2 CC f1  f2](Q ) = B CA : @ @ v2 v ; @ v1 v @ 1 @ 2 We compute rst the same Lie brackets as in the 3-dimensional case: 



b g f] = l  Q+ l  l    1 g g f]] = 0  l  l 2   1 g g f]] g f]] = 0 l  (l  l) + (l  l)  l : 2 Further, for any vector eld X 2 Vec(SO(3)  3) of the form 0 X= x  x | a constant vector eld on 3 we have  b X f] = Q x :

R

R



R

(6.23) (6.24)

To study the orbit of the 6-dimensional system (6:20) through a point (Q ) 2 SO(3)  3, we follow the dierent cases for the 3-dimensional system (6:19) in Subsect. 6.4.1. Case 1. l 2= , l 2= ei. We can choose 3 linearly independent vector elds in Lie(f g) of the form (6:23):

R

6.4 Controlled Rigid Body: Orbits

X1 = g

X2 = 12 g g f]]

95

X3 = 12 g g f] g f]]:

By the commutation rule (6:24), we have 6 linearly independent vectors in Lie(Q ) (f g): X1 ^ X2 ^ X3 ^ X1 f] ^ X2  f] ^ X3 f] 6= 0: Thus the orbit through (Q ) is 6-dimensional. Case 2. l 2  n e2. Case 2.1.  2= . First of all, Lie(f g) contains 2 linearly independent vector elds of the form (6:23): X1 = g X2 = 21 g g f]]: Since the trajectory of the free body in 3 through  is not at, we can assume that the vector v =    is linearly independent of l and l  l. Now our aim is to show that Lie(f g) contains 2 vector elds of the form  QM   QM  1 Y1 = v1  Y2 = v2 2  M1 ^ M2 6= 0 (6.25) where the vector elds v1 and v2 vanish at the point . If this is the case, then Lie(Q ) (f g) contains 6 linearly independent vectors: X1 (Q ) X2 (Q ) f(Q )   1  Y (Q ) = QM2  Y1 (Q ) = QM 2 0 0

R

R





Y1  Y2](Q ) = QM10 M2 ]  and the orbit through the point (Q ) is 6-dimensional. Now we construct 2 vector elds of the form (6:25) in Lie(f g). Taking appropriate linear combinations with the elds X1 , X2 , we project the second component of the elds g f] and 21 f g g f]] to the line v, thus we obtain the vector elds  b  b  Q l  Q (l  l) 2 Lie(f g): (6.26) k1 v k2 v If both k1 and k2 vanish at , these vector elds can be taken as Y1, Y2 in (6:25). And if k1 or k2 does not vanish at , we construct such vector elds Y1, Y2 taking appropriate linear combinations of elds (6:26) and f with the elds g, g g f]]. So in Case 2.1 the orbit is 6-dimensional. Case 2.2.  2 . There are 5 linearly independent vectors in the space Lie(Q ) (f g):

R

96

6 Rotations of the Rigid Body

X1 = g X2 = 12 g g f]] X1  f] X2 f] X1 f] X2 f]]:

RRR

R

Since the orbit in 3 is 2-dimensional, the orbit in SO(3)  3 is 5-dimensional. Case 3. l 2 e1 n f0g. Case 3.1.  2= e1. The argument is similar to that of Case 2.1. We can assume that the vectors l and v =    are linearly independent. The orbit in 3 is 2-dimensional and the vectors l, v span the tangent space to this orbit, thus we can nd vector elds in Lie(f g) of the form:

R





b b Y1 = g f] ; C1g ; C2f = Q l + 0C3Q    b b  b Q  l  ] + C Q   4 Y2 = Y1  f] = 0

for some real functions Ci , i = 1 : : :  4. Then we have 5 linearly independent vectors in Lie(Q ) (f g): g f Y1  Y2 Y1 Y2 ]: So the orbit of the 6-dimensional system (6:20) is 5-dimensional (it cannot have dimension 6 since the 3-dimensional system (6:19) has a 2-dimensional orbit). Case 3.2.  2 e1. The vectors

R

 b f(Q ) = Q0  

 b g f](Q ) = Q0 l 

are linearly dependent, thus dimLie(Q ) (f g) = dim span(f g)j(Q ) = 2. So the orbit is 2-dimensional. The cases l 2 ei n f0g, i = 1 2, are similar to Case 3. We completed the study of orbits of the controlled rigid body (6:20) and now summarize it. Proposition 6.8. Let (Q ) be a point in SO(3)  3. If the orbit O of the 3dimensional system (6:19) through the point  is 3- or 2-dimensional, then the orbit of the 6-dimensional system (6:20) through the point (Q ) is SO(3) O, i.e., respectively 6- or 5-dimensional. If dim O = 1, then the 6-dimensional system has a 2-dimensional orbit. We will describe attainable sets of this system in Sect. 8.4 after acquiring some general facts on attainable sets.

R

R

7 Control of Congurations

In this chapter we apply the Orbit Theorem to systems which can be controlled by the change of their conguration, i.e., of relative position of parts of the systems. A falling cat exhibits a well-known example of such a control. If a cat is left free over ground (e.g. if it falls from a tree or is thrown down by a child), then the cat starts to rotate its tail and bend its body, and nally falls to the ground exactly on its paws, regardless of its initial orientation over the ground. Such a behavior cannot be demonstrated by a mechanical system less skillful in turning and bending its parts (e.g. a dog or just a rigid body), so the crucial point in the falling cat phenomenon seems to be control by the change of conguration. We present a simple model of systems controlled in such a way, and study orbits in several simplest examples.

7.1 Model

R R

R

A system of mass points, i.e., a mass distribution in n, is described by a nonnegative measure  in n. We restrict ourselves by measures with compact support. For example, a system of points x1  : : : P xk 2 n with masses 1 : : :  k > 0 is modeled by the atomic measure  = ki=1 i xi , where xi is the Dirac function concentrated at xi. One can consider points xi free or restricted by constraints in n. More generally, mass can be distributed along segments or surfaces of various dimensions. So the state space M of a system to be considered is a reasonable class of measures in n. A controller is supposed to sit in the construction and change its conguration. The system is conservative, i.e., impulse and angular momentum are conserved. Our goal is to study orbits of systems subject to such constraints. Mathematically, conservation laws of a system come from N#other theorem due to symmetries of the system. Kinetic energy of our system is Z 1 (7.1) L = 2 jx_ j2 d

R

R

98

7 Control of Congurations

P in particular, for an atomic measure  = ki=1 i xi , k X L = 12 i jx_ ij2:

R

i=1

By N#other theorem (see e.g. 135]), if the ow of a vector eld V 2 Vec n preserves a Lagrangian L, then the system has an integral of the form @ L V (x) = const : @ x_ In our case, Lagrangian (7:1) is invariant w.r.t. isometries of the Euclidean space, i.e., translations and rotations in n. Translations in n are generated by constant vector elds: V (x) = a 2 n and our system is subject to the conservation laws

R

R

Z

That is,

R

R

hx_ ai d = const 8 a 2 n:

Z

xd _ = const

i.e., the center of mass of the system moves with a constant velocity (the total impulse is preserved). We choose the inertial frame of reference in which the center of mass is xed: Z xd _ = 0:

P For an atomic measure  = ki=1 i xi , this equality takes the form k X i=1

i xi = const

R

which is reduced by a change of coordinates in n to k X

i xi = 0: i=1 n. Let a vector eld

R

R

Now we pass to rotations in V (x) = Ax x 2 n preserve the Euclidean structure in n, i.e., its ow etV (x) = etA x

R

7.1 Model

preserve the scalar product: hetA x etAyi = hx yi

R 2R

x y 2 n:

Dierentiation of this equality at t = 0 yields

hAx yi + hx Ayi = 0

99

x y

i.e., the matrix A is skew-symmetric:

n

A = ;A: Conversely, if the previous equality holds, then

;etA = etA = e;tA = ;etA;1 

R

i.e., the matrix etA is orthogonal. We proved that the ow etA preserves the Euclidean structure in n if and only if A = ;A. Similarly to the 3dimensional case considered in Sect. 6.1, the group of orientation-preserving linear orthogonal transformations of the Euclidean space n is denoted by SO(n), and the corresponding Lie algebra of skew-symmetric transformations in n is denoted by so(n). In these notations, etA 2 SO(n) , A 2 so(n): Return to derivationR of conservation laws for our system of mass points. The Lagrangian L = 21 jx_ j2 d is invariant w.r.t. rotations in n, so N#other theorem gives integrals of the form @ L V (x) = Z hx_ Axi d = const A 2 so(n): @ x_

R

R

R

P For an atomic measure  = ki=1 i xi , we obtain k X i=1

i hx_ i  Axii = const

A 2 so(n)

(7.2)

and we restrict ourselves by the simplest case where the constant in the righthand side is just zero. Summingup, we have the following conservation laws for a system of points x1 : : :  xk 2 n with masses 1 : : :  k :

R

k X

i=1 k X i=1

i xi = 0 i hx_ i  Axii = 0

(7.3) A 2 so(n):

(7.4)

100

7 Control of Congurations

The state space is a subset

M

R|  {z  R} n

k

n

and admissible paths are piecewise smooth curves in M that satisfy constraints (7:3), (7:4). The rst equality (7:3) determines a submanifold in M in fact, this equality can obviously be resolved w.r.t. any variable xi , and one can get rid of this constraint by decreasing dimension of M. The second equality (7:4) is a linear constraint for velocities x_ i , it determines a distribution on M. So the admissibility conditions (7:3), (7:4) dene a linear in control, thus symmetric, control system on M. Notice that a more general condition (7:2) determines an \ane distribution", and control system (7:3), (7:2) is control-ane, thus, in general, not symmetric. We consider only the symmetric case (7:3), (7:4). Then orbits coincide with attainable sets. We compute orbits in the following simple situations: (1) Two free points: k = 2, (2) Three free points: k = 3, (3) A broken line with 3 links in 2.

R

7.2 Two Free Points

R

We have k = 2, and the rst admissibility condition (7:3) reads 1x1 + 2x2 = 0 x1 x2 2 n: We eliminate the second point: x1 = x x2 = ; 1 x 2

and exclude collisions of the points: x 6= 0: So the state space of the system is M = n n f0g: The second admissibility condition (7:4) 1hx_ 1  Ax1i + 2hx_ 2  Ax2i = 0 A 2 so(n) is rewritten as ; + 2=  hx_ Axi = 0 A 2 so(n) 1 1 2 i.e., hx_ Axi = 0 A 2 so(n): This equation can easily be analyzed via the following proposition.

R

(7.5)

R

7.3 Three Free Points

R

101

Exercise 7.1. If A 2 so(n), then hAx xi = 0 for all x 2 n. Moreover, for any vector x 2 n n f0g, the space fAx j A 2 so(n)g coincides with the whole orthogonal complement x? = fy 2 n j hy xi = 0g. So restriction (7:5) means that

R

x_ ^ x = 0 i.e., velocity of an admissible curve is proportional to the state vector. The distribution determined by this condition is one-dimensional, thus integrable. So admissible curves have the form x(t) = (t)x(0)

(t) > 0:

R

The orbit and admissible set through any point x 2 n n f0g is the ray

Ox =

R

+x = f x j > 0g:

R

The points x1, x2 can move only along a xed line in n, and orientation of the system cannot be changed. In order to have a more sophisticated behavior, one should consider more complex systems.

7.3 Three Free Points Now k = 3, and we eliminate the third point via the rst admissibility condition (7:3): x = 1 x1 y = 2 x2 x3 = ; 1 (x + y): 3

In order to exclude the singular congurations where the points x1, x2, x3 are collinear, we assume that the vectors x, y are linearly independent. So the state space is M = f(x y) 2 n  n j x ^ y 6= 0g: Introduce the notation i = 1  i = 1 2 3: i Then the second admissibility condition (7:4) takes the form:

R R

hx_ A((1 + 3 )x + 3 y)i + hy_ A((2 + 3 )y + 3 x)i = 0

A 2 so(n):

It turns out then that admissible velocities x,_ y_ should belong to the plane span(x y). This follows by contradiction from the following proposition.

102

7 Control of Congurations

R

Lemma 7.2. Let vectors v w   2 n satisfy the conditions v^w = 6 0 span(v w  ) 6= span(v w): Then there exists A 2 so(n) such that hAv i + hAw i =6 0: Proof. First of all, we may assume that

hv wi = 0:

(7.6) Indeed, choose a vector wb 2 span(v w) such that hv wbi = 0. Then w = wb + v and hAv i + hAw i = hAv + i + hAwb i thus we can replace w by w. b Second, we can renormalize vectors v, w and assume that jvj = jwj = 1: (7.7) Now let 62 span(v w), we can assume this since the hypotheses of the lemma are symmetric w.r.t. , . Then

= v + w + l for some vector l ? span(v w): Choose an operator A 2 so(n) such that Aw = 0 A : span(v l) ! span(v l) is invertible. Then hAv i + hAw i = hAv li 6= 0 i.e., the operator A is the required one. ut This lemma means that for any pair of initial points (x y) 2 M, all adn. So we missible curves xt and yt are contained in the plane span(x y) can reduce our system to such a plane and thus assume that x y 2 2. Thus we obtain the following system: hx_ A((1 + 3 )x + 3 y)i + hy_ A((2 + 3 )y + 3 x)i = 0 A 2 so(2) (7.8) (x y) 2 M = f(v w) 2 2  2 j v ^ w 6= 0g: Consequently,  0 1 A = const ;1 0 

RR

R R

7.3 Three Free Points

103

i.e., equality (7:8) denes one linear equation on velocities, thus a rank 3 distribution on a 4-dimensional manifold M. Using Exercise 7.1, it is easy to see that this distribution is spanned by the following 3 linear vector elds:  ( +  )x +  y  x @ 1 3 3 V1 = ((1 + 3 )x + 3 y) @ x = = B1 y  0   x @ 0 V2 = ((2 + 3 )y + 3 x) @ y = 3 x + (2 + 3 )y = B2 y  x x @ @ V3 = x @ x + y @ y = y = Id y  where  +    0 0  1 0 1 3 3 B1 = 0 0  B2 = 3 2 + 3  Id = 0 1 : In order to simplify notations, we write here 4-dimensional vectors as 2-dimensional columns: e.g., 0 +  )x +  y 1 1 3 1 3 1  ( +  )x +  y  B ( ( +  )x +  3 y2 C 1 3 3 1 3 2 CA  B V1 = = @ 0 0 0 where     y = yy1 : x = xx1  2 2 The rank 3 distribution in question can have only orbits of dimensions 3 or 4. In order to nd out, which of these possibilities are realized, compute the Lie bracket:   V1 V2 ] = B1  B2 ] xy    2 + 3 : 3 B1  B2] = 3 ;(1+ 3 ) ;3 It is easy to check that V1 ^ V2 ^ V3 ^ V1  V2] 6= 0 , B1 ^ B2 ^ Id ^B1  B2] 6= 0: We write 2  2 matrices as vectors in the standard basis of the space gl(2):

1 0 0 1 0 0 0 0     00

then

00

10

01

104

7 Control of Congurations

1 1 + 3 0 3 0  0  + 3 3 2 det(Id B1  B2 B1 B2 ]) = 0 0 3 ;(1 + 3 ) 1 0 2 + 3 ;3 = 23 (1 2 + 1 3 + 2 3 ) > 0: Consequently, the elds V1  V2  V3 V1 V2] are linearly independent everywhere on M, i.e., the control system has only 4-dimensional orbits. So the orbits coincide with connected components of the state space. The manifold M is decomposed into 2 connected components corresponding to positive or negative orientation of the frame (x y): M = M+  M;  M = f(x y) 2 2  2 j det(x y) ? 0g: So the system on M has 2 orbits, thus 2 attainable sets: M+ and M; . Given any pair of linearly independent vectors (x y) 2 2  2, we can reach any other nonsingular conguration (~x y~) 2 2  2 with x~ y~ 2 span(x y) and the frame (~x y~) oriented in the same way as (x y). Returning to the initial problem for 3 points x1 x2 x3 2 n: the 2-dimensional linear plane of the triangle (x1  x2 x3) should be preserved, as well as orientation and center of mass of the triangle. Except this, the triangle (x1 x2 x3) can be rotated, deformed or dilated as we wish. Congurations of 3 points that dene distinct 2-dimensional planes (or dene distinct orientations in the same 2-dimensional plane) are not mutually reachable: attainable sets from these congurations do not intersect one with another. Although, if two congurations dene 2-dimensional planes having a common line, then intersection of closures of attainable sets from these congurations is nonempty: it consists of collinear triples lying in the common line. Theoretically, one can imagine a motion that steers one conguration into another: rst the 3 points are made collinear in the initial 2-dimensional plane, and then this collinear conguration is steered to the nal one in the terminal 2-dimensional plane.

R R R RR R

R

7.4 Broken Line Consider a system of 4 mass points placed at vertices of a broken line of 3 segments in a 2-dimensional plane. We study the most symmetric case, where all masses are equal to 1 and lengths of all segments are also equal to 1, see Fig. 7.1. The holonomic constraints for the points x0  x1 x2 x3 2 2 = have the form

R C

7.4 Broken Line

105

x3

x1

x2

x0

Fig. 7.1. Broken line 3 X

j =0

xj = 0

jxj ; xj ;1j = 1

j = 1 2 3:

(7.9)

Thus

xj ; xj ;1 = eij 

j 2 S 1  j = 1 2 3: Position of the system is determined by the 3-tuple of angles ( 1  2  3), so the state space is the 3-dimensional torus: M = S 1  S 1  S 1 = 3 = f( 1  2  3) j j 2 S 1  j = 1 2 3g: The nonholonomic constraints on velocities reduce to the equality

T

3 X

hixj  x_ j i = 0:

j =0

In order to express this equality in terms of the coordinates j , denote rst yj = xj ; xj ;1 j = 1 2 3:

P

Taking into account the condition 3j =0 xj = 0, we obtain: x0 = ; 3y41 ; y22 ; y43  x1 = y41 ; y22 ; y43  x2 = y41 + y22 ; y43  x3 = y41 + y22 + 3y43 :

Now compute the dierential form:

106

7 Control of Congurations

!=

3 X

j =0

hixj  dxj i = hi ((3=4)y1 + (1=2)y2 + (1=4)y3 )  dy1i + hi ((1=2)y1 + y2 + (1=2)y3 )  dy2i + hi ((1=4)y1 + (1=2)y2 + (3=4)y3)  dy3i :

Since hiyj  dyk i = heij  eik d k i = cos( j ; k )d k , we have ! = ((3=4) + (1=2) cos( 2 ; 1 ) + (1=4) cos( 3 ; 1 )) d 1 + ((1=2) cos( 1 ; 2 ) + 1 + (1=2) cos( 3 ; 2 )) d 2 + ((1=4) cos( 1 ; 3 ) + (1=2) cos( 2 ; 3 ) + 3=4) d 3: Consequently, the system under consideration is the rank 2 distribution  = Ker ! on the 3-dimensional manifold M = 3. The orbits can be 2- or 3-dimensional. To distinguish these cases, we can proceed as before: nd a vector eld basis and compute Lie brackets. But now we study integrability of  in a dual way, via techniques of dierential forms. Assume that the distribution  has a 2-dimensional integral manifold N M. Then !jN = 0 consequently, 0 = d (!jN ) = (d!)jN  thus 0 = d!q j q = d!q jKer !q  q 2 N: In terms of exterior product of dierential forms, (! ^ d!)q = 0 q 2 N: We compute the dierential and exterior product: d! = sin( 2 ; 1 )d 1 ^ d 2 + sin( 3 ; 2 )d 2 ^ d 3 + 12 sin( 3 ; 1 )d 1 ^ 3  ! ^ d! = 21 (sin( 2 ; 1 ) + sin( 3 ; 2 ))d 1 ^ d 2 ^ d 3:

T

Thus ! ^ d! = 0 if and only if sin( 2 ; 1 ) + sin( 3 ; 2 ) = 0 i.e.,

3 = 1 or ( 1 ; 2 ) + ( 3 ; 2 ) = 

(7.10) (7.11)

7.4 Broken Line

Fig. 7.2. Hard to control conguration: 1 = 2

107

Fig. 7.3. Hard to control conguration: (1 ; 2 ) + (3 ; 2 ) = 

see Figs. 7.2, 7.3. Congurations (7:10) and (7:11) are hard to control: if neither of these equalities is satised, then ! ^ d! 6= 0, i.e., the system has 3-dimensional orbits through such points. If we choose basis vector elds X1 , X2 of the distribution , then already the rst bracket X1 X2 ] is linearly independent of X1 , X2 at points where both equalities (7:10), (7:11) are violated. Now it remains to study integrability of  at points of surfaces (7:10), (7:11). Here X1  X2](q) 2 q , but we may obtain nonintegrability of  via brackets of higher order. Consider rst the two-dimensional surface P = f 3 = 1 g: If the orbit through a point q 2 P is two-dimensional, then the distribution  should be tangent to P in the neighborhood of q. But it is easy to see that  is everywhere transversal to P: e.g., Tq P 3 @@ 2= q  q 2 P: 2 q So the system has 3-dimensional orbits through any point of P . In the same way one can see that the orbits through points of the second surface (7:11) are 3-dimensional as well. The state space M is connected, thus there is the only orbit (and attainable set) | the whole manifold M. The system is completely controllable.

8 Attainable Sets

In this chapter we study general properties of attainable sets. We consider families of vector elds F on a smooth manifold M that satisfy the property Lieq F = Tq M

8 q 2 M:

(8.1)

In this case the system F is called bracket-generating , or full-rank . By the analytic version of the Orbit Theorem (Corollary 5.17), orbits of a bracketgenerating system are open subsets of the state space M. If a family F Vec M is not bracket-generating, and M and F are real analytic, we can pass from F to a bracket-generating family FjO , where O is an orbit of F . Thus in the analytic case requirement (8:1) is not restrictive in essence.

8.1 Attainable Sets of Full-Rank Systems For bracket-generating systems both orbits and attainable sets are full-dimensional. Moreover, there holds the following important statement. Theorem 8.1 (Krener). If F Vec M is a bracket-generating system, then Aq0 int Aq0 for any q0 2 M . Remark 8.2. In particular, attainable sets for arbitrary time have nonempty interior: int Aq0 6= : Attainable sets may be:  open sets, Fig. 8.1,  manifolds with smooth boundary, Fig. 8.2,  manifolds with boundary having singularities (corner or cuspidal points), Fig. 8.3, 8.4.

110

8 Attainable Sets

q0

q0 Aq0

Fig. 8.1. Orbit an open set

q0

Aq0

M

Fig. 8.3. Orbit a manifold with nonsmooth boundary

Aq0

M

M

Fig. 8.2. Orbit a manifold with smooth boundary

Aq0

q0

M

Fig. 8.4. Orbit a manifold with nonsmooth boundary

One can easily construct control systems (e.g. in the plane) that realize these possibilities. On the other hand, Krener's theorem prohibits an attainable set Aq0 of a bracket-generating family to be:  a lower-dimensional subset of M, Fig. 8.5,  a set where boundary points are isolated from interior points, Fig. 8.6. Now we prove Krener's theorem. Proof. Fix an arbitrary point q0 2 M and take a point q0 2 Aq0 . We show that q0 2 int Aq0 : (8.2) (1) There exists a vector eld f1 2 F such that f1(q0 ) 6= 0, otherwise Lieq0 (F ) = 0 and dimM = 0. The curve s1 7! q0  es1 f1  0 < s1 < "1  (8.3) is a 1-dimensional submanifold of M for small enough "1 > 0.

8.1 Attainable Sets of Full-Rank Systems

M Fig. 8.5. Prohibited orbit: subset of non-full dimension

111

M Fig. 8.6. Prohibited orbit: subset with isolated boundary points

If dimM = 1, then q0  es1 f1 2 int Aq0 for suciently small s1 > 0, and inclusion (8:2) follows. (2) Assume that dimM > 1. Then arbitrarily close to q0 we can nd a point q1 on curve (8:3) and a eld f2 2 F such that the vector f2 (q1) is not tangent to curve (8:3):

q1 = q0  et11 f1  t11 suciently small (q1  f1 ) ^ (q1  f2 ) 6= 0 otherwise dimLieq F = 1 for q on curve (8:3) with small s1 . Then the mapping (s1  s2) 7! q0  es1 f1  es2 f2  (8.4) 1 1 t1 < s1 < t1 + "2  0 < s2 < "2  is an immersion for suciently small "2 , thus its image is a 2-dimensional submanifold of M. If dimM = 2, inclusion (8:2) is proved. (3) Assume that dimM > 2. We can nd a vector f3 (q), f3 2 F , not tangent to surface (8:4) suciently close to q0 : there exist t12 t22 > 0 and f3 2 F 1such that the vector eld f3 is not tangent to surface (8:4) at a point q2 = q0  et2 f1  et22 f2 . Otherwise the family F is not bracket-generating. The mapping (s1  s2 s3 ) 7! q0  es1 f1  es2 f2  es3 f3  ti2 < si < ti2 + "3  i = 1 2 0 < s3 < "3  is an immersion for suciently small "3 , thus its image is a smooth 3dimensional submanifold of M. If dimM = 3, inclusion (8:2) follows. Otherwise we continue this procedure. (4) For dimM = n, inductively, we nd a point

112

8 Attainable Sets

R

(t1n;1 t2n;1 : : : tnn;;11) 2 n;1 tin;1 > 0 and elds f1 : : :  fn 2 F such that the mapping (s1  : : :  sn ) 7! q0  es1 f1   esn fn  tin;1 < si < tin;1 + "n  i = 1 : : : n ; 1 0 < sn < "n  n ; 1 1 2 f 0 t f t t f qn;1 = q  e n;1 1  e n;1 2   e n;1 n;1  is an immersion. The image of this immersion is an n-dimensional submanifold of M, thus an open set. This open set is contained in Aq0 and can be chosen as close to the point q0 as we wish. Inclusion (8:2) is proved, and the theorem follows. ut We obtain the following proposition from Krener's theorem. Corollary 8.3. Let F Vec M be a bracket-generating system. If Aq0 (F ) = M for some q0 2 M , then Aq0 (F ) = M . Proof. Take an arbitrary point q 2 M. We show that q 2 Aq0 (F ). Consider the system ;F = f;V j V 2 Fg Vec M: This system is bracket-generating, thus by Theorem 8.1 Aq (;F ) int Aq (;F ) 8q 2 M: Take any point qb 2 int Aq (;F ) and a neighborhood of this point Oqb Aq (;F ): Since Aq0 (F ) is dense in M, then Aq0 (F ) \ Oqb 6= : That is why Aq0 (F ) \ Aq (;F ) 6= , i.e., there exists a point q0 2 Aq0 (F ) \ Aq (;F ): In other words, the point q0 can be represented as follows: q0 = q0  et1 f1   etk fk  fi 2 F  ti > 0 q0 = q  e;s1 g1   e;sl gl  gi 2 F  si > 0: We multiply both decompositions from the right by esl gl   es1g1 and obtain q = q0  et1 f1   etk fk  esl gl   es1 g1 2 Aq0 (F ) q.e.d. ut The sense of the previous proposition is that in the study of controllability, we can replace the attainable set of a bracket-generating system by its closure. In the following section we show how one can add new vector elds to a system without change of the closure of its attainable set.

8.2 Compatible Vector Fields and Relaxations

8.2 Compatible Vector Fields and Relaxations

113

Denition 8.4. A vector eld f 2 Vec M is called compatible with a system F Vec M if Aq (F  f) Aq (F ) 8q 2 M: Easy compatibility condition is given by the following statement. Proposition 8.5. Let F Vec M . For any vector elds f1  f2 2 F , and any functions a1 a2 2 C 1 (M), a1 a2  0, the vector eld a1f1 +a2f2 is compatible with F .

In view of Corollary 5.11, the following proposition holds. Corollary 8.6. If F Vec M is a bracket-generating system such that the positive convex cone generated by F

cone(F ) =

(X k i=1

aifi j fi 2 F  ai 2 C 1 (M) ai  0 k 2

N

)

Vec M

is symmetric, then F is completely controllable.

Proposition 8.5 is a corollary of the following general and strong statement. Theorem 8.7. Let X  Y , 2 0 t1], be nonautonomous vector elds with a common compact support. Let 0  ( )  1 be a measurable function. Then there exists a sequence of nonautonomous vector elds Zn 2 fX  Y g, i.e., Zn = X or Y for any and n, such that the ow

Zt Zt ;! ;! n exp Z d ! exp ( ( )X + (1 ; ( ))Y ) d  0

0

n ! 1

uniformly w.r.t. (t q) 2 0 t1]  M and uniformly with all derivatives w.r.t. q 2 M.

Now Proposition 8.5 follows. In the case a1 (q) + a2 (q) = 1 it is a corollary of Theorem 8.7. Indeed, it is easy to show that the curves q(t) = q0  et(a1 f1 +a2 f2 ) and

;! Z t ( 1( )f1 + 2( )f2 ) d  q0 exp 0

i(t) = ai(q(t))

coincide one with another (hint: prove that the curve

; q0  et(a1 f1 +a2 f2 )  exp

Zt 0

(; 1 ( )f1 ; 2 ( )f2 ) d

114

8 Attainable Sets

Y

Zn X

Fig. 8.7. Approximation of ow, Th. 8.7 is constant). For the case a1 (q) a2(q) > 0 we generalize by multiplication of control parameters by arbitrary positive function (this does not change attainable set for all nonnegative times), and the case a1(q) a2 (q)  0 is obtained by passage to limit. Remark 8.8. If the elds X , Y are piecewise continuous w.r.t. , then the approximating elds Zn in Theorem 8.7 can be chosen piecewise constant. Theorem 8.7 follows from the next two lemmas. Lemma 8.9. Under conditions of Theorem 8:7, there exists a sequence of nonautonomous vector elds Zn 2 fX  Y g such that

Zt 0

Zn d

!

Zt 0

( ( )X + (1 ; ( ))Y ) d

uniformly w.r.t. (t q) 2 0 t1]  M and uniformly with all derivatives w.r.t. q 2 M. Proof. Fix an arbitrary positive integer n. We can choose a covering of the

segment 0 t1] by subsets such that

N  i=1

Ei = 0 t1]

8i = 1 : : :  N 9Xi  Yi 2 Vec M s.t. kX ;Xi kn K  n1  kY ;Yi kn K  n1 

where K is the compact support of X , Y . Indeed, the elds X , Y are bounded in the norm k kn+1K , thus they form a precompact set in the topology induced by k kn K .

8.2 Compatible Vector Fields and Relaxations

115

Then divide Ei into n subsets of equal measure: Ei =

n

j =1

Eij  jEij j = n1 jEij

In each Eij pick a subset Fij so that

jFij j =

Fij Eij 

i j = 1 : : :  n:

Z Eij

( ) d :

Finally, dene the following vector eld: X  2 F  n Z = Y  2 Eij n F :  ij ij Then the sequence of vector elds Zn is the required one. ut Now we prove the second part of Theorem 8.7. Lemma 8.10. Let Zn , n = 1 2 : : : , and Z , 2 0 t1], be nonautonomous vector elds on M , bounded w.r.t. , and let these vector elds have a compact support. If

Zt 0

then

;! exp

Zn d !

Zt 0

Zt 0

Z d 

;! Zn d ! exp

Zt 0

n ! 1

Z d 

n ! 1

the both convergences being uniform w.r.t. (t q) 2 0 t1]  M and uniform with all derivatives w.r.t. q 2 M . Proof. (1) First we prove the statement for the case Z = 0. Denote the ow

;! Ptn = exp Then Ptn = Id +

Zt 0

Zt 0

0

Zn d :

Pn  Zn d

integrating by parts

Since

Zt

= Id +Ptn 

Zt 0

Zn d ;

Z t 0

Pn  Zn 

Z 0



Z d d :

Zn d ! 0, the last two terms above tend to zero, thus

116

8 Attainable Sets

Ptn ! Id and the statement of the lemma in the case Z = 0 is proved. (2) Now we consider the general case. Decompose vector elds in the sequence as follows: Zn = Z + Vn 

;! Denote Ptn = exp ;! exp

Zt 0

Zt 0

Zt 0

Vn d ! 0

n ! 1:

Vn d . From the variations formula, we have

;! Zn d = exp

Zt 0

;! (Vn + Z ) d = exp

Zt 0

Ad Pn Z d  Ptn:

Since Ptn ! Id by part (1) of this proof and thus Ad Ptn ! Id, we obtain the

required convergence:

;! exp

Zt 0

;! Zn d ! exp

Zt 0

Z d :

ut

So we proved Theorem 8.7 and thus Proposition 8.5.

8.3 Poisson Stability Denition 8.11. Let f 2 Vec M be a complete vector eld. A point q 2 M is called Poisson stable for f if for any t > 0 and any neighborhood Oq of q 0f 0 0 0 t there exists a point q 2 Oq and a time t > t such that q  e 2 Oq . In other words, all trajectories cannot leave a neighborhood of a Poisson stable point forever, some of them must return to this neighborhood for arbitrarily large times. Remark 8.12. If a trajectory q  etf is periodic, then q is Poisson stable for f. Denition 8.13. A complete vector eld f 2 Vec M is Poisson stable if all points of M are Poisson stable for f . The condition of Poisson stability seems to be rather restrictive, but nevertheless there are surprisingly many Poisson stable vector elds in applications, see Poincar"e's theorem below. But rst we prove a consequence of Poisson stability for controllability. Proposition 8.14. Let F Vec M be a bracket-generating system. If a vector

eld f 2 F is Poisson stable, then the eld ;f is compatible with F .

8.3 Poisson Stability

117

Proof. Choose an arbitrary point q0 2 M and a moment of time t > 0. To

prove the statement, we should approximate the point q0  e;tf by reachable points. Since F is bracket-generating, we can choose an open set W int Aq0 (F ) arbitrarily close to q0 . Then the set W  e;tf is close enough to q0  e;tf . By Poisson stability, there exists t0 > t such that ;   6= W  e;tf  et0 f \ W  e;tf = W  e(t0 ;t)f \ W  e;tf :

But W  e(t0 ;t)f

Aq0 (F ), thus Aq0 (F ) \ W  e;tf 6= : So in any neighborhood of q0  e;tf there are points of the attainable set Aq0 (F ), i.e., q0  e;tf 2 Aq0 (F ). ut Theorem 8.15 (Poincare). Let M be a smooth manifold with a volume form Vol. Let a vector eld f 2 Vec M be complete and its ow etf preserve volume. Let W

M , W intW , be a subset of nite volume, invariant for f : Vol(W ) < 1 W  etf W 8t > 0: Then all points of W are Poisson stable for f . Proof. Take any point q 2 W and any its neighborhood O M of nite volume. The set V = W \ O contains an open nonempty subset intW \ O, thus Vol(V ) > 0. In order to prove the theorem, we show that

V  et0 f \ V 6=  for some large t0: Fix any t > 0. Then all sets V  entf  n = 0 1 2 : : :  have the same positive volume, thus they cannot be disjoint. Indeed, if V  entf \ V  emtf =  8n m = 0 1 2 : ::  then Vol(W ) = 1 since all these sets are contained in W. Consequently, there exist nonnegative integers n > m such that V  entf \ V  emtf 6= : We multiply this inequality by e;mtf from the right and obtain V  e(n;m)tf \ V 6= : Thus the point q is Poisson stable for f. Since q 2 W is arbitrary, the theorem ut follows.

118

8 Attainable Sets

R

A vector eld that preserves volume is called conservative . Recall that a vector eld on n = fR(x1  : : :  xn)g is conservative, i.e., preserves the standard volume Vol(V ) = V dx1 : : : dxn i it is divergencefree: n n X X divx f = @@ xfi = 0 f = fi @@x : i=1

i

i=1

i

8.4 Controlled Rigid Body: Attainable Sets We apply preceding general results on controllability to the control system that governs rotations of the rigid body, see (6:20):  Q_  3 (8.5) _ = f(Q )  g(Q ) (Q ) 2 SO(3)    b      g= 0 : f = Q l By Proposition 8.5, the vector eld f = 21 (f +g)+ 12 (f ; g) is compatible with system (8:5). We show now that this eld is Poisson stable on SO(3)  3. Consider rst the vector eld f(Q ) on the larger space 9Q  3 , where 9Q is the space of all 3  3 matrices. Since div(Q ) f = 0, the eld f is conservative on 9Q  3 . Further, since the rst component of the eld f is linear in Q, it has the following left-invariant property in Q: Q Q   PQ   PQ  t tf tf e  =  ) e  = t t  (8.6) t Q Qt  P 2 9Q  t 2 3 : In view of this property, the eld f has compact invariant sets in 9Q  3 of the form W = (SO(3) K)  f ( )  C g K b 9Q K intK C > 0 so that W intW. By Poincar"e's theorem, the eld f is Poisson stable on all such sets W, thus on 9Q  3 . In view of the invariance property (8:6), the eld f is Poisson stable on SO(3)  3. Since f is compatible with (8:5), then ;f is also compatible. The vector elds g = (f  g) ; f are compatible with (8:5) as well. So all vector elds of the symmetric system span(f g) = faf + bg j a b 2 C 1g are compatible with the initial system. Thus closures of attainable sets of the initial system (8:5) and the extended system span(f g) coincide one with another.

R

R

R RR

R R

R R R

R R

R

R R

8.4 Controlled Rigid Body: Attainable Sets

119

Let the initial system be bracket-generating. Then the symmetric system span(f g) is bracket-generating as well, thus completely controllable. Hence the initial system (8:5) is completely controllable in the bracket-generating case. In the non-bracket-generating cases, the structure of attainable sets is more complicated. If l is a principal axis of inertia, then the orbits of system (8:5) coincide with attainable sets. If l 2  n e2, they do not coincide. This is easy to see from the phase portrait of the vector eld f() =    in the plane  : the line e2 consists of equilibria of f, and in the half-planes  n e2 trajectories of f are semicircles centered at the origin, see Fig. 8.8.

R

R

R

Re2

f

Fig. 8.8. Phase portrait of f j

R R



in the case l 2  n Re2

The eld f is not Poisson stable in the planes . The case l 2  n e2 diers from the bracket-generating case since the vector eld f preserves volume in 3, but not in  . A detailed analysis of the controllability problem in the non-bracket-generating cases was performed in 66].

9 Feedback and State Equivalence of Control Systems

9.1 Feedback Equivalence Consider control systems of the form q_ = f(q u) q 2 M u 2 U: (9.1) We suppose that not only M, but also U is a smooth manifold. For the righthand side, we suppose that for all xed u 2 U, f(q u) is a smooth vector eld on M, and, moreover, the mapping (u q) 7! f(q u) is smooth. Admissible controls are measurable locally bounded mappings t 7! u(t) 2 U (for simplicity, one can consider piecewise continuous controls). If such a control u(t) is substituted to control system (9:1), one obtains a nonautonomous ODE q_ = f(q u(t)) (9.2) with the right-hand side smooth in q and measurable, locally bounded in t. For such ODEs, there holds a standard theorem on existence and uniqueness of solutions, at least local. Solutions q( ) to ODEs (9:2) are Lipschitzian curves in M (see Subsect. 2.4.1). In Sect. 5.7 we already considered state transformations of control systems, i.e., dieomorphisms of M. State transformations map trajectories of control systems to trajectories, with the same control. Now we introduce a new class of feedback transformations, which also map trajectories to trajectories, but possibly with a new control. Denote the space of new control parameters by Ub . We assume that it is a smooth manifold.

122

9 Feedback and State Equivalence of Control Systems

Denition 9.1. Let ' : M  Ub ! U be a smooth mapping. A transformation

of the form

f(q u) 7! f(q '(q ub)) q 2 M u 2 U ub 2 Ub  is called a feedback transformation. Remark 9.2. A feedback transformation reparametrizes control u in a way depending on q. It is easy to see that any admissible trajectory q( ) of the system q_ = f(q '(q ub)) corresponding to a control ub( ) is also admissible for the system q_ = f(q u) with the control u( ) = '(q( ) ub( )), but, in general, not vice versa. In order to consider feedback equivalence, we consider invertible feedback transformations with Ub = U 'jq U 2 Di U: Such mappings ' : M  U ! U generate feedback transformations f(q u) 7! f(q '(q u)): The corresponding control systems q_ = f(q u) and q_ = f(q '(q u)) are called feedback equivalent . Our aim is to simplify control systems with state and feedback transformations. Remark 9.3. In mathematical physics, feedback transformations are often called gauge transformations. Consider control-ane systems q_ = f(q) +

k X i=1

R

u = (u1 : : :  uk ) 2 k q 2 M:

uigi (q)

R R

(9.3)

To such systems, it is natural to apply control-ane feedback transformations: ' = ('1  : : :  'k ) : M  k ! k 'i (q u) = ci (q) +

k X j =1

dij (q)uj 

i = 1 : : :  k:

(9.4)

Our aim is to characterize control-ane systems (9:3) which are locally equivalent to linear controllable systems w.r.t. state and feedback transformations (9:4) and to classify them w.r.t. this class of transformations.

9.2 Linear Systems

9.2 Linear Systems First we consider linear controllable systems x_ = Ax +

k X i=1

R

R R

x 2 n u = (u1  : : :  uk) 2 k

ui bi 

123

(9.5)

where A is an n  n matrix and bi , i = 1 : : :  k, are vectors in n. We assume that the vectors b1 : : :  bk are linearly independent: dimspan(b1 : : :  bk ) = k: If this is not the case, we eliminate some bi's. We nd normal forms of linear systems w.r.t. linear state and feedback transformations. To linear systems (9:5) we apply feedback transformations which have the form (9:4) and, moreover, preserve the linear structure: ci (x) = hci  xi ci 2 n i = 1 : : :  k (9.6) dij (x) = dij 2  i j = 1 : : :  k: Denote by D : span(b1 : : :  bk ) ! span(b1  : : :  bk) the linear operator with the matrix (dij ) in the base b1 : : :  bk . Linear feedback transformations (9:4), (9:6) map the vector elds in the right-hand side of the linear system (9:5) as follows:

R R



(Ax b1 : : :  bk ) 7! Ax +

k X i=1

!

hci  xibi Db1 : : :  Dbk :

(9.7)

Such mapping should be invertible, so we assume that the operator D (or, equivalently, its matrix (dij )) is invertible. Linear state transformations act on linear systems as follows: ;  (Ax b1 : : :  bk) 7! CAC ;1x Cb1 : : :  Cbk  (9.8) where C : n ! n is an invertible linear operator. State equivalence of linear systems means that these systems have the same coordinate representation in suitably chosen bases in the state space n.

R R

R

9.2.1 Linear Systems with Scalar Control Consider a simple model linear control system | scalar high-order control: x(n) +

R

nX ;1 i=0

i x(i) = u

R R

u2  x2 

(9.9)

where 0 : : :  n;1 2 . We rewrite this system in the standard form in the variables xi = x(i;1), i = 1 : : :  n:

124

9 Feedback and State Equivalence of Control Systems

8x_ = x  > > < 1 2 > xn > :x_ n;1 = P n;1

R

R

u 2  x = (x1  : : :  xn) 2 n: (9.10)

x_ n = ; i=0 ixi+1 + u

P

;1 ixi+1 +u as a new control, i.e., apply It is easy to see that if we take ; ni=1 the feedback transformation (9:4), (9:6) with k = 1 c = (; 0  : : :  ; n;1) d = 1 then system (9:10) maps into the system

8x_ = x  > > < 1 2 > > :x_ n;1 = xn

R

R

u 2  x = (x1 : : :  xn) 2 n

x_ n = u

R R

which is written in the scalar form as x(n) = u

u2  x2 :

(9.11)

(9.12)

So system (9:10) is feedback equivalent to system (9:11). It turns out that the simple systems (9:10) and (9:11) are normal forms of linear controllable systems with scalar control under state transformations and state-feedback transformations respectively.

Proposition 9.4. Any linear controllable system with scalar control x_ = Ax + ub u 2  x 2 n (9.13) n span(b Ab : : :  A ;1b) = n (9.14)

RRR

is state equivalent to a system of the form (9:10), thus state-feedback equivalent to system (9:11).

R

R

Proof. We nd a basis e1  : : :  en in n in which system (9:13) is written in

the form (9:10). Coordinates y1  : : :  yn of a point x 2 n in a basis e1  : : :  en are found from the decomposition x=

n X i=1

yi ei :

In view of the desired form (9:10), the vector b should have coordinates b = (0 : : :  0 1), thus the n-th basis vector is uniquely determined: en = b:

9.2 Linear Systems

125

Now we nd the rest basis vectors e1  : : :  en;1. We can rewrite our linear system (9:13) as follows: x_ = Ax mod b then we obtain in coordinates:

R

x_ = thus

nX ;1 i=1

n X i=1

n X

y_i ei =

y_i ei =

nX ;1 i=0

i=1

R R

yi Aei mod b

yi+1 Aei+1 mod b:

The required dierential equations: y_i = yi+1  i = 1 : : :  n ; 1 are fullled in a basis e1 : : :  en if and only if the following equalities hold: Aei+1 = ei + i b i = 1 : : :  n ; 1 (9.15) Ae1 = 0 b (9.16) for some numbers 0  : : :  n;1 2 . So it remains to show that we can nd basis vectors e1  : : :  en;1 which satisfy equalities (9:15), (9:16). We rewrite equality (9:15) in the form ei = Aei+1 ; i b i = 1 : : :  n ; 1 (9.17) and obtain recursively: en = b en;1 = Ab ; n;1 b en;2 = A2b ; n;1 Ab ; n;2 b (9.18)

R

e1 = An;1b ; n;1 An;2b ; ; 1 b: So equality (9:16) yields Ae1 = An b ; n;1 An;1b ; ; 1 Ab = 0 b: The equality

R

An b =

nX ;1 i=0

i Ai b

(9.19)

is satised for a unique n-tuple (0  : : :  n;1) since the vectors b, Ab, : : : , An;1b form a basis of n (in fact, i are coecients of the characteristic polynomial of A). With these numbers i , the vectors e1  : : :  en given by (9:18) form the required basis. Indeed, equalities (9:15), (9:16) hold by construction. The vectors e1 : : :  en are linearly independent by the controllability condition (9:14).

ut

126

9 Feedback and State Equivalence of Control Systems

Remark 9.5. The basis e1  : : :  en constructed in the previous proof is unique,

thus the state transformation that maps a controllable linear system with scalar control (9:13) to the normal form (9:10) is also unique.

9.2.2 Linear Systems with Vector Control Now consider general controllable linear systems: x_ = Ax +

k X i=1

uibi 

R

R

x 2 n u = (u1 : : :  uk ) 2 k

R

spanfAj bi j j = 0 : : :  n ; 1 i = 1 : : :  kg = n:

R

(9.20) (9.21)

Recall that we assume vectors b1  : : :  bk linearly independent. In the case k = 1, all controllable linear systems in n are state-feedback equivalent to the normal form (9:11), thus there are no state-feedback invariants in a given dimension n. If k > 1, this is not the case, and we start from description of state-feedback invariants.

Kronecker Indices

R

Consider the following subspaces in n: Dm = spanfAj bi j j = 0 : : :  m ; 1 i = 1 : : :  kg

m = 1 : : :  n: (9.22)

Invertible linear state transformations (9:8) preserve dimension of these subspaces, thus the numbers dimDm 

m = 1 : : :  n

are state invariants. Now we show that invertible linear feedback transformations (9:7) preserve the spaces Dm . Any such transformation can be decomposed into two feedback transformations of the form: (Ax b1 : : :  bk) 7! (Ax +

k X i=1

hci  xibi b1 : : :  bk )

(Ax b1 : : :  bk) 7! (Ax Db1  : : :  Dbk ):

(9.23) (9.24)

Transformations (9:24), i.e., changes of bi , obviously preserve the spaces Dm . Consider transformations (9:23). Denote the new matrix:

b = Ax + Ax

k X i=1

hci  xibi:

9.2 Linear Systems

127

We have:

Abj x = Aj x mod Dj  j = 1 : : :  n ; 1: But Dm;1 Dm , m = 2 : : :  n, thus feedback transformations (9:23) preserve the spaces Dm , m = 1 : : :  n. So the spaces Dm , m = 1 : : :  n, are invariant under feedback transformations, and their dimensions are state-feedback invariants. Now we express the numbers dimDm , m = 1 : : :  n, through other integers | Kronecker indices. Construct the following nk matrix whose elements are n-dimensional vectors: 0 b b 1 BB Ab11 Abkk CC (9.25) [email protected] ... ... ... CA : An;1b1 An;1bk Replace each vector Aj bi, j = 0 : : :  n ; 1, i = 1 : : :  k, in this matrix by a sign: cross  or circle , by the following rule. We go in matrix (9:25) by rows, i.e., order its elements as follows: b1  : : :  bk  Ab1 : : :  Abk  : : :  An;1b1 : : :  An;1bk : (9.26) A vector Aj bi in matrix (9:25) is replaced by  if it is linearly independent of the previous vectors in chain (9:26), otherwise it is replaced by . After this procedure we obtain a matrix of the form:

0    1 BB      CC =B BB . . . . . . CCC : @ .. .. .. .. .. .. A     

Notice that there are some restrictions on appearance of crosses and circles in matrix . The total number of crosses in this matrix is n (by the controllability condition (9:21)), and the rst row is lled only with crosses (since b1  : : :  bk are linearly independent). Further, if a column of  contains a circle, then all elements below it are circles as well. Indeed, if a vector Aj bi in (9:25) is replaced by circle in , then Aj bi 2 spanfAj b j  < ig + spanfA b j  < j  = 1 : : :  kg: Then the similar inclusions hold for all vectors Aj +1 bi : : :  An;1bi, i.e., below circles are only circles. So each column in the matrix  consists of a column of crosses over a column of circles (the column of circles can be absent). Denote by n1 the height of the highest column of crosses in the matrix , by n2 the height of the next highest column of crosses, : : : , and by nk the height of the lowest column of crosses in . The positive integers obtained:

128

9 Feedback and State Equivalence of Control Systems

n1  n2   nk are called Kronecker indices of the linear control system (9:20). Since the total number of crosses in matrix  is equal to dimension of the state space, then k X i=1

ni = n:

Moreover, by the construction, we have

R

span(b1 Ab1 : : :  An1;1 b1  : : :  bk  Abk  : : :  Ank ;1bk ) = n:

(9.27)

Now we show that Kronecker indices ni are expressed through the numbers dimDi . We have: dimD1 = k = number of crosses in the rst row of  dimD2 = number of crosses in the rst 2 rows of 

dimDi = number of crosses in the rst i rows of  so that (i) def = dimDi ; dimDi;1 = number of crosses in the i-th row of : Permute columns in matrix , so that the rst column become the highest one, the second column becomes the next highest one, etc. We obtain an n  kmatrix in the \block-triangular" form. This matrix rotated at the angle =2 gives the subgraph of the function  : f1 : : :  ng ! f1 : : :  kg. It is easy to see that the values of the Kronecker indices is equal to the points of jumps of the function , and the number of Kronecker indices for each value is equal to the height of the corresponding jump of . So Kronecker indices are expressed through dimDi , i = 1 : : :  k, thus are state-feedback invariants.

Brunovsky Normal Form Now we nd normal forms of linear systems under state and state-feedback transformations. In particular, we show that Kronecker indices form a complete set of state-feedback invariants of linear systems. Theorem 9.6. Any controllable linear system (9:20), (9:21) with k control parameters is state equivalent to a system of the form

9.2 Linear Systems

129

8y_1 = y1  8y_k = yk  > > 1 2 1 2 > > > > : : : : : :

j 1 k 1 > > y_n = ; ij yi+1 + u1  y_nk = ; kij yij+1 + uk  > > : : 1j k 1j k 0in ;1 0in ;1 1

1

1

j

j

where

x=

X 1ik 1j ni

yji eij 

(9.28)

(9.29)

and state-feedback equivalent to a system of the form

8 (n ) >

:yk(nk ) = uk 1

(9.30)

where ni, i = 1 : : :  k, are Kronecker indices of system (9:20). System (9:30) is called the Brunovsky normal form of the linear sys-

tem (9:20). We prove Theorem 9.6. Proof. We show rst that any linear controllable system (9:20) can be written, in a suitable basis in n: e11  : : :  e1n1  : : :   ek1  : : :  eknk (9.31) in the canonical form (9:28). We proceed exactly as in the scalar-input case (Subsect. 9.2.1). The required canonical form (9:28) determines uniquely the last basis vectors in all k groups: e1n1 = b1  : : :  eknk = bk : (9.32) Denote the space B = span(b1 : : :  bk). Then our system x_ = Ax mod B reads in coordinates as follows: X ii X i i x_ = y_j ej = yj Aej mod B:

R

1ik 1j ni

In view of the required equations

1ik 1j ni

130

9 Feedback and State Equivalence of Control Systems

y_ji = yji +1 

X

we have or, equivalently,

1ik 1j 0. Remark 10.4. The condition of convexity of the velocity sets fU (q) is natural in view of Theorem 8.7: the ow of the ODE q_ = (t)fu1 (q) + (1 ; (t))fu2 (q) 0  (t)  1 can be approximated by ows of the systems of the form q_ = fv (q) where v(t) 2 fu1(t) u2 (t)g: Now we give a sketch of the proof of Theorem 10.3. Proof. Notice rst of all that all nonautonomous vector elds fu (q) with admissible controls u have a common compact support, thus are complete. Further, under hypotheses of the theorem, velocities fu (q), q 2 M, u 2 U, are uniformly bounded, thus all trajectories q(t) of control system (10:1) starting at q0 are Lipschitzian with the same Lipschitz constant. Thus the set of admissible trajectories is precompact in the topology of uniform convergence. (We can embed the manifold M into a Euclidean space N , then the space of continuous curves q(t) becomes endowed with the uniform topology of continuous mappings from 0 t1] to N .) For any sequence qn(t) of admissible trajectories: q_n(t) = fun (qn(t)) 0  t  t1  qn(0) = q0  there exists a uniformly converging subsequence, we denote it again by qn(t): qn( ) ! q( ) in C0 t1] as n ! 1: Now we show that q(t) is an admissible trajectory of control system (10:1). Fix a suciently small " > 0. Then in local coordinates 1 (q (t + ") ; q (t)) = 1 Z t+" f (q ( )) d n " n " t un n  2 conv fU (qn( )) conv fU (q)

R

R

 2 tt+"]

R

q2Oq(t) (c")

142

10 Optimal Control Problem

where c is the doubled Lipschitz constant of admissible trajectories. Then we pass to the limit n ! 1 and obtain 1 (q(t + ") ; q(t)) 2 conv  f (q): U " q2Oq(t) (c")

Now let " ! 0. If t is a point of dierentiability of q(t), then q(t) _ 2 fU (q) since fU (q) is convex. In order to show that q(t) is an admissible trajectory of control system (10:1), we should nd a measurable selection u(t) 2 U that generates q(t). We do this via the lexicographic order on the set U = f(u1  : : :  um )g m. The set Vt = fv 2 U j q(t) _ = fv (q(t))g m is a compact subset of U, thus of . There exists a vector vmin (t) 2 Vt minimal in the sense of lexicographic order. To nd vmin (t), we minimize the rst coordinate on Vt : v1min = minf v1 j v = (v1  : : :  vm ) 2 Vt g then minimize the second coordinate on the compact set found at the rst step: v2min = minf v2 j v = (v1min  v2 : : :  vm ) 2 Vt g etc., vmmin = minf vm j v = (v1min  : : :  vmmin;1  vm ) 2 Vt g: The control vmin (t) = (v1min (t) : : :  vmmin (t)) is measurable, thus q(t) is an admissible trajectory of system (10:1) generated by this control. The proof of compactness of the attainable set Aq0 (t) is complete. Compactness of Atq0 is proved by a slightly modied argument. ut Remark 10.5. In Filippov's theorem, the hypothesis of common compact support of the vector elds in the right-hand side is essential to ensure the uniform boundedness of velocities and completeness of vector elds. On a manifold, sucient conditions for completeness of a vector eld cannot be given in terms of boundedness of the vector eld and its derivatives: a constant vector eld is not complete on a bounded domain in n. Nevertheless, one can prove compactness of attainable sets for many systems without the assumption of common compact support. If for such a system we have a priori bounds on solutions, then we can multiply its right-hand side by a cut-o function, and obtain a system with vector elds having compact support. We can apply Filippov's theorem to the new system. Since trajectories of the initial and new systems coincide in a domain of interest for us, we obtain a conclusion on compactness of attainable sets for the initial system.

R

R

R

R

10.5 Relaxations

143

For control systems on M = n, there exist well-known sucient conditions for completeness of vector elds: if the right-hand side grows at innity not faster than a linear eld, i.e., jfu (x)j  C(1 + jxj) x 2 n u 2 U (10.13) for some constant p C, then the nonautonomous vector elds fu(x) are complete (here jxj = x21 + + x2n is the norm of a point x = (x1 : : :  xn) 2 n). These conditions provide an a priori bound for solutions: any solution x(t) of the control system x_ = fu (x) x 2 n u 2 U (10.14) with the right-hand side satisfying (10:13) admits the bound jx(t)j  e2Ct (jx(0)j + 1)  t  0: So Filippov's theorem plus the previous remark imply the following sucient condition for compactness of attainable sets for systems in n. Corollary 10.6. Let system (10:14) have a compact space of control parameters U b m and convex velocity sets fU (x), x 2 n. Suppose moreover that the right-hand side of the system satis es a bound of the form (10:13). Then the attainable sets Ax0 (t) and Atx0 are compact for all x0 2 n, t > 0.

R

R

R

R

R

R

R

10.4 Time-Optimal Problem Given a pair of points q0 2 M and q1 2 Aq0 , the time-optimal problem consists in minimizing the time of motion from q0 to q1 via admissible controls of control system (10:1): min ft1 j qu (t1) = q1g: (10.15) u

That is, we consider the optimal control problem described in Sect. 10.1 with the integrand '(q u)  1 and free terminal time t1 . Reduction of optimal control problems to the study of attainable sets and Filippov's Theorem yield the following existence result. Corollary 10.7. Under the hypotheses of Theorem 10:3, time-optimal problem (10:1), (10:15) has a solution for any points q0 2 M , q1 2 Aq0 .

10.5 Relaxations Consider a control system of the form (10:1) with a compact set of control parameters U. There is a standard procedure called relaxation of control system (10:1), which extends the velocity set fU (q) of this system to its convex hull conv fU (q).

144

10 Optimal Control Problem

Recall that the convex hull conv S of a subset S of a linear space is the minimal convex set that contains S. A constructive description of convex hull is given by the following classical proposition: any point in the convex hull of a set S in the n-dimensional linear space is contained in the convex hull of some n + 1 points in S. Lemma 10.8 (Caratheodory). For any subset S n, its convex hull has the form

conv S =

(X n i=0

R

ixi j xi 2 S i  0

n X i=0

)

i = 1 :

For the proof of this lemma, one can consult e.g. 143]. Relaxation of control system (10:1) is constructed as follows. Let n = dimM be dimension of the state space. The set of control parameters of the relaxed system is V = n  U|  {z  U} where

(

n+1 times

n = ( 0 : : :  n) j i  0

n X i=0

i = 1

)

R

n+1

is the standard n-dimensional simplex. So the control parameter of the new system has the form v = (  u0 : : :  un) 2 V

= ( 0  : : :  n) 2 n ui 2 U:

If U is compact, then V is compact as well. The relaxed system is q_ = gv (q) =

n X i=0

i fui (q)

v = (  u0 : : :  un) 2 V q 2 M:

(10.16)

By Carath"eodory's lemma, the velocity set gV (q) of system (10:16) is convex, moreover, gV (q) = conv fU (q): If all vector elds in the right-hand side of (10:16) have a common compact support, we obtain by Filippov's theorem that attainable sets for the relaxed system are compact. By Theorem 8.7, any trajectory of relaxed system (10:16) can be uniformly approximated by families of trajectories of initial system (10:1). Thus attainable sets of the relaxed system coincide with closure of attainable sets of the initial system.

11 Elements of Exterior Calculus and Symplectic Geometry

In order to state necessary conditions of optimality for optimal control problems on smooth manifolds | Pontryagin Maximum Principle, see Chap. 12 | we make use of some standard technique of Symplectic Geometry. In this chapter we develop such a technique. Before this we recall some basic facts on calculus of exterior dierential forms on manifolds. The exposition in this chapter is rather explanatory than systematic, it is not a substitute to a regular textbook. For a detailed treatment of the subject, see e.g. 146], 135], 137].

11.1 Dierential 1-Forms 11.1.1 Linear Forms

R

Let E be a real vector space of nite dimension n. The set of linear forms on E, i.e., of linear mappings : E ! , has a natural structure of a vector space called the dual space to E and denoted by E  . If vectors e1  : : :  en form a basis of E, then the corresponding dual basis of E  is formed by the covectors e1  : : :  en such that

hei  ej i = ij 

i j = 1 : : :n

(we use the angle brackets to denote the value of a linear form 2 E  on a vector v 2 E: h  vi = (v)). So the dual space has the same dimension as the initial one: dimE  = n = dim E:

11.1.2 Cotangent Bundle Let M be a smooth manifold and Tq M its tangent space at a point q 2 M. The space of linear forms on Tq M, i.e., the dual space (Tq M) to Tq M, is

146

11 Elements of Exterior Calculus and Symplectic Geometry

called the cotangent space to M at q and is denoted as Tq M. The disjoint union of all cotangent spaces is called the cotangent bundle of M: T  M def =

  Tq M:

q 2M

The set T  M has a natural structure of a smooth manifold of dimension 2n, where n = dimM. Local coordinates on T  M are constructed from local coordinates on M. Let O M be a coordinate neighborhood and let  : O ! n (q) = (x1 (q) : : :  xn(q)) be a local coordinate system. Dierentials of the coordinate functions dxijq 2 Tq M i = 1 : : :  n q 2 O form a basis in the cotangent space Tq M. The dual basis in the tangent space Tq M is formed by the vectors @ 2 T M i = 1 : : :  n q 2 O q @x  [email protected]  dxi @ x  ij  i j = 1 : : :  n: j  Any linear form 2 Tq M can be decomposed via the basis forms:

R

=

n X i=1

i dxi:

So any covector 2 T  M is characterized by n coordinates (x1 : : :  xn) of the point q 2 M where is attached, and by n coordinates ( 1  : : :  n) of the linear form in the basis dx1 : : :  dxn. Mappings of the form

7! ( 1  : : :  n x1  : : :  xn) dene local coordinates on the cotangent bundle. Consequently, T  M is an 2n-dimensional manifold. Coordinates of the form (  x) are called canonical coordinates on T  M. If F : M ! N is a smooth mapping between smooth manifolds, then the dierential F : Tq M ! TF (q) N has the adjoint mapping F  def = (F ) : TF (q) N ! Tq M

dened as follows:

11.2 Di erential k-Forms

147

F  =  F 2 TF (q) N hF   vi = h  Fvi v 2 Tq M: A vector v 2 Tq M is pushed forward by the dierential F to the vector Fv 2 TF (q) N, while a covector 2 TF (q) N is pulled back to the covector F  2 Tq M. So a smooth mapping F : M ! N between manifolds induces a smooth mapping F  : T  N ! T  M between their cotangent bundles.

11.1.3 Dierential 1-Forms

A di erential 1-form on M is a smooth mapping q 7! !q 2 Tq M q 2 M i.e, a family ! = f!q g of linear forms on the tangent spaces Tq M smoothly depending on the point q 2 M. The set of all dierential 1-forms on M has a natural structure of an innite-dimensional vector space denoted as 1 M. Like linear forms on a vector space are dual objects to vectors of the space, dierential forms on a manifold are dual objects to smooth curves in the manifold. The pairing operation is the integral of a dierential 1-form ! 2 1M along a smooth oriented curve  : t0 t1] ! M, dened as follows:

Z



! def =

Zt

1

t0

h! (t)  _ (t)i dt:

The integral of a 1-form along a curve does not change under orientationpreserving smooth reparametrizations of the curve and changes its sign under change of orientation.

11.2 Dierential k-Forms A dierential k-form on M is an object to integrate over k-dimensional surfaces in M. Innitesimally, a k-dimensional surface is presented by its tangent space, i.e., a k-dimensional subspace in Tq M. We thus need a dual object to the set of k-dimensional subspaces in the linear space. Fix a linear space E. A k-dimensional subspace is dened by its basis v1  : : :  vk 2 E. The dual objects should be mappings (v1  : : :  vk ) 7! !(v1  : : :  vk ) 2 such that !(v1  : : :  vk ) depend only on the linear hull spanfv1  : : :  vk g and the oriented volume of the k-dimensional parallelepiped generated by v1 : : :  vk . Moreover, the dependence on the volume should be linear. Recall that ratio of volumes of the parallelepipeds generated by vectors wi = Pk the v ij j , i = 1 : : :  k, and the vectors v1  : : :  vk , equals det( ij )kij =1, and j =1 that determinant of a k  k matrix is a multilinear skew-symmetric form of the columns of the matrix. This is why the following denition of the \dual objects" is quite natural.

R

148

11 Elements of Exterior Calculus and Symplectic Geometry

11.2.1 Exterior k-Forms

N

Let E be a nite-dimensional real vector space, dimE = n, and let k 2 . An exterior k-form on E is a mapping

R

!:E |  {z  E} !  k times

which is multilinear: !(v1  : : :  1vi1 + 2vi2  : : :  vk ) = 1!(v1  : : :  vi1 : : :  vk ) + 2!(v1  : : :  vi2  : : :  vk )

R

1 2 2 

and skew-symmetric:

!(v1  : : :  vi : : :  vj  : : :  vk ) = ;!(v1  : : :  vj  : : :  vi  : : :  vk ) i j = 1 : : :  k: The set of all exterior k-forms on E is denoted by k E. By the skew-symmetry, any exterior form of order k > n is zero, thus k E = f0g for k > n. Exterior forms can be multiplied by real numbers, and exterior forms of the same order k can be added one with another, so each k E is a vector space. We construct a basis of k E after we consider another operation between exterior forms | the exterior product. The exterior product of two forms !1 2 k1 E, !2 2 k2 E is an exterior form !1 ^ !2 of order k1 + k2. Given linear 1-forms !1 !2 2 1E, we have a natural (tensor) product for them: v1  v2 2 E: !1 !2 : (v1  v2) 7! !1 (v1 )!2(v2 ) The result is a bilinear but not a skew-symmetric form. The exterior product is the anti-symmetrization of the tensor one: !1 ^ !2 : (v1  v2 ) 7! !1 (v1 )!2(v2 ) ; !1(v2 )!2 (v1 )

v1  v2 2 E:

Similarly, the tensor and exterior products of forms !1 2 k1 E and !2 2 k2 E are the following forms of order k1 + k2: !1 !2 : (v1  : : :  vk1+k2 ) 7! !1 (v1  : : :  vk1 )!2(vk1 +1  : : :  vk1 +k2 ) !1 ^ !2 : (v1  : : :  vk1+k2 ) 7! 1 X(;1) () ! (v  : : :  v )! (v 1 (1) (k1 ) 2 (k1 +1)  : : :  v(k1 +k2 ) ) (11.1) k1! k2!  where the sum is taken over all permutations  of order k1 + k2 and () is parity of a permutation . The factor k1 !1k2 ! normalizes the sum in (11:1) since it contains k1! k2! identically equal terms: e.g., if permutations  do not mix the rst k1 and the last k2 arguments, then all terms of the form

11.2 Di erential k-Forms

149

(;1) () !1(v(1)  : : :  v(k1) )!2(v(k1 +1)  : : :  v(k1 +k2 ) ) are equal to !1 (v1  : : :  vk1 )!2(vk1 +1  : : :  vk1 +k2 ): This guarantees the associative property of the exterior product: !i 2 ki E

!1 ^ (!2 ^ !3) = (!1 ^ !2 ) ^ !3 Further, the exterior product is skew-commutative: !2 ^ !1 = (;1)k1 k2 !1 ^ !2

!i 2 ki E:

Let e1  : : :  en be a basis of the space E and e1 ;: ::  en the corresponding dual basis of E  . If 1  k  n, then the following nk elements form a basis of the space k E: ei1 ^ : : : ^ eik 

1  i1 < i2 < < ik  n:

The equalities

(ei1 ^ : : : ^ eik )(ei1  : : :  eik ) = 1 (ei1 ^ : : : ^ eik )(ej1  : : :  ejk ) = 0 if (i1  : : :  ik ) 6= (j1  : : :  jk )

for 1  i1 < i2 < < ik  n imply that any k-form ! 2 k E has a unique decomposition of the form !=

X

1i1 0 and " 2 (0 t1). Then there exists a Lipschitzian curve t 2 Tq~(t)M t 6= 0

0  t  t1 

such that

_ t = ~hu~(t) (t ) hu~(t) (t ) = max h ( ) u2U u t hu~(t) (t ) = 0

(12.15)

for almost all t 2 0 t1]. Remark 12.9. In problems with free time, there appears one more variable,

the terminal time t1 . In order to eliminate it, we have one additional condition | equality (12:15). This condition is indeed scalar since the previous two equalities imply that hu~(t)(t ) = const, see remark after formulation of Theorem 12.1. Proof. We reduce the case of free time to the case of xed time by extension of the control system via substitution of time. Admissible trajectories of the extended system are reparametrized admissible trajectories of the initial system (the positive direction of time on trajectories is preserved). Let a new time be a smooth function

178

R! R

12 Pontryagin Maximum Principle

':

 '_ > 0:

We nd an ODE for a reparametrized trajectory: d q ('(t)) = '(t)f _ u('(t)) (qu('(t))) dt u so the required equation is q_ = '(t)f _ u('(t)) (q): Now consider along with the initial control system q_ = fu (q)

u 2 U

an extended system of the form q_ = vfu (q)

u 2 U jv ; 1j < 

(12.16)

where  = "=t1 2 (0 1). Admissible controls of the new system are w(t) = (v(t) u(t)) and the reference control corresponding to the control u~( ) of the initial system is w(t) ~ = (1 u~(t)): ;  It is easy to see that since q~(t1 ) 2 @ jt;t1j >

> : _1 = 0

2 = ; 1 : For the variable x1 we obtain the boundary value problem x(4) 1 = 0 x1 (0) = x01 x_ 1 (0) = x02 x1 (t1) = 0 x_ 1(t1 ) = 0: (13.19) For any (x01 x02), there exists exactly one solution x1 (t) of this problem | a cubic spline. The function x2 (t) is found from the equation x2 = x_ 1. So through any initial point x0 2 2 passes a unique extremal trajectory arriving at the origin. It is a curve (x1 (t) x2(t)), t 2 0 t1], where x1(t) is a cubic polynomial that satises the boundary conditions (13:19), and x2(t) = x_ 1(t). In view of existence, this is an optimal trajectory.

R

13.4 Control of a Linear Oscillator with Cost We control a linear oscillator, say a pendulum with a small amplitude, by an unbounded force R u, but take into account expenditure of energy measured by the integral 21 0t1 u2(t) dt. The optimal control problem reads

(

x_ 1 = x2 x_ 2 = ;x1 + u

x 

x = x1 2 2

x(0) = x0 x(t1 ) = 0 t1 xed 1 Z t1 u2 dt ! min: 2 0

R 2R 2

u



200

13 Examples of Optimal Control Problems

Existence of optimal control can be proved by the same argument as in the previous section. The Hamiltonian function of PMP is hu (  x) = 1x2 ; 2 x1 + 2 u + 2 u2: The corresponding Hamiltonian system yields (_

1 = 2 

_2 = ; 1 :

In the same way as in the previous problem, we show that there are no abnormal extremals, thus we can assume  = ;1. Then the maximality condition yields u~(t) = 2 (t): In particular, optimal control is a harmonic: u~(t) = sin(t + )   = const : The system of ODEs for extremal trajectories

(

x_ 1 = x2  x_ 2 = ;x1 + sin(t + )

is solved explicitly: x1(t) = ; 2 t cos(t + ) + a sin(t + b) x2(t) = 2 t sin(t + ) ; 2 cos(t + ) + a cos(t + b) a b 2 :

R

(13.20)

Exercise 13.1. Show that exactly one extremal trajectory of the form (13:20) satises the boundary conditions. In view of existence, these extremal trajectories are optimal.

13.5 Dubins Car In this section we study a time-optimal problem for a system called Dubins car , see equations (13:21) below. This system was rst considered by A.A. Markov back in 1887 109]. Consider a car moving in the plane. The car can move forward with a xed linear velocity and simultaneously rotate with a bounded angular velocity. Given initial and terminal position and orientation of the car in the plane, the problem is to drive the car from the initial conguration to the terminal one for a minimal time.

13.5 Dubins Car

201

Admissible paths of the car are curves with bounded curvature. Suppose that curves are parametrized by length, then our problem can be stated geometrically. Given two points in the plane and two unit velocity vectors attached respectively at these points, one has to nd a curve in the plane that starts at the rst point with the rst velocity vector and comes to the second point with the second velocity vector, has curvature bounded by a given constant, and has the minimal length among all such curves. Remark 13.2. If curvature is unbounded, then the problem, in general, has no solutions. Indeed, the inmum of lengths of all curves that satisfy the boundary conditions without bound on curvature is the distance between the initial and terminal points: the segment of the straight line through these points can be approximated by smooth curves with the required boundary conditions. But this inmum is not attained when the boundary velocity vectors do not lie on the line through the boundary points and are not collinear one to another. After rescaling, we obtain a time-optimal problem for a nonlinear system:

8 >

: _ 2= u

R

(13.21)

x = (x1 x2) 2 2 2 S 1  juj  1 x(0) (0) x(t1) (t1 ) xed t1 ! min: Existence of solutions is guaranteed by Filippov's Theorem. We apply Pontryagin Maximum Principle. We have (x1 x2 ) 2 M = 2x  S1 , let ( 1  2 ) be the corresponding coordinates of the adjoint vector. Then  = (x   ) 2 T  M

R

and the control-dependent Hamiltonian is hu() = 1 cos + 2 sin + u: The Hamiltonian system of PMP yields

_ = 0 _ = 1 sin ; 2 cos 

(13.22) (13.23)

and the maximality condition reads (t)u(t) = jmax (t)u: uj1

(13.24)

202

13 Examples of Optimal Control Problems

Equation (13:22) means that is constant along optimal trajectories, thus the right-hand side of (13:23) can be rewritten as

1 sin ; 2 cos = sin( + )

q

  = const = 12 + 22  0: (13.25)

So the Hamiltonian system of PMP (13:21){(13:23) yields the following system: ( _ = sin( + )

_ = u: Maximality condition (13:24) implies that u(t) = sgn (t) if (t) 6= 0:

(13.26)

If = 0, then ( 1  2)  0 and  = const 6= 0, thus u = const = 1. So the curve x(t) is an arc of a circle of radius 1. Let 6= 0, then in view of (13:25), we have > 0. Conditions (13:22), (13:23), (13:24) are preserved if the adjoint vector (  ) is multiplied p by any positive constant. Thus we can choose (  ) such that = 12 + 22 = 1. That is why we suppose in the sequel that = 1: Condition (13:26) means that behavior of sign of the function (t) is crucial for the structure of optimal control. We consider several possibilities for (t). (0) If the function (t) does not vanish on the segment 0 t1], then the optimal control is constant: u(t) = const = 1 t 2 0 t1] (13.27) and the optimal trajectory x(t), t 2 0 t1], is an arc of a circle. Notice that an optimal trajectory cannot contain a full circle: a circle can be eliminated so that the resulting trajectory satisfy the same boundary conditions and is shorter. Thus controls (13:27) can be optimal only if t1 < 2. In the sequel we can assume that the set N = f 2 0 t1] j ( ) 6= 0g does not coincide with the whole segment 0 t1]. Since N is open, it is a union of open intervals in 0 t1], plus, may be, semiopen intervals of the form 0 1), ( 2 t1]. (1) Suppose that the set N contains an interval of the form ( 1  2) 0 t1] 1 < 2 : (13.28) We can assume that the interval ( 1  2) is maximal w.r.t. inclusion:

13.5 Dubins Car

203

( 1 ) = ( 2 ) = 0 j(1 2 ) 6= 0: From PMP we have the inequality hu(t)((t)) = cos( (t) + ) + (t)u(t)  0: Thus cos( ( 1 ) + )  0: This inequality means that the angle

b = ( 1 ) +  satises the inclusion  h i 

b 2 0 2  3  2 : 2

 i

b 2 0 2 : Then ( _ 1 ) = sin b > 0, thus at 1 control switches from ;1 to +1, so _ = u(t)  1 t 2 ( 1  2):

(t) We evaluate the distance 2 ; 1 . Since Consider rst the case

( 2 ) = then 2 ; 1 = 2( ; b), thus

Z

2

1

sin( b + ; 1) d = 0

2 ; 1 2  2):

(13.29)

 

b 2 3  2 2 inclusion (13:29) is proved similarly,and in the case b = 0 we obtain no optimal controls (the curve x(t) contains a full circle, which can be eliminated). Inclusion (13:29) means that successive roots 1 , 2 of the function (t) cannot be arbitrarily close one to another. Moreover, the previous argument shows that at such instants i optimal control switches from one extremal value to another, and along any optimal trajectory the distance between any successive switchings i , i+1 is the same. So in case (1) an optimal control can only have the form  " t 2 ( 2k;1 2k) (13.30) u(t) = ;" t 2 ( 2k  2k+1) " = 1 i+1 ; i = const 2  2) i = 1 : : :  n ; 1 (13.31) 1 2 (0 2) In the case

204

13 Examples of Optimal Control Problems

here we do not indicate values of u in the intervals before the rst switching, t 2 (0 1), and after the last switching, t 2 ( n  t1). For such trajectories, control takes only extremal values 1 and the number of switchings is nite on any compact time segment. Such a control is called bang-bang . Controls u(t) given by (13:30), (13:31) satisfy PMP for arbitrarily large t, but they are not optimal if the number of switchings is n > 3. Indeed, suppose that such a control has at least 4 switchings. Then the piece of trajectory x(t), t 2  1  4], is a concatenation of three arcs of circles corresponding to the segments of time  1  2],  2 3],  3 4 ] with 4 ; 3 = 3 ; 2 = 2 ; 1 2  2): Draw the segment of line x~(t) t 2 ( 1 + 2 )=2 ( 3 + 4)=2]  dd x~t  1 the common tangent to the rst and third circles through the points x (( 1 + 2 )=2) and x (( 3 + 4 )=2) see Fig. 13.3. Then the curve  x(t) t 2= ( + )=2 ( + )=2]  1 2 3 4 y(t) = x~(t) t 2 ( 1 + 2 )=2 ( 3 + 4)=2]  is an admissible trajectory and shorter than x(t). We proved that optimal bang-bang control can have not more than 3 switchings. x(t)

x ~(t)

Fig. 13.3. Elimination of 4 switchings (2) It remains to consider the case where the set N does not contain intervals of the form (13:28). Then N consists of at most two semiopen intervals:

13.5 Dubins Car

205

N = 0 1)  ( 2  t1] 1  2  where one or both intervals may be absent. If 1 = 2 , then the function (t) has a unique root on the segment 0 t1], and the corresponding optimal control is determined by condition (13:26). Otherwise 1 < 2  and j 01 ) 6= 0 j 1 2 ]  0 j(2 t1] 6= 0: (13.32) In this case the maximality condition of PMP (13:26) does not determine optimal control u(t) uniquely since the maximum is attained for more than one value of control parameter u. Such a control is called singular . Nevertheless, singular controls in this problem can be determined from PMP. Indeed, the following identities hold on the interval ( 1  2): _ = sin( + ) = 0 ) +  = k ) = const ) u = 0: Consequently, if an optimal trajectory x(t) has a singular piece, which is a line, then 1 and 2 are the only switching times of the optimal control. Then uj(01) = const = 1 uj(2 t1) = const = 1 and the whole trajectory x(t), t 2 0 t1], is a concatenation of an arc of a circle of radius 1 t 2 0 1] x(t) u(t) = 1 a line x(t) u(t) = 0 t 2  1  2] and one more arc of a circle of radius 1 x(t) u(t) = 1 t 2  2 t1]: So optimal trajectories in the problem have one of the following two types: (1) concatenation of a bang-bang piece (arc of a circle, u = 1), a singular piece (segment of a line, u = 0), and a bang-bang piece, or (2) concatenation of bang-bang pieces with not more than 3 switchings, the arcs of circles between switchings having the same central angle 2  2). If boundary points x(0), x(t1 ) are suciently far one from another, then they can be connected only by trajectories containing singular piece. For such boundary points, we obtain a simple algorithm for construction of an optimal trajectory. Through each of the points x(0) and x(t1), construct a pair of circles of radius 1 tangent respectively to the velocity vectors x(0) _ = (cos (0) sin (0)) and x(t _ 1 ) = (cos (t1 ) sin (t1 )). Then draw common tangents to the circles at x(0) and x(t1 ) respectively, so that direction of motion along these tangents was compatible with direction of rotation along the circles determined by the boundary velocity vectors x(0) _ and x(t _ 1 ), see Fig. 13.4. Finally, choose the shortest curve among the candidates obtained. This curve is the optimal trajectory.

206

13 Examples of Optimal Control Problems x _ (t1 ) x _ (0) x(t1 ) x(0)

Fig. 13.4. Construction of the shortest motion for far boundary points

14 Hamiltonian Systems with Convex Hamiltonians

A well-known theorem states that if a level surface of a Hamiltonian is convex, then it contains a periodic trajectory of the Hamiltonian system 142], 147]. In this chapter we prove a more general statement as an application of optimal control theory for linear systems. Theorem 14.1. Let S be a strongly convex compact subset of n, n even, and let the boundary of S be a level surface of a Hamiltonian H 2 C 1 ( n). Then for any vector v 2 n there exists a chord in S parallel to v such that ~ passing through there exists a trajectory of the Hamiltonian system x_ = H(x)

R R

R R

the endpoints of the chord.

We assume here that n is endowed with the standard symplectic structure  0 Id  (x x) = hx Jxi J = ; Id 0  i.e., the Hamiltonian vector eld corresponding to a Hamiltonian H has the form H~ = J grad H. The theorem on periodic trajectories of Hamiltoniansystems is a particular case of the previous theorem with v = 0. Now we prove Th. 14.1. Proof. Without loss of generality, we can assume that 0 2 intS. Consider the polar of the set S:

R

S = fu 2 n j suphu xi  1g: x2S

It follows from the separation theorem that (S ) = S

0 2 int S 

R

and that S is a strongly convex compact subset of n. Introduce the following linear optimal control problem:

208

R

14 Hamiltonian Systems with Convex Hamiltonians

x_ = u u 2 S  x 2 n x(0) = a x(1) = b

Z1

hx Jui dt ! min: (14.1) Here a and b are any points in S suciently close to the origin and such that the vector J(b ; a) is parallel to v. By Filippov's theorem, this problem has 0

optimal solutions. We use these solutions in order to construct the required trajectory of the Hamiltonian system on @S. The control-dependent Hamiltonian of PMP has the form: hu (p x) = pu +  hx Jui:

We show rst that abnormal trajectories cannot be optimal. Let  = 0. Then the adjoint equation is p_ = 0, thus p = p0 = const : The maximality condition of PMP reads p0u(t) = vmax 2S  p0v:

Since the polar S is strictly convex, then u(t) = const

u(t) 2 @S :

Consequently, abnormal trajectories are lines with velocities separated from zero. For points a, b suciently close to the origin, abnormal trajectories cannot meet the boundary conditions. Thus optimal trajectories are normal, so we can set  = ;1. The normal Hamiltonian is hu(p x) = pu ; hx Jui and the corresponding Hamiltonian system reads

(

p_ = Ju x_ = u:

The normal Hamiltonian can be written as hu (p x) = hy ui y = p + Jx where the vector y satises the equation y_ = 2Ju:

14 Hamiltonian Systems with Convex Hamiltonians

209

Along a normal trajectory hu(t)(p(t) x(t)) = hy(t) u(t)i = vmax 2S  hy(t) vi = C = const :

(14.2)

Consider rst the case C 6= 0, thus C > 0. Then z(t) = C1 y(t) 2 (S ) = S i.e., z(t) 2 S. Moreover, z(t) 2 @S and the vector u(t) is a normal to @S at the point z(t). Consequently, the curve z(t) is, up to reparametrization, a trajectory of the Hamiltonian eld H~ = J grad H. Compute the boundary conditions: p(1) ; p(0) = J(x(1) ; x(0)) y(1) ; y(0) = 2J(x(1) ; x(0)) = 2J(b ; a) z(1) ; z(0) = C2 J(b ; a):

Thus z(t) is the required trajectory: the chord z(1) ; z(0) is parallel to the vector v. In order to complete the proof, we show now that the case C = 0 in (14:2) is impossible. Indeed, if C = 0, then y(t)  0, thus u(t)  0. If a 6= b, then the boundary conditions for x are not satised. And if a = b, then the pair (u(t) x(t))  (0 0) does not realize minimum of functional (14:1), which can take negative values: for any admissible 1-periodic trajectory x(t), the trajectory x^(t) = x(1 ; t) is periodic with the cost

Z1 0

hx^ J x^_ i dx = ;

Z1 0

hx Jui dx: ut

15 Linear Time-Optimal Problem

15.1 Problem Statement

R

RR

In this chapter we study the following optimal control problem: m x_ = Ax + Bu x 2 n u 2 U n x(0) = x0 x(t1 ) = x1  x0 x1 2 xed (15.1) t1 ! min where U is a compact convex polytope in m, and A and B are constant matrices of order n  n and n  m respectively. Such problem is called linear time-optimal problem . The polytope U is the convex hull of a nite number of points a1  : : :  ak in m: U = convfa1  : : :  ak g: We assume that the points ai do not belong to the convex hull of all the rest points aj , j 6= i, so that each ai is a vertex of the polytope U. In the sequel we assume the following General Position Condition : For any edge ai  aj ] of U, the vector eij = aj ; ai satises the equality span(Beij  ABeij  : : :  An;1Beij ) = n: (15.2) This condition means that no vector Beij belongs to a proper invariant subspace of the matrix A. By Theorem 3.3, this is equivalent to controllability of the linear system x_ = Ax+Bu with the set of control parameters u 2 eij. Condition (15:2) can be achieved by a small perturbation of matrices A B. We considered examples of linear time-optimal problems in Sects. 13.1, 13.2. Here we study the structure of optimal control, prove its uniqueness, evaluate the number of switchings. Existence of optimal control for any points x0, x1 such that x1 2 A(x0) is guaranteed by Filippov's theorem. Notice that for the analogous problem with an unbounded set of control parameters, optimal control may not exist: it is easy to show this using linearity of the system.

R

R

R

R

212

15 Linear Time-Optimal Problem

Before proceeding with the study of linear time-optimalproblems, we recall some basic facts on polytopes.

15.2 Geometry of Polytopes

R

The convex hull of a nite number of points a1  : : :  ak 2 m is the set def

U = convfa1  : : :  ak g =

R

(X k i=1

iai j i  0

k X i=1

i = 1 :

An ane hyperplane in m is a set of the form

R

)

R

R

 = fu 2 m j h  ui = cg 2 m n f0g c 2 : A hyperplane of support to a polytope U is a hyperplane  such that

h  ui  c 8u 2 U for the covector and number c that dene , and this inequality turns into equality at some point u 2 @U, i.e.,  \ U 6= .

a1

a2

a3 a4

U

Fig. 15.1. Polytope U with hyperplane of support  A polytope U = convfa1  : : :  akg intersects with any its hyperplane of support  = fu j h  ui = cg by another polytope: U \  = convfai1 : : :  ail g h  ai1i = = h  ail i = c h  aj i < c j 2= fi1  : : :  il g:

15.3 Bang-Bang Theorem

213

Such polytopes U \  are called faces of the polytope U. Zero-dimensional and one-dimensional faces are called respectively vertices and edges. A polytope has a nite number of faces, each of which is the convex hull of a nite number of vertices. A face of a face is a face of the initial polytope. Boundary of a polytope is a union of all its faces. This is a straightforward corollary of the separation theorem for convex sets (or the Hahn-Banach Theorem).

15.3 Bang-Bang Theorem Optimal control in the linear time-optimal problem is bang-bang, i.e., it is piecewise constant and takes values in vertices of the polytope U. Theorem 15.1. Let u(t), 0  t  t1 , be an optimal control in the linear time-optimal control problem (15:1). Then there exists a nite subset

T 0 t1]

#T < 1

such that

u(t) 2 fa1 : : :  ak g

t 2 0 t1] n T 

(15.3)

and restriction u(t)jt2 0t1 ]nT is locally constant. Proof. Apply Pontryagin Maximum Principle to the linear time-optimal prob-

lem (15:1). State and adjoint vectors are 0x 1 1 B . . [email protected] . C A 2 n = ( 1 : : :  n) 2 n xn and a point in the cotangent bundle is

R

R

R R R

 = (  x) 2 n  n = T  n: The control-dependent Hamiltonian is hu(  x) = Ax + Bu

(we multiply rows by columns). The Hamiltonian system and maximality condition of PMP take the form:

(

x_ = Ax + Bu

_ = ; A

(t) 6= 0

(t)Bu(t) = max

(t)Bu: u2U

(15.4)

214

15 Linear Time-Optimal Problem

The Hamiltonian system implies that adjoint vector

(t) = (0)e;tA 

(0) 6= 0

(15.5)

is analytic along the optimal trajectory. Consider the set of indices corresponding to vertices where maximum(15:4) is attained:





J(t) = 1  j  k j (t)Baj = max

(t)Bu = maxf (t)Bai j i = 1 : : :  kg : u2U At each instant t the linear function (t)B attains maximum at vertices of the polytope U. We show that this maximum is attained at one vertex always except a nite number of moments. Dene the set T = ft 2 0 t1] j #J(t) > 1g: By contradiction, suppose that T is innite: there exists a sequence of distinct moments f 1 : : :  n : : : g T : Since there is a nite number of choices for the subset J( n ) f1 : : :  kg, we can assume, without loss of generality, that J( 1) = J( 2 ) = = J( n ) = : Denote J = J( i ). Further, since the convex hull convfaj j j 2 J g is a face of U, then there exist indices j1 j2 2 J such that the segment aj1  aj2 ] is an edge of U. We have

( i )Baj1 = ( i )Baj2 

i = 1 2 : : : :

For the vector e = aj2 ; aj1 we obtain

( i )Be = 0

i = 1 2 : : : :

But ( i ) = (0)e;i A by (15:5), so the analytic function t 7! (0)e;tA Be

has an innite number of zeros on the segment 0 t1], thus it is identically zero:

(0)e;tA Be  0: We dierentiate this identity successively at t = 0 and obtain

15.4 Uniqueness of Optimal Controls and Extremals

215

(0)Be = 0 (0)ABe = 0 : : :  (0)An;1Be = 0: By General Position Condition (15:2), we have (0) = 0, a contradiction to (15:5). So the set T is nite. Out of the set T , the function (t)B attains maximum on U at one vertex aj (t), fj(t)g = J(t), thus the optimal control u(t) takes value in the vertex aj (t). Condition (15:3) follows. Further,

(t)Baj (t) > (t)Bai  i 6= j(t): But all functions t 7! (t)Bai are continuous, so the preceding inequality preserves for instants close to t. The function t 7! j(t) is locally constant on 0 t1] n T , thus the optimal control u(t) is also locally constant on 0 t1] n T .

ut

In the sequel we will need the following statement proved in the preceding argument. Corollary 15.2. Let (t), t 2 0 t1], be a nonzero solution of the adjoint equation _ = ; A. Then everywhere in the segment 0 t1], except a nite number of points, there exists a unique control u(t) 2 U such that (t)Bu(t) = max

(t)Bu. u2U

15.4 Uniqueness of Optimal Controls and Extremals Theorem 15.3. Let the terminal point x1 be reachable from the initial point x0 : x1 2 A(x0):

Then linear time-optimal control problem (15:1) has a unique solution. Proof. As we already noticed, existence of an optimal control follows from

Filippov's Theorem. Suppose that there exist two optimal controls: u1(t), u2 (t), t 2 0 t1]. By Cauchy's formula:



x(t1) = et1 A x0 + we obtain



et1 A x0 + thus

Zt

1

0

1

0



e;tA Bu(t) dt 





e;tA Bu1 (t) dt = et1 A x0 +

Zt

1

0

Zt

e;tA Bu1 (t) dt =

Zt

1

0

Zt

1

0



e;tA Bu2 (t) dt 

e;tA Bu2 (t) dt:

(15.6)

216

15 Linear Time-Optimal Problem

Let 1 (t) = 1 (0)e;tA be the adjoint vector corresponding by PMP to the control u1 (t). Then equality (15:6) can be written in the form

Zt 0

1

1(t)Bu1 (t) dt =

Zt

1

0

1 (t)Bu2 (t) dt:

(15.7)

By the maximality condition of PMP

1 (t)Bu1 (t) = max

(t)Bu u2U 1 thus

1 (t)Bu1 (t)  1 (t)Bu2 (t): But this inequality together with equality (15:7) implies that almost everywhere on 0 t1]

1 (t)Bu1 (t) = 1 (t)Bu2 (t): By Corollary 15.2, u1(t)  u2(t) almost everywhere on 0 t1]. ut So for linear time-optimal problem, optimal control is unique. The standard procedure to nd the optimal control for a given pair of boundary points x0, x1 is to nd all extremals ( (t) x(t)) steering x0 to x1 and then to seek for the best among them. In the examples considered in Sects. 13.1, 13.2, there was one extremal for each pair x0, x1 with x1 = 0. We prove now that this is a general property of linear time-optimal problems. Theorem 15.4. Let x1 = 0 2 A(x0 ) and 0 2 U n fa1 : : :  ak g. Then there exists a unique control u(t) that steers x0 to 0 and satis es Pontryagin Maximum Principle. Proof. Assume that there exist two controls

u1 (t) t 2 0 t1] and u2(t) t 2 0 t2] that steer x0 to 0 and satisfy PMP. If t1 = t2 , then the argument of the proof of preceding theorem shows that u1(t)  u2(t) a.e., so we can assume that t1 > t2: Cauchy's formula gives et1 A et2 A



Zt



Z0 t

x0 + x0 +

1

2

0



e;tA Bu1 (t) dt = 0



e;tA Bu2 (t) dt = 0

15.4 Uniqueness of Optimal Controls and Extremals

thus

Zt

1

0

e;tA Bu1 (t) dt =

Zt

2

0

e;tA Bu2 (t) dt:

217

(15.8)

According to PMP, there exists an adjoint vector 1 (t), t 2 0 t1], such that

1 (t) = 1 (0)e;tA 1 (0) 6= 0 (15.9)

1 (t)Bu1 (t) = max

(t)Bu: (15.10) u2U 1

Since 0 2 U, then

1 (t)Bu1 (t)  0 Equality (15:8) can be rewritten as

Zt

1

0

1(t)Bu1 (t) dt =

Zt

2

0

t 2 0 t1]:

(15.11)

1 (t)Bu2 (t) dt:

(15.12)

Taking into account inequality (15:11), we obtain

Zt 0

2

1(t)Bu1 (t) dt 

Zt

2

0

1 (t)Bu2 (t) dt:

(15.13)

But maximality condition (15:10) implies that t 2 0 t2]: (15.14)

1(t)Bu1 (t)  1 (t)Bu2 (t) Now inequalities (15:13) and (15:14) are compatible only if

1(t)Bu1 (t) = 1 (t)Bu2 (t) t 2 0 t2] thus inequality (15:13) should turn into equality. In view of (15:12), we have

Zt

2

t1

1 (t)Bu1 (t) dt = 0:

Since the integrand is nonnegative, see (15:11), then it vanishes identically:

1 (t)Bu1 (t)  0 t 2 t1 t2]: By the argument of Theorem 15.1, the control u1 (t) is bang-bang, so there exists an interval I t1 t2] such that u1(t)jI  aj 6= 0: Thus

1 (t)Baj  0 t 2 I: But 1 (t)0  0, this is a contradiction with uniqueness of the control for which maximum in PMP is obtained, see Corollary 15.2. ut

218

15 Linear Time-Optimal Problem

15.5 Switchings of Optimal Control Now we evaluate the number of switchings of optimal control in linear timeoptimal problems. In the examples of Sects. 13.1, 13.2 we had respectively one switching and an arbitrarily large number of switchings, although nite on any segment. It turns out that in general there are two cases: non-oscillating and oscillating, depending on whether the matrix A of the control system has real spectrum or not. Recall that in the example with one switching, Sect. 13.1, we had 0 1 A = 0 0  Sp(A) = f0g  and in the example with arbitrarily large number of switchings, Sect. 13.2,   A = ;01 10  Sp(A) = fig 6 : We consider systems with scalar control: x_ = Ax + ub u 2 U =   ]  x 2 n under the General Position Condition span(b Ab : : :  An;1b) = n: Then attainable set of the system is full-dimensional for arbitrarily small times. We can evaluate the minimal number of switchings necessary to ll a full-dimensional domain. Optimal control is piecewise constant with values in f   g. Assume that we start from the initial point x0 with the control . Without switchings we ll a piece of a 1-dimensional curve e(Ax+ b)tx0, with 1 switching we ll a piece of a 2-dimensional surface e(Ax+b)t2  e(Ax+ b)t1 x0 , with 2 switchings we can attain points in a 3-dimensional surface, etc. So the minimal number of switchings required to reach an n-dimensional domain is n ; 1. We prove now that in the non-oscillating case we never need more than n ; 1 switchings of optimal control. Theorem 15.5. Assume that the matrix A has only real eigenvalues: Sp(A) : Then any optimal control in linear time-optimal problem (15:1) has no more than n ; 1 switchings. Proof. Let u(t) be an optimal control and (t) = (0)e;tA the corresponding solution of the adjoint equation _ = ; A. The maximality condition of PMP reads

(t)bu(t) = umax 2 ] (t)bu

R R

R R R

R

15.5 Switchings of Optimal Control

219

(

thus

u(t) =  if (t)b > 0 if (t)b < 0: So the number of switchings of the control u(t), t 2 0 t1], is equal to the number of changes of sign of the function y(t) = (t)b t 2 0 t1]: We show that y(t) has not more than n ; 1 real roots. Derivatives of the adjoint vector have the form

(k)(t) = (0)e;tA (;A)k : By Cayley Theorem, the matrix A satises its characteristic equation: An + c1 An;1 + + cn Id = 0 where det(t Id ;A) = tn + c1tn;1 + + cn  thus (;A)n ; c1 (;A)n;1 + + (;1)ncn Id = 0: Then the function y(t) satises an n-th order ODE: y(n) (t) ; c1 y(n;1) (t) + + (;1)n cny(t) = 0: (15.15) It is well known (see e.g. 136]) that any solution of this equation is a quasipolynomial: k X y(t) = e;i tPi (t) i=1

Pi (t) a polynomial i 6= j for i 6= j where i are eigenvalues of the matrix A and degree of each polynomial Pi is less than multiplicity of the corresponding eigenvalue i , thus k X i=1

deg Pi  n ; k:

Now the statement of this theorem follows from the next general lemma. ut

Lemma 15.6. A quasipolynomial y(t) =

k X

ei tPi (t)

i=1 i 6= j for i 6= j

has no more than n ; 1 real roots.

k X i=1

deg Pi  n ; k

(15.16)

220

15 Linear Time-Optimal Problem

Proof. Apply induction on k.

If k = 1, then a quasipolynomial y(t) = et P(t) deg P  n ; 1 has no more than n ; 1 roots. We prove the induction step for k > 1. Denote ni = deg Pi i = 1 : : :  k: Suppose that the quasipolynomial y(t) has n real roots. Rewrite the equation kX ;1

y(t) = as follows:

kX ;1 i=1

i=1

ei t Pi (t) + ek tPk (t) = 0

e(i ;k )t Pi(t) + Pk (t) = 0:

(15.17)

The quasipolynomial in the left-hand side has n roots. We dierentiate this quasipolynomial successively (nk + 1) times so that the polynomial Pk (t) disappear. After (nk + 1) dierentiations we obtain a quasipolynomial kX ;1 i=1

e(i ;k )t Qi(t)

deg Qi  deg Pi 

which has (n ; nk ; 1) real roots by Rolle's Theorem. But by induction assumption the maximal possible number of real roots of this quasipolynomial is kX ;1 ni + k ; 2 < n ; nk ; 1: i=1

The contradiction nishes the proof of the lemma. ut So we completed the proof of Theorem 15.5: in the non-oscillating case an optimal control has no more than n ; 1 switchings on the whole domain (recall that n ; 1 switchings are always necessary even on short time segments since the attainable sets Aq0 (t) are full-dimensional for all t > 0). For an arbitrary matrix A, one can obtain the upper bound of (n ; 1) switchings for suciently short intervals of time. Theorem 15.7. Consider the characteristic polynomial of the matrix A: det(t Id ;A) = tn + c1tn;1 + + cn  and let

15.5 Switchings of Optimal Control

c = 1max in jcij:

221

R

Then for any time-optimal control u(t) and any t% 2 , the real segment





1 

%t %t + ln 1 + c contains not more than (n ; 1) switchings of an optimal control u(t). In the proof of this theorem we will require the following general proposition, which we learned from S. Yakovenko.

Lemma 15.8. Consider an ODE

y(n) + c1 (t)y(n;1) + + cn(t)y = 0

with measurable and bounded coecients:

ci = t2max jci (t)j:

t t +] If

n i X ci i! < 1

(15.18)

i=1

then any nonzero solution y(t) of the ODE has not more than n ; 1 roots on the segment t 2 t% %t + ]. Proof. By contradiction, suppose that the function y(t) has at least n roots

on the segment t 2 t% %t + ]. By Rolle's Theorem, derivative y(t) _ has not less than n ; 1 roots, etc. Then y(n;1) (t) has a root tn;1 2 t% %t + ]. Thus y(n;1) (t) =

Zt

tn;1

y(n) ( ) d :

Let tn;2 2 t% %t + ] be a root of y(n;2) (t), then y(n;2) (t) =

Zt

tn;2

d 1

Z

1

tn;1

y(n) ( 2 ) d 2:

We continue this procedure by integrating y(n;i+1) (t) from a root tn;i 2 t% %t + ] of y(n;i) (t) and obtain y(n;i) (t) =

Zt

tn;i

d 1

There holds a bound:

Z

1

tn;i+1

d 2

Z i;

1

tn;1

y(n) ( i ) d i

i = 1 : : :  n:

222

15 Linear Time-Optimal Problem

y(n;i) (t) 



Zt

Ztnt +;i t

i

d 1

Z

d 1

 i! sup t2 t t +] Then

n X i=1

ci (t)y(n;i) (t) 

i.e.,

n X i=1

1

t

Z n;i 1

+1

d 2

d 2

t

y(n) (t)

1

t ;1

Z i;n

1

t

y(n) ( i ) d i

y(n) ( i ) d i

:

jci(t)j y(n;i) (t) 

y(n) (t) 

Z i;

n i X ci i! sup y(n) (t) 

i=1

t2 tt+]

n i X ci i! sup y(n) (t) 

i=1

t2 tt+]

a contradiction with (15:18). The lemma is proved. ut Now we prove Theorem 15.7. Proof. As we showed in the proof of Theorem 15.5, the number of switchings of u(t) is not more than the number of roots of the function y(t) = (t)b, which satises ODE (15:15). We have n i X jcij i! < c(e ; 1) 8 > 0: By Lemma 15.8, if

i=1

c(e ; 1)  1

(15.19)

then the function y(t) has not more than n ; 1 real roots on any interval of length . But inequality (15:19) is equivalent to the following one:     ln 1 + 1c 

;



so y(t) has not more than n ; 1 roots on any interval of the length ln 1 + 1c .

ut

16.1 Problem Statement In this chapter we study a class of optimal control problems very popular in applications, linear-quadratic problems . That is, we consider linear systems with quadratic cost functional:

R R

x_ = Ax + Bu x 2 n u 2 m (16.1) x(0) = x0 x(t1 ) = x1  x0 x1 t1 xed Z t1 J(u) = 12 hRu(t) u(t)i + hPx(t) u(t)i + hQx(t) x(t)i dt ! min: 0 Here A, B, R, P, Q are constant matrices of appropriate dimensions, R and Q are symmetric: R = R Q = Q and angle brackets h  i denote the standard inner product in m and n. One can show that the condition R  0 is necessary for existence of optimal control. We do not touch here the case of degenerate R and assume that R > 0. The substitution of variables u 7! v = R1=2u transforms the functional J(u) to a similar functional with the identity matrix instead of R. That is why we assume in the sequel that R = Id. A linear feedback transformation kills the matrix P (exercise: nd this transformation). So we can write the cost functional as follows: Z t1 ju(t)j2 + hQx(t) x(t)i dt: J(u) = 21 0 For dynamics of the problem, we assume that the linear system is controllable:

R R

rank(B AB : : :  An;1B) = n:

(16.2)

224

16.2 Existence of Optimal Control

R

Since the set of control parameters U = m is noncompact, Filippov's Theorem does not apply, and existence of optimal controls in linear-quadratic problems is a nontrivial problem. In this chapter we assume that admissible controls are square-integrable: u 2 Lm2 0 t1] and use the Lm2 norm for controls:

kuk =

Z t

1

0

1=2 Z t

ju(t)j2 dt

=

1

0

1=2

u21(t) + + u2m (t) dt

:

Consider the set of all admissible controls that steer the initial point to the terminal one: U (x0  x1) = fu 2 Lm2 0 t1] j x(t1  u x0) = x1 g : We denote by x(t u x0) the trajectory of system (16:1) corresponding to an admissible control u 2 Lm2 starting at a point x0 2 n. By Cauchy's formula, the endpoint mapping u 7! x(t1 u x0) = et1 A x0 +

Zt

RR

1

0

R

e(t1 ; )A Bu( ) d

is an ane mapping from Lm2 0 t1] to n. Controllability of the linear system (16:1) means that for any x0 2 n, t1 > 0, the image of the endpoint mapping is the whole n. The subspace U (x0 x1) Lm2 0 t1] is ane, the subspace U (0 0) Lm2 0 t1] is linear, moreover,

R

U (x0 x1) = u + U (0 0) for any u 2 U (x0  x1): Thus it is natural that existence of optimal controls is closely related to behavior of the cost functional J(u) on the linear subspace U (0 0). Proposition 16.1. (1) If there exist points x0 x1 2 n such that inf

u2U (x0 x1 )

then

J(u) > ;1

J(u)  0 8u 2 U (0 0):

R

(16.3)

16.2 Existence of Optimal Control

(2) Conversely, if

225

J(u) > 0 8u 2 U (0 0) n 0

then the minimum is attained:

R

9 u2Umin J(u) 8x0  x1 2 n: (x0 x1 )

Remark 16.2. That is, the inequality

J jU (00)  0 is necessary for existence of optimal controls at least for one pair (x0 x1), and the strict inequality J jU (00)n0 > 0 is sucient for existence of optimal controls for all pairs (x0  x1). In the proof of Proposition 16.1, we will need the following auxiliary proposition. Lemma 16.3. If J(v) > 0 for all v 2 U (0 0) n 0, then J(v)  kvk2 for some > 0 and all v 2 U (0 0) or, which is equivalent,

inf fJ(v) j kvk = 1 v 2 U (0 0)g > 0: Proof. Let vn be a minimizing sequence of the functional J(v) on the sphere fkvk = 1g \ U (0 0). Closed balls in Hilbert spaces are weakly compact, thus we can nd a subsequence weakly converging in the unit ball and preserve the notation vn for its terms, so that vn ! vb weakly as n ! 1 kbvk  1 bv 2 U (0 0) J(vn ) ! inf fJ(v) j kvk = 1 v 2 U (0 0)g n ! 1: (16.4) We have Z t1 1 1 J(vn ) = 2 + 2 hQxn( ) xn( )i d : 0 Since the controls converge weakly, then the corresponding trajectories converge strongly: xn( ) ! xbv ( ) n ! 1 thus Z t1 1 1 n ! 1: J(vn ) ! 2 + 2 hQxbv ( ) xbv ( )i d  0 In view of (16:4), the inmum in question is equal to 1 + 1 Z t1 hQx ( ) x ( )i d = 1 ;1 ; kbv k2 + J(bv ) > 0: bv bv 2 2 0 2

ut

226

Now we prove Proposition 16.1. Proof. (1) By contradiction, suppose that there exists v 2 U (0 0) such that J(v) < 0. Take any u 2 U (x0  x1), then u + sv 2 U (x0 x1) for any s 2 . Let y(t), t 2 0 t1], be the solution to the Cauchy problem

R

y_ = Ay + Bv and let

y(0) = 0

Z t1 J(u v) = 12 hu( ) v( )i + hQx( ) y( )i d : 0

R

Then the quadratic functional J on the family of controls u + sv, s 2 , is computed as follows: J(u + sv) = J(u) + 2sJ(u v) + s2 J(v): Since J(v) < 0, then J(u + sv) ! ;1 as s ! 1. The contradiction with hypothesis (16:3) proves item (1). (2) We have

Z t1 J(u) = 12 kuk2 + 12 hQx( ) x( )i d : 0

The norm kukR is lower semicontinuous in the weak topology on Lm2 , and the functional 0t1 hQx( ) x( )i d is weakly continuous on Lm2 . Thus J(u) is weakly lower semicontinuous on Lm2 . Since balls are weakly compact in Lm2 and the ane subspace U (x0 x1) is weakly compact, it is enough to prove that J(u) ! 1 when u ! 1, u 2 U (x0 x1). Take any control u 2 U (x0 x1). Then any control from U (x0  x1) has the form u + v for some v 2 U (0 0). We have   J(u + v) = J(u) + 2kvkJ u kvvk + J(v):





Denote J(u) = C0. Further, J u kvvk  C1 = const for all v 2 U (0 0) n 0. Finally, by Lemma 16.3, J(v)  kvk2, > 0, for all v 2 U (0 0) n 0. Consequently, J(u + v)  C0 ; 2kvkC1 + kvk2 ! 1

v ! 1 v 2 U (0 0):

Item (2) of this proposition follows. ut So we reduced the question of existence of optimal controls in linear-quadratic problems to the study of the restriction J jU (00). We will consider this restriction in detail in Sect. 16.4.

16.3 Extremals

16.3 Extremals

227

We cannot directly apply Pontryagin Maximum Principle to the linearquadratic problem since we have conditions for existence of optimal controls in Lm2 only, while PMP requires controls from Lm1 . Although, suppose for a moment that PMP is applicable to the linear-quadratic problem. It is easy to write equations for optimal controls and trajectories that follow from PMP, moreover, it is natural to expect that such equations should hold true. Now we derive such equations, and then substantiate them. So we write PMP for the linear-quadratic problem. The control-dependent Hamiltonian is hu(  x) = Ax + Bu ; 2 (juj2 + hQx xi) x 2 n 2 n:

R R

Consider rst the abnormal case:  = 0. By PMP, adjoint vector along an extremal satises the ODE _ = ; A, thus (t) = (0)e;tA . The maximality condition implies that 0  (t)B = (0)e;tA B. We dierentiate this identity n ; 1 times, take into account the controllability condition (16:2) and obtain

(0) = 0. This contradicts PMP, thus there are no abnormal extremals. In the normal case we can assume  = 1. Then the control-dependent Hamiltonian takes the form hu(  x) = Ax + Bu ; 12 (juj2 + hQx xi) x 2 n 2 n: The term Bu ; 12 juj2 depending on u has a unique maximum in u 2 m at the point where @ hu = B ; u = 0 @u thus u = B  : (16.5) So the maximized Hamiltonian is 1 hQx xi + 1 jB   j2 H(  x) = umax h (  x) =

Ax ; u m 2R 2 2 1 1 2 = Ax ; 2 hQx xi + 2 jB j :

R R

R

The Hamiltonian function H(  x) is smooth, thus normal extremals are solutions of the corresponding Hamiltonian system x_ = Ax + BB    (16.6) _ = x Q ; A: (16.7) Now we show that optimal controls and trajectories in the linear-quadratic problem indeed satisfy equations (16:5){(16:7). Consider the extended system

228

x_ = Ax + Bu y_ = 12 (juj2 + hQx xi)

R R

and the corresponding endpoint mapping: F : u 7! (x(t1 u x0) y(t1  u 0)) F : Lm2 0 t1] ! n  : This mapping can be written explicitly via Cauchy's formula:



Zt

1

x(t1 u x0) = et1 A x0 +

0



e;tA Bu(t) dt 

(16.8)

Z t1 1 y(t1  u 0) = 2 ju(t)j2 + hQx(t) x(t)i dt: (16.9) 0 Let u~( ) be an optimal control and x~( ) = x(  u~ x0) the corresponding optimal trajectory, then F(~u) 2 @ ImF: By implicit function theorem, the dierential Du~ F : Lm2 0 t1] ! n ! is not surjective, i.e., there exists a covector (  ) 2 n ! , (  ) 6= 0, such that (16.10) (  ) ? Du~ Fv v 2 Lm2 0 t1]: The dierential of the endpoint mapping is found from the explicit formulas (16:8), (16:9):

R R

Du~ Fv =

Z t

1

0

e(t1 ;t)A Bv(t) dt

Zt  1

0

Zt

u~(t) +

1

t

R R

Then the orthogonality condition (16:10) reads:

Zt  1

0

B  e(t1 ;t)A + ~u(t) + 

that is, B  e(t1 ;t)A + ~u(t) + 

Zt

1

t

Zt

1

t

 

B  e( ;t)A Q~x( ) d  v(t) dt :



B  e( ;t)A Q~x( ) d  v(t) dt = 0 v 2 Lm2 0 t1]

B  e( ;t)A Q~x( ) d  0

t 2 0 t1]: (16.11)

16.4 Conjugate Points

229

The case  = 0 is impossible by condition (16:2). Denote  = ; =, then equality (16:11) reads u~(t) = B   (t) where

(t) =   e(t1 ;t)A ;

Zt

1

t

x~ ( )Qe( ;t)A dt:

(16.12)

So we proved equalities (16:5), (16:6). Dierentiating (16:12), we arrive at the last required equality (16:7). So we proved that optimal trajectories in the linear-quadratic problem are projections of normal extremals of PMP (16:6), (16:7), while optimal controls are given by (16:5). In particular, optimal trajectories and controls are analytic.

16.4 Conjugate Points Now we study conditions of existence and uniqueness of optimal controls depending upon the terminal time. So we write the cost functional to be minimized as follows: Zt 1 Jt (u) = 2 ju( )j2 + hQx( ) x( )i d : 0 Denote

Ut(0 0) = fu 2 Lm2 0 t] j x(t u x0) = x1g  (t) def = inf fJt(u) j u 2 Ut (0 0) kuk = 1g:

(16.13)

We showed in Proposition 16.1 that if (t) > 0 then the problem has solution for any boundary conditions, and if (t) < 0 then there are no solutions for any boundary conditions. The case (t) = 0 is doubtful. Now we study properties of the function (t) in detail. Proposition 16.4. (1) The function t 7! (t) is monotone nonincreasing and continuous. (2) For any t > 0 there hold the inequalities 2

1  2(t)  1 ; t2 e2tkAk kB k2kQk:

(16.14)

(3) If 1 > 2(t), then the in mum in (16:13) is attained, i.e., it is mini-

mum.

230

Proof. (3) Denote

Zt 1 It (u) = 2 hQx( ) x( )i d  0 the functional It (u) is weakly continuous on Lm2 . Notice that Jt(u) = 21 + It (u) on the sphere kuk = 1: Take a minimizing sequence of the functional It (u) on the sphere fkuk = 1g \ Ut (0 0). Since the ball fkuk  1g is weakly compact, we can nd a weakly converging subsequence: un ! ub weakly as n ! 1 kubk  1 ub 2 Ut (0 0) It (un) ! It (ub) = inf fIt(u) j kuk = 1 u 2 Ut (0 0)g n ! 1: If ub = 0, then It (ub) = 0, thus (t) = 21 , which contradicts hypothesis of item (3).   So ub 6= 0, It (ub) < 0, and It kuubbk  It (ub). Thus kubk = 1, and Jt (u) attains minimum on the sphere fkuk = 1g \ Ut(0 0) at the point ub. (2) Let kuk = 1 and x0 = 0. By Cauchy's formula, x(t) = thus

jx(t)j 

Zt 0

Zt 0

e(t; )A Bu( ) d 

e(t; )kAk kB k ju( )j d

by Cauchy-Schwartz inequality

 kuk =

Z t

Z t 0

0

e(t; )2kAk kB k2 d

e(t; )2kAk kB k2 d

1=2

1=2

:

We substitute this estimate of x(t) into Jt and obtain the second inequality in (16:14). The rst inequality in (16:14) is obtained by considering a weakly converging sequence un ! 0, n ! 1, in the sphere kunk = 1, un 2 Ut (0 0). (1) Monotonicity of (t). Take any ^t > t. Then the space Ut(0 0) is isometrically embedded into Ut^(0 0) by extending controls u 2 Ut(0 0) by zero: u 2 Ut (0 0) ) ub 2 Ut^(0 0)  t ub( ) = u( ) 0 > t:

16.4 Conjugate Points

Moreover, Thus

231

Jt^(ub) = Jt(u):

(t) = inf fJt(u) j u 2 Ut (0 0) kuk = 1g  inf fJt^(u) j u 2 Ut^(0 0) kuk = 1g = (t^): Continuity of (t): we show separately continuity from the right and from the left. Continuity from the right. Let tn & t. We can assume that (tn ) < 12 (otherwise (tn) = (t) = 12 ), thus minimum in (16:13) is attained: (tn) = 12 + Itn (un ) un 2 Utn (0 0) kunk = 1: Extend the functions un 2 Lm2 0 tn] to the segment 0 t1] by zero. Choosing a weakly converging subsequence in the unit ball, we can assume that un ! u weakly as n ! 1 u 2 Ut(0 0) kuk  1 thus Itn (un) ! It (u)  inf fIt(v) j v 2 Ut (0 0) kvk = 1g tn & t: Then (t)  12 + tlim Itn (un) = tlim (tn ): n &t n &t By monotonicity of , (t) = tlim &t (tn ) n

i.e., continuity from the right is proved. Continuity from the left. We can assume that (t) < 21 (otherwise ( ) = (t) = 21 for < t). Thus minimum in (16:13) is attained: (t) = 12 + It (ub) ub 2 Ut (0 0) kubk = 1: For the trajectory xb( ) = x(  ub 0) we have Z xb( ) = e( ;)A B ub( ) d : Denote and notice that

0

(") = k ubj 0"] k (") ! 0

" ! 0:

232

Denote the ball B = fu 2 Lm2 j kuk   u 2 U (0 0)g: Obviously,

R

x(" B (")  0) 3 xb("): The mapping u 7! x(" u 0) from Lm2 to n is linear, and the system x_ = Ax + Bu is controllable, thus x(" B (") 0) is a convex full-dimensional set in n such that the positive cone generated by this set is the whole n. That is why x(" 2B (")  0) = 2x(" B (")  0)  Ox("B(") 0) for some neighborhood Ox("B(") 0) of the set x(" B (")  0). Further, there exists an instant t" > " such that xb(t" ) 2 x(" 2B (") 0) consequently, xb(t" ) = x(" v"  0) kv" k  2 ("): Notice that we can assume t" ! 0 as " ! 0. Consider the following family of controls that approximate ub:  0   " u" ( ) = vub"( ( ) + t" ; ") " <  t + " ; t" : We have u" 2 Ut+";t" (0 0) kub ; u" k ! 0 " ! 0: But t + " ; t" < t and  is nonincreasing, thus it is continuous from the left. Continuity from the right was already proved, hence  is continuous. tu Now we prove that the function  can have not more than one root. Proposition 16.5. If (t) = 0 for some t > 0, then ( ) < 0 for all > t. Proof. Let (t) = 0, t > 0. By Proposition 16.4, inmum in (16:13) is attained at some control ub 2 Ut (0 0), kubk = 1: (t) = minfJt(u) j u 2 Ut (0 0) kuk = 1g = Jt(ub) = 0: Then Jt(u)  Jt(ub) = 0 8u 2 Ut(0 0) i.e., the control ub is optimal, thus it satises PMP. There exists a solution ( ( ) x( )), 2 0 t], of the Hamiltonian system

R

R

(_

= x Q ; A x_ = Ax + BB   

16.4 Conjugate Points

233

with the boundary conditions x(0) = x(t) = 0 and

u( ) = B   ( ) 2 0 t]: We proved that for any root t of the function , any control u 2 Ut (0 0), kuk = 1, with Jt(u) = 0 satises PMP. Now we prove that ( ) < 0 for all > t. By contradiction, suppose that the function  vanishes at some instant t0 > t. Since  is monotone, then j tt0 ]  0: Consequently, the control  b( )  t u0 ( ) = u0 2 t t0] satises the conditions: u0 2 Ut0 (0 0) ku0k = 1 Jt0 (u0 ) = 0: Thus u0 satises PMP, i.e.,

u0 ( ) = B   ( ) 2 0 t0] is an analytic function. But u0j tt0]  0, thus u0  0, a contradiction with ku0k = 1. ut It would be nice to have a way to solve the equation (t) = 0 without performing the minimization procedure in (16:13). This can be done in terms of the following notion. Denition 16.6. A point t > 0 is conjugate to 0 for the linear-quadratic problem in question if there exists a nontrivial solution ( ( ) x( )) of the 0

Hamiltonian system

(_

= x Q ; A x_ = Ax + BB  

such that x(0) = x(t) = 0. Proposition 16.7. The function  vanishes at a point t > 0 if and only if t is the closest to 0 conjugate point.

234

Proof. Let (t) = 0, t > 0. First of all, t is conjugate to 0, we showed this in

the proof of Proposition 16.5. Suppose that t0 > 0 is conjugate to 0. Compute the functional Jt0 on the corresponding control u( ) = B   ( ), 2 0 t0]: 1 Z t hB   ( ) B   ( )i + hQx( ) x( )i d 2 0 Z t0 = 12 hBB   ( )  ( )i + hQx( ) x( )i d 0 Z t0 1 = 2 ( )(x( ) _ ; Ax( )) + x ( )Qx( ) d Z0t0 1 _ d = 2 ( x_ + x) 0 = 12 ( (t0 )x(t0 ) ; (0)x(0)) = 0: 0

Jt0 (u) =



 Thus (t0)  Jt0 kuuk = 0. Now the result follows since  is nonincreasing.

ut

The rst (closest to zero) conjugate point determines existence and uniqueness properties of optimal control in linear-quadratic problems. Before the rst conjugate point, optimal control exists and is unique for any boundary conditions (if there are two optimal controls, then their dierence gives rise to a conjugate point). At the rst conjugate point, there is existence and nonuniqueness for some boundary conditions, and nonexistence for other boundary conditions. And after the rst conjugate point, the problem has no optimal solutions for any boundary conditions.

17 Su cient Optimality Conditions, Hamilton-Jacobi Equation, and Dynamic Programming

17.1 Sucient Optimality Conditions Pontryagin Maximum Principle is a universal and powerful necessary optimality condition, but the theory of sucient optimality conditions is not so complete. In this section we consider an approach to sucient optimality conditions that generalizes elds of extremals of the Classical Calculus of Variations. Consider the following optimal control problem: q_ = fu (q) q 2 M u 2 U q(0) = q0 q(t1 ) = q1 q0 q1 t1 xed

Zt

1

0

'(q(t) u(t)) dt ! min:

(17.1) (17.2) (17.3)

The control-dependent Hamiltonian of PMP corresponding to the normal case is hu() = h fu (q)i ; '(q u)

 2 T  M q = () 2 M u 2 U:

Assume that the maximized Hamiltonian H() = max h () u2U u

(17.4)

is dened and smooth on T  M. We can assume smoothness of H on an open domain O T  M and modify respectively the subsequent results. But for simplicity of exposition we prefer to take O = T  M. Then trajectories of the Hamiltonian system ~ _ = H() are extremals of problem (17:1){(17:3). We assume that the Hamiltonian vector eld H~ is complete.

236

17 Sucient Optimality Conditions

17.1.1 Integral Invariant First we consider a general construction that will play a key role in the proof of sucient optimality conditions. Fix an arbitrary smooth function a 2 C 1 (M):

Then the graph of dierential da is a smooth submanifold in T  M:

L0 = fdq a j q 2 M g T  M dim L0 = dimM = n:

Translations of L0 by the ow of the Hamiltonian vector eld

Lt = etH~ (L0 ) are smooth n-dimensional submanifolds in T  M, and the graph of the mapping t 7! Lt , L = f( t) j  2 Lt 0  t  t1g T  M  is a smooth (n + 1)-dimensional submanifold in T  M  .

R

Consider the 1-form

R

R

s ; H dt 2 1 (T  M  ): Recall that s is the tautological 1-form on T  M, s =  , and its dierential is the canonical symplectic structure on T  M, ds = . In mechanics, the form s ; H dt = p dq ; H dt is called the integral invariant of Poincare-Cartan on the extended phase space T  M  . Proposition 17.1. The form (s ; H dt)jL is exact. Proof. First we prove that the form is closed: 0 = d(s ; H dt)jL = ( ; dH ^ dt)jL : (17.5)

R

(1) Fix Lt = L\ft = constg and consider restriction of the form  ; dH ^ dt to Lt . We have ( ; dH ^ dt)jLt = jLt tH~  = , thus since dtjLt = 0. Recall that ed

d 

jLt = etH~  But sjL0 = d(a  )jL0 , hence

L0

= jL0 = dsjL0 :

17.1 Sucient Optimality Conditions

237

dsjL0 = d  d(a  )jL0 = 0: We proved that ( ; dH ^ dt)jLt = 0. (2) The manifold L is the image of the smooth mapping   ( t) 7! etH~  t  thus the tangent vector to L transversal to Lt is ~ + @ 2 T(t)L: H() @t So   ~ [email protected] : T(t)L = T(t)Lt ! H() @t @ ~ To complete the proof, we substitute the vector H()+ @ t as the rst argument to  ; dH ^ dt and show that the result is equal to zero. We have: iH~  = ;dH i @@t  = 0 ;  ;  iH~ (dH ^ dt) = iH~ dH ^ dt ; dH ^ iH~ dt = 0

R

| {z } | {z } =0 =0    i @@t (dH ^ dt) = i @@t dH ^ dt ; dH ^ i @@t dt = ;dH | {z } | {z } =0

consequently,

=1

iH~ + @@t ( ; dH ^ dt) = ;dH + dH = 0: We proved that the form (s ; H dt)jL is closed. (3) Now we show that this form is exact, i.e.,

Z



s ; H dt = 0

(17.6)

for any closed curve  : 7! (( ) t( )) 2 L 2 0 1]: The curve  is homotopic to the curve 0 : 7! (( ) 0) 2 L0 2 0 1]: Since the form (s ; H dt)jL is closed, Stokes' theorem yields that

Z



s ; H dt =

Z

0

s ; H dt:

But the integral over the closed curve 0 L0 is easily computed:

Z

0

s ; H dt =

Z

0

s=

Z

0

d(a  ) = 0:

Equality (17:6) follows, i.e., the form (s ; H dt)jL is exact.

ut

238

17 Sucient Optimality Conditions

17.1.2 Problem with Fixed Time Now we prove sucient optimality conditions for problem (17:1){(17:3). Theorem 17.2. Assume that the restriction of projection jLt is a di eomorphism for any t 2 0 t1]. Then for any 0 2 L0, the normal extremal trajectory

q~(t) =   etH~ (0 )

0  t  t1  R realizes a strict minimum of the cost functional 0t1 '(q(t) u(t)) dt among all admissible trajectories q(t), 0  t  t1, of system (17:1) with the same bound-

ary conditions:

q(0) = q~(0) q(t1) = q~(t1 ): (17.7) Remark 17.3. (1) Under the hypotheses of this theorem, no check of existence of optimal control is required. (2) If all assumptions (smoothness of H, extendibility of trajectories of H~ to the time segment 0 t1], dieomorphic property of jLt ) hold in a proper open domain O T  M, then the statement can be modied to give local optimality of q~( ) in (O). These modications are left to the reader. Now we prove Theorem 17.2. Proof. The curve q~(t) is projection of the normal extremal ~t = etH~ (0 ): Let u~(t) be an admissible control that maximizes the Hamiltonian along this extremal: H(~t ) = hu~(t) (~t ): On the other hand, let q(t) be an admissible trajectory of system (17:1) generated by a control u(t) and satisfying the boundary conditions (17:7). We compare costs of the pairs (~q u~) and (q u). Since  : Lt ! M is a dieomorphism, the trajectory fq(t) j 0  t  t1g M can be lifted to a smooth curve f(t) j 0  t  t1g T  M: 8t 2 0 t1] 9! (t) 2 Lt such that ((t)) = q(t): Then

Zt 0

1

'(q(t) u(t)) dt =

 = =

Zt

1

Zt 0

1

Zt 0

Z

0



1

h(t) fu(t) (q(t))i ; hu(t)((t)) dt h(t) q(t) _ i ; H((t)) dt _ i ; H((t)) dt hs(t)  (t)

s ; H dt

(17.8)

17.1 Sucient Optimality Conditions

239

where

 : t 7! ((t) t) 2 L t 2 0 t1]: By Proposition 17.1, the form (s ; H dt)jL is exact. Then integral of the form (s ; H dt)jL along a curve depends only upon endpoints of the curve. The curves  and ~ : t 7! (~t  t) 2 L ~t = etH~ (0 ) t 2 0 t1] have the same endpoints (see Fig. 17.1), thus

Z



s ; H dt = = =

So

Zt

1

0

Z

~

s ; H dt =

Zt

1

Zt 0 0

1

Zt

1

0

h~t  fu~(t) (~q(t))i ; hu~(t)(~t ) dt '(~q(t) u~(t)) dt:

'(q(t) u(t)) dt 

i.e., the trajectory q~(t) is optimal.

Zt

1

0

(t)

(0  0)

'(~q(t) u~(t)) dt

(17.9)

~ t ) ( t1 1

L

~ (t)

q(t)

q~(0)

h~t  q~_ (t)i ; H(~t ) dt

q~(t1)

q~(t)

M

Fig. 17.1. Proof of Th. 17.2 It remains to prove that the minimum of the pair (~q(t) u~(t)) is strict, i.e, that inequality (17:9) is strict.

240

17 Sucient Optimality Conditions

For a xed point q 2 M, write cotangent vectors as  = (p q), where p are coordinates of a covector  in Tq M. The control-dependent Hamiltonians hu(p q) are ane w.r.t. p, thus their maximum H(p q) is convex w.r.t. p. Any vector 2 Tq M such that

hp i = max hp fu (q)i u2U

denes a hyperplane of support to the epigraph of the mapping p 7! H(p q). Since H is smooth in p, such a hyperplane of support is unique and maximum in (17:4) is attained at a unique velocity vector. If q(t) 6 q~(t), then inequality (17:8) becomes strict, as well as inequality (17:9). The theorem is proved. ut Sucient optimality condition of Theorem 17.2 is given in terms of the manifolds Lt, which are in turn dened by a function a and the Hamiltonian ~ One can prove optimality of a normal extremal trajectory q~(t), ow of H. t 2 0 t1], if one succeeds to nd an appropriate function a 2 C 1 (M) for which the projections  : Lt ! M, t 2 0 t1], are dieomorphisms. For t = 0 the projection  : L0 ! M is a dieomorphism. So for small t > 0 any function a 2 C 1 (M) provides manifolds Lt projecting dieomorphically to M, at least if we restrict ourselves by a compact K b M. Thus the sucient optimality condition for small pieces of extremal trajectories follows. Corollary 17.4. For any compact K b M that contains a normal extremal trajectory

q~(t) =   etH~ (0 ) there exists t01 2 (0 t1] such that the piece q~(t)

0  t  t1 

0  t  t01

is optimal w.r.t. all trajectories contained in K and having the same boundary conditions.

In many problems, one can choose a suciently large compact K  q~ such that the functional J is separated from below from zero on all trajectories leaving K (this is the case, e.g., if '(q u) > 0). Then small pieces of q~ are globally optimal.

17.1.3 Problem with Free Time For problems with integral cost and free terminal time t1, a sucient optimality condition similar to Theorem 17.2 is valid, see Theorem 17.6 below. Recall that all normal extremals of the free time problem lie in the zero level H ;1 (0) of the maximized Hamiltonian H. First we prove the following auxiliary proposition.

17.1 Sucient Optimality Conditions

241

Proposition 17.5. Assume that 0 is a regular value of the restriction H jL0 , i.e. d H jT L0 6= 0 for all  2 L0 \ H ;1(0). Then the mapping  : L0 \ H ;1(0)  ! T  M (0 t) = etH~ (0 )

R

is an immersion and bs is an exact form. Proof. First of all, regularity of the value 0 for H jL0 implies that L0 \ H ;1(0)

is a smooth manifold. Then, the exactness of bs easily follows from Proposition 17.1. To prove that  is an immersion, it is enough to show that the vector @  (0  t) = H( ~ t ), t = (0  t), is not tangent to the image of L0 \ H ;1(0) @t under the dieomorphism etH~ : T  M ! T  M for all 0 2 L0 \ H ;1(0). Note that etH~ (L0 \ H ;1(0)) = Lt \ H ;1(0). We are going to prove a little bit more ~ t ) is not tangent to Lt. than we need, namely, that H( Indeed, Proposition 17.1 implies that jLt = dsjLt = 0. Hence it is enough to show that the form (iH~ )jLt does not vanish at the point t . Recall that ~ In particular, the Hamiltonian ow etH~ preserves both  and H.

;



tH~ (i ~ )jL0 = ;ed tH~ (dH jL0 ) : (iH~ )jLt = ed H

tH~ is invertible. So it is enough to prove that dH jL0 does not The mapping ed vanish at 0 . But the last statement is our assumption! ut Now we obtain a sucient optimality condition for the problem with free time. Theorem 17.6. Let W be a domain in L0 \ H ;1 (0)  such that   jW : W ! M is a di eomorphism of W onto a domain in M , and let ~t = etH~ (~0 ) t 2 0 t1] be a normal extremal such that (~0  t) 2 W for all t 2 0 t1]. Then the extremal trajectory q~(t) = (~tR) (with the corresponding control u~(t)) realizes a strict minimum of the cost 0 '(q(t) u(t)) dt among all admissible trajectories such that q(t) 2   (W) for all t 2 0 ], q(0) = q~(0), q( ) = q~(t1 ), > 0. Proof. Set L = (W), then  : L ! (L) is a dieomorphism and sjL is an exact form. Let q(t), t 2 0 ], be an admissible trajectory generated by a control u(t) and contained in (L), with the boundary conditions q(0) = q~(0), q( ) = q~(t1). Then q(t) = ((t)), 0  t  , where t 7! (t) is a smooth curve in L such that R(0) = ~0 , ( ) = ~t1 . R We have s = s. Further,

R

(  )

~ 

242

17 Sucient Optimality Conditions

Z

s=

~ 

Zt D 1

0

E

~ t q~_ (t) dt =

Zt D 1

E

~t  fu~(t) (~q(t)) dt =

0

Zt

1

'(~q(t) u~(t)) dt:

0

The last equality follows from the fact that D~ E (t) fu~(t) (~q(t)) ; '(~q (t) u~(t)) = H(~(t)) = 0: On the other hand,

Z

(  )

Z

Z

0

0

s = h(t) q(t) _ i dt =

Z

Z

hu(t)((t)) dt + '(q(t) u(t)) dt 0

 '(q(t) u(t)) dt: 0

The last inequality follows since max h ((t)) = H((t)) = 0. Moreover, the u2U u inequality is strict if the curve t 7! (t) is not a solution of the equation ~ _ = H(), i.e., if it does not coincide with ~(t). Summing up,

Zt

1

0

Z

'(~q (t) u~(t)) dt  '(q(t) u(t)) dt 0

ut

and the inequality is strict if q diers from q~.

17.2 Hamilton-Jacobi Equation Suppose that conditions of Theorem 17.2 are satised. As we showed in the proof of this theorem, the form (s ; H dt)jL is exact, thus it coincides with dierential of some function:

R

(s ; H dt)jL = dg

R

R

g : L! :

R

(17.10)

Since the projection  : Lt ! M is one-to-one, we can identify ( t) 2 L with (q t) 2 M  and dene g as a function on M  :

Lt 

g = g(q t):

In order to understand the meaning of the function g for our optimal control problem, consider an extremal ~ t = etH~ (0 )

17.2 Hamilton-Jacobi Equation

243

and the curve

~ L ~ : t 7! (~t  t) as in the proof of Theorem 17.2. Then

Z

~

s ; H dt =

Zt

1

0

'(~q( ) u~( )) d 

(17.11)

where q~(t) = (~t ) is an extremal trajectory and u~(t) is the control that maximizes the Hamiltonian hu () along ~t . Equalities (17:10) and (17:11) mean that Zt g(~q(t) t) = g(q0 0) + '(~q ( ) u~( )) d  0

i.e., g(q t) ; g(q0  0) is the optimal cost of motion between points q0 and q for the time t. Initial value for g can be chosen of the form g(q0 0) = a(q0)

q0 2 M:

(17.12)

Indeed, at t = 0 denition (17:11) of the function g reads dgjt=0 = (s ; H dt)jL0 = sjL0 = da

which is compatible with (17:12). We can rewrite equation (17:10) as a partial dierential equation on g. In local coordinates on M and T  M, we have q = x 2 M  = (  x) 2 T  M g = g(x t):

Then equation (17:10) reads i.e.,

( dx ; H(  x) dt)jL = dg(x t)

[email protected] g > < =  @x > : @@ gt = ;H(  x):

This system can be rewritten as a single rst order nonlinear partial dierential equation: @ g + H  @ g  x = 0 (17.13) @t @x which is called Hamilton-Jacobi equation . We showed that the optimal cost g(x t) satises Hamilton-Jacobiequation (17:13) with initial condition (17:12). Characteristic equations of PDE (17:13) have the form

244

17 Sucient Optimality Conditions

8 @H > x_ =  > < @@ H

_ = ; @ x  > > :d

d t g(x(t) t) = x_ ; H: ~ for normal The rst two equations form the Hamiltonian system _ = H() extremals. Thus solving our optimal control problem (17:1){(17:3) leads to the method of characteristics for the Hamilton-Jacobi equation for optimal cost.

17.3 Dynamic Programming One can derive the Hamilton-Jacobi equation for optimal cost directly, without Pontryagin MaximumPrinciple, due to an idea going back to Huygens and constituting a basis for Bellman's method of Dynamic Programming , see 3]. For this, it is necessary to assume that the optimal cost g(q t) exists and is C 1-smooth. Let an optimal trajectory steer a point q0 to a point q for a time t. Apply a constant control u on a time segment t t + t] and denote the trajectory starting at the point q by qu ( ), 2 t t + t]. Since qu (t + t) is the endpoint of an admissible trajectory starting at q0, the following inequality for optimal cost holds: g(qu (t + t) t + t)  g(q t) +

Z t+t t

'(qu ( ) u) d :

Divide by t: 1 (g(q (t + t) t + t) ; g(q t))  1 Z t+t '(q ( ) u) d u t u t t and pass to the limit as t ! 0: @ g  @g @ q  fu(q) + @ t  '(q u): So we obtain the inequality   @ g + h @ g  q  0 u 2 U: (17.14) @t u @q Now let (~q(t) u~(t)) be an optimal pair. Let t > 0 be a Lebesgue point of the control u~. Take any t 2 (0 t). A piece of an optimal trajectory is optimal, thus q~(t ; t) is the endpoint of an optimal trajectory, as well as q~(t). So the optimal cost g satises the equality:

17.3 Dynamic Programming

g(~q(t) t) = g(~q(t ; t) t ; t) +

Zt

t;t

245

'(~q ( ) u~( )) d :

We repeat the above argument: 1 (g(~q (t) t) ; g(~q(t ; t) t ; t)) = 1 Z t '(~q( ) u~( )) d  t t t;t take the limit t ! 0: @ g  @g +h (17.15) @ t u~(t) @ q  q = 0: This equality together with inequality (17:14) means that @ g  @ g  h q : hu~(t) @ q  q = max u2U u @ q We denote H(  q) = max h (  q) u2U u

and write (17:15) as Hamilton-Jacobi equation:   @ g + H @ g  q = 0: @t @q Thus derivative of the optimal cost @@ gq is equal to the impulse along the optimal trajectory q~(t). We do not touch here a huge theory on nonsmooth generalized solutions of Hamilton-Jacobi equation for smooth and nonsmooth Hamiltonians.

18 Hamiltonian Systems for Geometric Optimal Control Problems

18.1 Hamiltonian Systems on Trivialized Cotangent Bundle 18.1.1 Motivation Consider a control system described by a nite set of vector elds on a manifold M: q_ = fu (q)

u 2 f1 : : :  kg q 2 M:

(18.1)

We construct a parametrization of the cotangent bundle T  M adapted to this system. First, choose a basis in tangent spaces Tq M of the elds fu (q) and their iterated Lie brackets: Tq M = span(f1 (q) : : :  fn (q))

we assume that the system is bracket-generating. Then we have special coordinates in the tangent spaces:

8 v 2 Tq M

v=

R

( 1  : : :  n ) 2 n:

n X i=1

i fi (q)

R

Thus any tangent vector to M can be represented as an (n + 1)-tuple ( 1  : : :  n q)

( 1  : : :  n ) 2 n q 2 M

i.e., we obtain a kind of parametrization of the tangent bundle T M = q2M Tq M. One can construct coordinates on TM by choosing local coordinates in M, but such a choice is extraneous to our system, and we stay without any coordinates in M.

248

18 Hamiltonian Systems for Geometric Optimal Control Problems

Having in mind the Hamiltonian system of PMP, we pass to the cotangent bundle. Construct the dual basis in T  M: choose dierential forms !1  : : :  !n 2 1 M such that h!i fj i = ij  i j = 1 : : :  n: Then the cotangent spaces become endowed with special coordinates:

8  2 Tq M

=

R

n X i=1

i !iq 

(1 : : :  n) 2 n: So we obtain a kind of parametrization of the cotangent bundle:  7! (1  : : :  n q) (1  : : :  n) 2 n q 2 M: In notation of Sect. 11.5, i = fi () = h fi (q)i is the linear on bers Hamiltonian corresponding to the eld fi . Canonical coordinates on T  M arise in a similar way from commuting vector elds fi = @@xi , i = 1 : : :  n, corresponding to local coordinates (x1  : : :  xn) on M. Consequently, in the (only interesting in control theory) case where the elds fi do not commute, the \coordinates" (1 : : :  n q) on T  M are not canonical. Now our aim is to write Hamiltonian system in these nonstandard coordinates on T  M, or in other natural coordinates adapted to the control system in question.

R

18.1.2 Trivialization of T M

Let M be a smooth manifold of dimension n, and let E be an n-dimensional vector space. Suppose that we have a trivialization of the cotangent bundle T  M, i.e., a dieomorphism  : E  M ! T M such that: (1) the diagram  T M E  M ;;;;!

?? y

M is commutative, i.e.,   (e q) = q

?? y

M

e 2 E q 2 M

18.1 Hamiltonian Systems on Trivialized Cotangent Bundle

249

(2) for any q 2 M the mapping e 7! (e q) e 2 E is a linear isomorphism of vector spaces: (  q) : E ! Tq M: So the space E is identied with any vertical ber Tq M, it is a typical ber of the cotangent bundle T  M. For a xed vector e 2 E, we obtain a dierential form on M: e def = (e ) 2 1 M: In the previous section we had E = f(1 : : :  n)g = n (e q) =

n X i=1

i!iq 

R

but now we do not x any basis in E.

18.1.3 Symplectic Form on E M

In order to write a Hamiltonian system on E  M  = T  M, we compute the symplectic form b on E  M. We start from the Liouville form s 2 1 (T  M) and evaluate its pull-back bs 2 1(E  M): The tangent and cotangent spaces are naturally identied with the direct products: T(eq) (E  M)  = Te E ! Tq M  = E ! Tq M     T(eq) (E  M) = Te E ! Tq M  = E  ! Tq M: Any vector eld V 2 Vec(E  M) is a sum of its vertical and horizontal parts: V = Vv + Vh  Vv (e q) 2 E Vh (e q) 2 Tq M: Similarly, any dierential form ! 2 1(E  M) is decomposed into its vertical and horizontal parts:

250

18 Hamiltonian Systems for Geometric Optimal Control Problems

! = !v + !h 

!v (eq) 2 E   !h (eq) 2 Tq M:

The vertical part !v vanishes on horizontal tangent vectors, while the horizontal one !h vanishes on vertical tangent vectors. In particular, vector elds and dierential forms on M (possibly depending on e 2 E) can be considered as horizontal vector elds and dierential forms on E  M: Tq M = 0 ! Tq M T(eq) (E  M) Tq M = 0 ! Tq M T(eq) (E  M):

Compute the action of the form bs on a tangent vector (  v) 2 Te E ! Tq M:

hbs (  v)i = hs(eq)  (  v)i = hs(eq)  (  v)i = h(e q) vi:

Thus

(bs)(eq) = (e q)

(18.2)

where  in the right-hand side of (18:2) is considered as a horizontal form on E  M. We go on and compute the pull-back of the standard symplectic form: b = bds = dbs = d:

Recall that dierential of a form ! 2 1 (N) can be evaluated by formula (11:15): d!(W1  W2) = W1 h! W2i ; W2 h! W1 i ; h! W1 W2]i W1  W2 2 Vec N: (18.3) In our case N = E  M we take test vector elds of the form Wi = ( i  Vi) 2 Vec(E  M)

i = 1 2

where i = const 2 E are constant vertical vector elds and Vi 2 Vec M are horizontal vector elds. By (18:3), d(( 1 V1 ) ( 2 V2)) = ( 1  V1)h(  ) V2i ; ( 2  V2)h(  ) V1i ; h(  ) V1 V2]i since ( 1  V1) ( 2 V2 )] = V1 V2]. Further, (( 1  V1)h(  ) V2i)(e ) = ( 1 h(  ) V2i + V1 h(  ) V2i)(e ) and taking into account that  is linear w.r.t. e

18.1 Hamiltonian Systems on Trivialized Cotangent Bundle

251

= h1  V2i + V1 he  V2i: Consequently, d(( 1 V1 ) ( 2 V2))(e ) = h1  V2i ; h2  V1 i + V1he  V2i ; V2 he  V1i ; he V1 V2 ]i: We denote the rst two terms e(( 1  V1) ( 2 V2 )) = h1  V2i ; h2  V1i and apply formula (18:3) to the horizontal form e: de(V1  V2 ) = V1 he  V2i ; V2 he V1 i ; he V1 V2]i: Finally, we obtain the expression for pull-back of the symplectic form: b(e ) (( 1  V1) ( 2  V2)) = e(( 1  V1 ) ( 2 V2)) + de (V1  V2) (18.4) i.e., b(e ) = e + de: Remark 18.1. In the case of canonical coordinates we can take test vector elds Vi = @@xi , then it follows that de = 0.

18.1.4 Hamiltonian System on E M

Formula (18:4) describes the symplectic structure b on E  M. Now we compute the Hamiltonian vector eld corresponding to a Hamiltonian function h 2 C 1 (E  M): One can consider h as a family of functions on M parametrized by vectors from E: he = h(e ) 2 C 1 (M) e 2 E: Decompose the required Hamiltonian vector eld into the sum of its vertical and horizontal parts: ~h = X + Y X = X(e q) 2 E Y = Y (e q) 2 Tq M: By denition of a Hamiltonian eld, (18.5) iX +Y b = ;dh: Transform the both sides of this equality:

252

18 Hamiltonian Systems for Geometric Optimal Control Problems

;dh = ; @@ he ; |{z} dhe  | {z } 2T  M 2E b iX +Y je  = iX +Y je (e + de ) = i(XY ) e e + i(XY ) e de de} : = h|X{z }i ; h| {z Y i} + i|Y {z   2T  M 2T M 2E

Now we equate the vertical parts of (18:5): h  Y i = @@ he  (18.6) from this equation we can nd the horizontal part Y of the Hamiltonian eld ~h. Indeed, the linear isomorphism (  q) : E ! Tq M has a dual mapping  (  q) : Tq M ! E  : Then equation (18:6) can be written as (  q)Y = @@ he (e q) and then solved w.r.t. Y : Y = ;1 @@ he : To nd the vertical part X of the eld ~h, we equate the horizontal parts of (18:5): X + iY de = ;dhe  rewrite as X = ;iY de ; dhe  and solve this equation w.r.t. X: X = ;;1 (iY de + dhe): Thus the Hamiltonian system on E  M corresponding to a Hamiltonian h has the form: 8

q _ = < i=1 @ ui Vi n X > > u _ = ; V h + uk ckij @@uh  i = 1 : : :  n: i u :i j jk=1 Remark 18.2. This system becomes especially simple (triangular) when the

Hamiltonian does not depend upon the point in the base: @ h = 0: @q The vertical subsystem simplies even more when ckij = const i j k = 1 : : :  n: Both these conditions are satised for invariant problems on Lie groups discussed in subsequent sections.

18.2 Lie Groups

255

The structural constants ckij can easily be expressed in terms of Lie brackets of basis vector elds. Proposition 18.3. Let the frame of vector elds V1  : : :  Vn 2 Vec M be dual to the frame of 1-forms !1  : : :  !n 2 1(M):

h!i  Vj i = ij  Then

d!k =

if and only if

n 1 X

ij =1 2

Vi Vj ] = ;

i j = 1 : : :  n:

ckij !i ^ !j 

n X k=1

ckij Vk 

k = 1 : : :  n

i j = 1 : : :  n:

Proof. The equality for d!k can be written as

hd!k  (Vi  Vj )i = ckij 

i j k = 1 : : :  n:

The left-hand side is computed by formula (18:3):

hd!k  (Vi  Vj )i = V| i h!{zk  Vj i} ; V| j h!{zk  Vii} ;h!k  Vi Vj ]i =0

=0

and the statement follows. ut If the coecients ckij are constant, then the vector elds V1  : : :  Vn span a nite-dimensional Lie algebra, and the numbers ckij are called structural constants of this Lie algebra. As we mentioned above, for general vector elds ckij 6 const.

18.2 Lie Groups State spaces for many interesting problems in geometry, mechanics, and applications are often not just smooth manifolds but Lie groups, in particular, groups of transformations. A manifold with a group structure is called a Lie group if the group operations are smooth. The cotangent bundle of a Lie group has a natural trivialization. We develop an approach of the previous section and study optimal control problems on Lie groups.

256

18 Hamiltonian Systems for Geometric Optimal Control Problems

18.2.1 Examples of Lie Groups The most important examples of Lie groups are given by groups of linear transformations of nite-dimensional vector spaces. The group of all nondegenerate linear transformations of n is called the general linear group : GL(n) = fX : n ! n j det X 6= 0g: Linear volume-preserving transformations of n form the special linear group :

R

R R R f R !R j g R R f R !R j  g f R !R j  g C C R C C R  C  ;  j R ! R 6 R   j R !R  ;   ;  C\ C    C ; j R !R R

n det X = 1 : SL(n) = X : n Another notation for these groups is respectively GL( n) and SL( n). The orthogonal group is formed by linear transformations preserving Euclidean structure: n X X = Id  O(n) = X : n and orthogonal orientation-preserving transformations form the special orthogonal group : n X X = Id det X = 1 : SO(n) = X : n One can also consider the complex and Hermitian versions of these groups: GL( n ) SL( n ) U(n) SU(n) for this one should replace in the denitions above n by n . Each of these groups realizes as a subgroup of the corresponding real or orthogonal group. Namely, the general linear group GL( n ) and the unitary group U(n) can be considered respectively as the subgroups of GL( 2n) or O(2n) commuting with multiplication by the imaginary unit:

GL( n ) =

U(n) =

A B BA

A B BA

A B : n

n

det2 A + det2 B = 0

GL( 2n)

A B : n

AA + BB = Id BA

n

AB = 0

GL( n ) O(2n):

The special linear group SL( n ) and the special unitary group SU(n) realize as follows: A B A B : n n det(A + iB) = 1 SL( n ) = SL( 2n) BA

  SU(n) = ;AB B A j A B :

18.2 Lie Groups

R !R n

n

257



AA + BB  = Id BA ; AB  = 0 det(A + iB) = 1

C

= U(n) \ SL( n ) SO(2n): Lie groups of linear transformations are called linear Lie groups . These groups often appear as a state space of a control system: e.g., SO(n) arises in the study of rotating congurations. For such systems, one can consider, as usual, the problems of controllability and optimal control.

18.2.2 Lie's Theorem for Linear Lie Groups Consider a control system of the form X_ = XA X 2 M = GL(N) A 2 A gl(N)

(18.8)

where A is an arbitrary subset of gl(N), the space of all real N  N matrices. We compute orbits of this system. Systems of the form (18:8) are called leftinvariant : they are preserved by multiplication from the left by any constant matrix Y 2 GL(N). Notice that the ODE with a constant matrix A X_ = XA is solved by the matrix exponential: X(t) = X(0)etA : Lie bracket of left-invariant vector elds is left-invariant as well: XA XB] = XA B]

(18.9)

this follows easily from the coordinate expression for commutator (exercise). Remark 18.4. Instead of left-invariant systems X_ = XA, we can consider right-invariant ones: X_ = CX. These forms are equivalent and transformed one into another by the inverse of matrix. Although, the Lie bracket for rightinvariant vector elds is CX DX] = D C]X which is less convenient than (18:9). Return to control system (18:8). By the Orbit Theorem, the orbit through identity OId (A) is an immersed submanifold of GL(N). Moreover, by denition, the orbit admits the representation via composition of ows:

258

R

N

18 Hamiltonian Systems for Geometric Optimal Control Problems

OId (A) = fId et1 A1   etk Ak j ti 2  Ai 2 A k 2 g thus via products of matrix exponentials

R

N

= fet1 A1 etk Ak j ti 2  Ai 2 A k 2 g: Consequently, the orbit OId (A) is a subgroup of GL(N). Further, in the proof of the Orbit Theorem we showed that the point q  et1 A1   etk Ak depends continuously on (t1  : : :  tk ) in the \strong" topology of the orbit, thus it depends smoothly. To summarize, we showed that the orbit through identity has the following properties: (1) OId (A) is an immersed submanifold of GL(N), (2) OId (A) is a subgroup of GL(N), (3) the group operations (X Y ) 7! XY , X 7! X ;1 in OId (A) are smooth. In other words, the orbit OId (A) is a Lie subgroup of GL(N). The tangent spaces to the orbit are easily computed via the analytic version of the Orbit theorem (system (18:8) is real analytic): TId OId (A) = Lie(A) (18.10) TX OId (A) = X Lie(A): The orbit of the left-invariant system (18:8) through any point X 2 GL(N) is obtained by left translation of the orbit through identity: OX (A) = fXet1 A1 etk Ak j ti 2  Ai 2 A k 2 g = X OId (A): We considered before system (18:8) dened by an arbitrary subset A gl(N). Restricting to Lie subalgebras A = Lie(A) gl(N) we see that the following proposition was proved: to any Lie subalgebra A gl(N), there corresponds a connected Lie subgroup M GL(N) such that TId M = A. Here M = OId A. Now we show that this correspondence is invertible. Let M be a connected Lie subgroup of GL(N), i.e.: (1) M is an immersed connected submanifold of GL(N), (2) M is a group w.r.t. matrix product, (3) the group operations (X Y ) 7! XY , X 7! X ;1 in M are smooth mappings. Then Id 2 M. Consider the tangent space   d TId M = A = d t ;t j ;t 2 M ;t smooth ;0 = Id : t=0

R

N

18.2 Lie Groups

Since M

GL(N) gl(N), then TId M

Further, since

259

gl(N):

A 2 TId M X 2 M ) XA 2 TX M

XA = ddt X;t  t=0 the velocity of the curve X;t , where A = ;_0. Consequently, for any A 2 TId M the vector eld XA is identically tangent to M. So the following control system is well-dened on M: X_ = XA X 2 M A 2 TId M: This system has a full rank. Since the state space M is connected, it coincides with the orbit OId of this system through identity. We have already computed the tangent space to the orbit of a left-invariant system, see (18:10), thus TId M = TId (OId ) = Lie(TId M): That is, TId M is a Lie subalgebra of gl(N). We proved the following classical proposition.

Theorem 18.5 (Lie). There exists a one-to-one correspondence between Lie

subalgebras A TId M = A.

gl(N) and connected Lie subgroups M

GL(N) such that

We showed that Lie's theorem for linear Lie algebras and Lie groups follows from the Orbit Theorem: connected Lie subgroups are orbits of left-invariant systems dened by Lie subalgebras, and Lie subalgebras are tangent spaces to Lie subgroups at identity.

18.2.3 Abstract Lie Groups An abstract Lie group is an abstract smooth manifold (not considered embedded into any ambient space) which is simultaneously a group, with smooth group operations. There holds Ado's theorem 139] stating that any nitedimensional Lie algebra is isomorphic to a Lie subalgebra of gl(N). A similar statement for Lie groups is not true: a Lie group can be represented as a Lie subgroup of GL(N) only locally, but, in general, not globally. Although, the major part of properties of linear Lie groups can be generalized for abstract Lie groups. In particular, let M be a Lie group. For any point q 2 M, the left product by q: q% : M ! M q%(x) = qx x 2 M

260

18 Hamiltonian Systems for Geometric Optimal Control Problems

is a dieomorphism of M. Any tangent vector v 2 TId M can be shifted to any point q 2 M by the left translation q%: V (q) = q%v 2 Tq M q 2 M thus giving rise to a left-invariant vector eld on M: V 2 Vec M q% V = V q 2 M: There is a one-to-one correspondence between left-invariant vector elds on M and tangent vectors to M at identity: V 7! V (Id) = v: Left translations in M preserve ows of left-invariant vector elds on M, thus ows of their commutators. Consequently, left-invariant vector elds on a Lie group M form a Lie algebra, called the Lie algebra of the Lie group M. The tangent space TId M is thus also a Lie algebra. Then, similar to the linear case, one can prove Lie's theorem on one-to-one correspondence between Lie subgroups of a Lie group M and Lie subalgebras of its Lie algebra A.

18.3 Hamiltonian Systems on Lie Groups

18.3.1 Trivialization of the Cotangent Bundle of a Lie Group Let M GL(N) be a Lie subgroup. Denote by M the corresponding Lie subalgebra: M = TId M gl(N):

The cotangent bundle of M admits a trivialization of the form  : M  M ! T  M where M is the dual space to the Lie algebra M. We start from describing the dual mapping  : TM ! M  M: Recall that Tq M = qTId M = qM for any q 2 M. We set  : qa 7! (a q) a 2 M q 2 M qa 2 Tq M: (18.11) I.e., the value of a left-invariant vector eld qa at a point q is mapped to the pair consisting of the value of this eld at identity and the point q. Then the trivialization  has the form:  : (x q) 7! x%q  x 2 M  q 2 M x%q 2 Tq M (18.12) where x% is the left-invariant 1-form on M coinciding with x at identity: hx%q  qai def = hx ai:

18.3 Hamiltonian Systems on Lie Groups

18.3.2 Hamiltonian System on M

261

M

The Hamiltonian system corresponding to a Hamiltonian h = h(x q) 2 C 1 (M  M)

was computed in Sect. 18.1, see (18:7): 8

:x_ = ad @@ hx x ; ;1dhx: Recall that dhx is the horizontal part of dh, thus (dhx )q 2 Tq M

and

(x q) 2 M  M

;1dhx 2 M :

262

18 Hamiltonian Systems for Geometric Optimal Control Problems

System (18:14) describes the Hamiltonian system for an arbitrary Lie group and any Hamiltonian function h. In the case of commutative Lie groups (which arise in trivialization of T  M generated by local coordinates in M), the rst term in the second equation (18:14) vanishes, and we obtain the usual form of Hamiltonian equations in canonical coordinates: 8 @h

:x_ = ad @@ xh x: Here the second equation does not contain q. So in left-invariant control problems, where the Hamiltonian h of PMP is left-invariant, one can solve the equation for vertical coordinates x independently, and then pass to the horizontal equation for q.

18.3.3 Compact Lie Groups The Hamiltonian system (18:15) simplies even more in the case of compact Lie groups. Let M be a compact Lie subgroup of GL(N). Then M can be considered as a Lie subgroup of the orthogonal group O(N). Indeed, one can choose a Euclidean structure h  i in N invariant w.r.t. all transformations from M: hAv Awi = hv wi v w 2 N  A 2 M GL(N): Such a structure can be obtained from any Euclidean structure g(  ) on N by averaging over A 2 M using a volume form !1 ^ : : : ^ !n, where !i are basis left-invariant forms on M:

R

hv wi =

R

R

Z

vw !1 ^ : : : ^ !n vw (A) = g(Av Aw) A 2 M: So we will assume that elements of M are orthogonal N  N matrices, and the tangent space to M at identity consists of skew-symmetric matrices: M = TId M TId O(N) = so(N) = fa : N ! N j a + a = 0g: M

R R

18.3 Hamiltonian Systems on Lie Groups

263

There is an invariant scalar product on so(N) dened as follows: ha bi = ; tr ab a b 2 so(N): This product is invariant in the sense that het ad c a et ad c bi = ha bi a b c 2 so(N) t 2  (18.16) i.e., the operator Ad etc = et ad c : so(N) ! so(N) is orthogonal w.r.t. this product. Equality (18:16) is a corollary of the invariance of trace: het ad c a et ad cbi = h(Ad etc )a (Ad etc)bi = hetc ae;tc  etcbe;tci = ; tr(etc ae;tc etc be;tc) = ; tr(etc abe;tc ) = ; tr(ab) = ha bi: The sign minus in the denition of the invariant scalar product on so(N) provides positive-deniteness of the product. This can be easily seen in coordinates: if a = (aij ) b = (bij ) 2 so(N) aij = ;aji bij = ;bji  i j = 1 : : :  N then N N X X ; tr(ab) = ; aij bji = aij bij :

R

ij =1

The norm on so(N) is naturally dened:

p

jaj = ha ai

ij =1

a 2 so(N): The innitesimal version of the invariance property (18:16) is easily obtained by dierentiation at t = 0: hc a] bi + ha c b]i = 0 a b c 2 so(N): (18.17) That is, all operators ad c : so(N) ! so(N) c 2 so(N) are skew-symmetric w.r.t. the invariant scalar product. Equality (18:17) is a multidimensional generalization of a property of vector and scalar products in 3  = so(3). Since M so(N), there is an invariant scalar product in the Lie algebra M. Then the dual space M can be identied with the Lie algebra M via the scalar product h  i:

R

264

18 Hamiltonian Systems for Geometric Optimal Control Problems

M ! M 

a 7! ha i:

In terms of this identication, the operator (ad a) , a 2 M, takes the form: (ad a) : M ! M

In the case of a compact Lie group M, Hamiltonian system (18:15) for an invariant Hamiltonian h = h(a) becomes dened on M  M and reads 8 @h >

:a_ = a @@ ha : We apply this formula in the next chapter for solving several geometric optimal control problems.

19 Examples of Optimal Control Problems on Compact Lie Groups

19.1 Riemannian Problem Let M be a compact Lie group. The invariant scalar product h  i in the Lie algebra M = TId M denes a left-invariant Riemannian structure on M:

hqu qviq def = hu vi

u v 2 M q 2 M qu qv 2 Tq M:

So in every tangent space Tq M there is a scalar product h  iq . For any Lipschitzian curve q : 0 1] ! M its Riemannian length is dened as integral of velocity: l=

Z1 0

jq(t) _ j dt

p

jq_j = hq_ q_i:

The problem is stated as follows: given any pair of points q0  q1 2 M, nd the shortest curve in M that connects q0 and q1. The corresponding optimal control problem is as follows: q_ = qu q 2 M u 2 M q(0) = q0 q(1) = q1 q0 q1 2 M xed l(u) =

Z1 0

(19.1) (19.2) (19.3)

ju(t)j dt ! min:

First of all, we prove existence of optimal controls. Parametrizing trajectories of control system (19:1) by arc length, we see that the problem with unbounded admissible control u 2 M on the xed segment t 2 0 1] is equivalent to the problem with the compact space of control parameters U = fjuj = 1g and free terminal time. Obviously, afterwards we can extend the set of control

266

19 Examples of Optimal Control Problems on Compact Lie Groups

parameters to U = fjuj  1g so that the set of admissible velocities fU (q) become convex. Then Filippov's theorem implies existence of optimal controls in the problem obtained, thus in the initial one as well. By Cauchy-Schwartz inequality, (l(u))2 =

Z 1 0

2 Z 1

ju(t)j dt 

0

ju(t)j2 dt

moreover, the equality occurs only if ju(t)j  const. Consequently, the Riemannian problem l ! min is equivalent to the problem Z1 1 J(u) = 2 ju(t)j2 dt ! min: (19.4) 0 The functional J is more convenient than l since J is smooth and its extremals are automatically curves with constant velocity. In the sequel we consider the problem with the functional J: (19:1){(19:4). The Hamiltonian of PMP for this problem has the form: hu(a q) = ha%q  qui + 2 juj2 = ha ui + 2 juj2: The maximality condition of PMP is:  2 hu(t) (a(t) q(t)) = vmax 2M(ha(t) vi + 2 jvj )   0: (1) Abnormal case:  = 0. The maximality condition implies that a(t)  0. This contradicts PMP since the pair ( a) should be nonzero. So there are no abnormal extremals. (2) Normal case:  = ;1. The maximality condition gives u(t)  a(t), thus the maximized Hamiltonian is smooth: H(a) = 12 jaj2: Notice that the Hamiltonian H is invariant (does not depend on q), which is a corollary of left-invariance of the problem. Optimal trajectories are projections of solutions of the Hamiltonian system corresponding to H. This Hamiltonian system has the form (see (18:18)):

(

q_ = qa a_ = a a] = 0: Thus optimal trajectories are left translations of one-parameter subgroups in M: q(t) = q0eta  a 2 M recall that an optimal solution exists. In particular, for the case q0 = Id, we obtain that any point q1 2 M can be represented in the form

19.2 A Sub-Riemannian Problem

267

q1 = ea  a 2 M: That is, any element q1 in a compact Lie group M has a logarithm a in the Lie algebra M.

19.2 A Sub-Riemannian Problem Now we modify the previous problem. As before, we should nd the shortest path between xed points q0, q1 in a compact Lie group M. But now admissible velocities q_ are not free: they should be tangent to a left-invariant distribution (of corank 1) on M. That is, we dene a left-invariant eld of tangent hyperplanes on M, and q(t) _ should belong to the hyperplane attached at the point q(t). A problem of nding shortest curves tangent to a given distribution is called a sub-Riemannian problem , see Fig. 19.1. _( )

q t

()

q t q

(0)

q(t)

_( ) 2 q(t) ( 1 ) ( ()) ! min q t

q t

l q

Fig. 19.1. Sub-Riemannian problem To state the problem as an optimal control one, choose any element b 2 M, jbj = 1. Then the set of admissible velocities at identity is the hyperplane U = b? = fu 2 M j hu bi = 0g: Remark 19.1. In the case M = SO(3), this restriction on velocities means that

we x an axis b in a rigid body and allow only rotations of the body around any axis u orthogonal to b. The optimal control problem is stated as follows. q_ = qu q 2 M u 2 U q(0) = q0 q(1) = q1 q0  q1 2 M xed l(u) =

Z1 0

ju(t)j dt ! min:

Similarly to the Riemannian problem, Filippov's theorem guarantees existence of optimal controls, and the length minimization problem is equivalent to the problem

268

19 Examples of Optimal Control Problems on Compact Lie Groups

Z1 J(u) = 12 ju(t)j2 dt ! min: 0

The Hamiltonian of PMP is the same as in the previous problem: hu (a q) = ha ui + 2 juj2 but the maximality condition diers since now the set U is smaller: hu(t)(a(t) q(t)) = max? (ha(t) vi + 2 jvj2): v 2b Consider rst the normal case:  = ;1. Then the Lagrange multipliers rule implies that the maximum max? h;v 1 (a q) v2b

is attained at the vector

vmax = a ; ha bib the orthogonal projection of a to U = b? . The maximized Hamiltonian is smooth: H(a) = 21 (jaj2 ; ha bi2) and the Hamiltonian system for normal extremals reads as follows:

(

q_ = q(a ; ha bib) a_ = ha bib a]:

The second equation has an integral of the form ha bi = const this is easily veried by dierentiation w.r.t. this equation: d d t ha bi = ha bihb a] bi by invariance of the scalar product = ;ha biha b b]i = 0: Consequently, the equation for a can be written as a_ = ha0  bib a] = ad(ha0  bib)a where a0 = a(0). This linear ODE is easily solved: a(t) = et ad(ha0 bib)a0 :

19.2 A Sub-Riemannian Problem

269

Now consider the equation for q:

  q_ = q et ad(ha0 bib) a0 ; ha0  bib

since et ad(ha0 bib) b = b = qet ad(ha0 bib)(a0 ; ha0 bib): (19.5) This ODE can be solved with the help of the Variations formula. Indeed, we have (see (2:29)): ;! Z t e ad f g d  etf  et(f +g) = exp i.e.,

0

Zt ;! exp e ad f g d = et(f +g)  e;tf

(19.6)

0

for any vector elds f and g. Taking f = ha0 bib g = a0 ; ha0  bib we solve ODE (19:5): q(t) = q0 eta0 e;tha0 bib : (19.7) Consequently, normal trajectories are products of two one-parameter subgroups. Consider the abnormal case:  = 0. The Hamiltonian h0u (a q) = ha ui u ? b attains maximum only if a(t) = (t)b (t) 2 : (19.8) But the second equation of the Hamiltonian system reads a_ = a u] (19.9) thus ha_  ai = ha u] ai = ;hu a a]i = 0: That is, a_ ? a. In combination with (19:8) this means that a(t) = const = b 6= 0 2 : (19.10) Notice that 6= 0 since the pair ( a(t)) should be nonzero. Equalities (19:9) and (19:10) imply that abnormal extremal controls u(t) satisfy the relation

R

R

270

19 Examples of Optimal Control Problems on Compact Lie Groups

u(t) b] = 0: That is, u(t) belong to the Lie subalgebra Hb = fc 2 M j c b] = 0g M: For generic b 2 M the subalgebra Hb is a Cartan subalgebra of M, thus Hb is Abelian. In this case the rst equation of the Hamiltonian system q_ = qu contains only mutually commuting controls: u( ) 2 Hb ) u( 1) u( 2)] = 0 and the equation is easily solved: R q(t) = q0 e 0t u( ) d : (19.11) Conversely, any trajectory of the form (19:11) with u( ) 2 Hb, 2 0 t] is abnormal: it is a projection of abnormal extremal (q(t) a(t)) with a(t) = b for any 6= 0. We can give an elementary explanation of the argument on Cartan subalgebra in the case M = so(n). Any skew-symmetric matrix b 2 so(n) can be transformed by a change of coordinates to the diagonal form: 0 i 1 1 BB ;i 1 CC B CC ; 1 i 2 TbT = B (19.12) [email protected] ;i 2 C A ...

C

for some T 2 GL(n ). But changes of coordinates (even complex) do not aect commutativity: c b] = 0 , TcT ;1 T bT ;1] = 0 thus we can compute the subalgebra Hb using new coordinates: Hb = T ;1 HTbT ;1 T: Generic skew-symmetric matrices b 2 so(n) have distinct eigenvalues, thus for generic b the diagonal matrix (19:12) has distinct diagonal entries. For such b the Lie algebra HTbT ;1 is easily found. Indeed, the commutator of a diagonal matrix 0 1 1 B 2 CC b=B [email protected] ... C A n

19.3 Control of Quantum Systems

271

with any matrix c = (cij ) is computed as follows: (ad b) c = ((i ; j )cij ): If a diagonal matrix b has simple spectrum: i ; j 6= 0

i 6= j

then the Lie algebra Hb consists of diagonal matrices of the form (19:12), consequently Hb is Abelian. So for a matrix b 2 so(n) with mutually distinct eigenvalues (i.e., for generic b 2 so(n)) the Lie algebra HTbT ;1 is Abelian, thus Hb is Abelian as well. Returning to our sub-Riemannian problem, we conclude that we described all normal extremal curves (19:7), and described abnormal extremal curves (19:11) for generic b 2 M. Exercise 19.2. Consider a more general sub-Riemannian problem stated in the same way as in this section, but with the space of control parameters U M any linear subspace such that its orthogonal complement U ? w.r.t. the invariant scalar product is a Lie subalgebra: U ?  U ?] U ? :

(19.13)

Prove that normal extremals in this problem are products of two oneparameter groups (as in the corank one case considered above): a? = const aU (t) = et ad a? a0U  q(t) = q0 eta e;ta? 

a0

U

= aU (0)

(19.14) (19.15) (19.16)

where a = aU + a? is the decomposition of a vector a 2 M corresponding to the splitting M = U ! U ? . We apply these results in the next problem.

19.3 Control of Quantum Systems This section is based on the paper of U. Boscain, T. Chambrion, and J.P. Gauthier 104]. Consider a three-level quantum system described by the Schr#odinger equation (in a system of units such that = 1): i_ = H (19.17)

RC

~

where  : ! 3 ,  = (1  2 3), is a wave function and

272

19 Examples of Optimal Control Problems on Compact Lie Groups

0E  0 1 1 1 H = @  1 E2 2 A

(19.18) 0  2 E3 is the Hamiltonian. Here E1 < E2 < E3 are constant energy levels of the system and i : ! are controls describing the inuence of the external pulsed eld. The controls are connected to the physical parameters by j (t) = j Fj (t)=2, j = 1 2, with Fj the external pulsed eld and j the couplings (intrinsic to the quantum system) that we have restricted to couple only levels j and j + 1 by pairs. This nite-dimensional problem can be thought as the reduction of an innite-dimensional problem in the following way. We start with a Hamiltonian which is the sum of a drift-term H0, plus a time dependent potential V (t) (the control term, i.e., the lasers). The drift term is assumed to be diagonal, with eigenvalues (energy levels) E1 < E2 < E3 < . Then in this spectral resolution of H0, we assume the control term V (t) to couple only the energy levels E1, E2 and E2 , E3. The projected problem in the eigenspaces corresponding to E1, E2, E3 is completely decoupled and is described by Hamiltonian (19:18). The problem is stated as follows. Assume that at the initial instant t = 0 the state of the system lies in the eigenspace corresponding to the ground eigenvalue E1 . The goal is to determine controls 1 , 2 that steer the system at the terminal instant t = t1 to the eigenspace corresponding to E3, requiring that these controls minimize the cost (energy in the following):

RC

J=

Z T;

C RR



j1 (t)j2 + j2(t)j2 dt:

0

From the physical viewpoint, this problem may be considered either with arbitrary controls i (t) 2 , or with controls \in resonance": j (t) = uj (t)ei(!j t+ j )  !j = Ej +1 ; Ej  (19.19) uj : !  j 2 ; ] j = 1 2: (19.20) In the sequel we call this second problem of minimizing the energy J, which in this case reduces to

Zt ;  u21(t) + u22 (t) dt 1

0

(19.21)

the \real-resonant" problem. The rst problem (with arbitrary complex controls) will be called the \general-complex" problem. Since Hamiltonian (19:18) is self-adjoint: H  = H, it follows that Schr#odinger equation (19:17) is well-dened on the unit sphere " # SC = S 5 =  = (1  2 3) 2 3 j jj2 = j1 j2 + j2j2 + j3j2 = 1 : The source and the target, i.e., the initial and the terminal manifolds in the general-complex problem are respectively the circles

C

R

R

19.3 Control of Quantum Systems

SCd = f(ei' 0 0) j ' 2 g

273

TCd = f(0 0 ei') j ' 2 g:

The meaning of the label (d ) here will be claried later. Summing up, the general-complex problem is stated as follows: i_ = H  2 S 5  1  2 2  (0) 2 SCd  (t1 ) 2 TCd 

C

Zt ;  j1j2 + j2j2 dt ! min 1

0

with the Hamiltonian H dened by (19:18). For the real-resonant case, the control system is (19:17) with Hamiltonian (19:18), admissible controls (19:19), (19:20), and cost (19:21). The natural state space, source, and target in this problem will be found later.

19.3.1 Elimination of the Drift We change variables in order to transfer the ane in control system (19:17), (19:18) to a system linear in control, both in the general-complex and realresonant cases. For  2 , denote by Mj () and Nj () the n  n matrices:

C

Mj ()kl = jk j +1l  + j +1k jl  Nj ()kl = jk j +1l  ; j +1k  jl 

j = 1 2

(19.22)

where  is the Kronecker symbol: ij = 1 if i = j, ij = 0 if i 6= j. Let  = diag(E1  E2 E3), !j = Ej +1 ; Ej , j = 1 2. We will consider successively the general-complex problem: i_ = H

H = +

2 X

j =1

C

Mj (j ) j 2 

and the real-resonant problem: i_ = H

H = +

2 X

j =1

Mj (ei(!j t+ j ) uj )

R

uj  j 2 :

In both cases, we rst make the change of variable  = e;it  to get: i_ =

2 ; X

j =1



Ad eit Mj (j )  =

2 X ; Mj e;it!j j ):

j =1

The source S and the target T are preserved by this rst change of coordinates.

274

19 Examples of Optimal Control Problems on Compact Lie Groups

The General-Complex Case In that case, we make the time-dependent cost preserving change of controls: e;it!j j = i~j : Hence our problem becomes (after the change of notation  ! , ~j ! uj ):

X _ = Nj (uj ) = He C  2

j =1

Zt ; ju1j2 + ju2j2 dt ! min

C

uj 2 

(19.23)

1

0

(0) 2 SCd  where

(19.24)

(t1 ) 2 TCd 

(19.25)

0 0 u (t) 0 1 1 He C = @ ;u%1 (t) 0 u2 (t) A : 0

(19.26)

;u%2(t) 0

Notice that the matrices Nj (1) Nj (i) generate su(3) as a Lie algebra. The cost and the relation between controls before and after elimination of the drift are: J=

Zt ;  ju1(t)j2 + ju2(t)j2 dt 1

(19.27)

0

1 (t) = u1 (t)ei (E2 ;E1 )t+=2]  2 (t) = u2 (t)ei (E3 ;E2 )t+=2] :

(19.28) (19.29)

The Real-Resonant Case In this case j = uj ei(!j t+ j ) , and we have: i_ =

2 X

j =1

;  Mj ei j uj 

R

uj 2 :

We make another diagonal, linear change of coordinates:  = eiL ! which gives:

i!_ =

R

L = diag(1  2  3) j 2  2  i( X

j =1

Mj e

j +j+1 ;j ) uj



!:

19.3 Control of Quantum Systems

Choosing the parameters j such that ei(

X !_ = Nj (uj )! 2

j =1

j +j+1 ;j )

R

275

= i, we get:

uj 2 :

(19.30)

The source and the target are also preserved by this change of coordinates. Notice that the matrices N1 (1), N2 (1) in (19:30) generate so(3) as a Lie algebra. This means that the orbit of system (19:30) through the points (1 0 0) is the real sphere S 2 . Hence (by multiplication on the right by ei' ), the orbit through the points (ei'  0 0) is the set S 2 ei' . Therefore (after the change of notation ! ! ) the real-resonant problem is well-dened on the real sphere " # SR= S 2 =  = (1  2 3) 2 3 j jj2 = 12 + 22 + 32 = 1  as follows:

R

X _ = Nj (uj ) = H~ R 2

j =1

Zt;  u21 + u22 dt ! min 1

0

(0) 2 f(1 0 0)g where

R

2 S 2  uj 2 

(t1 ) 2 f(0 0 1)g

(19.31) (19.32) (19.33)

0 0 u (t) 0 1 1 H~ R= @ ;u1 (t) 0 u2(t) A :

(19.34) 0 ;u2 (t) 0 The cost is given again by formula (19:27) and the relation between controls before and after elimination of the drift is: j (t) = uj (t) ei(!j t+ j )  !j = Ej +1 ; Ej  uj : !  j 2 ; ] j = 1 2: In the following we will use the labels (C ) and (R) to indicate respectively the general-complex problem and the real-resonant one. When these labels are dropped in a formula, we mean that it is valid for both the real-resonant and the general-complex problem. With this notation: SCd = f(ei'  0 0)g TCd = f(0 0 ei')g d = f(1 0 0)g d = f(0 0 1)g: SR TR

RR

19.3.2 Lifting of the Problems to Lie Groups The problems (19:23){(19:25) and (19:31){(19:33) on the spheres SC = S 5 and SR= S 2 are naturally lifted to right-invariant problems on the Lie groups MC = SU(3) and MR= SO(3) respectively. The lifted systems read

276

19 Examples of Optimal Control Problems on Compact Lie Groups

e q_ = Hq

q 2 M:

(19.35)

Denote the projections C : SU(3) ! S 5  R: SO(3) ! S 2 both dened as 011 q 7! q @ 0 A  0 i.e., a matrix maps to its rst column. We call problems (19:35) on the Lie groups M problems upstairs, and problems (19:23), (19:31) on the spheres S problems downstairs. We denote the problem upstairs by the label (u ) in parallel with the label (d ) for the problem downstairs. Now we compute boundary conditions for the problems upstairs. Dene the corresponding sources and targets: S u = ;1 (S d ) T u = ;1 (T d ): The source SCu consists of all matrices q 2 SU(3) with the rst column in SCd : 0 0 1 C B q=B @ 0 A CA  2 U(1) A 2 U(2) det q = 1: We denote the subgroup of SU(3) consisting of such matrices by S(U(1)  U(2)). So the source upstairs in the general-complex problem is the subgroup SCu = S(U(1)  U(2)): Further, the matrix 00 1 01 qb = @ 0 0 1 A 100 d d maps SC into TC , thus TCu = qb SCu = qb S(U(1)  U(2)): Similarly, in the real case the source upstairs is u = S(O(1)  O(2)) SR the subgroup of SO(3) consisting of the matrices 0 0 1 C B q=B @ 0 A CA  2 O(1) A 2 O(2) det q = 1

19.3 Control of Quantum Systems

and the target is

u = qb S u = qb S(O(1)  O(2)): TR R

277

R

Summingup, we state the lifted problems. The real problem upstairs reads: q_ = He Rq = (u1 X1 + u2X2 ) q q 2 SO(3) u1 u2 2  (19.36) u u q(0) 2 SR q(t1 ) 2 TR

Zt;  u21 + u22 dt ! min 1

where

0

0 0 1 01 X1 = @ ;1 0 0 A 

00 0 01 X2 = @ 0 0 1 A :

(19.37) 0 ;1 0 0 00 Notice that the real problem upstairs is a right-invariant sub-Riemannian problem on the compact Lie group SO(3) with a corank one set of control parameters U so(3), i.e., a problem already considered in Sect. 19.2. We have 0 0 0 11 U = span(X1  X2) U ? = span(X3 ) X3 = @ 0 0 0 A : ;1 0 0 Moreover, the frame (19:37) is orthonormal w.r.t. the invariant scalar product hX Y i = ; 21 tr(XY ) X Y 2 so(3): The complex problem upstairs is stated as follows: q_ = He C q = (u1 X1 + u2X2 + u3Y1 + u4Y2 ) q q 2 SU(3) uj 2 

R

q(0) 2 SCu  Z t1 ; 0

q(t1 ) 2 T u  C

 u2 + u2 + u2 + u2 dt ! min: 1

2

3

4

Here X1 and X2 are given by (19:37) and 00 i 01 00 0 01 Y1 = @ i 0 0 A  Y2 = @ 0 0 i A : 000 0i0 The set of control parameters is U = span(X1  X2  Y1 Y2):

(19.38)

278

19 Examples of Optimal Control Problems on Compact Lie Groups

Notice that its orthogonal complement is

U ? = span(Z1  Z2  Z3 Z4)

where

0 0 0 11 Z1 = @ 0 0 0 A  0 ;i 100 001 Z3 = @ 0 ;i 0 A  0 0 0

00 0 i 1 Z2 = @ 0 0 0 A  0 0i 00 00 1 Z4 = @ 0 i 0 A  0 0 ;i

and it is easy to check that U ? is a Lie subalgebra. So the general-complex problem is of the form considered in Exercise 19.2. Again the distribution is right-invariant and the frame (X1  X2  Y1 Y2) is orthonormal for the metric hX Y i = ; 21 tr(XY ) X Y 2 su(3): The problems downstairs and upstairs are related as follows. For any trajectory upstairs q(t) 2 M satisfying the boundary conditions in M, its projection (t) = (q(t)) 2 S is a trajectory of the system downstairs satisfying the boundary conditions in S. And conversely, any trajectory downstairs (t) with the boundary conditions can be lifted to a trajectory upstairs q(t) with the corresponding boundary conditions (such q(t) is a matrix fundamental solution of the system downstairs). The cost for the problems downstairs and upstairs is the same. Thus solutions of the optimal control problems downstairs are projections of the solutions upstairs.

19.3.3 Controllability The set of control parameters U in the both problems upstairs (19:38), (19:36) satises the property U U] = U ? , thus U + U U] = M = TId M:

(19.39)

The systems upstairs have a full rank and are symmetric, thus they are completely controllable on the corresponding Lie groups SU(3), SO(3). Passing to the projections , we obtain that the both systems downstairs (19:23), (19:31) are completely controllable on the corresponding spheres S 5 , S 2 .

19.3.4 Extremals The problems upstairs are of the form considered in Sect. 19.2 and Exercise 19.2, but right-invariant not left-invariant ones. Thus normal extremals

19.3 Control of Quantum Systems

279

are given by formulas (19:14){(19:16), where multiplication from the left is replaced by multiplication from the right: a? = const aU (t) = e;t ad a? a0U  a0U = aU (0) q(t) = e;ta? eta q0 (19.40) for any a? 2 U ? , a0U 2 U. Geodesics are parametrized by arclength i ha0U  a0U i = 1: (19.41) Equality (19:39) means that in the problems upstairs, vector elds in the right-hand sides and their rst order Lie brackets span the whole tangent space. Such control systems are called 2-generating. In Chap. 20 we prove that for such systems strictly abnormal geodesics (i.e., trajectories that are projections of abnormal extremals but not projections of normal ones) are not optimal, see the argument before Example 20.19. Thus we do not consider abnormal extremals in the sequel.

19.3.5 Transversality Conditions In order to select geodesics meeting the boundary conditions, we analyze transversality conditions upstairs. Transversality conditions of PMP on T  M corresponding to the boundary conditions q(0) 2 S  q(t1 ) 2 T  S  T M read as follows: h0  Tq(0)Si = ht1  Tq(t1 ) T i = 0: (19.42) Via trivialization (18:12) of T  M, transversality conditions (19:42) are rewritten for the extremal (x(t) q(t)) 2 M  M in the form: x(0) q(0);1 T S  = x(t ) q(t );1 T T  = 0: 1 1 q(0) q(t1 ) Here the brackets h  i denote action of a covector on a vector. The transversality conditions for the extremal (a(t) q(t)) 2 M  M read as follows: a(0) q(0);1 T S  = a(t ) q(t );1 T T  = 0 1 1 q(0) q(t1 ) where the brackets denote the invariant scalar product in M. For the right-invariant problem, transversality conditions are written in terms of right translations: a(0) (T S ) q(0);1  = a(t ) (T T ) q(t );1  = 0: (19.43) 1 1 q(0) q(t1 ) The following features of transversality conditions for our problems upstairs simplies their analysis.

280

19 Examples of Optimal Control Problems on Compact Lie Groups

Lemma 19.3. (1) Transversality conditions at the source are required only at the identity.

(2) Transversality conditions at the source imply transversality conditions at the target.

Proof. Item (1) follows since the problem is right-invariant and the source S u

is a subgroup. Item (2). Let t 2 Tq(t) M be a normal extremal for the problem upstairs such that q(0) = Id. We assume the transversality conditions at the source:

h0  TId S u i = 0 and prove the transversality conditions at the target:

ht1  Tq(t1 ) T u i = 0: (19.44) Notice rst of all that since q(t1 ) 2 T u = qb S u , then qb;1q(t1 ) 2 S u and T u = qb S u = qb (qb;1 q(t1))S u = q(t1 )S u : Then transversality conditions at the target (19:44) read

ht1  Tq(t1 ) (q(t1 )S u )i = 0: In order to complete the proof, we show that the function I(t) = ht  Tq(t) (q(t)S u )i

t 2 0 t1]

is constant. Denote the tangent space S = TId S u . Then we have:

I(t) = ht  q(t)S i = x(t) q(t)Sq(t);1 = hx(t) (Adq(t))S i = ha(t) (Ad q(t))S i D E = (Ad e;ta? )a(0) (Ad e;ta? )(Ad e;ta(0) )S

by invariance of the scalar product

D

E D

E

= a(0) (Ad e;ta(0))S = (Ad e;ta(0) )a(0) S = ha(0) S i = I(0): That is, I(t)  const, and item (2) of this lemma follows.

19.3.6 Optimal Geodesics Upstairs and Downstairs Similarly to N1 , N2 (see formula (19:22)), let us dene N13 by:

ut

0 N13(a3 ei ) = @ 3

19.3 Control of Quantum Systems

1 0 0 a3 ei 0 0 0 A: ;a3 e;i 0 0

281

3

3

Let us set, in the real-resonant case a0U = a1 N1 (1) + a2N2 (1) a? = a3 N13(1): In the general complex case, set a0U = N1 (a1 ei1 ) + N2 (a2 ei2 ) a? = a4 Z3 + a5 Z4 + N13 (a3 ei3 ): Here ai 2 and i 2 ; ].

R

The Real-Resonant Case Proposition 19.4. For the real-resonant problem, transversality condition at u i = 0 means that a2 = 0. the identity in the source ha TId SR Proof. We have

9 80 0 0 0 1 = < u = @ 0 0 ; A   2 TId SR & : 0 0

R

R

u i = 0 is satised for every  2 if and only if thus the equation ha TId SR a2 = 0. ut From Proposition 19.4 and condition (19:41), one gets the covectors to be used in formula (19:40): 0 0 1 a 1 3 a = @ 1 0 0 A : (19.45) ;a3 0 0 Proposition 19.5. Geodesics (19:40) with the initial condition q(0) = Id and u for the smallest time (armatrix a given by (19:45) reach the ptarget TR clength) jtj, if and only if a3 = 1= 3. Moreover, the 4 geodesics (corresponding to a and to the signs  in a3) have the same length and reach the

target at the time

p

t1 = 23 : Proof. Computing q(t) = e;a? te(a? +a0U )t, with a given by formula (19:45), and recalling that 011 (t) = q(t) @ 0 A  0

282

19 Examples of Optimal Control Problems on Compact Lie Groups

one gets for the square of the third component of the wave function: (3 (t))2 =

q

;cos(t a ) sin(t ) a  ; cos (t ) sin(t a ) 2 2 3 3 3  4

(19.46)

 = 1 + a23 : Then the following lemma completes the proof of this proposition. ut p 2 pa p 2 Lemma 19.6. Set fa = cos(ta) sin(t 1 + a ) 1+a2 ; cos(t 1 + a ) sin(ta), jaj 2 = 1 + k0 < 1, k 6= 0 and then jfa j  1. Moreover, jfa j = 1 i p1+ 2k k a k . In particular, the smallest jtj is obtained for k = 1, a =  p1 , t = p1+ 2 3 pa t = 2 3 . p Proof. Set  = p1+a a2 , = t 1 + a2, then:

fa (t) =  cos( ) sin( ) ; cos( ) sin( ) = h( cos( ) sin( )) (sin( ) ; cos( ))i = hv1 v2 i: Both v1  v2 have norm  1 and jfa j  1. Hence, for jfa j = 1, we must have jv1j = jv2j = 1, v1 = v2 . It follows that cos( ) = 0 and cos( ) = 1. Hence

= k,  = 2 + k0,  = 21k + kk0 . Therefore, 21k + kk0 =  < 1. Conversely, choose k k0 meeting this condition, and = k. Then cos( ) = 1, sin( ) = k , and the smallest jtj is obtained for k = 1 1, fa (t) = 1. Now, jtj = p1+ a2 (if k = 0, = 0 and fa (t) = 0). Moreover, 21k + kk0 < 1 is possible only for (k k0) = (1 0) orp(1 ;1) or (;1 0) or (;1 ;1). In all cases, jj = 21 , a =  p13 , and t =   2 3 . ut

p

Let us x for instance the sign ; in (19:45) and a3 = +1= 3. The expressions of the three components of the wave function are:  3 1 (t) = cos pt  p 3  2 (t) = 23 sin p2 t   t 33 3 (t) = ;sin p : 3 Notice that this curve is not a circle on S 2 . Controls can be obtained with the following expressions: u1 = (qq _ ;1)12 u2 = (qq _ ;1)23 :

19.3 Control of Quantum Systems

We get:

283

t

u1(t) = ; cos p   t 3 u2(t) = sin p : 3 Using conditions (19:19){(19:20) (resonance hypothesis), we get for the external elds:

 p 1 (t) = ; cos t= 3 ei(!1 t+ 1 ) 

 p 2 (t) = sin t= 3 ei(!2 t+ 2 ) : Notice that the phases 1 2 are arbitrary.

The General-Complex Case Proposition 19.7. For the general-complex problem, transversality condition at the identity in the source ha TId SCu i = 0 means that a2 = a4 = a5 = 0. Proof. We have:

9 80 i 1 0 0 = < 1 TId SCu = :@ 0 i( 2 ; 1) 1 + i2 A  1 2 1  2 2 & : 0 ;1 + i2 ;i 2

R

R

The equation ha TId SCu i = 0 is satised for every 1 2 1 2 2 if and only if a2 = a4 = a5 = 0. ut The covector to be used in formula (19:40) is then: 0 0 ei1 a ei3 1 3 a(1 3) = @ ;e;i1 0 0 A : (19.47) ;a3 e;i3 0 0

Proposition 19.8. The geodesics (19:40), with a given by formula (19:47) (for which q(0) = Id), reach p the target TCu for the smallest time (arclength) jtj, if and only if a3 = 1= 3. Moreover, all the geodesics of the two parameter family corresponding to 1  3 2 ; ], have the same length: p t1 = 23 : Proof. The explicit expression for j3j2 is given by the right-hand side of formula (19:46). The conclusion follows as in the proof of Proposition 19.5.

ut

284

19 Examples of Optimal Control Problems on Compact Lie Groups

The expressions of the three components of the wave function and of optimal controls are:  t 3 1(t) = cos p  p 3  2(t) = ; 23 sin p2 t e;i1   t 3 3 3(t) = ;sin p e;i3  3 and

 p u1(t) = cos t= 3 ei1 

 p u2(t) = ; sin t= 3 ei(3 ;1) :

Notice that all the geodesics of the family described by Proposition 19.8 have the same length as the 4 geodesics described by Proposition 19.5. This proves that the use of the complex Hamiltonian (19:26) instead of the real one (19:34) does not allow to reduce the cost (19:27). We obtain the following statement.

Proposition 19.9. For the three-level problem with complex controls, optimality implies resonance. More precisely, controls 1 , 2 are optimal if and only if they have the following form:

p

1 (t) = cos(t= 3)ei (E2 ;E1 )t+'1 ]  p 2 (t) = sin(t= 3)ei (E3 ;E2 )t+'2 ] 

where '1  '2 are two arbitrary phases. Here the nal time t1 is xed in such a way sub-Riemannian geodesics are parametrized by arclength, and it is given p by t1 = 23 .

19.4 A Time-Optimal Problem on SO(3)

R

R

Consider a rigid body in 3. Assume that the body can rotate around some axis xed in the body. At each instant of time, orientation of the body in 3 denes an orthogonal transformation q 2 SO(3). We are interested in the length of the curve in SO(3) corresponding to the motion of the body. Choose a natural parameter (arc length) t, then the curve q = q(t) satises the ODE q_ = qf where

19.4 A Time-Optimal Problem on SO(3)

285

f 2 so(3) jf j = 1 is the unit vector of angular velocity corresponding to the xed axis of rotation in the body. The curve is a one-parameter subgroup in SO(3): q(t) = q(0)etf  and we obviously have no controllability on SO(3). In order to extend possibilities of motion in SO(3), assume now that there are two linearly independent axes in the body: f g 2 so(3)

jf j = jgj = 1

f ^ g 6= 0

and we can rotate the body around these axes in certain directions. Now we have a control system ( q_ = qf  qg which is controllable on SO(3): Lie(qf qg) = span(qf qg qf g]) = q so(3) = Tq SO(3): In order to simplify notation, choose vectors a b 2 so(3) such that

f = a + b Then the control system reads

g = a ; b:

q_ = q(a  b): We are interested in the shortest rotation of the body steering an initial orientation q0 to a terminal orientation q1. The corresponding optimal control problem is q(0) = q(0) l=

Zt

1

0

q(t1 ) = q1

jq_j dt ! min:

Since jq_j = ja  bj = 1, this problem is equivalent to the time-optimal problem: t1 ! min: Notice that

ha bi = h(f + g)=2 (f ; g)=2i = 0:

(19.48)

286

19 Examples of Optimal Control Problems on Compact Lie Groups

Moreover, by rescaling time we can assume that

jaj = 1:

(19.49) Passing to convexication, we obtain the following nal form of the problem: q_ = q(a + ub) u 2 ;1 1] q 2 SO(3) q(0) = q0 q(t1) = q1 t1 ! min where a b 2 so(3) are given vectors that satisfy equalities (19:48), (19:49). Now we study this time-optimal problem. By PMP, if a pair (u( ) q( )) is optimal, then there exists a Lipschitzian curve x(t) 2 so(3) such that:

(

q_ = q(a + u(t)b) x_ = x a + u(t)b] hu(t)(x(t)) = hx(t) a + u(t)bi = jmax hx(t) a + vbi  0 vj1 moreover, hu(t)(x(t)) = const : The maximality condition for the function v 7! hx(t) a + vbi = hx(t) ai + vhx(t) bi

v 2 ;1 1]

is easily resolved if the switching function x 7! hx bi

x 2 M

does not vanish at x(t). Indeed, in this case optimal control can take only extremal values 1:

hx(t) bi 6= 0 ) u(t) = sgnhx(t) bi: If the switching function has only isolated roots on some real segment, then the corresponding control u(t) takes on this segment only extremal values. Moreover, the instants where u(t) switches from one extremal value to another are isolated. Such a control is called bang-bang . Now we study the structure of optimal controls. Take an arbitrary extremal with the curve x(t) satisfying the initial condition

hx(0) bi 6= 0: Then the ODE

19.4 A Time-Optimal Problem on SO(3)

287

x_ = x a  b]  = sgnhx(0) bi is satised for t > 0 until the switching function hx(t) bi remains nonzero. Thus at such a segment of time x(t) = e;t ad(ab)x(0):

We study the switching function hx(t) bi. Notice that its derivative does not depend upon control: d hx(t) bi = hx(t) a + u(t)b] bi = ;hx(t) a b]i: dt If the switching function vanishes:

hx(t) bi = 0 at a point where

hx(t) a b]i 6= 0

then the corresponding control switches, i.e., changes its value from +1 to ;1 or from ;1 to +1. In order to study, what sequences of switchings of optimal controls are possible, it is convenient to introduce coordinates in the Lie algebra M. In view of equalities (19:48), (19:49), the Lie bracket a b] satises the conditions a b] ? a a b] ? b ja b]j = jbj this follows easily from properties of cross-product in 3. Thus we can choose an orthonormal basis: so(3) = span(e1  e2 e3 ) such that a = e2  b = e3 a b] = e1  > 0: In this basis, switching points belong to the horizontal plane span(e1  e2). Let x( 0) be a switching point, i.e., t = 0 is a positive root of hx(t) bi. Assume that at this point control switches from +1 to ;1 (the case of switching from ;1 to +1 is completely similar, we show this later). Then

R

hx( _ 0 ) bi = ;hx( _ 0 ) a b]i  0 thus

hx( 0 ) e1 i  0:

Further, since the Hamiltonian of PMP is nonnegative, then hu(0 ) (x( 0 )) = hx( 0) ai = hx( 0) e2 i  0: So the point x( 0 ) lies in the rst quadrant of the plane span(e1  e2 ):

288

19 Examples of Optimal Control Problems on Compact Lie Groups

x( 0) 2 cone(e1  e2): Let x( 1 ) be the next switching point after 0 . The control has the form

(

u(t) = 1 t 2  0 ; " 0 ] ;1 t 2  0 1] and the curve x(t) between the switchings is an arc of the circle obtained by rotation of the point x( 0 ) around the vector a ; b = e2 ; e3: x(t) = e;t ad(a;b)x( 0 ) t 2  0  1]: The switching points x( 0 ), x( 1 ) satisfy the equalities: hx( 0 ) e3i = hx( 1 ) e3i = 0 hx( 0 ) e2i = hx( 1 ) e2i = hu(0 ;")(x( 0 ; ")) jx( 0) = jx( 1)j: Consequently, hx( 0 ) e1i = ;hx( 1 ) e1i i.e., x( 1 ) is the reection of x( 0 ) w.r.t. the plane span(e2  e3 ). Geometrically it is easy to see that the angle of rotation from x( 0) to x( 1 ) around a ; b is bounded as follows:

2  2] see Fig. 19.2. The extremal values of are attained when the point x( 0 ) is on the boundary of cone(e1  e2): x( 0 ) 2 +e1 ) =  x( 0 ) 2 +e2 ) = 2: In the second case the point x(t), as well as the point q(t), makes a complete revolution at the angle 2. Such an arc cannot be a part of an optimal trajectory: it can be eliminated with decrease of the terminal time t1 . Consequently, the angle between two switchings is

2  2): Let x( 2) be the next switching after x( 1 ). The behavior of control after the switching x( 1) from ;1 to +1 is similar to the behavior after x( 0 ). Indeed, our time-optimal problem admits the symmetry b 7! ;b: After the change of basis e3 7! ;e3  e1 7! ;e1  e2 7! e2

RR

19.4 A Time-Optimal Problem on SO(3)

289

b = e3 e3

e2

=a

x(0)

e1

a b] =

x(1 )

e1 

a;b

Fig. 19.2. Estimate of rotation angle  the curve x(t) is preserved, but now it switches at x( 1 ) from +1 to ;1. This case was already studied, thus the angle of rotation from x( 1) to x( 2 ) is again , moreover, x( 2) = x( 0 ). The next switching point is x( 3 ) = x( 1 ), and so on. Thus the structure of bang-bang optimal trajectories is quite simple. Such trajectories contain a certain number of switching points. Between these switching points the vector x(t) rotates alternately around the vectors a + b and a ; b at an angle 2  2) constant along each bang-bang trajectory. Before the rst switching and after the last switching the vector x(t) can rotate at angles 0 and 1 respectively, 0 < 0  1  . The system of all optimal bang-bang trajectories is parametrized by 3 continuous parameters 0 , , 1 , and 2 discrete parameters: the number of switchings and the initial control sgnhx(0) bi. An optimal trajectory can be not bang-bang only if the point x( 0 ) corresponding to the rst nonnegative root of the equation hx(t) bi = 0 satises the equalities hx( 0 ) bi = hx( 0 ) a b]i = 0: Then x( 0 ) = e2   6= 0: There can be two possibilities: (1) either the switching function hx(t) bi takes nonzero values for some t > 0 and arbitrarily close to 0 , (2) or

290

19 Examples of Optimal Control Problems on Compact Lie Groups

hx(t) bi  0

t 2  0 0 + "]

(19.50)

for some " > 0. We start from the rst alternative. From the analysis of bang-bang trajectories it follows that switching times cannot accumulate to 0 from the right: the angle of rotation between two consecutive switchings  . Thus in case (1) we have hx(t) bi > 0 t 2  0 0 + ] for some  > 0. That is, 0 is a switching time. Since x( 0) 2 e1, then the angle of rotation until the next switching point is = 2, which is not optimal. So case (1) cannot occur for an optimal trajectory. Consider case (2). We dierentiate identity (19:50) twice w.r.t. t: d hx(t) bi = ;hx(t) a b]i  0 dt d = hx(t) a + u(t)b a b]i = u(t)hx(t) b] a b]i d t hx(t) a b]i = 0: Then x(t) = (t)e2 , t 2  0 0 + "], thus u(t)ha b] a b]i = 0 i.e., u(t)  0 t 2  0 0 + "]: This control is not determined directly from PMP (we found it with the help of dierentiation). Such a control is called singular . Optimal trajectories containing a singular part (corresponding to the control u(t)  0) can have an arc with u  1 before the singular part, with the angle of rotation around a  b less then 2 such an arc can also be after the singular one. So there can be 4 types of optimal trajectories containing a singular arc: + 0 + + 0 ; ; 0 + ; 0 ;: The family of such trajectories is parametrized by 3 continuous parameters (angles of rotation at the corresponding arcs) and by 2 discrete parameters (signs at the initial and nal segments). So we described the structure of all possible optimal trajectories: the bangbang one, and the strategy with a singular part. The domains of points in SO(3) attained via these strategies are 3-dimensional, and the union of these domains covers the whole group SO(3). But it is easy to see that a suciently long trajectory following any of the two strategies is not optimal: the two domains in SO(3) overlap. Moreover, each of the strategies overlaps with itself.

R

19.4 A Time-Optimal Problem on SO(3)

291

In order to know optimal trajectory for any point in SO(3), one should study the interaction of the two strategies and intersections of trajectories that follow the same strategy. This interesting problem remains open. Notice that the structure of optimal trajectories in this left-invariant timeoptimal problem on SO(3) is similar to the structure of optimal trajectories for Dubins car (Sect. 13.5). This resemblance is not accidental: the problem on Dubins car can be formulated as a left-invariant time-optimal problem on the group of isometries of the plane.

20 Second Order Optimality Conditions

20.1 Hessian In this chapter we obtain second order necessary optimality conditions for control problems. As we know, geometrically the study of optimality reduces to the study of boundary of attainable sets (see Sect. 10.2). Consider a control system m q_ = fu (q) q 2 M u 2 U = int U (20.1) where the state space M is, as usual, a smooth manifold, and the space of control parameters U is open (essentially, this means that we study optimal controls that do not come to the boundary of U, although a similar theory for bang-bang controls can also be constructed). The attainable set Aq0 (t1 ) of system (20:1) is the image of the endpoint mapping

R

Z t1 ;! Ft1 : u( ) 7! q0 exp fu(t) dt: 0

We say that a trajectory q(t), t 2 0 t1], is geometrically optimal for system (20:1) if it comes to the boundary of the attainable set for the terminal time t1 : q(t1 ) 2 @ Aq0 (t1 ): Necessary conditions for this inclusion are given by Pontryagin Maximum Principle. A part of the statements of PMP can be viewed as the rst order optimality condition (we see this later). Now we seek for optimality conditions of the second order. Consider the problem in a general setting. Let F : U !M be a smooth mapping, where U is an open subset in a Banach space and M is a smooth n-dimensional manifold (usually in the sequel U is the space of

294

20 Second Order Optimality Conditions

admissible controls L1 (0 t1] U) and F = Ft1 is the endpoint mapping of a control system). The rst dierential Du F : Tu U ! TF (u) M is well dened independently on coordinates. This is not the case for the second dierential. Indeed, consider the case where u is a regular point for F, i.e., the dierential Du F is surjective. By implicit function theorem, the mapping F becomes linear in suitably chosen local coordinates in U and M, thus it has no intrinsic second dierential. In the general case, well dened independently of coordinates is only a certain part of the second dierential. The dierential of a smooth mapping F : U ! M can be dened via the rst order derivative (20.2) Du F v = dd" F('(")) "=0 along a curve ' : (;"0  "0) ! U with the initial conditions '(0) = u 2 U 

'(0) _ = v 2 Tu U :

In local coordinates, this derivative is computed as d F '_ '_ = '(0): _ du In other coordinates q~ in M, derivative (20:2) is evaluated as

d Fe '_ = d q~ d F ':_ du dq du Coordinate representation of the rst order derivative (20:2) transforms under changes of coordinates as a tangent vector to M | it is multiplied by the Jacobian matrix dd qq~. The second order derivative d2 (20.3) d "2 "=0 F('(")) '(0) = u 2 U  '(0) _ = v 2 Tu U 

is evaluated in coordinates as d F '#: d2 F ('_ ') _ + 2 du du Transformation rule for the second order directional derivative under changes of coordinates has the form:

 d q~ d2 F

20.1 Hessian

 e d2 Fe ('_ ') d F d F _ + d u '# d u2 _ + d u '# = d q d u2 ('_ ')  d2 q~ d F d F 

295

+ d q2 d u '_ d u '_ : (20.4) The second order derivative (20:3) transforms as a tangent vector in TF (u) M only if '_ = v 2 Ker Du F , i.e., if term (20:4) vanishes. Moreover, it is determined by u and v only modulo the subspace ImDu F , which is spanned by # the term dd Fu '. Thus intrinsically dened is the quadratic mapping Ker Du F ! TF (u) M= ImDu F 2 (20.5) v 7! dd"2 F('(")) mod ImDu F: "=0 After this preliminary discussion, we turn to formal denitions. The Hessian of a smooth mapping F : U ! M at a point u 2 U is a symmetric bilinear mapping Hessu F : Ker Du F  Ker Du F ! Coker Du F = TF (u) M= ImDu F: (20.6) In particular, at a regular point Coker Du F = 0, thus Hessu F = 0. Hessian is dened as follows. Let v w 2 Ker Du F and  2 (Im Du F)? TF (u) M: In order to dene the value  Hess u F(v w) take vector elds V W 2 Vec U  and a function Then

a 2 C 1 (M)

V (u) = v W(u) = w dF (u)a = :

 Hess u F (v w) def = V  W (a  F)ju :

(20.7)

We show now that the right-hand side does not depend upon the choice of V , W , and a. The rst Lie derivative is W(a  F) = hdF (  ) a F W( )i

296

20 Second Order Optimality Conditions

and the second Lie derivative V  W (a  F )ju does not depend on second derivatives of a since FW(u) = 0. Moreover, the second Lie derivative obviously depends only on the value of V at u but not on derivatives of V at u. In order to show the same for the eld W, we prove that the right-hand side of the denition of Hessian is symmetric w.r.t. V and W : (W  V (a  F) ; V  W(a  F))ju = W V ] (a  F)ju = dF (u) a Du F W V ](u) = 0

| {z } =

since  ? ImDu F . We showed that the mapping Hessu F given by (20:7) is intrinsically dened independently of coordinates as in (20:6). Exercise 20.1. Show that the quadratic mapping (20:5) dened via the second order directional derivative coincides with Hessu F (v v). If we admit only linear changes of variables in U , then we can correctly dene the full second di erential Du2 F : Ker Du F  Ker Du F ! TF (u) M in the same way as Hessian (20:7), but the covector is arbitrary:  2 TF (u) M

and the vector elds are constant: V  v

W  w:

The Hessian is the part of the second dierential independent on the choice of linear structure in the preimage. Exercise 20.2. Compute the Hessian of the restriction F jf ;1 (0) of a smooth mapping F to a level set of a smooth function f. Consider the restriction of a smooth mapping F : U ! M to a smooth hypersurface S = f ;1 (0), f : U ! , df 6= 0, and let u 2 S be a regular point of F. Prove that the Hessian of the restriction is computed as follows:

R

 Hessu (F jS ) = Du2 F ; d2uf

 ? ImDu F jS   2 TF (u) M n f0g

and the covector  is normalized so that Du F = du f:

20.2 Local Openness of Mappings

20.2 Local Openness of Mappings

297

A mapping F : U ! M is called locally open at a point u 2 U if F(u) 2 int F(Ou) for any neighborhood Ou U of u. In the opposite case, i.e., when F(u) 2 @F(Ou ) for some neighborhood Ou, the point u is called locally geometrically optimal for F. A point u 2 U is called locally nite-dimensionally optimal for a mapping F if for any nite-dimensional smooth submanifold S U , u 2 S, the point u is locally geometrically optimal for the restriction F jS .

20.2.1 Critical Points of Corank One Corank of a critical point u of a smooth mapping F is by denition equal to corank of the dierential Du F: corank Du F = codimImDu F: In the sequel we will often consider critical points of corank one. In this case the Lagrange multiplier  2 (ImDu F )?  6= 0

R

is dened uniquely up to a nonzero factor, and  Hess u F : Ker Du F  Ker Du F ! is just a quadratic form (in the case corank Du F > 1, we should consider a family of quadratic forms). Now we give conditions of local openness of a mapping F at a corank one critical point u in terms of the quadratic form  Hessu F. Theorem 20.3. Let F : U ! M be a continuous mapping having smooth restrictions to nite-dimensional submanifolds of U . Let u 2 U be a corank one critical point of F , and let  2 (Im Du F)? ,  6= 0. (1) If the quadratic form  Hess u F is sign-inde nite, then F is locally open at u. (2) If the form  Hess u F is negative (or positive), then u is locally nitedimensionally optimal for F . Remark 20.4. A quadratic form is locally open at the origin i it is signindenite.

298

20 Second Order Optimality Conditions

Proof. The statements of the theorem are local, so we x local coordinates

R

in U and M centered at u and F (u) respectively, and assume that U is a Banach space and M = n. (1) Consider the splitting into direct sum in the preimage: Tu U = E ! Ker Du F

dimE = n ; 1

(20.8)

and the corresponding splitting in the image: TF (u) M = ImDu F ! V

dimV = 1:

(20.9)

The quadratic form  Hessu F is sign-indenite, i.e., it takes values of both signs on Ker Du F . Thus we can choose vectors v w 2 Ker Du F such that

Fu00 (v v) = 0 Fu00(v w) 6= 0 we denote by F 0, F 00 derivatives of the vector function F in local coordinates. Indeed, let the quadratic form Q = Fu00 take values of opposite signs at some v0 w 2 Ker Du F . By continuity of Q, there exists a nonzero vector v 2 span(v0  w) at which Q(v v) = 0. Moreover, it is easy to see that Q(v w) 6= 0. Since the rst dierential is an isomorphism: Du F = Fu0 : E ! ImDu F = ? 

there exists a vector x0 2 E such that

Fu0 x0 = ; 12 Fu00 (v v):

Introduce the following family of mappings:

R

R

" : E  ! M "2  2 3 " (x y) = F(" v + " yw + "4 x0 + "5 x)

R

x 2 E y 2 

notice that

Im" ImF for small ". Thus it is sucient to show that " is open. The Taylor expansion "(x y) = "5 (Fu0 x + yFu00(v w)) + O("6 )

" ! 0

implies that the family "15 " is smooth w.r.t. parameter " at " = 0. For " = 0 this family gives a surjective linear mapping. By implicit function theorem, the mappings "15 " are submersions, thus are locally open for small " > 0. Thus the mapping F is also locally open at u.

20.2 Local Openness of Mappings

299

(2) Take any smooth nite-dimensional submanifold S U , u 2 S. Similarly to (20:8), (20:9), consider the splittings in the preimage: S = Tu S = L ! Ker Du F jS  and in the image: M = TF (u) M = ImDu F jS ! W dimW = k = corank Du F jS  1: Since the dierential Du F : E ! ImDu F is an isomorphism, we can choose, by implicit function theorem, coordinates (x y) in S and coordinates in M such that the mapping F takes the form  x  F(x y) = '(x y)  x 2 L y 2 Ker Du F jS : Further, we can choose coordinates ' = ('1  : : :  'k ) in W such that F(x y) = '1 (x y): Now we write down hypotheses of the theorem in these coordinates. Since ImDu F jS \ W = f0g, then D(00)'1 = 0: Further, the hypothesis that the form  Hess u F is negative reads @ 2 '1 @ y2 (00) < 0: Then the function

'1 (0 y) < 0 for small y. Thus the mapping F jS is not locally open at u. ut There holds the following statement, which is much stronger than the previous one. Theorem 20.5 (Generalized Morse's lemma). Suppose that u 2 U is a corank one critical point of a smooth mapping F : U ! M such that Hessu F is a nondegenerate quadratic form. Then there exist local coordinates in U and M in which F has only terms of the rst and second orders:

F (x v) = Du F x + 21 Hessu F(v v) (x v) 2 U  = E ! Ker Du F: We do not prove this theorem since it will not be used in the sequel.

300

20 Second Order Optimality Conditions

20.2.2 Critical Points of Arbitrary Corank The necessary condition of local openness given by item (1) of Theorem 20.3 can be generalized for critical points of arbitrary corank. Recall that positive (negative) index of a quadratic form Q is the maximal dimension of a positive (negative) subspace of Q:

n n

o o

ind+ Q = max dimL j QjLnf0g > 0  ind; Q = max dimL j QjLnf0g < 0 :

Theorem 20.6. Let F : U ! M be a continuous mapping having smooth restrictions to nite-dimensional submanifolds. Let u 2 U be a critical point of F of corank m. If

ind;  Hess u F  m

8  ? ImDu F  6= 0

then the mapping F is locally open at the point u.

R

Proof. First of all, the statement is local, so we can choose local coordinates

and assume that U is a Banach space and u = 0, and M = n with F(0) = 0. Moreover, we can assume that the space U is nite-dimensional, now we prove this. For any  ? Im Du F,  6= 0, there exists a subspace E U 

dimE = m

such that

 Hessu F jE nf0g < 0: We take  from the unit sphere

n

o

S m;1 =  2 (ImDu F )? j jj = 1 :

For any  2 S m;1 , there exists a neighborhood O S m;1 ,  2 O, such that E0 = E for any 0 2 O, this easily follows from continuity of the form 0 Hess u F on the unit sphere in E . Choose a nite covering: S m; 1 =

N 

i=1

Oi :

P

Then restriction of F to the nite-dimensional subspace Ni=1 Ei satises the hypothesis of the theorem. Thus we can assume that U is nite-dimensional. Then the theorem is a consequence of the following Lemmas 20.7 and 20.8. ut Lemma 20.7. Let F : N ! n be a smooth mapping, and let F(0) = 0.

R R

Assume that the quadratic mapping

20.2 Local Openness of Mappings

301

Q = Hess 0 F : Ker D0 F ! Coker D0 F has a regular zero:

9 v 2 Ker D0 F s.t. Q(v) = 0 Dv Q surjective:

R

Then the mapping F has regular zeros arbitrarily close to the origin in N . Proof. We modify slightly the argument used in the proof of item (1) of The-

R

orem 20.3. Decompose preimage of the rst dierential: N = E ! Ker D0 F

dimE = n ; m

then the restriction

D0 F : E ! ImD0 F is one-to-one. The equality Q(v) = Hess0 F(v) = 0 means that F000 (v v) 2 ImD0 F:

Then there exists x0 2 E such that

F00 x0 = ; 12 F000 (v v):

Dene the family of mappings

x 2 E y 2 Ker D0 F:

" (x y) = F("2 v + "3 y + "4 x0 + "5 x)

The rst four derivatives of " vanish at " = 0, and we obtain the Taylor expansion 1 0 00 "5 "(x y) = F0x + F0 (v y) + O(") " ! 0: Then we argue as in the proof of Theorem 20.3. The family "15 " is smooth and linear surjective at " = 0. By implicit function theorem, the mappings "15 " are submersions for small " > 0, thus they have regular zeros in any neighborhood of the origin in N . Consequently, the mapping F also has regular zeros arbitrarily close to the origin in N . ut Lemma 20.8. Let Q : N ! m be a quadratic mapping such that

R

R R R

ind; Q  m

8 2

Then the mapping Q has a regular zero.

R m

  6= 0:

Proof. We can assume that the quadratic form Q has no kernel:

Q(v ) 6= 0

8 v 6= 0:

(20.10)

302

20 Second Order Optimality Conditions

If this is not the case, we factorize by kernel of Q. Since Dv Q = 2Q(v ), condition (20:10) means that Dv Q 6= 0 for v 6= 0. Now we prove the lemma by induction on m. In the case m = 1 the statement is obvious: a sign-indenite quadratic form has a regular zero. Induction step: we prove the statement of the lemma for any m > 1 under the assumption that it is proved for all values less than m. (1) Suppose rst that Q;1 (0) 6= f0g. Take any v 6= 0 such that Q(v) = 0. If v is a regular point of Q, then the statement of this lemma follows. Thus we assume that v is a critical point of Q. Since Dv Q 6= 0, then 0 < k < m: rank Dv Q = k Consider Hessian of the mapping Q: Hess v Q : Ker Dv Q ! m;k: The second dierential of a quadratic mapping is the doubled mapping itself, thus  Hessv Q = 2 QjKer Dv Q : Further, since ind; Q  m and codimKer Dv Q = k, then ind;  Hessv Q = ind; QjKer Dv Q  m ; k: By the induction assumption, the quadratic mapping Hessv Q has a regular zero. Then Lemma 20.7 applied to the mapping Q yields that Q has a regular zero as well. The statement of this lemma in case (1) follows. (2) Consider now the second case: Q;1 (0) = f0g. (2.a) It is obvious that ImQ is a closed cone. (2.b) Moreover, we can assume that Im Q n f0g is open. Indeed, suppose that there exists x = Q(v) 2 @ ImQ x 6= 0: Then v is a critical point of Q, and in the same way as in case (1) the induction assumption for Hessv Q yields that Hessv Q has a regular zero. By Lemma 20.7, Q is locally open at v and Q(v) 2 intImQ. Thus we assume in the sequel that ImQ n f0g is open. Combined with item (a), this means that Q is surjective. (2.c) We show now that this property leads to a contradiction which proves the lemma. The smooth mapping Q : S N ;1 ! S m;1  v 7! Q(v)  v 2 S N ;1  jQj jQ(v)j is surjective. By Sard's theorem, it has a regular value. Let x 2 S m;1 be a regular value of the mapping Q=jQj.

R

20.2 Local Openness of Mappings

303

Now we proceed as follows. We nd the minimal a > 0 such that v 2 S N ;1 

Q(v) = ax

and apply optimality conditions at the solution v0 to show that ind; Q  m ; 1, a contradiction. So consider the following nite-dimensional optimization problem with constraints: a ! min Q(v) = ax a > 0 v 2 S N ;1 :

(20.11)

This problem obviously has a solution, let a pair (v0  a0) realize minimum. We write down rst- and second-order optimality conditions for problem (20:11). There exist Lagrange multipliers ( ) 6= 0

R

R

2   2 Ta0 x m

such that the Lagrange function

L(  a v) = a + (Q(v) ; ax) satises the stationarity conditions: @ L =  ; x = 0 @a @L = Dv0 QjS N ;1 = 0: @v

(20.12)

(v0a0 )

Since v0 is a regular point of the mapping Q=jQj, then  6= 0, thus we can set  = 1: Then second-order necessary optimality condition for problem (20:11) reads  Hess v0 QjS N ;1  0:

(20.13)

Recall that Hessian of restriction of a mapping is not equal to restriction of Hessian of this mapping, see Exercise 20.2 above. Exercise 20.9. Prove that  (Hess v QjS N ;1 ) (u) = 2(Q(u) ; juj2Q(v)) v 2 S N ;1  u 2 Ker Dv QjS N ;1 : That is, inequality (20:13) yields Q(u) ; juj2Q(v0 )  0

u 2 Ker Dv0 QjS N ;1 

304

20 Second Order Optimality Conditions

thus i.e.,

Q(u)  juj2Q(v0 ) = juj2a0 x = juj2a0 = juj2a0 > 0

Q(u)  0 u 2 Ker Dv0 QjS N ;1 : Moreover, since v0 2= Tv0 S N ;1 , then QjL  0

R

L = Ker Dv0 QjS N ;1 ! v0:

Now we compute dimension of the nonnegative subspace L of the quadratic form Q. Since v0 is a regular value of jQ Qj , then dimImDv0 jQ Qj = m ; 1: Thus Im Dv0 QjS N ;1 can have dimension m or m ; 1. But v0 is a critical point of QjS N ;1 , thus dimImDv0 QjS N ;1 = m ; 1 and dimKer Dv0 QjS N ;1 = N ; 1 ; (m ; 1) = N ; m: Consequently, dimL = N ; m + 1, thus ind; Q  m ; 1, which contradicts the hypothesis of this lemma. So case (c) is impossible, and the induction step in this lemma is proved.

tu

Theorem 20.6 is completely proved.

20.3 Dierentiation of the Endpoint Mapping In this section we compute dierential and Hessian of the endpoint mapping for a control system

R

m U = int U q 2 M q_ = fu (q) u2U q(0) = q0 u( ) 2 U = L1 (0 t1] U)

(20.14)

with the right-hand side fu (q) smooth in (u q). We study the endpoint mapping Ft1 : U ! M

Z t1 ;! Ft1 : u( ) 7! q0  exp fu(t) dt 0

20.3 Di erentiation of the Endpoint Mapping

305

in the neighborhood of a xed admissible control u~ = u~( ) 2 U : In the same way as in the proof of PMP (see Sect. 12.2), the Variations formula yields a decomposition of the ow:

;! Z t1 gtu(t) dt  Pt  Ft1 (u) = q0  exp 1 0

where

Zt ;! Pt = exp fu~( ) d  0

gtu = Pt; 1(fu ; fu~(t) ): Further, introduce an intermediate mapping Gt1 : U ! M

;! Z t1 g dt: Gt1 : u 7! q0  exp tu(t) 0

Then

Ft1 (u) = Pt1 (Gt1 (u))

consequently,

Du~ Ft1 = Pt1 Du~ Gt1  Hessu~ Ft1 = Pt1 Hessu~ Gt1  so dierentiation of Ft1 reduces to dierentiation of Gt1 . We compute derivatives of the mapping Gt1 using the asymptotic expansion of the chronological exponential: a(Gt1 (u))

0 Z ZZ t = q0  @Id + gu( ) d + 1

0

1 g u( )  g u( ) d 1 d 2A a 0  t ;  + O ku ; u~k3 : (20.15) 2

1

2

2

1

1

1

L1

Introduce some more notations:

306

20 Second Order Optimality Conditions

g0 = @@u gu u~( ) 2 g00 = @@u2 gu u~( ) hu () = h fu (q)i hu  h0 = @@u u~( ) 2 hu : h00 = @@u2 u~( )

 2 Tq M

Then dierential (the rst variation) of the mapping Gt1 has the form: (Du~ Gt1 )v = q0 

Zt

1

0

gt0 v(t) dt

v = v( ) 2 Tu~ U :

The control u~ is a critical point of Ft1 (or, which is equivalent, of Gt1 ) if and only if there exists a Lagrange multiplier 0 2 Tq0 M 0 6= 0 such that 0 (Du~ Gt1 )v = 0 8 v 2 Tu~ U  i.e., 0 gt0 (q0 ) = 0 t 2 0 t1]: Translate the covector 0 along the reference trajectory q(t) = q0  Pt we obtain the covector curve t = Pt;10 = 0 Pt; 1 2 Tq(t) M which is a trajectory of the Hamiltonian system _ t = ~hu~(t)(t ) t 2 0 t1] see Proposition 11.14. Then 0 gt0 (q0 ) = 0 Pt; 1 @@u fu (q(t)) = h0t(t ): u~(t) We showed that u~ is a critical point of the endpoint mapping Ft1 if and only if there exists a covector curve t 2 Tq(t) M t 6= 0 t 2 0 t1]

20.3 Di erentiation of the Endpoint Mapping

307

such that _ t = ~hu~(t)(t ) @ hu ( ) = 0 @ u u~(t) t

(20.16) t 2 0 t1]:

(20.17)

In particular, any Pontryagin extremal is a critical point of the endpoint mapping. Pontryagin Maximum Principle implies rst order necessary optimality conditions (20:16), (20:17). Notice that PMP contains more than these conditions: by PMP, the Hamiltonian hu (t ) is not only critical, as in (20:17), but attains maximum along the optimal u~(t). We go further to second order conditions. Asymptotic expansion (20:15) yields the expression for the second dierential: Du2~ Gt1 (v w) a

1 0Z ZZ t (g0 v( 2 ))  g0 w( 1 ) d 1 d 2A a = q0@ g00 (v( ) w( )) d + 2 1

0

2

02 1 t1

1

where a 2 C 1 (M) and v w 2 Ker Du~ Gt1 = Ker Du~ Ft1  i.e.,

q0 

Zt

1

0

gt0 v(t) dt = q0 

Zt

1

0

gt0 w(t) dt = 0:

Now we transform the formula for the second variation via the following decomposition into symmetric and antisymmetric parts. Exercise 20.10. Let X be a nonautonomous vector eld on M. Then

ZZ

02 1 t

X2  X1 d 1 d 2

Zt Zt 1 = 2 X d  X d + 12 0 0

ZZ

X2  X1 ] d 1d 2:

02 1 t

Choosing Xt = gt0 v(t) and taking into account that q0 

we obtain: q0 

ZZ

02 1 t1

X2  X1 d 1 d 2 = 21 q0 

ZZ

02 1 t1

Zt

1

0

gt0 v(t) dt = 0,

X2  X1 ] d 1d 2 

308

20 Second Order Optimality Conditions

thus Du2~ Gt1 (v w)a

0Z 1 ZZ t 00 0 0 = q0  @ g (v( ) w( )) d + g v( 2 ) g w( 1 )] d 1 d 2A a 0 0  t Z t   Z t Z  1

2

1

= q0 

0

g00(v( ) w( )) d +

2

1

1

1

1

0

0

1

g0 2 v( 2 ) d 2 g0 1 w( 1 ) d 1 a:

The rst term can conveniently be expressed in Hamiltonian terms since 0 gu = 0 P;1(fu ; fu~( ) ) = hu ( ) ; hu~( ) ( ):

Then t1 Du2~ Ft1 (v w) = 0 Du2~ Gt1 (v w)

Zt

1

h00 ( )(v( ) w( )) d +

Zt

Z 



g0 2 v( 2 ) d 2 g0 1 w( 1 ) d 1: 0 0 0 (20.18) In order to write also the second term in this expression in the Hamiltonian form, compute the linear on bers Hamiltonian corresponding to the vector eld g0 v:     0g0 v = 0  P;1 @@u fu v = P;10 @@u fu v   = @@u P;10  fu v = @@u hu  P;1(0 )v =

1

0

1

where derivatives w.r.t. u are taken at u = u~( ). Introducing the Hamiltonian hu () = hu (P;1())

we can write the second term in expression (20:18) for the second variation as follows:

Zt Z 1

0

1

0

= =

' ( 0 g0 2 v( 2 ) g0 1 w( 1 ) d 2d 1

 Zt Z @ @ h w( ) ( ) d d h v( ) u 2 u 1 0 2 1  Z0t Z0 @ u @ ;! @ u @ ;! 1

1

2

1

1

1

0 @ u hu2 v( 2 ) @ u hu1 w( 1) d 2 d 1 : (20.19) ;! Here the derivatives @@u hui and @@u hui are evaluated at u = u~( i ). 0

0

20.4 Necessary Optimality Conditions

309

20.4 Necessary Optimality Conditions

Now we apply our results on second variation and obtain necessary conditions for geometric optimality of an extremal trajectory of system (20:1).

20.4.1 Legendre Condition Fix an admissible control u~ which is a corank m  1 critical point of the endpoint mapping Ft1 . For simplicity, we will suppose that u~( ) is piecewise smooth. Take any Lagrange multiplier 0 2 (ImDu~ Ft1 )? n f0g then Zt ;! ; 1 t = Pt 0 = 0  exp ~hu~( ) d  t 2 0 t1] 0

is a trajectory of the Hamiltonian system of PMP. Denote the corresponding quadratic form that evaluates Hessian of the endpoint mapping in (20:18): Q : Tu~ U !  Q(v) =

Zt

1

0

R00

h ( )(v( ) v( )) d +

Zt

1

0

0

Z  0

1



g0 2 v( 2 ) d 2 g0 1 v( 1 ) d 1 :

Then (20:18) reads t1 Hess u~ Ft1 (v v) = Q(v) v 2 Ker Du~ Ft1 : By Theorem 20.6, if a control u~ is locally geometrically optimal (i.e., the endpoint mapping Ft1 is not locally open at u~), then there exists a Lagrange multiplier 0 such that the corresponding form Q satises the condition ind; QjKer Du~ Ft1 < m = corank Du~ Ft1 : (20.20) The kernel of the dierential Du~ Ft1 is dened by a nite number of scalar linear equations:



Ker Du~ Ft1 = v 2 Tu~ U j q0 

Zt

1

0



gt0 v(t) dt = 0 

i.e., it has a nite codimension in Tu~ U . Thus inequality (20:20) implies that ind; Q < +1 for the corresponding extremal t . If we take the extremal ;t projecting to the same extremal curve q(t), then we obtain a form Q with a nite positive index. So local geometric optimality of u~ implies niteness of positive index of the form Q for some Lagrange multiplier 0.

310

20 Second Order Optimality Conditions

Proposition 20.11. If the quadratic form Q has a nite positive index, then

R

there holds the following inequality along the corresponding extremal t :

h00t (t )(v v)  0 t 2 0 t1] v 2 m: (20.21) Inequality (20:21) is called Legendre condition . In particular, if a trajectory q(t) is locally geometrically optimal, then Legendre condition holds for some extremal t , (t ) = q(t). However, necessity of Legendre condition for optimality follows directly from the maximality condition of PMP (exercise). But we will need in the sequel the stronger statement related to index of Q as in Proposition 20.11. Notice once more that in the study of geometric optimality,all signs may be reversed: multiplying t by ;1, we obtain a quadratic form with ind; Q < +1 and the reversed Legendre condition h00t (t )(v v)  0. Of course, this is true also for subsequent conditions related to geometric optimality. Now we prove Proposition 20.11. Proof. Take a smooth vector function v : ! m supp v 0 1] and introduce a family of variations of the form:  ; %  v "( ) = v "  % 2 0 t1) " > 0: Notice that the vector function v " is concentrated at the segment %  % + "]. Compute asymptotics of the form Q on the family introduced:

RR

Q(v " ) = "

="

Z1 0

Z1 0

h00 +"s( +"s )(v(s) v(s)) ds + "2

Z 1 Z 1 0

0

0



g0 +"s2 v(s2 ) ds2  g0 +"s1 v(s1 ) ds1 (20.22)

h00 ( )(v(s) v(s)) ds + O("2 )

where O("2) is uniform w.r.t. v in the L1 norm. Suppose, by contradiction, that h00 ( )(v v) > 0 for some % 2 0 t1), v 2 m. In principal axes, the quadratic form becomes a sum of squares: m X h00 ( )(v v) = i (vi )2

R

with at least one coecient

i=1

20.4 Necessary Optimality Conditions

311

i > 0: Choose a vector function v of the form 0 v1 (s) 1

0 0 1 BB CC BB CC CC = BB vi (s) CC v(s) = B [email protected] v i(s) A @ A vm (s)

0

with the only nonzero component vi (s). For suciently small " > 0, Q(v ") > 0. But for any xed % and ", the space of vector functions v " is innitedimensional. Thus the quadratic form Q has an innite positive index. By contradiction, the proposition follows. ut

20.4.2 Regular Extremals We proved that Legendre condition is necessary for niteness of positive index of the quadratic form Q. The corresponding sucient condition is given by the strong Legendre condition : h00t (t )(v v) < ; jvj2 t 2 0 t1] v 2 m (20.23) > 0: An extremal that satises the strong Legendre condition is called regular (notice that this denition is valid only in the case of open space of control parameters U, where Legendre condition is related to maximality of hu ). Proposition 20.12. If t, t 2 0 t1], is a regular extremal, then: (1) For any 2 0 t1) there exists " > 0 such that the form Q is negative on the space Lm 1   + "], (2) The form Q has a nite positive index on the space Tu~ U = Lm1 0 t1]. Proof. (1) We have: Q(v) = Q1 (v) + Q2 (v)

R

Q1(v) =

Zt

1

Zt 0

h00 ( )(v( ) v( )) d 

Z 



Q2(v) = 0 g0 2 v( 2 ) d 2  g0 1 v( 1 ) d 1 0 0 Z t1 Z 1 @ ;! @ ;! 1

1



@ u hu2 v( 2 ) @ u hu1 v( 1 ) d 2 d 1 : By continuity of h00 ( ) w.r.t. , the strong Legendre condition implies that   Q1 vj  +"] < ; 2 " kvk2L2 =

0

0

0

312

20 Second Order Optimality Conditions

for small " > 0. It follows by the same argument as in (20:22) that the term Q1 dominates on short segments:





Q2 vj  +"] = O("2 )kvk2L2 



thus

" ! 0



Q vj  +"] < 0 for suciently small " > 0 and all v 2 Lm1 0 t1], v 6= 0. (2) We show that the form Q is negative on a nite codimension subspace in Lm1 0 t1], this implies that ind+ Q < 1. By the argument used in the proof of item (1), any point 2 0 t1] can be covered by a segment  ; " + "] such that the form Q is negative on the space Lm1  ; " +"]. Choose points 0 = 0 < 1 < < N = t1 such that Q is negative on the spaces Lm1  i;1 i], i = 1 : : :  N. Dene the following nite codimension subspace of Lm1 0 t1]: L=

(

v 2 Lm 0 t1] j 0 

1

) Z i @ ;! @ u hu v( ) d = 0 i = 1 : : :  N : i;1

For any v 2 L, v 6= 0, Q(v) =

N  X i=1



Q vj i;1 i ] < 0:

Thus L is the required nite codimension negative subspace of the quadratic form Q. Consequently, the form Q has a nite positive index. ut Propositions 20.11 and 20.12 relate sign-deniteness of the form h00 (t ) with sign-deniteness of the form Q, thus, in the corank one case, with local geometric optimality of the reference control u~ (via Theorem 20.3). Legendre condition is necessary for niteness of ind+ Q, thus for local geometric optimality of u~. On the other hand, strong Legendre condition is sucient for negativeness of Q on short segments, thus for local nite-dimensional optimality of u~ on short segments. Notice that we can easily obtain a much stronger result from the theory of elds of extremals (Sect. 17.1). Indeed, under the strong Legendre condition the maximized Hamiltonian of PMP is smooth, and Corollary 17.4 gives local optimality on short segments (in C(0 t1] M) topology, thus in L1 (0 t1] U) topology and in topology of convergence on nite-dimensional submanifolds in U ).

20.4.3 Singular Extremals Now we consider the case where the second derivative of the Hamiltonian hu vanishes identically along the extremal, in particular, the case of control-ane

20.4 Necessary Optimality Conditions

313

P systems q_ = f0 (q) + mi=1 ui fi (q). So we assume that an extremal t satises

the identity

h00t (t )  0 t 2 0 t1]: (20.24) Such an extremal is called totally singular . As in the case of regular extremals, this denition is valid only if the set of control parameters U is open. For a totally singular extremal, expression (20:18) for the Hessian takes the form: t1 Hessu~ Ft1 (v1  v2) = 0

Z t Z  1

0

1

0



g0 2 v1 ( 2 ) d 2 g0 1 v2 ( 1) d 1 :

In order to nd the dominating term of the Hessian (concentrated on the diagonal 1 = 2), we integrate by parts. Denote

Zt

1

wi ( ) = vi (s) ds  g_ 0 = dd g0 :

Then t1 Hessu~ Ft1 (v1  v2) = 0

Z t 

Z

 0Z t

0

1

= 0 ;

;g0 1 w1( 1 ) + g00 w1(0) +

1



0

0:

Zt  1

0



g0 w1( ) g0 v2 ( )] d + g00 w1(0) g00 w2(0)]

+ g00 w1 (0) +



g_ 0 2 w1( 2 ) d 2  g0 1 v2 ( 1 ) d 1

1

Zt

1

0

 Zt

g_ 0 w2( ) d +

g_ 0 2 w1 ( 2 )

Zt 2

1

1

0

_g0 w1( ) g0 w2 ( )] d





g_ 0 1 w2( 1 ) d 1 d 2 :

We integrate by parts also the admissibility condition q0  q0 

Z t

1

0



g_ t0 wi (t) dt + g00 wi(0) = 0:

(20.25)

Zt 0

1

gt0 vi (t) dt = (20.26)

In the sequel we take variations vi subject to the restriction wi(0) =

Zt

1

0

vi (t) dt = 0

i = 1 2:

We ; assume  that functions v(s) used in construction of the family v "( ) = v  ;"  satisfy the equality

314

20 Second Order Optimality Conditions

Z1

v(s) ds = 0

0

then the primitive

w(s) =

Zs 0

v(s0 ) ds0

is also concentrated at the segment 0 1]. Then the last term in expression (20:25) of the Hessian vanishes, and equality (20:26) reduces to q0 

Zt

1

0

g_ t0 wi(t) dt = 0:

Asymptotics of the Hessian on the family v " has the form: t1 Hessu~ Ft1 (v "  v " ) = Q(v " ) = "2 0

Z1 0

g0 w(s) g0 v(s)] ds + O("3 ):

The study of this dominating term provides necessary optimality conditions. Proposition 20.13. Let t, t 2 0 t1], be a totally singular extremal. If the quadratic form Q = t1 Hessu~ Ft1 has a nite positive index, then 0 gt0 v1  gt0 v2] = 0

8 v1  v2 2

R

m

t 2 0 t1]:

(20.27)

Equality (20:27) is called Goh condition . It can be written also as follows:   t @@fuu v1  @@fuu v2 = 0 or in Hamiltonian form: @ h @ h  @  @ ~h = 0 u  u ( ) =  ~ h  t  u u t @u @ ui @ uj @ uj i i j = 1 : : :  m t 2 0 t1]: As before, derivatives w.r.t. u are evaluated at u = u~(t). Now we prove Proposition 20.13. Proof. Take a smooth Zvector function v : ! m concentrated at the seg2 ment 0 2] such that v(s) ds = 0, and construct as before the variation 0 of controls  ; %  v " ( ) = v " :

R R

Then

Q(v ") = "2

Z 2 0

0 g0 w(s) g0 v(s)] ds + O("3 )kvk2L2 

where w(s) =

Zs

Z 2

0

20.4 Necessary Optimality Conditions

315

v(s0 ) ds0. The leading term is the integral

Z 2

0 g0 w(s) g0 v(s)] ds = !(w(s) v(s)) ds 0 0 !(x y) = 0 g0 x g0 y] x y 2 m

R

(20.28)

notice that the bilinear skew-symmetric form ! enters Goh condition (20:27). In order to prove the proposition, we show that if ! 6 0, then the leading term (20:28) of Hessian has a positive subspace of arbitrarily large dimension. Let ! 6 0 for some % 2 0 t1], then rank ! = 2l > 0, and there exist coordinates in m in which the form ! reads

R

!(x y) =

Xl

0 x1 i1=1 x = @ A

(xiyi+l ; xi+l yi )

xm Take a vector function v of the form 0 v1 (s) 1 BB 0 CC

BB CC B 0 CC v(s) = B BB vl+1 (s) CC  BB 0 CC @ A 0 X 1 v (s) =

k>0

k cos ks

0 y1 1 y = @ A : ym

vl+1 (s) =

Substituting v(s) to (20:28), we obtain:

Z 2 0

!(w(s) v(s)) ds = ;2

X k>0

X1 k>0 k

k sin ks:

k k :

This form obviously has a positive subspace of innite dimension. For an arbitrarily great N, we can nd an N-dimensional positive space LN for form (20:28). There exists "N > 0 such that Q(v "N ) > 0 for any v 2 LN . ut Thus ind+ Q = 1. By contradiction, Goh condition follows. Exercise 20.14. Show that Goh condition is satised not only for piecewise smooth, but also for measurable bounded extremal control u~ at Lebesgue points.

316

20 Second Order Optimality Conditions

Goh condition imposes a strong restriction on a totally singular optimal control u~. For a totally singular extremal, the rst two terms in (20:25) vanish by Goh condition. Moreover, under the condition w(0) = 0, the third term in (20:25) vanishes as well. Thus the expression for Hessian (20:25) reduces to the following two terms: t1 Hessu~ Ft1 (v v) = Q(v)

Z t

1

_g0 w( ) g0 w( )] d +

Zt  1

g_ 0 2 w( 2 )



Zt



g_ 0 1 w( 1 ) d 1 d 2 : 0 0 2 (20.29) Suppose that the quadratic form Q has a nite positive index. Then by the same argument as in Proposition 20.11 we prove one more pointwise condition: 0 _gt0 v gt0 v]  0 8 v 2 m t 2 0 t1]: (20.30) This inequality is called generalized Legendre condition . Notice that generalized Legendre condition can be rewritten in Hamiltonian terms: ""h  h0 v#  h0 v# ( )+"h00(u~_ (t) v) h0 v# ( )  0 v 2 m t 2 0 t ]: t t 1 u~(t) t t t t This easily follows from the equalities: gt0 v = Pt; 1 @@fuu v = Ad Pt @@fuu v ;! Z t adf d @ fu v g_ t0 v = ddt exp u~( )  0 @ f  @ [email protected] f = Pt; 1 fu~(t)  @ uu v + Pt; 1 @ u2u (u~_ (t) v): The strong version (20:31) of generalized Legendre condition plays in the totally singular case the role similar to that of the strong Legendre condition in the regular case. Proposition 20.15. Let an extremal t be totally singular, satisfy Goh condition, the strong generalized Legendre condition: ""h  h0 v#  h0 v# ( ) + "h00(u~_ (t) v) h0 v# ( )  ; jvj2 t t u~(t) t t t t v 2 m t 2 0 t1] (20.31) for some > 0, and the following nondegeneracy condition: 0) m ! Tq M is injective: the linear mapping @ [email protected] (q (20.32) 0 u u~(0) : Then the quadratic form QjKer Du~ Ft is negative on short segments and has a

nite positive index on Lm 1 0 t1]. = 0

1

R

R

R

R

20.4 Necessary Optimality Conditions

317

Proof. This proposition is proved similarly to Proposition 20.12. In decom-

position (20:25) the rst two terms vanish by Goh condition, and the fourth term is negative and dominates on short segments. The third term is small on short segments since 0) q0  g00 w1 (0) = @ [email protected] (q u u~(0) w1(0)

R

and condition (20:32) allows to express w1 (0) through the integral 0t1 w1 ( ) d on the kernel of Du~ Ft1 , which is dened by equality (20:26). ut We call an extremal that satises all hypotheses of Proposition 20.15 a nice singular extremal .

20.4.4 Necessary Conditions Summarizing the results obtained in this section, we come to the following necessary conditions for the quadratic form Q to have a nite positive index. Theorem 20.16. Let a piecewise smooth control u~ = u~(t), t 2 0 t1], be a critical point of the endpoint mapping Ft1 . Let a covector t1 2 TFt1 (~u) M be a Lagrange multiplier:

t1 Du~ Ft1 = 0

t1 6= 0:

If the quadratic form Q has a nite positive index, then: (I) The trajectory t of the Hamiltonian system of PMP

_ t = ~hu~(t)(t ) hu () = h fu (q)i

satis es the equality

h0t(t ) = 0

t 2 0 t1]

R R 2R

(II.1) Legendre condition is satis ed: h00t (t )(v v)  0 v 2 m t 2 0 t1]: (II.2) If the extremal t is totally singular: h00t (t )(v v)  0 v 2 m t 2 0 t1] then there hold Goh condition:

fh0t v1  h0tv2g (t )  0

v1 v2

and generalized Legendre condition:

m

t 2 0 t1]

""h  h0 v#  h0 v# ( ) + "h00(u~_ (t) v) h0 v# ( )  0 t t u~(t) t t t t

R

(20.33)

v 2 m t 2 0 t1]: (20.34)

318

20 Second Order Optimality Conditions

Remark 20.17. If the Hamiltonian hu is ane in u (for control-ane systems),

then the second term in generalized Legendre condition (20:34) vanishes. Recall that the corresponding sucient conditions for niteness of index of the second variation are given in Propositions 20.12 and 20.15. Combining Theorems 20.16 and 20.6, we come to the following necessary optimality conditions. Corollary 20.18. If a piecewise smooth control u~ = u~(t) is locally geometrically optimal for control system (20:14), then rst-order conditions (I) and second-order conditions (II.1), (II.2) of Theorem 20:16 hold along the corresponding extremal t.

20.5 Applications In this section we apply the second order optimality conditions obtained to particular problems.

20.5.1 Abnormal Sub-Riemannian Geodesics Consider the sub-Riemannian problem: q_ =

m X i=1

ui fi (q)

R

q 2 M u = (u1  : : :  um ) 2 m

q(0) = q0  q(1) = q1  Z 1X Z1 m J(u) = 12 u2i dt = 12 juj2 dt ! min : 0 i=1 0

The study of optimality is equivalent to the study of boundary of attainable set for the extended system:

The Hamiltonian is hu ( ) =

m X i=1

8 X m >

:y_ = 1 juj2 y2 : 2

R

uih fi (q)i + 2 juj2

R R

 2 T  M  2  = :

The parameter  is constant along any geodesic (extremal). If  6= 0 (the normal case), then extremal control can be recovered via PMP. In the sequel we consider the abnormal case:

20.5 Applications

319

= 0:

Then

hu () = hu ( 0) =

m X i=1

ui hi()

hi () = h fi (q)i i = 1 : : :  m: The maximality condition of PMP does not determine controls in the abnormal case directly (abnormal extremals are totally singular). What that condition implies is that abnormal extremals t satisfy, in addition to the Hamiltonian system m X _ t = ui(t)~hi (t ) the following identities:

i=1

hi (t )  0 i = 1 : : :  m: We apply second order conditions. As we already noticed, Legendre condition degenerates. Goh condition reads: fhi  hj g(t )  0 i j = 1 : : :m: If an abnormal extremal t projects to an optimal trajectory q(t), then at any point q of this trajectory there exists a covector  2 Tq M  6= 0 such that h fi (q)i = 0 i = 1 : : :  m h fi  fj ](q)i = 0 i j = 1 : : :  m: Consequently, if span(fi (q) fi fj ](q)) = Tq M (20.35) then no locally optimal strictly abnormal trajectory passes through the point q. An extremal trajectory is called strictly abnormal if it is a projection of an abnormal extremal and it is not a projection of a normal extremal. Notice that in the case corank > 1 extremal trajectories can be abnormal but not strictly abnormal (i.e., can be abnormal and normal simultaneously), there can be two Lagrange multipliers ( 0) and (0   6= 0). Small arcs of such trajectories are always local minimizers since the normal Hamiltonian P H = 21 mi=1 h2i is smooth (see Corollary 17.4). Distributions span(fi (q)) that satisfy condition (20:35) are called 2-generating . E.g., the left-invariant bracket-generating distributions appearing in the sub-Riemannian problem on a compact Lie group in Sect. 19.2 and Exercise 19.2 are 2-generating, thus there are no optimal strictly abnormal trajectories in those problems.

320

20 Second Order Optimality Conditions

Example 20.19. Consider the following left-invariant sub-Riemannian problem

on GL(n) with a natural cost: Q_ = QV Q 2 GL(n) V = V   (20.36) Z 1 J(V ) = 21 tr V 2 dt ! min: (20.37) 0 Exercise 20.20. Show that normal extremals in this problem are products of 2 one-parameter subgroups. (Hint: repeat the argument of Sect. 19.2.) Then it follows that any nonsingular matrix can be represented as a product of two exponentials eV e(V ;V  )=2. Notice that not any nonsingular matrix can be represented as a single exponential eV . There are many abnormal extremals in problem (20:36), (20:37), but they are never optimal. Indeed, the distribution dened by the right-hand side of the system is 2-generating. We have QV1 QV2] = QV1 V2 ] and if matrices Vi are symmetric then their commutator V1 V2 ] is antisymmetric. Moreover, any antisymmetric matrix appears in this way. But any n  n matrix is a sum of symmetric and antisymmetric matrices. Thus the distribution fQV j V  = V g is 2-generating, and strictly abnormal extremal trajectories are not optimal.

20.5.2 Local Controllability of Bilinear System

R R

Consider a bilinear control system of the form x_ = Ax + uBx + vb u v 2  x 2 n: (20.38) We are interested, when the system is locally controllable at the origin, i.e., 0 2 int A0 (t) 8 t > 0: Negation of necessary conditions for geometric optimality gives sucient conditions for local controllability.Now we apply second order conditions of Corollary 20.18 to our system. Suppose that 0 2 @ A0 (t) for some t > 0: Then the reference trajectory x(t)  0 is geometrically optimal, thus it satises PMP. The control-dependent Hamiltonian is huv (p x) = pAx + upBx + vpb  = (p x) 2 T  n = n  n: The vertical part of the Hamiltonian system along the reference trajectory x(t) reads:

R R R

p_ = ;pA

R

20.6 Single-Input Case

p 2 n:

321

(20.39)

It follows from PMP that p( )b = p(0)e;A b  0 2 0 t] i.e., p(0)Aib = 0 i = 0 : : :  n ; 1 (20.40) for some covector p(0) 6= 0, thus span(b Ab : : :  An;1b) 6= n: We pass to second order conditions. Legendre condition degenerates since the system is control-ane, and Goh condition takes the form: p( )Bb  0 2 0 t]: Dierentiating this identity by virtue of Hamiltonian system (20:39), we obtain, in addition to (20:40), new restrictions on p(0): p(0)Ai Bb = 0 i = 0 : : :  n ; 1: Generalized Legendre condition degenerates. Summing up, the inequality span(b Ab : : :  An;1b Bb ABb : : :  An;1Bb) 6= n is necessary for geometric optimality of the trajectory x(t)  0. In other words, the equality span(b Ab : : :  An;1b Bb ABb : : :  An;1Bb) = n is sucient for local controllability of bilinear system (20:38) at the origin.

R

R R

20.6 Single-Input Case In this section we apply rst- and second-order optimality conditions to the simplest (and the hardest to control) case with scalar input: q_ = f0 (q) + uf1 (q) u 2   ]  q 2 M: (20.41) Since the system is control-ane, Legendre condition automatically degenerates. Further, control is one-dimensional, thus Goh condition is trivial. Although, generalized Legendre condition works (we write it down later). We apply rst Pontryagin Maximum Principle. Introduce the following Hamiltonians linear on bers of the cotangent bundle:

R

322

20 Second Order Optimality Conditions

hi () = h fi (q)i then the Hamiltonian of the system reads

i = 0 1

hu () = h0() + uh1 (): We look for extremals corresponding to a control u(t) 2 (  ): The Hamiltonian system of PMP reads _ t = ~h0 (t ) + u(t)~h1(t ) and maximality condition reduces to the identity

(20.42) (20.43)

h1 (t )  0: (20.44) Extremals t are Lipschitzian, so we can dierentiate the preceding identity: h_ 1(t ) = ddt h1 (t ) = fh0 + u(t)h1  h1g(t ) = fh0  h1g(t)  0: (20.45) Equalities (20:44), (20:45), which hold identically along any extremal t that satises (20:42), do not allow us to determine the corresponding control u(t). In order to obtain an equality involving u(t), we proceed with dierentiation: #h1 (t ) = fh0 + u(t)h1  fh0 h1gg(t ) = fh0 fh0 h1gg(t ) + u(t)fh1  fh0 h1gg(t)  0: Introduce the notation for Hamiltonians: hi1 i2 :::ik = fhi1  fhi2  : : :  fhik;1  hik g : : : gg ij 2 f0 1g: then any extremal t with (20:42) satises the identities h1 (t ) = h01 (t )  0 (20.46) h001(t ) + u(t)h101(t )  0: (20.47) If h101(t ) 6= 0, then extremal control u = u(t) is uniquely determined by t : t) : (20.48) u(t) = ; hh001( 101(t ) Notice that the regularity condition h101(t ) 6= 0 is closely related to generalized Legendre condition. Indeed, for the Hamiltonian hu = h0 + uh1 generalized Legendre condition takes the form ffh0 + uh1 h1g h1g(t ) = ;h101(t )  0

20.6 Single-Input Case

323

i.e.,

h101(t )  0: And if this inequality becomes strong, then the control is determined by relation (20:48). Assume that h101(t ) 6= 0 and plug the control u() = ;h001()=h101() given by (20:48) to the Hamiltonian system (20:43): _ = ~h0 () + u()~h1 (): (20.49) Any extremal with (20:42) and h101(t ) 6= 0 is a trajectory of this system.

Lemma 20.21. The manifold f 2 T  M j h1() = h01() = 0 h101() 6= 0g

(20.50)

is invariant for system (20:49). Proof. Notice rst of all that the regularity condition h101() 6= 0 guarantees that conditions (20:50) determine a smooth manifold since dh1 and dh01 are linearly independent. Introduce a Hamiltonian

'() = h0 () + u()h1 (): The corresponding Hamiltonian vector eld '~ () = ~h0() + u() ~h1() + h1() ~u() coincides with eld (20:49) on the manifold fh1 = h01 = 0g, so it is sucient to show that '~ is tangent to this manifold. Compute derivatives by virtue of the eld '~ : h_ 1 = fh0 + uh1 h1g = h01 ; (~h1 u)h1 h_ 01 = fh0 + uh1  h01g = h| 001 +{zuh101} ;(~h01 u)h1 = ;(~h01 u)h1:

0

The linear system with variable coecients for h1(t) = h1 (t), h01(t) = h01(t ) (_ h1 (t) = h01 (t) ; (~h1u)(t ) h1 (t) h_ 01(t) = ;(~h01u)(t ) h1 (t) has a unique solution. Thus for the initial condition h1(0) = h01(0) = 0 we obtain the solution h1 (t) = h01(t)  0. So manifold (20:50) is invariant for the eld '~ (), thus for eld (20:49). ut Now we can describe all extremals of system (20:41) satisfying the conditions (20:42) and h101 6= 0. Any such extremal belongs to the manifold

324

20 Second Order Optimality Conditions

fh1 = h01 = 0g, and through any point 0 of this manifold with the bound-

ary restrictions on control satised: 0 ) 2 (  ) u(0) = ; hh001( ( 101 0 ) passes a unique such extremal | the trajectory t of system (20:49). In problems considered in Chaps. 13 and 18 (Dubins car, rotation around 2 axes in SO(3)), all singular extremals appeared exactly in this way. Generically, h101 6= 0, thus all extremals with (20:42) can be studied as above. But in important examples the hamiltonian h101 can vanish. E.g., consider a mechanical system with a controlled force: y# = g(y) + ub y b 2 n u 2   ]  or, in the standard form:

R

R

(

y_1 = y2  y_2 = g(y1 ) + ub:

The vector elds in the right-hand side are f0 = y2 @@y + g(y1 ) @@y  1 2 @ f1 = b @ y  2

thus

h101() = h f | 1 f{z0 f1]]}i  0:

0

More generally, h101 vanishes as well for systems of the form

(

x_ = f(x y) y# = g(x y) + ub

R

x 2 M y b 2 n u 2   ]

R

: (20.51)

An interesting example of such systems is Dubins car with control of angular acceleration : 8x_ = cos  > >

:y_ = u Having such a motivation in mind, we consider now the case where h101()  0: (20.52)

R

R

20.6 Single-Input Case

325

Then equality (20:47) does not contain u(t), and we continue dierentiation in order to nd an equation determining the control: h(3) 1 (t ) = h_ 001(t ) = h0001(t ) + u(t)h1001(t )  0: It turns out that the term near u(t) vanishes identically under condition (20:52): h1001 = fh1  fh0 fh0 h1ggg = ff | h1 h0g{z fh0 h1gg} +fh0 fh1 fh0 h1ggg =0

= fh0  h101g = 0: So we obtain, in addition to (20:46), (20:47), and (20:52), one more identity without u(t) for extremals: h0001(t )  0: Thus we continue dierentiation: h(4) (20.53) 1 (t ) = h_ 0001(t ) = h00001(t ) + u(t)h10001(t )  0: In Dubins car with angular acceleration control h10001(t ) 6= 0, and generically (in the class of systems (20:51)) this is also the case. Under the condition h10001(t ) 6= 0 we can express control as u = u() from equation (20:53) and nd all extremals in the same way as in the case h101(t ) 6= 0. Exercise 20.22. Show that for Dubins car with angular acceleration control, singular trajectories are straight lines in the plane (x1  x2): x1 = x01 + t cos 0  x2 = x02 + t sin 0  = 0  y = 0: Although, now geometry of the system is new. There appears a new pattern for optimal control, where control has an innite number of switchings on compact time intervals. For the standard Dubins car (with angular velocity control) singular trajectories can join bang trajectories as follows: u(t) = 1 t < t% u(t) = 0 t > t% (20.54) or u(t) = 0 t < t% u(t) = 1 t > t%: (20.55) We show that such controls cannot be optimal for the Dubins car with angular acceleration control. The following argument shows how our methods can be applied to problems not covered directly by the formal theory. In this argument we prove Proposition 20.23 stated below at p. 330.

326

20 Second Order Optimality Conditions

Consider the time-optimal problem for our single-input system (20:41). We prove that there do not exist time-optimal trajectories containing a singular piece followed by a bang piece. Suppose, by contradiction, that such a trajectory q(t) exists. Consider restriction of this trajectory to the singular and bang pieces: q(t) t 2 0 t1] u(t) 2 (  ) t 2 0 %t ] u(t) =  2 f   g t 2 t% t1]: Let t be an extremal corresponding to the extremal trajectory q(t). We suppose that such t is unique up to a nonzero factor (generically, this is the case). Reparametrizing control (i.e., taking u ; u(t%; 0) as a new control), we obtain u(t% ; 0) = 0 < 0 <  without any change of the structure of Lie brackets. Notice that now we study a time-optimal trajectory, not geometrically optimal one as before. Although, the Hamiltonian of PMP hu = h0 + uh1 for the time-optimal problem is the same as for the geometric problem, thus the above analysis of singular extremals applies. In fact, we prove below that a singular piece and a bang piece cannot follow one another not only for a time-minimal trajectory, but also for a time-maximal trajectory or for a geometrically optimal one. We suppose that the elds f0 , f1 satisfy the identity f1 f0 f1 ]]  0 and the extremal t satises the inequality h10001(t ) 6= 0: Since u(t% ; 0) = 0, then equality (20:53) implies that h00001(t ) = 0. It follows from the maximality condition of PMP that hu(t)(t ) = h0 (t ) + u(t)h1(t )  h0(t ) i.e., along the whole extremal u(t)h1 (t )  0 t 2 0 t1]: But along the singular piece h1 (t )  0, thus u(t)h1 (t )  0 t 2 0 %t ]: The rst nonvanishing derivative of u1 (t)h1(t ) at t = t%+ 0 is positive. Keeping in mind that u(t)   at the singular piece t 2 t% t1], we compute this derivative. Since h1 (t ) = h01 (t ) = h001(t ) = h0001(t ) = h1001(t ) = 0, then the rst three derivatives vanish:

20.6 Single-Input Case

327

dk dtk t=t +0 u(t)h1 (t ) = 0 k = 0 1 2 3: Thus the fourth derivative is nonnegative: d4 dt4 t=t +0 u(t)h1(t ) = (h00001(t ) + h10001(t )) =  2 h10001(t )  0: Since  2 > 0, then h10001(t ) > 0: (20.56) Now we apply this inequality in order to obtain a contradiction via the theory of second variation. Recall expression (20:29) for Hessian of the endpoint mapping: t Hessu Ft(v) =

Zt 0

0 _g0  g0 ] w2( ) d +

Here

Z tZ 

' ( 0 g_ 0 2  g_ 0 1 w( 2 )w( 1 ) d 2d 1: (20.57)

1

0

0

Zt

w( ) = v( ) d   0g = P;1f1  g_ 0 = P;1f0 f1 ]

;! P = exp

Z 0

w(0) = 0

fu() d :

The rst term in expression (20:57) for the Hessian vanishes: 0 _g0  g0 ] = ;h101( )  0: Integrating the second term by parts twice, we obtain: t Hessu Ft(v) = where

Zt 0

0 #g0  g_ 0 ] 2( ) d +

Z tZ 

1

0

0

' ( 0 g#0 2  g#0 1 ( 2 )( 1 ) d 2 d 1 (20.58)

g#0 = P;1f0  f0 f1 ]] ( ) =

Z 0

w( 1 ) d 1

(t) = 0:

328

20 Second Order Optimality Conditions

The rst term in (20:58) dominates on needle-like variations v = vt " : t Hessu Ft (vt " ) = "4 0  g#t0  g_ t0 ] + O("5 ) we compute the leading term in the Hamiltonian form: 0  g#t0  g_ t0 ] = t f0 f0 f1]] f0 f1 ]] = fh001 h01g(t ) = ffh1 h0g h001g(t ) = fh1 fh0 h001gg(t ) ; fh0 f| h1{zh001g}g(t ) = h10001(t ): =h1001 0

By virtue of inequality (20:56), t Hess u Ft (v0) > 0 where v0 = vt " for small enough " > 0. This means that d2 d s2 s=0 a  Ft (u + sv0 ) = t Hessu Ft (v0 ) > 0 for any function a 2 C 1 (M), a(q(t%)) = 0, dq(t ) a = t . Then 2 a  Ft (u + sv0 ) = s2 t Hessu Ft (v0 ) + O(s3 ) s ! 0 i.e., the curve Ft (u + psv0 ) is smooth at s = +0 and has the tangent vector p d d s s=+0 Ft (u + sv0 ) = 0  ht  0i > 0: (20.59) That is, variation of the optimal control u in direction of v0 generates a tangent vector 0 to the attainable set Aq0 (t%) that belongs to the half-space ht  i > 0 in Tq(t ) M. Since the extremal trajectory q(t) is a projection of a unique, up to a scalar factor, extremal t , then the control u is a corank one critical point of the endpoint mapping: dimImDu Ft = dimM ; 1 = n ; 1: This means that there exist variations of control that generate a hyperplane of tangent vectors to Aq0 (t%): 9 v1  : : :  vn;1 2 Tu U such that d d s s=0 Ft (u + svi ) = i  i = 1 : : :  n ; 1 span( 1 : : :  n;1) = ImDu Ft :

20.6 Single-Input Case

329

Summing up, the variations v0, v1 , : : : , vn;1 of the control u at the singular piece generate a nonnegative half-space of the covector t : us = u + ps0 v0 +

nX ;1 i=1

s = (s0  s1  : : :  sn;1) 2

si vi 

R R ; +

n 1

@ i = 0 1 : : :  n ; 1 @ si s=0 Ft (us ) = i  + 0 + span( 1  : : :  n;1) = fht  i  0g: Now we add a needle-like variation on the bang piece. Since the control u(t), t 2 t% t1], is nonsingular, then the switching function h1(t ) 6 0, t 2 t% t1]. Choose any instant %t1 2 (t% t1) such that h1(t 1 ) 6= 0: Add a needle-like variation concentrated at small segments near %t1: 8 > t 2 0 %t ] 0, thus ht1  ni < 0: Now we translate the tangent vectors i , i = 0 : : :  n;1, from q(t%) to q(t1 ): @ @ t1 F t 1 (us" ) = @ " ("s)=(00) @ " ("s)=(00) Pt (Ft (us )) ;  = Pt t1  i = i  i = 0 : : :  n ; 1:

330

20 Second Order Optimality Conditions

Inequality (20:59) translates to

ht1  0i = ht  0i > 0 and, of course,

ht1  ii = ht  i i = 0 i = 1 : : :  n ; 1: The inequality ht1  ni < 0 means that the needle-like variation on the bang piece generates a tangent vector in the half-space ht1  i < 0 complementary to the half-space ht1  i  0 generated by variations on the singular piece.

R R ; R ! R R ; R R

Summing up, the mapping F:

+

n 1

+

M

F(s ") = Ft1 (us" )

satises the condition D(00)F(

+

n 1

+) =

R

+0 + span(1 : : :  n;1) + +n = Tq(t1 ) M:

By Lemma 12.4 and remark after it, the mapping F is locally open at (s ") = (0 0). Thus the image of the mapping Ft1 (us") contains a neighborhood of the terminal point q(t1 ). By continuity, q(t1 ) remains in the image of Ft1 ; (us") for suciently small  > 0. In other words, the point q(t1) is reachable from q0 at t1 ; instants of time, i.e., the trajectory q(t), t 2 0 t1], is not time-optimal, a contradiction. We proved that a time-optimal trajectory q(t) cannot have a singular piece followed by a bang piece. Similarly, a singular piece cannot follow a bang piece. We obtain the following statement on the possible structure of optimal control.

Proposition 20.23. Assume that vector elds in the right-hand side of system (20:41) satisfy the identity

f1  f0 f1]] = 0:

(20.60)

Let a time-optimal trajectory q(t) of this system be a projection of a unique, up to a scalar factor, extremal t , and let h10001(t ) 6= 0. Then the trajectory q(t) cannot contain a singular piece and a bang piece adjacent one to another. Remark 20.24. In this proposition, time-optimal (i.e., time-minimal) control

can be replaced by a time-maximal control or by a geometrically optimal one. What happens near singular trajectories under hypothesis (20:60)? Assume that a singular trajectory is optimal (as straight lines for Dubins car with angular acceleration control). Notice that optimal controls exist, thus the cost function is everywhere dened. For boundary conditions suciently close to the singular trajectory, there are two possible patterns of optimal control:

20.6 Single-Input Case

331

(1) either it makes innite number of switchings on a compact time segment adjacent to the singular part, so that the optimal trajectory \gets o" the singular trajectory via innite number of switchings, (2) or optimal control is bang-bang, but the number of switchings grows innitely as the terminal point approaches the singular trajectory. Pattern (1) of optimal control is called Fuller's phenomenon . It turns out that Fuller's phenomenon takes place in Dubins car with angular acceleration control, see Fig. 20.1. As our preceding arguments suggest, this phenomenon is not a pathology, but is ubiquitous for certain classes of systems (in particular, in applications). One can observe this phenomenon trying to stop a pingpong ball jumping between the table and descending racket. The theory of Fuller's phenomenon is described in book 18]. It follows from this theory that possibility (1) is actually realized for Dubins car with angular acceleration control.

Fig. 20.1. Singular arc adjacent to arc with Fuller phenomenon

21 Jacobi Equation

In Chap. 20 we established that the sign of the quadratic form t Hessu~ Ft is related to optimality of the extremal control u~. Under natural assumptions, the second variation is negative on short segments. Now we wish to catch the instant of time where this quadratic form fails to be negative. We derive an ODE (Jacobi equation) that allows to nd such instants (conjugate times). Moreover, we give necessary and sucient optimality conditions in these terms. Recall expression (20:18) for the quadratic form Q with t Hessu~ Ft = QjKer Du~ Ft obtained in Sect. 20.3: Q(v) =

Zt 0

h00 (v( )) d +

Z t Z  0

0

1

0



g0 2 v( 2 ) d 2 g0 1 v( 1 ) d 1 :

We extend the form Q from L1 to L2 by continuity. We will consider a family of problems on segments 0 t], t 2 0 t1], so we introduce the corresponding sets of admissible controls: Ut = fu 2 L2 (0 t1] U) j u( ) = 0 for > tg and spaces of variations of controls:

Vt = Tu~ Ut = fv 2 Lm2 0 t1] j v( ) = 0 for > tg = Lm2 0 t]: We denote the second variation on the corresponding segment as Qt = QjVt : Notice that the family of spaces Vt is ordered by inclusion: t0 < t00 ) Vt0 Vt00 

334

21 Jacobi Equation

and the family of forms Qt respects this order: Qt0 = Qt00 jVt0 :

In particular,

Qt00 < 0 ) Qt0 < 0: Denote the instant of time where the forms Qt lose their negative sign:

"

#

t def = sup t 2 (0 t1] j Qt jKt < 0 



where

Kt = v 2 Vt j q0 

Zt 0

g0 v( ) d = 0



is the closure of the space Ker Du~ Ft in L2 . If Qt jKt is negative for all t 2 (0 t1], then, by denition, t = +1.

21.1 Regular Case: Derivation of Jacobi Equation Proposition 21.1. Let t be a regular extremal with t 2 (0 t1]. Then the quadratic form Qt jKt is degenerate. Proof. By the strong Legendre condition, the norm

kvkh00 =

Z t 0

;h00 (v( )) d

1=2

is equivalent to the standard Lm2 -norm. Then Qt =

Z t 0

h00 (v( )) d +

= ;kvk2h00 + hRv vi

Z t Z  0

0

1

0



g0 2 v( 2 ) d 2 g0 1 v( 1 ) d 1

where R is a compact operator in Lm2 0 t]. First we prove that the quadratic form Qt is nonpositive on the kernel Kt . Assume, by contradiction, that there exists v 2 Vt such that Qt (v) > 0 v 2 Kt : The linear mapping Du~ Ft has a nite-dimensional image, thus Vt = Kt ! E dimE < 1: The family Du~ Ft is weakly continuous in t, hence Du~ Ft ;"jE is invertible and Vt = Kt ;" ! E

21.1 Regular Case: Derivation of Jacobi Equation

335

for small " > 0. Consider the corresponding decomposition v = v" + x" 

v" 2 Kt ;"  x" 2 E:

Then x" ! 0 weakly as " ! 0, so x" ! 0 strongly since E is nite-dimensional. Consequently, v" ! v strongly as " ! 0. Further, Qt ;" (v" ) = Qt (v" ) ! Qt (v) as " ! 0 since the quadratic forms Qt are continuous. Summing up, Qt;" (v" ) > 0 for small " > 0, a contradiction with denition of t . We proved that Now we show that

Qt jKt  0:

(21.1)

9 v 2 Kt  v 6= 0 such that Qt (v) = 0: By the argument similar to the proof of Proposition 16.4 (in the study of conjugate points for the linear-quadratic problem), we show that the function (t) = sup fQt(v) j v 2 Kt  kvkh00 = 1g (21.2) satises the following properties: (t) is monotone nondecreasing, the supremum in (21:2) is attained, and (t) is continuous from the right. Inequality (21:1) means that (t )  0. If (t ) < 0, then (t + ") < 0 for small " > 0, which contradicts denition of the instant t . Thus (t ) = 0, moreover, there exists v 2 Kt  kvkh00 = 1 such that Qt (v) = 0: Taking into account that the quadratic form Qt is nonpositive, we conclude that the element v 6= 0 is in the kernel of Qt jKt . ut Proposition 21.1 motivates the introduction of the following important notion. An instant tc 2 (0 t1] is called a conjugate time (for the initial instant t = 0) along a regular extremal t if the quadratic form Qtc jKtc is degenerate. Notice that by Proposition 20.12, the forms Qt jKt are negative for small t > 0, thus short arcs of regular extremals have no conjugate points: for them t > 0. Proposition 21.1 means that the instant t where the quadratic forms QtjKt lose their negative sign is the rst conjugate time. We start to derive a dierential equation on conjugate time for a regular extremal pair (~u(t) t). The symplectic space  = T0 (T  M)

will be the state space of that ODE. Introduce the family of mappings

336

R

21 Jacobi Equation

Jt : m !  ;! Jt v = @@u hut v: u~(t) In these terms, the bilinear form Qt reads Qt (v1  v2) =

Zt 0

ZZ

h00 (v1 ( ) v2( )) dt +

02 1 t

(J2 v1 ( 2 ) J1 v2 ( 1 )) d 1 d 2 (21.3)

see (20:18), (20:19). We consider the form Qt on the subspace



Kt = Ker Du~ Ft = vi 2 Vt j where

Zt 0



J vi ( ) d 2 0 

(21.4)

0 = T0 (Tq0 M)

is the vertical subspace. A variation of control v 2 Vt satises the inclusion ;  v 2 Ker Qt jKt i the linear form Qt (v ) annihilates the subspace Kt Vt . Since the vertical subspace 0  is Lagrangian, equality (21:4) can be rewritten as follows:



Kt = vi 2 Vt j 

Z t 0





J vi ( ) d   = 0 8 2 0 :

That is, the annihilator of the subspace Kt Vt coincides with the following nite-dimensional space of linear forms on Vt :

Z t 0



(J  ) d j  2 0 :

;

(21.5)



Summing up, we obtain that v 2 Ker QtjK;t i the form Qt(v ) on Vt  belongs to subspace (21:5). That is, v 2 Ker Qt jKt i there exists  2 0 such that Qt(v ) =

Zt 0

(J  ) d :

We transform equality of forms (21:6):

Zt 0

(J  ) d = =

Zt 0

Zt 0

h00 (v( ) ) d +

ZZ

02 1 t

h00 (v( ) ) d +

(J2 v( 2 ) J1 ) d 1d 2

Z t Z  0



(21.6)

0



J v( ) d  J d :

21.1 Regular Case: Derivation of Jacobi Equation

337

This equality of forms means that the integrands coincide one with another: (J  ) = h00 (v( ) ) + 

Z  0



J v( ) d  J 

2 0 t]:

(21.7)

In terms of the curve in the space   =

Z 0

J v( ) d + 

2 0 t]

(21.8)

equality of forms (21:7) can be rewritten as follows: h00 (v( ) ) + (  J ) = 0 2 0 t]: The strong Legendre condition implies that the linear mapping h00 : m ! m

(21.9)

R R

is nondegenerate (we denote here and below the linear mapping into the dual space by the same symbol as the corresponding quadratic form), thus the inverse mapping is dened:

R R

(h00 );1 : m ! m:

Then equality (21:9) reads

v( ) + (h00 );1 (  J ) = 0 2 0 t]: (21.10) We come to the following statement. Theorem 21.2. Let t, t 2 0 t1], be a regular extremal. An instant t 2 (0 t1] is a conjugate time i there exists a nonconstant solution  to Jacobi equation _ = J (h00 );1 (J   )

2 0 t]

(21.11)

that satis es the boundary conditions

0 2 0

t 2 0:

(21.12) Jacobi equation (21:11) is a linear nonautonomous Hamiltonian system on  : _ = ~b ( ) (21.13) with the quadratic Hamiltonian function

b () = ; 12 (h00 );1 ((J  ) (J  )) 

R

where (h00 );1 is a quadratic form on m.

 2 

338

21 Jacobi Equation

Proof. We already proved that existence of v 2 Ker QtjKt is equivalent to

existence of a solution  to Jacobi equation that satises the boundary conditions (21:12). If v  0, then   const by virtue of (21:8). Conversely, if   const, then J v( ) = _  0. By (21:3), the second variation takes the form Qt(v) =

Zt 0

h00 (v( )) d < ; kvk2L2

for some > 0:

But v 2 Ker Qt, so Qt(v) = 0, consequently v  0. Thus nonzero v correspond to nonconstant  and vice versa. It remains to prove that b is the Hamiltonian function for Jacobi equation (21:11). Denote

R

A () = (h00 );1 (J  ) 2 m

then Jacobi equation reads

 2 

_ = J A ( ) so we have to prove that J A () = ~b ()

 2 :

(21.14)

Since then

D E b () = ; 12 (J  ) (h00 );1 (J  ) = ; 12 h(J  ) A ()i  hd b  i = ;h(J  ) A ()i = (  J A ()):

Thus equality (21:14) follows and the proof is complete.

ut

21.2 Singular Case: Derivation of Jacobi Equation In this section we obtain Jacobi equation for a nice singular extremal pair (~u(t) t ). In contrast to the regular case, the second variation in the singular case can be nondegenerate at the instant t where it loses its negative sign. In order to develop the theory of conjugate points for the singular case, we introduce a change of variables in the form Qt. We denote, as before, the integrals wi ( ) =

Zt 

vi (s) ds

i = 1 2

and denote the bilinear form that enters generalized Legendre condition:

R

21.2 Singular Case: Derivation of Jacobi Equation

lt (w1 w2) = (J_t w1  Jtw2)

339

wi 2 m:

For a nice singular extremal, expression (20:25) for the second variation reads Qt(v1  v2 ) =

Zt 0

l (w1 ( ) w2( )) d +

Zt 

 J_ w1 ( )



0



Zt

+  J0w1 (0)

J_ w2 ( ) d d



Zt 0



J_ w2( ) d :

Admissibility condition (20:26) for variations of control vi ( ) can be written as follows:

Zt 0

The mapping

J_ w( ) d + J0 w(0) 2 0:

(21.15)

R

R R

v( ) 7! (w( ) w(0)) 2 Lm2  m has a dense image in Lm2  m, and the Hessian Qt and admissibility condition (21:15) are extended to Lm2  m by continuity. Denote  = J0 w(0) 2 ;0 and consider the extended form Qt(w1  w2) =

Zt 0

l (w1 ( ) w2( )) d +

Zt  0

 J_ w1( )

Zt

 Z t

+  1  on the space

Zt 0

0



J_ w2 ( ) d d J_ w2 ( ) d

J_ w( ) d +  2 0:



(21.16)

Then in the same way as in the regular case, it follows that the restriction of the quadratic form Qt(w) to the space (21:16) is degenerate at the instant t = t . An instant t that satises such a property is called a conjugate time for the nice singular extremal t . Similarly to the regular case, we derive now a Hamiltonian Jacobi equation on conjugate times for nice singular extremals, although the Hamiltonian function and boundary conditions dier from the ones for the regular case. Let t 2 (0 t1] be a conjugate time, i.e., let the form Qt (w1 w2) have a nontrivial kernel on the space (21:16). That is, there exists a pair (w ) 2 Lm2 0 t]  ;0

Zt 0

J_ w( ) d +  2 0

340

21 Jacobi Equation

such that the linear form on the space Lm2 0 t]  ;0 Qt(  w) =

Zt 0

l ( L2  w( )) d +

Zt  0

Zt

 J_ L2 





+  ;0 



J_ w( ) d d

Zt 0

J_ w( ) d



(21.17)

annihilates the admissible space (21:16). In turn, the annihilator of the admissible space (21:16) is the space of linear forms

Zt    J_ L   d +  ( ;  )

2 0 :

0

2

0

Thus, similarly to the regular case, there exists  2 0 such that Qt (  w) =

Zt 



 J_ L2   d +  ( ;0  ):

0

By virtue of (21:17), the previous equality of forms splits:



l ( Rm  w( )) +  J_ Rm 



 ;0 

Zt 0



Zt 







J_ w( ) d =  J_ Rm   

2 0 t]

J_ w( ) d =  ( ;0  ):

That is,



l w( ) = ; J_ Rm 



 ;0 

Zt 0



Zt 

J_ w( ) d ;  



J_ w( ) d ;  = 0:

(21.18) (21.19)

In terms of the curve in the space  = T0 (T  M)  =

Zt 

J_ w( ) d ; 

equalities (21:18), (21:19) take the form





l w( ) = ; J_ Rm     (;0 0) = 0:

2 0 t] 2 0 t]

(21.20) (21.21)

The last equality means that 0 belongs to the skew-orthogonal complement ;0\. On the other hand, 0 2 ;0 + 0, compare denition (21:20) with (21:16). That is,

21.2 Singular Case: Derivation of Jacobi Equation

341

0 2 (0 + ;0) \ ;0\ = 0;0 :

Recall that 0;0 is a Lagrangian subspace in the symplectic space  containing the isotropic subspace ;0, see denition (11:28). Notice that Goh condition (Jt v1  Jtv2 )  0 v1 v2 2 m t 2 0 t1]

R

means that the subspaces

R

;t = spanfJtv j v 2 mg  are isotropic. We obtain boundary conditions for the curve  : 0 2 0;0 

t 2 0 :

(21.22)

Moreover, equality (21:21) yields an ODE for  : _ = ;J_ w( ) = J_ l;1((J_   )) 2 0 t]:

(21.23)

Similarly to the regular case, it follows that this equation is Hamiltonian with the Hamiltonian function ^b () = ; 1 l;1 ((J_  ) (J_  ))  2 : 2 The linear nonautonomous equation (21:23) is Jacobi equation for the totally singular case. Now the next statement follows in the same way as in the regular case. Theorem 21.3. Let t be a nice singular extremal. An instant t 2 (0 t1] is a conjugate time i there exists a nonconstant solution  to Jacobi equation   _ = J_ l;1 (J_   )  2 0 t] (21.24) with the boundary conditions

0 2 0;0 

t 2 0:

(21.25)

Jacobi equation (21:24) is Hamiltonian:

_ = ~^b ( )

(21.26)

with the nonautonomous quadratic Hamiltonian function

  ^b () = ; 1 l;1 (J_  ) (J_  )  2

 2 :

The following statement provides a rst integral of equation (21:23), it can be useful in the study of Jacobi equation in the singular case.

342

21 Jacobi Equation

Lemma 21.4. For any constant vector v 2 integral of Jacobi equation (21:23). Proof. We have to show that

R

m, the function

( J v) is an

(_  J v) + (  J_ v)  0 (21.27) for a solution  to (21:23). The rst term can be computed via Jacobi equation: (_  J v) = ;hd ^b  J vi   = l;1 (J_  J v) (J_   ) where l;1 is a bilinear form

D

E

= (J_   ) l;1(J_  J v) where l;1 is a linear mapping to the dual space

D

E

= (J_   ) v = ;(  J_ v) and equality (21:27) follows. ut In particular, this lemma means that 0 2 ;0\ ,  2 ;\  i.e., the ow of Jacobi equation preserves the family of spaces ;\. Since this equation is Hamiltonian, its ow preserves also the family ; . Consequently, boundary conditions (21:22) can equivalently be written in the form 0 2 0 t 2 0;t :

21.3 Necessary Optimality Conditions Proposition 21.5. Let (~u t) be a corank one extremal pair. Suppose that t is regular or nice singular. Let t 2 (0 t1]. Then: (1) Either for any nonconstant solution t, t 2 0 t], to Jacobi equation

(21:13) or (21:26) that satis es the boundary conditions (21:12) or (21:25) the continuation

(

%t = t t  satis es Jacobi equation on 0 t1],

t 2 0 t] t 2 t t1]

(21.28)

21.4 Regular Case: Transformation of Jacobi Equation

343

(2) Or the control u~ is not locally geometrically optimal on 0 t1]. Proof. Assume that condition (1) does nothold, weprove that condition (2) is then satised. Take any nonzero v 2 Ker Qt jKt and let t, t 2 0 t], be the corresponding nonconstant solution to Jacobi equation with the boundary conditions. Consider the continuation of v by zero:

(

v%(t) = v(t) 0

t 2 0 t] t 2 t  t1]:

and the corresponding continuation by constant %t as in (21:28). Since %t does not satisfy Jacobi equation on 0 t1], then v% 2= Ker(Qt1 jKt1 ). Notice that Qt1 (%v ) = Qt (v) = 0. On the other hand, there exists w 2 Kt1 such that Qt1 (%v  w) 6= 0. Then the quadratic form Qt1 takes values of both signs in the plane span(%v  w). In the singular case, since the extended form Qt is sign-indenite, then the initial form is sign-indenite as well. Summing up, the form Qt1 is sign-indenite on Kt1 . By Theorem 20.3, the control u~(t) is not optimal on 0 t1]. ut Notice that case (1) of Proposition 21.5 imposes a strong restriction on an extremal t. If this case realizes, then the set of conjugate points coincides with the segment t  t1]. Assume that the reference control u~(t) is analytic, then solutions t to Jacobi equation are analytic as well. If t is constant on some segment, then it is constant on the whole domain. Thus in the analytic case alternative (1) of Proposition 21.5 is impossible, and the rst conjugate time t provides a necessary optimality condition: a trajectory cannot be locally geometrically optimal after t . Absence of conjugate points implies nite-dimensional local optimality in the corank one case, see Theorem 20.3. In the following two sections, we prove a much stronger result for the regular case: absence of conjugate points is sucient for strong optimality.

21.4 Regular Case: Transformation of Jacobi Equation Let t be a regular extremal, and assume that the maximized Hamiltonian H() is smooth in a neighborhood of t. The maximality condition of PMP yields the equation @ hu () = 0 @u which can be resolved in the neighborhood of t : @ hu () = 0 , u = u(): @u

344

21 Jacobi Equation

The mapping  7! u() is smooth near t and satises the equality u(t) = u~(t): The maximized Hamiltonian of PMP is expressed in the neighborhood of t as H() = hu()() see Proposition 12.3. Consider the ow on T  M: etH~ 

Zt ; exp ;~hu~( ) d = etH~  P : t

0

By the variations formula in the Hamiltonian form, see (2:27) and (11:22), this ow is Hamiltonian:

;! etH~  Pt = exp

Zt 0

'~  d

(21.29)

with the Hamiltonian function

't() = (H ; hu~(t))(Pt;1 ()):

Notice that

0  etH~  Pt = t  Pt = 0  i.e., 0 is an equilibrium point of the eld '~ t . In other words, 0 is a critical point of the Hamiltonian function: 't ()  0 = 't (0 ) ) d0 't = 0: It is natural to expect that the corresponding Hessian is related to optimality of the extremal t . The following statement relates two Hamiltonian systems: Jacobi equation on  and the maximized Hamiltonian system on T  M. We will use this relation in the proof of sucient optimality conditions in Sect. 21.5. Proposition 21.6. The Hamiltonian bt of Jacobi equation coincides with one half of Hessian of the Hamiltonian 't at 0 : bt = 12 Hess 0 't: Proof. Recall that Hamiltonian of Jacobi equation for the regular case is   bt() = ; 12 (Jt  ) (h00t );1 (Jt  ) : Transform the linear form:

21.4 Regular Case: Transformation of Jacobi Equation

 @ ;!

(Jt  ) =  @ u hut   where hut() = hu (Pt;1())



 @ h  ;   dt @ uu Pt;1    ;  ;  where Pt;1  is dierential of the dieomorphism Pt;1 : T  M ! T  M  @ hu  



345

= ; d0 @@u hut   = ;

0

= ; dt @ u   ; 

= Pt;1 0  2 Tt (T  M): Then the Hamiltonian bt can be rewritten as     bt () = ; 21 dt @@huu   (h00t );1 dt @@huu  : Now we compute Hessian of the Hamiltonian 't () = (hu() ; hu~(t) )(Pt;1()): We have Hess 0 't() = Hess t (hu() ; hu~(t))( ): Further, du + (dhu )ju() ; d hu~(t)  d(hu() ; hu~(t)) = @@huu u()

| {z }  0

!

@ hu dt u: t (hu() ; hu~(t) ) = dt @ u u(t) The dierential dt u can be found by dierentiation of the identity @ hu @ u u()  0 D2

at  = t . Indeed, we have @ 2 hu d u + d @ hu = 0  @u @ u2  thus dt u = ;(h00t );1 dt @@huu :

346

21 Jacobi Equation

Consequently, i.e.,

D2 t (hu() ; hu~(t)) = ;dt @@huu (h00t );1 dt @@huu 

Hess0 't () = Hesst (hu() ; hu~(t) )( ) = 2bt() and the statement follows. ut Since the Hamiltonian 't attains minimum at 0, the quadratic form bt is nonnegative: bt  0: Denote by Ct the space of constant vertical solutions to Jacobi equation at the segment 0 t]: n o Ct =  2 0 j ~b ()  0 2 0 t] : (21.30) Now we can give the following simple characterization of this space: \;  Ct = 0 \ 2 0t] Ker b : Indeed, equilibrium points of a Hamiltonian vector eld are critical points of the Hamiltonian, and critical points of a nonnegative quadratic form are elements of its kernel.

21.5 Sucient Optimality Conditions In this section we prove sucient conditions for optimality in the problem with integral cost: m q_ = fu (q) q 2 M u 2 U = int U q(0) = q0  q(t1 ) = q1

Zt

1

0

R

'(q(t) u(t)) dt ! min

with xed or free terminal time. Notice that now we study an optimal problem, not a geometric one as before. Although, the theory of Jacobi equation can be applied here since Jacobi equation depends only on a Hamiltonian hu () and an extremal pair (~u(t) t ). For the normal Hamiltonian of PMP hu() = h fu (q)i ; '(q u)  2 T  M and a regular extremal pair (~u(t) t) of the optimal control problem, consider Jacobi equation _ = ~bt()  2  = T0 (T  M): In Sect. 21.3 we showed that absence of conjugate points at the interval (0 t1) is necessary for geometric optimality (at least in the corank one analytic case).

21.5 Sucient Optimality Conditions

347

Exercise 21.7. Show that absence of conjugate points on (0 t1) is necessary

also for optimality (in the analytic case) reducing the optimal control problem to a geometric one. Now we can show that absence of conjugate points is also sucient for optimality (in the regular case). A trajectory q(t), t 2 0 t1], is called strongly optimal for an optimal control problem if it realizes a local minimum of the cost functional w.r.t. all trajectories of the system close to q(t) in the uniform topology C(0 t1] M) and having the same endpoints as q(t). If the minimum is strict, then the trajectory q(t) is called strictly strongly optimal . Theorem 21.8. Let t , t 2 0 t1], be a regular normal extremal in the problem with integral cost and xed time, and let the maximized Hamiltonian H() be smooth in a neighborhood of t . If the segment (0 t1] does not contain conjugate points, then the extremal trajectory q(t) = (t ), t 2 0 t1], is strictly strongly optimal. Proof. We apply the theory of elds of extremals (see Sect. 17.1) and embed t

into a family of extremals well projected to M. The maximized Hamiltonian H() = max h ()  2 T  M u2U u

is dened and smooth. Then by Theorem 17.2, it is enough to construct a function a 2 C 1 (M) such that the family of manifolds

Lt = etH~ (L0) T  M t 2 0 t1] L0 = f = dq ag T  M 0 2 L0 

has a good projection to M:  : Lt ! M is a dieomorphism near t  t 2 0 t1]: In other words, we require that the tangent spaces Tt Lt = etH~ (T0 L0) have zero intersection with the vertical subspaces t = Tt (Tq(t) M): etH~ (T0 L0 ) \ t = f0g t 2 0 t1]: This is possible due to the absence of conjugate points (a typical picture for a conjugate point | fold for projection onto M | is shown at Fig. 21.1). Below we show that such a manifold L0 exists by passing to its tangent space L0 | a Lagrangian subspace in  (see denition in Subsect. 11.5.3). For any Lagrangian subspace L0  transversal to 0, one can nd a function a 2 C 1 (M) such that the graph of its dierential L0 = f = dq ag T  M satises the conditions:

348

21 Jacobi Equation etH~ 0

L0 \ Tq0 M

q0

t

t

q(t) M

Fig. 21.1. Conjugate point as a fold (1) 0 2 L0 , (2) T0 L0 = L0: Indeed, in canonical coordinates (p q) on T  M, take a function of the form 0 = (p0  0) a(q) = hp0 qi + 12 qT Sq with a symmetric n  n matrix S. Then L0 = f = (p q) j p = p0 + Sqg T0 L0 = f(dp dq) j dp = Sdqg and it remains to choose the linear mapping S with the graph L0 . Notice that the symmetry of the matrix S corresponds to the Lagrangian property of the subspace L0 . Below we use a similar construction for parametrization of Lagrangian subspaces by quadratic forms. To complete the proof, we have to nd a Lagrangian subspace L0  such that  tH~  e L0 \ t = f0g t 2 0 t1]: By (21:29), the ow of the maximized Hamiltonian decomposes: etH~ = t  Pt;1

;! t = exp

Zt 0

'~  d :

21.5 Sucient Optimality Conditions

349

Notice that the ow Pt;1 on T  M is induced by the ow Pt on M, thus it preserves the family of vertical subspaces: ;P ;1  =  : t  0 t So it remains to show that there exists a Lagrangian subspace L0 which (tL0 ) \ 0 = f0g

t 2 0 t1]:

for

(21.31)

Proposition 21.6 relates the Hamiltonian bt of Jacobi equation to the Hamiltonian 't : 1 Hess ' = b : 0 t t 2 Thus the eld ~bt is the linearization of the eld '~ t at the equilibrium point 0 : the Hamiltonian bt and the Hamiltonian eld ~bt are respectively the main terms in Taylor expansion of 't and '~ t at 0 . Linearization of a ow is the ow of the linearization, thus

 ;! Z t exp

0

'~  d



0

;! = exp

Zt 0

~b d :

Introduce notation for the ow of Jacobi equation:

Zt ;! Bt = exp ~b d  0

then and equality (21:31) reads

t0 = Bt 

(Bt L0 ) \ 0 = f0g t 2 0 t1]: (21.32) It remains to prove existence of a Lagrangian subspace L0 that satises this equality. Recall that the segment (0 t1] does not contain conjugate points: (Bt 0) \ 0 = Ct t 2 (0 t1] where Ct is the space of constant vertical solutions to Jacobi equation on 0 t], see (21:30). In order to make the main ideas of the proof more clear, we consider rst the simple case where Ct = f0g t 2 (0 t1] (21.33) i.e.,

350

21 Jacobi Equation

(Bt 0) \ 0 = f0g t 2 (0 t1]: Fix any " 2 (0 t1). By continuity of the ow Bt , there exists a neighborhood of the vertical subspace 0 such that for any Lagrangian subspace L0 from this neighborhood (Bt L0 ) \ 0 = f0g t 2 " t1]: In order to complete the proof, it remains to nd such a Lagrangian subspace L0 satisfying the condition (Bt L0 ) \ 0 = f0g t 2 0 "]: We introduce a parametrization of the set of Lagrangian subspaces L0  suciently close to 0 . Take any Lagrangian subspace H  which is horizontal, i.e., transversal to the vertical subspace 0. Then the space  splits:  = 0 ! H: Introduce Darboux coordinates (p q) on  such that 0 = f(p 0)g H = f(0 q)g: Such coordinates can be chosen in many ways. Indeed, the symplectic form  denes a nondegenerate pairing of the mutually transversal Lagrangian subspaces 0 and H: H = 0 hf ei = (e f) e 2 0 f 2 H: Taking any basis e1  : : :  en in 0 and the corresponding basis f1  : : :  fn in H dual w.r.t. this pairing, we obtain a Darboux basis in . In Darboux coordinates the symplectic form reads ((p1  q1) (p2 q2)) = hp1  q2i ; hp2 q1i: Any n-dimensional subspace L  transversal to H is a graph of a linear mapping S : 0 ! H i.e., L = f(p Sp) j p 2 0g: A subspace L is Lagrangian i the corresponding mapping S has a symmetric matrix in a symplectic basis (exercise): S = S :

21.5 Sucient Optimality Conditions

351

Introduce the quadratic form on 0 with the matrix S: S(p p) = hp Spi: So the set of Lagrangian subspaces L  transversal to the horizontal space H is parametrized by quadratic forms S on 0. We call such parametrization of Lagrangian subspaces L , L \ H = f0g, a (0 H)parametrization. Consider the family of quadratic forms St that parametrize a family of Lagrangian subspaces of the form Lt = Bt L0  i.e., Lt = f(p Stp) j p 2 0g:

Lemma 21.9.

S_t (p p) = 2bt(p St p):

Proof. Take any trajectory (p q) = (pt  qt) of the Hamiltonian eld ~bt. We

have

q = St p

thus

q_ = S_ t p + St p_

i.e.,





~bt (p q) = p_ S_ t p + St p_ : Since the Hamiltonian bt is quadratic, we have





 (p q) ~bt(p q) = 2bt (p q): But the left-hand side is easily computed:





 (p q) ~bt(p q) =  ((p q) (p_ q)) _



 D

E

=  (p Stp) (p_ S_ tp + St p)_ = p S_ tp + St p_ ; hp_ Stpi D E = p S_ t p by symmetry of St . Since the Hamiltonian 't attains minimum at 0 , then bt  0, thus S_t  0:

ut

352

21 Jacobi Equation

The partial order on the space of quadratic forms induced by positive forms explains how one should choose the initial subspace L0 . Taking any Lagrangian subspace L0  with the corresponding quadratic form S0 > 0 suciently close to the zero form, we obtain St > 0

t 2 0 "]:

That is,

Lt \ 0 = f0g on 0 "], thus on the whole segment 0 t1]. We proved equality (21:32) in the simple case (21:33). Now we consider the general case. The intersection (Bt 0) \ 0 = Ct is nonempty now, but we can get rid of it by passing to Jacobi equation on the quotient Ct\ =Ct. The family of constant vertical solutions Ct is monotone nonincreasing: Ct0  Ct00 for t0 < t00 :

We have C0 = 0 and set, by denition, Ct1+0 = f0g. The family Ct is continuous from the left, denote its discontinuity points: 0  s1 < s2 < < sk  t1 (notice that in the simple case (21:33), we have k = 1, s1 = 0). The family Ct is constant on the segments (si  si+1 ]. Construct subspaces Ei 0 , i = 1 : : :  k, such that Ct = Ei+1 ! Ei+2 ! ! Ek 

t 2 (si  si+1 ]:

Notice that for t = 0, we obtain a splitting of the vertical subspace: 0 = C0 = E1 ! ! Ek : For any horizontal Lagrangian subspace H , one can construct the corresponding splitting of H: H = F1 ! ! Fk 

(Ei  Fj ) = 0 i 6= j:

(21.34)

Fix any initial horizontal subspace H0 , H0 \ 0 = f0g. The following statement completes the proof of Theorem 21.8 in the general case. Lemma 21.10. For any i = 1 : : :  k, there exist a number "i > 0 and a Lagrangian subspace Hi  , Hi \ 0 = f0g, such that any Lagrangian subspace L0  , L0 \ H0 = f0g, with a (0  H0)-parametrization S0 (p p) = "hp pi, 0 < " < "i , satis es the conditions:

21.5 Sucient Optimality Conditions

353

(1) Lt \ 0 = f0g t 2 0 si], (2) Lt \ Hi = f0g, t 2 0 si], and the Lagrangian subspace Lt has a (0  Hi)parametrization St > 0. Proof. We prove this lemma by induction on i. Let i = 1. For s1 = 0, the statement is trivial, so we assume that s1 > 0. Take any "1 > 0 and any Lagrangian subspace L0  with a quadratic form "hp pi, 0 < " < "1 , in the (0 H0)-parametrization. Notice that Ct = 0, i.e., Bt j0 = Id, for t 2 (0 s1]. We have Lt \ 0 = Bt L0 \ Bt 0 = Bt (L0 \ 0) = f0g

t 2 0 s1]:

By continuity of the ow Bt , there exists a horizontal Lagrangian subspace H1 with a (0 H0)-parametrization ; hp pi,  > 0, such that Lt \ H1 = f0g, t 2 0 s1]. One can easily check that the subspace L0 in (0 H1)-parametrization is given by the quadratic form S0 (p p) = "0 hp pi > 0, "0 = "=(1 + "=) < ". We already proved that S_ t  0, thus St > 0

t 2 0 s1]

in the (0  H1)-parametrization. The induction basis (i = 1) is proved. Now we prove the induction step. Fix i  1, assume that the statement of Lemma 21.10 is proved for i, and prove it for i + 1. Let t 2 (si  si+1 ], then Ct = Ei+1 ! ! Ek . Introduce a splitting of the horizontal subspace Hi as in (21:34): Hi = F1 ! ! Fk : Denote

E10 = E1 ! ! Ei  F10 = F1 ! ! Fi  L10 = L0 \ (E10 ! F10 )

E20 = Ct = Ei+1 ! ! Ek  F20 = Fi+1 ! ! Fk  L20 = L0 \ (E20 ! F20 ):

Since Bt E20 = E20 , then the skew-orthogonal complement (E20 )\ = E10 ! 0 E2 ! F10 is also invariant for the ow of Jacobi equation: Bt (E20 )\ = (E20 )\ . In order to prove that Lt \ 0 = f0g, compute this intersection. We have 0 (E20 )\ , thus Bt L0 \ 0 = Bt L0 \ Bt (E20 )\ \ 0 = Bt (L0 \ (E20 )\ ) \ 0 = Bt L10 \ 0:

(21.35)

So we have to prove that Bt L10 \ 0 = f0g, t 2 (si  si+1 ]. Since the subspaces E20 and (E20 )\ are invariant w.r.t. the ow Bt , the quotient ow is well-dened:

354

21 Jacobi Equation

e e = (E20 )\=E20 : Bet : e !  In the quotient, the ow Bet has no constant vertical solutions: Bet e 0 \ e 0 = f0g t 2 (si  si+1 ] e 0 = 0=E20 : By the argument already used in the proof of the simple case (21:33), it follows that Bet Le10 \ e 0 = f0g t 2 (si  si+1 ] Le10 = L10 =E20  for L0 suciently close to 0, i.e., for " suciently small. That is, Bt L10 \ 0 E20  t 2 (si  si+1 ]: Now it easily follows that this intersection is empty: Bt L10 \ 0 Bt L10 \ E20 = Bt L10 \ BtE20 = Bt (L10 \ E20 ) = f0g t 2 (si  si+1 ]: In view of chain (21:35), Lt \ 0 = f0g t 2 (si  si+1] that is, we proved condition (1) in the statement of Lemma 21.10 for i + 1. Now we pass to condition (2). In the same way as in the proof of the induction basis, it follows that there exists a horizontal Lagrangian subspace Hi+1  such that the curve of Lagrangian subspaces Lt , t 2 0 si+1], is transversal to Hi+1 . In the (0  Hi+1)-parametrization, the initial subspace L0 is given by a positive quadratic form S0 (p p) = "0 hp pi, 0 < "0 < ". Since S_ t  0, then St > 0 t 2 0 si+1]: Condition (2) is proved for i + 1. The induction step is proved, and the statement of this lemma follows. ut By this lemma, Lt \ 0 = f0g t 2 0 t1] for all initial subspaces L0 given by quadratic forms S0 = "hp pi, 0 < " < "k , for some "k > 0, in a (0 H0)-parametrization. This means that we constructed a family of extremals containing t and having a good projection to M. By Theorem 17.2, the extremal t, t 2 0 t1], is strongly optimal. Theorem 21.8 is proved. ut For the problem with integral cost and free terminal time t1, a similar argument and Theorem 17.6 yield the following sucient optimalitycondition. Theorem 21.11. Let t , t 2 0 t1], be a regular normal extremal in the problem with integral cost and free time, and let H() be smooth in a neighborhood of t . If there are no conjugate points at the segment (0 t1], then the extremal trajectory q(t) = (t ), t 2 0 t1], is strictly strongly optimal.

22 Reduction

In this chapter we consider a method for reducing a control-ane system to a nonlinear system on a manifold of a less dimension.

22.1 Reduction Consider a control-ane system q_ = f(q) +

m X i=1

R

ui 2  q 2 M

ui gi (q)

(22.1)

with pairwise commuting vector elds near controls: gi gj ]  0 i j = 1 : : :  m: The ow of the system can be decomposed by the variations formula:



!

m ;! Z t f + X ;! Z t ePmi=1 wi ( ) ad gi f d  ePmi=1 wi (t)gi  exp ui ( )gi d = exp 0

wi (t) =

Zt 0

i=1

0

(22.2)

ui( ) d :

P

Here we treat mi=1 ui ( )gi as a nonperturbed ow and take into account that the elds gi mutually commute. Introduce the partial system corresponding to the second term in composition (22:2): P q_ = e mi=1 wi ad gi f(q) wi 2  q 2 M (22.3) where wi are new controls. Attainable sets A1 (t) of the initial system (22:1) and A2 (t) of the partial system (22:3) for time t from a point q0 2 M are closely related one to another:

R

356

22 Reduction

n Pm

i=1 wi gi

A1 (t) A2 (t)  e

j wi 2

Ro

cl(A1 (t)):

(22.4)

Indeed, the rst inclusion follows directly from decomposition (22:2). To prove the second inclusion in (22:4), notice that the mapping

;! Z t ePmi=1 wi ( ) ad gi f d w( ) 7! q0  exp 0

is continuous in L1 topology, this follows from the asymptotic expansion of the chronological exponential. Thus the mapping

;! Z t ePmi=1 wi ( ) ad gi f d  ePmi=1 vigi (w( ) v) 7! q0  exp

R0 

0

is continuous in topology of L1  m. Finally, the mapping

1 u( ) 7! (w( ) v) = @ u( ) d  u( ) d A

R

Z

Zt

0

0

has a dense image in L1  inclusion in (22:4). The partial system (22:3) is invariant w.r.t. the elds gi :

m. Then decomposition (22:2) implies the second

 Pm e

i=1 vi gi

 Pm i e

=1

wi ad gi f

Pmi

=e

=1

(wi ;vi) ad gi f:

(22.5)

Thus chain (22:4) and equality (22:5) mean that the initial system (22:1) can be considered as a composition of the partial system (22:3) with the ow of the elds gi: any time t attainable set of the initial system is (up to closure) the time t attainable set of the partial system plus a jump along gi, moreover, the jump along gi is possible at any instant. Let (u(t) t ) be an extremal pair of the initial control-ane system. The extremal t is necessarily totally singular, moreover the maximality condition of PMP is equivalent to the identity ht  gi i  0: It is easy to see that P  t = e mi=1 wi (t)gi t is an extremal of system (22:3) corresponding to the control w(t) = moreover,

Zt 0

u( ) d 

ht  gii  0:

(22.6)

22.1 Reduction

357

(Here, we use the term extremal as a synonym of a critical point of the endpoint mapping, i.e., we require that the extremal control be critical, not necessarily minimizing, for the control-dependent Hamiltonian of PMP.) Conversely, if t is an extremal of (22:3) with a Lipschitzian control w(t), and if identity (22:6) holds, then

 P  t = e; mi=1 wi (t)gi t

is an extremal of the initial system (22:1) with the control u(t) = w(t): _ Moreover, the strong generalized Legendre condition for an extremal t of the initial system coincides with the strong Legendre condition for the corresponding extremal t of the partial system. In other words, the passage from system (22:1) to system (22:3) transforms nice singular extremals t into regular extremals t . Exercise 22.1. Check that the extremals t and t have the same conjugate times. Since system (22:3) is invariant w.r.t. the elds gi , this system can be considered on the quotient manifold of M modulo action of the elds gi if the quotient manifold is well-dened. Consider the following equivalence relation on M: q0  q , q0 2 Oq (g1 : : :  gm ): Suppose that all orbits Oq (g1 : : :  gm ) have the same dimension and, moreover, the following nonrecurrence condition is satised: for each point q 2 M there exist a neighborhood Oq 3 q and a manifold Nq M, q 2 Nq , transversal to Oq (g1  : : :  gm ), such that any orbit Oq0 (g1  : : :  gm ), q0 2 Oq , intersects Nq at a unique point. In particular, these conditions hold if M = n and gi are constant vector elds, or if m = 1 and the eld g1 is nonsingular and nonrecurrent. If these conditions are satised, then the space of orbits M= is a smooth manifold. Then system (22:3) is well-dened on the quotient manifold M=:

R

Pmi

q_ = e

=1

wi ad gi f(q)

R

wi 2  q 2 M=:

(22.7)

The passage from the initial system (22:1) ane in controls to the reduced system (22:7) nonlinear in controls decreases dimension of the state space and transforms singular extremals into regular ones. Let  : M ! M= be the projection. For the attainable set A3 (t) of the reduced system (22:7) from the point (q0), inclusions (22:4) take the form

A1 (t) ;1 (A3 (t)) cl(A1 (t)):

(22.8)

358

22 Reduction

It follows from the analysis of extremals above that q(t) is an extremal curve of the initial system (22:1) i its projection (q(t)) is an extremal curve of the reduced system (22:7). The rst inclusion in (22:8) means that if (q( )), 2 0 t], is geometrically optimal, then q( ), 2 0 t], is also geometrically optimal. One can also dene a procedure of inverse reduction. Given a control system

R

q 2 M w 2 n

q_ = f(q w)

(22.9)

we restrict it to Lipschitzian controls w( ) and add an integrator:

(

R R

q_ = f(q w) w_ = u

(q w) 2 M  n u 2 n:

(22.10)

Exercise 22.2. Prove that system (22:9) is the reduction of system (22:10).

22.2 Rigid Body Control Consider the time-optimal problem for the system that describes rotations of a rigid body, see Sect. 19.4:

R

q 2 SO(3) u 2 

q_ = q(a + ub) where

(22.11)

a b 2 so(3) ha bi = 0 jbj = 1 a 6= 0: Notice that in Sect. 19.4 we assumed jaj = 1, not jbj = 1 as now, but one case is obtained from another by dividing the right-hand side of the system by a constant. We construct the reduced system for system (22:11). The state space SO(3) factorizes modulo orbits qesb , s 2 , of the eld qb. The corresponding equivalence relation is: q  qesb 

R

R

s2 

and the structure of the factor space is described in the following statement.

Proposition 22.3. the canonical projection is

q 7! q

SO(3)= ' S 2  q 2 SO(3)  2 S 2 :

(22.12)

Here  2 S 2

R

22.3 Angular Velocity Control

359

3 is the unit vector corresponding to the matrix b 2 so(3):

0b 1 1  = @ b2 A  b3

0 0 ;b b 1 3 2 b = @ b3 0 ;b1 A : ;b2 b1 0

Proof. The group SO(3) acts transitively on the sphere S 2 . The subgroup of

SO(3) leaving a point  2 S 2 xed consists of rotations around the line , i.e., it is eRb = fesb j s 2 g: Thus the quotient SO(3)=eRb = SO(3)= is dieomorphic to S 2 , projection SO(3) ! S 2 is given by (22:12), and level sets of this mapping coincide with orbits of the eld qb. ut The partial system (22:3) in this example takes the form q_ = qew ad ba q 2 SO(3) w 2  and the reduced system (22:7) is d w ad b q 2 S 2 : (22.13) d t (q) = qe a The right-hand side of this symmetric control system denes a circle of radius jaj in the tangent plane (q)? = Tq S 2 . In other words, system (22:13) determines a Riemannian metric on S 2 . Since the vector elds in the righthand side of system (22:13) are constant by absolute value, then the timeoptimal problem is equivalent to the Riemannian problem (time minimization is equivalent to length minimization if velocity is constant by absolute value). Extremal curves (geodesics) of a Riemannian metric on S 2 are arcs of great circles, they are optimal up to semicircles. And the antipodal point is conjugate to the initial point. Conjugate points for the initial and reduced systems coincide, thus for both systems extremal curves are optimal up to the antipodal point.

R

R

22.3 Angular Velocity Control Consider the system that describes angular velocity control of a rotating rigid body, see (6:19): _ =    + ul u 2   2 3: (22.14) Here  is the vector of angular velocity of the rigid body in a coordinate system connected with the body, and l 2 3 is a unit vector in general position along which the torque is applied. Notice that in Sect. 6.4 we allowed only torques u = 1, while now the torque is unbounded. In Sect. 8.4 we proved

R

R R

360

22 Reduction

that the system with bounded control is completely controllable (even in the six-dimensional space). Now we show that with unbounded control we have complete controllability in 3 for an arbitrarily small time. We apply the reduction procedure to the initial system (22:14). The partial system reads now

R

_ = ew ad l (  ) = ( + wl)  ( + wl)

R R R R

R R

w2  2

3:

The quotient of 3 modulo orbits of the constant eld l can be realized as the plane 2 passing through the origin and orthogonal to l. Then projection 3 ! 2 is the orthogonal projection along l, and the reduced system reads x_ = (x + wl)  (x + wl) ; hx  (x + wl) li l

R

x2

R 2R 2

w

: (22.15)

Introduce Cartesian coordinates in 3 corresponding to the orthonormal frame with basis vectors collinear to the vectors l, l  l, l  (l  l). In these coordinates x = (x1  x2) and the reduced system (22:15) takes the form: x_ 1 = b13x22 + ((b11 ; b33)x2 ; b23x1)w ; b13w2 x_ 2 = ;b13x1x2 + ((b22 ; b11)x1 + b23x2)w

(22.16) (22.17)

where b = (bij ) is the matrix of the operator  in the orthonormal frame. Direct computation shows that b13 < 0 and b22 ; b11 6= 0. In polar coordinates (r ') in the plane (x1  x2), the reduced system (22:16), (22:17) reads r_ = rF (cos ' sin ')w ; b13 cos ' w2  '_ = ;b13r sin ' ; (1=r) sin ' w2 + G(cos ' sin')w where F and G are homogeneous polynomials of degree 2 with G(1 0) = b22 ; b11. Choosing appropriate controls, one can construct trajectories of the system in 2 of the following two types: (1) \spirals", i.e., trajectories starting and terminating at the positive semiaxes x1, not passing through the origin (r 6= 0), and rotating counterclockwise ('_ > 0), (2) \horizontal" trajectories almost parallel to the axis x1 (x_ 1 # x_ 2 ). Moreover, we can move fast along these trajectories. Indeed, system (22:16), (22:17) has an obvious self-similarity | it is invariant with respect to the changes of variables x1 7! x1, x2 7! x2, w 7! w, t 7! ;1t ( > 0). Consequently, one can nd \spirals" arbitrarily far from the origin and with an arbitrarily small time of complete revolution. Further, it is easy to see from equations (22:16), (22:17) that taking large in absolute value controls w

R

22.3 Angular Velocity Control

361

one obtains arbitrarily fast motions along the \horizontal" trajectories in the positive direction of the axis x1 . Combining motions of types (1) and (2), we can steer any point x 2 2 to any point x^ 2 2 for any time " > 0, see Fig. 22.1. Details of this argument are left to the reader as an exercise, see also 41].

R

R

x2

x

x1

x^

Fig. 22.1. Complete controllability of system (22:15) That is, time t attainable sets A3x (t) of the reduced system (22:15) from a point x satisfy the property:

A3x (") =

R R

R 2R

8 x 2 2 " > 0: By virtue of chain (22:8), attainable sets A1 (t) of the initial system (22:14) satisfy the equality

2

cl(A1 (")) =

3

8

3

" > 0:

Since the vector l is in general position, the 3-dimensional system (22:14) has a full rank (see Proposition 6.3), thus it is completely controllable for arbitrarily small time: A1 (") = 3 8  2 3 " > 0:

R

R

23 Curvature

23.1 Curvature of 2-Dimensional Systems Consider a control system of the form q_ = fu (q) q 2 M u 2 U (23.1) where dimM = 2 U = or S 1 : We suppose that the right-hand side fu (q) is smooth in (u q). A well-known example of such a system is given by a two-dimensional Riemannian problem: locally, such a problem determines a control system q_ = cos u f1 (q) + sin u f2(q) q 2 M u 2 S 1  where f1  f2 is a local orthonormal frame of the Riemannian structure. For control systems (23:1), we obtain a feedback-invariant form of Jacobi equation and construct the main feedback invariant | curvature (in the Riemannian case this invariant coincides with Gaussian curvature). We prove comparison theorem for conjugate points similar to those in Riemannian geometry. We assume that the curve of admissible velocities of control system (23:1) satises the following regularity conditions: fu (q) ^ @ [email protected](q) 6= 0 @ fu (q) ^ @ 2 fu (q) 6= 0 q 2 M u 2 U: (23.2) @u @ u2

R

Condition (23:2) means that the curve ffu (q) j u 2 U g Tq M is strongly convex, it implies strong Legendre condition for extremals of system (23:1). Introduce the following control-dependent Hamiltonian linear on bers of the cotangent bundle: hu() = h fu (q)i

364

23 Curvature

and the maximized Hamiltonian H() = max h (): u2U u

(23.3)

We suppose that H() is dened in a domain in T  M under consideration. Moreover, we assume that for any  in this domain maximum in (23:3) is attained for a unique u 2 U, this means that any line of support touches the curve of admissible velocities at a unique point. Then the convexity condition (23:2) implies that H() is smooth in this domain and strongly convex on bers of T  M. Moreover, H is homogeneous of order one on bers, thus we restrict to the level surface H = H ;1 (1) T  M: Denote the intersection with a ber Hq = H \ Tq M:

23.1.1 Moving Frame We construct a feedback-invariant moving frame on the 3-dimensional manifold H in order to write Jacobi equation in this frame. Notice that the maximized Hamiltonian H is feedback-invariant since it depends on the whole admissible velocity curve fU (q), not on its parametrization by u. Thus the level surface H and the ber Hq are also feedback-invariant. We start from a vertical eld tangent to the curve Hq . Introduce polar coordinates in a ber: p = (r cos ' r sin ') 2 Tq M then Hq is parametrized by angle ': Hq = fp = p(')g: Since the curve Hq does not pass through the origin: p(') 6= 0, it follows that p(') ^ dd 'p (') 6= 0: (23.4) Decompose the second derivative in the frame p, dd 'p : d2 p (') = a (')p(') + a (') d p ('): 1 2 d '2 d' The curve Hq is strongly convex, thus a1 (') < 0:

23.1 Curvature of 2-Dimensional Systems

365

A change of parameter = (') gives   d2 p = a (') d ' 2 p( ) + ~a ( ) d p ( ) 2 d

d 2 1 d

thus there exists a unique (up to translations and orientation) parameter

on the curve Hq such that d2 p = ;p( ) + b( ) d p ( ): d 2 d

We x such a parameter and dene the corresponding vertical vector eld on H: v = @@ : In invariant terms, v is a unique (up to multiplication by 1) vertical eld on H such that L2v s = ;s + b Lv s (23.5) where s = p dq is the tautological form on T  M restricted to H. We dene the moving frame on H as follows: ~ V3 = H: ~ V1 = v V2 = v H] Notice that these vector elds are linearly independent since v is vertical and the other two elds have linearly independent horizontal parts:  H~ = f d u 6= 0: ~ = @ fu d u   v H] @u d

d

Here we denote by u( ) the maximizing control on Hq : hp( ) fu() i  hp( ) fu i u 2 U: Dierentiating the identity * + @ f u p( ) @ u 0 u() w.r.t. , we obtain dd u 6= 0. In order to write Jacobi equation along an extremal t, we require Lie brackets of the Hamiltonian eld H~ with the vector elds of the frame: ~ V1] = ;V2  H ~ V2] = ? H ~ V3] = 0: H The missing second bracket is given by the following proposition.

366

23 Curvature

Theorem 23.1.

~ H~  v]] = ;"v: H (23.6) The function " = "(),  2 H, is called the curvature of the twodimensional control system (23:1). The Hamiltonian eld H~ is feedbackinvariant, and the eld v is feedback-invariant up to multiplication by 1. Thus the curvature " is a feedback invariant of system (23:1). Now we prove Theorem 23.1. Proof. The parameter provides an identication H = f g  M (23.7) thus tangent spaces to H decompose into direct sum of the horizontal and vertical subspaces. By duality, any dierential form on H has a horizontal and vertical part. Notice that trivialization (23:7) is not feedback invariant since the choice of the section = 0 is arbitrary, thus the form d and the property of a subspace to be horizontal are not feedback-invariant. For brevity, we denote in this proof s = sjH  a horizontal form on H. Denote the Lie derivative: Lv = L @@ = 0 and consider the following coframe on H: d  s s0 : (23.8) It is easy to see that these forms are linearly independent: d is vertical, while the horizontal forms s, s0 are linearly independent by (23:4). Now we construct a frame on H dual to coframe (23:8). Decompose H~ into the horizontal and vertical parts: H~ = |{z} Y + @@  = (  q): (23.9) horizontal |{z} vertical

We prove that the elds

  @  Y Y 0 = @  Y @

@

give a frame dual to coframe (23:8). We have to show only that the pair of horizontal elds Y , Y 0 is dual to the pair of horizontal forms s, s0 . First, hs  Y i = hs  H~ i = h fui = H() = 1:

23.1 Curvature of 2-Dimensional Systems

Further,

367

 @f   @f  du 0 0 ~ hs  Y i = hs  H i =  @ u =  @ uu d = 0: | {z } =0

Consequently, i.e.,

0 = hs Y i0 = hs0  Y i + hs Y 0 i

hs0  Y i = 0:

Finally,

0 = hs0  Y i0 = hs00  Y i + hs0  Y 0i: Equality (23:5) can be written as s00 = ;s + bs0 , thus hs0  Y 0 i = ;hs00 Y i = hs ; bs0  Y i = 1: So we proved that the frame is dual to the coframe

@  Y Y 0 2 Vec H @

d  s s0 2 1(H): ~ H~  v]] using We complete the proof of this theorem computing the bracket H these frames. First consider the standard symplectic form: jH = d (sjH ) = ds = d ^ s0 + dq s where dq s is the dierential of the form s w.r.t. horizontal coordinates. The horizontal 2-form dq s decomposes: dq s = c s ^ s0  c = c(  q) thus Since then

jH = d ^ s0 + cs ^ s0 : iH~ jH = 0 ~ ) = jH (Y + @  ) jH (H @

= s0 ; hs0  Y i d + c hs Y is0 ; c hs0  Y is = s0 + cs0 = 0

368

23 Curvature

i.e., = ;c, thus

H~ = Y ; c @@ : Now we can compute the required Lie bracket.   H~ 0 = @@  H~ = Y 0 ; c0 @@  consequently,    @  h i  ~ ;H~ 0 = Y ; c @  ;Y 0 + c0 @ ~ H ~ = H H @

@

@

~ 0 ~0  @ 0 00 0 0 = Hc ; H c @ + Y |  Y ] + {zcY ; c Y } :

|

{z

vertical part

}

horizontal part

In order to complete the proof, we have to show that the horizontal part of ~ H ~ @@ ]] vanishes. the bracket H The equality s00 = ;s + bs0 implies, by duality of the frames Y , Y 0 and s, s0 , that Y 00 = ;Y ; bY 0 : Further, ds = d ^ s0 + cs ^ s0  d(s0) = (ds)0 = d ^ s00 + c0 s ^ s0 + cs ^ s00 = ;d ^ s ; b d ^ s0 + (c0 + cb)s ^ s0  and we can compute the bracket Y 0  Y ] using duality of the frames and Proposition 18.3: Y 0 Y ] = cY + (c0 + cb)Y 0: ~ H ~ v]] is Summing up, the horizontal part of the eld H Y 0  Y ] + cY 00 ; c0 Y 0 = cY + (c0 + cb)Y 0 ; cY ; cbY 0 ; c0 Y 0 = 0: We proved that   @  @ ~ H ~ H @ = ;" @  where the curvature has the form ~ 0 + H~ 0 c: " = ;Hc

ut

Remark 23.2. Recall that the vertical vector eld v that satises (23:5) is

unique up to a factor 1. On the other hand, the vertical eld v that satises (23:6) is unique, up to a factor constant along trajectories of H~ (so this factor does not aect "). Consequently, any vertical vector eld v for which an equality of the form (23:6) holds can be used for computation of curvature ".

23.1 Curvature of 2-Dimensional Systems

369

So now we know all brackets of the Hamiltonian vector eld X = H~ with the vector elds of the frame V1 , V2, V3 : ~ V1] = ;V2  H (23.10) ~ V2] = "V1  H (23.11) ~ H V3] = 0: (23.12)

23.1.2 Jacobi Equation in Moving Frame We apply the moving frame constructed to derive an ODE on conjugate time of our two-dimensional system | Jacobi equation in the moving frame. As in Chap. 21, consider Jacobi equation along a regular extremal t , t 2 0 t1], of the two-dimensional system (23:1): _ = ~bt()  2  = T0 (T  M) and its ow

;! Bt = exp

Zt 0

~b d :

Recall that 0 = T0 (Tq0 M) is the vertical subspace in  and Ct 0 is the subspace of constant vertical solutions to Jacobi equation at 0 t], see (21:30). The intersection Bt 0 \ 0 always contains the subspace Ct. An instant t 2 (0 t1] is a conjugate time for the extremal t i that intersection is greater than Ct : Bt 0 \ 0 6= Ct: In order to complete the frame V1 , V2 , V3 to a basis in T0 (T  M), consider a vector eld transversal to H | the vertical Euler eld E 2 Vec(T  M) with the ow   etE = et   2 T  M t 2 : In coordinates (p q) on T  M, this eld reads E = p @@p : The vector elds V1  V2 V3 E form a basis in T (T  M),  2 H. The elds V1 = @@ and E are vertical: 0 = span(V1 (0 ) E(0)): To compute the constant vertical subspace Ct, evaluate the action of the ow Bt on these elds. In the proof of Theorem 21.8, we decomposed the ow of Jacobi equation: Bt () = (Pt ) etH~ (): Thus

R

370

23 Curvature

Bt E(0 ) = (Pt ) etH~ E(0 ): The Hamiltonian H is homogeneous of order one on bers, consequently the ow of H~ is homogeneous as well:





(k)  etH~ = k   etH~ 

k > 0

and the elds H~ and E commute. That is, the Hamiltonian vector eld H~ preserves the vertical Euler eld E. Further, the ow Pt is linear on bers, thus it also preserves the eld E. Summing up, the vector E(0 ) is invariant under the action of the ow of Jacobi equation, i.e.,

R

E(0) Ct:

It is easy to see that this inclusion is in fact an equality. Indeed, in view of bracket (23:10), etH~ V1 (0 ) = t  e;t ad H~ V1 = t  (V1 + tV2 + o(t)) 2= Tt (Tq(t) M) thus

Bt V1 (0 ) 2= 0 for small t > 0. This means that

R

t 2 (0 t1]:

Ct = E(0)

R 2R 2R

Thus an instant t is a conjugate time i Bt 0 \ 0 6= E(0) i.e., or, equivalently,

etH~ V1(0 ) 0  et ad H~ V1

V1(t ) (0  V1 ):

(23.13)

Now we describe the action of the ow of a vector eld on a moving frame. Lemma 23.3. Let N be a smooth manifold, dimN = m, and let vector elds V1 : : :  Vm 2 Vec N form a moving frame on N . Take a vector eld X 2 Vec N . Let the operator ad X have a matrix A = (aij ) in this frame: (ad X) Vj =

m X i=1

aij Vi 

aij 2 C 1 (N):

Then the matrix ;(t) = (ij (t)) of the operator et ad H~ in the moving frame:

23.1 Curvature of 2-Dimensional Systems

et ad X Vj =

m X i=1

ij (t) 2 C 1 (N)

ij (t)Vi 

371

(23.14)

is the solution to the following Cauchy problem:

_ = ;(t)A(t) ;(t) ;(0) = Id

(23.15) (23.16)

where A(t) = (etX aij ). Proof. Initial condition (23:16) is obvious. In order to derive the matrix equa-

tion (23:15), we dierentiate identity (23:14) w.r.t. t: m X i=1

_ ij (t)Vi = et ad X X Vj ] = et ad X m ; X  = etX akj ik Vi 

X m

k=1

! X m ;  akj Vk = etX akj et ad X Vk k=1

ki=1

and the ODE follows. ut In view of inclusion (23:13), an instant t is a conjugate time i the coecients in the decomposition 0  et ad H~ Vj =

3 X

i=1

ij (t)(0  Vi )

satisfy the equalities:

21 (t) = 31(t) = 0: By the previous lemma, the matrix ;(t) = (ij (t)) is the solution to Cauchy problem (23:15), (23:16) with the matrix 0 0 " 01 t A(t) = @ ;1 0 0 A  "t = "(t ) 0 00 see Lie bracket relations (23:10){(23:12). Summing up, an instant t 2 (0 t1] is a conjugate time i the solutions to the Cauchy problems

(

and

21 (0) = 0 22 (0) = 1

_ 31 = ;32  _ 32 = "t 31

31 (0) = 0 32 (0) = 0

(

_ 21 = ;22  _ 22 = "t 21

372

23 Curvature

satisfy the equalities

21 (t) = 31(t) = 0: But Cauchy problem for 31, 32 has only trivial solution. Thus for a conjugate time t, we obtain the linear nonautonomous system for (x1 x2) = (21 22 ):

(

x_ 1 = ;x2  x_ 2 = "t x1

x1(0) = x1(t) = 0:

(23.17)

We call system (23:17), or, equivalently, the second order ODE x# + "t x = 0 x(0) = x(t) = 0 (23.18) Jacobi equation for system (23:1) in the moving frame. We proved the following statement. Theorem 23.4. An instant t 2 (0 t1] is a conjugate time for the two-dimensional system (23:1) i there exists a nontrivial solution to boundary problem (23:18). Sturm's comparison theorem for second order ODEs (see e.g. 136]) implies the following comparison theorem for conjugate points. Theorem 23.5. (1) If " < C 2 for some C > 0 along an extremal t, then there are no conjugate points at the time segment 0 C ]. In particular, if "  0 along t , then there are no conjugate points. (2) If "  C 2 along t , then there is a conjugate point at the segment  0 C ]. A typical behavior of extremal trajectories of the two-dimensional system (23:1) in the cases of negative and positive curvature is shown at Figs. 23.1 and 23.2 respectively.

q q

Fig. 23.1.  < 0

Fig. 23.2.  > 0

23.2 Curvature of 3-Dimensional Control-Ane Systems

373

Example 23.6. Consider the control system corresponding to a Riemannian

problem on a 2-dimensional manifold M:

q_ = cos u f1 (q) + sin u f2(q)

q 2 M u 2 S 1 

where f1  f2 is an orthonormal frame of the Riemannian structure h  i:

hfi  fj i = ij 

i j = 1 2:

In this case, " is the Gaussian curvature of the Riemannian manifold M, and it is evaluated as follows: " = ;c21 ; c22 + f1 c2 ; f2 c1 where ci are structure constants of the frame f1 , f2 : f1  f2] = c1 f1 +c2 f2 . We prove this formula for " in Chap. 24. For the Riemannian problem, the curvature " = "(q) depends only on the base point q 2 M, not on the coordinate in the ber. Generally, this is not the case: the curvature is a function of (q ) 2 H. Optimality conditions in terms of conjugate points obtained in Chap. 21 can easily be applied to the two-dimensional system (23:1) under consideration. Assume rst that tc 2 (0 t1) is a conjugate time for an extremal t , t 2 0 t1], of system (23:1). We verify hypotheses of Proposition 21.5. Condition (23:2) implies that the extremal is regular. The corresponding control u~ has corank one since the Lagrange multiplier t is uniquely determined by PMP (up to a scalar factor). Further, Jacobi equation cannot have solutions of form (21:28): if this were the case, Jacobi equation in the moving frame x# + "t x = 0 would have a nontrivial solution with the terminal conditions x(tc) = x(t _ c ) = 0, which is impossible. Summing up, the extremal t satises hypotheses of Proposition 21.5, and alternative (1) of this proposition is not realized. Thus the corresponding extremal trajectory is not locally geometrically optimal. If the segment 0 t1] does not contain conjugate points, then by Theorem 21.11 the corresponding extremal trajectory is time-optimal compared with all other admissible trajectories suciently close in M.

23.2 Curvature of 3-Dimensional Control-Ane Systems

R

In this section we consider control-ane 3-dimensional systems: q_ = f0 (q) + uf1 (q) dimM = 3:

u 2  q 2 M

(23.19)

374

23 Curvature

We reduce such a system to a 2-dimensional one as in Chap. 22 and compute the curvature of the 2-dimensional system obtained | a feedback invariant of system (23:19). We assume that the following regularity conditions hold on M: f0 ^ f1 ^ f0 f1 ] 6= 0 (23.20) f1 ^ f0 f1 ] ^ f1  f0 f1]] 6= 0: (23.21) Any extremal t of the control-ane system (23:19) is totally singular, it satises the equality h1(t ) = ht  f1i  0 (23.22) and the corresponding extremal control cannot be found immediately from this equality. Dierentiation of (23:22) w.r.t. t yields h01(t ) = ht  f0 f1 ]i  0 and one more dierentiation leads to an equality containing control: h001(t ) + u(t)h101(t ) = ht  f0 f0 f1]]i + u(t)ht  f1 f0 f1 ]]i  0: Then the singular control is uniquely determined:  h1 () = h01() = 0: u = u~(q) = ; hh001() 101() We apply a feedback transformation to system (23:19): u 7! u ; u~(q): This transformation aects the eld f0 , but preserves regularity conditions (23:20), (23:21). After this transformation the singular control is u = 0: In other words, f1 = f0 f1 ] = 0 ) f0  f0 f1]] = 0: So we assume below that f0 f0 f1]] 2 span(f1  f0 f1]): (23.23) In a tubular neighborhood of a trajectory of the eld f0 , consider the reduction of the three-dimensional system (23:19): d q~ = ew ad f1 f (~q) f = M=eRf1 w 2 (;" ") q~ 2 M (23.24) 0 dt

23.2 Curvature of 3-Dimensional Control-Ane Systems

375

for a small enough ". This system has the same conjugate points as the initial one (23:19). If system (23:24) has no conjugate points, then the corresponding singular trajectory of system (23:19) is strongly geometrically optimal, i.e., comes, locally, to the boundary of attainable set. f. A tangent space to Mf Describe the cotangent bundle of the quotient M consists of tangent vectors to M modulo f1 : f = Tq M= f1(q) Tq~M (23.25) R f f q 2 M q~ = q  e 1 2 M identication (23:25) is given by the mapping

R

f v 7! v~ v 2 Tq M v~ 2 Tq~M v~ = ddt q~(t): v = ddt q(t) t=0 t=0 f consists of covectors on M orthogonal to f1: Thus a cotangent space to M f = TqM \ fh1 = 0g Tq~ M f  7! ~  2 Tq M \ fh1 = 0g ~ 2 Tq~ M f: h~  v~i = h vi v 2 Tq M v~ 2 Tq~M Taking into account that the eld f1 is the projection of the Hamiltonian eld ~h1 , it is easy to see that

f = fh1 = 0g=eR~h1 T M where the mapping  7! ~ is dened above (exercise: show that ~ 1 = ~ 2 , f is obtained 2 2 1  eR~h1 ). Summing up, cotangent bundle to the quotient M  from T M via Hamiltonian reduction by ~h1: restriction to the level surface of h1 with subsequent factorization by the ow of ~h1 . Further, regularity condition (23:21) implies that the eld ~h1 is transversal to the level surface fh1 = h01 = 0g, so this level surface gives another realization of the cotangent bundle to the quotient: f = fh1 = h01 = 0g: T M In this realization, ~h0 is the Hamiltonian eld corresponding to the maximized Hamiltonian | generator of extremals (H~ in Sect. 23.1). The level surface of the maximized Hamiltonian (H in Sect. 23.1) realizes as the submanifold fh1 = h01 = 0 h0 = 1g T  M: Via the canonical projection  : T  M ! M, this submanifold can be identied with M, so the level surface H of Sect. 23.1 realizes now as M. We use

376

23 Curvature

this realization to compute curvature of the three-dimensional system (23:19) as the curvature " of its two-dimensional reduction (23:24). The Hamiltonian eld H~ of Sect. 23.1 is now f0 , and f1 is a vertical eld. It remains to normalize f1 , i.e., to nd a vertical eld af1 , a 2 C 1 (M), such that f0 f0 af1]] = ;"af1  (23.26) see (23:6). The triple f0  f1  f2 = f0  f1] forms a moving frame on M, consider the structure constants of this frame: fi  fj ] =

2 X

k=0

ckjifk 

i j = 0 1 2:

Notice that inclusion (23:23) obtained after preliminary feedback transformation reads now as c002 = 0. That is why f0 f0 f1 ]] = ;c102f1 ; c202f0  f1]: Now we can nd the normalizing factor a for f1 such that (23:26) be satised. We have f0  f0 af1]] = f0  (f0a) + af0  f1]] = (f02 a)f1 + 2(f0 a)f0 f1 ] + af0 f0 f1 ]] = (f02 a ; c102a)f1 + (2f0 a ; c201)f0  f1]: Then the required function a is found from the rst order PDE 2f0 a ; c202a = 0 and the curvature is computed: 2 1 " = ; f0 a ;a c02 a : Summing up, curvature of the control-ane 3-dimensional system (23:19) is expressed through the structure constants as " = c102 ; 14 (c202 )2 ; 12 f0 c202 a function on the state space M. Bounds on curvature " along a (necessarily singular) extremal of a 3dimensional control-ane system allow one to obtain bounds on conjugate time, thus on segments where the extremal is locally optimal. Indeed, by construction, " is the curvature of the reduced 2-dimensional system. As we know from Chap. 22, reduction transforms singular extremals into regular ones, and the initial and reduced systems have the same conjugate times. Thus Theorem 23.5 can be applied, via reduction, to the study of optimality of singular extremals of 3-dimensional control-ane systems.

24 Rolling Bodies

We apply the Orbit Theorem and Pontryagin Maximum Principle to an intrinsic geometric model of a pair of rolling rigid bodies. We solve the controllability problem: in particular, we show that the system is completely controllable i the bodies are not isometric. We also state an optimal control problem and study its extremals.

24.1 Geometric Model Consider two solid bodies in the 3-dimensional space that roll one on another without slipping or twisting.

Mc

M Fig. 24.1. Rolling bodies Rather than embedding the problem into geometric model of the system.

R

3,

we construct an intrinsic

378

24 Rolling Bodies

c be two-dimensional connected manifolds | surfaces of the Let M and M c, we suppose rolling bodies. In order to measure lengths of paths in M and M that each of these manifolds is Riemannian , i.e., endowed with a Riemannian structure | an inner product in the tangent space smoothly depending on the point in the manifold: hv1  v2iM  hbv1  bv2iMc

vi 2 Tx M bvi 2 TxbMc:

R

c are oriented (which is natural since Moreover, we suppose that M and M 3 surfaces of solid bodies in are oriented by the exterior normal vector). c, their tangent spaces are At contact points of the bodies x 2 M and xb 2 M identied by an isometry (i.e., a linear mapping preserving the Riemannian structures) c q : Tx M ! TxbM see Fig. 24.2. We deal only with orientation-preserving isometries and omit

M x

TxM

+

xb

M

TxbMc =

Mc

Fig. 24.2. Identication of tangent spaces at contact point

Mc

the words \orientation-preserving" in order to simplify terminology. An isometry q is a state of the system, and the state space is the connected 5dimensional manifold

c j x 2 M xb 2 Mc q an isometry g: Q = f q : Tx M ! TxbM

c: Denote the projections from Q to M and M

c (q) = x b(q) = xb q : Tx M ! TxbM c: q 2 Q x 2 M xb 2 M

Local coordinates on Q can be introduced as follows. Choose arbitrary local c: orthonormal frames e1 , e2 on M and eb1 , be2 on M

hei  ej iM = ij 

hbei  bej iMc = ij 

i j = 1 2:

24.2 Two-Dimensional Riemannian Geometry

379

For any contact conguration of the bodies q 2 Q, denote by the angle of rotation from the frame be1 , eb2 to the frame qe1 , qe2 at the contact point: qe1 = cos eb1 + sin eb2  qe2 = ; sin be1 + cos eb2 : Then locally points q 2 Q are parametrized by triples (x xb ), x = (q) 2 M, c, 2 S1 . Choosing local coordinates (x1 x2) on M and (xb1 xb2) xb = b(q) 2 M c on M , we obtain local coordinates (x1  x2 xb1 xb2 ) on Q. Let q(t) 2 Q be a curve corresponding to a motion of the rolling bodies, then x(t) = (q(t)) and xb(t) = b(q(t)) are trajectories of the contact points c respectively. The condition of absence of slipping means that in M and M q(t)x(t) _ = xb_ (t) (24.1) and the condition of absence of twisting is geometrically formulated as follows: q(t) (vector eld parallel along x(t)) = (vector eld parallel along xb(t)) : (24.2) Our model ignores the state constraints that correspond to admissibility of contact of the bodies embedded in 3. Notice although that if the surfaces M c have respectively positive and nonnegative Gaussian curvatures at a and M point, then their contact is locally admissible. The admissibility conditions (24:1) and (24:2) imply that a curve x(t) 2 M determines completely the whole motion q(t) 2 Q. That is, velocities of admissible motions determine a rank 2 distribution  on the 5-dimensional space Q. We show this formally and compute the distribution  explicitly below. Before this, we recall some basic facts of Riemannian geometry.

R

24.2 Two-Dimensional Riemannian Geometry Let M be a 2-dimensional Riemannian manifold. We describe Riemannian geodesics, Levi-Civita connection and parallel translation on T  M  = TM. Let h  i be the Riemannian structure and e1 , e2 a local orthonormal frame on M.

24.2.1 Riemannian Geodesics For any xed points x0  x1 2 M, we seek for the shortest curve in M connecting x0 and x1: x_ = u1e1 (x) + u2e2 (x) x 2 M (u1 u2) 2 x(0) = x0 x(t1) = x1 l=

Zt 0

1

hx_ x_ i1=2 dt =

Zt

1

0

R

(u21 + u22)1=2 dt ! min:

2

380

24 Rolling Bodies

In the same way as in Sect. 19.1, it easily follows from PMP that arc-length parametrized extremal trajectories in this problem (Riemannian geodesics) are projections of trajectories of the normal Hamiltonian eld: x(t) =   etH~ ()  2 H = fH = 1=2g T  M H = 12 (h21 + h22 ) hi () = h ei i i = 1 2: The level surface H is a spherical bundle over M with a ber Hq = fh21 + h22 = 1g \ Tq M = S 1 parametrized by angle ': h1 = cos ' h2 = sin ': Cotangent bundle of a Riemannian manifold can be identied with the tangent bundle via the Riemannian structure: TM  = T  M v 7!  = hv i: Then H T  M is identied with the spherical bundle S = fv 2 TM j kvk = 1g TM

of unit tangent vectors to M. After this identication, etH~ can be considered as a geodesic ow on S .

24.2.2 Levi-Civita Connection A connection on the spherical bundle S ! M is an arbitrary horizontal distribution D: D = fDv Tv S j v 2 Sg Dv ! Tv (Sx ) = Tv S  Sx = S \ Tx M: Any connection D on M denes a parallel translation of unit tangent vectors along curves in M. Let x(t), t 2 0 t1], be a curve in M, and let v0 2 Tx(0) M be a unit tangent vector. The curve x(t) has a unique horizontal lift on S starting at v0: v(t) 2 S    v(t) = x(t) v_ (t) 2 Dv(t) v(0) = v0 :

24.2 Two-Dimensional Riemannian Geometry

381

Indeed, if the curve x(t) satises the nonautonomous ODE x_ = u1(t) e1 (x) + u2(t) e2 (x) then its horizontal lift v(t) is a solution to the lifted ODE v_ = u1(t) 1 (v) + u2(t) 2 (v)

(24.3)

where i are horizontal lifts of the basis elds ei : Dv = span( 1 (v) 2 (v))

 i = ei :

Notice that solutions of ODE (24:3) are continued to the whole time segment 0 t1] since the bers Sx are compact. The vector v(t1 ) is the parallel translation of the vector v0 along the curve x(t). A vector eld v(t) along a curve x(t) is called parallel if it is preserved by parallel translations along x(t). Levi-Civita connection is the unique connection on the spherical bundle S ! M such that: (1) velocity of a Riemannian geodesic is parallel along the geodesic (i.e., the geodesic eld H~ is horizontal), (2) parallel translation preserves angle, i.e., horizontal lifts of vector elds on the base M commute with the vector eld @@ ' that determines the element of length (or, equivalently, the element of angle) in the ber Sx . Now we compute the Levi-Civita connection as a horizontal distribution on H  = S . In Chap. 23 we constructed a feedback-invariant frame on the manifold H:  @  @  0 0 ~ ~ ~ T H = span H @ '  H  H = @ '  H~ : We have     ~H = h1 e1 + c1 @ + h2 e2 + c2 @  (24.4)  @ '@   @ '@  H~ 0 = ;h2 e1 + c1 @ ' + h1 e2 + c2 @ '  (24.5) where ci are structure constants of the orthonormal frame on M:

e1  e2] = c1 e1 + c2 e2  ci 2 C 1 (M): Indeed, the component of the eld H~ = h1~h1 + h2~h2 in the tangent space of the manifold M is equal to h1e1 +h2 e2 . In order to nd the component of the ~ i in two dierent ways: eld H~ in the ber, we compute the derivatives Hh

382

24 Rolling Bodies

~ 1 = (h1~h1 + h2~h2 )h1 = h2(~h2 h1 ) = h2 fh2 h1g = h2 (;c1 h1 ; c2 h2) Hh ~ 1 = H~ cos ' = ; sin '(H') ~ = ;h2(H') ~ Hh similarly thus Consequently,

~ 2 = h1(c1 h1 + c2 h2) = h1(H') ~ Hh ~ = c1h1 + c2h2 : H' H~ = h1e1 + h2e2 + (c1 h1 + c2 h2) @@' 

and equality (24:4) follows. Then equality (24:5) is obtained by straightforward dierentiation. Notice that using decompositions (24:4), (24:5), we can easily compute Gaussian curvature k of the Riemannian manifold M via the formula of Theorem 23.1:   @  @ : ~ H ~ H = ; k @' @' Since ~ H~ 0] = (c21 + c22 ; e1 c2 + e2 c1) @  H @' then k = ;c21 ; c22 + e1 c2 ; e2 c1 : (24.6) Properties (1) and (2) of the horizontal distribution D @on H that determines the Levi-Civita connection mean that H~ 2 D and es @ ' D = D, thus



@ D = span es @ ' H~ j s 2

Since





R

:





@ es @ ' H~ = h1 (' ; s) e1 + c1 @@ ' + h2(' ; s) e2 + c2 @@ ' 

we obtain





~ H~ 0 : D = span H The 1-form of the connection D: ! 2 1(H) D = Ker ! reads ! = c1 !1 + c2 !2 ; d' where (!1 !2) is the dual coframe to (e1  e2 ): !i 2 1(M) h!i ej i = ij  i j = 1 2:

383

We return to the rolling bodies problem and write down admissibility conditions (24:1), (24:2) for a curve q(t) 2 Q as restrictions on velocity q(t). _ c in the orthonormal Decompose velocities of the contact curves in M and M frames: x_ = a1 e1 (x) + a2 e2 (x) xb_ = ba1 be1 (xb) + ba2 eb2 (xb):

(24.7) (24.8)

Then the nonslipping condition (24:1) reads:

ba1 = a1 cos ; a2 sin 

ba2 = a1 sin + a2 cos :

(24.9)

Now we consider the nontwisting condition (24:2). Denote the structure constants of the frames: e1 e2 ] = c1e1 + c2 e2  be1 be2 ] = bc1eb1 + bc2 be2 

ci 2 C 1(M) bci 2 C 1(Mc):

c be the mapping induced by the isometry q via identiLet q~ : Tx M ! Txb M cation of tangent and cotangent spaces: q~ !1 = cos !b1 + sin !b2 q~ !2 = ; sin !b1 + cos !b2 : In the cotangent bundle, the nontwisting condition means that if (t) = (x(t) '(t)) 2 H is a parallel covector eld along a curve x(t) 2 M, then b(t) = q~(t)(t) = (xb(t) 'b(t)) 2 Hb

c. is a parallel covector eld along the curve xb(t) 2 M Since the isometry q(t) rotates the tangent spaces at the angle (t), then the mapping q~(t) rotates the cotangent spaces at the same angle: 'b(t) = '(t) + (t), thus _ = 'b_ (t) ; '(t):

(t) _ (24.10) A covector eld (t) is parallel along the curve in the base x(t) i _ 2 Ker !, i.e., '_ = hc1!1 + c2 !2  x_ i = c1 a1 + c2 a2: Similarly, b(t) is parallel along xb(t) i

384

24 Rolling Bodies

D

E

'b_ = bc1 !b1 + bc2 !b2  xb_ = bc1 ba1 + bc2 ba2 = a1(bc1 cos + bc2 sin ) + a2(;bc1 sin + bc2 cos ): In view of (24:10), the nontwisting condition reads

_ = bc1 ba1 + bc2ba2 ; (c1 a1 + c2 a2) = a1 (;c1 + bc1 cos + bc2 sin ) + a2 (;c2 ; bc1 sin + bc2 cos ):

(24.11)

Summing up, admissibility conditions (24:1) and (24:2) for rolling bodies determine constraints (24:9) and (24:11) along contact curves (24:7), (24:8), i.e., a rank two distribution  on Q spanned locally by the vector elds X1 = e1 + cos be1 + sin be2 + (;c1 + bc1 cos + bc2 sin ) @@  (24.12) X2 = e2 ; sin eb1 + cos be2 + (;c2 ; bc1 sin + bc2 cos ) @@ : (24.13)

R

Admissible motions of the rolling bodies are trajectories of the control system q_ = u1X1 (q) + u2X2 (q) q 2 Q u1  u2 2 :

(24.14)

24.4 Controllability

c by Denote the Gaussian curvatures of the Riemannian manifolds M and M c to Q: k and bk respectively. We lift these curvatures from M and M k(q) = k((q)) bk(q) = bk(b(q)) q 2 Q:

Theorem 24.1. (1) The reachable set O of system (24:14) from a point q 2 Q is an immersed smooth connected submanifold of Q with dimension equal to 2 or 5. Speci cally : (k ; bk)jO  0 ) dimO = 2 (k ; bk)jO 6 0 ) dimO = 5:

(2) There exists an injective correspondence between isometries i : M ! c and 2-dimensional reachable sets O of system (24:14). In particular, if M c are isometric, then system (24:14) is not completely the manifolds M and M

controllable. c are complete and simply con(3) Suppose that both manifolds M and M c and 2nected. Then the correspondence between isometries i : M ! M dimensional reachable sets O of system (24:14) is bijective. In particular, sysc are not tem (24:14) is completely controllable i the manifolds M and M isometric.

24.4 Controllability

385

Proof. (1) By the Orbit theorem, the reachable set of the symmetric sys-

tem (24.14), i.e., the orbit of the distribution  through any point q 2 Q, is an immersed smooth connected submanifold of Q. Now we show that any orbit O of  has dimension either 2 or 5. Fix an orbit O and assume rst that at some point q 2 O the manifolds M c have dierent Gaussian curvatures: k(q) 6= bk(q). In order to construct and M a frame on Q, compute iterated Lie brackets of the elds X1 , X2 : X3 = X1 X2 ] = c1X1 + c2 X2 + (bk ; k) @@  (24.15) X4 = X1 X3 ]  @  @ b b = (X1 c1 )X1 + (X1 c2)X2 + c2 X3 + (X1 (k ; k)) @ + (k ; k) X1  @  (24.16) X5 = X2 X3 ]  @  @ b b = (X2 c1 )X1 + (X2 c2)X2 ; c1 X3 + (X2 (k ; k)) @ + (k ; k) X2  @  (24.17)



 X1  @@ = sin be1 ; cos eb2 + ( ) @@  (24.18)  @  X2  @ = cos eb1 + cos be2 + ( ) @@ : (24.19) In the computation of bracket (24:15) we used expression (24:6) of Gaussian curvature through structure constants. It is easy to see that   @ Lie(X1  X2)(q) = span (X1  X2  X3 X4  X5) (q) = span e1  e2 be1  be2 @ (q) = Tq Q:

System (24.14) has the full rank at the point q 2 O where k 6= bk, thus dimO = 5. On the other hand, if k(q) = bk(q) at all points q 2 O, then equality (24:15) implies that the distribution  is integrable, thus dimO = 2. c be an isometry. Its graph (2) Let i : M ! M

n

c x 2 M xb = i(x) 2 Mc ; = q 2 Q j q = ix : Tx M ! Txb M

o

is a smooth 2-dimensional submanifold of Q. We prove that ; is an orbit of . Locally, choose an orthonormal frame e1 , e2 in M and take the corresponding c. Then j; = 0. Since bc1 = c1, orthonormal frame be1 = i e1 , be2 = i e2 in M b bc2 = c2, and k(q) = k(q), restrictions of the elds X1, X2 read

386

24 Rolling Bodies

X1 j; = e1 + eb1  X2 j; = e2 + eb2 : Then it follows that the elds X1 , X2 are tangent to ;. Lie bracket (24:15) yields X1 X2 ]j; = c1X1 + c2 X2  thus ; is an orbit of . Distinct isometries i1 6= i2 have distinct graphs ;1 6= ;2, i.e., the correspondence between isometries and 2-dimensional orbits is injective. c are complete and simply (3) Now assume that the manifolds M and M connected. Let O be a 2-dimensional orbit of . We construct an isometry c with the graph O. i : M !M Notice rst of all that for any Lipschitzian curve x(t) 2 M, t 2 0 t1], and any point q0 2 Q, there exists a trajectory q(t) of system (24:14) such that (q(t)) = x(t) and q(0) = q0. Indeed, a Lipschitzian curve x(t), t 2 0 t1], is a trajectory of a nonautonomous ODE x_ = u1 (t)e1 (x) + u2(t)e2 (x) for some ui 2 L1 0 t1]. Consider the lift of this equation to Q: q_ = u1(t)X1 (q) + u2(t)X2 (q)

q(0) = q0:

(24.20)

We have to show that the solution to this Cauchy problem is dened on the whole segment 0 t1]. Denote by R the Riemannian length of the curve x(t) and by B(x0  2R) M the closed Riemannian ball of radius 2R centered at x0. The curve x(t) is contained in B(x0  2R) and does not intersect with its boundary. Notice that the ball B(x0  2R) is a closed and bounded subset of the c of the maximal complete space M, thus it is compact. The projection xb(t) 2 M solution q(t) to Cauchy problem (24:20) has Riemannian length not greater c, xb0 = b(q0), and than R, thus it is contained in the compact B(xb0  2R) M does not intersect with its boundary. Summing up, the maximal solution q(t) to (24:20) is contained in the compact K = B(x0  2R)  B(xb0  2R)  S 1 and does not come to its boundary. Thus the maximal solution q(t) is dened at the whole segment 0 t1]. Now it easily follows that (O) = M for the two-dimensional orbit O. Indeed, let q0 2 O, then x0 = (q0 ) 2 (O). Take any point x1 2 M and connect it with x0 by a Lipschitzian curve x(t), t 2 0 t1]. Let q(t) be the lift of x(t) to the orbit O with the initial condition q(0) = q0. Then q(t1 ) 2 O and c. x1 = (q(t1 )) 2 (O). Thus (O) = M. Similarly, b(O) = M The projections  : O!M

c and b : O ! M

are local dieomorphisms since  (X1 ) = e1   (X2 ) = e2 

b (X1 ) = cos eb1 + sin eb2  b (X2 ) = ; sin be1 + cos be2 :

(24.21)

24.5 Length Minimization Problem

387

Moreover, it follows that projections (24:21) are global dieomorphisms. Indeed, let q 2 O. Any Lipschitzian curve x( ) on M starting from (q) has a unique lift to O starting from q and this lift continuously depends on x( ). Suppose that q0 2 O, q0 6= q, (q0 ) = (q), and q( ) is a path on O connecting q with q0 . Contracting the loop (q( )) and taking the lift of the contraction, we come to a contradiction with the local invertibility of jO . Hence jO is globally invertible, thus it is a global dieomorphism. The same is true for bjO . Thus we can dene a dieomorphism c: i = b  (jO );1 : M ! M Since i e1 = cos be1 + sin be2  i e2 = ; sin eb1 + cos eb2  the mapping i is an isometry. c are not isometric, then all reachable sets of If the manifolds M and M system (24:14) are 5-dimensional, thus open subsets of Q. But Q is connected, thus it is a single reachable set. ut

24.5 Length Minimization Problem 24.5.1 Problem Statement c, i.e., k ;bk 6= 0 on Q. Then, by Suppose that k(x) = 6 bk(xb) for any x 2 M, xb 2 M

item (1) of Theorem 24.1, system (24:14) is completely controllable. Consider the following optimization problem: given any two contact congurations of the system of rolling bodies, nd an admissible motion of the system that steers the rst conguration into the second one and such that the path of c) was the shortest possible. This the contact point in M (or, equivalently, in M geometric problem is stated as the following optimal control one: q_ = u1X1 + u2X2  q 2 Q u = (u1  u2) 2 2 (24.22) q(0) = q0 q(t1 ) = q1 t1 xed l=

Zt

1

0

R

(u21 + u22 )1=2 dt ! min:

c read respectively as Notice that projections of ODE (24:22) to M and M x_ = u1e1 + u2e2  x 2 M and c xb_ = u1(cos eb1 + sin eb2 ) + u2(; sin eb1 + cos be2 ) xb 2 M

388

24 Rolling Bodies

thus the sub-Riemannian length l of the curve q(t) is equal to the Riemannian length of the both curves x(t) and xb(t). As usual, we replace the length l by the action: Z t1 J = 12 (u21 + u22) dt ! min 0 and restrict ourselves to constant velocity curves: u21 + u22  const 6= 0:

24.5.2 PMP As we showed in the proof of Theorem 24.1, the vector elds X1  : : :  X5 form a frame on Q, see (24:15){(24:17). Denote the corresponding Hamiltonians linear on bers in T  Q: gi () = h Xii  2 T  Q i = 1 : : :  5: Then the Hamiltonian of PMP reads gu () = u1 g1() + u2g2 () + 2 (u21 + u22 ) and the corresponding Hamiltonian system is _ = u1~g1() + u2~g2()

 2 T  Q:

(24.23)

24.5.3 Abnormal Extremals Let  = 0. The maximality condition of PMP implies that g1(t ) = g2(t )  0 (24.24) along abnormal extremals. Dierentiating these equalities by virtue of the Hamiltonian system (24:23), we obtain one more identity: g3(t )  0: (24.25) The next dierentiation by virtue of (24:23) yields an identity containing controls: (24.26) u1(t)g4 (t ) + u2 (t)g5 (t)  0: It is natural to expect that conditions (24:23){(24:26) on abnormal exc. tremals on Q should project to reasonable geometric conditions on M and M This is indeed the case, and now we derive ODEs for projections of abnormal c. extremals to M and M

24.5 Length Minimization Problem

389

According to the splitting of the tangent spaces c ! T S1  Tq Q = Tx M ! TxbM the cotangent space split as well: c ! TS1  Tq Q = Tx M ! Txb M c d 2 TS1:  =  + b + d   2 Tq Q  2 Tx M b 2 Txb M Then   @ b g1 () = h X1 i =  +  + d  e1 + cos eb1 + sin be2 + b1 @

= h1 () + cos bh1 (b) + sin bh2 (b) + b1 (24.27)   g2 () = h X2 i =  + b + d  e2 ; sin be1 + cos be2 + b2 @@

= h2 () ; sin bh1(b) + cos bh2 (b) + b2 (24.28) where b1 = ;c1 + bc1 cos + bc2 sin , b2 = ;c2 ; bc1 sin + bc2 cos ,

 g3 () = h X3 i =  + b + d  c1X1 + c2 X2 + (bk ; k) @@



= c1 g1() + c2 g2() + (bk ; k): (24.29) Then identities (24:24) and (24:25) read as follows: = 0 h1 + cos bh1 + sin bh2 = 0 h2 ; sin bh1 + cos bh2 = 0: Under these conditions, taking into account equalities (24:16){(24:19), we have:



g4() =  + b (X1 c1 )X1 + (X2 c2 )X2 + c2 X3   +(X1 (bk ; k)) @@ + (bk ; k) X1  @@

= (bk ; k)(sin bh1 ; cos bh2 ) = (bk ; k)h2 



g5() =  + b (X2 c1 )X1 + (X2 c2 )X2 ; c1 X3 +   (X2 (bk ; k)) @@ + (bk ; k) X2  @@

= (bk ; k)(cos bh1 + sin bh2 ) = ;(bk ; k)h1:

390

24 Rolling Bodies

Then identity (24:26) yields u1h2 ; u2 h1 = 0: That is, up to reparametrizations of time, abnormal controls satisfy the identities: u1 = h1 ()

u2 = h2():

(24.30)

In order to write down projections of the Hamiltonian system (24:23) to c, we decompose the Hamiltonian elds ~g1, ~g2. In view of equalT  M and T  M ities (24:27), (24:28), we have ~g1 = ~h1 + cos ~bh1 + sin ~bh2 + (; sin bh1 + cos bh2)~ + ~a1 + a1 ~  ~g2 = ~h2 ; sin ~bh1 + cos ~bh2 + (; cos bh1 ; sin bh2)~ + ~a2 + a2 ~ : It follows easily that ~ = ; @@ . Since = 0 along abnormal extremals, projection to T  M of system (24:23) with controls (24:30) reads ~ _ = h1~h1 + h2~h2 = H H = 12 (h21 + h22 ): Consequently, projections x(t) = (q(t)) are Riemannian geodesics in M. c we obtain the equalities Similarly, for projection to M u1 = ; cos bh1 ; sin bh2 

thus

u2 = sin bh1 ; cos bh2 

  b_ = ; cos bh1 ; sin bh2 cos ~bh1 + sin ~bh2    b ~ ~ b b b + sin h1 ; cos h2 ; sin h1 + cos h2   ~b = ;bh1 ~bh1 ; bh2 ~hb2 = ;H Hb = 1 hb21 + bh22 

2 c. i.e., projections xb(t) = b(q(t)) are geodesics in M We proved the following statement.

Proposition 24.2. Projections of abnormal extremal curves x(t) = (q(t))

c. and xb(t) = b(q(t)) are Riemannian geodesics respectively in M and M

Abnormal sub-Riemannian geodesics q(t) are optimal at segments 0 ] at which at least one of Riemannian geodesics x(t), xb(t) is a length minimizer. In particular, short arcs of abnormal geodesics q(t) are optimal.

24.5 Length Minimization Problem

391

24.5.4 Normal Extremals Let  = ;1. The normal extremal controls are determined from the maximality condition of PMP:

u1 = g1 u2 = g2 and normal extremals are trajectories of the Hamiltonian system ~ _ = G()  2 T  Q (24.31) G = 12 (g12 + g22 ): The maximized Hamiltonian G is smooth, thus short arcs of normal extremal trajectories are optimal. c = 2. Consider the case where one of the rolling surfaces is the plane: M In this case the normal Hamiltonian system (24:31) can be written in a simple form. Choose the following frame on Q: Y1 = X1  Y2 = X2  Y3 = @@  Y4 = Y1 Y3 ] Y5 = Y2 Y3 ] and introduce the corresponding linear on bers Hamiltonians

R

mi () = h Yii

i = 1 : : :  5:

Taking into account that bc1 = bc2 = bk = 0, we compute Lie brackets in this frame: Y1 Y2] = c1 Y1 + c2 Y2 ; kY3  Y1 Y4] = ;c1 Y5  Y2  Y4] = ;c2Y5  Y1 Y5] = c1 Y4 Y2  Y5] = c2Y4 : Then the normal Hamiltonian system (24:31) reads as follows: m_ 1 = ;m2 (c1 m1 + c2m2 ; km3 ) m_ 2 = m1 (c1m1 + c2 m2 ; km3 ) m_ 3 = m1 m4 + m2 m5  m_ 4 = ;(c1 m1 + c2m2 )m5  m_ 5 = (c1m1 + c2 m2 )m4  q_ = m1 X1 + m2 X2 : Notice that, in addition to the Hamiltonian G = 12 (m21 + m22 ), this system has one more integral:  = (m24 + m25 )1=2. Introduce coordinates on the level surface G = 21 :

392

24 Rolling Bodies

m1 = cos  m2 = sin  m3 = m m4 =  cos( + ) m5 =  sin( + ): Then the Hamiltonian system simplies even more: _ = c1 cos  + c2 sin  ; km m_ =  cos  _ = km q_ = cos  X1 + sin  X2 : The case k = const, i.e., the sphere rolling on a plane, is completely integrable. This problem was studied in detail in book 12].

Fig. 24.3. Sphere on a plane

A Appendix

In this Appendix we prove several technical propositions from Chap. 2.

A.1 Homomorphisms and Operators in C 1(M ) Lemma A.1. On any smooth manifold M , there exists a function a 2 C 1(M) such that for any N > 0 exists a compact K b M for which a(q) > N 8 q 2 M n K: Proof. Let ek , k 2 , be a partition of unity on M: the functions ek 2 C 1 (M) haveP compact supports supp ek b M, which a locally nite covering of M, P1 form and 1 e  1. Then the function ke can be taken as a. ut k k k=1 k=1 Now we recall and prove Proposition 2.1. Proposition 2.1. Let ' : C 1(M) ! be a nontrivial homomorphism of algebras. Then there exists a point q 2 M such that ' = qb. Proof. For the homomorphism ' : C 1 (M) ! , the set Ker ' = ff 2 C 1 (M) j 'f = 0g is a maximal ideal in C 1 (M). Further, for any point q 2 M, the set of functions Iq = ff 2 C 1 (M) j f(q) = 0g is an ideal in C 1(M). To prove the proposition, we show that Ker ' Iq (A.1) for some q 2 M. Then it follows that Ker ' = Iq and ' = qb. By contradiction, suppose that Ker ' 6 Iq for any q 2 M. This means that

N

R

R

394

A Appendix

8 q 2 M 9 bq 2 Ker ' s. t. bq (q) 6= 0:

Changing if necessary the sign of bq , we obtain that 8 q 2 M 9 bq 2 Ker ' Oq M s. t. bq jOq > 0: (A.2) Fix a function a given by Lemma A.1. Denote '(a) = , then '(a ; ) = 0, i.e., (a ; ) 2 Ker ': Moreover, 9 K b M s. t. a(q) ; > 0 8 q 2 M n K: Take a nite covering of the compact K by the neighborhoods Oq as in (A.2): K

n

i=1

Oqi 

and let e0  e1 : : :  en 2 C 1(M) be a partition of unity subordinated to the covering of M: M n K Oq1  : : :  Oqn : Then we have a globally dened function on M: c = e0 (a ; ) +

n X i=1

ei bqi > 0:

    1 = ' c 1c = '(c) ' 1c 

Since then

'(c) 6= 0: But c 2 Ker ', a contradiction. Inclusion (A.1) is proved, and the proposition follows. ut Now we formulate and prove the theorem on regularity properties of composition of operators in C 1 (M), in particular, for nonautonomous vector elds or ows on M. Proposition A.2. Let At and Bt be continuous w.r.t. t families of linear continuous operators in C 1 (M). Then the composition At  Bt is also continuous w.r.t. t. If in addition the families At and Bt are di erentiable at t = t0 , then the family At  Bt is also di erentiable at t = t0, and its derivative is given by the Leibniz rule:



!



!

d (A  B ) = d A  B + A  d B : t0 t0 d t t0 t t d t t0 t d t t0 t

A.2 Remainder Term of the Chronological Exponential

395

Proof. To prove the continuity, we have to show that for any a 2 C 1 (M),

the following expression tends to zero as " ! 0:

(At+"  Bt+" ; At  Bt ) a = At+"  (Bt+" ; Bt ) a + (At+" ; At )  Bt a: By continuity of the family At , the second term (At+" ; At )  Bt a ! 0 as " ! 0. Since the family Bt is continuous, the set of functions (Bt+" ; Bt ) a lies in any preassigned neighborhood of zero in C 1 (M) for suciently small ". For any "0 > 0, the family At+" , j"j < "0 , is locally bounded, thus equicontinuous by the Banach-Steinhaus theorem. Consequently, At+"  (Bt+" ; Bt ) a ! 0 as " ! 0. Continuity of the family At  Bt follows. The dierentiability and Leibniz rule follow similarly from the decomposition 1 (A  B ; A  B ) a = A  1 (B ; B ) a + 1 (A ; A )  B a: t+" " t+" t " t+" t+" t t " t+" t t

ut

A.2 Remainder Term of the Chronological Exponential Here we prove estimate (2:13) of the remainder term for the chronological exponential. Lemma A.3. For any t1 > 0, complete nonautonomous vector eld Vt , compactum K b M , and integer s  0, there exist C > 0 and a compactum K 0 b M , K K 0 , such that

R kPtaksK  C eC 0t kV ksK0 d kaksK 0 

where

;! Pt = exp

Proof. Denote the compact set

a 2 C 1 (M) t 2 0 t1] (A.3)

Zt 0

V d :

Kt =  fP (K) j 2 0 t]g and introduce the function   (t) = sup kkaPktaksK j a 2 C 1 (M) kaks+1Kt 6= 0 s+1Kt = sup fkPtaksK j a 2 C 1 (M) kaks+1Kt = 1g :

(A.4)

Notice that the function (t), t 2 0 t1], is measurable since the supremum in the right-hand side of (A.4) may be taken over only an arbitrary countable dense subset of C 1 (M). Moreover, in view of inequalities (2:3) and

396

A Appendix

kaksPt(K )  kaks+1Kt  the function (t) is bounded on the segment 0 t1]. As in the denition of the seminorms k ksK in Sect. 2.2, we x a proper N and vector elds h1 : : :  hN 2 Vec M that span tangent embedding M spaces to M. Let q0 2 K be a point at which

R

kPtaksK

= sup fjhil   hi1 (Pt a)(q) j q 2 K 1  i1  : : :  il  N 1  l  sg attains its upper bound, and let pa = pa (x1 : : :  xN ) be the polynomial of degree  s whose derivatives of order up to and including s at the point qt = Pt(q0 ) coincide with the corresponding derivatives of a at qt. Then kPtaksK = jhil   hi1 (Pt pa )(q0)j  kPtpa ksK  (A.5) kpa ksqt  kaksKt : In the nite-dimensional space of all real polynomials of degree  s, all norms are equivalent, so there exists a constant C > 0 which does not depend on the choice of the polynomial p of degree  s such that kpksKt  C: (A.6) kpksqt Inequalities (A.5) and (A.6) give the estimate kPtaksK  kPtpa ksK  C kPtpa ksK = C kPtpa ksK  C (t): (A.7) kaksKt kpa ksqt kpa ksKt kpa ks+1Kt Since Zt Pta = a + P  V a d  0

then

kPtaksK  kaksK +

Zt 0

kP  V aksK d

by inequality (A.7) and denition (2:2)

 kaksK + C kaks+1Kt Dividing by kaks+1Kt , we arrive at

Zt 0

( ) kV ksKt d :

kPtaksK  1 + C Z t ( ) kV k d :  sKt kaks+1Kt 0

A.2 Remainder Term of the Chronological Exponential

Thus we obtain the inequality (t)  1 + C

Zt 0

( ) kV ksKt d 

from which it follows by Gronwall's lemma that Rt (t)  eC 0 kV ksKt d : Then estimate (A.7) implies that R kPtaksK  C eC 0t kV ksKt d kaksKt  and the required inequality (A.3) follows for any compactum K 0  Kt . Now we prove estimate (2:13). Proof. Decomposition (2:11) can be rewritten in the form

Z

397

ut

Z

Pt = Sm (t) + Pm  Vm   V1 d m : : : d 1 

m (t)

where Sm (t) = Id +

mX ;1 Z

Z

Vk   V1 d k : : : d 1 :

k=1 k (t)

Then Z Z k(Pt ; Sm (t))aksK  kPm  Vm   V1 aksK d m : : : d 1

m (t)

by Lemma A.3

Z

Z

R  CeC 0t kV ksK0 d kVm   V1 aksK 0 d m : : : d 1 :

m (t)

Now we estimate the last integral. By denition (2:2) of seminorms,

Z

Z

kVm   V1 aksK 0 d m : : : d 1

m (t)

Z

Z

 kVm ksK 0 kVm;1 ks+1K 0 kV1 ks+m;1K 0 kaks+mK 0 d m : : : d 1

m (t)  kaks+mK 0

Z

Z

kVm ks+m;1K 0 kVm;1 ks+m;1K 0 kV1 ks+m;1K 0 d m : : : d 1

m (t)

1 = kaks+mK 0 m!

Z t 0

kV ks+m;1K 0 d

m



398

A Appendix

and estimate (2:13) follows:

k(Pt ; Sm (t)) aksK

C eC R0t kV ksK0 d  m!

Z t 0

kV ks+m;1K 0 d

m

kaks+mK 0 : ut

References

The bibliography contains the references mentioned in the text and some other works helpful for deeper acquaintance with the subject it does not claim to be complete or systematic.

Books on control theory and nonholonomic geometry 1] Alekseev, V.M., Tikhomirov, V.M., Fomin S.V.: Optimal Control. Consultants Bureau, New York (1987) 2] Bellaiche, A., Risler, J., Eds.: Sub-Riemannian Geometry. Birkhaser, Progress in Math., 144 (1996) 3] Bellman, R.: Introduction to the Mathematical Theory of Control Processes, vol. 1, 2. Academic Press, New York (1967) 4] Bloch, A.: Nonholonomic Mechanics and Control, Interdisciplinary Applied Mathematics, Springer, 24, (2003) 5] Boscain, U., Piccoli, B.: Optimal Synthesis for Control Systems on 2-D Manifolds. Springer SMAI, 43, to appear 6] Brockett, R.W., Millman, R.S., Sussmann, H.J., Eds.: Di erential Geometric Conrol Theory. Birkhauser, Boston (1983) 7] Davydov, A.A.: Qualitative Theory of Control Systems. Translations of Mathematical Monographs. American Mathematical Society (1994) 8] Gamkrelidze, R.V.: Principles of Optimal Control Theory. Plenum Publishing Corporation, New York (1978) 9] Gauthier, J.-P., Kupka, I.A.K.: Deterministic Observation Theory and Applications. Cambridge University Press (2001) 10] Isidori, A.: Nonlinear Control Systems: an Introduction. Springer-Verlag (1985) 11] Jakubczyk, B., Respondek, W., Eds.: Geometry of Feedback and Optimal Control. Marcel Dekker (1998) 12] Jurdjevic, V.: Geometric Control Theory. Cambridge University Press (1997) 13] Montgomery, R.: A Tour of Subriemannian Geometries, Their Geodesics and Applications. American Mathematical Society (2002) 14] Nijmeijer, H., van der Schaft, A.: Nonlinear Dynamical Control Systems. Springer-Verlag (1990)

400

References

15] Pontryagin, L.S., Boltyanskij, V.G., Gamkrelidze, R.V., Mishchenko, E.F.: The Mathematical Theory of Optimal Processes. Pergamon Press, Oxford (1964) 16] Sontag, E.D.: Mathematical Control Theory: Deterministic Finite Dimensional Systems. Springer-Verlag (1990) 17] Sussmann, H.J., Ed.: Nonlinear Controllability and Optimal Control. Marcel Dekker (1990) 18] Zelikin, M.I., Borisov, V.F.: Theory of Chattering Control with Applications to Astronautics, Robotics, Economics, and Engineering. Birkhauser, Basel (1994)

Papers Chronological calculus and related topics 19] Agrachev, A.A., Gamkrelidze, R.V.: The exponential representation of ows and the chronological calculus. Matem. sbornik, 107 (149), 467{532 (1978) 20] Agrachev, A.A., Gamkrelidze, R.V.: The chronological algebras and nonstationary vector elds. Itogi nauki. VINITI. Problemy geometrii, 11, 135{176 (1980) 21] Agrachev, A.A., Gamkrelidze, R.V.: Volterra series and groups of permutations. Itogi nauki. VINITI. Sovremennye problemy matematiki. Novejshie dostideniya, 39, 3{40 (1991) 22] Agrachev, A.A., Gamkrelidze, R.V., Sarychev, A.V.: Local invariants of smooth control systems. Acta Appl. Math., 14, 191{237 (1989) 23] Agrachev, A.A., Marigo, A.: Nonholonomic tangent spaces: intrinsic construction and rigid dimensions. Electronic Research Announcements of the American Mathematical Society, 9, 111{120 (2003) 24] Agrachev, A.A., Sarychev, A.V.: Filtrations of a Lie algebra of vector elds and the nilpotent approximation of controlled systems. Dokl. Acad. Nauk SSSR, 295, 777{781 (1987) 25] Agrachev, A.A., Vakhrameev, S.: Chronological series and Cauchy { Kovalevska theorem. Itogi nauki. VINITI. Problemy geometrii, 12, 165{189 (1981) 26] Chen, K.T.: Integration of paths, geometric invariants and a generalized Baker{ Hausdorf formula. Annals of Mathematics, 65, 163{178 (1957) 27] Fliess, M.: Fonctionelles causales nonlineaires et indeterminees noncommutatives, Bull. Soc. Math. France, 109, 3{40 (1981) 28] Gamkrelidze, R., Agrachev, A., Vakhrameev, S.: Ordinary di erential equations on vector bundles and the chronological calculus. Itogi nauki. VINITI. Sovremennye problemy matematiki. Novejshie dostigeniya, 35, 3{107 (1989) 29] Kawski, M.: Nilpotent Lie algebras of vector elds. J. Reine & Angewandte Mathematic, 188, 1{17 (1988) 30] Kawski, M.: Combinatorics of nonlinear controllability and noncommuting ows. In: Mathematical control theory, ICTP Lecture Notes Series, 8, 222{ 311 (2002) 31] Kawski, M., Sussmann, H.J.: Noncommutative power series and formal Lie algebraic Techniques in nonlinear control theory. In: Operators, Systems and Linear Algebra, U. Helmke, D. Pratzel-Wolters and E. Zers, Eds., Teubner 111{128 (1997)

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List of Figures

1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11

Coordinate system in smooth manifold M . . . . . . . . . . . . . . . . . . . Tangent vector (0) _ ...................................... Tangent space Tq M . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solution to ODE q_ = V (q) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Dierential Dq  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Flow P t of vector eld V . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Attainable set Aq0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Initial and nal congurations of the car . . . . . . . . . . . . . . . . . . . . Steering the car from q0 to q1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Lie bracket of vector elds V1 , V2 . . . . . . . . . . . . . . . . . . . . . . . . . . . Lie bracket for a moving car . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3 5 5 6 7 11 15 16 16 17 19

4.1 Lattice generated by vectors e1 , e2 . . . . . . . . . . . . . . . . . . . . . . . . . . 62 5.1 5.2 5.3 5.4 5.5

Attainable set Aq0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Orbit Oq0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Immersed manifold . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Complete controllability of the family F . . . . . . . . . . . . . . . . . . . . . Integral manifold Nq of distribution  . . . . . . . . . . . . . . . . . . . . . .

64 64 65 71 75

6.1 6.2 6.3 6.4

Fixed and moving frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Phase portrait of system (6:12). . . . . . . . . . . . . . . . . . . . . . . . . . . . . Orbits in the case l 2 e1 n f0g . . . . . . . . . . . . . . . . . . . . . . . . . . . . Orbits in the case l 2 e2 n f0g . . . . . . . . . . . . . . . . . . . . . . . . . . . .

82 89 93 93

RR

7.1 Broken line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 7.2 Hard to control conguration: 1 = 2 . . . . . . . . . . . . . . . . . . . . . . . 107 7.3 Hard to control conguration: ( 1 ; 2 ) + ( 3 ; 2 ) =  . . . . . . . 107 8.1 Orbit an open set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 8.2 Orbit a manifold with smooth boundary. . . . . . . . . . . . . . . . . . . . . 110

408

List of Figures

8.3 8.4 8.5 8.6 8.7 8.8

Orbit a manifold with nonsmooth boundary . . . . . . . . . . . . . . . . . 110 Orbit a manifold with nonsmooth boundary . . . . . . . . . . . . . . . . . 110 Prohibited orbit: subset of non-full dimension . . . . . . . . . . . . . . . . 111 Prohibited orbit: subset with isolated boundary points . . . . . . . . 111 Approximation of ow, Th. 8.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 Phase portrait of f j in the case l 2  n e2 . . . . . . . . . . . . . . 119

R

10.1 Optimal trajectory qu~ (t) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 12.1 Hyperplane of support and normal covector to attainable set Aq0 (t1 ) at the point q1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 12.2 Transversality conditions (12:34) . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 13.1 13.2 13.3 13.4

Optimal synthesis in problem (13:1){(13:3) . . . . . . . . . . . . . . . . . . 194 Optimal synthesis in problem (13:9){(13:11) . . . . . . . . . . . . . . . . . 197 Elimination of 4 switchings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 Construction of the shortest motion for far boundary points . . . 206

15.1 Polytope U with hyperplane of support  . . . . . . . . . . . . . . . . . . . 212 17.1 Proof of Th. 17.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 19.1 Sub-Riemannian problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267 19.2 Estimate of rotation angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289 20.1 Singular arc adjacent to arc with Fuller phenomenon . . . . . . . . . . 331 21.1 Conjugate point as a fold . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 348 22.1 Complete controllability of system (22:15) . . . . . . . . . . . . . . . . . . . 361 23.1 " < 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372 23.2 " > 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372 24.1 Rolling bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377 24.2 Identication of tangent spaces at contact point . . . . . . . . . . . . . . 378 24.3 Sphere on a plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392

Index

Symbols

Du2 F , 296 Dq , 7 F , 146 L\ , 165 Lf , 154 P t, 10 T  M , 146 Tq M , 5 Tx M , 4 V1 V2 ], 17 Ad B , 37

Aq0 , 14, 63, 139 Aqt 0 (t), 14, 138 Aq0 , 138 Oq10 , 63 C (M ), 21 Di M , 4  V , 8 GL(n), 256 1M , 147 k M , 150 ; , 166 Lie F , 66 Lieq F , 66 O(n), 256 SL(C n ), 256 SL(n), 256 SO(3), 81, 275 SO(n), 99, 256 SU(3), 275 b, 4 Vec M , 6

ad V , 38 kaksK , 25 Fb, 22, 150 ; 39 exp, !1 ^ !2 , 148 GL(C n ), 256 Hessu F , 295 SU(n), 257 U(n), 256 ind+ Q, 300 ind; Q, 300

, 158 so(n), 99 ~h, 159 ; 34 exp, ;! 30, 39 exp, fa bg, 161 d!, 152 etV , 35 et ad V , 40 if , 154 s, 157

A

attainable set, 14, 138

C

Cartan's formula, 154 Cauchy's formula, 47 chronological exponential, left, 34 chronological exponential, right, 30 conjugate point, 233 conjugate time, 335, 339 connection, 380

410

Index

connection, Levi-Civita, 381 control system, control-ane, 47 control parameter, 12 control system, 12 control system, bilinear, 320 control system, completely controllable, 49 control system, left-invariant, 257 control system, linear, 47 control system, linear, Brunovsky normal form, 129 control system, linear, Kronecker indices, 128 control system, right-invariant, 257 control system, single-input, 321 control, bang-bang, 204, 286 control, singular, 205, 290 controllability, local, 320 convex hull, 144 cost functional, 137 cotangent bundle, 146 cotangent bundle, canonical coordinates, 146 cotangent bundle, canonical symplectic structure, 158 cotangent bundle, trivialization, 248 cotangent space, 146 critical point, corank, 297 curvature, 2-dimensional systems, 366 curvature, 3-dimensional control-ane systems, 376 curvature, Gaussian, 373

D

derivation of algebra, 24 di eomorphism, 3 di erential, 7 di erential form, 147, 150 di erential form, exterior di erential, 152 di erential form, integral, 147, 150 di erential form, interior product, 154 di erential form, Lie derivative, 153 di erential, second, 296 distribution, 74 distribution, 2-generating, 319 distribution, integrable, 74 distribution, integral manifold, 74 dual basis, 145

dual space, 145 Dubins car, 200 Dubins car, with angular acceleration control, 324 Dynamic Programming, 244 dynamical system, 6

E

embedding, 4 endpoint mapping, 224, 293 endpoint mapping, di erential, 306 endpoint mapping, second di erential, 308 equivalence, feedback, 122 equivalence, local state, 77 equivalence, local state-feedback, 131 exponential of vector eld, 35 exterior k-form, 148 exterior product, 148 extremal, 168 extremal trajectory, 168 extremal, abnormal, 181 extremal, nice singular, 317 extremal, normal, 181 extremal, regular, 311 extremal, strictly abnormal, 319 extremal, totally singular, 313

F

family of functionals, regularity properties, 27 family of functions, regularity properties, 26 family of vector elds, bracket-generating, 67, 109 family of vector elds, completely nonholonomic, 67 family of vector elds, full-rank, 109 family of vector elds, symmetric, 63 feedback transformation, 122 Filippov's theorem, 141 ow, 27 Frobenius condition, 75 Fuller's phenomenon, 331

G

General Position Condition, 211 group, general linear, 256 group, orthogonal, 256

Index group, special linear, 256 group, special orthogonal, 256 group, special unitary, 256 group, unitary, 256

H

Hamilton-Jacobi equation, 243 Hamiltonian, 159 Hamiltonian system, 160 Hessian, 295

I

immersion, 64 index of quadratic form, 300 integral invariant of Poincare-Cartan, 236 isotropic space, 166

J

Jacobi equation, in moving frame, 372 Jacobi equation, regular case, 337 Jacobi equation, singular case, 341 Jacobi identity, 38

L

Lagrangian subspace, 166 Lebesgue point, 174 Lie algebra, 39 Lie algebra, structural constants, 255 Lie bracket of vector elds, 17 Lie bracket, formula in coordinates, 17 Lie group, 255 Lie group, linear, 257 Lie subgroup, 258 Lie's theorem, 259 Liouville 1-form, 157

M

manifold, 2 manifold, Riemannian, 378 manifolds, di eomorphic, 3 mapping, locally open, 297 mapping, proper, 4 mapping, smooth, 3 module, 72 module, nitely generated, 72 module, locally nitely generated, 73 Morse's lemma, 299

411

N

needle-like variations, 175

O

ODE, 6 ODE, nonautonomous, 28 optimal control, 138 optimal control problem, 138 optimal trajectory, 138 optimality condition, Goh, 314, 317 optimality condition, Legendre, 310, 317 optimality condition, Legendre, generalized, 316, 317 optimality condition, Legendre, generalized, strong, 316 optimality condition, Legendre, strong, 311 optimality conditions, necessary, 167, 318 optimality conditions, sucient, 238, 241, 346, 347, 354 optimality, geometric, 293 optimality, local nite-dimensional, 297 optimality, local geometric, 297 optimality, strict strong, 347 optimality, strong, 347 orbit, 63

P

parallel translation, 380 point, Poisson stable, 116 Poisson bracket, 161 Pontryagin Maximum Principle, 167, 177, 181, 183, 188 problem, linear time-optimal, 211 problem, linear-quadratic, 223 problem, Riemannian, 265 problem, sub-Riemannian, 267, 318 problem, time-optimal, 143

R

reachable set, 14 reduction, 355 relaxation, 143 relaxed system, 144 Riemannian structure, 378 rigid body, 81 rigid body, angular velocity, 82 rigid body, Euler equations, 87

412

Index

S

space of control parameters, 12 state, 12 state space, 12 submanifold, 1 submanifold, immersed, 64 switching function, 286 symplectic manifold, 159 symplectic space, 165 symplectic structure, 159

T

tangent space, 4, 5 tangent vector, 4 tautological 1-form, 157 topology, C 1 (M ), 26 transversality conditions, 182

V

variations formula, 42 vector eld, 6, 27 vector eld, autonomous, 35 vector eld, compatible, 113 vector eld, complete, 9, 29 vector eld, conservative, 118 vector eld, ow, 10 vector eld, Hamiltonian, 159 vector eld, Hamiltonian lift, 164 vector eld, nonautonomous, 27 vector eld, parallel, 381 vector eld, Poisson stable, 116 vector eld, with compact support, 29 vector elds, commutator, 17