This book presents some facts and methods of the Mathematical Control Theory treated from the geometric point of view. T
380 111 2MB
English Pages 412 [424] Year 2010
Table of contents :
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 FullRank 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 StateFeedback 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 TimeOptimal Problem 143 10.5 Relaxations 143 11 Elements of Exterior Calculus and Symplectic Geometry 145 11.1 Differential 1Forms 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 TimeOptimal Problem 211 15.1 Problem Statement 211 15.2 Geometry of Polytopes 212 15.3 BangBang Theorem 213 15.4 Uniqueness of Optimal Controls and Extremals 215 15.5 Switchings of Optimal Control 218 16 LinearQuadratic 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, HamiltonJacobi Equation,Dynamic Programming 235 17.1 Sufficient Optimality Conditions 235 17.2 HamiltonJacobi Equation 242 17.3 Dynamic Programming 244
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 Dierential 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 dierential equations. In this book we deal only with nite dimensional systems: they are governed by ordinary dierential equations on nite dimensional smooth manifolds. A control system for us is thus a family of ordinary dierential equations. The family is parametrized by control parameters. All dierential equations of the family are dened 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 dierential 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
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state) is just a starting point of the trajectory dierent controls provide, generally speaking, dierent admissible trajectories started from a xed state. All these trajectories ll the attainable (reachable) set of the given initial state. To characterize the states reachable from a given initial one is the rst natural problem to study in Control Theory: the Controllability Problem. As soon as the possibility to reach a certain state is established, we try to do it in the best way. Namely, we try to steer the initial state to the nal one as fast as possible, or try to nd the shortest admissible trajectory connecting the initial and the nal states, or to minimize some other cost. This is the Optimal Control Problem. These two problems are our leading lights throughout the book. Why Geometry? The righthand side of the ordinary dierential equation is a vector eld and the dynamical system governed by the equation is the ow generated by this vector eld. Hence a control system is a family of vector elds. The features of control systems we study do not change under transformations induced by smooth transformations of the state space. Moreover, our systems admit a wide class of reparametrizations of the family of vector elds, which are called feedback transformations in Control Theory and gauge transformations in Geometry and Mathematical Physics. This is a formal reason why the intrinsic geometric language and geometric methods are relevant to Control Theory. There is another more fundamental reason. As we mentioned, a dynamical system is a ow (a oneparametric group of transformations of the state space) generated by a vector eld. An admissible trajectory of the control system associated to a constant control is a trajectory of the corresponding ow. Admissible trajectories associated with a piecewise constant control are realized by the composition of elements of the ows corresponding to the values of the control function. The arbitrary control case is realized via an approximation by piecewise constant controls. We see that the structure of admissible trajectories and attainable sets is intimately related to the group of transformations generated by the dynamical systems involved. In turn, groups of transformations form the heart of Geometry. Now, what could be the position of Control techniques and the Control way of thinking in Geometry and, more generally, in the study of basic structures of the world around us? A naive innitesimal version of attainable set is the set of admissible velocities formed by velocities of all admissible trajectories passing through the given state. It is usual in Control Theory for the dimension of attainable sets to be essentially greater than the dimension of the sets of admissible velocities. In particular, a generic pair of vector elds on an ndimensional manifold provides ndimensional attainable sets, where n is as big as we want. In other words, constraints on velocities do not imply state constraints. Such a situation is traditionally indicated by saying that constraints are \nonholonomic". Control theory is a discipline that systematically studies various types of behavior under nonholonomic constraints and
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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 dierential equations on manifolds are, and dene 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 eective 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 classication of linear systems, and eectively 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 dierential 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 timeoptimal 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 denition 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 sucient 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
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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 (PereslavlZalessky, 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 { PereslavlZalessky { 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 Dierential Equations and Flows on Manifolds . . . . . . . Control Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1 4 8 12
2 Elements of Chronological Calculus : : : : : : : : : : : : : : : : : : : : : : : : 21 2.1 Points, Dieomorphisms, 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 Dieomorphisms 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 Congurations :: : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 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 FullRank 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 StateFeedback 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 TimeOptimal Problem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 Relaxations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
11 Elements of Exterior Calculus and Symplectic Geometry : : 145
11.1 Dierential 1Forms. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 11.2 Dierential kForms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 11.3 Exterior Dierential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 11.4 Lie Derivative of Dierential 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
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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 TimeOptimal Problem : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 211 15.1 15.2 15.3 15.4 15.5
Problem Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 Geometry of Polytopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212 BangBang Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 Uniqueness of Optimal Controls and Extremals. . . . . . . . . . . . . . 215 Switchings of Optimal Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218
16 LinearQuadratic 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, HamiltonJacobi Equation, and Dynamic Programming : : : : : : : : : : : : : : : : : : : : : 235
17.1 Sucient Optimality Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 17.2 HamiltonJacobi 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 SubRiemannian Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267 19.3 Control of Quantum Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271 19.4 A TimeOptimal 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 Dierentiation of the Endpoint Mapping . . . . . . . . . . . . . . . . . . . 304 Necessary Optimality Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . 309
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20.5 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318 20.6 SingleInput 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 Sucient 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 2Dimensional Systems . . . . . . . . . . . . . . . . . . . . . . . 363 23.2 Curvature of 3Dimensional ControlAne Systems . . . . . . . . . . 373
24 Rolling Bodies : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 377 24.1 Geometric Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377 24.2 TwoDimensional 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
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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 . Denition 1.1. A subset M n is called a smooth kdimensional 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 vectorfunction 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 vectorfunction
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 vectorfunction
: 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 dierent onedimensional manifolds: the line 1 and the circle S 1 . The sphere S 2 and the torus 2 = S 1 S 1 are twodimensional manifolds. The torus can be viewed as a sphere with a handle. Any compact orientable twodimensional 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 dened 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 dened as follows. Denition 1.4. A smooth kdimensional 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 Denition 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 denitions. Denition 1.5. Let M and N be k and ldimensional 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
Denition 1.6. A smooth manifold M is called dieomorphic 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 dieomorphism.
4
1 Vector Fields and Control Systems on Smooth Manifolds
The set of all dieomorphisms 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 dieomorphism. 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 kdimensional 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. Denition 1.8. Let M be a smooth kdimensional submanifold of n and x 2 M its point. Then the tangent space to M at the point x is a kdimensional 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 coordinateinvariant 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. Denition 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 1st order Taylor polynomial as ( ), for any coordinate system in a neighborhood of q. _ (0) (t) (0)
Fig. 1.2. Tangent vector _ (0)
Denition 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 kdimensional manifold and q 2 M. Show that the tangent space Tq M has a natural structure of a linear kdimensional space.
6
1 Vector Fields and Control Systems on Smooth Manifolds
Denition 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. Denition 1.14. A smooth dynamical system, or an ordinary dierential 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
Denition 1.15. Let : M ! N be a smooth mapping between smooth manifolds M and N . The dierential 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 dieomorphism : M ! N : q 7! x = (q) on the vector eld V (q), take a solution q(t) of (1.1) and compute the ODE satised 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 righthand side here is the transformed vector eld on N induced by the dieomorphism :
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 dierential 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 dieomorphism.
1.3 Smooth Dierential 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 suciently 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 suciently 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 Denition 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 oneparameter 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 dieomorphisms. 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
Denition 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 sucient 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 inmum 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 dened for t 2 (;" "). But V jM nsupp V = 0, and all trajectories of V starting outside of supp V are constant, thus dened 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 aect 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 2dimensional 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 simplied 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 3dimensional 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 __ ==u02:
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. Denition 1.31. The attainable set (or reachable set) of control system (1.8) with piecewiseconstant controls from a point q0 2 M for a time t 0 is de ned as follows:
Aq0 (t) = f Pukk : : : Pu11 (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.
t0
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 P2k P1k;1 : : : P22 P11 (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 P2k P1k;1 : : : P22 P11 = P2k +:::+2 P1k;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 twodimensional 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 kdimensional 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 dened by two vector elds (1:11) on the 3dimensional manifoldM = 2xS1 .
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 congurations 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 conguration q0 = (x0 0) 2 M we can drive the car to any terminal conguration q1 = (x1 1) 2 M by alternating linear motions and rotations (both with xed velocities), see Fig. 1.9. So any point in the 3dimensional 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 lowerorder 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 rstorder derivatives @ @ = 0 @ t s=t=0 @ s s=t=0 = V2 (q) are obviously useless, as well as the pure secondorder 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 secondorder 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 eective 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 dene 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 ecient 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, innitesimally, 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 innitesimal 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 innitesimal direction of motion with a suciently 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(x1x2 ) S1 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 denition 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 innitedimensional one: the commutative algebra of smooth functions on M (for details, see 19], 22]). For basic denitions and facts of functional analysis used in this chapter, one can consult e.g. 144].
2.1 Points, Dieomorphisms, and Vector Fields In this section we identify points, dieomorphisms, and vector elds on the manifold M with functionals and operators on the algebra C 1 (M) of all smooth realvalued functions on M. Addition, multiplication, and product with constants are dened 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 denes 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.
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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 denition": 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 dieomorphism P : M ! M denes 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
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qb A = P(q) = qb Pb
8q 2 M
i.e.,
b A = P and P is the required dieomorphism. 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 identied 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) satises the multiplicative rule
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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 dierentiate this equality at t = 0 and obtain that the velocity vector to the curve of functionals
def = dd qbt : C 1 (M) ! t=0 satises the Leibniz rule:
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(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
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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 satises 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:
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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
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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
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ut
So tangent vectors v 2 Tq M can be identied 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 satises 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.
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We show that the linear functional dq def = qb D : C 1 (M) ! is a directional derivative at the point q, i.e., satises 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):
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ut
So we can identify points q 2 M, dieomorphisms 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 dieomorphism, and P(q) is the value of the dieomorphism 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).
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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 suciently 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 dene the family of seminorms k ksK on the space C 1 (M) in the following way:
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kaksK = 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 denes a topology on C 1 (M). A local base of this topology is given by the subsets
N
a 2 C 1(M) j kaknKn < n1
N
n2
where Kn, n 2 , is a chained system of compacta that cover M:
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2 Elements of Chronological Calculus
Kn Kn+1
1
n=1
Kn = M:
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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 ; aksK = 0 8 s 0 K b M: k!1 k For vector elds V 2 Vec M, we dene the seminorms
kV ksK = sup fkV aksK j kaks+1K = 1g s 0 K b M: (2.2) One can prove that any vector eld V 2 Vec M has nite seminorms kV ksK , and that there holds an estimate of the action of a dieomorphism P 2 Di M on a function a 2 C 1 (M): kPaksK CsP kaksP (K )
s 0 K b M:
(2.3)
Thus vector elds and dieomorphisms 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 oneparameter families of points, dieomorphisms, and vector elds that satisfy various regularity properties (e.g. dierentiability or absolute continuity) with respect to the parameter. Since we treat points as functionals, and dieomorphisms 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 oneparameter families of functions t 7! at
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at 2 C 1 (M) t 2 :
So we start from denitions for families of functions. Continuity and di erentiability of a family of functions at w.r.t. parameter t are dened 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 katksK dt < 1 8 s 0 K b M t0 t1 2 : t0
A family at is called absolutely continuous w.r.t. t if
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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 ksK CsK jt ; j 8s 0 K b M t 2 and locally bounded w.r.t. t if katksK CsKI 8 s 0 K b M I b t 2 I where CsK and CsKI are some constants depending on s, K, and I. Now we can dene 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 dieomorphisms 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 dened in the weak sense:
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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 dierentiable at t0, then the family At Bt is continuous, moreover, dierentiable at t0 , and satises the Leibniz rule:
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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 RightHand 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 dierential equations in n with righthand 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
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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:
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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) dierentiable in x with locally bounded partial derivatives: @ vi (t x) C t 2 I b x 2 K b Ox0 i = 1 : : : n: IK @x By the classical Carath"eodory Theorem (see e.g. 8]), the Cauchy problem (2:6) has a unique solution, i.e., a vectorfunction x(t x0), Lipschitzian w.r.t. t and smooth w.r.t. x0 , and such that: (1) ODE (2:6) is satised for almost all t, (2) initial condition holds: x(0 x0) = x0. Then the pullback of this solution from n to M
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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 satises 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 dened 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 ).
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2.4.2 Denition 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
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q(0) = q0
is satised for the family of functionals
q(t q0) : C 1 (M) !
q0 2 M t 2
(2.7)
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constructed in the previous subsection. We prove later that this Cauchy problem has no other solutions (see Proposition 2.9). Thus the ow dened as P t : q0 7! q(t q0) is a unique solution of the operator Cauchy problem
(2.8)
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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 justies such a notation.
2.4.3 Formal Series Expansion We rewrite dierential 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 righthand side = q0 +
Z t
q0 +
0 Z t
= q0 Id +
0
Z
1
0
q( 2 ) V2 d 2 V1 d 1
ZZ
V dt +
02 1 t
q( 2 ) V2 V1 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 Vn V d n : : : d 1C A+
n (t) Z Z 2
1
2
1
q( n+1 ) Vn+1 V1 d n+1 : : : d 1: (2.11)
n+1 (t)
Here
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n(t) = f( 1 : : : n ) 2 n j 0 n 1 tg is the ndimensional 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 Vn V d n : : : d 1C @Id + A
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1
n=1 n (t)
thus for the solution P t to problem (2:9): Id +
1Z X
Z
Vn V1 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 dened 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 + (;Vn ) (;V1 ) 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 mth partial sum of series (2:12) as Sm (t) = Id +
mX ;1 Z
Z
Vn V1 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 sK Z t R C t kV ksK0 d 1 Ce
m
m! 0 kV ks+m;1K 0 d kaks+mK 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 sK
" ! 0
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2 Elements of Chronological Calculus
where Sm" (t) is the mth 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
Vn V1 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
02 1 t
V2 V1 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 welldened 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
Vn V1 a d n : : : d 1:
n=1 n (t)
(2.16)
Z Z Vn V a d n : : : d 1 n(t) Z Z 1
0n 1 t
kVn k kV1 k d n : : : d 1 kak
by symmetry w.r.t. permutations of indices : f1 : : : ng ! f1 : : : ng
Z
Z
kVn k kV1 k d n : : : d 1 kak 0(n) (1) t
=
passing to the integral over cube
Zt Zt = n!1 : : : kVn k kV1 k d n : : : d 1 kak Z0 t 0 n 1 = n! kV k d kak: 0
2.4 Chronological Exponential
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So series (2:16) is majorized by the exponential series, thus the operator series (2:14) converges on L. Series (2:16) can be dierentiated termwise, thus it satises 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)
Vn V1 a d n : : : d 1:
So in the case (2:15) the asymptotic series converges to the chronological exponential and there holds the bound
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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:
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Notice that conditions (2:15) are satised for any nitedimensional 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 suciently 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 dierentiation 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
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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 righthand side: Qt = Id + = Id + = Id +
Zt Z0t 0
(;V ) Q d (;V ) d +
mX ;1 Z
Z
ZZ
2 (t)
(;V1 ) (;V2 ) Q2 d 2 d 1 =
(;V1 ) (;Vn ) d n : : : d 1
n=1 n (t)
Z
Z
+ (;V1 ) (;Vm ) Qm 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
(;V1 ) (;Vn ) 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 (;Vn ), n : : : 1 . (2) The right and left chronological exponentials satisfy the corresponding dierential 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 righthand 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!
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2 Elements of Chronological Calculus
i.e, it is the standard exponential series. The exponential of an autonomous vector eld satises 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 ij =1
Similarly,
q etV esW e;tV = q (Id +tV + ) (Id +sW + ) (Id ;tV + ) = q (Id +sW + tsV 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 Dieomorphisms on Vector Fields
37
We have already found counterparts to points, dieomorphisms, and vector elds among functionals and operators on C 1 (M). Now we consider action of dieomorphisms on vector elds. Take a tangent vector v 2 Tq M and a dieomorphism 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 Pv 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) PV = (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 dieomorphisms act on vector elds as similarities. In particular, dieomorphisms 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 = PV 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 innitesimal 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: Dierentiation 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) skewsymmetry:
R
X Y ] = ;Y X]
X Y 2 Vec M
2.5 Action of Di eomorphisms on Vector Fields
39
(3) Jacobi identity (2:20). In other words, the set Vec M of all smooth vector elds on a smooth manifold M forms a Lie algebra . ;! Z t V d of a nonautonomous vector eld Vt . Consider the ow P t = exp 0 We nd an ODE for the family of operators Ad P t = (P t); 1 on the Lie algebra Vec M. d (Ad P t)X = d ;P t X (P t);1 dt dt = P t Vt X (P t);1 ; P t X Vt (P t);1 = (Ad P t)Vt X] = (Ad P t) ad Vt X X 2 Vec M: Thus the family of operators Ad P t satises the ODE d t t (2.21) d t Ad P = (Ad P ) ad Vt with the initial condition Ad P 0 = Id : (2.22) So the family Ad P t is an invertible solution for the Cauchy problem A_ t = At adVt A0 = Id for operators At : Vec M ! Vec M. We can apply the same argument as for the analogous problem (2:9) for ows to derive the asymptotic expansion Ad P t Id +
Zt 0
adV d +
Z
Z
+ ad Vn ad V1 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 V1 ) (; ad Vn ) 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 Vn ad V1 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 sK Z t R C t kV ks K0 d 1 C1e
1 0
+1
m!
0
kV ks+mK 0 d
m
kX ks+mK 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. PtW = 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 sucient. 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 Vthed commutativity t = eR0t V d dened 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 righthand side satises the same ODE as the chronological exponential in the lefthand 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 righthand 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 veried by dierentiation. In view of uniqueness, it gives the solution to the Cauchy problem. Linear system (3:1) is a particular case of a controlane 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 ane. Thus the attainable set Ax0 (t) of linear system (3:1) for a xed time t > 0 is an ane subspace in n. Denition 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 denition 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 ane 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 CayleyHamilton 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. Suciency. 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 ith 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 dierentiate this equality repeatedly at s = 0 and obtain pAk bi = 0
k = 0 1 : : : i = 1 : : : m
a contradiction with (3:10). Suciency 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 dieomorphism (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 dieomorphisms: 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 = f0 :{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 satised 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 ndimensional 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 dieomorphism 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 ndimensional 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 dieomorphisms. Suciency. 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 dened on a suciently 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 dieomorphism of a neighborhood of 0 in n and a neighborhood of q0 in M, according to the implicit function theorem. Now we prove that recties 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 dieomorphisms, and for controllable linear systems conditions (4:2), (4:3) hold. Suciency. 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 dieomorphism: : Oq0 ! O0 Y = @ @x = 1 : : : n: We show that is the required dieomorphism. (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 ndimensional 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 kdimensional 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
dieomorphic to n, i.e., k = 0. (2) If, on the contrary, M is compact, i.e., dieomorphic 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. Suciency. 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 + aX 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 dene the \inverse" of the required dieomorphism 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 dierential 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 dieomorphism. 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
qe 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 xY 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 xY = 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 dieomorphism. (7) The mapping is welldened 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 welldened. (8) The mapping (4:11) is onetoone: if x y 2 n (x) = (y) then P P q0 e n=1 x Y = q0 e n=1 yY 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 dieomorphism. 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 dieomorphism = ;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 welldened on the quotient k n;k, that is why equalities (4:6) hold. The suciency 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 satised. 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 ndimensional 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 satises equality (4:12) since any element that satises (4:12) can be translated to the strip ;1 (I) by elements of the lattice ;1. That is why a suciently 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 denitions 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 dieomorphisms of M generated by ows in F : P = fet1 f1 etk fk j ti 2 fi 2 F k 2 g Di M: We dene 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 Denition 5.2. A subset W of a smooth ndimensional manifold is called an immersed kdimensional submanifold of M , k n, if there exists a onetoone immersion : N ! M Ker q = 0 8 q 2 N of a kdimensional smooth manifold N such that
W = (N): Remark 5.3. An immersed submanifold W of M can also be dened as a manifold contained in M such that the inclusion mapping i : W ! M i : q 7! q is an immersion. Suciently 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 Denition 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 onetoone immersion of the line into the plane such that limt!+1 (t) = (0). Then W = ( ) is an immersed onedimensional 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 Denition 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_ = iw w2 and their solutions have the form z(t) = ei tz(0) w(t) = eit 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 = Imq : 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 = spanff1 : : :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) Denition 5.10. A family F Vec M that satis es property (5:3) is called
completely nonholonomic or bracketgenerating. 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 bracketgenerating 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. dimq = dimq0 for all q 2 Oq0 , q0 2 M . Proof. If q 2 Oq0 , then q = q0 Q for some dieomorphism 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 dimq0 dimq . But q0 and q can be switched, that is why dimq dimq0 . Finally, dimq = 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 = dimq 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 suciently 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 mdimensional 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 Gt(Tt m) = G(t), t 2 O0. Since rank GtjO0 = 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 dimq^ = 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 suciently 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 suciently 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) suciently 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 dened 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 dierent 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 nonanalytic. 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 coecients from C 1(M). We dene 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, dene 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 denition 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. Denition 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. Denition 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 dimq = 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 onedimensional submanifolds with tangent spaces q . But in the general case (dimq > 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 dimq = 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 dimq = 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 + af 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 nonconstant dimq . Such a distribution is dened 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, dierent, 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 dimq > 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 bracketgenerating: Liex0 fU = Tx0 M Liey0 gU = Ty0 N:
5.7 State Equivalence of Control Systems
77
Denition 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: dene 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 conguration of vectors f u1:::uk g to the conguration fu1:::uk g. It turns out that in the analytic case this condition is sucient. 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 bracketgenerating 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 = fu1:::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 bracketgenerating: 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 conguration 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 conguration should be expressed through the base vectors with the same coecients as the conguration , 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 congurations and as in (5:15) and (5:17) such that decompositions (5:16) and (5:18) with the same coecients ciu1 :::uk hold, then there exists a linear isomorphism A with (5:14). Indeed, we dene 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. Congurations f u1 :::uk g and fu1:::uk g are called equivalent if the sets of relations K(fU ) and K(gU ) between elements of these congurations coincide: K(fU ) = K(gU ). We denote here by K(fU ) the set of all systems of coecients 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 bracketgenerating families of analytic vector elds is contained in Lie brackets.
Notice, although, that the conguration 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 suciency 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, dene their direct product f g 2 Vec(M N) as the derivation (f g)aj(xy) = (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 dened 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 dene 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 conguration to as in (5:14), and construct the local equivalence (fU x0) ' (gU y0). In view of denition (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 ndimensional 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 dieomorphisms since the dierentials 1j(x0 y0 ) : (v Av) 7! v v 2 Tx0 M 2j(x0 y0 ) : (v Av) 7! Av v 2 Tx0 M are onetoone. Now = 2 (1jO );1 is a local dieomorphism 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 threedimensional 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 simplied 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 threedimensional space with positive orientation. This set can be identied with SO(3), the group of linear orthogonal orientationpreserving 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 dierentiate 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 (Qtt x Qty) + (Qt x Qtty) = 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 = Qtt 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
ij =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 hAt 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 = Qtt 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(:" ) = hAt Uti dt 0 "=0
by (6:3) = by (6:8) =
Zt
1
0
Zt
1
0
h(Ad Qt)At (Ad Qt)Ut i dt
h(Ad Qt)At 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 )At = 0 t 2 0 t1]: Hence Ad Qt(t At] + A_ t ) = 0 t 2 0 t1] that is why A_ t = At t] t 2 0 t1]: (6.10) Introduce the operator Mt = At
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 orientationpreserving n n matrices with an arbitrary n > 0. Then we come to equations of rotations of a generalized ndimensional rigid body. Now we rewrite Euler equations via isomorphism (6:2) of so(3) and 3, which is essentially 3dimensional 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 = Qtbt 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 denite 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)1132:
(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 121 + 2 22 + 3 23 = const
(6.13)
i.e., belong to intersection of spheres with ellipsoids. Moreover, since the dierential 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 2dimensional 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 bt
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 6dimensional control system (6:20). But before that we study orbits of the 3dimensional system (6:19).
6.4.1 Orbits of the 3Dimensional 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 3dimensional 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 3dimensional. We proved the following statement. Proposition 6.4. Assume that l 2 + n e2. Then system (6:19) has one 2dimensional orbit, the plane + , and two 3dimensional 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 2dimensional. 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 1dimensional orbit  the line e1 (C = 0), and an in nite number of 2dimensional 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 1dimensional orbit  the line e3, and an in nite number of 2dimensional 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 1dimensional orbit  the line e2, and an in nite number of 2dimensional orbits of the following
R
R
form: (1) connected components of hyperbolic cylinders (6:22) for C 6= 0 (2) halfplanes  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 3dimensional system (6:19). Now we study orbits of the full 6dimensional system (6:20).
6.4.2 Orbits of the 6Dimensional System The vector elds in the righthand side of the 6dimensional system (6:20) are
Qb 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 ) = Qv w() i @ w 1 0 @ w 2 1 b b b BB Qw1 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 3dimensional 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 6dimensional system (6:20) through a point (Q ) 2 SO(3) 3, we follow the dierent cases for the 3dimensional 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 6dimensional. 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 ) = QM10 M2 ] and the orbit through the point (Q ) is 6dimensional. 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 6dimensional. 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 2dimensional, the orbit in SO(3) 3 is 5dimensional. 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 2dimensional 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 6dimensional system (6:20) is 5dimensional (it cannot have dimension 6 since the 3dimensional system (6:19) has a 2dimensional 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 2dimensional. 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 2dimensional, then the orbit of the 6dimensional system (6:20) through the point (Q ) is SO(3) O, i.e., respectively 6 or 5dimensional. If dim O = 1, then the 6dimensional system has a 2dimensional 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 Congurations
In this chapter we apply the Orbit Theorem to systems which can be controlled by the change of their conguration, i.e., of relative position of parts of the systems. A falling cat exhibits a wellknown 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 conguration. 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 conguration. 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 Congurations
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:
Dierentiation of this equality at t = 0 yields
hAx yi + hx Ayi = 0
99
x y
i.e., the matrix A is skewsymmetric:
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 orientationpreserving linear orthogonal transformations of the Euclidean space n is denoted by SO(n), and the corresponding Lie algebra of skewsymmetric 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 Congurations
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) dene a linear in control, thus symmetric, control system on M. Notice that a more general condition (7:2) determines an \ane distribution", and control system (7:3), (7:2) is controlane, 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 onedimensional, 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 congurations 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 Congurations
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) denes one linear equation on velocities, thus a rank 3 distribution on a 4dimensional 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 4dimensional vectors as 2dimensional 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 Congurations
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 = 23 (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 4dimensional 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 conguration (~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 2dimensional 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. Congurations of 3 points that dene distinct 2dimensional planes (or dene distinct orientations in the same 2dimensional plane) are not mutually reachable: attainable sets from these congurations do not intersect one with another. Although, if two congurations dene 2dimensional planes having a common line, then intersection of closures of attainable sets from these congurations is nonempty: it consists of collinear triples lying in the common line. Theoretically, one can imagine a motion that steers one conguration into another: rst the 3 points are made collinear in the initial 2dimensional plane, and then this collinear conguration is steered to the nal one in the terminal 2dimensional 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 2dimensional 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 = eij
j 2 S 1 j = 1 2 3: Position of the system is determined by the 3tuple of angles ( 1 2 3), so the state space is the 3dimensional 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 dierential form:
106
7 Control of Congurations
!=
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 = heij eik 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 3dimensional manifold M = 3. The orbits can be 2 or 3dimensional. 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 dierential forms. Assume that the distribution has a 2dimensional 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 dierential forms, (! ^ d!)q = 0 q 2 N: We compute the dierential 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 conguration: 1 = 2
107
Fig. 7.3. Hard to control conguration: (1 ; 2 ) + (3 ; 2 ) =
see Figs. 7.2, 7.3. Congurations (7:10) and (7:11) are hard to control: if neither of these equalities is satised, then ! ^ d! 6= 0, i.e., the system has 3dimensional 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 twodimensional surface P = f 3 = 1 g: If the orbit through a point q 2 P is twodimensional, 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 3dimensional 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 3dimensional 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 bracketgenerating , or fullrank . 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 bracketgenerating, and M and F are real analytic, we can pass from F to a bracketgenerating 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 FullRank Systems For bracketgenerating systems both orbits and attainable sets are fulldimensional. Moreover, there holds the following important statement. Theorem 8.1 (Krener). If F Vec M is a bracketgenerating 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 bracketgenerating family to be: a lowerdimensional 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 1dimensional submanifold of M for small enough "1 > 0.
8.1 Attainable Sets of FullRank Systems
M Fig. 8.5. Prohibited orbit: subset of nonfull dimension
111
M Fig. 8.6. Prohibited orbit: subset with isolated boundary points
If dimM = 1, then q0 es1 f1 2 int Aq0 for suciently 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 suciently 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 suciently small "2 , thus its image is a 2dimensional 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) suciently 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 bracketgenerating. 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 suciently 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 ndimensional 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 bracketgenerating 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 bracketgenerating, 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 bracketgenerating 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
Denition 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 bracketgenerating 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 Zn 2 fX Y g, i.e., Zn = 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 Zn 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 Zn 2 fX Y g such that
Zt 0
Zn 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+1K , 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, dene the following vector eld: X 2 F n Z = Y 2 Eij n F : ij ij Then the sequence of vector elds Zn is the required one. ut Now we prove the second part of Theorem 8.7. Lemma 8.10. Let Zn , 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
Zn d !
Zt 0
Zt 0
Z d
;! Zn 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
Zn d :
Pn Zn d
integrating by parts
Since
Zt
= Id +Ptn
Zt 0
Zn d ;
Z t 0
Pn Zn
Z 0
Z d d :
Zn 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: Zn = Z + Vn
;! Denote Ptn = exp ;! exp
Zt 0
Zt 0
Zt 0
Vn d ! 0
n ! 1:
Vn d . From the variations formula, we have
;! Zn d = exp
Zt 0
;! (Vn + Z ) d = exp
Zt 0
Ad Pn 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
;! Zn d ! exp
Zt 0
Z d :
ut
So we proved Theorem 8.7 and thus Proposition 8.5.
8.3 Poisson Stability Denition 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. Denition 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 bracketgenerating 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 bracketgenerating, 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 (Poincare). 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 leftinvariant 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 bracketgenerating. Then the symmetric system span(f g) is bracketgenerating as well, thus completely controllable. Hence the initial system (8:5) is completely controllable in the bracketgenerating case. In the nonbracketgenerating 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 halfplanes 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 diers from the bracketgenerating case since the vector eld f preserves volume in 3, but not in . A detailed analysis of the controllability problem in the nonbracketgenerating 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 righthand 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., dieomorphisms 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
Denition 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 controlane 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 controlane 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 controlane 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 righthand 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 highorder 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 statefeedback 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 statefeedback 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 nth 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 dierential equations: y_i = yi+1 i = 1 : : : n ; 1 are fullled 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 satised for a unique ntuple (0 : : : n;1) since the vectors b, Ab, : : : , An;1b form a basis of n (in fact, i are coecients 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 statefeedback equivalent to the normal form (9:11), thus there are no statefeedback invariants in a given dimension n. If k > 1, this is not the case, and we start from description of statefeedback 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 statefeedback invariants. Now we express the numbers dimDm , m = 1 : : : n, through other integers  Kronecker indices. Construct the following nk matrix whose elements are ndimensional 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 ith 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 \blocktriangular" 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 statefeedback invariants.
Brunovsky Normal Form Now we nd normal forms of linear systems under state and statefeedback transformations. In particular, we show that Kronecker indices form a complete set of statefeedback 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 > > : : 1j k 1j k 0in ;1 0in ;1 1
1
1
j
j
where
x=
X 1ik 1j ni
yji eij
(9.28)
(9.29)
and statefeedback 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 scalarinput 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
1ik 1j ni
In view of the required equations
1ik 1j ni
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9 Feedback and State Equivalence of Control Systems
y_ji = yji +1
X
we have or, equivalently,
1ik 1j 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 C0 t1] as n ! 1: Now we show that q(t) is an admissible trajectory of control system (10:1). Fix a suciently 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 tt+"]
R
q2Oq(t) (c")
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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 dierentiability 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 modied argument. ut Remark 10.5. In Filippov's theorem, the hypothesis of common compact support of the vector elds in the righthand side is essential to ensure the uniform boundedness of velocities and completeness of vector elds. On a manifold, sucient 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 righthand side by a cuto 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 wellknown sucient conditions for completeness of vector elds: if the righthand side grows at innity 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 righthand 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 sucient 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 righthand 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 TimeOptimal Problem Given a pair of points q0 2 M and q1 2 Aq0 , the timeoptimal 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, timeoptimal 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).
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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 ndimensional linear space is contained in the convex hull of some n + 1 points in S. Lemma 10.8 (Caratheodory). 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 ndimensional 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 righthand 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 dierential 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 Dierential 1Forms 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 e1 : : : en such that
hei 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. Dierentials 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) dene local coordinates on the cotangent bundle. Consequently, T M is an 2ndimensional 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 dierential F : Tq M ! TF (q) N has the adjoint mapping F def = (F ) : TF (q) N ! Tq M
dened as follows:
11.2 Di erential kForms
147
F = F 2 TF (q) N hF vi = h Fvi v 2 Tq M: A vector v 2 Tq M is pushed forward by the dierential F to the vector Fv 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 Dierential 1Forms
A di erential 1form 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 dierential 1forms on M has a natural structure of an innitedimensional vector space denoted as 1 M. Like linear forms on a vector space are dual objects to vectors of the space, dierential forms on a manifold are dual objects to smooth curves in the manifold. The pairing operation is the integral of a dierential 1form ! 2 1M along a smooth oriented curve : t0 t1] ! M, dened as follows:
Z
! def =
Zt
1
t0
h! (t) _ (t)i dt:
The integral of a 1form along a curve does not change under orientationpreserving smooth reparametrizations of the curve and changes its sign under change of orientation.
11.2 Dierential kForms A dierential kform on M is an object to integrate over kdimensional surfaces in M. Innitesimally, a kdimensional surface is presented by its tangent space, i.e., a kdimensional subspace in Tq M. We thus need a dual object to the set of kdimensional subspaces in the linear space. Fix a linear space E. A kdimensional subspace is dened 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 kdimensional 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 )kij =1, and j =1 that determinant of a k k matrix is a multilinear skewsymmetric form of the columns of the matrix. This is why the following denition of the \dual objects" is quite natural.
R
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11 Elements of Exterior Calculus and Symplectic Geometry
11.2.1 Exterior kForms
N
Let E be a nitedimensional real vector space, dimE = n, and let k 2 . An exterior kform 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 skewsymmetric:
!(v1 : : : vi : : : vj : : : vk ) = ;!(v1 : : : vj : : : vi : : : vk ) i j = 1 : : : k: The set of all exterior kforms on E is denoted by k E. By the skewsymmetry, 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 1forms !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 skewsymmetric form. The exterior product is the antisymmetrization 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 kForms
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 skewcommutative: !2 ^ !1 = (;1)k1 k2 !1 ^ !2
!i 2 ki E:
Let e1 : : : en be a basis of the space E and e1 ;: :: en the corresponding dual basis of E . If 1 k n, then the following nk elements form a basis of the space k E: ei1 ^ : : : ^ eik
1 i1 < i2 < < ik n:
The equalities
(ei1 ^ : : : ^ eik )(ei1 : : : eik ) = 1 (ei1 ^ : : : ^ eik )(ej1 : : : ejk ) = 0 if (i1 : : : ik ) 6= (j1 : : : jk )
for 1 i1 < i2 < < ik n imply that any kform ! 2 k E has a unique decomposition of the form !=
X
1i1 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 satises 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 hu ( 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) satises the boundary conditions. In view of existence, these extremal trajectories are optimal.
13.5 Dubins Car In this section we study a timeoptimal 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 conguration 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 inmum 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 inmum 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 timeoptimal 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 S1 , let ( 1 2 ) be the corresponding coordinates of the adjoint vector. Then = (x ) 2 T M
R
and the controldependent 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: uj1
(13.24)
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13 Examples of Optimal Control Problems
Equation (13:22) means that is constant along optimal trajectories, thus the righthand 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 ) + satises 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 bangbang . 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 bangbang 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 01 ) 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(01) = 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 bangbang piece (arc of a circle, u = 1), a singular piece (segment of a line, u = 0), and a bangbang piece, or (2) concatenation of bangbang 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 suciently 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 wellknown 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 suciently 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 controldependent Hamiltonian of PMP has the form: hu (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 suciently 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 satises 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 satised. 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 1periodic 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 TimeOptimal 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 timeoptimal 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 satises 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 timeoptimal 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 TimeOptimal Problem
Before proceeding with the study of linear timeoptimalproblems, 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 ane 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 dene , 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 BangBang Theorem
213
Such polytopes U \ are called faces of the polytope U. Zerodimensional and onedimensional 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 HahnBanach Theorem).
15.3 BangBang Theorem Optimal control in the linear timeoptimal problem is bangbang, 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 timeoptimal 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 0t1 ]nT is locally constant. Proof. Apply Pontryagin Maximum Principle to the linear timeoptimal 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 controldependent 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 TimeOptimal 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. Dene the set T = ft 2 0 t1] j #J(t) > 1g: By contradiction, suppose that T is innite: 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 innite number of zeros on the segment 0 t1], thus it is identically zero:
(0)e;tA Be 0: We dierentiate 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 timeoptimal 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 TimeOptimal 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 timeoptimal 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 timeoptimal 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 bangbang, 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 TimeOptimal 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: nonoscillating 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) = fig 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 fulldimensional for arbitrarily small times. We can evaluate the minimal number of switchings necessary to ll a fulldimensional 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 1dimensional curve e(Ax+ b)tx0, with 1 switching we ll a piece of a 2dimensional surface e(Ax+b)t2 e(Ax+ b)t1 x0 , with 2 switchings we can attain points in a 3dimensional surface, etc. So the minimal number of switchings required to reach an ndimensional domain is n ; 1. We prove now that in the nonoscillating 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 timeoptimal 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 satises 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) satises an nth 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
ei 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 TimeOptimal Problem
Proof. Apply induction on k.
If k = 1, then a quasipolynomial y(t) = et 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
ei t Pi (t) + ek tPk (t) = 0
e(i ;k )t Pi(t) + Pk (t) = 0:
(15.17)
The quasipolynomial in the lefthand side has n roots. We dierentiate this quasipolynomial successively (nk + 1) times so that the polynomial Pk (t) disappear. After (nk + 1) dierentiations 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 nonoscillating 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 fulldimensional for all t > 0). For an arbitrary matrix A, one can obtain the upper bound of (n ; 1) switchings for suciently 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 in jcij:
221
R
Then for any timeoptimal 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 coecients:
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 TimeOptimal 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 tt+]
n i X ci i! sup y(n) (t)
i=1
t2 tt+]
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 satises 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 LinearQuadratic Problem
16.1 Problem Statement In this chapter we study a class of optimal control problems very popular in applications, linearquadratic 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)
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16 LinearQuadratic Problem
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 linearquadratic problems is a nontrivial problem. In this chapter we assume that admissible controls are squareintegrable: 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 ane 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 ane, 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 (00) 0 is necessary for existence of optimal controls at least for one pair (x0 x1), and the strict inequality J jU (00)n0 > 0 is sucient 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 inmum 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
16 LinearQuadratic Problem
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 ane 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 linearquadratic problems to the study of the restriction J jU (00). 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 linearquadratic 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 linearquadratic problem. The controldependent 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 satises the ODE _ = ; A, thus (t) = (0)e;tA . The maximality condition implies that 0 (t)B = (0)e;tA B. We dierentiate 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 controldependent 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 linearquadratic problem indeed satisfy equations (16:5){(16:7). Consider the extended system
228
16 LinearQuadratic Problem
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 dierential 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 dierential 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). Dierentiating (16:12), we arrive at the last required equality (16:7). So we proved that optimal trajectories in the linearquadratic 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
16 LinearQuadratic Problem
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 CauchySchwartz 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
16 LinearQuadratic Problem
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 fulldimensional 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, inmum 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 satises 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 satises 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 tt0 ] 0: Consequently, the control b( ) t u0 ( ) = u0 2 t t0] satises the conditions: u0 2 Ut0 (0 0) ku0k = 1 Jt0 (u0 ) = 0: Thus u0 satises PMP, i.e.,
u0 ( ) = B ( ) 2 0 t0] is an analytic function. But u0j tt0] 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. Denition 16.6. A point t > 0 is conjugate to 0 for the linearquadratic 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
16 LinearQuadratic Problem
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 linearquadratic problems. Before the rst conjugate point, optimal control exists and is unique for any boundary conditions (if there are two optimal controls, then their dierence 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, HamiltonJacobi Equation, and Dynamic Programming
17.1 Sucient Optimality Conditions Pontryagin Maximum Principle is a universal and powerful necessary optimality condition, but the theory of sucient optimality conditions is not so complete. In this section we consider an approach to sucient 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 controldependent 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 dened 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 Sucient Optimality Conditions
17.1.1 Integral Invariant First we consider a general construction that will play a key role in the proof of sucient optimality conditions. Fix an arbitrary smooth function a 2 C 1 (M):
Then the graph of dierential 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 ndimensional 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 1form
R
R
s ; H dt 2 1 (T M ): Recall that s is the tautological 1form on T M, s = , and its dierential 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 PoincareCartan 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 Sucient 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 Sucient Optimality Conditions
17.1.2 Problem with Fixed Time Now we prove sucient 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], dieomorphic property of jLt ) hold in a proper open domain O T M, then the statement can be modied to give local optimality of q~( ) in (O). These modications 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 dieomorphism, 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 Sucient 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 Sucient 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 controldependent Hamiltonians hu(p q) are ane 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
denes 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 Sucient optimality condition of Theorem 17.2 is given in terms of the manifolds Lt, which are in turn dened 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 dieomorphisms. For t = 0 the projection : L0 ! M is a dieomorphism. So for small t > 0 any function a 2 C 1 (M) provides manifolds Lt projecting dieomorphically to M, at least if we restrict ourselves by a compact K b M. Thus the sucient 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 suciently 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 sucient 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 Sucient 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 dieomorphism 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 sucient 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 dieomorphism 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 Sucient 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 diers from q~.
17.2 HamiltonJacobi Equation Suppose that conditions of Theorem 17.2 are satised. As we showed in the proof of this theorem, the form (s ; H dt)jL is exact, thus it coincides with dierential of some function:
R
(s ; H dt)jL = dg
R
R
g : L! :
R
(17.10)
Since the projection : Lt ! M is onetoone, we can identify ( t) 2 L with (q t) 2 M and dene 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 HamiltonJacobi 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 denition (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 dierential 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 dierential equation: @ g + H @ g x = 0 (17.13) @t @x which is called HamiltonJacobi equation . We showed that the optimal cost g(x t) satises HamiltonJacobiequation (17:13) with initial condition (17:12). Characteristic equations of PDE (17:13) have the form
244
17 Sucient 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 HamiltonJacobi equation for optimal cost.
17.3 Dynamic Programming One can derive the HamiltonJacobi 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 1smooth. 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 satises 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 HamiltonJacobi 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 HamiltonJacobi 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 bracketgenerating. 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 dierential 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 ndimensional vector space. Suppose that we have a trivialization of the cotangent bundle T M, i.e., a dieomorphism : 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 identied 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 dierential 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 pullback bs 2 1(E M): The tangent and cotangent spaces are naturally identied with the direct products: T(eq) (E M) = Te E ! Tq M = E ! Tq M T(eq) (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 dierential 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 (eq) 2 E !h (eq) 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 dierential forms on M (possibly depending on e 2 E) can be considered as horizontal vector elds and dierential forms on E M: Tq M = 0 ! Tq M T(eq) (E M) Tq M = 0 ! Tq M T(eq) (E M):
Compute the action of the form bs on a tangent vector ( v) 2 Te E ! Tq M:
hbs ( v)i = hs(eq) ( v)i = hs(eq) ( v)i = h(e q) vi:
Thus
(bs)(eq) = (e q)
(18.2)
where in the righthand side of (18:2) is considered as a horizontal form on E M. We go on and compute the pullback of the standard symplectic form: b = bds = dbs = d:
Recall that dierential 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
= h1 V2i + V1 he V2i: Consequently, d(( 1 V1 ) ( 2 V2))(e ) = h1 V2i ; h2 V1 i + V1he V2i ; V2 he V1i ; he V1 V2 ]i: We denote the rst two terms e(( 1 V1) ( 2 V2 )) = h1 V2i ; h2 V1i and apply formula (18:3) to the horizontal form e: de(V1 V2 ) = V1 he V2i ; V2 he V1 i ; he V1 V2]i: Finally, we obtain the expression for pullback of the symplectic form: b(e ) (( 1 V1) ( 2 V2)) = e(( 1 V1 ) ( 2 V2)) + de (V1 V2) (18.4) i.e., b(e ) = e + de: Remark 18.1. In the case of canonical coordinates we can take test vector elds Vi = @@xi , then it follows that de = 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 denition 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 + de ) = i(XY ) e e + i(XY ) e de de} : = hX{z }i ; h {z Y i} + iY {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 de = ;dhe rewrite as X = ;iY de ; dhe and solve this equation w.r.t. X: X = ;;1 (iY de + 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 jk=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 simplies even more when ckij = const i j k = 1 : : : n: Both these conditions are satised 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 1forms !1 : : : !n 2 1(M):
h!i Vj i = ij Then
d!k =
if and only if
n 1 X
ij =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 lefthand 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 coecients ckij are constant, then the vector elds V1 : : : Vn span a nitedimensional 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 nitedimensional 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 volumepreserving 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 orientationpreserving 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 denitions 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 congurations. 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 leftinvariant vector elds is leftinvariant as well: XA XB] = XA B]
(18.9)
this follows easily from the coordinate expression for commutator (exercise). Remark 18.4. Instead of leftinvariant systems X_ = XA, we can consider rightinvariant 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 denition, 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 leftinvariant 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) dened 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 welldened 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 leftinvariant 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 onetoone 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 leftinvariant systems dened 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 dieomorphism 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 leftinvariant vector eld on M: V 2 Vec M q% V = V q 2 M: There is a onetoone correspondence between leftinvariant vector elds on M and tangent vectors to M at identity: V 7! V (Id) = v: Left translations in M preserve ows of leftinvariant vector elds on M, thus ows of their commutators. Consequently, leftinvariant 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 onetoone 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 leftinvariant 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 leftinvariant 1form 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 leftinvariant control problems, where the Hamiltonian h of PMP is leftinvariant, 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) simplies 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 leftinvariant forms on M:
R
hv wi =
R
R
Z
vw !1 ^ : : : ^ !n vw (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 skewsymmetric 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) dened 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 denition of the invariant scalar product on so(N) provides positivedeniteness 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
ij =1
The norm on so(N) is naturally dened:
p
jaj = ha ai
ij =1
a 2 so(N): The innitesimal version of the invariance property (18:16) is easily obtained by dierentiation at t = 0: hc 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 skewsymmetric 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 identied 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 identication, the operator (ad a) , a 2 M, takes the form: (ad a) : M ! M
(ada) = ; ad a:
In the case of a compact Lie group M, Hamiltonian system (18:15) for an invariant Hamiltonian h = h(a) becomes dened 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 denes a leftinvariant 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 dened 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 CauchySchwartz 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: hu(a q) = ha%q qui + 2 juj2 = ha ui + 2 juj2: The maximality condition of PMP is: 2 hu(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 leftinvariance 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 oneparameter 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 SubRiemannian 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 SubRiemannian 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 leftinvariant distribution (of corank 1) on M. That is, we dene a leftinvariant 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 subRiemannian 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. SubRiemannian 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: hu (a q) = ha ui + 2 juj2 but the maximality condition diers since now the set U is smaller: hu(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 bib a]:
The second equation has an integral of the form ha bi = const this is easily veried by dierentiation w.r.t. this equation: d d t ha bi = ha bihb 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 bib 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 SubRiemannian 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 oneparameter 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 = ha 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 skewsymmetric 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 aect 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 skewsymmetric 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 subRiemannian 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 subRiemannian 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 threelevel 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 inuence 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 nitedimensional problem can be thought as the reduction of an innitedimensional problem in the following way. We start with a Hamiltonian which is the sum of a driftterm 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
j1 (t)j2 + j2(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 \realresonant" problem. The rst problem (with arbitrary complex controls) will be called the \generalcomplex" problem. Since Hamiltonian (19:18) is selfadjoint: H = H, it follows that Schr#odinger equation (19:17) is welldened on the unit sphere " # SC = S 5 = = (1 2 3) 2 3 j jj2 = j1 j2 + j2j2 + j3j2 = 1 : The source and the target, i.e., the initial and the terminal manifolds in the generalcomplex 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 claried later. Summing up, the generalcomplex problem is stated as follows: i_ = H 2 S 5 1 2 2 (0) 2 SCd (t1 ) 2 TCd
C
Zt ; j1j2 + j2j2 dt ! min 1
0
with the Hamiltonian H dened by (19:18). For the realresonant 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 ane in control system (19:17), (19:18) to a system linear in control, both in the generalcomplex and realresonant cases. For 2 , denote by Mj () and Nj () the n n matrices:
C
Mj ()kl = jk j +1l + j +1k jl Nj ()kl = jk j +1l ; j +1k jl
j = 1 2
(19.22)
where is the Kronecker symbol: ij = 1 if i = j, ij = 0 if i 6= j. Let = diag(E1 E2 E3), !j = Ej +1 ; Ej , j = 1 2. We will consider successively the generalcomplex problem: i_ = H
H = +
2 X
j =1
C
Mj (j ) j 2
and the realresonant 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 GeneralComplex Case In that case, we make the timedependent 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 RealResonant 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 realresonant problem is welldened on the real sphere " # SR= S 2 = = (1 2 3) 2 3 j jj2 = 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 generalcomplex problem and the realresonant one. When these labels are dropped in a formula, we mean that it is valid for both the realresonant and the generalcomplex 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 rightinvariant 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 dened 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. Dene 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 generalcomplex 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 rightinvariant subRiemannian 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 generalcomplex problem is of the form considered in Exercise 19.2. Again the distribution is rightinvariant 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) satises 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 rightinvariant not leftinvariant 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 righthand sides and their rst order Lie brackets span the whole tangent space. Such control systems are called 2generating. 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: h0 Tq(0)Si = ht1 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 rightinvariant 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 simplies their analysis.
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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 rightinvariant 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:
h0 TId S u i = 0 and prove the transversality conditions at the target:
ht1 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
ht1 Tq(t1 ) (q(t1 )S u )i = 0: In order to complete the proof, we show that the function I(t) = ht 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) = ht 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 dene N13 by:
ut
0 N13(a3 ei ) = @ 3
19.3 Control of Quantum Systems
1 0 0 a3 ei 0 0 0 A: ;a3 e;i 0 0
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3
3
Let us set, in the realresonant case a0U = a1 N1 (1) + a2N2 (1) a? = a3 N13(1): In the general complex case, set a0U = N1 (a1 ei1 ) + N2 (a2 ei2 ) a? = a4 Z3 + a5 Z4 + N13 (a3 ei3 ): Here ai 2 and i 2 ; ].
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The RealResonant Case Proposition 19.4. For the realresonant 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
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u i = 0 is satised 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
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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, jj = 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)12 u2 = (qq _ ;1)23 :
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 GeneralComplex Case Proposition 19.7. For the generalcomplex 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 + i2 A 1 2 1 2 2 & : 0 ;1 + i2 ;i 2
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The equation ha TId SCu i = 0 is satised 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 ei1 a ei3 1 3 a(1 3) = @ ;e;i1 0 0 A : (19.47) ;a3 e;i3 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 j3j2 is given by the righthand side of formula (19:46). The conclusion follows as in the proof of Proposition 19.5.
ut
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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;i1 t 3 3 3(t) = ;sin p e;i3 3 and
p u1(t) = cos t= 3 ei1
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 threelevel 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 subRiemannian geodesics are parametrized by arclength, and it is given p by t1 = 23 .
19.4 A TimeOptimal Problem on SO(3)
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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 denes 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) satises the ODE q_ = qf where
19.4 A TimeOptimal 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 oneparameter 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 qf 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 timeoptimal problem: t1 ! min: Notice that
ha bi = h(f + g)=2 (f ; g)=2i = 0:
(19.48)
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19 Examples of Optimal Control Problems on Compact Lie Groups
Moreover, by rescaling time we can assume that
jaj = 1:
(19.49) Passing to convexication, 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 timeoptimal 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 vj1 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 bangbang . 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 TimeOptimal Problem on SO(3)
287
x_ = x a b] = sgnhx(0) bi is satised for t > 0 until the switching function hx(t) bi remains nonzero. Thus at such a segment of time x(t) = e;t ad(ab)x(0):
We study the switching function hx(t) bi. Notice that its derivative does not depend upon control: d hx(t) bi = hx(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] satises the conditions a b] ? a a b] ? b ja b]j = jbj this follows easily from properties of crossproduct 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
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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 ):
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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 reection 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 timeoptimal 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 TimeOptimal 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 bangbang 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 bangbang 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 bangbang 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 bangbang only if the point x( 0 ) corresponding to the rst nonnegative root of the equation hx(t) bi = 0 satises 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
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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 bangbang 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 dierentiate identity (19:50) twice w.r.t. t: d hx(t) bi = ;hx(t) a b]i 0 dt d = hx(t) a + u(t)b a b]i = u(t)hx(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)ha 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 dierentiation). 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 3dimensional, and the union of these domains covers the whole group SO(3). But it is easy to see that a suciently 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.
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19.4 A TimeOptimal 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 leftinvariant 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 leftinvariant timeoptimal 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 bangbang 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 ndimensional 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 dierential Du F : Tu U ! TF (u) M is well dened independently on coordinates. This is not the case for the second dierential. Indeed, consider the case where u is a regular point for F, i.e., the dierential 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 dierential. In the general case, well dened independently of coordinates is only a certain part of the second dierential. The dierential of a smooth mapping F : U ! M can be dened 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 dened 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 denitions. 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 dened as follows. Let v w 2 Ker Du F and 2 (Im Du F)? TF (u) M: In order to dene 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 righthand 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
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20 Second Order Optimality Conditions
and the second Lie derivative V W (a F )ju does not depend on second derivatives of a since FW(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 righthand side of the denition 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 dened independently of coordinates as in (20:6). Exercise 20.1. Show that the quadratic mapping (20:5) dened 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 dene 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 dierential 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:
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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 nitedimensionally optimal for a mapping F if for any nitedimensional 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 denition equal to corank of the dierential 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 dened 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 nitedimensional 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 signinde 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 signindenite.
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 signindenite, 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 dierential 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 sucient 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 nitedimensional 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 dierential 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(00)'1 = 0: Further, the hypothesis that the form Hess u F is negative reads @ 2 '1 @ y2 (00) < 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 nitedimensional 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 nitedimensional, 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 jj = 1 :
For any 2 S m;1 , there exists a neighborhood O S m;1 , 2 O, such that E0 = 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
Oi :
P
Then restriction of F to the nitedimensional subspace Ni=1 Ei satises the hypothesis of the theorem. Thus we can assume that U is nitedimensional. 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 dierential: N = E ! Ker D0 F
dimE = n ; m
then the restriction
D0 F : E ! ImD0 F is onetoone. 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):
Dene 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 signindenite 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 dierential 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 nitedimensional 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 secondorder optimality conditions for problem (20:11). There exist Lagrange multipliers ( ) 6= 0
R
R
2 2 Ta0 x m
such that the Lagrange function
L( a v) = a + (Q(v) ; ax) satises the stationarity conditions: @ L = ; x = 0 @a @L = Dv0 QjS N ;1 = 0: @v
(20.12)
(v0a0 )
Since v0 is a regular point of the mapping Q=jQj, then 6= 0, thus we can set = 1: Then secondorder 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) ; juj2Q(v)) v 2 S N ;1 u 2 Ker Dv QjS N ;1 : That is, inequality (20:13) yields Q(u) ; juj2Q(v0 ) 0
u 2 Ker Dv0 QjS N ;1
304
20 Second Order Optimality Conditions
thus i.e.,
Q(u) juj2Q(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 Dierentiation of the Endpoint Mapping In this section we compute dierential 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 righthand 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 gtu(t) dt Pt Ft1 (u) = q0 exp 1 0
where
Zt ;! Pt = exp fu~( ) d 0
gtu = Pt; 1(fu ; fu~(t) ): Further, introduce an intermediate mapping Gt1 : U ! M
;! Z t1 g dt: Gt1 : u 7! q0 exp tu(t) 0
Then
Ft1 (u) = Pt1 (Gt1 (u))
consequently,
Du~ Ft1 = Pt1 Du~ Gt1 Hessu~ Ft1 = Pt1 Hessu~ Gt1 so dierentiation of Ft1 reduces to dierentiation 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 + gu( ) 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
g0 = @@u gu u~( ) 2 g00 = @@u2 gu u~( ) hu () = h fu (q)i hu h0 = @@u u~( ) 2 hu : h00 = @@u2 u~( )
2 Tq M
Then dierential (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 Tq0 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;10 = 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 dierential: Du2~ Gt1 (v w) a
1 0Z ZZ t (g0 v( 2 )) g0 w( 1 ) d 1 d 2A a = q0@ g00 (v( ) w( )) d + 2 1
0
2
02 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
02 1 t
X2 X1 d 1 d 2
Zt Zt 1 = 2 X d X d + 12 0 0
ZZ
X2 X1 ] d 1d 2:
02 1 t
Choosing Xt = gt0 v(t) and taking into account that q0
we obtain: q0
ZZ
02 1 t1
X2 X1 d 1 d 2 = 21 q0
ZZ
02 1 t1
Zt
1
0
gt0 v(t) dt = 0,
X2 X1 ] 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
g00(v( ) w( )) d +
2
1
1
1
1
0
0
1
g0 2 v( 2 ) d 2 g0 1 w( 1 ) d 1 a:
The rst term can conveniently be expressed in Hamiltonian terms since 0 gu = 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
g0 2 v( 2 ) d 2 g0 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 g0 v: 0g0 v = 0 P;1 @@u fu v = P;10 @@u fu v = @@u P;10 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 g0 2 v( 2 ) g0 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 hu2 v( 2 ) @ u hu1 w( 1) d 2 d 1 : (20.19) ;! Here the derivatives @@u hui and @@u hui 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
g0 2 v( 2 ) d 2 g0 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 satises the condition ind; QjKer Du~ Ft1 < m = corank Du~ Ft1 : (20.20) The kernel of the dierential Du~ Ft1 is dened 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
g0 +"s2 v(s2 ) ds2 g0 +"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 coecient
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 suciently small " > 0, Q(v ") > 0. But for any xed % and ", the space of vector functions v " is innitedimensional. Thus the quadratic form Q has an innite 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 sucient 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 satises the strong Legendre condition is called regular (notice that this denition 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 g0 2 v( 2 ) d 2 g0 1 v( 1 ) d 1 0 0 Z t1 Z 1 @ ;! @ ;! 1
1
@ u hu2 v( 2 ) @ u hu1 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 suciently 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. Dene 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 signdeniteness of the form h00 (t ) with signdeniteness 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 sucient for negativeness of Q on short segments, thus for local nitedimensional 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 nitedimensional 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 controlane
20.4 Necessary Optimality Conditions
313
P systems q_ = f0 (q) + mi=1 ui fi (q). So we assume that an extremal t satises
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 denition 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
g0 2 v1 ( 2 ) d 2 g0 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 g0 :
Then t1 Hessu~ Ft1 (v1 v2) = 0
Z t
Z
0Z t
0
1
= 0 ;
;g0 1 w1( 1 ) + g00 w1(0) +
1
0
0:
Zt 1
0
g0 w1( ) g0 v2 ( )] d + g00 w1(0) g00 w2(0)]
+ g00 w1 (0) +
g_ 0 2 w1( 2 ) d 2 g0 1 v2 ( 1 ) d 1
1
Zt
1
0
Zt
g_ 0 w2( ) d +
g_ 0 2 w1 ( 2 )
Zt 2
1
1
0
_g0 w1( ) g0 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
g0 w(s) g0 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 g0 w(s) g0 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 g0 w(s) g0 v(s)] ds = !(w(s) v(s)) ds 0 0 !(x y) = 0 g0 x g0 y] x y 2 m
R
(20.28)
notice that the bilinear skewsymmetric 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 innite dimension. For an arbitrarily great N, we can nd an Ndimensional 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 satised 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
_g0 w( ) g0 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 dened by equality (20:26). ut We call an extremal that satises 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 TFt1 (~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 ane in u (for controlane systems),
then the second term in generalized Legendre condition (20:34) vanishes. Recall that the corresponding sucient 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 rstorder conditions (I) and secondorder 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 SubRiemannian Geodesics Consider the subRiemannian 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 2generating . E.g., the leftinvariant bracketgenerating distributions appearing in the subRiemannian problem on a compact Lie group in Sect. 19.2 and Exercise 19.2 are 2generating, thus there are no optimal strictly abnormal trajectories in those problems.
320
20 Second Order Optimality Conditions
Example 20.19. Consider the following leftinvariant subRiemannian 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 oneparameter 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 dened by the righthand side of the system is 2generating. We have QV1 QV2] = QV1 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 2generating, 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 sucient 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 satises PMP. The controldependent Hamiltonian is huv (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 SingleInput 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 controlane, and Goh condition takes the form: p( )Bb 0 2 0 t]: Dierentiating 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 sucient for local controllability of bilinear system (20:38) at the origin.
R
R R
20.6 SingleInput Case In this section we apply rst and secondorder 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 controlane, Legendre condition automatically degenerates. Further, control is onedimensional, 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 dierentiate 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 satises (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 dierentiation: #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) satises 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 SingleInput 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 dh1 and dh01 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 sucient 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 coecients 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 satised: 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 righthand 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 SingleInput Case
325
Then equality (20:47) does not contain u(t), and we continue dierentiation 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 dierentiation: 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 innite 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 timeoptimal problem for our singleinput system (20:41). We prove that there do not exist timeoptimal 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 timeoptimal trajectory, not geometrically optimal one as before. Although, the Hamiltonian of PMP hu = h0 + uh1 for the timeoptimal 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 timeminimal trajectory, but also for a timemaximal 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 satises 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 SingleInput 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 _g0 g0 ] 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;1f0 f1 ]
;! P = exp
Z 0
w(0) = 0
fu() d :
The rst term in expression (20:57) for the Hessian vanishes: 0 _g0 g0 ] = ;h101( ) 0: Integrating the second term by parts twice, we obtain: t Hessu Ft(v) = where
Zt 0
0 #g0 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;1f0 f0 f1 ]] ( ) =
Z 0
w( 1 ) d 1
(t) = 0:
328
20 Second Order Optimality Conditions
The rst term in (20:58) dominates on needlelike 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 ht 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 halfspace ht 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 SingleInput Case
329
Summing up, the variations v0, v1 , : : : , vn;1 of the control u at the singular piece generate a nonnegative halfspace 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) = fht i 0g: Now we add a needlelike 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 needlelike variation concentrated at small segments near %t1: 8 > t 2 0 %t ] 0, thus ht1 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)=(00) @ " ("s)=(00) Pt (Ft (us )) ; = Pt t1 i = i i = 0 : : : n ; 1:
330
20 Second Order Optimality Conditions
Inequality (20:59) translates to
ht1 0i = ht 0i > 0 and, of course,
ht1 ii = ht i i = 0 i = 1 : : : n ; 1: The inequality ht1 ni < 0 means that the needlelike variation on the bang piece generates a tangent vector in the halfspace ht1 i < 0 complementary to the halfspace ht1 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" )
satises the condition D(00)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 suciently 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 timeoptimal, a contradiction. We proved that a timeoptimal 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 righthand side of system (20:41) satisfy the identity
f1 f0 f1]] = 0:
(20.60)
Let a timeoptimal 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, timeoptimal (i.e., timeminimal) control
can be replaced by a timemaximal 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 dened. For boundary conditions suciently close to the singular trajectory, there are two possible patterns of optimal control:
20.6 SingleInput Case
331
(1) either it makes innite number of switchings on a compact time segment adjacent to the singular part, so that the optimal trajectory \gets o" the singular trajectory via innite number of switchings, (2) or optimal control is bangbang, but the number of switchings grows innitely 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 sucient 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
g0 2 v( 2 ) d 2 g0 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
g0 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 denition, 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
g0 2 v( 2 ) d 2 g0 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 nitedimensional 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 nitedimensional. 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 denition 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 linearquadratic problem), we show that the function (t) = sup fQt(v) j v 2 Kt kvkh00 = 1g (21.2) satises 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 denition 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 dierential equation on conjugate time for a regular extremal pair (~u(t) t). The symplectic space = T0 (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 hut v: u~(t) In these terms, the bilinear form Qt reads Qt (v1 v2) =
Zt 0
ZZ
h00 (v1 ( ) v2( )) dt +
02 1 t
(J2 v1 ( 2 ) J1 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 = T0 (Tq0 M)
is the vertical subspace. A variation of control v 2 Vt satises 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 nitedimensional 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
02 1 t
h00 (v( ) ) d +
(J2 v( 2 ) J1 ) 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 dened:
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 satises 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 satises 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 dier 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 = T0 (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 skeworthogonal complement ;0\. On the other hand, 0 2 ;0 + 0, compare denition (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 denition (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 nothold, weprove that condition (2) is then satised. 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 signindenite, then the initial form is signindenite as well. Summing up, the form Qt1 is signindenite 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 nitedimensional 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 sucient 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 satises 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 ) ) d0 '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 sucient 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 hut where hut() = hu (Pt;1())
@ h ; dt @ uu Pt;1 ; ; where Pt;1 is dierential of the dieomorphism Pt;1 : T M ! T M @ hu
345
= ; d0 @@u hut = ;
0
= ; dt @ u ;
= Pt;1 0 2 Tt (T M): Then the Hamiltonian bt can be rewritten as bt () = ; 21 dt @@huu (h00t );1 dt @@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, du + (dhu )ju() ; d hu~(t) d(hu() ; hu~(t)) = @@huu u()
 {z } 0
!
@ hu dt u: t (hu() ; hu~(t) ) = dt @ u u(t) The dierential dt u can be found by dierentiation of the identity @ hu @ u u() 0 D2
at = t . Indeed, we have @ 2 hu d u + d @ hu = 0 @u @ u2 thus dt u = ;(h00t );1 dt @@huu :
346
21 Jacobi Equation
Consequently, i.e.,
D2 t (hu() ; hu~(t)) = ;dt @@huu (h00t );1 dt @@huu
Hess0 't () = Hesst (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 0t] 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 Sucient Optimality Conditions In this section we prove sucient 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 = T0 (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 Sucient 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 sucient 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 dened 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 dieomorphism near t t 2 0 t1]: In other words, we require that the tangent spaces Tt Lt = etH~ (T0 L0) have zero intersection with the vertical subspaces t = Tt (Tq(t) M): etH~ (T0 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 denition 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 dierential L0 = f = dq ag T M satises 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) T0 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 T0 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 Sucient 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 (tL0 ) \ 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
t0 = Bt
(Bt L0 ) \ 0 = f0g t 2 0 t1]: (21.32) It remains to prove existence of a Lagrangian subspace L0 that satises 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 suciently 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 denes 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 ndimensional 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 Sucient 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 lefthand 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 suciently 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 denition, 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 Sucient 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 j0 = 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 skeworthogonal 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 welldened:
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 suciently close to 0, i.e., for " suciently 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 sucient 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 controlane system to a nonlinear system on a manifold of a less dimension.
22.1 Reduction Consider a controlane 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 controlane system. The extremal t is necessarily totally singular, moreover the maximality condition of PMP is equivalent to the identity ht 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
ht 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 controldependent 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 welldened. 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 satised: 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 satised, then the space of orbits M= is a smooth manifold. Then system (22:3) is welldened 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) ane 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 dene 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 timeoptimal 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 righthand 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 dieomorphic 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 righthand side of this symmetric control system denes 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 sixdimensional 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 selfsimilarity  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 3dimensional 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 2Dimensional 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 righthand side fu (q) is smooth in (u q). A wellknown example of such a system is given by a twodimensional 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 feedbackinvariant 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) satises 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 controldependent 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 dened 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 feedbackinvariant moving frame on the 3dimensional manifold H in order to write Jacobi equation in this frame. Notice that the maximized Hamiltonian H is feedbackinvariant 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 feedbackinvariant. 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 2Dimensional 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 dene 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 dene 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: Dierentiating 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 feedbackinvariant 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 identication H = f g M (23.7) thus tangent spaces to H decompose into direct sum of the horizontal and vertical subspaces. By duality, any dierential 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 feedbackinvariant. 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 2Dimensional 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 dierential of the form s w.r.t. horizontal coordinates. The horizontal 2form 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 satises (23:5) is
unique up to a factor 1. On the other hand, the vertical eld v that satises (23:6) is unique, up to a factor constant along trajectories of H~ (so this factor does not aect "). 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 2Dimensional 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 twodimensional 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 twodimensional system (23:1): _ = ~bt() 2 = T0 (T M) and its ow
;! Bt = exp
Zt 0
~b d :
Recall that 0 = T0 (Tq0 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 T0 (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 ) etH~ (): Thus
R
370
23 Curvature
Bt E(0 ) = (Pt ) etH~ 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), etH~ V1 (0 ) = t e;t ad H~ V1 = t (V1 + tV2 + o(t)) 2= Tt (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,
etH~ 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 2Dimensional 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 dierentiate 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
ki=1
and the ODE follows. ut In view of inclusion (23:13), an instant t is a conjugate time i the coecients 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 twodimensional 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 twodimensional 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 3Dimensional ControlAne Systems
373
Example 23.6. Consider the control system corresponding to a Riemannian
problem on a 2dimensional 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 twodimensional 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 satises 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 timeoptimal compared with all other admissible trajectories suciently close in M.
23.2 Curvature of 3Dimensional ControlAne Systems
R
In this section we consider controlane 3dimensional 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 2dimensional one as in Chap. 22 and compute the curvature of the 2dimensional 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 controlane system (23:19) is totally singular, it satises the equality h1(t ) = ht f1i 0 (23.22) and the corresponding extremal control cannot be found immediately from this equality. Dierentiation of (23:22) w.r.t. t yields h01(t ) = ht f0 f1 ]i 0 and one more dierentiation leads to an equality containing control: h001(t ) + u(t)h101(t ) = ht f0 f0 f1]]i + u(t)ht 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 aects 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 threedimensional 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 3Dimensional ControlAne 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 identication (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 = TqM \ 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 dened 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 identied 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 threedimensional system (23:19) as the curvature " of its twodimensional 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 ; c202f0 f1]: Now we can nd the normalizing factor a for f1 such that (23:26) be satised. We have f0 f0 af1]] = f0 (f0a) + af0 f1]] = (f02 a)f1 + 2(f0 a)f0 f1 ] + af0 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 controlane 3dimensional 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 controlane 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 2dimensional 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 3dimensional controlane 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 3dimensional 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 twodimensional 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 identied 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 orientationpreserving isometries and omit
M x
TxM
+
xb
M
TxbMc =
Mc
Fig. 24.2. Identication of tangent spaces at contact point
Mc
the words \orientationpreserving" 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 TwoDimensional Riemannian Geometry
379
For any contact conguration 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 5dimensional 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 TwoDimensional Riemannian Geometry Let M be a 2dimensional Riemannian manifold. We describe Riemannian geodesics, LeviCivita 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 arclength 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 identied with the tangent bundle via the Riemannian structure: TM = T M v 7! = hv i: Then H T M is identied with the spherical bundle S = fv 2 TM j kvk = 1g TM
of unit tangent vectors to M. After this identication, etH~ can be considered as a geodesic ow on S .
24.2.2 LeviCivita 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 denes 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 TwoDimensional Riemannian Geometry
381
Indeed, if the curve x(t) satises 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). LeviCivita 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 LeviCivita connection as a horizontal distribution on H = S . In Chap. 23 we constructed a feedbackinvariant 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 dierent 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 dierentiation. 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 LeviCivita connection mean that H~ 2 D and es @ ' D = D, thus
@ D = span es @ ' H~ j s 2
Since
R
:
@ es @ ' H~ = h1 (' ; s) e1 + c1 @@ ' + h2(' ; s) e2 + c2 @@ '
we obtain
~ H~ 0 : D = span H The 1form 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:
24.3 Admissible Velocities
24.3 Admissible Velocities
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 identiLet 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 2dimensional 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 dierent 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 = ix : Tx M ! Txb M
o
is a smooth 2dimensional 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 2dimensional orbits is injective. c are complete and simply (3) Now assume that the manifolds M and M connected. Let O be a 2dimensional 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 dened 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 dened at the whole segment 0 t1]. Now it easily follows that (O) = M for the twodimensional 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 dieomorphisms 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 dieomorphisms. 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 dieomorphism. The same is true for bjO . Thus we can dene a dieomorphism 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 5dimensional, 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 congurations of the system of rolling bodies, nd an admissible motion of the system that steers the rst conguration 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 subRiemannian 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. Dierentiating these equalities by virtue of the Hamiltonian system (24:23), we obtain one more identity: g3(t ) 0: (24.25) The next dierentiation 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 ! TS1 Tq Q = Tx M ! Txb M c d 2 TS1: = + 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 subRiemannian 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 simplies 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 dened 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 suciently small ". For any "0 > 0, the family At+" , j"j < "0 , is locally bounded, thus equicontinuous by the BanachSteinhaus theorem. Consequently, At+" (Bt+" ; Bt ) a ! 0 as " ! 0. Continuity of the family At Bt follows. The dierentiability 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 kPtaksK C eC 0t kV ksK0 d kaksK 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 kkaPktaksK j a 2 C 1 (M) kaks+1Kt 6= 0 s+1Kt = sup fkPtaksK j a 2 C 1 (M) kaks+1Kt = 1g :
(A.4)
Notice that the function (t), t 2 0 t1], is measurable since the supremum in the righthand 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
kaksPt(K ) kaks+1Kt the function (t) is bounded on the segment 0 t1]. As in the denition of the seminorms k ksK 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
kPtaksK
= 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 kPtaksK = jhil hi1 (Pt pa )(q0)j kPtpa ksK (A.5) kpa ksqt kaksKt : In the nitedimensional 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 kpksKt C: (A.6) kpksqt Inequalities (A.5) and (A.6) give the estimate kPtaksK kPtpa ksK C kPtpa ksK = C kPtpa ksK C (t): (A.7) kaksKt kpa ksqt kpa ksKt kpa ks+1Kt Since Zt Pta = a + P V a d 0
then
kPtaksK kaksK +
Zt 0
kP V aksK d
by inequality (A.7) and denition (2:2)
kaksK + C kaks+1Kt Dividing by kaks+1Kt , we arrive at
Zt 0
( ) kV ksKt d :
kPtaksK 1 + C Z t ( ) kV k d : sKt kaks+1Kt 0
A.2 Remainder Term of the Chronological Exponential
Thus we obtain the inequality (t) 1 + C
Zt 0
( ) kV ksKt d
from which it follows by Gronwall's lemma that Rt (t) eC 0 kV ksKt d : Then estimate (A.7) implies that R kPtaksK C eC 0t kV ksKt d kaksKt 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) + Pm Vm V1 d m : : : d 1
m (t)
where Sm (t) = Id +
mX ;1 Z
Z
Vk V1 d k : : : d 1 :
k=1 k (t)
Then Z Z k(Pt ; Sm (t))aksK kPm Vm V1 aksK d m : : : d 1
m (t)
by Lemma A.3
Z
Z
R CeC 0t kV ksK0 d kVm V1 aksK 0 d m : : : d 1 :
m (t)
Now we estimate the last integral. By denition (2:2) of seminorms,
Z
Z
kVm V1 aksK 0 d m : : : d 1
m (t)
Z
Z
kVm ksK 0 kVm;1 ks+1K 0 kV1 ks+m;1K 0 kaks+mK 0 d m : : : d 1
m (t) kaks+mK 0
Z
Z
kVm ks+m;1K 0 kVm;1 ks+m;1K 0 kV1 ks+m;1K 0 d m : : : d 1
m (t)
1 = kaks+mK 0 m!
Z t 0
kV ks+m;1K 0 d
m
398
A Appendix
and estimate (2:13) follows:
k(Pt ; Sm (t)) aksK
C eC R0t kV ksK0 d m!
Z t 0
kV ks+m;1K 0 d
m
kaks+mK 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.: SubRiemannian Geometry. Birkhaser, 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 2D Manifolds. Springer SMAI, 43, to appear 6] Brockett, R.W., Millman, R.S., Sussmann, H.J., Eds.: Di erential Geometric Conrol Theory. Birkhauser, 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. SpringerVerlag (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. SpringerVerlag (1990)
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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) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Dierential Dq . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Flow P t of vector eld V . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Attainable set Aq0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Initial and nal congurations 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 conguration: 1 = 2 . . . . . . . . . . . . . . . . . . . . . . . 107 7.3 Hard to control conguration: ( 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 nonfull 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 SubRiemannian 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 Identication 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 kaksK , 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, LeviCivita, 381 control system, controlane, 47 control parameter, 12 control system, 12 control system, bilinear, 320 control system, completely controllable, 49 control system, leftinvariant, 257 control system, linear, 47 control system, linear, Brunovsky normal form, 129 control system, linear, Kronecker indices, 128 control system, rightinvariant, 257 control system, singleinput, 321 control, bangbang, 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, 2dimensional systems, 366 curvature, 3dimensional controlane 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, 2generating, 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 statefeedback, 131 exponential of vector eld, 35 exterior kform, 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, bracketgenerating, 67, 109 family of vector elds, completely nonholonomic, 67 family of vector elds, fullrank, 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
HamiltonJacobi equation, 243 Hamiltonian, 159 Hamiltonian system, 160 Hessian, 295
I
immersion, 64 index of quadratic form, 300 integral invariant of PoincareCartan, 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 1form, 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
needlelike 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, sucient, 238, 241, 346, 347, 354 optimality, geometric, 293 optimality, local nitedimensional, 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 timeoptimal, 211 problem, linearquadratic, 223 problem, Riemannian, 265 problem, subRiemannian, 267, 318 problem, timeoptimal, 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 1form, 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