Optimal Space Flight Navigation

Control Engineering

Ashish Tewari

Optimal Space Flight Navigation An Analytical Approach

Control Engineering Series Editor William S. Levine Department of Electrical and Computer Engineering University of Maryland College Park, MD USA Editorial Advisory Board Richard Braatz Massachusetts Institute of Technology Cambridge, MA USA Graham Goodwin University of Newcastle Australia Davor Hrovat Ford Motor Company Dearborn, MI USA Zongli Lin University of Virginia Charlottesville, VA USA

Mark Spong University of Texas at Dallas Dallas, TX USA Maarten Steinbuch Technische Universiteit Eindhoven Eindhoven, The Netherlands Mathukumalli Vidyasagar University of Texas at Dallas Dallas, TX USA Yutaka Yamamoto Kyoto University Kyoto, Japan

More information about this series at http://www.springer.com/series/4988

Ashish Tewari

Optimal Space Flight Navigation An Analytical Approach

Ashish Tewari Department of Aerospace Engineering Indian Institute of Technology, Kanpur IIT-Kanpur, India

ISSN 2373-7719 ISSN 2373-7727 (electronic) Control Engineering ISBN 978-3-030-03788-8 ISBN 978-3-030-03789-5 (eBook) https://doi.org/10.1007/978-3-030-03789-5 Library of Congress Control Number: 2018961705 Mathematics Subject Classification: 70F05, 70F07, 70F10, 49N05, 49N25, 49J15, 49K15, 49J30, 37N05, 34B10 © Springer Nature Switzerland AG 2019 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, express or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This book is published under the imprint Birkhäuser, www.birkhauser-science.com by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Preface

This book is primarily written to consolidate optimal space flight navigation theory as a discipline. There are several excellent textbooks on optimal control theory and equally good books on orbital space mechanics, but the application of optimal control theory to orbital mechanics is mainly confined to research articles. Over the last several decades, the research literature on optimal space flight navigation has become voluminous and has fragmented into many esoteric subdisciplines. However, a monograph which organizes this content and vitalizes it for contemporary application is hitherto lacking. Optimal space flight navigation is a problem of practical interest and has been so ever since humans imagined journeying into space. While visionaries such as Tsiolkovsky and Goddard helped in taking the first steps by developing rocketry, it required enthusiasts to devise fuel-optimal trajectories for practical space missions. The vision of Walter Hohmann, a civil engineer, revealed in his 1925 book Die Erreichbarkeit der Himmelskörper (The Accessibility of Celestial Bodies) is one of the most useful and commonly applied space flight maneuvers—the Hohmann transfer. As recently as in 2014, the Hohmann transfer was successfully applied to efficiently send India’s first mission to Mars. While arising as an elegant idea in Hohmann’s mind, the optimality of the Hohmann transfer can now be rigorously proved in various ways, including by optimal control theory. Similarly, the straightforward and insightful application by D.F. Lawden of the erstwhile nascent optimal control theory to the space navigation problem produced in 1963 in the monograph entitled Optimal Trajectories for Space Navigation has guided many researchers since then. The first footsteps on the Moon would not have appeared if NASA’s engineers (including Richard Battin who passed away in 2014) had not devised simple but efficient techniques to guide the lunar module on a flat thrusting trajectory to a safe landing. Battin in his treatise An Introduction to the Mathematics and Methods of Astrodynamics reminisces about how he came up with the velocity-to-be-gained and cross-product steering as simple but practical guidance strategies for space flight in the era of primitive computer technology. He writes of the enormous buildings required to house the computer mainframes of the day with only very modest computing power. Perhaps the total computational v

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Preface

resources present on board Apollo 11 could be surpassed by those of today’s pocket calculator. The feat performed by those pioneering guidance engineers can only be appreciated by the complexity of their task—to send people to the Moon and to bring them successfully back to the Earth. The design and analysis of practical lunar and interplanetary trajectories could be conducted in the analytical framework established by Victor Szebehely in his 1967 masterpiece, Theory of Orbits: The Restricted Problem of Three Bodies. Rather than highlighting the computational (and often unrealizable) solutions of mathematically complex problems, this book emphasizes the analytical approach to optimal space navigation. The need for simplicity and practicability of guidance methods is to be contrasted with the unfortunate emphasis which has been placed these days on increasing the complexity of control schemes. Multilevel iterative algorithms must be solved in real-time computations and therefore have unresolved convergence issues which prevent applications to actual missions. It should thus be asked: Is more sophistication always better, and does it lead to any advantage in the practical sense? I think a simpler guidance solution must be sought, wherever possible, due to its reliable implementation in a realistic situation, which has been amply demonstrated by the analytical guidance methods. That is why even the latest space flight missions are based upon analytical techniques. The complex interplanetary and asteroid (or comet) flyby and rendezvous missions are analytically derived from the application of optimal control theory to multi-body dynamics, such as the optimal transfers between halo and quasihalo orbits and Lissajous trajectories using the stable and unstable manifolds of the restricted three-body problem. Studying such optimal paths often imparts new insights into the problem of multi-body transfers, which would be lacking in a purely numerical search for suboptimal solutions. In a similar vein, the problem of orbiting and landing on an irregularly shaped body with an uncertain gravity field (such as an asteroid or a comet) can be solved out by time-optimal and fuel-optimal methods, as has been demonstrated in many practical missions. The thought of writing this book arose from the course AE-777 (Optimal Space Flight Control), which I have taught for the past several years and for which a single textbook containing all the relevant material was unavailable. Therefore, the book is designed to be used as a textbook in a course on optimal space flight at the graduate and senior undergraduate levels. The first three chapters give a basic introduction to the topic, namely, optimal control theory (Chaps. 1 and 2) and orbital mechanics with impulsive transfers (Chap. 3). Chapter 4 covers optimal transfers in a spherical gravity field with continuous, unconstrained acceleration inputs, whereas Chap. 5 extends the treatment to trajectory optimization with bounded acceleration inputs. Finally, Chap. 6 introduces the reader to the advanced topics of optimal transfer in time-varying gravity fields, including multi-body transfers. Unfortunately, due to the vast scope of these topics, it is not possible to do them full justice in an introductory text. Hence, references to research articles are provided for the advanced topics. The end-of-chapter exercises test the understanding of the basic concepts, whereas several references are provided for undertaking research and for advanced applications. A basic knowledge of control systems and dynamics is

Preface

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assumed of the reader, which can be supplemented by textbooks on these topics, such as those previously published by me (e.g., Modern Control Design, and Atmospheric and Space Flight Dynamics, Automatic Control of Atmospheric and Space Flight Vehicles). It is a pleasure to offer this manuscript for publication with Birkhäuser. I would like to thank the editorial and production staff at Birkhäuser for their invaluable assistance, especially Bill Levine, Benjamin Levitt, Samuel DiBella, and Christopher Tominich. I would finally like to thank my wife Prachi and daughter Manya for their patience during the preparation of this manuscript. Kanpur, India September 2018

Ashish Tewari

Contents

1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Optimal Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Space Vehicle Guidance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 3

2

Analytical Optimal Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Optimization of Static Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Static Equality Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2 Inequality Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Dynamic Equality Constraints and Unbounded Inputs . . . . . . . . . . . . . . 2.4 Special Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 Fixed Terminal Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.2 Free Terminal Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Sufficient Condition for Optimality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Pontryagin’s Minimum Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Hamilton–Jacobi–Bellman Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8 Time-Invariant Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9 Linear Systems with Quadratic Cost Functions . . . . . . . . . . . . . . . . . . . . . . 2.10 Illustrative Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.11 Singular Optimal Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.11.1 Generalized Legendre–Clebsch Necessary Condition . . . . . 2.11.2 Jacobson’s Necessary Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.12 Numerical Solution Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.12.1 Shooting Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.12.2 Collocation Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7 7 8 11 18 20 26 26 28 31 31 32 34 35 43 55 58 67 69 70 71 71

3

Orbital Mechanics and Impulsive Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Keplerian Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.1 Reference Frames of Keplerian Motion. . . . . . . . . . . . . . . . . . . . .

77 77 78 80

ix

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Contents

3.2.2 Time Equation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.3 Lagrange’s Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Impulsive Orbital Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 Minimum Energy Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Lambert’s Transfer. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.1 Stumpff Function Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.2 Hypergeometric Function Method . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Optimal Impulsive Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.1 Coasting Arc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.2 Hohmann Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.3 Outer Bi-elliptical Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

83 86 88 92 95 96 99 101 102 105 110 112

4

Two-Body Maneuvers with Unbounded Continuous Inputs . . . . . . . . . . . 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 A Motivating Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Optimal Low-Thrust Orbital Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.1 Coplanar Orbital Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.2 Plane Change Maneuver . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.3 General Orbital Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Variational Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Optimal Regulation of Circular Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6.1 Coplanar Regulation with Radial Thrust. . . . . . . . . . . . . . . . . . . . 4.6.2 Coplanar Regulation with Tangential Thrust . . . . . . . . . . . . . . . 4.7 General Orbital Tracking . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8 Basic Guidance with Continuous Inputs. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.9 Line-of-Sight Guidance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.10 Cross-Product Steering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.11 Energy-Optimal Guidance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.12 Hill–Clohessy–Wiltshire Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

115 115 115 117 120 122 122 123 124 127 128 131 132 138 139 145 148 151 155

5

Optimal Maneuvers with Bounded Inputs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Optimal Thrust Direction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.1 Constant Acceleration Bound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.2 Bounded Exhaust Rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Time-Invariant Gravity Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Null-Thrust Arc in Central Gravity Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.1 Inverse-Square Gravity Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Intermediate-Thrust Arc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Lawden’s Spiral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7 Powered Arcs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

159 159 159 166 166 168 169 172 173 174 180

Contents

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5.8

Linearization Relative to a Circular Orbit . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8.1 Out-of-Plane Rendezvous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8.2 Coplanar Rendezvous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

189 193 196 197

Flight in Non-spherical Gravity Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Gravity Field of a Non-spherical Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.1 Gravity of an Axisymmetric Planet . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.2 Gravity Field of an Oblate Planet . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.3 Gravity of an Irregular Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Circular Restricted Three-Body Problem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.1 Lagrangian Points and Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.2 Hamiltonian Formulation and Jacobi’s Integral . . . . . . . . . . . . 6.4 Special Three-Body Trajectories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.1 Perturbed Orbits About a Primary. . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.2 Free-Return Trajectories Between the Primaries . . . . . . . . . . . 6.5 Optimal Low-Thrust Three-Body Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6 Missions Around Collinear Lagrangian Points . . . . . . . . . . . . . . . . . . . . . . . 6.6.1 Motion About the Collinear Points. . . . . . . . . . . . . . . . . . . . . . . . . . 6.6.2 Lindstedt–Poincaré Method for Halo Orbits. . . . . . . . . . . . . . . . 6.7 Numerical Computation of Halo Orbits and Manifolds . . . . . . . . . . . . . 6.7.1 Monodromy Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7.2 Stable and Unstable Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.8 Optimal Station Keeping around Collinear Points . . . . . . . . . . . . . . . . . . . 6.8.1 Impulsive Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.8.2 Continuous Thrust Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

199 199 201 201 207 208 208 213 216 219 219 220 222 232 234 237 242 243 246 249 249 250

Answers to Selected Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267

Chapter 1

Introduction

Navigating a space vehicle consists of first determining and then steering a desired path under the gravitational influence of several large bodies. This results in two problems to be solved: (a) determination of a suitable flight path which takes the vehicle from an initial to a final state, and (b) guiding the spacecraft such that it always remains close to the desired optimal trajectory. Either of these problems could lead to infinitely many solutions, each of which can have a different energy expenditure and a time of flight. However, a unique solution is obtained when either the energy or the time (or any other performance measure) of flight is minimized for a given set of initial and final conditions, and with realistic physical constraints. Such an approach—called the optimal control—thus provides a closure to the spacecraft navigation problem, which would otherwise remain both arbitrary and intractable. The application of the systematic framework supplied by the optimal control theory applied to practical problems in space flight navigation is the objective of this monograph.

1.1 Optimal Control A general control problem is solved by determining the control history for changing the state of a given system as required. Such a determination can be carried out only if a mathematical model of the actual system is available. Deriving a suitable math model in terms of differential, integro-differential, and algebraic equations requires various idealizations and simplifying assumptions to be applied to the actual system. The solution to such equations subject to given initial conditions and prescribed control history must be unique if it has to be applied for controlling the actual system. The cases where either the governing equations are uncertain or have non-unique solutions to a prescribed control history and a set of initial conditions © Springer Nature Switzerland AG 2019 A. Tewari, Optimal Space Flight Navigation, Control Engineering, https://doi.org/10.1007/978-3-030-03789-5_1

1

2

1 Introduction

are termed non-deterministic and must be handled by stochastic (probabilistic) techniques. For our present purposes, we shall assume that the mathematical model is a deterministic one. Furthermore, the system of interest in spacecraft navigation is completely described at any instant by a finite number of variables x(t) : R → Rn×1 called the state variables, where n is the finite order of the system. The set of state variables is said to uniquely determine its state. An additional set of variables u(t) : R → Rm×1 called the control variables is applied externally to the system in order to manipulate its state. The state vector x(t) and the control input vector u(t) are mutually related by a set of ordinary differential equations governing the system. Consider a system governed by the following set of first-order ordinary differential equations called the state equation: x˙ = f (x, u, t)

(1.1)

where x(t) ∈ Rn×1 is the state vector and u(t) ∈ Rm×1 is the control vector, subject to the initial condition x(t0 ) = x0

(1.2)

The optimal control problem refers to finding a set of control variables, u(t), which minimizes a performance index,  J = φ[x(tf ), tf ] +

tf

L(x, u, t)dt

(1.3)

t0

such that the dynamic equality constraint given by the state equation is satisfied, while taking the terminal state, x(tf ), to a specified set, x(tf ) ∈ Xf ⊂ Rn×1 . Both the Lagrangian, L(x, u), representing a cost on the transient behavior, and the terminal cost function, φ[x(tf ), tf ], denoting the penalty at the terminal time, tf , are continuously differentiable functions up to an indefinite order. The control interval, t ∈ (t0 , tf ), could be either open or closed, depending upon whether the terminal time, tf , is specified a priori. There could be additional inequality constraints to be specified for the state and control variables, given by g(x, u, t) ≤ 0

(1.4)

It is clear from the Weierstrass’ extreme value theorem that a continuous, real-valued functional, J (x, u), would possess both a maximum and a minimum with respect to u either inside or on the boundary of any bounded region of space, u ∈ U ⊂ Rm×1 . Thus by imposing constraints on the state and control variables in the form of equality and inequality constraints, as well as the initial and terminal boundary conditions, the existence of a minimum (or maximum) point of J relative to u is guaranteed. The optimal control theory is a mathematical framework for deriving the necessary and sufficient conditions of the existence of a minimum (or maximum) point, from which the optimizing control variables could be determined uniquely.

1.2 Space Vehicle Guidance

3

The solution to the optimal control problem can be derived either by analytically through the framework of calculus of variations or by a direct transcription method where the solution is directly sought by nonlinear programming methods. The direct solution methods involve a “blind” search over a feasible space by computationally intensive schemes, which either may or may not produce a solution. The latter invariably focus on sophisticated algorithms, without offering any insight into the physical problem being solved. This book focuses on the analytical methods of investigating the extremal trajectories, which drive the state from x0 to x(tf ) ∈ Xf while minimizing (or maximizing) J relative to u, subject to the constraints, Eqs. (1.1) and (1.4). This optimal control theoretic method has been developed over the last century [15, 19, 68]. The formulation of the analytical method results in a two-point boundary value problem which must be solved iteratively, and has a guaranteed solution when the sufficient conditions of optimality are met. The analytical derivation of the necessary and sufficient conditions of optimality produces an insight into the actual problem being solved, which can be employed to guide the direct search methods. Hence, the two approaches can be combined to yield an efficient procedure. For the sake of brevity, we shall not discuss the direct search methods any further, and instead refer the reader to the excellent survey paper by Betts [14].

1.2 Space Vehicle Guidance All spacecraft require a precise control of their position for a successful mission. The flight control tasks can be broadly classified into two distinct categories: 1. Navigation (or guidance) consisting of a control of the spatial position, r(t) ∈ R3×1 , and linear velocity, v(t) ∈ R3×1 (i.e., translation) of the craft’s center of mass relative to a fixed reference frame. 2. Attitude control, which involves a control of the rigid vehicle’s angular orientation (or attitude) represented by the kinematic parameters, q(t) ∈ R4×1 , called the quaternion, and angular velocity, ω(t) ∈ R3×1 , about the center of mass, relative to a non-rotating frame of reference. The reference frame for the navigation problem is regarded as being nonaccelerating (or inertial), with its origin conveniently chosen to be at the center of a large body (such as a planet, the Moon, or the Sun). During the time of flight, the motion of the spacecraft is resolved relative to that of the body using the translational dynamic variables, r(t), v(t). The space vehicle’s translational dynamics relative to the large body is referred to as its orbital motion, and can be controlled by applying external thrust via rocket engines. In contrast, the reference frame for attitude control has its origin at the spacecraft’s center of mass (thus translates with it), and has its axes fixed with respect to the distant stars. The angular velocity, ω(t), and the attitude, q(t), of the rigid spacecraft are measured relative to this reference frame, and are resolved in a frame called the body axes, which is rigidly

4

1 Introduction

attached to the spacecraft. The attitude is controlled by externally applied torque inputs and/or internally variable angular momentum via rotors. The desired attitude of the spacecraft is dependent on the vehicle’s spatial position in most space missions, which invariably involve pointing one of the body axes in a particular direction. Furthermore, the vehicle’s rotation about its own center of mass causes the thrust of its engines to produce time-varying acceleration components in the inertial space, thereby affecting the translational motion. Thus the two motions are inherently coupled. However, since the characteristic time scale of translational dynamics is typically much larger than that of attitude dynamics, the two control problems are addressed separately by dedicated systems. Due to its much shorter time scale, the attitude controller acts as a slave (or actuator) for the navigation system. Another control subsystem that is subservient to the navigational system is the spacecraft’s propulsion system, whose time scale is much shorter than even the attitude subsystem. The propulsion system generates a thrust vector, T (t) ∈ R3×1 , from the engines, as demanded by the navigational system. A spacecraft of mass, m(t), has the following equations of translational motion: r˙ = v

(1.5)

v˙ = g(r, t) + T /m

(1.6)

where g(r, t) represents the acceleration due to gravity, and the mass variation is related to the net mass exhaust rate, β(t), of all rocket engines as follows: m ˙ = −β

(1.7)

If the spacecraft is assumed to be rigid with a constant inertia tensor, J ∈ R3×3 , its rotational (or attitude) dynamics equations resolved in the body axes are the following: q˙ = A(ω)

(1.8)

ω˙ = J−1 τ − J−1 (ω × Jω)

(1.9)

where A(.) : R3×1 → R4×1 is a nonlinear attitude kinematics functional and τ (t) ∈ R3×1 is the applied torque input. The thrust vector, T (t), in the inertial space is related to a changing attitude as follows: T = C(q)T¯

(1.10)

where T¯ is the thrust vector resolved in the body axes, and C(.) : R4×1 → R3×3 is the orthogonal rotation matrix representing the coordinate transformation from the body axes to the inertial space, with the property det (C) = 1. Therefore, | T (t) |=| T¯ (t) |. The spacecraft’s engines can be rotated relative to the body axes

1.2 Space Vehicle Guidance

5

r(t0)=r0 , v(t0)=v0 F[r(tf), v(tf), tf]=0 Navigational Feedforward Controller

rd(t), vd(t) −

Navigational Feedback Controller

Attitude Feedforward Controller

qd(t),

ωd(t)

Attitude Feedback Controller

−

Td(t)

q(t),

τ (t)

Spacecraft Propulsion System

Spacecraft Attitude Dynamics

ω (t)

Spacecraft Orbital Dynamics

r(t), v(t)

(t)

Fig. 1.1 Spacecraft control system comprising the navigational, propulsion, and attitude subsystems

using two gimbal angles per engine, and can also be throttled by varying the mass exhaust rate, β, thereby producing a variable, T¯ , as required. The inter-relationship of the various spacecraft control systems is schematically depicted in Fig. 1.1, wherein the outer navigation (or guidance) loop takes reference inputs, rd (t), vd (t), from either a ground station or an on-board flight computer (called the navigational feedforward controller). The navigational feedforward controller generates the trajectory waypoints, rd (t), vd (t), to meet the boundary conditions specified by the initial conditions, r(t0 ) = r0 , v(t0 ) = v0 , and a terminal condition, F [r(tf ), v(tf ), tf ] = 0, where F (.) : R3×1 ×R3×1 → R. A navigational feedback controller compares the actual trajectory, r(t), v(t), with the specific waypoints to issue corrective control inputs. The navigational control inputs are the commands issued to the propulsion subsystem (its internal structure is not shown here) for producing a desired thrust vector, Td (t), and to the attitude control system to rotate the vehicle for achieving the desired rotational state, qd (t), ωd (t), at any time, t. A fully automated (autonomous) spacecraft has navigational, propulsion, and attitude control subsystems functioning without any manual intervention. Although the three control systems work in tandem, they can be designed and studied independently of each other due to their much different time scales. For this reason, the present text considers only the design and analysis of optimal navigation systems, and assumes the tasks requiring the generation of the required thrust as well as pointing of the vehicle in desired directions are instantly achieved by separate propulsion and attitude control systems, respectively. Such an assumption involves neglecting the time scales of engine control system dynamics and attitude motion in comparison with that of the translational dynamics. Figure 1.2 is the schematic diagram of the navigational system assuming instantaneous vehiclerotation and

6

1 Introduction

r(t0)=r0 , v(t0)=v0 F[r(tf), v(tf), tf]=0

Navigational Feedforward Controller

rd(t), vd(t)

-

Navigational Feedback Controller

a(t)

Spacecraft Orbital Dynamics

r(t), v(t)

Fig. 1.2 Spacecraft navigational system with propulsion and attitude dynamics neglected

thrust vector generation, such that T¯ (t) = Td (t), q(t) = qd (t), and ω(t) = ωd (t) at any given time, t. The acceleration input, a(t) = T (t)/m(t) ∈ R3×1 , which actually arises out of T¯ (t), q(t), ω(t), β(t), is thus assumed to be instantly variable, and hence regarded as the sole control input vector for the navigation problem. Therefore we have no further need to model the propulsion and the attitude control subsystems in order to study the navigational system.

Chapter 2

Analytical Optimal Control

2.1 Introduction Optimization refers to the process of achieving the best possible result (objective), given the circumstances (constraints). When applied to determine a control strategy for fulfilling a desired task, such an optimization is called optimal control. Although optimization has been studied for millennia, optimal control received a fillip in the 1950s with the publication of two treatises, one by Richard Bellman [13], and the other by Pontryagin [68]. Bellman essentially introduced the terminology we find in usage today and proposed the principle that a trajectory which minimizes an objective for a part of it continues to minimize the same objective through the remainder of the trajectory. Pontryagin’s minimum principle on the other hand states that any trajectory which fulfills the necessary (but not sufficient) conditions for the minimization of an objective function will always have a value of the objective function greater than that on the optimal trajectory. Both of these observations may seem trivial, but are the basic foundations of the modern optimal control theory. For example, a small but arbitrary variation of the control input from its optimal history can be used to derive the variations in the state variables that are required for satisfying the conditions of optimality. This chapter briefly presents such a variational method for obtaining the necessary and sufficient conditions for minimizing a specific objective function along a trajectory which is the solution to a dynamic constraint equation. The dynamic optimization problem involves determining a solution to the governing equations of a dynamic system, which satisfy the necessary conditions for minimizing a prescribed objective function, subject to boundary conditions and interior constraints. Such a trajectory (and the corresponding control history) is said to be extremal, and all quantities evaluated on the extremal trajectory are indicated by the superscript ∗ in this book. While an extremal trajectory satisfies the necessary conditions of optimality, it may or may not be the optimal one. In order to be the optimal trajectory, an extremal trajectory must satisfy additional conditions, called © Springer Nature Switzerland AG 2019 A. Tewari, Optimal Space Flight Navigation, Control Engineering, https://doi.org/10.1007/978-3-030-03789-5_2

7

8

2 Analytical Optimal Control

the sufficient conditions for optimality. We will briefly consider in this chapter how an extremal trajectory can be determined, and then tested for optimality. The process of analytically deriving the necessary and sufficient conditions for optimization involves the calculus of variations. An alternative method can also be employed where computational techniques directly search for an optimal trajectory without first deriving the conditions for optimality. Such direct search techniques are computationally intensive and fraught with convergence issues, hence their study is a separate topic in itself. Our focus in this book is therefore on the analytical methods of dynamic optimization, whose solution can be derived and implemented much more easily. Notation is kept simple in this chapter due to the many variables involved. Whereas in the later chapters, we will distinguish between scalars and vectors with (or without) the use of bold symbols, in the present chapter, all the variables are in normal font but their dimensions are specified. For example, a vector spanning (n × 1) real space is denoted by x ∈ Rn×1 .

2.2 Optimization of Static Systems In order to motivate the reader, it is necessary to introduce the concepts of optimization of a system that does not depend upon time. Such an optimization consists of time-invariant objective functions and constraints, hence it is considered to be a static problem. To generalize the theory to dynamic systems, time is merely added to the set of independent variables. Optimization refers to either the minimization or maximization of a given function. In the book, we shall pose the optimization problem as a minimization problem without any loss of generality, since changing the sign of the given objective function can easily transform it into a maximization problem. Consider the minimization of a scalar objective function, L(x, u), called the performance index, with respect to a set of mutually independent, unconstrained scalar variables, u = (u1 , u2 , . . . , um )T . The set u ∈ Rm×1 is referred to as control vector, while the set of additional variables, x ∈ Rn×1 , necessary to describe the static system is called the state vector. Clearly, control variables are those with respect to which the performance index is minimized. It is assumed that L(x, u) is continuous and possesses continuous partial derivatives with respect to both x and u up to an order which is sufficient to validate all the arguments presented here. This holds for all the performance indexes contained in this book. In order that a given candidate point (x, u) minimizes a continuously differentiable performance index L(x, u) with respect to u, it must satisfy the following necessary condition:

2.2 Optimization of Static Systems



9

∂L ∂u

T =

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

∂L ∂u1 ∂L ∂u2

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬

⎪ .. ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ . ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ∂L ⎪ ⎭

=0

(2.1)

∂um

where the extreme right-hand side is a null vector of dimension m × 1 and all the derivatives are evaluated at the candidate point. Any point (x ∗ , u∗ ) that satisfies the necessary condition for a minimum is called an extremal point (or a stationary point) of L. However, an extremal point is not guaranteed to be a minimum point, because the necessary condition of Eq. (2.1) is satisfied also by maximum points and saddle points. In order to unambiguously identify a minimum point, a separate condition must be satisfied. This condition requires an increase of L whenever departing from the stationary point in any direction, hence it must be determined from the sign of the second-order partial derivatives of L with respect to u. A stationary point is necessarily a minimum of L(x, u) with respect to u, if the following Hessian is positive definite (that is, has all real, positive eigenvalues) at the given point: ⎛ ⎜ ⎜ ⎜ ⎜ 2 ∂ L ⎜ ⎜ =⎜ ⎜ ∂u2 ⎜ ⎜ ⎜ ⎝

∂2L ∂u1 2

∂2L ∂u1 ∂u2

···

∂2L ∂u1 ∂um

∂2L

∂2L ∂u2 2

···

∂2L

∂u2 ∂u1

∂u2 ∂um

.. .

.. .

.. .

.. .

···

∂2L ∂um ∂um

∂2L ∂2L ∂um ∂u1 ∂um ∂u2

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

(2.2)

The positive definiteness of the Hessian at an extremal point is therefore a sufficient condition for the existence of a minimum of L with respect to u and is denoted by: ∂ 2L >0 ∂u2

(2.3)

On the other hand, if the stationary point satisfies ∂ 2L <0 ∂u2 then it is a maximum point of L(x, u) with respect to u.

(2.4)

10

2 Analytical Optimal Control

If the Hessian has both positive and negative real eigenvalues at a stationary point, then that point is said to be a saddle point of L(x, u) with respect to u. If any of the eigenvalues of the Hessian vanishes at a stationary point, then it is called a singular point. Since a singular point does not strictly satisfy the sufficient condition inequality, it is not an optimal point. Additional condition/s must be imposed to convert a singular point into an optimal point. Any point u satisfying both the necessary and sufficient conditions for a minimum is called a minimum point (or optimal point) of L with respect to u. If L(x, u) ˆ ≤ L(x, u) for all u = u, ˆ then the optimal point uˆ is said to be a global minimum of L(x, u) with respect to u. If a minimum point is not a global minimum, then it is a local minimum. Example 2.1 Determine the minimum points of the following function of real variables:    1 −1 u1 L = {u1 , u2 } u2 −1 4 Since the performance index is a quadratic function of u = (u1 , u2 )T , its only stationary point is the origin: Lu =

   1 −1 ∂L  ∗ = u1 , u∗2 = {0 , 0} −1 4 ∂u

or, u∗1 = u∗2 = 0. The Hessian is the following: Luu

∂ 2L = = ∂u2



1 −1 −1 4

 ,

whose both eigenvalues are real and positive, (0.7, 4.3). As the sufficient condition is satisfied by the Hessian, the origin is the only minimum point of L, hence L = 0 is its global minimum value. Example 2.2 Does the following function of real variables have a minimum point: L = −u21 + 2u1 u2 + 3u22 ? Since the performance index is a quadratic function of u = (u1 , u2 )T , its only stationary point is the origin: Lu =

   −1 1 ∂L  ∗ = u1 , u∗2 = {0 , 0} 1 3 ∂u

or, u∗1 = u∗2 = 0. The Hessian is the following:

2.2 Optimization of Static Systems

11

Luu

∂ 2L = = ∂u2



−1 1 1 3



√ whose eigenvalues are 1 ± 5. Since an eigenvalue of the Hessian is negative while the other is positive, the only stationary point (the origin) is a saddle point of L. Therefore, there are no minimum points of L. Example 2.3 Determine whether the origin is a minimum point of the following function: L = (u1 − u22 )(u1 − 3u22 ) It is easily seen that u∗ = (0, 0)T satisfies the stationary condition, Lu = 0: 

∂L ∂u

T

 =

2u∗1 − 4(u∗2 )2 −8u∗1 u∗2 + 12(u∗2 )3



  0 = 0

and is hence an extremal point of L. The Hessian at the extremal point is the following:  Luu =

∂ 2L ∂u2

T

 =

−8u∗2 2 − 8u∗2 ∗ ∗ −8u2 −8u1 + 36(u∗2 )2



 =

2 0 0 0



whose eigenvalues are (0, 2). As the sufficient condition is not satisfied by the Hessian, the origin is not a minimum point of L. Since one of the eigenvalues of the Hessian vanishes at the origin, the latter is a singular point of L.

2.2.1 Static Equality Constraints Consider the minimization of L(x, u) with respect to u and subject to the vector constraint: f (x, u) = 0

(2.5)

with f : Rn×1 ×Rm×1 → Rn×1 having continuous partial derivatives with respect to both the state variables, x, and control variables, u. The equality constraint must be such that it uniquely determines x from u. This fact translates into the requirement that a small and arbitrary variation in the control, du, must cause a corresponding variation in the state, dx, in order to satisfy the equality constraint. The variations being small allow one to use the following first-order approximation: df = fx dx + fu du = 0

(2.6)

12

2 Analytical Optimal Control

or dx = −fx−1 fu du

(2.7)

with fx = ∂f/∂x ∈ Rn×n and fu = ∂f/∂u ∈ Rn×m . Equation (2.7) requires the square matrix, fx (called the Jacobian) must be non-singular. Since a stationary point must also have a stationary L in the presence of small and arbitrary variation in u, we must have to a first-order approximation: dL = Lx dx + Lu du = 0

(2.8)

while holding df = 0, with Lu = ∂L/∂u ∈ R1×m and Lx = ∂L/∂x ∈ R1×n . This implies that the following must be true:   dL = Lu − Lx fx−1 fu du = 0

(2.9)

Since du is arbitrary, we have the following m scalar equations relating the partial derivatives of L at the stationary point: Lu − Lx fx−1 fu = 0

(2.10)

An alternative necessary condition for minimization in the presence of the equality constraints can be derived by adjoining the constraints with the function being minimized. Define the following Hamiltonian function in which the function to be minimized, L(x, u) (called the Lagrangian), is adjoined to the vector constraint f (x, u) = 0 through a set of additional variables λ ∈ Rn×1 called the Lagrange multipliers or the adjoint variables: H (x, u, λ) = L(x, u) + λT f (x, u)

(2.11)

The stationary condition in the presence of a small but arbitrary variation in u can now be expressed as follows: dH = Hx dx + Hu du = 0

(2.12)

with Hu = ∂H /∂u ∈ R1×m and with Hx = ∂H /∂x ∈ R1×n . For the convenience of determining the Lagrange multipliers, λ (which are completely arbitrary), the following condition is imposed at the stationary point: Hx = Lx + λT fx = 0

(2.13)

(λ∗ )T = −Lx fx−1

(2.14)

This implies that

2.2 Optimization of Static Systems

13

Upon substituting Hx = 0 in Eq. (2.12), the following stationary condition is derived: Hu = Lu + (λ∗ )T fu = 0

(2.15)

Equations (2.13)–(2.15) are the necessary conditions for the minimization of L(x, u) with respect to u, subject to f (x, u) = 0. The sufficient conditions for the minimization of L(x, u) with respect to u, subject to f (x, u) = 0, are derived by taking the variations in L and f up to terms of the second order. Consider the following Taylor series expansions about the stationary point:  dL = {Lx , Lu }  df = {fx , fu }

   L 1 T dx dx xx Lxu dx , duT + Lux Luu du du 2

(2.16)

   f 1 T dx dx xx fxu dx , duT + fux fuu du du 2

(2.17)

where Lxu =

∂ ∂ 2L = ∂x∂u ∂u



∂L ∂x

T

∈ Rn×m , etc., are the second-order partial derivatives. The Hamiltonian variation can similarly be expanded into a second-order Taylor series, and substituting the necessary conditions for optimality, Hx = 0 and Hu = 0, we have dH =

  H 1 T dx xx Hxu dx , duT Hux Huu du 2

(2.18)

The substitution of Eqs. (2.17)–(2.18) into dL = dH − λT df ,

(2.19)

along with the following Taylor series variation in x, dx = −fx−1 fu du + terms involving higher powers of (dx, du)

(2.20)

results in the following variation in L up to second-order terms: ⎫ ⎛ ⎞⎧ −1  Hxx Hxu ⎨ −fx fu ⎬ 1 T  T T −1 ⎠ dL = du −fu (fx ) , I ⎝ du ⎩ ⎭ 2 Hux Huu I

(2.21)

14

2 Analytical Optimal Control

The sufficient condition for the minimization of L(x, u) with respect to u, subject to f (x, u) = 0, is thus the following: 

∂ 2L ∂u2

 f =0

⎫ ⎛ ⎞⎧ −1   Hxx Hxu ⎨ −fx fu ⎬ ⎠ = −fuT (fxT )−1 , I ⎝ >0 ⎩ ⎭ Hux Huu I

(2.22)

or Huu − Hux fx−1 fu − fuT (fxT )−1 Hxu + fuT (fxT )−1 Hxx fx−1 fu > 0

(2.23)

This form of the sufficient condition for optimality in the presence of equality constraints is referred to as the Legendre–Clebsch sufficient condition. Example 2.4 Determine the extremal and minimum points for   1 y2 u2 L= + 2 a2 b2 with respect to u, and subject to the following constraint: y + mu − c = 0 where a, b, c, m are real constants. We begin with defining the following Hamiltonian function in which the function to be minimized, L, and the equality constraint are combined: H =

1 2



y2 u2 + 2 2 a b

 + λ (y + mu − c)

The necessary conditions for minimization are the following: ∂H y∗ = 2 +λ=0 ∂y a ∂H u∗ = 2 + mλ = 0 ∂u b solving which along with the equality constraint, y ∗ + mu∗ − c = 0, results in the following extremal point: y∗ =

a2c a 2 + m2 b 2

b2 mc + m2 b 2 c λ∗ = − 2 a + m2 b 2

u∗ =

a2

2.2 Optimization of Static Systems

15

The sufficient condition for minimization requires that (Luu )f =0 > 0, which is seen to be satisfied at the extremal point: (Luu )f =0 =

1 m2 + >0 b2 a2

Thus, the extremal point is a minimum point, with the following minimal value of L: Lmin =

2(a 2

c2 + m2 b 2 )

Example 2.5 Determine the extremal and minimum points for L = 3z2 + u21 ez + u22 y 2 + (u3 − 2)2 with respect to u = (u1 , u2 , u3 )T ∈ R3×1 , subject to the following constraints: z − u1 y 2 = 0 z + 2u3 = 0 where x T = (z, y) ∈ R1×3 . We begin with defining the following Hamiltonian function: H = 3z2 + u21 ez + u22 y 2 + (u3 − 2)2 + λ1 (z − u1 y 2 ) + λ2 (y + 2u3 ) Here, (λ1 , λ2 ) ∈ R1×2 are the Lagrange multipliers corresponding to the two constraints, respectively. The problem of minimizing L relative to u while satisfying the two equality constraints is thus translated into the problem of minimizing the Hamiltonian function, H , relative to u. The extremal control, u∗ , satisfying the stationary condition is derived as follows: ⎧ ⎫ ⎧ ⎫ ∗ ⎪ ⎪ 0⎪ 2u∗1 ez − λ∗1 (y ∗ )2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪  T ⎬ ⎨ ⎨ ⎬ ∂H ∗ ∗ 2 = 0 = 2u2 (y ) ⎪ ⎪ ⎪ ⎪ ∂u ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ ⎩ ⎭ ∗ ∗ 0 2(u3 − 2) + 2λ2 which is identically satisfied by: λ∗1 (y ∗ )2 ∗ 2ez u∗2 = 0 u∗1 =

u∗3 = 2 − λ∗2

16

2 Analytical Optimal Control

In order to determine the Lagrange multipliers, the following condition is imposed: ⎫ ⎧ ∂H ⎪ ⎬ ⎨ ∂z ⎪ ⎪ ⎭ ⎩ ∂H ⎪

=

⎧ ⎨ ⎩

∂y

∗

6z∗ + (u∗1 )2 ez + λ∗1 2(u∗2 )2 y ∗

− 2λ∗1 u∗1 y ∗

⎫ ⎬

⎭ + λ∗2

=

⎧ ⎫ ⎨0⎬ ⎩ ⎭ 0

which are satisfied by: λ∗1 = −6x ∗ − (u∗1 )2 ez

∗

λ∗2 = 2u∗1 λ∗1 y ∗ When the extremal control values are substituted into the Lagrange multipliers, we have a quadratic equation for u∗1 which is solved to yield the following stationary value of u∗1 :  u∗1

1 =− ∗ 2 ± (y )

1 6z∗ − z∗ , ∗ 4 e (y )

(y ∗ = 0)

For the extremal control variable u∗1 to have a real value, it is necessary to impose the restriction: ∗

ez ≥ 6z∗ (y ∗ )4 ,

(y ∗ = 0)

The stationary control variable u∗1 can now be substituted into the expressions derived above for λ∗1 , λ∗2 , and u∗3 in order to determine their values. Hence, the stationary points are obtained. In order to investigate the minimum points, we derive the Hessian for the sufficient condition in the following steps: Hu =

 ∂H = 2u1 ez − λ1 y 2 , ∂u ⎛

2u2 y 2 ,

2(u3 − 2) + 2λ2

⎞ 2ez 0 0 Huu = ⎝ 0 2y 2 0 ⎠ 0 0 2   z 0 0 2e u1 T = Hux −2yλ1 4yu2 0

 Hx = 6z + ez u21 + λ1 ,  Hxx =

2yu22 − 2λ1 u1 y + λ2

0 6 + u21 ez 0 2u22 − 2λ1 u1







2.2 Optimization of Static Systems

17

⎛

T Hxu

⎞ 2u21 ez −2yλ1 =⎝ 0 4yu2 ⎠ 0 0 

1 −2yu1 0 1

fx = fx−1

 = 

fu = fx−1 fu =



1 2yu1 0 1

−y 2 0 0 0 0 2







−y 2 0 4u1 y 0 0 2



⎛

⎞ −2y 2 u1 ez 0 8ez yu21 − 4λ1 y ⎠ Hux fx−1 fu = ⎝ 0 0 8yu2 0 0 0 ⎛

(fx−1 fu )T Hxu

⎞ −2y 2 u1 ez 0 0 =⎝ 0 0 0⎠ 2 z 8e yu1 − 4λ1 y 8yu2 0

fuT (fxT )−1 Hxx fx−1 fu = ⎞ 6y 4 + u21 y 4 ez 0 −24u1 y 3 − 4u31 y 3 ez ⎠ ⎝ 0 0 0 3 2 4 2 3 3 z 2 2 z −24u1 y − 4y u1 e 0 96u1 y + 16y u1 e + 8u2 − 8u1 λ1 ⎛

∗

Substitution of λ∗1 = −6x ∗ − (u∗1 )2 ez and u∗2 = 0 for the stationary points results in the following expression for the required Hessian: (Luu )f =0 = Huu − (fx−1 fu )T Hxu − Hux fx−1 fu + fuT (fxT )−1 Hxx fx−1 fu ⎛ ⎞∗  4  6y + u21 y 4 ez 0 − 24u1 y 3 + 4u31 y 3 ez  ⎟ ⎜ +2ez + 4y 2 u ez  +24xy + 12ez yu21 1 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 2 =⎜ −8yu2 0 2y ⎟ ⎜ ⎟ ⎜ ⎟  ⎜  ⎟ ⎝ − 24u1 y 3 + 4u31 y 3 ez −8yu2 2 + 96u21 y 2 + 16y 2 u41 ez ⎠   +8u22 − 8u31 ez − 48xu1 +24xy + 12ez yu21

18

2 Analytical Optimal Control

where the stationary expression for u1 , calculated above, must be substituted. The minimum points of L are the ones for which (Luu )f =0 has real and positive eigenvalues in the (z, y) space. Of course, this has to be determined by a numerical search. As the previous example illustrates, often the solution for the minimum points cannot be obtained in a closed form, and instead requires a numerical solution. The search for extremal points that satisfy the sufficient condition of minimization in the presence of implicit, nonlinear constraints in terms of the control and state variables must be carried out by a nonlinear programming algorithm. Such iterative computational schemes form the crux of the current research on multivariable optimization methods.

2.2.2 Inequality Constraints Consider general inequality constraints of the form: f (x, u) ≤ 0 ,

(2.24)

while minimizing L(x, u) with respect to u. Extending the concept of Lagrange multipliers, λ, we adjoin L with the inequality constraint function into the following augmented objective function: J (x, u) = L(x, u) + λT f (x, u) .

(2.25)

Noting the two possibilities for an optimal point (if it exists), i.e., either f (x, ˆ u) ˆ <0,

(2.26)

f (x, ˆ u) ˆ =0,

(2.27)

or

we say that in the first case (Eq. (2.26)), the constraint is inactive and thus it can be ignored by putting λ = 0 in Eq. (2.25), which is treated as unconstrained minimization we discussed above. In the second case (Eq. (2.27)), the minimum point (if it exists) lies on the constraining boundary, and a small control variation, δu, about the minimum point, (x, ˆ u), ˆ would result in either an increase, or no change of L: δL =

∂L δu ≥ 0 , ∂u

(2.28)

2.2 Optimization of Static Systems

19

where δu must satisfy the constraint, Eq. (2.24), with λ > 0: λT δf = λT

∂f δu ≤ 0 ∂u

(2.29)

Since δu is arbitrary, there are only the following two possibilities of reconciling Eqs. (2.28) and (2.29), i.e., either we should have (with λ > 0) ∂L ∂f = −λT ∂u ∂u

(2.30)

or ∂L =0; ∂u

∂f =0. ∂u

(2.31)

The three possibilities represented by Eqs. (2.26), (2.30), and (2.31) can be expressed in the following compact form: ∂f ∂L + λT =0, ∂u ∂u

(2.32)

where  λ

= 0, f (x, u) < 0 > 0, f (x, u) = 0

(2.33)

Equation (2.32) is the necessary condition for optimality and must be solved in an iterative manner to yield an optimal point (if it exists). Whenever inequality constraints are specified, they denote a bounded (or feasible) region in which the search for any optimal points is to be performed. If a point falls at the boundary of the feasible region, then it may not satisfy the necessary and sufficient conditions for minimization (or maximization) of the performance index, yet it could be an optimal point simply because it has the smallest (or the largest) numerical value of the performance index in the feasible region. Thus, we can categorize the search for a minimum in the presence of inequality constraints as follows: 1. The minimum point lies inside the feasible region, where both the necessary and sufficient conditions of optimality are satisfied. 2. The minimum point lies on a boundary of the feasible region and satisfies the necessary (but not sufficient) conditions of optimality. An example of this case is a singular control problem in which an inequality constraint supplies the additional information for determining the minimality of the given point. 3. The minimum point lies on a boundary of the feasible region, but does not satisfy the necessary conditions of optimality.

20

2 Analytical Optimal Control

2.3 Dynamic Equality Constraints and Unbounded Inputs We begin the treatment of the optimal control problem with the case of continuously applied, smooth control inputs which are unbounded by any physical constraints. Consider the problem of finding an admissible set of control inputs u∗ (t) ∈ Rm×1 to follow an extremal trajectory x ∗ (t) ∈ Rn×1 , which is the solution to the following state equation: x˙ = f (x, u, t)

(2.34)

x(t0 ) = x0

(2.35)

subject to the initial condition:

such that the following performance index is minimized with respect to the control inputs: 



J = φ[x(tf ), tf ] +

tf

L(x, u, t)dt

(2.36)

t0

where f : Rn×1 × Rm×1 × R → Rn×1 , φ : Rn×1 × R → R, and L : Rn×1 × Rm×1 × R → R are continuous and possess continuous partial derivatives with respect to their arguments, up to an indefinite order. While φ is the terminal cost to be minimized, the integral term containing the Lagrangian, L(x, u), represents the transient cost to be minimized over the control interval, t ∈ [t0 , tf ]. The general form of the performance index given in Eq. (2.36) is termed the Bolza problem. A special optimal control problem in which φ[x(tf ), tf ] = 0 is called the Lagrange problem, whereas the one with L(x, u, t) = 0 is termed the Mayer problem. We begin by writing the terminal cost as follows: 

tf

φ[x(tf ), tf ] =

˙ φ(x, t)dt + φ[x(t0 ), t0 ]

(2.37)

t0

which substituted into Eq. (2.36) results in J = φ[x(t0 ), t0 ] +



tf

˙ [L(x, u, t) + φ(x, t)]dt

(2.38)

t0

Since x(t0 ) and t0 are both fixed, they do not affect the minimization, and only the following performance index needs to be minimized:  J =

tf

t0

˙ [L(x, u, t) + φ(x, t)]dt

(2.39)

2.3 Dynamic Equality Constraints and Unbounded Inputs

21

which is also expressed as follows:  J =

tf

t0

  ∂φ ∂φ L(x, u, t) + (x, t) + (x, t)x˙ dt ∂t ∂x

(2.40)

Let x ∗ (t) be an extremal trajectory satisfying the necessary conditions for the minimization of J subject to Eq. (2.34), and let u∗ (t) be the corresponding control history. We augment J by adjoining the state equation as the dynamic equality constraint through Lagrange multipliers, λ(t) ∈ Rn×1 (also called the adjoint, or costate variables):  Ja =

tf

[L +

t0

∂φ ∂φ + x˙ + λT (f − x)]dt ˙ ∂t ∂x

(2.41)

Introducing the following augmented Lagrangian: ∂φ ∂φ ˙ + x˙ + λT (f − x) ∂t ∂x

La = L +

we have the following augmented performance index to be minimized:  tf La (x, u, λ, x, ˙ t)dt Ja =

(2.42)

(2.43)

t0

Next, consider infinitesimal and arbitrary variations in x(t), u(t), λ(t), x(t) ˙ about the extremal trajectory represented by δx(t), δu(t), δλ(t), δ x(t), ˙ respectively.1 The net change in the augmented performance index over the extremal trajectory is then given by:  dJa =  +

tf

[La (x ∗ + δx, u∗ + δu, λ∗ + δλ, x˙ ∗ + δ x, ˙ t) − La (x ∗ , u∗ , λ∗ , x˙ ∗ , t)]dt

t0 tf +δtf

La (x, u, λ, x, ˙ t)dt

(2.44)

tf

The first integrand on the right-hand side of Eq. (2.44) is expanded in a first-order Taylor series to yield  dJa =

tf

t0

 +



 ∂La ∂La ∂La ∂La δx + δu + δ x˙ + δλ dt ∂x ∂u ∂ x˙ ∂λ

tf +δtf

La (x, u, λ, x, ˙ t)dt

(2.45)

tf

1 Since the dynamic equality constraint of Eq. (2.34) must be enforced separately, x˙ is taken to be an independent variable in the minimization of Ja .

22

2 Analytical Optimal Control

where all coefficients are evaluated on the admissible trajectory. Now, the second integral on the right-hand side of Eq. (2.44) is expressed as follows: 

tf +δtf

La (x, u, λ, x, ˙ t)dt = La (x(tf ), u(tf ), λ(tf ), x(t ˙ f ), tf )δtf

tf

+ higher order terms involving δtf

(2.46)

Neglecting the second- (and higher-) order terms in δtf in Eq. (2.46), and expanding the augmented Lagrangian in a first-order Taylor series about the extremal trajectory at the final time, we have La (x(tf ), u(tf ), λ(tf ), x(t ˙ f ), tf ) = La [x ∗ (tf ), u∗ (tf ), λ∗ (tf ), x˙ ∗ (tf ), tf ]     ∂La ∂La δx(tf ) + δu(tf ) + ∂x t=tf ∂u t=tf     ∂La ∂La + δλ(tf ) + δ x(t ˙ f) ∂λ t=tf ∂ x˙ t=tf (2.47) When Eq. (2.47) is substituted into Eq. (2.46), we have the following first-order expansion of the second integral on the right-hand side of Eq. (2.44): 

tf +δtf

La (x, u, λ, x, ˙ t)dt = La [x ∗ (tf ), u∗ (tf ), λ∗ (tf ), x˙ ∗ (tf ), tf ]δtf

(2.48)

tf

In the first integral on the right-hand side of Eq. (2.45), the term involving δ x˙ is expanded by integration by parts as follows: 

tf

t0

 tf  tf   ∂La d ∂La δx δxdt − ∂ x˙ ∂ x˙ t0 dt t0   ∂La = δx(tf ) ∂ x˙ t=tf      tf d ∂La ∂La δxdt δx(t0 ) − − ∂ x˙ t=t0 ∂ x˙ t0 dt

∂La δ xdt ˙ = ∂ x˙



(2.49)

However, since δx(t0 ) = 0 (the initial condition being fixed), we have 

tf

t0

∂La δ xdt ˙ = ∂ x˙



∂La ∂ x˙



 δx(tf ) −

t=tf

tf

t0

d ∂La δxdt dt ∂ x˙

(2.50)

2.3 Dynamic Equality Constraints and Unbounded Inputs

23

The substitution of Eqs. (2.48) and (2.49) into Eq. (2.45) leads to the following necessary condition for the minimization of Ja :  dJa = 0 =  +

tf

t0

 ∂La δx(tf ) + La [x ∗ (tf ), u∗ (tf ), λ∗ (tf ), x˙ ∗ (tf ), tf ]δtf ∂ x˙ t=tf     ∂La ∂La d ∂La ∂La δx + − δu + δλ dt (2.51) ∂x dt ∂ x˙ ∂u ∂λ

Before proceeding further, it is necessary to express the variation in the terminal state by the following first-order approximation: ∂x(tf ) = δxf − x˙ ∗ (tf )δtf

(2.52)

This is tantamount to a linear extrapolation around the extremal trajectory at the terminal time, as indicated by the schematic diagram of Fig. 2.1. Clearly, when the terminal condition is not fixed, the variation of the terminal state depends linearly upon the variation in the terminal time. When Eq. (2.52) is substituted into Eq. (2.51), we have the following necessary condition for minimization:     ∂La ∂La ∗ δ x˙ dJa = 0 = δxf + La − δtf ∂ x˙ t=tf ∂ x˙ t=tf     tf  ∂La d ∂La ∂La ∂La − δu + δλ dt (2.53) δx + + ∂x dt ∂ x˙ ∂u ∂λ t0 Fig. 2.1 Extrapolation of the terminal state in the neighborhood of the extremal trajectory for a small, arbitrary variation in the terminal time

.*

(tf)δtf

x

δx(tf)

δ xf

xf

x*(t) x(t)

xi

ti

tf

tf +δ tf

t

24

2 Analytical Optimal Control

It is now possible to derive the individual necessary conditions for optimality from Eq. (2.53). The terms involving the terminal cost, φ, in the augmented Lagrangian (see Eq. (2.42)) must satisfy the following condition inside the integral term of Eq. (2.53): ∂ ∂x



∂φ ∂x



δ x˙ ∗ +

∂φ ∂t

 x=x ∗

    d ∂ ∂φ x˙ ∗ (2.54) dt ∂ x˙ ∂x x=x ∗  2    ∂ φ d ∂φ ∂ 2φ ∗ = x ˙ − + ∂t∂x dt ∂x ∂x 2    2  ∂ 2φ ∂ φ ∂ 2φ ∂ 2φ ∗ x ˙ x˙ ∗ − − = + 2 2 ∂t∂x ∂x∂t ∂x ∂x

−

Since the partial derivatives of φ must necessarily be continuous, we must have ∂ 2 φ/∂t∂x = ∂ 2 φ/∂x∂t, hence the term on the right-hand side of Eq. (2.54) vanishes identically. Thus, all the terms involving φ inside the integral term of Eq. (2.53) cancel out, thereby yielding the following necessary condition for optimality arising out of that particular term:    ∂L ∂L d ∂f ∂f + (λ∗ )T − (−λ∗ )T δx + + (λ∗ )T δu ∂x ∂x dt ∂u ∂u t0 T   + f − x˙ ∗ δλ dt (2.55) 

0=

tf



Equation (2.55) must hold irrespective of any boundary conditions on x and λ. Furthermore, the equation of dynamic constraint (Eq. (2.34)) must be satisfied along the extremal trajectory: x˙ ∗ = f (x ∗ , u∗ , t)

(2.56)

which implies that the coefficient of δλ must vanish in Eq. (2.55), thereby leaving the following necessary condition to be satisfied for optimality:     ∂L ∂L d ∗ T ∗ T ∂f ∗ T ∂f + (λ ) + (λ ) δx + + (λ ) δu dt = 0 ∂x ∂x dt ∂u ∂u t0 (2.57) Because Eq. (2.57) must hold irrespective of variations on x and u, the coefficients of both δx and δu must independently vanish along the extremal trajectory. Thus, we have 

tf



(λ˙ ∗ )T = −

∂f ∂L − (λ∗ )T ∂x ∂x

(2.58)

and ∂L ∂f + (λ∗ )T =0 ∂u ∂u

(2.59)

2.3 Dynamic Equality Constraints and Unbounded Inputs

25

These last two equations, along with the state equation, Eq. (2.34), are the necessary conditions which must be satisfied by the extremal trajectory, x ∗ (t) and the extremal control history u∗ (t), in the control interval t ∈ [t0 , tf ]. Of these, Eq. (2.58) is termed the costate equation (or the adjoint equation), while Eq. (2.59) is called the stationary condition. Furthermore, the term containing δxf and δtf in Eq. (2.53), representing the terminal boundary variation, must also vanish, resulting in the following transversality condition: 

∂La ∂ x˙

  ∂La ∗ x˙ δxf + La − δtf = 0 ∂ x˙ t=tf

 t=tf

(2.60)

In order to combine the requirements of optimality with the dynamic constraint, a Hamiltonian function is defined as follows: H (x, u, λ, t) = L(x, u, t) + λT f (x, u, t)

(2.61)

Then, the necessary conditions for optimality are expressed as follows in terms of the Hamiltonian function:   ∂H T ∗ ∗ ∗ ∗ ∗ ∗ x˙ = f (x , u , t) = (x , u , λ , t) (2.62) ∂λ λ˙ ∗ = −



∂L ∂x

T

 −

∂f ∂x

T

∗



λ =

∂H ∂x

T

(x ∗ , u∗ , λ∗ , t)

(2.63)

and 

∂L ∂u

T

 +

∂f ∂x

T

λ∗ = 0 =

∂H ∗ ∗ ∗ (x , u , λ , t) ∂u

(2.64)

The performance index of optimization can be expressed as follows in terms of the Hamiltonian:  tf  H (x, u, λ, t) − λT x˙ dt (2.65) J = φ[x(tf ), tf ] + t0

It is to be noted that the Hamiltonian is related to the augmented Lagrangian, Eq. (2.42), as follows: La = H +

∂φ ∂φ + x˙ − λT x˙ ∂t ∂x

(2.66)

Then, the transversality condition can be derived from Eq. (2.60) to be the following: !

∂φ ∂x

T

"T ∗

− λ (tf ) t=tf

  ∂φ δxf + H + δtf = 0 ∂t t=tf

(2.67)

26

2 Analytical Optimal Control

Equations (2.62)–(2.64), along with Eq. (2.67), are collectively termed the Euler– Lagrange equations, which must be satisfied on any trajectory (the extremal) minimizing the performance index. In addition to the scalar boundary condition of Eq. (2.67) imposed on the terminal time tf , boundary conditions on x ∗ (t), λ∗ (t) must be specified in order to satisfy the Euler–Lagrange equations. For a system of order n, the total number of boundary conditions necessary to pose the optimal control problem is thus (2n + 1). Of these, n boundary conditions are already specified by the initial condition, Eq. (2.35). Hence, only (n+1) additional boundary conditions are necessary to completely pose the optimal control problem.

2.4 Special Boundary Conditions Optimal control formulations can be classified according to the boundary conditions imposed on the feasible space and control interval. These are now taken up in detail depending upon how the terminal time and state are specified.

2.4.1 Fixed Terminal Time When the terminal time is fixed, we cannot have any variation in tf , therefore δtf = 0, and the total number of boundary conditions to be imposed on the system is reduced to 2n. A further classification occurs due to the boundary conditions imposed upon the terminal state, x(tf ), as follows.

Fixed Terminal State Here, the terminal state is specified to be a fixed point in the state space: x(tf ) = xf

(2.68)

which implies that there can be no variation on the terminal state at all (δxf = 0). Consequently, Eq. (2.67) is identically satisfied, and the initial condition, Eq. (2.35), along with the final boundary condition, Eq. (2.68), furnishes the required 2n boundary conditions for the problem.

Free Terminal State When no boundary condition is imposed at all on the terminal state, the transversality condition, Eq. (2.67), results in the following n boundary conditions on the adjoint (or costate) variables:

2.4 Special Boundary Conditions

27

λ∗ (tf ) =



∂φ ∂x

T (2.69) t=tf

Terminal State on a Hypersurface Sometimes instead of fixing the terminal state x(tf ) to be at a point, it is given to lie anywhere in a region of the state space specified by: F [x(tf )] = 0

(2.70)

where F : Rn×1 → Rk×1 , 1 ≤ k ≤ (n − 1). Note that the requirement n < 1 is necessary to preclude the case n = 1, which is tantamount to fixing the final state (δxf = 0). The scalar components of Eq. (2.70) can be expressed as follows: Fi [x(tf )] = 0 ,

i = 1, . . . , k

(2.71)

The hypersurface is thus represented by contours, Fi = 0, i = 1, . . . , k, and the condition that x(tf ) must lie on the hypersurface is enforced by ensuring that the variation of x(tf ) is tangential to each contour. Thus, we have k #

 ci

i=1

∂Fi ∂x

 δx(tf ) = 0 ,

i = 1, . . . , k

(2.72)

t=tf

However, since the final state x(tf ) should also be on an extremal trajectory minimizing J with respect to u, the variation in the terminal state, δxf , is exactly specified by the normal δx(tf ) to each of the hypersurfaces, Fi [x(tf )] = 0, 1 ≤ k ≤ (n − 1), bounding the terminal region at the final extremal point, x ∗ (tf ), represented by Eq. (2.67) as follows: !

∂φ ∂x

"T

T

∗

− λ (tf )

δx(tf ) = 0

(2.73)

t=tf

where x(tf ) satisfies Eq. (2.72). Hence, the requirements that x(tf ) should be on the hypersurface and that it should also fall on the extremal trajectory are combined into the following n terminal boundary conditions for the state variables: !

∂φ ∂x

"

T −λ



∗

= c1 t=tf

∂F1 ∂x



 + c2 t=tf

∂F2 ∂x



 + · · · + ck t=tf

∂Fk ∂x

 t=tf

(2.74) where c1 , c2 , . . . , ck are the constant Lagrange multipliers to be determined. This implies that apart from the initial condition, Eq. (2.35), there are a total of (n + k)

28

2 Analytical Optimal Control

terminal boundary conditions of which n are specified by Eq. (2.74), and the remaining k conditions are given by: F [x ∗ (tf )] = 0

(2.75)

Equation (2.74) is equivalent to replacing the terminal cost φ in Eq. (2.69) with the following: Φ = φ + cT F [x(tf )]

(2.76)

where c = (c1 , c2 , . . . , ck )T . Thus, the boundary condition for free terminal state is used to represent the case of a terminal state constrained to be on a hypersurface using the modified terminal cost, which adjoins the terminal constraint with the original terminal cost using the additional adjoint parameters, c1 , c2 , . . . , ck , which are constants due to the static nature of the equality constraint, Eq. (2.75).

2.4.2 Free Terminal Time When the terminal time tf is not specified while stating the optimal control problem, the variations on the terminal time are allowed, (δtf = 0), and Eq. (2.67) provides the boundary condition to determine tf .

Fixed Terminal State When the terminal state is fixed, x(tf ) = xf

(2.77)

there cannot be any variation on the terminal state (δxf = 0). Consequently, Eq. (2.67) leads to the following scalar boundary condition:   ∂φ H+ =0 ∂t t=tf

(2.78)

Free Terminal State Since both δxf and δtf are arbitrary and nonzero, their respective coefficients in Eq. (2.67) must independently vanish. Thus, we have the following boundary conditions:

2.4 Special Boundary Conditions

29

λ∗ (tf ) =



∂φ ∂x

T (2.79) t=tf

and   ∂φ H+ =0 ∂t t=tf

(2.80)

Time-Varying Terminal State When the terminal boundary condition is a moving point in the state space, described by a time-varying function h(t) : R → Rn×1 , the variation in the terminal state is expressed by a first-order time difference as follows:  δxf =

dh dt

 (2.81)

δtf t=tf

Substituted into Eq. (2.67), this results in the following scalar boundary condition: 

∂φ H+ ∂t

!

 + t=tf

∂φ ∂x

"T

T −λ



∗ t=tf

dh dt

 =0

(2.82)

t=tf

Terminal State on a Moving Hypersurface The terminal state lies anywhere in a region of the state space specified by: F [x(tf ), tf ] = 0

(2.83)

where F : R(n+1)×1 → Rk×1 , 1 ≤ k ≤ (n − 1). On such a surface, δxf and δtf are independent of each other, and both, being arbitrary variations, must have independently vanishing coefficients in Eq. (2.67). Thus, we have the following boundary conditions:   ∂φ H+ =0 ∂t t=tf and !

∂φ ∂x

"

T −λ



∗

= c1 t=tf

∂F1 ∂x



 + c2 t=tf

∂F2 ∂x

(2.84)



 + · · · + ck t=tf

∂Fk ∂x

 t=tf

(2.85)

30

2 Analytical Optimal Control

where c1 , c2 , . . . , ck are time-varying coefficients to be treated as additional variables for adjoining the dynamic terminal constraint, Eq. (2.83), to the performance index. Hence, there are in total (2n+k+1) boundary conditions for an equal number of variables, x ∗ , λ∗ , tf , c1 , c2 , . . . , ck .

Terminal State Partially Fixed The case where some state variables are either fixed or constrained to lie on a specified hypersurface, while other state variables are free at the unspecified terminal time, has to be specially treated. Let the partially specified state variables be referred by the following boundary condition: xi (tf ) = xf ,

i = 1, 2, . . . , q

(2.86)

where q ≤ n. This can be regarded as a special case of final state lying on a hypersurface, where F [x(tf ), tf ] = Gx(tf ) − xf

(2.87)

% $ G = Iq×q , 0q×(n−q)

(2.88)

with

with Iq×q and 0q×(n−q) being the identity and null matrices, respectively, of the assigned dimensions. When substituted into Eq. (2.85), the resulting boundary condition can be stated as follows: λ∗ (tf ) =



∂F ∂φ + c ∂x ∂x

T

 =

t=tf

∂φ ∂x

T + GT c

(2.89)

t=tf

where c ∈ Rq×1 is a vector of constant coefficients (Lagrange multipliers), and 

∂φ ∂x

T t=tf

⎧ 0q×1 ⎪ ⎪ ⎪ ⎨ ∂φ/∂xq+1 T = .. ⎪ ⎪ . ⎪ ⎩ (∂φ/∂xn )T

⎫T ⎪ ⎪ ⎪ ⎬ (2.90)

⎪ ⎪ ⎪ ⎭ t=tf

Therefore, the terminal boundary conditions for the costate variables can be expressed as follows: & ∗

λ (tf ) =



c T ∂φ/∂xj t=t

, f

j = (q + 1), . . . , n

(2.91)

2.6 Pontryagin’s Minimum Principle

31

2.5 Sufficient Condition for Optimality Let [x ∗ (t), u∗ (t), λ∗ (t)] be an extremal trajectory and associated control history satisfying Hu = 0 as well as the other Euler–Lagrange necessary conditions for optimality. The weak form of the Legendre–Clebsch condition given by: Huu ≥ 0

(2.92)

is a necessary condition for minimization of H relative to u. However, an extremal trajectory satisfying all the necessary conditions of optimality may not be the optimal trajectory. In addition to satisfying the necessary conditions, if the given extremal trajectory satisfies the following Legendre–Clebsch condition of a positive definite Hessian of the Hamiltonian with respect to the control variables: Huu =

∂ 2H >0 ∂u2

(2.93)

then the extremal trajectory is the optimal one. Hence, the condition given by the strict inequality of Eq. (2.93) (called the strong Legendre–Clebsch condition) is the sufficient condition for optimality on an extremal trajectory. In cases where the Hamiltonian is a linear function of the control variables, the Legendre–Clebsch sufficient condition is not satisfied, and instead we have Huu = 0 on an extremal trajectory. Such a problem is said to be singular in control, and requires additional conditions to determine the optimal trajectory.

2.6 Pontryagin’s Minimum Principle Pontryagin’s minimum principle states that if u(t) ˆ minimizes the performance index J given by Eq. (2.40), subject to the dynamics state equality constraint, Eq. (2.34), then any variation δ uˆ in the control applied on the corresponding optimal trajectory, ˆ [x(t), ˆ λ(t)], produces either an increase, or an invariance of the Hamiltonian, H . In other words, an optimal control u(t) ˆ is the one that minimizes the Hamiltonian among all admissible controls, u ∈ U ⊂ Rm×1 , at any point on the optimal trajectory. The minimum principle can thus be expressed as the following inequality: H (x, ˆ uˆ + δ u, ˆ λˆ , t) ≥ H (x, ˆ u, ˆ λˆ , t) ,

∀u ∈ U ,

t ∈ [t0 , tf ]

(2.94)

Since the minimum principle does not require a Hamiltonian to be a smooth function of the control variables, it provides a more general necessary condition of optimality than Euler–Lagrange equations. Hence, optimal control problems with a non-smooth Hamiltonian can be solved using the minimum principle.

32

2 Analytical Optimal Control

It is a trivial matter to show that Euler–Lagrange equations lead to the minimum principle, if the Hamiltonian function possesses continuous partial derivatives up to the second order with respect to u. The necessary conditions for the minimization of a smooth performance index are the following: Hu (x, ˆ u, ˆ λˆ , t) = 0 ˆ t) ≥ 0 Huu (x, ˆ u, ˆ λ,

(2.95)

A variation in the control about the given optimal trajectory while holding (x, ˆ λˆ ) unchanged produces the following second-order expansion of the Hamiltonian: 

H (x, ˆ uˆ + δ u, ˆ λˆ , t) = H (x, ˆ u, ˆ λˆ , t) +

∂H ∂u



1 δu + δ uˆ T 2



∂ 2H ∂u2

 δ uˆ

(2.96)

which, after substitution of Eq. (2.95), produces Eq. (2.94).

2.7 Hamilton–Jacobi–Bellman Formulation An alternative formulation for the sufficient conditions of optimality is possible through the Hamilton–Jacobi–Bellman approach. For an optimal trajectory, x(t), ˆ minimizing the performance index: 

tf

J = φ[x(tf ), tf ] +

L[x(t), u(t), t]dt ,

(2.97)

t0

subject to: x˙ = f (x, u, t) ,

x(t0 ) = x0 ,

(2.98)

and corresponding to the optimal control history, u(t), ˆ one can define the following optimal return function for t0 ≤ t ≤ tf : Vˆ [x(t), ˆ t] = φ[x(t ˆ f ), tf ] +



tf

L[x(τ ˆ ), u(τ ˆ ), τ ]dτ ,

(2.99)

t

or Vˆ [x(t), ˆ t] = φ[x(t ˆ f ), tf ] −



t tf

Differentiating Vˆ with time, we have

L[x(τ ˆ ), u(τ ˆ ), τ ]dτ .

(2.100)

2.7 Hamilton–Jacobi–Bellman Formulation

33

dVˆ = −L[x(t), ˆ u(t), ˆ t] , dt

(2.101)

and also dVˆ ∂ Vˆ ∂ Vˆ = + x˙ dt ∂t ∂x ∂ Vˆ ∂ Vˆ + f . = ∂t ∂x

(2.102)

By equating Eqs. (2.101) and (2.102), we have the following partial differential equation, called the Hamilton–Jacobi–Bellman (HJB) equation, which must be satisfied by the optimal return function: −

∂ Vˆ ∂ Vˆ = L[x(t), ˆ u(t), ˆ t] + f . ∂t ∂x

(2.103)

The boundary condition for the optimal return function is the following: Vˆ [x(t ˆ f ), tf ] = φ[x(t ˆ f ), tf ] .

(2.104)

A solution for the HJB equation in the optimal trajectory space, x, ˆ and time, t, for a set of initial states, x0 , results in a field of optimal solutions. A finite set of small initial perturbations about a given optimal path thus produces a set of neighboring optimal solutions by the solution of HJB equation. However, a solution of the HJB equation is a formidable task requiring simultaneous integration in space and time for neighboring extremal trajectories with finite-difference (or finitevolume) methods. A more practical use of the HJB equation is to provide a sufficient condition for testing the optimality of a given return function. On comparing Eqs. (2.103) and (2.61), we note that −

∂ Vˆ = H [x(t), ˆ λˆ (t), u(t), ˆ t] ∂t

(2.105)

where λˆ T =

∂ Vˆ . ∂x

(2.106)

With these identities, the HJB equation can be used to provide optimality conditions, but its main application lies in deriving nonlinear feedback control laws of the form: u(t) = uˆ [t, x(t)] , from the arguments of the optimal return function, Vˆ [x(t), ˆ t].

(2.107)

34

2 Analytical Optimal Control

For a system described by: x(t0 ) = x0 ,

x˙ = f (x, t) ;

(2.108)

ˆ t], whose equilibrium point is xe = 0, if the optimal return function, Vˆ [x(t), satisfies Lyapunov’s theorem [74] for global asymptotic stability for all optimal trajectories, x(t), ˆ then the given system is globally asymptotically stable at the origin, x = 0. In this regard, the optimal return function is regarded as a Lyapunov function. Hence, the task of a designer is to find a return function, V (x, t), that is continuously differentiable with respect to the time and the state variables of the system, and satisfies the following sufficient conditions for global asymptotic stability about the equilibrium point at origin: V (0, t) = 0 ,

V (x, t) > 0 ;

dV (x, t) < 0 ; dt

| x |→ ∞ implies V (x, t) → ∞ .

for all x = 0 (2.109) (2.110)

2.8 Time-Invariant Systems Time-invariant systems are those whose properties do not change with time. They are also referred to as autonomous systems in control parlance. The state equation of such a system does not depend explicitly on time, and hence can be expressed as follows: x˙ = f (x, u)

(2.111)

with x(t) : R → Rn×1 and u(t) : R → Rm×1 , and the initial condition: x(t0 ) = x0

(2.112)

If the Lagrangian to be optimized is also time invariant, then the performance index is given by:  J = φ[x(tf ), tf ] +

tf

L(x, u)dt

(2.113)

t0

resulting in the following Hamiltonian: H = L(x, u) + λT f (x, u)

(2.114)

To show that the Hamiltonian of a time-invariant system is constant along an extremal, [x ∗ (t), u∗ (t)], let us consider the following necessary conditions of optimality derived above:

2.9 Linear Systems with Quadratic Cost Functions

35

 T  ∂H ∗ λ˙ ∗ = − ∂x x˙ ∗ = f (x ∗ , u∗ ) =



(2.115)

∂H ∂λ

∗ T (2.116)

and 

∂H ∂u

∗

=0

(2.117)

Substituting Eqs. (2.115)–(2.117) into the following time derivative of the Hamiltonian evaluated along the extremal trajectory: H˙ =



∂H ∂x

∗

x˙ ∗ +



∂H ∂λ

∗

λ˙ ∗ +



∂H ∂u

∗

u˙ ∗

(2.118)

we have H˙ =



∂H ∂x

∗ 

∂H ∂λ

∗ T

 −

∂H ∂λ

∗ 

∂H ∂x

∗ T

=0

(2.119)

This result is useful in deriving relationships among the variables along an extremal trajectory.

2.9 Linear Systems with Quadratic Cost Functions The special case of linear systems with a cost function to be optimized which contains only quadratic terms of state and input variables can be solved analytically. Let the state equation in such a case be given by: x˙ = A(t)x(t) + B(t)u(t)

(2.120)

x(t0 ) = x0

(2.121)

with the initial condition:

where A(t) : R →R n × n and B(t) : R →R n × m are the state-space coefficient matrices.

36

2 Analytical Optimal Control

The objective function for minimization is then expressed as follows:  J = φ[x(tf ), tf ] +

tf

L(x, u, t)dt

(2.122)

1 T x (tf )Qf (tf )x(tf ) 2

(2.123)

t0

where the terminal cost is given by: φ[x(tf ), tf ] = and the Lagrangian is given by: L(x, u, t) =

1 T 1 x (t)Q(t)x(t) + x T (t)S(t)u(t) + uT (t)R(t)u(t) 2 2

(2.124)

where Q(t) : R →R n × n, Qf (tf ) : R →R n × n, S(t) : R →R m × n, and R(t) : R →R m × m are the specified cost coefficient matrices. Equations (2.123) and (2.124) substituted into Eq. (2.122) can also be expressed as follows: 1 J = x T (tf )Qf (tf )x(tf ) + 2



tf

t0



Q(t) S(t) {x (t), u (t)} S T (t) R(t) T

T



 x(t) dt u(t) (2.125)

From the foregoing discussion, it is evident that the matrices Q, Qf , and R must be chosen to be symmetric. Since J ≥ 0, it is also requires that Qf (tf ), Q(t), and R(t) be at least positive semi-definite. Other requirements will be imposed on the cost coefficients from the sufficient conditions of optimality. While the focus of the present section is on linear systems, it is important to realize that an actual physical system is often nonlinear which can be linearized in the neighborhood of an extremal trajectory. A fuller insight into linearized systems’ optimization with a quadratic cost function can be obtained from the expansion of the actual nonlinear system about an extremal trajectory. Consider a dynamic plant with state vector, ξ(t) : R →R n × 1, and control input vector, η(t) : R →R m × 1, governed by the following state equation: ξ˙ = f (ξ, η, t) .

(2.126)

Let an extremal trajectory, ξ ∗ (t), and the corresponding extremal control history, η∗ (t), both satisfying Eq. (2.126), be available from the necessary conditions for the solution of the optimal control problem minimizing an objective function:  J (ξ, η) = φ[ξ(tf ), tf ] +

tf

L[ξ(t), η(t), t]dt ,

(2.127)

t0

subject to certain specific constraints. The extremal control and trajectory are the nominal functions satisfying Eq. (2.126):

2.9 Linear Systems with Quadratic Cost Functions

37

ξ˙ ∗ = f (ξ ∗ , η∗ , t) ,

(2.128)

about which small control and state deviations: u(t) = η(t) − η∗ (t) ;

t 0 ≤ t ≤ tf

(2.129)

x(t) = ξ(t) − ξ ∗ (t) ;

t 0 ≤ t ≤ tf

(2.130)

are to be minimized. A first-order Taylor series expansion about the extremal trajectory leads to the following linearized state equation governing small deviations, x(t), u(t): x(t) ˙ = A(t)x(t) + B(t)u(t) ;

x(t0 ) = x0 ,

(2.131)

' ' ∂f '' ∂f '' A(t) = ; B(t) = ∂ξ 'ξ ∗ ,η∗ ∂η 'ξ ∗ ,η∗

(2.132)

where

are the Jacobian matrices of the expansion. Similarly, the objective function can be expanded about the extremal solution, (ξ ∗ , η∗ ), up to the second-order terms: J (ξ ∗ + x, η∗ + u) J (ξ ∗ , η∗ ) + Δ2 J (x, u) ,

(2.133)

noting that the first variation of J about the extremal trajectory is identically zero: ' ' ∂J '' ∂J '' x+ u=0. dJ (x, u) = ∂ξ 'ξ ∗ ,η∗ ∂η 'ξ ∗ ,η∗

(2.134)

The second variation of J about the extremal trajectory is given by: 1 T x (tf )Qf x(tf ) 2     1 tf T x(t) Q(t) S(t) {x (t), uT (t)} T + , u(t) S (t) R(t) 2 t0

d2 J (x, u) =

(2.135)

which is a quadratic form with the following cost coefficient matrices: ' ' ∂ 2 φ '' ∂ 2 L '' ; Q(t) = Qf = ∂ξ 2 'ξ ∗ (tf ) ∂ξ 2 'ξ ∗ ,η∗

(2.136)

' ' ∂ 2 L '' ∂ 2 L '' S(t) = ; R(t) = . ∂ξ ∂η 'ξ ∗ ,η∗ ∂η2 'ξ ∗ ,η∗

(2.137)

and

38

2 Analytical Optimal Control

Thus, the second variation of the objective function, d2 J , about the extremal trajectory is a quadratic cost function that must be minimized subject to a linearized dynamic equation for a neighboring extremal trajectory. This forms the basis of the linear, quadratic, optimal control problem for neighboring extremal trajectories. The Hamiltonian corresponding to the quadratic cost function, d2 J , and subject to linear dynamic constraint of Eq. (2.131) is the following: H =

1 T 1 x (t)Q(t)x(t) + x T (t)S(t)u(t) + uT (t)R(t)u(t) 2 2

+ λT (t)[A(t)x(t) + B(t)u(t)] .

(2.138)

The necessary conditions for optimality with a fixed terminal time, tf , are then given by Euler–Lagrange equations as follows: λ˙ = −



∂H ∂x

T = −Q(t)x(t) − S(t)u(t) − AT (t)λ(t) ,  λ(tf ) =

∂φ ∂x

(2.139)

T = Qf x(tf ) ,

(2.140)

t=tf

∂H = 0 = S T (t)x(t) + R(t)u(t) + B T (t)λ(t) , ∂u

(2.141)

the last of which is solved for the optimal control as follows:  u(t) = −R −1 (t) S T (t)x(t) + B T (t)λ(t) .

(2.142)

This is the well-known linear, quadratic regulator (LQR) control law. The sufficient condition for optimality requires a positive definite Hessian: ∂ 2H = R(t) > 0 , ∂u2

(2.143)

which is satisfied by selecting a positive definite control cost matrix, R(t), and ensures the existence of its inverse in Eq. (2.142). Substitution of Eq. (2.142) into Eqs. (2.131) and (2.139) results in the following set of linear state and costate equations for the LQR:  x˙ = A(t) − B(t)R −1 (t)S T (t) x(t) − B(t)R −1 (t)B T (t)λ(t) ,

(2.144)

  λ˙ = − AT (t) − S(t)R −1 (t)B T (t) λ(t) + S(t)R −1 (t)S T (t) − Q(t) x(t) , (2.145)

2.9 Linear Systems with Quadratic Cost Functions

39

that must be solved subject to the boundary conditions: x(t0 ) = x0 ;

λ(tf ) = Qf x(tf ) .

(2.146)

The linear, two-point boundary value problem comprising Eqs. (2.144)–(2.146) must be integrated in time by an iterative, numerical method such that the boundary conditions are satisfied. However, since the state and costate vectors of the linearized system are related by Eq. (2.140) at the final time, we look for a solution to Eqs. (2.144)–(2.146) that satisfies a state transition matrix, thereby ensuring linear independence of solutions, x(t) and λ(t). Thus, we write the combined state and costate equations as follows:    x˙ A − BR −1 S T = λ˙ SR −1 S T − Q

−BR −1 B T −AT + SR −1 B T

  x λ

(2.147)

which must satisfy the boundary conditions of Eq. (2.146). To begin the solution, we integrate backward in time from t = tf , for which we have 

x(t) λ(t)





X(t) 0 = 0 Λ(t)



x(tf ) λ(tf )

 ,

(2.148)

where X(t) and Λ(t) are the state transition matrices corresponding to the backward evolution in time of x(t) and λ(t), respectively. Clearly, we must have λ(t) = Λ(t)λ(tf ) = Λ(t)Qf x(tf ) ,

(2.149)

x(t) = X(t)x(tf ) ,

(2.150)

x(t0 ) = X(t0 )x(t) .

(2.151)

and

Inversion of Eq. (2.150) and substitution into Eq. (2.149) yields λ(t) = Λ(t)Qf X−1 (t)x(t) .

(2.152)

Since both the state transition matrices must satisfy X(tf ) = I ;

Λ(tf ) = I.

(2.153)

Equation (2.152) represents the adjoint relationship between the solutions, x(t) and λ(t), written as follows: λ(t) = P (t)x(t) ,

(2.154)

40

2 Analytical Optimal Control

where P (t) = Λ(t)Qf X−1 (t). Substituting Eq. (2.154) into Eq. (2.142), we have the following linear, optimal feedback control law: u(t) = −R −1 (t)[B T (t)P (t) + S T (t)]x(t) .

(2.155)

Taking the time derivative of Eq. (2.154) and substituting into Eqs. (2.144), (2.145), and (2.155), we have the following matrix Riccati equation (MRE) to be satisfied by the matrix, P (t): − P˙ = Q + (A − BR −1 S T )T P + P (A − BR −1 S T ) − P BR −1 B T P − SR −1 S T ,

(2.156)

which must be solved subject to the boundary condition: P (tf ) = Qf .

(2.157)

An alternative derivation of the matrix Riccati equation is via the HJB formulation, where the following quadratic value function is proposed with a positive definite, symmetric matrix, P (t): V [x(t), t] = x T (t)P (t)x(t) ,

(2.158)

whose optimal value must satisfy the HJB equation with the terminal boundary condition: Vˆ [x(t ˆ f ), tf ] = x T (tf )Qf x(tf ) .

(2.159)

Thus, we have λT =

∂V = 2x T (t)P (t) , ∂x

(2.160)

and the following Hamiltonian: H = x T (t)Q(t)x(t) + 2x T (t)S(t)u(t) + uT (t)R(t)u(t) + 2x T (t)P (t)[A(t)x(t) + B(t)u(t)] .

(2.161)

For deriving the optimal control, we differentiate H with respect to u and equate the result to zero Hu = 0: xˆ T (t)S(t) + uˆ T (t)R(t) + xˆ T (t)Pˆ (t)B(t) = 0 ,

(2.162)

or ˆ . u(t) ˆ = −R −1 (t)[B T (t)Pˆ (t) + S T (t)]x(t)

(2.163)

2.9 Linear Systems with Quadratic Cost Functions

41

which is the same as Eq. (2.155). Since the matrix P (t) is chosen to be positive definite, we have V [x(t), t] > 0 for all x(t), and by substituting the optimal feedback control law into ∂ Vˆ Hˆ = − = −xˆ T P˙ˆ xˆ , ∂t

(2.164)

we have the Legendre–Clebsch sufficient condition, Huu > 0, implying a negative definite matrix, P˙ˆ (t) (i.e., V˙ [x(t), t] < 0 for all x(t)). Therefore, the optimal return function given by Eq. (2.158) with a positive definite Pˆ (t) ensures a globally asymptotically stable system by Lyapunov’s theorem [74]. We again note that the Legendre–Clebsch sufficient condition implies that R(t) is positive definite. By substituting Eq. (2.163) into Eq. (2.161), we have the optimal Hamiltonian which satisfies the following HJB equation, Eq. (2.103) for the LQR problem: xˆ T P˙ˆ xˆ = −xˆ T [(A − BR −1 S T )T Pˆ + Pˆ (A − BR −1 S T ) − Pˆ BR −1 B T Pˆ + Q − SR −1 S T ]xˆ ,

(2.165)

which yields the following matrix Riccati equation to be satisfied by the optimal matrix, Pˆ : − P˙ˆ = Q + (A − BR −1 S T )T Pˆ + Pˆ (A − BR −1 S T ) − Pˆ BR −1 B T Pˆ − SR −1 S T ,

(2.166)

which must be solved subject to the boundary condition: Pˆ (tf ) = Qf .

(2.167)

Since the matrix Riccati equation (MRE) is derived both from the necessary conditions (Euler–Lagrange equations) and the sufficient condition (Hamilton– Jacobi–Bellman equation) of optimality, it must reflect both necessary and sufficient conditions for the existence of an optimal control law for linear systems with a quadratic performance index. As discussed earlier, a sufficient condition for optimality is the existence of a positive definite solution to MRE, P (t), for all times in the control interval, t0 ≤ t ≤ tf . The MRE is at the heart of modern control theory, and can be regarded as a simplified form of the nonlinear, two-point boundary value problem (2PBVP) in time represented by the Euler–Lagrange equations, or the nonlinear partial differential equations in space and time derived through Hamilton–Jacobi–Bellman (HJB) formulation. However, being nonlinear in character, the solution procedure to a general MRE is only slightly simpler, often requiring iterative numerical methods similar to the nonlinear 2PBVP. Although simple iterative schemes based upon repeated linear, algebraic solutions are usually applied when the coefficient matrices

42

2 Analytical Optimal Control

are either slowly or periodically varying, the convergence to a positive definite (or even positive semi-definite) solution is not always guaranteed. Given the complexity of an MRE solution, it is much easier to directly solve the linear state and costate equations, Eqs. (2.144)–(2.146), by an iterative 2PBVP procedure like the shooting, or collocation methods. Linearity of the adjoint system of equations, Eq. (2.147), assures the existence of a state transition matrix, Φ(t, t0 ), such that 

x(t) λ(t)



 = Φ(t, t0 )

x(t0 ) λ(t0 )

 (2.168)

,

with the boundary conditions: x(t0 ) = x0 ;

λ(tf ) = Qf x(tf ) .

(2.169)

The state transition matrix has the usual properties of Φ(t0 , t) = Φ −1 (t, t0 ) ,

Φ(t0 , t0 ) = I ;

(2.170)

and ˙ t0 ) = Φ(t,



A − BR −1 S T SR −1 S T − Q

−BR −1 B T −AT + SR −1 B T

 Φ(t, t0 ) ,

(2.171)

as well as the special property of being a symplectic matrix2 , that is: 

0 Φ (t, t0 ) −I T

I 0



 Φ(t, t0 ) =

0 −I

I 0

 (2.172)

.

On partitioning Φ(t, t0 ) as follows: 

2A

x(t) λ(t)





Φxx (t, t0 ) = Φλx (t, t0 )

Φxλ (t, t0 ) Φλλ (t, t0 )



x(t0 ) λ(t0 )

 ,

(2.173)

symplectic matrix, A, satisfies AT J A = J , where   0 I J = −I 0

is analogous to the imaginary number, j = matrix, A, has the following properties:

√

−1, in complex algebra since J 2 = −I . A symplectic

1. Its determinant is unity, i.e., det(A) = 1. 2. If λ is an eigenvalue of A, then 1/λ is also an eigenvalue of A.

2.10 Illustrative Examples

43

the symplectic nature of the state transition matrix implies that ⎡ Φ −1 (t, t0 ) = ⎣

T (t, t ) Φλλ 0

T (t, t ) −Φxλ 0

T (t, t ) −Φλx 0

T (t, t ) Φxx 0

⎤ ⎦ ,

(2.174)

which is very useful in carrying out the matrix operations required for the solution of the 2PBVP. In order to prove that Φ(t, t0 ) is indeed symplectic, we note that Φ(t0 , t0 ) = I satisfies Φ T J Φ = J . Therefore, if it can be shown that d (Φ T J Φ) = 0 , dt

(2.175)

then the identity Φ T J Φ = J is satisfied for all times. Making use of Eq. (2.171), which we re-write as follows:   E F Φ, (2.176) Φ˙ = G −E T where E = A − BR −1 S T F = −BR −1 B T

(2.177)

G = SR −1 S T − Q , we have d (Φ T J Φ) = Φ˙ T J Φ + Φ T J Φ˙ dt   G − GT E T − E T T Φ=0, =Φ E−E F − FT

(2.178)

because F, G are symmetric matrices. Thus, we have proved that Φ(t, t0 ) is symplectic.

2.10 Illustrative Examples Example 2.6 Consider a second-order, linear time-invariant system with the following state equations: x˙1 = x2 x˙2 = x1 + u

(2.179)

44

2 Analytical Optimal Control

where the scalar input u(t) is unconstrained. The control task is to move the system from initial state x1 (0) = x2 (0) = 0 to x1 (5) = 1 while minimizing 1 1 J = x22 (5) + 2 2



5

u2 (t)dt

(2.180)

0

Note that the terminal time is specified to be tf = 5, whereas the terminal cost function is given as φ = 12 x22 (5). The Lagrangian, L = 12 u2 , represents the total control effort (or energy) spent during the control interval, 0 ≤ t ≤ 5. We begin by writing the Hamiltonian of the system as follows: H = L + λT f =

1 2 u + λ1 x2 + λ2 (x1 + u) 2

(2.181)

The costate equations of the system are next derived as follows:  T  ∂H ∗ ˙λ∗ = − = Λλ∗ ∂x

(2.182)

where  Λ=

0 −1 −1 0

 (2.183)

The solution to the costate equations subject to the initial condition, λ∗ (0) = (c1 , c2 )T , is expressed by: ⎞⎧ ⎫ ⎨ c1 ⎬ 1 ⎠ λ∗ (t) = eΛt λ∗ (0) = L−1 (sI − Λ)−1 λ∗ (0) = ⎝ ⎩ ⎭ 2 −et + e−t et + e−t c2 (2.184) Note that the constants c1 , c2 (thus the extremal costate trajectory) are to be determined from the boundary conditions. The necessary conditions for optimality include the following stationary condition of the Hamiltonian with respect to the control history, u(t): ⎛

 Hu =

∂H ∂u

∗ T

et + e−t

−et + e−t

= u∗ (t) + λ∗2 (t) = 0

(2.185)

which results in the following extremal control history:   % 1 $  −t c1 e − et + c2 et + e−t 2 % 1$ = (0 ≤ t ≤ 5) (c1 − c2 ) et − (c1 + c2 ) e−t , 2

u∗ (t) = −λ∗2 (t) = −

(2.186)

2.10 Illustrative Examples

45

In order to determine the constants c1 , c2 , the state equations must be solved, subject to the boundary conditions. The extremal trajectory resulting from the extremal control history (Eq. (2.186)) and zero initial state is expressed as follows: x ∗ (t) =



t

eA(t−τ ) Bu∗ (τ )dτ ,

(0 ≤ t ≤ 5)

(2.187)

0

where  A=

 0 1 , 1 0

  0 B= 1

(2.188)

The state transition matrix is determined to be the following: ⎛ eAt = L−1 (sI − A)−1 =

1⎝ 2

et + e−t et

− e−t

et − e−t et

⎞ ⎠

(2.189)

+ e−t

After Eqs. (2.186) and (2.189) are substituted into Eq. (2.187), and the integration is performed, we have the following extremal trajectory:  $ %⎫ ⎧  −t ⎨ c1 e − et + t (c1 − c2 ) et + (c1 + c2 ) e−t ⎬ 1 x ∗ (t) = ,  $ %⎭ 4 ⎩  −t c2 e − et + t (c1 − c2 ) et − (c1 + c2 ) e−t

(0 ≤ t ≤ 5) (2.190)

The boundary conditions are given by: x1∗ (5) = 1

(2.191)

and λ∗2 (5) =



∂φ ∂x2

∗

= x2∗ (5)

(2.192)

t=5

Substituting the boundary conditions into Eqs. (2.184) and (2.190) results in the following set of linear algebraic equations:   1 5 5  −5 2e + 3e−5 c1 + e − e5 c2 = 1 2 4     3 −5 7 5 −5 5 e −e e + 2e c2 = 0 c1 − 4 2

(2.193)

whose solution is easily derived to be the following: c1 = −0.072 ,

c2 = −0.063

(2.194)

46

2 Analytical Optimal Control 1

0.8

x2∗

x∗1

0.6 0.5

0.4 0.2

0

0

1

2

t

3

4

0

5

0

1

2

−0.5 0

1

2

0

4

5

3

4

5

0.5

λ ∗2

λ ∗1

3

1

−0.2 −0.4

0

−0.6 −0.8 0

t

1

2

t

3

4

5

t

Fig. 2.2 The extremal trajectory and the corresponding costate trajectory for the optimal control problem of Example 2.6

Thus, the optimal control problem is solved, and the extremal trajectory and the corresponding control history are obtained. The results are plotted in Figs. 2.2 and 2.3. It can be verified that x2∗ (5) = λ∗2 (5) = 0.6689 and x1∗ (5) = 1. It is quite easily verified that the extremal trajectory is also the optimal one, because  Huu =

∂ 2H ∂u2

∗ T =1>0

(2.195)

hence the sufficient condition for optimality is satisfied on the extremal trajectory. Example 2.7 Let the optimizing conditions for the system in Example 2.6 be changed such that the terminal time, tf , is unspecified, and the system is to be moved from a zero initial state, x1 (0) = x2 (0) = 0 to x1 (tf ) = 1, while minimizing J =

1 1 2 x2 (tf ) + 2 2



tf

u2 (t)dt

(2.196)

0

While the state and costate solutions are still given by Eqs. (2.190) and (2.184), respectively, and the stationary condition:

2.10 Illustrative Examples

47

0

−0.1

−0.2

u∗

−0.3

−0.4

−0.5

−0.6

−0.7 0

0.5

1

1.5

2

2.5

t

3

3.5

4

4.5

5

Fig. 2.3 The extremal control history for the optimal control problem of Example 2.6

Hu = 0 = u∗ (t) + λ∗2 (t)

(2.197)

is unchanged, the boundary conditions are now changed to the following: x1∗ (tf ) = 1 λ∗2 (tf ) =



∂φ ∂x2

∗

(2.198) = x2∗ (tf )

(2.199)

t=tf

and   %2 ∂φ ∗ 1$ ∗ u (tf ) + λ∗1 (tf )x2∗ (tf ) + λ∗2 (tf )x1∗ (tf ) = 0 H+ = ∂t t=tf 2

(2.200)

Substituting Eqs. (2.197)–(2.199) into Eq. (2.200), we have −

%2 1$ ∗ λ (tf ) + λ∗1 (tf )λ∗2 (tf ) + λ∗2 (tf ) = 0 2 2

(2.201)

or $ % λ∗2 (tf ) = 2 λ∗1 (tf ) + 1

(2.202)

48

2 Analytical Optimal Control

The algebraic equations in terms of the unknowns, c1 , c2 , tf , resulting from the boundary conditions are the following: % c1  t  tf $ e f − e−tf = 1 (c1 − c2 ) etf + (c1 + c2 ) e−tf − 4 4  t    − c1 3e f + e−tf + c2 3etf − e−tf = 4

(2.203) (2.204)

and    c1 etf − e−tf tf + 2   c2 = t 3e f + e−tf + tf etf + e−tf

(2.205)

When Eqs. (2.204) and (2.205) are substituted into Eq. (2.203), the following necessary condition is derived for the terminal time, tf : 9e3tf + 45etf + 11e−tf − e−3tf + 66tf etf + 26tf e−tf + 6tf e3tf − 2tf e−3tf + 21tf2 etf + 11tf2 e−tf + tf2 e3tf − tf2 e−3tf = 0

(2.206)

Newton’s iterative solution method [77] yields the only real solution to Eq. (2.206) as tf = −0.7167, which is invalid because the terminal time must always be positive. Hence, the stated optimal control problem does not have a feasible solution. This example illustrates the fact that not every set of objective function and boundary conditions has a feasible solution. Example 2.8 Consider another variation for the system in Example 2.6 where the terminal time, tf , is unspecified, and the system is to be moved from a zero initial state, x1 (0) = x2 (0) = 0, to final state, x1 (tf ) = x2 (tf ) = 1, while minimizing 1 1 J = k 2 tf2 + 2 2



tf

u2 (t)dt

(2.207)

0

Since both the states are fixed at the terminal time, we have the following boundary conditions for determining the coefficients, c1 , c2 , in Eqs. (2.184) and (2.190): $ %   tf (c1 − c2 ) etf + (c1 + c2 ) e−tf − c1 etf − e−tf = 4

(2.208)

$ %   tf (c1 − c2 ) etf − (c1 + c2 ) e−tf − c2 etf − e−tf = 4

(2.209)

and

These are solved to yield $  % 4 + c2 etf − e−tf + tf etf + e−tf   c1 = tf etf − e−tf

(2.210)

2.10 Illustrative Examples

49

Table 2.1 The results of iterative solution in Example 2.8

k 0.1 1 10 100

tf 2.9026 1.4528 0.6490 0.2723

c1 −0.2495 −2.3939 −33.0767 −522.0565

c2 −0.2409 −1.7046 −11.3930 −73.7958

1.4

k k k k

1.2

= = = =

0.1 1 10 100

1

x1

0.8

0.6

0.4

0.2

0

0

0.5

1

1.5

t

2

2.5

3

Fig. 2.4 The extremal trajectory of the first state variable for different values of the terminal cost parameter, k, in Example 2.8

where   4 e−tf + etf + 2tf e−tf c2 =  2 etf − e−tf − 4tf2

(2.211)

The additional boundary condition required to solve for the terminal time, tf , is the following:   %2 ∂φ ∗ 1$ H+ = − λ∗2 (tf ) + λ∗1 (tf ) + λ∗2 (tf ) + k 2 tf = 0 ∂t t=tf 2

(2.212)

wherein the terminal boundary condition for the state variables has been substituted, and the costate variables, λ∗1 , λ∗2 , are given by Eq. (2.184). Equation (2.212) requires an iterative numerical solution, which is performed by Newton’s method [77]. The results when the cost coefficient k is varied from 0.1 to 100 are tabulated in Table 2.1 and Figs. 2.4, 2.5, and 2.6. The terminal time, tf , is seen to decrease almost exponentially with the increase of k whereas the coefficients, c1 , c2 , are

50

2 Analytical Optimal Control 6

k k k k

5

= = = =

0.1 1 10 100

x2

4

3

2

1

0

0

0.5

1

1.5

t

2

2.5

3

Fig. 2.5 The extremal trajectory of the second state variable for different values of the terminal cost parameter, k, in Example 2.8

almost linearly related to k. As k is increased, the state variable, x1∗ , reaches its final value in a smaller time in a continuously ascending curve (Fig. 2.4), whereas x2∗ encounters increasing overshoots in its trajectory for k > 1 (Fig. 2.5). The largest control input magnitude, | u∗ |, increases in proportion with k, its peak magnitude being roughly the same as k (Fig. 2.6). Hence, the cost parameter, k, offers a direct trade-off between the terminal time, tf , and the required peak input. Example 2.9 In order to illustrate the effect of the boundary conditions, consider a lunar landing problem in which a continuous vertical acceleration input, u(t), is applied to control the altitude, z(t), of the spacecraft’s center of mass. The lander is assumed to have approached sufficiently near the lunar surface in order for the acceleration due to gravity, g, to be considered a constant. The differential equation governing the spacecraft’s translational motion is thus simply given by: z¨ = −g + u

(2.213)

Let us choose the state variables to be the following: 1 x1 = z + gt 2 2 x2 = x˙1 = z˙ + gt

(2.214)

2.10 Illustrative Examples

51

0.3

2

k = 0.1

0.2

u

u

0.1

0

0 −0.1 0

1

t

2

−1 0

3

20

0.5

t

1

1.5

100

k = 10

k = 100

50 0

u

u

10

0

−10 0

k=1

1

−50

0.2

0.4

t

0.6

0.8

−100 0

0.1

0.2

t

0.3

0.4

Fig. 2.6 The extremal control history for different values of the terminal cost parameter, k, in Example 2.8

The state equations can then be expressed as follows: x˙1 = x2 x˙2 = u

(2.215)

This would appear to be no different than the simple linear acceleration problem in a zero gravity situation, but this is deceptive because although the acceleration due to gravity does not appear in the state equations, it is very much a part of the boundary conditions. For a smooth landing to take place, both vertical displacement and vertical speed must vanish at some terminal time, tf , implying 1 2 gt 2 f x2 (tf ) = gtf x1 (tf ) =

(2.216)

Therefore, a solution to the optimal landing problem must be sought where Eq. (2.216) is imposed as a terminal constraint. This must be ensured irrespective of the initial condition: x1 (0) = z(0) x2 (0) = z˙ (0)

(2.217)

52

2 Analytical Optimal Control

A possible choice of the objective function for the smooth vertical landing is the total control energy given by: J =

1 2



tf

u2 (t)dt

(2.218)

0

resulting in the following Hamiltonian: H =

1 2 u + λ1 x2 + λ2 u 2

(2.219)

The costate equations of the system are next derived as follows: λ˙ ∗ =



λ∗1 λ∗2



   T  ∂H ∗ 0 =− = −λ∗1 ∂x

(2.220)

whose solution subject to the initial condition, λ∗ (0) = (c1 , c2 )T , is simply the following: λ∗ (t) =

⎧ ⎨ ⎩

⎫ ⎬

c1 c2 − c1 t

(2.221)

⎭

The stationary condition of the Hamiltonian:  Hu =

∂H ∂u

∗ T

= u∗ (t) + λ∗2 (t) = 0

(2.222)

results in the following expression for the extremal control history: u∗ (t) = −λ∗2 (t) = c2 − c1 t,

(0 ≤ t ≤ tf )

(2.223)

This is substituted into the state equations, Eq. (2.215), to yield the extremal trajectory as follows: x ∗ (t) =



x1∗ x2∗

 =

⎧ ⎨ ⎩

c1 t 2 /2 − c2 t + c3 c1 t 3 /6 − c2 t 2 /2 + c3 t + c4

⎫ ⎬ ⎭

,

(0 ≤ t ≤ tf )

(2.224)

where x ∗ (0) = (c3 , c4 )T . The initial condition of the system, Eq. (2.217), requires that c3 = z˙ (0) and c4 = z(0). Finally, the terminal condition, Eq. (2.216), is applied to determine the coefficients, c1 , c2 , as follows:

2.10 Illustrative Examples

53

800

x∗1(m)

600 400 200 0

0

5

10

15

20

25

30

20

25

30

t(s)

x∗2 (m/s)

60 40 20 0

0

5

10

15

t(s) Fig. 2.7 The optimal state trajectory for the lunar landing problem of Example 2.9



c1 c2



⎛ ⎜ =⎝

6/tf2 2/tf

⎫ ⎞⎧ ⎪ ⎪ −˙z(0) + gtf ⎬ ⎨ ⎟ ⎠ ⎪ ⎪ −6/tf2 ⎩ −z(0) − z˙ (0)tf + gtf2 /2 ⎭

−12/tf3

(2.225)

Since Huu = 3/2 > 0 on the extremal trajectory, the latter is also the optimal trajectory. Figures 2.7, 2.8, and 2.9 are the plots of the optimal state trajectory, control history, and the vertical displacement and velocity, respectively, for g = 1.6 m/s2 , z(0) = 100 m, z˙ (0) = 10 m/s, and tf = 30 s. For this condition, the values of the coefficients are c1 = 1/9 and c2 = 2/5. The optimal control increases linearly from −0.4 m/s2 to 2.93 m/s2 (Fig. 2.8), whereas the vertical displacement (Fig. 2.9) has an overshoot of 28 m at t = 6 s before smoothly decreasing to zero at the terminal time. This overshoot is due to the fact that the spacecraft is initially traveling upward, thereby requiring a negative vertical acceleration input to be applied initially in order to make it land in the given time. By increasing the value of tf , one can decrease the magnitude of the control input as well as the vertical overshoot before landing. By keeping the terminal time open and adding a terminal cost to the objective function, one can optimally combine the requirements of minimizing both the control energy and the time. This is to be explored in one of the exercises at the end of the chapter. Caution must be exercised when choosing a too small value of tf , which would not only greatly increase the control input magnitudes, but more importantly, can

54

2 Analytical Optimal Control 3

2.5

2

u∗(m/s2)

1.5

1

0.5

0

−0.5 0

5

10

15

t(s)

20

25

30

Fig. 2.8 The optimal control history for the lunar landing problem of Example 2.9

z∗ (m)

150 100 50 0

0

5

10

15

20

25

30

20

25

30

t(s) 10

z˙ ∗ (m/s)

5 0 −5 −10 0

5

10

15

t(s)

Fig. 2.9 The optimal vertical displacement and velocity for the lunar landing problem of Example 2.9

2.11 Singular Optimal Control

55

also lead to an undershoot in the displacement near the terminal time. This implies a crash into the ground at a nonzero vertical velocity, and is automatically avoided by adding an inequality constraint to the optimization problem, such as x1 (t) ≥ 12 gt 2 , 0 ≤ t ≤ tf . However, the resulting problem would be a nonlinear one as opposed to the linear one solved in the present example.

2.11 Singular Optimal Control A singular control problem is the one in which the necessary conditions of optimality do not determine the optimal control profile. In such a case, additional conditions must be specified in order to define the optimal control. A large majority of optimal control problems in space navigation have a Hamiltonian which is a linear function of the control variables, and therefore the stationary condition, Hu = 0, cannot be used to solve for u(t). Furthermore, such a Hamiltonian does not satisfy the Legendre–Clebsch sufficient condition for optimality, Huu > 0, because the Hessian, Huu , is singular. Examples of singular control include the minimization of a performance index which is a linear function of the control variables, such as the time minimization and fuel minimization maneuvers. Other singular optimal control problems are the ones in which the state equations are linearized, the control cost in the performance index is quadratic, and the control variables are bounded by inequality constraints. Since the sufficient condition for optimality is unusable in singular control problems, additional conditions must be available to determine the minimizing input and trajectory. As an illustrative example of singular control, consider a control-affine system described by the following state equation: x˙ = f (x, t) + B(x, t)u(t)

(2.226)

and the following Lagrangian which is linear in control: L(x, u, t) = 1 (x, t) + 2 (t)u(t)

(2.227)

resulting in the following Hamiltonian: H (x, u, λ, t) = 1 + 2 u + λT (f + Bu)

(2.228)

Unless other conditions are specified, there is no solution to the optimal control problem in which H is minimized relative to u, because Huu = 0 on an extremal trajectory for which Hu = 0. Even though the non-convexity of H does not guarantee optimality, a practical minimum solution is obtained for the case of bounded control inputs, such as that given by the following inequality constraint: | u(t) |≤ um

(2.229)

56

2 Analytical Optimal Control

In such a case, the Pontryagin’s minimum principle given by Eq. (2.94) yields the ˆ following condition for minimality on an optimal trajectory, (x, ˆ u, ˆ λ): ˆ t) + 2 (t)u(t) ˆ + λˆ T [f (x, ˆ t) + B(x, ˆ t)u(t)] ˆ 1 (x, ≤ 1 (x, ˆ t) + 2 (t)u(t) + λˆ T [f (x, ˆ t) + B(x, ˆ t)u(t)]

(2.230)

or ˆ t)]u(t)] ˆ ≤ [2 (t) + λˆ T B(x, ˆ t)]u(t)] [2 (t) + λˆ T B(x,

(2.231)

This results in the following switching condition for the optimal control, u(t): ˆ ⎧ ⎪ −um , (2 + λˆ T B)i > 0 ⎪ ⎪ ⎪ ⎪ ⎨ uˆ i (t) = 0 , (2 + λˆ T B)i = 0 , ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ um , (2 + λˆ T B)i < 0

(i = 1, . . . , m)

(2.232)

where the index i refers to the i th element of the particular vector. The magnitude of each element of uˆ thus depends upon the sign of the switching function, (2 + λˆ T B)i , and can also be expressed as follows: uˆ i (t) = −um sgn(2 + λˆ T B)i ,

(i = 1, . . . , m)

(2.233)

Clearly, the optimality requires that the norm of the control input vector must be on the boundary, that is either the maximum or the minimum value. Such an optimal control law switching between the two extreme input values is typical of the singular problems and is referred to as bang–bang control. Note that the bound on the norm of the input vector, Eq. (2.229), implies that if the i th element is either uˆ i = −um or uˆ i = um , then uˆ j = 0 for all i = j . Example 2.10 Let us now take up an example of time-optimal control with bounded control magnitude as an example of singular control problems. For illustration, the following linear system is chosen: x˙1 = x2 x˙2 = u

(2.234)

which can represent a rigid spacecraft undergoing a single axis rotation with u(t) being the angular acceleration input and x1 (t) the rotation angle. Alternatively, these state equations also model the rectilinear displacement, x1 (t), of a mass moving in a frictionless groove when forced by a linear acceleration input, u(t). Consider moving the system from x(0) = (0, 0)T to x(tf ) = (1, 0)T in the minimum time, tf , with the following restriction on the input magnitude:

2.11 Singular Optimal Control

57

|u(t)| ≤ 1

(2.235)

Since the Lagrangian of a time-optimal control problem is L = 1, the Hamiltonian of the system is given by: H = 1 + λT f = 1 + λ1 x2 + λ2 u

(2.236)

while Pontryagin’s minimum principle applied to this singular problem results in the following switching condition:  u(t) ˆ =

−1 , λˆ 2 (t) > 0 1 , λˆ 2 (t) < 0

(2.237)

where λˆ 1 = c1 λˆ 2 = −c1 t + c2

(2.238)

The constants c1 , c2 must be determined in order to meet the boundary conditions in the minimum possible time, tf , after the switching is applied at t = tˆ. Since λˆ 2 (t) is a linear function of time, it can change sign only once. Therefore, a change in the system’s state from x(0) = (0, 0)T to x(tf ) = (1, 0)T requires that a positive control, u(t) ˆ = 1, must be applied before switching, t < tˆ, which implies λˆ 2 (t) < 0 for 0 ≤ t < tˆ. Clearly, we must have c2 < 0 and tˆ = c2 /c1 > 0, thereby implying c1 < 0. Thus, we have x(t) ˆ =

 t 0

1 t −τ 0 1

   2  0 t /2 (0 ≤ t < tˆ) dτ = t 1

(2.239)

After switching at t = tˆ, the optimal control changes to u(t) ˆ = −1, resulting in the following state response:   t   1 t −τ 0 tˆ2 /2 − dτ x(t) ˆ = ˆt 0 1 1 tˆ   2t tˆ − tˆ2 − t 2 /2 (tˆ < t ≤ tf ) = 2tˆ − t 

1 t − tˆ 0 1



(2.240)

In order to meet the terminal condition, we must have x(t ˆ f ) = (1, 0)T , which requires tˆ = tf /2 and tf = 2. The resulting bang–bang optimal trajectory is plotted in Fig. 2.10 with a time step of 1/100. Since there can be only one switching due to the time-linear profile of the costate variable, λˆ 2 (t), the resulting terminal time

58

2 Analytical Optimal Control

x

1 0.5 0

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

−1 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

x˙

1 0.5

0

u

1 0

t

Fig. 2.10 The optimal bang–bang trajectory and control history of the time-optimal control problem of Example 2.10

tf is the absolute minimum, with the switching performed at the midpoint of the control interval, tf /2. The velocity, x, ˙ reaches a maximum at the switching time and decreases to zero at the final time.

2.11.1 Generalized Legendre–Clebsch Necessary Condition Since a singular control history, u(t), satisfies the Legendre–Clebsch necessary condition, Huu ≥ 0, with an equality, it may (or may not) be the optimal control. In order to determine the optimality of a singular control solution, a generalized Legendre–Clebsch necessary condition must be derived. Most of the singular control problems in space navigation involve a scalar control variable, hence we first consider the optimal control problem of minimization of  J = φ[x(tf ), tf ] +

tf

L(x, t)dt

(2.241)

t0

with respect to a scalar control variable, u(t) ∈ R, which satisfies the control-affine dynamic constraint (state equation): x˙ = f (x, t) + b(x, t)u

(2.242)

2.11 Singular Optimal Control

59

with a specified initial state: x(t0 ) = x0

(2.243)

F [x(tf )] = 0

(2.244)

and the terminal state satisfying:

where tf is specified in advance. The Hamiltonian: H = L(x, t) + λT [f (x, t) + b(x, t)u]

(2.245)

is linear in control, u, thus the optimal control problem is singular. Theorem 2.11 If u is the optimal control solution of the problem given by Eqs. (2.241)–(2.244), then ∂ d2 ∂ d2 T Hu = λ b≤0 2 ∂u dt ∂u dt 2

(2.246)

for all t ∈ [t0 , tf ]. The proof of this theorem is derived by taking the second-order variation of the augmented performance index along the singular trajectory, and requiring it to be non-negative. See, for example, Kelley et al. [46]. Equation (2.246) is the modified necessary condition for optimality of singular control problem, that is when Hu = 0. If satisfied with a strict inequality, this condition gives the information to determine the optimal singular control. In case Eq. (2.246) is satisfied with an equality, it does not give any information regarding the optimal control, u. In that case, the following condition must be checked: ∂ d4 ∂ d4 T H = λ b≤0 u ∂u dt 4 ∂u dt 2

(2.247)

for all t ∈ [t0 , tf ]. Once again, if Eq. (2.247) is satisfied with an equality, it does not give any information regarding the optimal control, u. In that case, the next higher derivative must be considered. Kelley et al. [46] derived the following generalized necessary condition for optimality: (−1)q

∂ d2q Hu ≥ 0 ∂u dt 2q

(2.248)

for all t ∈ [t0 , tf ], where q = 1, 2, . . .. Thus, an odd time derivative of Hu cannot appear in the generalized necessary condition for the scalar input.

60

2 Analytical Optimal Control

Example 2.12 Consider the minimization of 

π

J =



0

 x22 − x12 dt

relative to u ∈ R, subject to: x˙1 = x2 ,

x1 (0) = 0

x˙2 = u ,

x2 (0) = 1

The Hamiltonian of the problem is the following:   H = x22 − x12 + λ1 x2 + λ2 u which results in the following costate equations: λ˙ 1 = 2x1 ,

λ1 (π ) = 0

λ˙ 2 = −2x2 − λ1 ,

λ2 (π ) = 0

The extremal solution of this singular problem is derived as follows: Hu = 0 = λ∗2 or λ∗1 = −2x2∗ which implies that the second state equation is the following along the extremal arc: x˙2∗ = u∗ = −x1∗ The extremal trajectory is thus represented by: 

x˙1∗ x˙2∗



 =

0 1 −1 0



x1∗ x2∗



 ,

x1∗ (0) x2∗ (0)



  0 = 1

whose solution is the following: 

x1∗ (t) x2∗ (t)



 =

sin t cos t



Thus, u∗ (t) = − sin t, 0 ≤ t ≤ π is the singular control history.

2.11 Singular Optimal Control

61

To investigate the optimality of the singular control profile, we apply Theorem 2.11 on the extremal arc as follows: d2 Hu = λ¨ ∗2 = −2x˙2∗ − λ˙ ∗1 = −2u∗ − 2x1∗ dt 2 ∂ d2 Hu = −2 < 0 ∂u dt 2 Thus, Theorem 2.11 is satisfied for all t, implying that u∗ (t) = − sin t, 0 ≤ t ≤ π could be an optimal solution of the problem. However, since it is not a sufficient condition for optimality, additional information is necessary to determine whether the singular solution is optimal. The generalized Legendre–Clebsch necessary condition for multi-input singular optimal control can be derived using a transformational approach [34], wherein the singular control problem is converted into an equivalent non-singular one by transforming the control variables. For the control-affine system with u ∈ Rm×1 and x ∈ Rn×1 , represented by: x˙ = f (x, t) + g(x, t)u

(2.249)

and with a specified initial state: x(t0 ) = x0

(2.250)

the minimization of 

tf

J = φ[x(tf ), tf ] +

L(x, t)dt

(2.251)

t0

subject to the dynamic equality constraint, Eq. (2.249), and a terminal state constraint: F [x(tf ), tf ] = 0

(2.252)

H = L(x, t) + λT [f (x, t) + g(x, t)u]

(2.253)

results in the Hamiltonian:

which is linear in control, u. The augmented cost function in terms of the Hamiltonian is expressed as follows:  Ja = φ[x(tf ), tf ] + c F [x(tf ), tf ] + T

tf

t0

H (x, u, λ, t) − λT xdt ˙

(2.254)

62

2 Analytical Optimal Control

The Euler–Lagrange necessary conditions for optimality along an extremal arc, [x ∗ (t), u∗ (t)], are based on the following first-order variation of Ja along the extremal trajectory. The first-order variations in the state equation and the boundary conditions due to δx, δu, δt are expressed as follows: δ x˙ = Aδx + Bδu

(2.255)

δx(t0 ) = 0

(2.256)

˙ [Ft δt + Fx (δx + xδt)] t=tf = 0

(2.257)

and

The first-order variation in Ja is the following:  ˙ + (φx + cT Fx )dx dJa = (φt + cT Ft + H − λT x)dt  +

tf



t=tf



Hx δx + Hu δu − λT δ x˙ dt

(2.258)

t0

As before, we integrate the term involving δ x˙ by parts, express the total derivative dx = xdt ˙ + δx, and note that δx(t0 ) = 0. This produces the following first-order variation:  dJa = (φt + cT Ft + H )δt + (φx + cT Fx − λT ) (xδt ˙ + δx)  +

t=tf

tf

  Hx + λ˙ T δx + Hu δu dt

(2.259)

t0

Since dJa = 0 in the presence of arbitrary variations δx, δu, δt, we have the following necessary conditions for minimization of Ja : λ˙ T = −



∂H ∂x

∗

= −Hx∗

(2.260)

∗  λT (tf ) = φx + cT Fx

(2.261)

 ∗ φt + c T F t + H

(2.262)

t=tf

t=tf

=0

and Hu∗ = 0. Here, A = (∂f/∂x)∗ is the Jacobian matrix, and B = g(x ∗ , t), both being evaluated on the extremal arc. The second variation of Ja is now derived by taking the total derivative of Eq. (2.258) as follows:

2.11 Singular Optimal Control

63

 d2 Ja = (φt + cT Ft + H − λT x)d ˙ 2 t + (φx + cT Fx )d2 x + (φtt + cT Ftt + H˙ − λT x)dt ¨ 2 + 2(φtx + cT Ftx )dtdx + dx T (φxx + cT Fxx )dx + 2Hx δxdt + 2Hu δudt − 2λT δ xdt ˙  +

t=tf

(2.263)

tf 

 Hx δ 2 x + Hu δ 2 u − λT δ 2 x+ ˙ δx T Hxx δx+ 2δuT Hux δx+ δuT Huu δu dt

t0

where the time derivative of the Hamiltonian is the following: H˙ = Ht + Hx x˙ + Hλ λ˙ + Hu u˙

(2.264)

H˙ = Ht + Hx x˙ ∗ − x˙ ∗ HxT + Hu u˙ ∗ = Ht

(2.265)

which becomes

on the extremal trajectory. Furthermore, integrating the term involving δ x˙ by parts, expressing the total derivatives of x by: dx = xdt ˙ + δx d2 x = xd ˙ 2 t + xdt ¨ 2 + 2δ xdt ˙ + δ2x

(2.266)

the terminal state variation by: ˙ f )δtf dx(tf ) = δx(tf ) + x(t

(2.267)

and substituting Eqs. (2.260)–(2.262) into Eq. (2.264), we have the following expression for the second variation:  ˙ + (δx + xδt) ˙ T Φxx (δx + xδt) ˙ d2 Ja = Φtt δt 2 + Φtx (δx + xδt)δt + (Ht − Hx x)δt ˙ 2 + Hx (δx + xδt)δt ˙  +

tf

t=tf

(2.268)

  δx T Hxx δx + 2δuT Hux δx + δuT Huu δu dt

t0

where Φ[x(tf ), tf ] = φ[x(tf ), tf ] + cT F [x(tf ), tf ]

(2.269)

When the terminal time is specified, we have δtf = 0, hence the terminal constraint variation becomes the following: (Fx δx)t=tf = 0

(2.270)

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2 Analytical Optimal Control

As before, a necessary condition for minimization of Ja is that of a positive semidefinite second variation, d2 Ja , which (as discussed earlier) requires that for the non-singular case (Huu > 0) the following matrix Riccati equation must be satisfied with a non-negative, finite square matrix, P (t), along the extremal arc for a specific terminal time, tf : −1 − P˙ = Hxx + P A + AT P − (Hxu + P B)Huu (Hux + B T P )

(2.271)

with the terminal boundary condition: P (tf ) = Φxx [x(tf ), tf ]

(2.272)

As before, we introduce the following notation for convenience: Hxx = Q ,

Hux = S ,

(Φxx )t=tf = Qf ,

Huu = R

(Fx )t=tf = D

The requirement of a non-negative second variation of Ja with a specific terminal time, tf , results in the following accessory problem of minimization of  J2 = x(tf ) Qf x(tf ) + T

tf

  x T Qx + 2uT Sx + uT Ru dt

(2.273)

t0

subject to: x˙ = A(t)x + B(t)u ,

x(t0 ) = 0

(2.274)

and Dx(tf ) = 0

(2.275)

For the non-singular case, R = Huu > 0, this is the standard linear, quadratic regulator (LQR) problem discussed above, and has the solution: u(t) = −R −1 (t)[B T (t)P (t) + S T (t)]x(t) .

(2.276)

where P (t) is a non-negative matrix satisfying − P˙ = Q + P A + AT P − (S T + P B)R −1 (S + B T P )

(2.277)

with the terminal boundary condition: P (tf ) = Qf

(2.278)

However, for the present singular case we have R = 0, hence there is no solution to the Riccati equation, Eq. (2.277).

2.11 Singular Optimal Control

65

In order to transform the singular problem to a non-singular one (for which the Riccati solution is available), consider the following definition: v˙ = u

(2.279)

v(t0 ) = 0

(2.280)

with initial condition:

Furthermore, a new state vector is defined by: z = x − Bv

(2.281)

Thus, the first-order variational state equation and boundary conditions, Eqs. (2.274) and (2.275), become the following: ˙ , z˙ = Az + (AB − B)v

z(t0 ) = 0

(2.282)

and D[z(tf ) + Bv(tf )] = 0

(2.283)

The objective function to be minimized associated with the second-order variation of the original performance index is now the following: (2.284) J2 = [z(tf ) + B(tf )v(tf )]T Qf [z(tf ) + B(tf )v(tf )]  tf  + zT Qz + 2v T B T Qz + v T B T QBv + 2v˙ T S(z + Bv) dt t0

resulting in the following Hamiltonian: ˙ H = zT Qz + 2v T B T Qz + v T B T QBv + 2v˙ T S(z + Bv) + λT [Az + (AB − B)v] (2.285) and the following costate equation: λ˙ = −HzT = −Qz − 2QBv − 2S T v˙ − AT λ

(2.286)

On integrating the term v˙ T Sz in Eq. (2.284) by parts, we have the following result: J2 = zT (tf )Qf z(tf ) + 2v T (tf )[S(tf ) + B T (tf )Qf ]z(tf )  tf  zT Qz + v T (tf )B T (tf )Qf B(tf )v(tf ) + t0

(2.287)

 ˙ + v T [B T QB − 2S(AB − B)]v ˙ + 2vSBv ˙ dt + 2v T (B T Q − SA − S)z

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2 Analytical Optimal Control

However, from Eq. (2.285) it can be shown that: ˙ − d (SB) = − ∂ H¨ u B T QB − 2S(AB − B) dt ∂u

(2.288)

A substitution of Eq. (2.288) into Eq. (2.287) and integration of the last term inside the integral by parts results in the following: J2 = zT (tf )Qf z(tf ) + 2v T (tf )[S(tf ) + B T (tf )Qf ]z(tf )  tf  zT Qz + v T (tf )[S(tf )B(tf ) + B T (tf )Qf B(tf )]v(tf ) +   ∂ ¨ T T T ˙ + 2v (B Q − SA − S)z + 2v − Hu v ∂u  + v˙ T (SB − B T S T )v dt

t0

(2.289)

The term (SB − B T S T ) in Eq. (2.289) is a skew-symmetric matrix, which can make J2 either positive or negative. But, this cannot be allowed if J2 has a minimum on the extremal arc. Hence, the following necessary condition for optimality must be enforced: SB − B T S T = 0

for all t ∈ [t0 , tf ]

(2.290)

which makes the v˙ term vanish inside the integral of Eq. (2.289). It is to be noted that SB − B T S T =

∂ ˙ Hu ∂u

(2.291)

which corresponds to the generalized necessary Legendre–Clebsch condition given by Theorem 2.11 (Eq. (2.248)) with q = 1 for the scalar case, and is trivially satisfied even for the vector case when the condition of Eq. (2.290) is imposed. Therefore, another transformation is necessary to derive the nontrivial generalized optimality condition by taking: w˙ = v ,

w(t0 ) = 0

for which the second variation becomes the following: ˙ f ), y(tf )] J2 = ψ[y(t  tf   + y˙ T N y˙ + 2y˙ T Uy + y T My dt t0

(2.292)

2.11 Singular Optimal Control

67

and must be minimized subject to the first-order state variation: ¯ =0, B¯ y˙ + Ay

y(t0 ) = 0

(2.293)

where y = (zT , w T )T , A¯ = (−A, 0) , ⎛ N =⎝

0

0

0

∂ ¨ − ∂u Hu

B¯ = [In , (B˙ − AB)] ⎛

⎞

U =⎝

⎠ ,

0

0

˙ (B T Q − SA − S)

0

⎞ ⎠

and ⎛ M=⎝

Q 0

⎞ ⎠

0 0 The classical Legendre–Clebsch condition of a non-negative Hessian applied to this accessory problem yields the following necessary condition to be satisfied along an optimal trajectory: r T Nr ≥ 0

(2.294)

¯ =0 Br

(2.295)

for all r ∈ R(n+m)×1 satisfying

This results in the following necessary condition: −

∂ ¨ Hu ≥ 0 ∂u

for all t ∈ [t0 , tf ]

(2.296)

If this condition is satisfied with an equality for all t ∈ [t0 , tf ], then the problem is totally singular in v, and therefore another transformation must be made to determine optimality. This process is repeated until a nontrivial condition is satisfied.

2.11.2 Jacobson’s Necessary Condition Another necessary condition for the optimality of singular problems is given by the following theorem [42]:

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2 Analytical Optimal Control

Theorem 2.13 A necessary condition for the second variation, J2 , to be nonnegative relative to u for the singular optimal control problem posed by Eqs. (2.273)–(2.275) is SB + B T W B ≥ 0

(2.297)

for all t ∈ [t0 , tf ], where − W˙ = Q + AT W + W A ,

W (tf ) = Qf

(2.298)

Proof Adjoin the dynamic state constraint Eq. (2.274) to the performance index given by Eq. (2.273) for the singular case (R = 0) via the costate vector, λ(t) = W T (t)x(t), with W (t) ∈ Rn×n being an arbitrary, symmetric matrix which is continuously differentiable with time. This results in the following augmented performance index (i.e., the second variation on the performance index of the accessory problem): J2 = x(tf )T Qf x(tf )  tf  x T Qx + 2uT Sx + x T W (Ax + Bu − x) + ˙ dt

(2.299)

t0

On integrating by parts the term involving x, ˙ we have 

tf

J2 = x(tf )T Qf x(tf ) +

 x T (W˙ + Q + AT W + W A)x

(2.300)

t0

+ 2uT (S + B T W )x dt + x(tf )T [Qf − W (tf )]x(tf ) The substitution of Eq. (2.298) into Eq. (2.300) yields 

tf

J2 = 2

uT (S + B T W )xdt

(2.301)

t0

Now, consider an arbitrary control variation, u, which vanishes over the entire control interval, t ∈ [t0 , tf ], except in a narrow interval, t ∈ [t1 , t1 + ], with t1 ∈ [t0 , tf ], where it has a constant and arbitrary value, k ∈ Rm×1 . Thus, the second variation becomes the following:  J2 = 2

t1 +

k T (S + B T W )xdt

(2.302)

t1

whose dominant term in the limit,  → 0, is  k T (S + B T W )Bk 2

t=t1

(2.303)

2.12 Numerical Solution Procedures

69

The necessary condition for this term to be non-negative for any arbitrary t1 ∈ [t0 , tf ] is therefore SB + B T W B ≥ 0, which completes the proof. Example 2.14 Let us revisit Example 2.12 in the light of Jacobson’s condition. Here, we have  Q = Hxx =

−2 0 0 2

 ,

S = Hux = (0, 0) ,

B = fu = (0, 1)T

from which it follows that  S + B T W = (0, 0) + (0, 1)

w11 w12 w12 w22

 = (w12 , w22 )

SB + B T W B = (w12 , w22 )(0, 1)T = w22 and Eq. (2.298) produces  −

w˙ 11 w˙ 12 w˙ 12 w˙ 22



 =

−2 w11 w11 2 + 2w12



 ,

w11 (π ) w12 (π ) w12 (π ) w22 (π )



 =

0 0 0 0



whose solution is easily derived to be the following: w11 = −2τ ,

w12 = −τ 2 ,

2 w22 = 2τ − τ 3 3

√ where τ = π − t is the time-to-go. Since w22 < 0 in the interval τ > 3, Jacobson’s necessary condition √ given by Theorem 2.13 fails to be satisfied for a part of the control interval, 3 < τ ≤ π . Thus, the singular solution is seen to be non-optimal. We note that while the generalized Legendre–Clebsch necessary condition was seen to be satisfied in Example 2.12, it did not guarantee the solution to be optimal because it was not a sufficient condition. Jacobson’s condition is the additional test which reveals the same solution to be actually a non-optimal one.

2.12 Numerical Solution Procedures As has been discussed in the previous sections of this chapter, any optimal control formulation without specific interior point constraints results in a twopoint boundary value problem (2PBVP), which can be expressed as the following set of first-order, ordinary differential equations comprising the state and costate equations: y˙ = F (y, t) ,

(2.304)

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2 Analytical Optimal Control

where  y(t) =

x(t) λ(t)

 ∈ R2n×1

and F (.) : R(2n+1)×1 → R2n×1 is a mapping vector functional. The equality boundary conditions applied at the ends of the control interval, t ∈ [t0 , tf ], are generally expressed as follows: g0 [y(t0 )] = 0 ;

gf [y(tf )] = 0 .

(2.305)

As discussed previously, the initial conditions are specified only for the state vector, x(t), whereas the terminal boundary conditions are to be satisfied by both the state and the costate (λ(t)) vectors. A solution procedure for Eq. (2.304) is thus necessary, which can march both forward and backward in time until all the boundary conditions, Eq. (2.305), are satisfied. If the boundary conditions were to be only specified at one end, the resulting problem is called an initial value problem (IVP) whose integration can be carried out by a variable-step time-marching scheme such as the Runge–Kutta methods [77] based upon a Taylor series approximation of Eq. (2.304). However, a 2PBVP is inherently more difficult to solve than an IVP since boundary conditions must be satisfied at both the ends. It is not necessary that there should exist a solution to a particular 2PBVP, or that the solution be unique. The existence and uniqueness of the solution to a 2PBVP depend upon the nature of the functional, F (y, t), at the interior and boundary points. The matters become complicated if F (y, t) has either singularities, or discontinuous partial derivatives. Difficulties of computing a solution to a 2PBVP are further compounded by errors in the numerical approximation employed for the purpose. In some cases, there may be spurious numerical solutions when none really exist, and in others a particular numerical scheme may be unable to find a solution when one does exist. Even when a unique solution does exist to a 2PBVP, iterative numerical schemes are often required and are divided into two broad categories: (a) shooting method, and (b) collocation method.

2.12.1 Shooting Method A shooting method is based upon posing the boundary value problem as if it were an initial value problem (IVP), and then changing the initial conditions by trial and error until the boundary conditions are satisfied with a given accuracy. Thus, a shooting method is based upon a repeated integration of the governing differential equations for various initial conditions, using a standard numerical scheme such as the Runge–Kutta method [77]. Simple shooting methods solve the IVP over the

2.12 Numerical Solution Procedures

71

entire control interval, t0 ≤ t ≤ tf , and can face convergence problems even if the governing system of equations is stable and well-conditioned. In order to remove such difficulties, multiple shooting methods have been developed [77] that partition the control interval into several grid points, and shooting is carried out in several stages.

2.12.2 Collocation Method In contrast with the shooting methods, a collocation method approximates the solution by a linear combination of a number of piecewise continuous polynomials that are usually defined on a mesh of collocation points [59, 72]. The approximate solution is then substituted into the system of ODEs such that the system is exactly satisfied at each collocation point. The number of collocation points plus the number of boundary conditions must equal the number of unknown coefficients in the approximate solution. The most common choice of approximation is a linear combination of spline functions. The collocation solution strategy is thus based upon dividing the control interval into a number of sub-intervals, and approximating the solution in each sub-interval by an appropriate spline interpolated at the midpoint. In order to achieve a given accuracy, the collocation points must be carefully selected. Commercially available software such as Maple and MATLAB contain inbuilt 2PBVP solvers based upon either the shooting or the collocation scheme. For a further description of the numerical solution procedures, a reader is referred to a textbook devoted to such methods [59, 72, 77].

Exercises 2.1 A scalar system with the state equation, x˙ = x + u, is to be moved from the initial state, x(0) = 1, in exactly five units of time, while minimizing the following performance index: J =

1 1 [x(5)]2 + 2 2



5

u2 (t)dt

0

Determine the extremal trajectory and the corresponding control history and find out whether they are optimal. 2.2 A block sliding in a straight, frictionless groove on a horizontal table has its equation of motion given as follows: y¨ = u

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2 Analytical Optimal Control

where y(t) is the displacement of the block measured from one end of the table, and u(t) is the applied acceleration input. The system is to be controlled such that beginning from rest at t = 0, the block should come to rest at a final displacement of 10 m after 1 s, and the total control effort, given by: 1 J = 2



1

u2 (t)dt

0

is a minimum. Solve for the extremal trajectory and the corresponding control history and determine whether the extremal control history is optimal. 2.3 Repeat Exercise 2.2 if the final displacement and velocity after 1 s are not specified, and the performance index is changed to the following: 2 J = k12 [y(1) − 10]2 + k22 [y(1)] ˙ +

1 2



1

u2 (t)dt

0

where k1 ∈ R, k2 ∈ R. Determine the value of the cost coefficients, k12 , k22 , such that the final displacement and the velocity of the block exactly after 1 s are within ±1% of 10 m and 0, respectively. Plot the optimal trajectory and control history for this set of the cost coefficients. 2.4 Repeat Exercise 2.2 if the terminal time tf is not specified, and the performance index to be minimized while moving the block from rest to rest by 10 m is the following: J =

900tf2

1 + 2



tf

u2 (t)dt

0

2.5 Does the optimal control problem for the system in Example 2.6 have any solution if the terminal time, tf , is unspecified, and the system is to be moved from a zero initial state, x1 (0) = x2 (0) = 0 to x1 (tf ) = 1, x2 (tf ) = 0, while minimizing J =

1 2



tf

u2 (t)dt ?

0

Explain your answer. 2.6 Consider a system with the following state equations: x˙1 = x2 − x1 x˙2 = −2x1 − 3x2 + u

2.12 Numerical Solution Procedures

73

Determine the extremal trajectory and control history for a fixed terminal time, tf = 1, if the system is to be taken from an initial state of x(0) = (0, 0)T , to a free final state such that the following performance index is minimized: 1 1 1 J = [x1 (tf ) − 1]2 + [x2 (tf ) − 1]2 + 2 2 2



1

u2 (t)dt

0

2.7 Solve the lunar landing problem of Example 2.9 with the revised initial conditions, z(0) = 100 m, z˙ (0) = −10 m/s, and for tf = 30 s. What is the value of the vertical acceleration input at the final time and why? 2.8 For the lunar landing problem of Example 2.9, revise the objective function to the following:  1 1 1 tf 2 J = k 2 tf2 + [x2 (tf ) − 1]2 + u (t)dt 2 2 2 0 and find the optimal trajectory and control history for a suitable value of the cost parameter k. 2.9 For a system with the following state equations: x˙1 = x2 x˙2 = −x2 + u Determine the extremal trajectory and control history for a fixed terminal time, tf = 2, if x(0) = (0, 0)T , x(2) is not specified, and the following performance index is minimized: J =

1 1 1 [x1 (2) − 5]2 + [x2 (2) − 2]2 + 2 2 2



2

u2 (t)dt

0

2.10 Repeat Exercise 2.9 for x(0) = (0, 0)T , x(2) = (5, 2)T , and J =

1 2



2

u2 (t)dt

0

2.11 Repeat Exercise 2.9 if the system is to be transferred from x(0) = (0, 0)T to the following straight line at tf = 2: x1 (t) + 5x2 (t) = 15 while minimizing 1 J = 2

 0

2

u2 (t)dt

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2 Analytical Optimal Control

2.12 Consider a particle moving in a plane acted upon by a known acceleration, a(t), which is produced by a force applied at an angle, β(t), measured from the direction of motion. The equations of motion with (x, y) as the displacement coordinates, and (u, v) as the velocity components are the following: x˙ = u y˙ = v u˙ = a cos β v˙ = a sin β It is required to move the particle in a given time T such that y(T ) = h and v(T ) = 0. Considering β(t) to be the control input, formulate the two-point boundary value problem to be solved for the extremal trajectory. 2.13 Consider a bead sliding on a frictionless wire stretched between two points with the following state equations: x˙ = v(y) cos θ y˙ = v(y) sin θ where x, y are the horizontal and vertical coordinates of the bead, θ (t) is the local angle made , by the wire with the horizontal, the speed of the bead is given by

v(y) = v02 + 2gy, with initial speed v0 , and g being the constant acceleration due to gravity. State the necessary conditions for the minimum-time path taken by the bead between two given points, if θ (t) is the control input. (This problem is called the brachistochrone problem.) 2.14 The state equations of a ship with the displacement coordinates (x, y), moving at a constant speed V with heading angle, β(t), between strong ocean currents of known velocity field, [u(x, y), v(x, y)] are the following: x˙ = V cos β + u(x, y) y˙ = V sin β + v(x, y) It is required to steer the ship in the minimum time between two specified positions, (xi , yi ), and (xf , yf ). (This is called Zermelo’s navigation problem.) (a) Considering β(t) to be the control input, formulate the two-point boundary value problem to be solved for the extremal trajectory. (b) From (a), derive the differential equation to be solved for β(t). (c) What is the solution to (b), if the ocean currents have constant velocity components, (u, v)?

2.12 Numerical Solution Procedures

75

2.15 Suppose the scalar system with the state equation, x˙ = u, has its input bounded by: |u(t)| ≤ 1 Derive the switching condition on u(t) for the time-optimal transfer from x(0) = 1 to x(tf ) = 0 and solve for the resulting optimal trajectory. 2.16 Does the minimum-time problem for the block sliding in a frictionless, horizontal, straight groove (Exercise 2.2) have any solution if the input is unconstrained? Explain your answer. 2.17 State the necessary conditions for the minimum-time transfer of the particle in Exercise 2.11. 2.18 Reformulate the lunar landing problem of Example 2.9 as a time-optimal control problem with the vertical acceleration input bounded by | u |≤ a, and the initial conditions, z(0) = 100 m, z˙ (0) = −10 m/s. (a) Can the upper bound, a, be smaller than g for a smooth landing (˙z(tf ) = 0)? (b) For a = 1.5g, solve the minimum-time problem and determine the trajectory and control history.

Chapter 3

Orbital Mechanics and Impulsive Transfer

3.1 Introduction Orbital mechanics refers to the study of the translational motion of bodies in mutual gravitational attraction, and the resulting spatial paths of the centers of mass of the respective bodies are called the orbits. The relative motion of two spherical bodies in mutual gravitational attraction is the fundamental problem of translational space dynamics, called the two-body problem (or the Keplerian motion), and possesses an analytical solution. The motion of a spacecraft under the influence of a celestial body is usually approximated as a two-body problem by neglecting the gravitational effects caused by the other bodies, the actual, aspherical shapes of the two bodies, and non-gravitational disturbances due to the atmosphere, continuous thrust, and solar radiation pressure. The gravity field in such a case is exactly spherical, governed by the inverse-squared law of Newton. However, when a spacecraft comes under the influence of two (or more) bodies whose gravitational pull is of comparable magnitudes in certain regions of the flight, then the approximation of Keplerian motion becomes invalid, and an n-body gravity model with n > 2 is required. Lunar and interplanetary flights are the examples of such problems. For a flight within the solar system, it is rarely necessary to take into account more than three bodies in mutual gravitational attraction. Thus the three-body problem (n = 3) is the limiting case of analytical space flight navigation problems. For n > 3, a numerical solution is invariably required, hence a theoretical orbital model is unavailable. Furthermore, the mass of the spacecraft is negligible in comparison with those of the bodies attracting it, resulting in what is referred to as a restricted two- (or three-) body problem. The present chapter introduces the reader to the two-body orbits and their impulsive changes (called impulsive orbital transfers). The following chapters will cover the more general cases of continuous transfers and non-spherical gravity fields. For a greater detail, a textbook on orbital mechanics can be consulted, such as Chobotov [23] or Battin [8]. © Springer Nature Switzerland AG 2019 A. Tewari, Optimal Space Flight Navigation, Control Engineering, https://doi.org/10.1007/978-3-030-03789-5_3

77

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3 Orbital Mechanics and Impulsive Transfer

3.2 Keplerian Motion Consider spherical masses m1 , m2 in mutual attraction given by Newton’s law of gravitation. The equation of motion of each mass is given by Newton’s second law of motion referred to an inertial frame. Subtracting the two equations of motion from one another, we have r¨ +

μ r=0 r3

(3.1)

where r(t) is the position of center of mass m2 relative to that of mass m1 , μ = G(m1 + m2 ), with G = 6.6726 × 10−11 m3 /kg/s2 is the universal gravitational constant, and overdot represents the time derivative, d/dt. Since the spacecraft’s mass, m2 , is negligible in comparison with that of the celestial body, m1 , we can approximate μ ≈ Gm1 . Take the right vector product of both the sides of Eq. (3.1) with r to show that the specific angular momentum of m2 relative to m1 , defined by h = r × r˙ = r × v

(3.2)

is conserved in Keplerian motion, where v = r˙ is the orbital velocity. Since h is a constant vector, it follows that (a) The direction of h is a constant, implying that the vectors r and v are always in the same plane—called orbital plane—and h is normal to that plane. (b) The magnitude of h is constant. Writing h in polar coordinates, (r, θ ), h =| h |=| r × v |= r 2 θ˙

(3.3)

which implies that the radius vector, r, sweeps out area at a constant rate, thereby confirming Kepler’s second law of planetary motion. The trajectories in the orbital plane—called orbits—are classified according to the magnitude and direction of a constant h. The case of h = 0 represents rectilinear motion along the line joining the two bodies, while h = 0 represents the more common orbits involving the mass m2 describing a curve about m1 . A scalar product of Eq. (3.1) with v = r˙ reveals that the specific energy of the two-body relative motion is also constant, given by v2 μ − = 2 r

(3.4)

The right vector product of both sides of Eq. (3.1) with h leads to another constant vector, e, called the eccentricity vector, and defined by μe = v × h −

μr r

(3.5)

3.2 Keplerian Motion

79

Since the eccentricity vector is normal to h, it always lies in the orbital plane, and can be used as a reference to specify the direction of the relative position vector, r(t). The angle, θ (t), made by r(t) with e (measured along the flight direction) is called the true anomaly, θ (t). Taking the scalar product of μe with r(t) produces the following orbit equation: r(θ ) =

p 1 + e cos θ

(3.6)

where p = h2 /μ is the parameter of the orbit and e is termed eccentricity of the orbit. The orbit equation defines the shape of the orbit in polar coordinates, (r, θ ), and indicates that the orbit is symmetrical about e. Furthermore, it shows that the orbit is a conic section, with the minimum separation of the two bodies (called periapsis) occurring for θ = 0. Hence, e points toward the periapsis. For e < 1, the orbit is an ellipse, whose special case is the circular orbit (e = 0). For e = 1, the orbit is a parabola, whereas for e > 1, it is a hyperbola. A time derivative of the orbit equation produces the radial velocity, vr = r˙ =

eh sin θ = p

-

μ sin θ p

(3.7)

while the circumferential velocity is derived from Eqs. (3.3) and (3.6) to be the following: vθ = r θ˙ =

h = r

-

μ (1 + e cos θ ) p

(3.8)

The ratio of the two velocity components gives the flight-path angle, φ, as follows: tan φ =

vr r˙ sin θ = = ˙ vθ 1 + e cos θ rθ

(3.9)

while their contribution to the velocity magnitude (orbital speed), v, is given by v=

,

vr2 + vθ2 =

μ (1 + 2e cos θ + e2 ) p

(3.10)

A substitution of Eq. (3.10) into Eq. (3.4) yields =

μ μ(1 − e2 ) μ v2 − =− =− 2 r 2p 2a

(3.11)

where a = p/(1 − e2 ) is defined to be the semi-major axis of the orbit. For an elliptical orbit, a > 0 has a geometric significance, whereas a = ∞ for a parabolic orbit, and a < 0 for a hyperbolic orbit. In all cases, the focus of the orbit is at

80

3 Orbital Mechanics and Impulsive Transfer

the center of the celestial body, and the smallest distance between the two bodies (periapsis radius) is rp = p/(1 + e).

3.2.1 Reference Frames of Keplerian Motion The velocity vector can be resolved in the local-horizon frame, (ir , iθ , ih ), with ir = r/r, and iθ being normal to both ir and ih , such that ih = ir × iθ : v = vr ir + vθ iθ =

μ [sin θ ir + (1 + e cos θ )iθ ] p

(3.12)

The local-horizon frame—being attached to the spacecraft—is a moving frame, and the plane (iθ , ih ), which is normal to the radius vector, r, is called the local horizon. A right-handed coordinate frame fixed to the orbital plane, with unit vectors ih = h/ h, ie = e/e, and ip = ih × ie , is used to specify the relative position and velocity vectors of the spacecraft in the orbital plane. This frame, (ie , ip , ih ), is called the perifocal frame. The position and velocity of the spacecraft in the perifocal frame are the following: r = r cos θ ie + r sin θ ip v = v(sin φ cos θ − cos φ sin θ )ie + v(sin φ sin θ + cos φ cos θ )ip

(3.13)

where φ is the flight-path angle (defined earlier) measured above the local horizon. On eliminating the flight path angle from the velocity expression, we have v=−

μ μ sin θ ie + (e + cos θ )ip h h

(3.14)

The spacecraft’s position and velocity can also be resolved in an inertial celestial frame, (I, J, K), which is fixed with respect to distant stars, with its origin, o, at the center of the celestial body (Fig. 3.1). The position vector, r, can be expressed in terms of spherical coordinates, (r, δ, λ), where r is the radius, δ is the declination (or latitude) measured above the (I, J) plane, and λ the right ascension (or celestial longitude) measured from the I axis to the projection of r in the (I, J) plane along the right-hand rotation about K (Fig. 3.1). Another orientation of the localhorizon frame is useful in representing the velocity vector in terms of the spherical coordinates. It consists of the unit vectors, (ir , iλ , iδ ), where iλ points toward the direction of increasing λ, and iδ toward that of increasing δ, such that ir × iλ = iδ (Fig. 3.1). The velocity vector, v, is then resolved in spherical coordinates, (v, φ, ψ), relative to the local-horizon frame, where v is the speed, φ the flight-path angle made by v with the local-horizon plane, (iλ , iδ ), measured positive toward ir , and ψ the azimuth angle between the projection of v on the local-horizon plane, (iλ , iδ ), and the axis iδ , measured positive toward iλ , as shown in Fig. 3.1.

3.2 Keplerian Motion

81

v

Fig. 3.1 Position and velocity vectors resolved in the celestial, (I, J, K), and local horizon, (ir , iλ , iδ ), frames, respectively

id K

il

f

ir

y

r

d

o

J

l I

K

Local Horizon

v

f

ih

ip i

r

ie Periapsis

q o I Vernal Equinox

J

w

W Orbit

n Ascending Node

Fig. 3.2 Orientation of the perifocal frame relative to celestial plane given by Euler angles, Ω, i, ω

Examples of Earth-based celestial frames are with the (I, J) plane being either the equatorial plane or the ecliptic plane, with the axis I pointing toward the vernal equinox. This is the point of intersection between the equatorial and ecliptic planes, where the Sun appears to cross the equator from the south to the north at mid-day (which usually happens around March 21). The intersection of the orbital plane with the reference plane, (I, J), yields the line of nodes, as shown in Fig. 3.2. The ascending node is the point on the line of nodes where the orbit crosses the plane (I, J) from the south to the north. A unit vector, n, pointing toward the ascending node makes an angle Ω with the axis I. The angle Ω is measured in the plane (I, J) in an anti-clockwise direction (right-handed rotation about K) (Fig. 3.2), and is termed the right ascension of the ascending node. The orbital inclination, i, is

82

3 Orbital Mechanics and Impulsive Transfer

the angle between the orbital plane and the (I, J) plane, and is the positive (righthanded) rotation about n required to produce ih from the axis K. The angle ω represents a positive rotation of n about ih to produce ie in the orbital plane, and is called the argument of periapsis. A coordinate transformation between the perifocal and celestial frames is necessary. The orientation of the perifocal frame relative to the celestial frame is classically described by the Ω, i, ω Euler angles, sequentially applied in the righthand sense about the third, first, and third axes, respectively, of the successively rotated reference frame, as shown in Fig. 3.2. This results in the following rotation matrix [80]: C = C3 (ω)C1 (i)C3 (Ω)

(3.15)

where ⎞ ⎞ ⎛ 1 0 0 cos(.) − sin(.) 0 C1 (.) = ⎝ 0 cos(.) − sin(.) ⎠ , C3 (.) = ⎝ sin(.) cos(.) 0 ⎠ 0 sin(.) cos(.) 0 0 1 ⎛

(3.16)

Hence the transformation between the perifocal and the celestial frames is given as follows: ⎧ ⎫ ⎧ ⎫ ⎨ ie ⎬ ⎨I⎬ T (3.17) = C i J ⎩ p⎭ ⎩ ⎭ ih K Similarly, a coordinate transformation between the local-horizon and celestial frames is described by the Euler angles, λ, −δ, sequentially applied in the righthand sense about the third and the second axes, respectively, of the successively rotated reference frame, as shown in Fig. 3.1: ⎧ ⎫ ⎧ ⎫ ⎨I⎬ ⎨ ir ⎬ J = CLH T iλ ⎩ ⎭ ⎩ ⎭ iδ K

(3.18)

CLH = C2 (−δ)C3 (λ)

(3.19)

⎞ cos(.) 0 sin(.) C2 (.) = ⎝ 0 1 0 ⎠ − sin(.) 0 cos(.)

(3.20)

where

where ⎛

3.2 Keplerian Motion

83

3.2.2 Time Equation While the vectors h and e completely determine the shape and orientation of a twobody orbit, they do not provide any information about the location of the spacecraft at a given time along the orbit. This missing information is expressed as a differential equation which relates the variation of true anomaly with time, θ (t), beginning with the time of periapsis, τ , for which θ = 0. On substituting the orbit equation, Eq. (3.6), into Eq. (3.8), we have θ˙ =

-

μ (1 + e cos θ )2 p3

(3.21)

The integration of Eq. (3.21) provides τ , thereby determining the function θ (t), and completing the solution to the two-body problem. However, such an integration is usually carried out by a numerical procedure, depending upon whether the orbit is elliptical, parabolic, or hyperbolic. Elliptical Orbit (0 ≤ e < 1) For an elliptical orbit, Eq. (3.21) becomes the following: 3

(1 − e2 ) 2 dθ = ndt (1 + e cos θ )2

(3.22)

where n=

μ a3

(3.23)

is the orbital mean motion. Now introduce an eccentric anomaly, E defined by cos E =

e + cos θ 1 + e cos θ

(3.24)

which substituted into Eq. (3.22), yields the following Kepler’s equation: E − e sin E = M

(3.25)

where M = n(t − τ ) is called the mean anomaly. The unambiguous relationship between the true and eccentric anomalies is given by tan

θ = 2

-

E 1+e tan 1−e 2

(3.26)

84

3 Orbital Mechanics and Impulsive Transfer

Note that E2 and θ2 are always in the same quadrant. Solution to Kepler’s equation is simply obtained by Newton’s method. Consider a first-order, finite-difference approximation of the function, f (E) = E − e sin E − M, as follows: f (E + ΔE) f (E) + f (E)ΔE

(3.27)

where f = df (E)/dE = 1 − e cos E and ΔE is the step size. Newton’s method is then based upon starting with an initial guess of the solution (usually E = M + e sin M), updating it to E + ΔE where ΔE = −f (E)/f (E) until f (E + ΔE) < δ, a given tolerance. The position and velocity vectors in an elliptical orbit can be expressed directly in terms of the eccentric anomaly as follows: √ r = a(cos E − e)ie + ap sin Eip √ √ μp μa sin Eie + cos Eip v=− r r

(3.28)

Example 3.1 A low-Earth satellite is to be de-orbited from a circular orbit of period 90 min. by firing a retro-rocket which applies tangential velocity impulse of Δv = −0.5iθ km/s. Calculate the radius, the speed, and the flight-path angle 5 min. after firing the rocket. (For Earth: μ = 398,600.4 km3 /s2 , mean radius, R0 = 6378.14 km.) We begin by calculating the spacecraft’s orbit following the velocity impulse. The initial orbital radius is calculated from the time period of the orbit as follows: 2π = 0.0013884 rad/s r1 = (n2 μ)1/3 = 6652.6 km 60 × 90 √ The initial speed in the circular orbit is therefore v1 = μ/r1 = 7.7406, from which the horizontal impulse is subtracted to yield the initial speed of the new orbit, v1 = v1 − 0.5 = 7.2406 km/s. The semi-major axis of the new orbit is calculated using Eq. (3.11) as follows: n=

a=

μ = 5913.3 km 2μ/r1 − v12

Since the semi-major axis of this elliptical orbit is smaller than Earth’s radius, a < R0 , the satellite will finally impact Earth. Furthermore, since a < r1 , the initial tangential impulse point is the apogee (the point of the largest radius) of the new orbit. Therefore, we put r1 = r(π ) in the orbit equation, Eq. (3.6), and obtain r1 = a(1 + e), from which the eccentricity is derived to be e = r1 /a − 1 = 0.125. (The same result can be obtained by calculating the angular momentum, h, of the new √ orbit, h = r1 v1 = 48,168.49 km2 s, giving p = h2 /μ = 5820.9 km, and e = 1 − p/a = 0.125.)

3.2 Keplerian Motion

85

Next, Kepler’s equation is solved by taking the initial anomalies for the apogee point, θ1 = E1 = M1 = π , and determining the final eccentric anomaly, E2 , after 5 min. The mean anomaly at the final position is M2 = π + n × 5 × 60 = 3.5581221 rad., from which the initial guess of E (0) = M2 is made. The steps of Newton’s iteration are then the following: (i) f (E (0) ) = E (0) − e sin E (0) − M2 = 0.05058, f (E (0) ) = 1 − e cos E (0) 1.1143, ΔE = −f (E (0) )/f (E (0) ) = −0.045391 rad. (ii) E (1) = E (0) + ΔE = 3.5127312 rad., f (E (1) ) = E (1) − e sin E (1) M2 = −5.03 × 10−5 , f (E (1) ) = 1 − e cos E (1) = 1.1165, ΔE −f (E (1) )/f (E (1) ) = 4.5 × 10−5 rad. (iii) E (2) = E (1) + ΔE = 3.5127763 rad., f (E (2) ) = E (2) − e sin E (2) − M2 −4.6 × 10−11 rad.

= − = =

Hence, the solution has converged to within a tolerance of 10−10 rad. in two iterations, resulting in E2 = E (2) = 3.5127763 rad. The radius at this point is r2 = a(1 − e cos E2 ) = 6602.21 km, the speed,  2μ μ − = 7.3034 km/s v2 = r2 a and the flight-path angle, φ2 = cos−1

h = −2.617◦ r2 v2

where the correct quadrant (π < θ < 2π ) is employed. Hyperbolic Orbit (e > 1) The equivalent form of Kepler’s equation can be obtained for hyperbolic orbits by introducing the hyperbolic anomaly, H. The parametric equation of a hyperbola centered at x = 0, y = 0, with semi-major axis a, can be written as, x = a cosh H , or r cos θ = a(cosh H − e). Hence the relationship between the hyperbolic anomaly and the true anomaly is given by cos θ =

cosh H − e 1 − e cosh H

(3.29)

whose substitution into Eq. (3.21) and integration from H = 0, t = τ yields e sinh H − H = n(t − τ )

(3.30)

with the hyperbolic mean motion, n given by μ n= − 3 a

(3.31)

86

3 Orbital Mechanics and Impulsive Transfer

Equation (3.30) can be solved numerically using Newton’s method in a manner similar to Kepler’s equation. However, since a hyperbolic orbit is an open trajectory, the hyperbolic anomaly, H , keeps on increasing with time, and hence a larger number of iterations are required for an increased difference between the initial and final times. Parabolic Escape Trajectory (e = 1) A parabolic orbit is the minimum energy escape trajectory ( = 0), thus it has zero speed at r → ∞. It is often a valuable mathematical aid for quickly determining the minimum fuel requirements for a given mission, as well as for beginning the iterations of Kepler’s equation in the case e 1 or e > 1. Equation (3.21) for a parabolic orbit is integrated to yield the following Barker’s equation: tan3

θ μ θ (t − τ ) + 3 tan = 6 2 2 p3

(3.32)

whose real, unique, and closed-form solution is obtained by substituting tan

θ 1 =α− 2 α

(3.33)

and solving the resulting quadratic equation for α 3 , resulting in tan

. . 1 1 θ = (C + 1 + C 2 ) 3 − (C + 1 + C 2 )− 3 2

(3.34)

where μ . C=3 (t − τ ) p3

(3.35)

is an equivalent mean anomaly.

3.2.3 Lagrange’s Coefficients A two-body trajectory expressed in the perifocal frame (ie , ip , and ih ) is the following (Eq. (3.13)): r = r cos θ ie + r sin θ ip v=−

μ μ sin θ ie + (e + cos θ )ip h h

(3.36) (3.37)

3.2 Keplerian Motion

87

Given the position and velocity at a time, t0 , one would like to determine the position and velocity at some other time, t. In order to do so, the known position and velocity are expressed as follows: &

r0

/

! =

v0

r0 cos θ0

"&

r0 sin θ0

− μh sin θ0

μ h (e

+ cos θ0 )

ie

/ (3.38)

ip

Since the square matrix in Eq. (3.38) is non-singular (its determinant is equal to h), it can be inverted to obtain ie and ip as follows: &

ie ip

⎡

/

=⎣

1 p (e

1 p

⎤&

− rh0 sin θ0

+ cos θ0 )

r0 h (e

sin θ0

+ cos θ0 )

⎦

r0

/ (3.39)

v0

On substituting Eq. (3.39) into Eqs. (3.36) and (3.37), we have & / r v

0 =

f

g

1&

f˙ g˙

r0

/ (3.40)

v0

where f = 1+

r [cos(θ − θ0 ) − 1] p

rr0 sin(θ − θ0 ) h h df = − 2 [sin(θ − θ0 ) + e(sin θ − sin θ0 )] f˙ = dt p g=

g˙ =

r0 dg = 1 + [cos(θ − θ0 ) − 1] dt p

(3.41)

The functions f and g, called Lagrange’s coefficients, are very useful in determining a two-body orbit from a known position and velocity. The matrix, 0 (t, t0 ) =

f

g

f˙

g˙

1 (3.42)

called state transition matrix, has a special significance, because it uniquely determines the current state, (r, v), from the initial state, (r0 , v0 ). Such a relationship between the initial and final states is rarely possible in the solution to a nonlinear differential equation, and is thus a valuable property of the two-body problem. The state transition matrix is seen to have the following properties:

88

3 Orbital Mechanics and Impulsive Transfer

1. From the conservation of angular momentum, h = r × v = (f g˙ − g f˙)r0 × v0 = r0 × v0

(3.43)

|  |= f g˙ − g f˙ = 1

(3.44)

it follows that

A consequence of the unity determinant is that the inverse of the state transition matrix is given by 0 

−1

. =

g˙

−g

−f˙

f

1 (3.45)

Such a matrix is said to be symplectic. (It can be shown [60] that the state transition matrix of any conservative Hamiltonian system is symplectic.) 2. Given any three points (t0 , t1 , t2 ) along the trajectory, it is true that (t2 , t0 ) = (t2 , t1 )(t1 , t0 )

(3.46)

3.3 Impulsive Orbital Transfer A spacecraft’s nominal trajectory relative to the central body is generally the solution to the two-body problem, unless significant orbital perturbations are present. Even if orbital perturbations are present, the two-body orbit can be taken to be a nominal flight path, which should be either followed despite small perturbations by the application of control inputs or used as an initial guess for obtaining the closedloop solution to the perturbed problem by iterative techniques [80]. Therefore, the problem of guiding the spacecraft from an initial to a desired final position in space can be generally treated as the problem of determining an optimal trajectory that passes through the two given positions. Consider the impulsive orbital transfer from an initial position vector, r1 (t1 ), to a final position, r2 (t2 ) (see Fig. 3.3). The transfer angle, Δθ = θ2 −θ1 , in Fig. 3.3 is the difference of the respective true anomalies, (θ1 , θ2 ), along the transfer orbit. If the time of flight, tf = t2 −t1 , is known a priori, the problem has a unique solution, and is referred to as Lambert’s problem. However—as is commonly the case—if we have a choice of varying the time of flight according to the requirement of minimizing an objective function (such as the time of flight, or the propellant mass) for the transfer, one arrives at the problem of optimal impulsive transfer that must be solved by optimal control techniques (Chap. 2). In either case, one has to solve a two-point, boundary value problem yielding the required orbit, r(t), taking the spacecraft from r1 to r2 , which can be used as a nominal flight path to be followed by a closed-loop guidance system despite perturbations.

3.3 Impulsive Orbital Transfer

89

Fig. 3.3 Geometry of impulsive transfer between two given positions, r1 , r2

Transfer Orbit

c

r1 r2 Dq

o

b1 v1¢

Dv1

b2 v2

a1-f1 v1

Df1

a2-f2¢ v2¢

Df2

a1

f1

f 2¢

(a)

Dv2

a2

(b)

Fig. 3.4 Geometry of the two impulse applications in Example 3.2

Example 3.2 Calculate the magnitudes and the directions of the two velocity impulses required for transferring a spacecraft around Earth from r1 = 7000 km, v1 = 7 km/s, φ1 = 25◦ to r2 = 7500 km, v2 = 8 km/s, φ2 = 30◦ via a transfer ellipse of a = 7000 km and e = 0.5. Let the first impulse of magnitude Δv1 be applied at an angle α1 from the local horizon on the given orbit. Figure 3.4a shows the geometry of impulse application, which yields the following relationships: φ1 + β1 − α1 = 0 v1

v1 Δv1 = = sin(π + φ1 − α1 ) sin β1 sin Δφ1 Δv1 =

,

v12 + (v1 )2 − 2v1 v1 cos Δφ1

Here, Δ indicates the change caused by a velocity impulse, and the prime denotes the velocities at the two extremities of the transfer orbit, (v1 , φ1 ) and (v2 , φ2 ). Thus we have φ1 = φ1 + Δφ1 and φ2 = φ2 − Δφ2 . The transfer orbit yields the following at r = r1 : v1 = 7.54605 km/s, φ1 = 30◦ , implying Δφ1 = 5◦ , from which the

90

3 Orbital Mechanics and Impulsive Transfer

relationships derived from Fig. 3.4a produce the first velocity impulse as Δv1 = 0.83677 km/s and α1 = 76.81◦ . Similarly, the second impulse produces the geometry shown in Fig. 3.4b, resulting in the following relationships: φ2 + β2 − α2 = 0 v2

Δv2 v2 = = sin(π + φ2 − α2 ) sin β2 sin Δφ2 Δv2 =

,

v22 + (v2 )2 − 2v2 v2 cos Δφ2

The transfer orbit yields the following at r = r2 : v2 = 7.025 km/s, φ2 = ±29.745◦ , implying either Δφ2 = −0.2551◦ or Δφ2 = −59.745◦ . The first option produces Δv2 = 0.97558 km/s and α2 = 212.1◦ , while the second option (φ2 = −29.745◦ ) yields Δv2 = 7.5311 km/s and α2 = 276.57◦ . Clearly, the second option is uneconomical in comparison with the first, because it requires a larger velocity impulse magnitude. It can be easily calculated that the second option also requires a larger transfer angle, θ2 − θ1 = 113.13◦ , compared to that of the first option, θ2 − θ1 = 6.87◦ , which translates into a much larger flight time. Therefore, the first option is selected, Δv2 = 0.97558 km/s and α2 = 212.1◦ . For a two-body orbital transfer with an arbitrary transfer time, t2 − t1 , we express the final position vector, r2 , in terms of the initial position, r1 , and initial velocity, v1 , through Lagrange’s coefficients, f, g, as follows: r2 = f r1 + gv1 ,

(3.47)

where f = 1+

r2 (cos Δθ − 1) p

r1 r2 sin Δθ h h df = − 2 [sin Δθ + e(sin θ2 − sin θ1 )] f˙ = dt p g=

g˙ =

r1 dg = 1 + (cos Δθ − 1) , dt p

(3.48)

and Δθ = θ2 − θ1 is the given transfer angle (Fig. 3.3). Furthermore, the initial position vector in terms of the final position and velocity is given by r1 = f r2 + g v2 ,

(3.49)

3.3 Impulsive Orbital Transfer

91

where r1 (cos Δθ − 1) = g˙ p r1 r2 g = − sin Δθ = −g . h

f = 1 +

(3.50)

Thus, we have r1 = gr ˙ 2 − gv2 .

(3.51)

The initial and final velocity vectors can be expressed as components along the chord, vc , and the local radius, vr . Note from Fig. 3.3 that the chord and local radius are not necessarily perpendicular to each other (vr = r˙ ). Thus we have v1 =

1 [(r2 − r1 ) + (1 − f )r1 ] = vc ic + vr ir1 g

v2 =

1 ˙ 2 ] = vc ic − vr ir2 , [(r2 − r1 ) − (1 − g)r g

(3.52)

where r2 − r1 c r1 = r1 r2 = r2

ic = ir1 ir2

vc =

ch c = g r1 r2 sin Δθ

vr =

h(1 − cos Δθ ) . p sin Δθ

(3.53)

The last two equations of Eq. (3.53) imply that p=

r1 r2 (1 − cos Δθ ) vc . c vr

(3.54)

Furthermore, we have vc vr =

cμ sec2 Δθ 2 , 2r1 r2

(3.55)

92

3 Orbital Mechanics and Impulsive Transfer

which is an important result showing that the product of the two velocity components depends only upon the geometry of transfer specified by (r1 , r2 , Δθ ) (and not on the shape or size of the transfer orbit). Thus, we can select various transfer orbits to meet other requirements (such as the time or fuel for transfer).

3.3.1 Minimum Energy Transfer A practical solution for the impulsive two-point, boundary value problem is the case of the minimum energy transfer, which corresponds to the smallest value of orbital energy  = −μ/2a for the transfer orbit. The minimum energy orbit involves the smallest positive value of the semi-major axis, a, of the orbit joining the two given radii, (r1 , r2 ), and would normally result in the smallest propellant expenditure. Let the minimum energy, elliptical orbit have the semi-major axis amin , as shown in Fig. 3.5, and the respective distances of the initial and final positions from the vacant focus, oV , be (r1∗ , r2∗ ). For an ellipse of major axis a, we have r1 + r1∗ = 2a ;

r2 + r2∗ = 2a

(3.56)

or r1 + r2 + r1∗ + r2∗ = 4a

(3.57)

Minimum Energy Transfer Orbit

r1* c r2*

r1

oV Dq

r2

o

2amin

Fig. 3.5 Geometry of minimum energy orbital transfer for spacecraft guidance

3.3 Impulsive Orbital Transfer

93

Then it follows that the smallest possible value for the distance, r1∗ +r2∗ , is the chord, c =| r2 − r1 |, thereby implying that the vacant focus, oV , must lie on the chord line, as shown in Fig. 3.5. Thus we have 2amin =

1 (r1 + r2 + c) 2

(3.58)

The semi-perimeter, s, of the transfer triangle with sides r1 , c, r2 is given by s=

r1 + r2 + c . 2

(3.59)

Hence, the major axis of the minimum energy transfer ellipse equals the semiperimeter, 2amin = s .

(3.60)

Note that the minimum energy transfer has the same orbit, am , if the focus, o, is shifted to the vacant focus, oV , the transfer takes place from r1∗ to r2∗ (i.e., Δθ = π and c = r1 + r2 ), which implies from Eq. (3.52) that s = c and vc /vr = 1. Since the ratio vc /vr is unity for the minimum energy transfer, the following is the corresponding parameter from Eq. (3.54): pm =

r1 r2 (1 − cos Δθ ) . c

(3.61)

The following method can be employed for determining the minimum energy transfer orbit (and the corresponding time of flight) between two given positions, (r1 , r2 ): (a) Determine the chord, the initial and final radii, and the semi-major axis of the transfer orbit as follows: c =| r2 − r1 | ; a=

r1 =| r1 | ;

r2 =| r2 |

1 (r1 + r2 + c) 4

(b) Find the unit vectors along the chord and the two radii, and the transfer angle, Δθ , as follows: ic =

r2 − r1 ; c

cos Δθ = ir1 · ir2

ir1 =

r1 ; r1

ir2 =

r2 r2

sin Δθ =| ir1 × ir2 |

94

3 Orbital Mechanics and Impulsive Transfer

(c) Calculate the parameter and the orbital eccentricity as follows: p=

r1 r2 (1 − cos Δθ ) c p e = 1− a

(d) Referring to Fig. 3.5, determine the initial and final eccentric anomalies, the time of periapsis, and the time of flight as follows (see Eqs. (3.6) and (3.24)): E1 = cos

−1



1 r1 − e ae

E2 = 2π − cos−1 n=

μ ; a3

t2 − t1 =





1 r2 − e ae

τ = t1 −



E1 − e sin E1 n

E2 − E1 − e(sin E2 − sin E1 ) n

Example 3.3 Find the minimum energy transfer trajectory and the corresponding time of flight for reaching a 500 km altitude above the Earth (μ = 398,600.4 km3 /s2 , surface radius, r0 = 6378.14 km), latitude 45◦ , and right ascension 250◦ , beginning from an initial altitude, latitude, and right ascension of 200 km, −10◦ , and 30◦ , respectively. The necessary computation is performed as follows: r1 = r0 + 200 = 6578.14 km ,

r2 = r0 + 500 = 6878.14 km

r1 = 5610.289I + 3239.102J − 1142.282K km r2 = −1663.442I − 4570.270J + 4863.579K km c =| r2 − r1 |= 12,245.971 km a= ic =

1 (r1 + r2 + c) = 6425.563 km 4

r2 − r1 ; c

cos Δθ =

ir1 =

r1 ; r1

ir2 =

r2 r2

r1 · r2 = −0.656233760652892 r1 r2

3.4 Lambert’s Transfer

95

p = r1 r2 (1 − cos Δθ )/c = 6119.311 km , E1 = cos

−1



1 − r1 /a e

E2 = 2π − cos−1 n= t2 − t1 =

,



e=

. 1 − p/a = 0.218315

 = 1.679778 rad.

1 − r2 /a e

 = 4.383888 rad.

μ/a 3 = 0.00122575 rad/s

E2 − E1 − e(sin E2 − sin E1 ) = 2551.719 s n

Hence the required transfer orbit has elements, a = 6425.563 km, e = 0.218315, and τ = −(E1 − e sin E1 )/n = −1193.357 s (measured from t1 = 0 at launch), and the required time of flight is 2551.719 s. Since the two foci must always lie on the chord for the minimum energy ellipse (Fig. 3.5), the corresponding time of flight for even a small transfer angle, Δθ , can be quite large. In fact, it can be shown that the minimum energy orbit yields the maximum time of flight between a given pair of positions. In any case, a different transfer strategy becomes necessary when the time of flight is fixed, such as in an orbital rendezvous between two spacecraft, or interception of a target in space by a free-flight interceptor.

3.4 Lambert’s Transfer For an impulsive two-body orbital transfer, Lambert’s theorem states that the transfer time, t2 − t1 , is a function of only the semi-major axis, a, of the transfer orbit, the sum of the two radii, r1 + r2 , and the chord, c, joining the two positions. Thus, we have t2 − t1 = f (a, r1 + r2 , c)

(3.62)

Since there is no dependence of the transfer time on the orbital eccentricity, one can choose any value of e for the orbit, provided (a, r1 + r2 , c) are unchanged. This is tantamount to changing the orbital shape for given initial and final positions by moving the two foci in such a way that both (a, r1 + r2 ) are unaffected. Lambert’s theorem is readily proved for parabolic, elliptic, and hyperbolic transfer orbits [8]. An outcome of Lambert’s theorem is Lambert’s problem which refers to the twopoint boundary value problem (2PBVP) resulting from an orbital transfer between two position vectors in a given time. Such a problem is commonly encountered in the

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3 Orbital Mechanics and Impulsive Transfer

guidance of spacecraft and ballistic missiles, as well as in the orbital determination of space objects from two observed positions separated by a specific time interval. For the initial and final positions, (r1 , r2 ), the specified time of flight, t12 = t2 − t1 , (or the transfer angle, Δθ , in Fig. 3.3) Lambert’s theorem indicates a unique value of the semi-major axis, a, which must be determined from the orbital plane, (r1 , r2 ), and the respective true anomalies, (θ1 , θ2 ). Since each value of semi-major axis produces two possible transfer times corresponding to either Δθ ≤ π or Δθ > π , a simple iterative procedure may not converge to a unique a. Furthermore, the derivative of t12 with respect to a is infinite for a minimum energy orbit, therefore a gradient-based scheme such as Newton’s method cannot be used for solving the time equation with respect to a, if a = am . For practical guidance, a robust algorithm is needed which converges to a unique, non-singular solution. Examples of Lambert’s algorithms are the universal variable methods by Battin [8], which can be applied to elliptical, parabolic, and hyperbolic transfer trajectories. These methods are formulated either by the evaluation of a power series (Stumpff functions) or via continued fractions (hypergeometric functions).

3.4.1 Stumpff Function Method Consider an auxiliary variable, x, given by √ x = ΔE a ,

(3.63)

where ΔE = E2 − E1 , and a is the semi-major axis of the transfer orbit. Lagrange’s coefficients, f, g, are expressed as functions of x as follows [7]: f = 1− f˙ =

x2 C(z) r1

√ x μ [zS(z) − 1] r1 r2

1 g = t12 − √ x 3 S(z) μ g˙ = 1 −

(3.64)

x2 C(z) r2

where z=

x2 , a

(3.65)

3.4 Lambert’s Transfer

97

and C(z), S(z) are the following Stumpff functions [76]: C(z) =

1 z z2 − + − ··· 2! 4! 6!

S(z) =

z z2 1 − + − ··· 3! 5! 7!

(3.66)

The elliptical transfer is represented by z > 0, and the hyperbolic transfer by z < 0. The parabolic case, z = 0, results in C(0) = 1/2, S(0) = 1/6, and can be solved analytically. Another universal variable, y, is now defined by y=

r1 r 2 (1 − cos Δθ ) p

(3.67)

where p is the parameter of the transfer orbit. The Lagrange’s coefficients are expressed for the orbital transfer as follows: r2 (1 − cos Δθ ) p   -  1 − cos Δθ 1 μ 1 − cos Δθ 1 ˙ f = − − p Δθ p r1 r2 r1 r2 sin Δθ g= √ μp r1 g˙ = 1 − (1 − cos Δθ ) p

f = 1−

(3.68)

whose comparison with Eq. (3.64) yields the following: x=

y C(z)

(3.69)

where x is the solution to the following cubic equation: . √ t12 μ = Ax C(z) + S(z)x 3 ,

(3.70)

and A is given by A = sin Δθ

r1 r2 1 − cos Δθ

(3.71)

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3 Orbital Mechanics and Impulsive Transfer

Thus we have f = 1−

y r1

y g=A μ y g˙ = 1 − r2

(3.72) (3.73)

while the remaining Lagrange’s coefficient, f˙, is obtained from the relationship f g˙ − g f˙ = 1. By expressing the transfer time in terms of the universal variables, (x, y), rather than the classical elements, (a, e) (or (a, p)), the singularity and convergence issues discussed earlier have been avoided. However, the convergence of the present formulation is guaranteed only for Δθ < π . In a terminal control application, it is required to find out the initial and final velocities of the transfer, which can be finally derived from Lambert’s solution, f, g, g, ˙ as follows: v1 =

1 (r2 − f r1 ) g

1 ˙ 1 = (gr v2 = f˙r1 + gv ˙ 2 − r1 ) g

(3.74)

The relationship among the transfer geometry, time, and Lagrange’s coefficients is a transcendental one, which requires an iterative solution for the universal variables, (x, y). Such an algorithm using Newton’s method is coded by Tewari [81]. For carrying out Newton’s iteration, we write . √ F (x) = Ax C(z) + S(z)x 3 − t12 μ

(3.75)

and take the derivative with respect to x,   . dz C (z) + S (z)x 3 F (x) = A C(z) + 3S(z)x 2 + Ax √ dx 2 C(z)

(3.76)

where dz 2x = dx a and C (z) = −

2z 3z2 1 + − − ··· 4! 6! 8!

(3.77)

3.4 Lambert’s Transfer

99

S (z) = −

2z 3z2 1 + − − ··· 5! 7! 9!

(3.78)

The correction in each step is then calculated by Δx = −

F (x) F (x)

(3.79)

3.4.2 Hypergeometric Function Method Battin [8] presents an alternative procedure for the general Lambert’s problem through the use of hypergeometric functions evaluated by continued fractions, instead of the Stumpff functions. Such a formulation improves convergence, even for transfer angles greater than 180◦ , and renders the method insensitive to the initial choice of z. Consider a universal variable x such that the semi-major axis, a, of the transfer orbit is given by a=

am , 1 − x2

(3.80)

where am is the semi-major axis of the minimum energy ellipse connecting the initial and final radii. Clearly, for an elliptical transfer we have −1 < x < 1 (x = 0 for minimum energy transfer), while for a parabola, x = 1. A hyperbolic transfer requires x > 1. Furthermore, for a given transfer, (r1 , r2 , Δθ ), we define the following parameter, λ: √ λ=

r1 r2 Δθ cos s 2

(3.81)

where s is the semi-perimeter, s = 2am =

r1 + r2 + c 2

(3.82)

Note that −1 < λ < 1. Additional variables, y, η, and z, are defined by y=

.

1 − λ2 (1 − x 2 )

η = y − xλ

(3.83) (3.84)

and z=

1 (1 − λ − xη) 2

(3.85)

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3 Orbital Mechanics and Impulsive Transfer

which transform the time equation to the following: μ 4 t12 = η3 F (z) + 4λη 3 3 am

(3.86)

Here F (z) is a hypergeometric function [8] that can be evaluated by a continued fraction as follows: F (z) =

1 1−

(3.87)

γ1 z

1−

γ2 z γ z 1− 3 1−

and γn =

⎧ ⎨ ⎩

(n+2)(n+5) (2n+1)(2n+3) n(n−3) (2n+1)(2n+3)

..

.

n odd (3.88) n even

When adopting Newton’s method to solve the time equation for x, we write [81] 4 3 μ f (x) = η F (z) + 4λη − t12 3 3 am

(3.89)

 dη 4 3 dz f (x) = 4 η2 F (z) + λ + η F (z) dx 3 dx

(3.90)

and

where prime denotes differentiation with respect to the argument of a function. From the definitions given above, we have dη λη =− dx y

(3.91)

η2 dz =− dx 2y

(3.92)

and F (z) =

6 − 3γ1 G(z) 2(1 − z)[1 − zγ1 G(z)]

(3.93)

3.5 Optimal Impulsive Transfer

101

where G(z) =

1 1−

(3.94)

γ2 z

1−

γ3 z γ z 1− 4 1−

..

.

3.5 Optimal Impulsive Transfer The problem of transferring a spacecraft from one set of position and velocity, [r(t0 ), r(t0 )], to another, [r(tf ), r(tf )], can be optimized either by minimizing the total velocity change magnitude, Δv or the total time of transfer, tf − t0 . The transfer is made by applying a number N of velocity impulses, Δvk , at different time instants, t0 ≤ tk ≤ tf , such that immediately after the application of the k th impulse, there is an instantaneous change in the velocity without affecting the position: v¯ k = v(tk− ) + Δvk ,

r¯ k = r(tk− ) (1 ≤ k ≤ N )

(3.95)

where r(tk− ), v(tk− ) denote the position and velocity, respectively, of the two-body orbit evolved from the initial state, [r(tk−1 ), v¯ (tk−1 )], to the time t = tk , after the application of the previous velocity impulse at t = tk−1 , as follows: r˙ (t) = v(t) v˙ (t) = −μ

r( tk−1 ) = r¯ k−1

r(t) , r 3 (t)

v( tk−1 ) = v¯ k−1 , (tk−1 ≤ t ≤ tk )

(3.96)

Clearly, the motion evolves from the first impulse applied at time t = t0 , with v(t0− ) = v(t0 ), to the final impulse at t = tf = tN , and the terminal boundary condition requires the final impulse to be the following: ΔvN = v(tf ) − v(tN − )

(3.97)

A fuel-optimal trajectory would minimize the following Hamiltonian: H =

N # k=1

| Δvk | +λTrk [v(tk− ) + Δvk ] − μλTvk

r(tk ) r 3 (tk )

(3.98)

relative to Δvk , k = 1, . . . , N , where λrk and λvk are the costate vectors corresponding to r(tk ) and v(tk ), respectively. A spacecraft equipped with a chemical rocket engine can adjust both the magnitude and the direction of each velocity impulse, Δvk , their time instants, tk , as well as the total number, N , of the impulses. In general, due to the large number of variables, such an optimization problem has many possible solutions, and is therefore indeterminate. However, solutions can be derived analytically for specific problems.

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3 Orbital Mechanics and Impulsive Transfer

3.5.1 Coasting Arc Since an impulsive transfer consists of a zero-thrust (also called null-thrust or coasting) arc between any two velocity impulses, it is possible to use the optimal control method to evolve the costate vectors such that the optimal magnitude and direction of the next impulse is obtained. Consider the following Hamiltonian along a coasting arc: H = λTr v + λTv g

(3.99)

where g = −μr/r 3 . The costate equations for the arc to be an extremal trajectory are the following:   T  ∂g(r) T ˙λr = − ∂H =− λv ∂r ∂r

(3.100)

T  ˙λv = − ∂H = −λr ∂v

(3.101)

where the gravity-gradient matrix, ∂g/∂r, is the following: μ ∂g(r) 3μ = − 3 I3×3 + 5 rrT ∂r r r

(3.102)

A substitution of Eq. (3.101) into Eq. (3.100) yields the following second-order differential equation for λv (t): λ¨ v =



∂g(r) ∂r

T λv

(3.103)

Resolving λv into components along the local-horizon axes, (ir , iθ , ih ), we have λv = λvr ir + λvθ iθ + λvn ih

(3.104)

and substituting into Eqs. (3.102) and (3.103), the latter becomes the following: μ λ¨ v = 3 (2λvr ir − λvθ iθ − λvn ih ) r

(3.105)

Now, take the time derivatives of λv as follows: ∂λv λ˙ v = + θ˙ ih × λv ∂t ∂ 2 λv + θ¨ih × λv λ¨ v = ∂t 2 ∂λv + θ˙ 2 ih × (ih × λv ) + 2θ˙ ih × ∂t

(3.106)

(3.107)

3.5 Optimal Impulsive Transfer

103

where ∂t∂ denotes taking the time derivative of a vector’s components. Introduce for convenience, λv = rq, with q = (q1 , q2 , q3 )T , and substitute the Keplerian equations of motion expressed as follows: r¨ − r θ˙ 2 = −

μ r2

r θ¨ + 2˙r θ˙ = 0

(3.108)

Then Eq. (3.107) becomes the following: ∂ 2q ∂q ˙h× λ¨ v = r 2 + 2r θi ∂t ∂t μ + r θ˙ 2 ih × (ih × q) − 2 q r

(3.109)

In order to integrate the equation, we change the independent variable from the time to the true anomaly, θ , which results in the following differential equation with the prime denoting the derivative with respect to θ :   μp r λ¨ v = 3 q

+ 2ih × q + ih × (ih × q) − q p r

(3.110)

On substituting Eq. (3.110) into Eq. (3.105) with λv = rq, we have the following scalar differential equations for the elements of q: q1

− 2q2 =

3r q1 p

q2

+ 2q1 = 0 q3

(3.111)

+ q3 = 0

Of these, the last is readily integrated to yield q3 = A3 cos θ + B3 sin θ

(3.112)

where A3 , B3 are the constants of integration. Furthermore, since the gravity field is time invariant, the Hamiltonian is constant on the coasting arc: T H = −λ˙ v v + λTv g

= eq1 sin θ (1 + e cos θ ) + q2 (1 + e cos θ )2 + q1 (2 + e2 + 3e cos θ ) = C = const.

(3.113)

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3 Orbital Mechanics and Impulsive Transfer

This integral is identically satisfied by the first of Eq. (3.111), which is therefore discarded from future consideration. The second of Eq. (3.111) is now integrated to q2 = Ae − 2q1 , where A is a constant, and this result is used to eliminate q2 from Eq. (3.113), which is then integrated to yield the following result [52]: (3.114) q1 = (1 + e cos θ )(A cos θ + Be sin θ + CI1 )   D − A sin θ + CI2 q2 = (1 + e cos θ ) −A sin θ + B(1 + e cos θ ) + 1 + e cos θ where B, D are constants, and 1 + e cos θ cot θ + I1 e(1 + e cos θ ) e sin θ  dθ I1 = sin θ sin2 θ (1 + e cos θ )2 I2 =

(3.115)

Analytical expressions for evaluating the integral, I1 , are provided by Lawden [50]. Finally, the substitution of the orbit equation, Eq. (3.6), gives the costate vector for velocity, λv , as follows: λvr = A cos θ + Be sin θ + CI2 λvθ = −A sin θ + B(1 + e cos θ ) + λvn =

D − A sin θ + CI2 1 + e cos θ

(3.116)

A3 cos θ + B3 sin θ 1 + e cos θ

The vector λ˙ v = −(λrr , λrθ , λrn )T (the negative of the costate vector for radius λr ) can now be obtained by the first equation of Eq. (3.106), thereby completing the optimal solution for the coasting arc:   μp A sin θ − D − B + CI 3 1 + e cos θ r2 μ = −(λvr + λ vθ )θ˙ = − [−A(e + cos θ ) + De sin θ + C cos θ ] p3 μ

˙ = −λvr θ = − (3.117) [−A3 sin θ + B3 (e + cos θ )] p3

λrr = −(λ vr − λvθ )θ˙ = λrθ λrn

√

where I3 =

e sin θ − cos θ I1 − 2 e sin θ e sin θ (1 + e cos θ )

(3.118)

3.5 Optimal Impulsive Transfer

105

Equations (3.116) and (3.117) cannot be applied for the special case of a circular orbit (e = 0), for which the required integrations can be carried out separately (and much more easily) by putting e = 0 in the differential equations. The results, only for e = 0, are the following: λvr = A cos θ + B sin θ + 2C λvθ = −2A sin θ + 2B cos θ + D − 3Cθ

(3.119)

λvn = A3 cos θ + B3 sin θ μ λrr = − 3 (A sin θ − B cos θ + 3Cθ − D) r μ λrθ = (A sin θ − B cos θ + 3Cθ − D) r3 μ λrn = (A3 sin θ − B3 cos θ ) r3

(3.120)

The necessary conditions of optimality on a coasting arc require that both λv and λ˙ v must be continuous across a velocity impulse, and that the direction of each impulse must be opposite to λv . These can be obtained by taking a derivative of the Hamiltonian, Eq. (3.98), with respect to Δvk , k = 1, . . . , N, and equating it to zero for the stationary condition. We shall return to the optimal thrust direction in Chap. 5.

3.5.2 Hohmann Transfer Hohmann showed that the optimal transfer between two circular orbits consists of two velocity impulses applied tangentially to the respective orbits. Hohmann transfer between two circular orbits of radii r1 and r2 , respectively, is depicted in Fig. 3.6. Since both the velocity impulses, Δv1 and Δv2 , are tangential to the orbits, the semi-major axis of the transfer ellipse is the average of the two radii, a = (r1 + r2 )/2. It can be shown that Hohmann transfer satisfies the necessary conditions for optimality as follows. Let the two velocity impulses be applied at angles φ1 and φ2 , respectively, from the local horizon, (ir , iθ ): Δv1 = Δv1θ iθ1 + Δv1r ir1 = (v1 cos φ1 − v1 )iθ1 + v1 sin φ1 ir1 Δv2 = Δv2θ iθ2 + Δv2r ir2 =

(v2 − v2 cos φ2 )iθ2

(3.121) − v2 sin φ2 ir2

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3 Orbital Mechanics and Impulsive Transfer

Fig. 3.6 Geometry of Hohmann transfer between two circular orbits

Dv2

r2

r1

Dv1

√ √ where v1 = μ/r1 and v2 = μ/r2 are the two circular orbital speeds, v1 denotes the speed immediately after the application of the first impulse, and v2 the speed immediately before the application of the second impulse. The impulse magnitudes are thus the following: Δv1 = | Δv1 |= Δv2 = | Δv2 |=

, ,

(v1 − v1 )2 + 2v1 v1 (1 − cos φ1 ) (v2 − v2 )2 + 2v2 v2 (1 − cos φ2 )

(3.122)

The total velocity impulse magnitude, J = Δv1 + Δv2 , is the performance index to be minimized with respect to the control variables, v1 , v2 , φ1 , φ2 , subject to the following constraints on the transfer orbit: (i) The specific energy, , is constant along the transfer ellipse. This implies that (v1 )2 − (v2 )2 + 2(v22 − v12 ) = 0

(3.123)

(ii) The specific angular momentum magnitude, h, is conserved along the transfer ellipse, thereby implying that v22 v1 cos φ1 − v12 v2 cos φ2 = 0

(3.124)

3.5 Optimal Impulsive Transfer

107

The Hamiltonian of the optimal control problem is thus given by H =

,

(v1 − v1 )2 + 2v1 v1 (1 − cos φ1 ) +

,

(v2 − v2 )2 + 2v2 v2 (1 − cos φ2 )

+ λ1 [(v1 )2 − (v2 )2 + 2(v22 − v12 )] + λ2 (v22 v1 cos φ1 − v12 v2 cos φ2 ) (3.125) where λ1 , λ2 are the Lagrange multipliers associated with the two equality constraints. The stationary conditions yield the following: v1 − v1 cos φ1 ∂H = + 2λ1 v1 + λ2 v22 cos φ1 = 0 ∂v1

Δv1 v − v2 cos φ2 ∂H = 2 − 2λ1 v2 − λ2 v12 cos φ2 = 0

∂v2 Δv2 v1 v1

∂H = sin φ1 − λ2 v22 v1 sin φ1 = 0 ∂φ1 Δv1

(3.126)

v2 v2

∂H = sin φ2 + λ2 v12 v2 sin φ2 = 0 ∂φ2 Δv2 the last two of which require φ1 = φ2 = 0, which substituted into the constraints, Eqs. (3.123) and (3.124), yield v1 = , v2 = ,

v12 v12 +v22 2

v22 v12 +v22 2

(3.127)

This proves that Hohmann transfer ellipse is indeed an extremal trajectory, satisfying the necessary conditions of optimality. The range of the ratio, r2 /r1 , for which Hohmann transfer is optimal can be derived by the continuity of the costate vector, λv , and its time derivative, λ˙ v , at both the points of application of the impulses (see Eq. (3.116)). Using e = 0 for the initial and final circular orbits, the costate solutions at the impulse application points can be expressed as follows [52]: λvr = R sin θ λvθ = 2R cos θ + D

(3.128)

√ where R = A2 + B 2 and θ is the true anomaly measured from the first impulse point. The magnitude of λv on a circular orbit is thus given by λ2v = λ2vr + λ2vθ = R 2 + D 2 + 4RD cos θ + 3R 2 cos2 θ

(3.129)

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3 Orbital Mechanics and Impulsive Transfer

For optimality, λ2v must be stationary with respect to θ , which happens either at the points θ = 0, π or when 2D + 3R cos θ = 0

(3.130)

However, for a real solution of Eq. (3.130), we must have | D |≤ 3R/2, in which range λ2v is stationary only for θ = 0 and θ = π . Therefore, in all cases (unless R = 0) the impulses must be applied either at the periapsis (θ = 0) or the apoapsis (θ = π ) of the transfer ellipse. The cases θ = 0, π require that λvr = 0, which implies the impulses should be applied along ±iθ , that is, tangentially to the circular orbit. However, to have a continuity of the time derivative λ˙ v , across each impulse applied on different circular orbits lying at either end of the coasting arc, one can set A = 0, B = (1 − D)/2e, resulting in the following: 1 (1 − D) sin θ 2 = (1 − D) cos θ ) ± D

λvr = λvθ

(3.131)

with 0 ≤ D ≤ 1. Here the positive sign is taken for the impulse applied at θ = 0, and the negative for that at θ = π . For the impulse applied in the direction of flight, the λ˙ v components are given by

λrr

⎧ , ⎨ 1 μ3 (1 + e)2 (2 − e) (θ = 0) 2 p = 1, μ 2 ⎩ 2 p3 (1 − e) (2 + e) (θ = π )

λrθ = 0

(3.132)

Applying Eq. (3.132) to the impulse at the periapsis, we have 1 2

-

μ 1 (1 + e)2 (2 − e) = − 3 2 p



μ r13



D−1 −D + 2

 (3.133)

or √ (2 − e) 1 + e = 1 + D

(3.134)

Similarly, for the impulse at the apoapsis, r = r2 , we have 

r2 p

3/2

√ (1 − e)2 (2 + e) = (2 + e) 1 − e = 1 + D¯

(3.135)

3.5 Optimal Impulsive Transfer

109

2 1.8 1.6 1.4

f(z)

1.2 1 0.8 0.6 0.4 0.2 0 −1 −z1

−0.5

0

0.5

z

1

z2 1.5

2

Fig. 3.7 Roots for satisfying the optimality constraints on the eccentricity of a Hohmann transfer ellipse

where D¯ is the value of D for the larger circle. Since both D and D¯ should have values between zero and unity, we have the following inequalities to be satisfied for optimality: √ 1 ≤ (2 − e) 1 + e ≤ 2 √ 1 ≤ (2 + e) 1 − e ≤ 2

(3.136)

To solve for the appropriate value of the orbital√eccentricity which satisfies these inequalities, consider the plot of f (z) = (2 − z) 1 + z for −1 ≤ z ≤ 2, shown in Fig. 3.7, wherein the range 1 ≤ f (z) ≤ 2 corresponds to −z1 ≤ z ≤ z2 , where z1 = 0.8794 and z2 = 1.3473. Therefore, the optimal value of e satisfying Eq. (3.136) must fall between zero and the smaller of z1 and z2 , that is, 0 ≤ e ≤ 0.8794. Now, since e = (r2 −r1 )/(r1 +r2 ) for the transfer ellipse, it follows that x = r2 /r1 must satisfy 0≤

x−1 ≤ 0.8794 x+1

(3.137)

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3 Orbital Mechanics and Impulsive Transfer

or 1 ≤ x ≤ 1.8794/0.1206 = 15.58. Hence, Hohmann transfer is optimal only for 1 ≤ r2 /r1 ≤ 15.58. For larger values of the radius ratio, r2 /r1 > 15.58, a transfer with more than two impulses would be optimal (instead of Hohmann transfer).

3.5.3 Outer Bi-elliptical Transfer A generalization of Hohmann transfer for the three impulse case is the outer bielliptical transfer depicted in Fig. 3.8. Here the transfer consists of two ellipses, and takes the spacecraft to an intermediate radius, rt > r2 > r1 , with all the three impulses applied tangentially. The third impulse, Δv3 , is applied opposite to the direction of the other two impulses, as shown in Fig. 3.8. The time of flight is determined by rt , which itself is a nonlinear function of r1 and r2 . Thus the sole design parameter for this transfer is the intermediate radius, rt . It has been indicated above that the outer bi-elliptical transfer can be more efficient than Hohmann transfer for r2 /r1 > 15.58. The reader can show this by minimizing the net velocity change, | Δv1 | + | Δv2 | + | Δv3 |, with respect to rt , and equating the result with that of Hohmann transfer. Consider x = r2 /r1 ≥ 1 and y = rt /r1 ≥ 1 as the two non-dimensional parameters of the outer bi-elliptical (OB) transfer. The total velocity change required Fig. 3.8 Geometry of the outer bi-elliptical transfer between two circular orbits

Dv2

rt r2

r1

Dv1

Dv3

3.5 Optimal Impulsive Transfer

111

for the Hohmann and OB transfers for the same value of r1 and r2 is then respectively expressed as follows:  2x 2 1 ΔvH = − −1 + √ + 1+x x(1 + x) x     - 0 μ 2y 2x 2 2y 1 + − + −1 − √ + ΔvB = r1 1+y y(x + y) y(1 + y) x(x + y) x -

μ r1

0

(3.138) The difference between the non-dimensional velocity change of the two transfers is then expressed by the following function: f (x, y) =

ΔvB − ΔvH , μ r1

(3.139)

which vanishes for x = y (that is, when the two transfers become exactly the same). The value of x = y for which the OB transfer becomes more efficient than Hohmann transfer is then obtained from the stationary condition of f (x, y) given by ∂f ∂f = =0 ∂x ∂y 3y + 1 = − (y + 1)3/2 =

-

x x+y

(3.140)

3x + 1 1 −√ (x + 1)3/2 2

or x 3 − 15x 2 − 9x − 1 = 0

(3.141)

whose only positive real root is x = 15.5817. Since the second derivative of f (x, y) is always negative, given by ∂ 2f 1 =− √ <0 ∂x∂y 2 2xy(x + y)3/2

(3.142)

it follows that f (x, y) = 0 is a maximum point for x = y. Therefore, f (x, y) < 0 for x > 15.5817, thereby implying that OB transfer is more efficient than Hohmann transfer for r2 /r1 > 15.5817.

112

3 Orbital Mechanics and Impulsive Transfer

Exercises 3.1 A spacecraft is detected to be moving at a speed 7 km/s relative to Earth, when its altitude is 500 km. What additional minimum velocity must be provided to the spacecraft by an impulsive rocket thrust in order to escape Earth’s gravity? (For Earth: μ = 398,600.4 km3 /s2 , mean surface radius, R0 = 6378.14 km). 3.2 A spacecraft is initially in an Earth orbit of perigee altitude 300 km and apogee altitude 1000 km. It is required to transfer the spacecraft to a new coplanar orbit of a constant altitude of 500 km. Determine the minimum total velocity change required, and the corresponding time of transfer in each of the following cases: (a) Only one impulse is applied. (b) Two impulses are applied. 3.3 Two astronauts called A and B have landed on Charon (μ = 101.25 km3 /s2 , surface radius = 605 km), while a third one called C is on a spacecraft orbiting Charon in a circular orbit of radius 650 km. The astronauts have decided to play a game of cricket. Astronaut B bowls to A, who hits the ball such that it reaches C in a minimum energy orbit when C is 10◦ ahead of the radial position of A. The atmosphere on Charon is negligible. (a) (b) (c) (d) (e)

Determine the magnitude and direction of the velocity provided by A to the ball. Determine the ball’s velocity (magnitude and direction) when it is caught by C. Determine the time of transfer of the ball from A to C. Calculate the position of C relative to A when A hits the ball. Use the hypergeometric function method to compute the transfer orbit, such that the transfer from A to C in the given position should take exactly 5 min. Verify that your answer is correct. (f) After catching the ball, astronaut C waits until he is directly overhead of A. At this time, he throws the ball vertically downwards toward A. But the ball instead of reaching A, returns to C in 5174 s. What is the velocity provided to the ball by C?

3.4 Repeat parts (a) and (b) of Exercise 3.3, if the transfer angle is not specified, but the ball is hit horizontally by A, and reaches C via a minimum energy transfer orbit. 3.5 Determine the minimum energy transfer orbit and the required velocities at both the ends, for the positions, r1 , r2 , given in Example 3.2. Is it a practical transfer trajectory? 3.6 Use the hypergeometric function method for solving Lambert’s problem to compute the transfer orbit for sending a spacecraft around Earth from a position in the Earth centered frame, (I, J, K), r1 = −3000I + 6000K km

3.5 Optimal Impulsive Transfer

113

to the position r2 = 500I − 7000J + 1000K km in exactly 20 min. Give your answer in terms of the classical orbital elements. 3.7 Calculate the velocity impulse vectors for transferring a spacecraft around Earth from a position r1 = −3000I + 6000K km and velocity v1 = 7I − 2J km/s to the position r2 = 7000I − 2000J km and velocity v2 = 8I + J − K km/s in exactly 30 min. 3.8 Estimate the smallest total velocity impulse magnitude required for sending a spacecraft from Earth (orbital radius = 1 astronomical unit (A.U.), orbital period = 365.256 mean solar days) to Mars (orbital period 687 mean solar days). (For Sun, μ = 1.3271244 × 1011 km3 /s2 , 1 mean solar day = 24 h, 1 A.U. = 1.495978 × 108 km). 3.9 Estimate the smallest total velocity impulse magnitude required for sending a spacecraft from Earth to Neptune (orbital radius 30.1 A.U.).

Chapter 4

Two-Body Maneuvers with Unbounded Continuous Inputs

4.1 Introduction A majority of spacecraft navigation problems involve transfers between two given positions and velocities. When continuous inputs are applied for the orbital transfer of a spacecraft around a central body, the perturbed two-body model can be used as a plant to derive the optimal trajectory and control history. In such a model, the central gravity field is regarded to be spherical and the gravitational perturbations caused by other bodies are neglected. The effects of any atmosphere and solar radiation pressure are also neglected, unless the nature of the trajectory demands that they be included. After an optimal trajectory has been determined, an optimal feedback loop performs automatic course corrections at regular intervals to keep the vehicle on the nominal trajectory despite small, unmodeled disturbances. In this chapter, we shall consider how the optimal trajectories can be devised using the perturbed Keplerian equations and unbounded acceleration inputs, and used to perform the task of optimal orbital regulation.

4.2 A Motivating Example Consider the problem of optimally controlling the displacement, y(t), of a block of unit mass sliding in a horizontal, frictionless track by the application of a force, u(t). Since the block faces no natural resistance to its motion for u(t) = 0, this system can be regarded as the simplest possible idealization of a spaceflight problem where neither the atmospheric drag nor any other dissipative natural forces are present. The second-order, linear time-invariant system can be represented by the following state equations: x˙1 = x2 x˙2 = u © Springer Nature Switzerland AG 2019 A. Tewari, Optimal Space Flight Navigation, Control Engineering, https://doi.org/10.1007/978-3-030-03789-5_4

(4.1) 115

116

4 Two-Body Maneuvers with Unbounded Continuous Inputs

where x1 (t) = y(t). The control task is to move the system from an initial state of rest, x1 (0) = x2 (0) = 0, to a final rest position of x1 (tf ) = 1 and x2 (tf ) = 0 at an unspecified time tf , while minimizing the total control effort given by  tf J = u2 (t)dt (4.2) 0

The extremal control history, u∗ (t), and the corresponding trajectory, are readily solved from the Euler–Lagrange necessary conditions (Chap. 2). The Hamiltonian of the problem is given by x1∗ (t), x2∗ (t),

H = L + λT f = u2 + λ1 x2 + λ2 u

(4.3)

which results in the following costate equations: λ˙ ∗ =



λ˙ ∗1 λ˙ ∗2



   T  ∂H ∗ 0 =− = −λ∗1 ∂x

(4.4)

The solution to the costate equations subject to the initial condition, λ∗ (0) = (c1 , c2 )T , is expressed by   ∗  c1 λ1 ∗ = (4.5) λ (t) = λ∗2 c2 − c1 t where the constants c1 , c2 are to be determined from the boundary conditions. The stationary condition of the Hamiltonian with respect to the control history, u(t),  T  ∂H ∗ Hu = = 2u∗ (t) + λ∗2 (t) = 0 (4.6) ∂u yields the following extremal control history: 1 u∗ (t) = −λ∗2 (t)/2 = − (−c2 + c1 t) , 2

(0 ≤ t ≤ tf )

(4.7)

In order to determine the constants c1 , c2 , the state equations are solved, subject to the boundary conditions. The extremal trajectory resulting from the extremal control history (Eq. (4.7)) and zero initial state is expressed as follows: x ∗ (t) =



x1∗ (t) x2∗ (t)



 =

 c1 t 3 /12 − c2 t 2 /4 , c1 t 2 /4 − c2 t/2

(0 ≤ t ≤ tf )

(4.8)

The terminal boundary conditions are now applied to determine the constants c1 , c2 , tf : x2∗ (tf ) = c1 tf2 /4 − c2 tf /2 = 0

(4.9)

4.3 Equations of Motion

117

which yields tf = 2c2 /c1 . x1∗ (tf ) = c1 tf3 /12 − c2 tf2 /4 = 1

(4.10)

into which the substitution of Eq. (4.9) yields tf3 = −24/c1 We must have c1 < 0 (and c2 < 0) for tf > 0. Finally, the terminal boundary condition for extremal control is applied,   ∂φ ∗ H+ = (u∗ (tf ))2 + λ∗1 (tf )x2∗ (tf ) + λ∗2 (tf )u∗ (tf ) = 0 ∂t t=tf

(4.11)

to produce the result u∗ (tf ) = 0, or tf = c2 /c1 . Clearly, there is a contradiction between Eqs. (4.9) and (4.11), both of which cannot be simultaneously satisfied. Hence, there does not exist any extremal trajectory satisfying the boundary condition x2 (tf ) = 0. This fact can be explained physically as follows. Since there is no resistance to the unforced motion, the velocity at the terminal time cannot be made zero by an extremal control, because the optimality condition, Eq. (4.11), demands the control force—which is initially positive (u∗ (0) = −c2 /2)—must vanish at the terminal time. In fact, to make the terminal velocity zero, a negative force must act for some time t < tf , and due to the time linearity of the control history, should become increasingly negative as the time increases. Thus u(tf ) < 0 is necessary for satisfying x2 (tf ) = 0, which conflicts with the necessary condition for optimality. Therefore, the boundary condition x2 (tf ) = 0 causes the optimal control problem with continuous inputs to be ill posed. Such a characteristic of the energyoptimal, continuous control history is valid for spaceflight problems wherein a given unforced terminal state can only be achieved by a suboptimal path. As discussed in Chap. 3, the application of discontinuous (or impulsive) thrust inputs can solve the problem of optimal transfer between orbits (which are the unforced states of the system). Alternatively, a well-posed spaceflight optimal navigation problem with continuous thrust is the one where the terminal velocity constraint is replaced by a suitable terminal cost, φ[x(tf ), tf ], in the objective function to be minimized. This will be further explored in this chapter for controlling spaceflight trajectories with continuous, unbounded inputs.

4.3 Equations of Motion The orbital motion of a spacecraft around a spherical body perturbed by a continuous acceleration input, a(t) ∈ R3×1 , can be represented by the following vector differential equations: v˙ + μ

r =a r3

(4.12)

118

4 Two-Body Maneuvers with Unbounded Continuous Inputs

and r˙ = v

(4.13)

where r(t) ∈ R3×1 (with r(t) = |r(t)|) is the position vector measured from the center of gravity of the two-body system, and μ is the gravitational constant. If the spacecraft’s mass is assumed to be negligible relative to the central body, the origin of the coordinate system shifts to the center of the body, which is then assumed to be at rest. Hence a coordinate frame with origin at the center of the large body can be regarded as an inertial reference frame, and r¨ (t) is the absolute acceleration of the spacecraft. Equations (4.12) and (4.13) can generally be expressed either in spherical or Cartesian coordinates. However, an alternative system of coordinates regards the plane of the unperturbed Keplerian orbit (i.e., when a = 0) as a reference plane. Here the polar coordinates, r(t), θ (t), can be used to resolve the spacecraft’s position vector, r(t), in the unperturbed orbital plane, while the angle, α(t), represents the plane change caused by the out-of-plane component of a(t). Such a representation is useful especially when the magnitude of the perturbing acceleration, a(t) = |a(t)|, is always small compared to the gravitational acceleration (i.e., a << rμ2 ), resulting in a slow departure from the Keplerian motion. In such a case, at any given instant, t0 , one can imagine an osculating orbit, which is the solution to Eq. (4.12) for a(t) = 0, t ≥ t0 . The osculating orbit is thus the imaginary two-body orbit with initial condition, r(t0 ), v(t0 ), and will be the trajectory of the spacecraft if the perturbing acceleration, a(t), suddenly vanishes at t = t0 and remains zero for t > t0 . If the applied perturbing acceleration is small1 at all times, then the osculating orbit is nearly tangential to the actual trajectory of the spacecraft at any given instant, t0 , thereby explaining the adjective “osculating” (or “touching”). Whenever continuous acceleration inputs are applied in practice, their magnitudes are always much smaller than the acceleration due to gravity, except in certain special situations such as lift-off and landing (which may be considered separately). Therefore, the approximation, a << rμ2 , is generally valid in orbital maneuvers involving continuous inputs. The osculating orbit is thus a useful reference trajectory for space navigation problems involving small, continuous inputs because it allows a simplification of the governing equations of motion.

1 An

osculating orbit is defined to be the tangential trajectory that will be followed if the perturbing acceleration were suddenly to be removed. If the perturbation is large and were suddenly to be removed, there would be an almost impulsive change in the acceleration at the instant of removal, which translates into a step change of velocity. Therefore, in that case the resulting orbit would not be tangential to the original (perturbed) orbit. In fact, there will be a “bump” in the trajectory. Hence the concept of osculating orbit is valid only if the applied perturbation is always small, so that at any instant, an abrupt vanishing of the acceleration would not cause an abrupt change in the velocity, thereby approximating the resulting orbit as being tangential (or osculating) to the original (perturbed) orbit.

4.3 Equations of Motion

119

Utilizing the set of nonlinear ordinary differential equations, Eqs. (4.12) and (4.13), expressed in Cartesian or spherical coordinates for optimal guidance application requires numerical integration, which has both computational and implementation difficulties. Instead, an approximate model based upon either linearization about the osculating orbit or a variation of Keplerian orbital parameters is employed in practical applications. The linearized model is useful when the reference orbit is fixed, such as in orbital regulation of and maneuvers around a given circular orbit. On the other hand, the variational model is more useful when apart from the thrust acceleration, other perturbations are also present, such as due to nonspherical gravity field, atmospheric drag, or solar radiation pressure. The classical variational model is via the Lagrange’s (or alternatively, Gauss’) perturbation equations, wherein the variation of the classical orbital elements of the osculating is used to represent the perturbed motion. A more general perturbation model utilizes the universal or canonical variables instead of the classical elements, in order to avoid the singularity of the latter in certain conditions. In any case, a set of six state variables is necessary to represent the position and velocity, r(t), r˙ (t), of the spacecraft in an inertial reference frame. When the osculating orbital model is used as a reference, then only four state variables are required to represent the motion of the spacecraft relative to the instantaneous osculating plane at t = t0 . In that case, the deviation of the spacecraft’s trajectory from the osculating orbit is represented by the respective changes in the orbital radius, r, the radial velocity, r˙ , the orbital angular momentum, h = r 2 θ˙ , and the instantaneous plane change angle, α. The remaining two variables required to complete the description are the two angles used to resolve the orbital angular momentum vector, h(t0 ) = r(t0 ) × v(t0 ), of the osculating orbit in the inertial frame. The osculating orbit model is depicted in Fig. 4.1. The vector differential equation, Eq. (4.12), results in the following scalar constituents when the deviation from a reference osculating orbit is considered: r¨ −

Fig. 4.1 The osculating orbit used as a reference for the actual trajectory of a spacecraft under the influence of small perturbing acceleration inputs, a

μ h2 + 2 = ar r3 r

(4.14)

v(t)

h(t) = r(t)´v(t)

Actual orbit

v(t0)

r(t)

a

a

h(t0) = r(t0)´v(t0) r(t0) Osculating orbit

120

4 Two-Body Maneuvers with Unbounded Continuous Inputs

h˙ = raθ

(4.15)

r an h

(4.16)

and α˙ =

where the input acceleration vector is resolved into the radial, circumferential, and normal components, ar , aθ , and an , respectively: a = (ar , aθ , an )T

(4.17)

4.4 Optimal Low-Thrust Orbital Transfer The slow departure from a Keplerian orbit when a small input acceleration is applied by a low-thrust ion/plasma engine is readily modeled by Eqs. (4.14)–(4.16). Let the state vector be the following: x = (r, r˙ , h, α)T

(4.18)

which results in the following state equation: x˙ = f(x, a) where

⎛

r˙

(4.19) ⎞

⎜ ⎟ ⎜ h2 ⎟ ⎜ r 3 − rμ2 + ar ⎟ ⎟ f=⎜ ⎜ ⎟ ⎜ ⎟ raθ ⎝ ⎠ r h an

(4.20)

The simplest optimal control problem relates to the low-thrust orbital transfer from the initial state x(ti ) = xi

(4.21)

x(tf ) = xf

(4.22)

L = aT Ra

(4.23)

to a final state

such that the net control effort,

4.4 Optimal Low-Thrust Orbital Transfer

121

is minimized for the control interval, t ∈ [ti , tf ], with R ∈ R3×3 being a symmetric cost coefficient matrix. The performance index to be minimized is thus the following: 

tf

J =

a(t)T R(t)a(t)dt

(4.24)

ti

and the Hamiltonian for the problem is given by H = L + λT f = aT Ra + λ1 r˙ + λ2 (

r h2 μ − 2 + ar ) + λ3 raθ + λ4 an 3 h r r

(4.25)

where λ = (λ1 , λ2 , λ3 , λ4 )T is the costate vector. The transverse condition for optimality, ∗ T  ˙λ∗ = − ∂H ∂x

(4.26)

yields the following costate equations: λ˙ ∗1 =



3(h∗ )2 2μ − ∗ 3 (r ∗ )4 (r )

λ˙ ∗3 = −



λ∗2 − λ∗3 aθ∗ − λ∗4

an∗ h∗

(4.27)

λ˙ ∗2 = −λ∗1

(4.28)

2h∗ ∗ r ∗a∗ λ2 + ∗ n2 λ∗4 ∗ 3 (r ) (h )

(4.29)

λ˙ ∗4 = 0

(4.30)

Finally, the stationary condition, Ha = 0 on the extremal trajectory, gives the following equation to be solved for the extremal control, a∗ : ⎛

⎞ λ∗2 2Ra∗ + ⎝ λ∗3 r ∗ ⎠ = 0 λ∗4 r ∗ / h∗

(4.31)

where the cost coefficient matrix R must be positive definite in order to satisfy the sufficient condition for optimality on the extremal trajectory: Haa = 2R > 0 Thus by choosing a positive definite R, we have aˆ (t) = a∗ (t), t ∈ [ti , tf ].

(4.32)

122

4 Two-Body Maneuvers with Unbounded Continuous Inputs

4.4.1 Coplanar Orbital Transfer The maneuvers which do not require a change of plane are termed coplanar transfers, and must have an∗ (t) = 0, t ∈ [ti , tf ], which implies λ∗4 (t) = 0 throughout the control interval. The fourth costate equation, Eq. (4.30), as well as the fourth state variable, α, is now redundant in solving the optimal control problem. Thus we have for the special case of coplanar maneuvers, x = (r, r˙ , h)T , a = (ar , aθ )T , λ = (λ1 , λ2 , λ3 )T , and ⎛ ⎞ r˙ ⎜ 2 ⎟ f = ⎝ h3 − μ2 + ar ⎠ (4.33) r r raθ which result in the following costate equations:   3(h∗ )2 2μ λ˙ ∗1 = λ∗2 − λ∗3 aθ∗ − (r ∗ )4 (r ∗ )3 λ˙ ∗2 = −λ∗1 λ˙ ∗3 = −

2h∗ ∗ λ (r ∗ )3 2

(4.34) (4.35) (4.36)

It is to be noted that the extremal control can be easily derived in the coplanar case by choosing   R1 0 R= (4.37) 0 R2 with R1 > 0, R2 > 0, resulting in the following: aˆ r = ar∗ = −

λ∗2 2R1

(4.38)

aˆ θ = aθ∗ = −

λ∗3 r ∗ 2R2

(4.39)

and

4.4.2 Plane Change Maneuver When changing the orbital plane without affecting the shape of the orbit, the normal acceleration, an , is the only control input, whereas the angular deviation, α, from the osculating orbit is the sole state variable. The formulation is greatly simplified owing to the fact that neither r nor r˙ are changed by the control, and evolve naturally by

4.4 Optimal Low-Thrust Orbital Transfer

123

Keplerian motion. The orbital angular momentum magnitude, h, remains constant since aθ = 0. Thus we have the following scalar state equation: α˙ =

r an h

(4.40)

which results in the following Hamiltonian: H = Ran2 + λ

ran h

(4.41)

The costate equation is now the following: λ˙ ∗ = 0

(4.42)

which implies that λ∗ is constant. The stationary condition produces the following optimal control input: aˆ n = an∗ = −

λ∗ r ∗ 2h∗ R

(4.43)

where R > 0. A practical choice for R is the following: R=

r(tf )∗ 2h∗

(4.44)

where r ∗ (tf ) is known from the Kepler’s equation for a specific value of terminal time, tf . Thus we have aˆ n = −

λ∗ r ∗ (t) r(tf )∗

(4.45)

The value of λ∗ can be chosen based upon the maximum available acceleration magnitude.

4.4.3 General Orbital Transfer A more general optimal transfer is for the minimization of  tf J = φ[x(tf ), tf ] + [ L[x(t), a(t)]dt

(4.46)

ti

such that the terminal state, x(tf ), lies on a hypersurface defined by F[x(tf ), tf ] = 0

(4.47)

This results in the change in the terminal boundary conditions to the following transversality condition on the costate vector (Chap. 2):

124

4 Two-Body Maneuvers with Unbounded Continuous Inputs

λ∗ (tf ) =



∂Φ ∂x

T (4.48) t=tf

where Φ = φ + cT F

(4.49)

with c = (c1 , c2 , . . . , ck )T being arbitrary constants. The terminal time, tf , must satisfy   ∂H Φ+ =0 (4.50) ∂t t=tf Since neither the costate vector, λ, nor the coefficients, c, have any physical significance, these boundary conditions can be satisfied in a variety of ways, which can cause difficulty in their implementation. The transversality conditions are common when a transfer is to be made either to a specific final orbit or to a specific relative position in a constellation of spacecraft (formation flying).

4.5 Variational Model The problem of Keplerian motion perturbed by a continuous, small acceleration, u, caused by the thrust inputs and external disturbances, can be modeled as a continuous, slow deviation of the trajectory from a two-body orbit. Such a variational model is developed by approximating the orbital elements by first-order time variations, while the others are averaged over an osculating orbit. The Gauss’ form of Lagrange’s planetary model subject to a perturbing acceleration, u = (ar , aθ , an )T , is expressed as the following set of variational equations [8] in terms of the classical orbital elements, x = (a, e, i, Ω, ω, θ )T : 2a 2 v da = aθ dt μ

(4.51)

de 1 r = [2(e + cos θ )aθ + ar sin θ ] dt v a

(4.52)

r di = cos(θ + ω)an dt h

(4.53)

r dΩ = sin(θ + ω)an dt h sin i

(4.54)

  1  r dω a 2aθ sin θ − 2e + cos θ ar − = an sin(θ + ω) cos i dt ev r h sin i

(4.55)

4.5 Variational Model

125

  dθ 1  a h 2aθ sin θ − 2e + cos θ ar = 2− dt ev r r

(4.56)

where θ is the true anomaly obtained from solving the Kepler’s equation for the osculating orbit from the time of periapsis, τ , to the present time, t, and (r, v, h) are the instantaneous radius, speed, and angular momentum of the osculating orbit. Equations (4.51)–(4.56) can be alternatively expressed in terms of the variation with respect to the eccentric . anomaly, E, instead of time, by dividing them by dE/dt = na/r, where n = μ/a 3 is the orbital mean motion. An averaging procedure is utilized to determine each element’s mean rate of change by calculating its variation over a single revolution, and then dividing by the orbital period, T = 2π/n. The averaged state vector is thus computed as follows: x¯ =

1 T



E2 E1

 r  dx dE na dt

(4.57)

where dx/dt is calculated from Eqs. (4.51)–(4.56) with the terms on the righthand side fixed at their mean values over a complete revolution. The integration limits of the averaging integral, E1 , E2 = E1 + 2π , are traditionally selected based upon astronomical observation issues, such as the points of entering and exiting the Earth’s shadow. Most of the low-thrust transfer problems are posed as time-optimal problems, because with a nearly constant mass exhaust rate of rocket engines, the minimum time also results in the minimum fuel consumption. Furthermore, a practical transfer between two orbits requires a terminal condition, Eq. (4.47), in which the final shape (a, e) and orbital inclination, i, are specified. For a time-optimal transfer involving only the changes in (a, e, i), the Hamiltonian of the variational model is given by H = 1 + λT x˙ = 1 + λa

da de di + λe + λi dt dt dt

(4.58)

Since the Hamiltonian is linear in control variables, (ar , aθ , an ), the time-optimal problem is singular in terms of the acceleration input, u. Its solution requires either imposing constraints on the magnitude of input acceleration, u, and solving the singular control problem using the method of Chap. 5 or by introducing new control variables which make the Hamiltonian non-singular. In order to do the latter, we resolve the thrust acceleration in spherical coordinates using the in-plane pitch angle, α, and out-of-plane yaw angle, β, as follows: u = u(sin α cos βir + cos α cos βiθ + sin βih )

(4.59)

where ir = r/r and ih = h/ h = ir × iθ (withh = r × v). This gives ar = a sin α cos β, aθ = a cos α cos β, and an = a sin β. Substitution of these components into Eqs. (4.51)–(4.56), and treating the thrust angles, (α, β), as the new control variables, produces the following stationary conditions:

126

4 Two-Body Maneuvers with Unbounded Continuous Inputs

∂H 2a 2 v = −uλ∗a sin α ∗ cos β ∗ ∂α μ u r + λ∗e [−2(e + cos θ ) sin α ∗ + sin θ cos α ∗ ] cos β ∗ = 0 v a

(4.60)

∂H 2a 2 v u r = −uλ∗a cos α ∗ sin β ∗ −λ∗e [2(e + cos θ ) cos α ∗ + sin θ sin α ∗ ] sin β ∗ ∂β μ v a r + uλ∗i cos(θ +ω) cos β ∗ = 0 (4.61) h The quadrant ambiguity in the optimal pitch and yaw angles is avoided by ensuring Hαα > 0 and Hββ > 0, which yields the following optimal angles: −λe ar sin θ sin αˆ = -  2 2 2 2 4 λ∗a a μv + λ∗e (e + cos θ ) + (λ∗e )2 ar 2 sin2 θ  2 2 −2 λ∗a a μv + λ∗e (e + cos θ )

cos αˆ = -  2 2 4 λ∗a a μv + λe (e + cos θ )

sin βˆ = − cos βˆ = −

2

(4.62)

2 + (λ∗e )2 ar 2

2

sin θ

λ∗i rv h cos(θ + ω) Δ

$ 2 2 2λ∗a a μv cos α ∗ + λ∗e 2(e + cos θ ) cos α ∗ +

r a

sin θ sin α ∗

Δ

% (4.63)

where  Δ = (λ∗i )2

2   r 2 v2 a2 v2 r 2 (θ + ω) + 2λ∗ ∗ + λ∗ 2(e + cos θ) cos α ∗ + ∗ cos cos α sin θ sin α a e μ a h2

The costate equations to be solved for the optimal steering laws of Eqs. (4.62) and (4.63) are the following: λ˙ ∗a = −

av r ∂H = −4uλ∗a cos α ∗ cos β ∗ + λ∗e 2 sin θ sin α ∗ cos β ∗ ∂a μ va λ˙ ∗e = −

(4.64)

2 ∂H = −λ∗e cos α ∗ cos β ∗ ∂e v

(4.65)

∂H =0 ∂i

(4.66)

λ˙ ∗i = −

4.6 Optimal Regulation of Circular Orbits

127

the last of which yields λ∗i (t) = λi (tf ) = const. The other two costate variables must be determined from the transversality conditions given by Eqs. (4.48) and (4.50). The resulting 2PBVP is to be solved numerically. Edelbaum [28] derived the low-thrust optimal profiles for transfers between circular orbits of different inclinations with Gauss perturbation equations, using ar = 0, and a constant thrust acceleration magnitude, u. The optimal paths √ with a low-thrust acceleration are nearly circular, i.e., e 0, r a, and v = μ/r. Putting these conditions into Eqs. (4.62) and (4.63), along with Edelbaum’s arbitrary selection of λa = 1/a, reduces them to the following time-optimal steering laws for a nearly circular transfer between inclined circular orbits [28]: tan α ∗ =

λ∗e sin θ 2(1 + λ∗e cos θ )

(4.67)

tan β ∗ =

λ∗i cos(θ + ω) 2

(4.68)

Chobotov [23] presents a Mayer formulation (see Chap. 2) using the equinoctial elements, which avoid the singularities of the classical elements for e = 0, i = 0, π/2 (but are singular for i = π instead).

4.6 Optimal Regulation of Circular Orbits Most of the maneuvering tasks in the life of a satellite involve the regulation of radial, in-track, and out-of-the plane position around a desired circular orbit. Here we consider the optimal station keeping maneuvers where the initial errors in position and velocity relative to a point in the circular orbit must be driven to zero with the minimum expenditure of the propellant. Let the constant radius and frequency of the desired orbit be r0 and n, respectively. The state variables of this problem can be chosen to be the deviation from the radius, δr = r − r0 , the radial velocity, r˙ , the in-track angular deviation, δθ , and out-of-plane angular displacement, α (both measured from a given point in the orbit), and the orbital angular momentum, h. The state equations can then be expressed as follows: δ r˙ = r˙ r¨ = δ θ˙ =

h2 μ − + ar 3 (r0 + δr) (r0 + δr)2 h (r0 + δr)2

h˙ = (r0 + δr)aθ α˙ =

(r0 + δr) an h

(4.69)

128

4 Two-Body Maneuvers with Unbounded Continuous Inputs

These nonlinear state equations couple the coplanar and out-of-the plane motion, whose solution requires numerical solution. However, a decoupling of the two types of maneuvers is possible by applying either coplanar or normal acceleration inputs, which simplifies their control. Furthermore, nearly all practical maneuvers with low thrust require only small displacements from a reference point in the given circular orbit, which allows a linearization of the governing equations.

4.6.1 Coplanar Regulation with Radial Thrust Regulating the orbital position is possible by radial acceleration input, ar , alone, which allows the angular momentum, h, to remain unchanged and simplifies the state equations, derived from Eq. (4.69) to be the following: δ r˙ = r˙ r¨ = δ θ˙ =

h2 μ − + ar 3 (r0 + δr) (r0 + δr)2

(4.70)

h (r0 + δr)2

√ where the angular momentum of the desired orbit, h = μr0 , is constant. Since the positional errors are usually small in comparison with the orbital radius, it is reasonable to linearize the state equations around a reference point in the circular orbit through the binomial approximations: −k

(r0 + δr)

r0−k

  kδr 1− , r0

(k = 2, 3)

(4.71)

Thus the following linearized state equations are derived by neglecting the second and higher-order terms involving δr and r0 δθ : δ r˙ = r˙

1 3h2 2μ − 3 δr + ar r¨ = − r04 r0

δ θ˙ = −

0

(4.72)

2h r03

By re-defining the state vector to be the following: x = (δr, r˙ , r0 δθ )T

(4.73)

4.6 Optimal Regulation of Circular Orbits

and noting that h/r02 =

,

129

μ/r03 = n and 3h2 2μ μ − 3 = 3 = n2 4 r0 r0 r0

(4.74)

the linearized state equations are expressed as follows: x˙ = Ax + Bar

(4.75)

where ⎛

⎞ 0 1 0 A = ⎝ −n2 0 0 ⎠ , −2n 0 0

⎛ ⎞ 0 B = ⎝1⎠

(4.76)

0

Example 4.1 An orbital regulation system is to be designed for a spacecraft in a circular orbit of radius r0 and frequency n around a planet, for maintaining the circular shape of the orbit with a continuous, radial acceleration control input, ar (t), in the presence of an initial perturbation, δ r˙ (0). The initial radial and in-track deviations, δr(0), δθ (0), are zeros. Use the linearized state equations to determine the extremal trajectory, δr ∗ (t), and the extremal control history for minimizing the following performance index:  1 1 tf 2 J = [δ r˙ (tf )]2 + a (t)dt . 2 2 0 r where tf is specified to be the orbital period. The final deviations, δr(tf ), δ r˙ (tf ), and δθ (tf ), are unspecified. The Hamiltonian of this problem is given by H = L + λT (Ax + Bar )

(4.77)

where L = 12 ar2 , x and (A, B) are given by Eqs. (4.73) and (4.76), respectively, and λ = (λr , λr˙ , λθ )T is the costate vector. The terminal cost function is φ = 1 2 2 [δ r˙ (tf )] . The costate equations are thus the following:   ∂H ∗T ∗ λ˙ = − = −AT λ∗ ∂x

(4.78)

Since the costate equations are linear and homogeneous, their solution to the initial condition, λ(0) = (c1 , c2 , c3 )T , can be expressed as follows: ⎛

λ∗ (t) = e

−AT t

cos nt n sin nt λ(0) = ⎝ − n1 sin nt cos nt 0 0

⎞⎧ ⎫ 2 sin nt ⎨ c1 ⎬ 2 ⎠ c2 (cos nt − 1) n ⎩ ⎭ c3 1

(4.79)

130

4 Two-Body Maneuvers with Unbounded Continuous Inputs

The transverse boundary condition for this case is the following: λ∗ (tf ) =



∂φ ∂x

∗T

⎧ ⎨

⎫ 0 ⎬ = δ r˙ (tf ) ⎩ ⎭ 0

(4.80)

Since we have tf = 2π/n, this condition results in λ∗r (tf ) = c1 = 0 and λ∗θ (tf ) = c3 = 0, therefore, the solution for the costate vector on the extremal trajectory is given by ⎧ ⎫ ⎨ n sin nt ⎬ λ∗ (t) = c2 cos nt ⎩ ⎭ 0

(4.81)

The stationary condition, Har = 0, produces ar∗ (t) = −λ∗r˙ (t) = −c2 cos nt, and the state solution is derived to be the following: x∗ (t) = eAt x(0) +

=

⎧ ⎪ ⎨ ⎪ ⎩



t 0

1 n

eA(t−τ ) Bar∗ (τ )dτ

$ % δ r˙ (0) −c2 2t sin nt

δ r˙ (0) cos nt −

c2 2

sin nt n

+ t cos nt

0

⎫ ⎬ ⎪ (4.82)

⎪ ⎭

Finally, the application of the terminal boundary condition gives λ∗r˙ (tf )

c2 = c2 = δ r˙ (tf ) = δ r˙ (0) cos ntf − 2 π = δ r˙ (0) − c2 n



sin ntf + tf cos ntf n



(4.83)

or c2 = δ r˙ (0)/(1 + πn ). Hence, the extremal control history is ar∗ (t) = −

δ r˙ (0) cos nt , 1 + πn

and the extremal trajectory is the following:   t δ r˙ (0) δr ∗ (t) = sin nt , 1+ n 1 + πn

(0 ≤ t ≤ 2π/n)

(0 ≤ t ≤ 2π/n)

The extremal control, ar∗ (t), is also optimal because Har ar = 1 > 0.

(4.84)

(4.85)

4.6 Optimal Regulation of Circular Orbits

131

4.6.2 Coplanar Regulation with Tangential Thrust When only a tangential thrust acceleration, aθ , is applied to the spacecraft the radial acceleration is applied only indirectly through a continuous change of the orbital angular momentum, h. The state equations are then expressed as follows: δ r˙ = r˙ r¨ = δ θ˙ =

h2 μ − (r0 + δr)3 (r0 + δr)2 h (r0 + δr)2

(4.86)

h˙ = (r0 + δr)aθ When a linearization is carried out based upon the assumption of small acceleration input and small displacements, the angular momentum can be expressed as follows: h = r 2 θ˙ = (r0 + δr)2 (n + δ θ˙ ) r02 n + 2nr0 δr + r02 δ θ˙

(4.87)

whose substitution into Eq. (4.86) results in the following approximate model: δ r˙ = r˙ r¨ = −n2 δr

(4.88)

r0 δ θ¨ = −2nr0 δ r˙ + r0 aθ By re-defining the state vector to be the following: x = (δr, r˙ , r0 δθ, r0 δ θ˙ )T

(4.89)

we express the linearized state equations as follows: x˙ = Ax + Bu

(4.90)

where ⎛

0 1 ⎜ −n2 0 A=⎜ ⎝ 0 0 0 −2nr0

⎞ 0 0 0 0⎟ ⎟ , 0 1⎠ 0

⎞ 0 ⎜0⎟ ⎟ B=⎜ ⎝0⎠ r0 ⎛

(4.91)

132

4 Two-Body Maneuvers with Unbounded Continuous Inputs

4.7 General Orbital Tracking Many space missions require a spacecraft to track a predetermined (nominal) trajectory in the presence of disturbing forces. This could either be a Keplerian orbit in an inertial frame (Chap. 3) or a multi-body trajectory in a rotating frame (Chap. 6). With the use of the appropriate motion coordinates, it is possible to linearize the governing equations close to the given trajectory, and then apply control inputs to track the latter in the presence of nonlinear perturbations. Here we shall consider a Keplerian nominal orbit for tracking, which can be extended to a more general case, such as a halo orbit around the collinear Lagrangian point of a three-body problem (Chap. 6). The equation governing perturbed two-body motion can be expressed as follows: r˙ = v μ v˙ + 3 r = u + ad r

(4.92)

where ad (t) represents a disturbing acceleration, and u(t) is the required corrective control input. Unless modeled by appropriate physical processes, ad (t) is treated as a stochastic disturbance called the process noise, ν(t). Furthermore, the measurement of the position and velocity vectors is subject to a random disturbance called the measurement noise, w(t). The nominal orbit being described by Keplerian motion is modeled by r˙ n = vn μ v˙ n + 3 rn = 0 rn

(4.93)

This results in a first-order Taylor series expansion of the position and velocity vectors about their nominal values as follows: r(t) rn (t) + δr(t) v(t) vn (t) + δv(t)

(4.94)

which is a valid approximation as long as | ad |

μ ; r2

| u |

μ r2

Writing the spherical gravitational acceleration as follows: μ μ g = − 3 r = − 3 rn + r rn



∂g ∂r

 δr = − r=rn

μ rn + δg rn 3

(4.95)

or, δg = Gn δr

(4.96)

4.7 General Orbital Tracking

133

where Gn is the gravity-gradient matrix evaluated on the nominal trajectory:   ∂g Gn = ∂r r=rn Similarly, the perturbing acceleration can be linearized about the nominal trajectory as follows: ad = Pn δr + Qn δv where

 Pn =

and

 Qn =

∂ad ∂r

∂ad ∂v

(4.97)

 r=rn

 v=vn

The design of a tracking system is thus based upon the following linearized plant derived above with the small perturbation approximation: r˙ = δv v˙ = (Gn + Pn ) δr + Qn δv + u

(4.98)

which can be expressed in the following state-space form: x˙ (t) = A(t)x(t) + B(t)u(t)

x(0) = x0

(4.99)

where  x=  A=

0 (Gn + Pn )

δr δv



   0 I ; B= Qn I

If the disturbance coefficient matrices Pn , Qn cannot be determined by a physical model, one has to include their effects by the stochastic process and measurement noise vectors, ν(t) and w(t), respectively. In such a case, the state vector is not determined directly, but instead estimated from the measurement of (noisy) output variables using an optimal observer (Kalman filter) [79]. Hence, the stochastic tracking plant is expressed by x˙ (t) = A(t)x(t) + B(t)u(t) + F(t)ν(t)

x(0) = x0

(4.100)

134

4 Two-Body Maneuvers with Unbounded Continuous Inputs

where  A=

   I 0 ; B= 0 I

0 Gn

along with the output equation, y(t) = C(t)x(t) + D(t)u(t) + w(t)

(4.101)

We note that the state transition matrix, (t, t0 ), corresponding to the homogeneous part of the state equation, Eq. (4.100), is a symplectic matrix (see Chap. 2), because the gravity-gradient matrix, Gn (t), is always symmetric. Thus the state transition matrix of the nominal plant [ν(t) = 0, w(t) = 0] is expressed as follows:  (t, t0 ) =

˜ R(t) R(t) ˜ V(t) V(t)

 (4.102)

with the property 0 

−1

(t, t0 ) = (t0 , t) =

VT (t) −RT (t) ˜ T (t) R ˜ T (t) −V

1 (4.103)

˜ V, V, ˜ must obey Eq. (4.103), which implies that Here the partition matrices, R, R, the following matrices are symmetric: ˜ T , VV ˜T, R ˜ T V, ˜ RT V RR ˜ V−1 V, ˜ V ˜R ˜ −1 VR−1 , R−1 R, ˜ −1 R, V ˜ −1 V, R ˜V ˜ −1 RV−1 , R Furthermore, the following identity must be satisfied: T

˜ − VR ˜ T =I VR

(4.104)

and the partition matrices should also satisfy the governing, homogeneous equations, and the initial conditions, that is, ˙˜ = V ˜ R ˙˜ = G R V n˜ ˜ 0) = I R(t ˜ 0) = 0 V(t

(4.105)

4.7 General Orbital Tracking

135

and ˙ =V R ˙ = Gn R V R(t0 ) = 0

(4.106)

V(t0 ) = I Battin [8] presents an algorithm for computing the state transition matrix based upon Lagrange’s coefficients and universal functions which require series (or continued fraction) evaluations. For practical guidance problems, it appears much simpler to solve the nonlinear 2PBVP by a Lambert algorithm, rather than determine a linearized solution in this manner. The general planar orbit tracking requires the solution of a linear, time varying regulator problem. Since the nominal trajectory is unenforced (that is, un = 0), there is no need for a feedforward control [81]. Linear optimal control (Chap. 2) is best suited to this problem. Such a controller can be later modified to include an observer to account for stochastic disturbances, ν(t), w(t), by a robust, multivariable control strategy, such as either the linear, quadratic Gaussian (LQG) or H∞ method [79]. Since the disturbance acceleration is unknown, it is not possible to make the trajectory converge to the nominal at any given time. Instead, the best that can be achieved is to drive the state vector, x(t), asymptotically to a zero steady state (t → ∞) by an infinite-time, linear, quadratic regulator (LQR) with the following cost function:  1 ∞ T J∞ (t) = x (τ )Q(τ )x(τ ) + uT (τ )R(τ )u(τ ) dτ (4.107) 2 t subject to the state constraint equation x˙ (t) = A(t)x(t) + B(t)u(t)

x(0) = x0

(4.108)

and the linear, full-state feedback control law (see Chap. 2): u(t) = −K∞ (t)x(t)

(4.109)

Since the plant is slowly time varying, the stabilizing steady-state solution for the feedback gain matrix at each time instant, K∞ (t), is given by K∞ (t) = R−1 (t)BT (t)P(t)

(4.110)

where P∞ (t) is the symmetric, positive semi-definite, quasi-steady solution of the following algebraic Riccati equation: P∞ (t)A(t) + AT (t)P∞ (t) − P∞ (t)B(t)R−1 (t)BT (t)P∞ (t) + Q(t) = 0

(4.111)

136

4 Two-Body Maneuvers with Unbounded Continuous Inputs

The quasi-steady approximation is tantamount to finding a steady-state solution to Eq. (4.111) at each time instant, provided the variation of A, B, Q, R with time is much slower than the settling time of all the transients of the closed-loop dynamics matrix, A − BK. For an observer-based application, a suitable Kalman filter can be designed for a practical LQG implementation, with loop transfer recovery (LTR) at either the plant’s input or its output [79]. Example 4.2 Consider, for example, a planar tracking problem, in which the perturbations always lie in the plane of the orbit. The orbital tracking plant can be expressed in a state-space form as follows: x˙ (t) = A(t)x(t) + B(t)u(t)

x(0) = x0

(4.112)

where ⎧ ⎫ ⎨ δr(t) ⎬ x(t) = δ r˙ (t) ⎩ ⎭ δh(t)  u(t) = ⎛ ⎜ A(t) = ⎝ −



ar aθ

0 3h2n rn4 (t)

−

2μ rn3 (t)





0

1

0

0

2hn rn3 (t)

0

0

⎞ ⎟ ⎠

and ⎛

0 B(t) = ⎝ 1 0

⎞ 0 0 ⎠ rn (t)

The stability of the linear, planar orbital plant is analyzed by finding the eigenvalues of A as follows:   2μ 3h2n 2 − |λI − A| = λ λ + 4 rn (t) rn3 (t)

(4.113)

Clearly, for stability we must have 3h2n 2μ > 3 4 rn (t) rn (t)

(4.114)

4.7 General Orbital Tracking

137

or, h2n >

2 μrn (t) 3

Since there will be points on any given orbit where the stability condition will be unsatisfied, we need a feedback controller for stability augmentation at such points. Even for those points where the stability condition is satisfied, the plant is not asymptotically stable, thereby requiring a feedback controller to provide active damping for bringing the orbital deviations to zero in the steady state. Since the plant is unconditionally controllable, i.e., the controllability test matrix [79], ⎡

0

0

1

⎢ ⎢1 0 0 ⎣ 0 rn (t) 0

2hn rn3 (t)



0 0

0 2μ rn3 (t)

−

3h2n rn4 (t)

 

⎤

0 4μhn rn6 (t)

0

−

6h3n rn7 (t)

⎥ ⎥ ⎦

0

is of rank three (the order of plant) at all times, we can satisfy the sufficient conditions for the existence of a unique, positive semi-definite solution to the algebraic Riccati equation by selecting suitable state and control cost coefficient matrices, Q(t), R(t). A possible choice is the following constant pair: ⎛

Q1 Q=⎝ 0 0

0 Q2 0

⎞ 0 0 ⎠ ; Q3

 R=

R1 0

0 R2



where Qi ≥ 0, Ri > 0 are selected based upon the relative importance of the state and control variables. The algebraic Riccati equation, Eq. (4.111), must be solved in real time with a changing set of state-space coefficient matrices. Because such a controller is rarely implementable, a constant feedback gain matrix based on a specific time instant is selected, and applied throughout the control interval. Another approach would be to use the averaged orbital parameters of a perturbation model over a complete orbit. A linear regulator based upon either a time-frozen or time-averaged plant ¯ A, ¯ can be designed to be stabilizing because of the slowly varying dynamics, A, plant dynamics. In a real application, a control interval of a few hundred seconds is quite small compared to the orbital period of a satellite. Therefore, we modify the control law by taking a constant gain matrix, ¯ −1 B¯ T P¯ ∞ ¯∞ =R K

(4.115)

where P¯ ∞ is the constant solution to ¯ =0 ¯ +A ¯ T P¯ ∞ − P¯ ∞ B¯ R ¯ −1 B¯ T P¯ ∞ + Q P¯ ∞ A

(4.116)

138

4 Two-Body Maneuvers with Unbounded Continuous Inputs

Referring to the orbital plant dynamics, the constant gain approximation consists of neglecting the variation in the nominal orbital radius with time, rn (t) ≈ rn (0).

4.8 Basic Guidance with Continuous Inputs Either positioning the spacecraft relative to a given point in an orbit or changing its position and velocity to achieve a new orbit, both require an ability to move the vehicle relative to a desired terminal state. Hence the term guidance refers to the process of changing the flight path in order to meet desired terminal conditions. Guiding a space vehicle involves modifying the translational motion of its center of mass (i.e., orbital mechanics) by the application of acceleration inputs. These inputs can be applied either impulsively (as seen in Chap. 3) or continuously. In a spacecraft—as in any other vehicle—there are two broad ways of applying continuous acceleration inputs for guidance purposes: (a) varying the flight direction via acceleration inputs applied normal to the flight path, and (b) changing the kinetic energy via tangential inputs. The adjectives “tangential” and “normal” refer to the instantaneous flight direction, which could be changing continuously. Method (a) is referred to as maneuvering (or steering) whereas method (b) is called energy control. Each of the two methods require a guidance logic whose derivation can be carried out by optimal control theory. While it is possible to devise multi-variable control laws for performing both maneuvering and energy control simultaneously, it is much more intuitive to develop guidance techniques for them separately and analytically. Being much simpler than a numerical 2PBVP solution, basic guidance methods have been practically used in even the most challenging missions, such as guiding the landers to the surface of the various planets and the moons, beginning with the Apollo program, as well as in making orbital rendezvous between two (or more) spacecraft. Guidance for changing the flight direction is referred to as the maneuvering guidance. Here the application of normal acceleration control inputs causes a variation in the flight path, which in turn results in a new orbit. The most common example of maneuvering guidance is for achieving a desired final position from a given current position and velocity, such as in a terminal interception problem. Since the terminal velocity is an unspecified target, the flight energy is not a part of maneuvering, and can be controlled separately as discussed in the previous section. The dependence of the optimal control input on the ever changing current position makes the maneuvering guidance a time-dependent (or non-autonomous) system. Maneuvering guidance is aimed at intercepting a specific target, which can either be another spacecraft or a specified set of position and velocity coordinates defining a moving point in space. Such an interception requires a continuous reduction in the distance between the target and the maneuvering spacecraft. The acceleration control input is applied normal to the relative velocity vector such that the spacecraft closes in on the target at any given instant, often maintaining a constant relative speed. Let us begin the treatment of guidance laws with a simple

4.9 Line-of-Sight Guidance

139

motivating method. Although the line-of-sight navigation was originally developed for air-to-air missiles, it can be applied to spacecraft maneuvers around an intended target.

4.9 Line-of-Sight Guidance Consider a target moving with a position rt (t) and velocity vt (t). A maneuvering spacecraft is required to intercept the target by applying an acceleration input u(t). Both the target and the interceptor are assumed to be sufficiently close to disregard the differences in the acceleration due to gravity between them. This is possible when the two points are roughly in the same orbit but separated by a small distance, and have a small relative velocity. The instantaneous position, rm (t), and velocity, vm (t), of the spacecraft are thus changed such that the relative position, r(t) = rm (t) − rt (t), decreases in magnitude r(t) =| r(t) | at every instant. Without knowing the instantaneous velocity and acceleration of the target, the simplest way in which an interception can be ensured is by applying the control u(t) normal to the relative velocity vector, v(t) = vm (t) − vt (t), such that component of v(t) in the direction of r(t) is always negative. Since r(t) is along the line of sight from the maneuvering spacecraft to the target, such a scheme is referred to as lineof-sight interception. This situation is depicted in Fig. 4.2 where the right-handed coordinate frame attached to the target, iθ , ir , in , with in = iθ × ir , is used to resolve the relative position and velocity, with angles θ (t) and φ(t) denoting the inertial directions of r(t) and v(t), respectively. The line-of-sight direction, given by ir (t), can be determined either by an optical or radar sensor on board the maneuvering spacecraft or from an inertial navigation computer if the target is not a physical object. vm

Fig. 4.2 The geometry of line-of-sight interception

v

u

f

Inertial reference

q r ir

iq vt

140

4 Two-Body Maneuvers with Unbounded Continuous Inputs

The angle (φ + θ ) is subtended by the relative velocity v with the line of sight, ir (t), in either the second or the third quadrant, otherwise the closure rate r˙ would be non-negative. Therefore, the relative velocity can be expressed as follows: v = v [ir cos(π − φ − θ ) + iθ sin(π − φ − θ )] = v [−ir cos(φ + θ ) + iθ sin(φ + θ )]

(4.117)

The kinematic relationship between r and v is the following: dr = r˙ ir + (−θ˙ in ) × r dt ˙θ = r˙ ir + r θi

v=

(4.118)

On comparing Eqs. (4.117) and (4.118), we have the following kinematic scalar equations of motion: r˙ = −v cos(φ + θ )

(4.119)

and θ˙ =

v sin(φ + θ ) r

(4.120)

For the kinetic equation of motion, we recall that the target and the interceptor are roughly in the same orbit and separated by a small distance compared to the orbital radius, which implies that the difference in their accelerations due to gravity can be neglected. We will consider a more accurate model later in the chapter where the difference in gravity is also accounted for. Thus the relative acceleration is supplied by the control acceleration, u, applied to the interceptor in a direction normal to v. Hence we write dv = v˙ [−ir cos(φ + θ ) + iθ sin(φ + θ )] dt ˙ [ir sin(φ + θ ) + iθ cos(φ + θ )] + v(φ˙ + θ) $     % + v − −θ˙ in × ir cos(φ + θ ) + −θ˙ in × iθ sin(φ + θ )

u=

(4.121)

or, noting that in × ir = −iθ and in × iθ = ir , we have u = v˙ [−ir cos(φ + θ ) + iθ sin(φ + θ )] + v φ˙ [ir sin(φ + θ ) + iθ cos(φ + θ )]

(4.122)

Since u is normal to v, we also have the following:  v u = u in × v = u [ir sin(φ + θ ) + iθ cos(φ + θ )]

(4.123)

4.9 Line-of-Sight Guidance

141

Therefore, on comparing Eqs. (4.122) and (4.123), it follows that u v

(4.124)

v˙ = 0

(4.125)

φ˙ = and

This implies that the relative speed v always remains a constant, which is not surprising since the input acceleration is normal to the relative velocity vector. A substitution of Eq. (4.120) into Eq. (4.124) gives the following rate of change of the net angle between ir and v: φ˙ + θ˙ =

u v + sin(φ + θ ) v r

(4.126)

The equations governing the relative motion are thus Eqs. (4.119) and (4.126) where the relative speed, v, is constant. The acceleration normal to the projected line of sight is given by Eq. (4.124) to be the following: z¨ = v(φ˙ + θ˙ ) = u +

v2 sin(φ + θ ) r

(4.127)

Equation (4.127) can be integrated twice to yield the projected miss distance, z(t), defined to be the lateral distance of closest approach between the target and the interceptor if the acceleration input, u, remains unchanged. Referring to Fig. 4.2, the projected miss distance is given by z = r sin(φ + θ )

(4.128)

whose time derivative is the following: z˙ = r˙ sin(φ + θ ) + r(φ˙ + θ˙ ) cos(φ + θ )

(4.129)

or, by Eqs. (4.119) and (4.126), we have z˙ = −v sin(φ + θ ) cos(φ + θ ) + =

ur cos(φ + θ ) + v sin(φ + θ ) cos(φ + θ ) v

ur cos(φ + θ ) v

(4.130)

For a successful interception, it must be ensured that z(t) → 0 as t → tf . A simple navigation law for accomplishing this is derived by treating the net angle, (φ + θ ), to be small at all times, resulting in the following approximations: z r(φ + θ )

(4.131)

142

4 Two-Body Maneuvers with Unbounded Continuous Inputs

θ˙

v zv (φ + θ ) = 2 r r

(4.132)

ur v

(4.133)

r˙ −v

(4.134)

r = rf − vt

(4.135)

z˙ and

It follows from Eq. (4.134) that

or, z˙ = u

r

f

v

−t

 (4.136)

If we regard the time, tf = rf /v, as the projected terminal time at which the interception takes place, that is z(tf ) = 0, then the control u(t) is applied in the remaining time, i.e., the time to go, (tf − t), before the projected interception. When Eq. (4.135) is substituted into Eqs. (4.132) and (4.133), the following linearized state equations are obtained for the line-of-sight navigation problem: z v(tf − t)2

(4.137)

z˙ = u(tf − t)

(4.138)

θ˙ = and

These can serve as the basis of designing an optimal line-of-sight navigation system. Equation (4.138) shows that the normal velocity, z˙ , vanishes at the terminal time, unless u(tf ) → ∞. Furthermore, if u(t) were to remain constant throughout the time to go, the integration of Eq. (4.138) results in the following:   t z = ut tf − 2

(4.139)

which implies that the actual miss distance, z(tf ) = utf2 /2 = 0, could be large. Clearly, the acceleration input, u(t), must not remain constant, but should evolve in time in order to make z(tf ) = 0. Example 4.3 Consider a spacecraft approaching a space station for docking in a circular orbit. When the two spacecraft are in the same orbit and separated by a small

4.9 Line-of-Sight Guidance

143

lateral distance, x(t), the spacecraft is guided by line-of-sight navigation, resulting in the following equation of relative motion:   x˙ = tf − t u ,

(4.140)

where u(t) is the normal acceleration input applied to the maneuvering spacecraft. It is assumed that the two spacecraft are traveling in approximately parallel straight lines with a constant relative speed called closure rate, which will bring them exactly abreast in time tf . A successful docking thus requires the final lateral separation, x(tf ), must vanish while requiring only small transient input magnitudes, u(t), t ≤ tf , both of which are satisfied by minimizing the following objective function:  J = φ[x(tf ), tf ] +

tf

L(x, u)dτ

(4.141)

t

where the terminal penalty function φ[x(tf ), tf ] = k 2 x 2 (tf ) ,

(4.142)

ensures that x(tf ) is driven to small values by appropriate selection of the cost parameter, k, and the minimization of the transient cost with Lagrangian, L = u2 (t) ,

(4.143)

begins at the current time, t ≤ tf . The Hamiltonian function for this scalar state variable, x(t), is thus the following: H = L(x, u) + λT f (x, u) = u2 + (tf − t)λu ,

(4.144)

with the costate equation  T  ∂H ∗ ˙λ∗ (t) = − =0 ∂x

(4.145)

λ∗ (tf ) = 2k 2 x ∗ (tf )

(4.146)

with boundary condition

The stationary condition for optimization,  Hu =

∂H ∂u

∗ T

= 2u∗ (t) + (tf − t)λ∗ (t) = 0

(4.147)

144

4 Two-Body Maneuvers with Unbounded Continuous Inputs

gives the following expression for the extremal control: 1 u∗ (t) = − (tf − t)λ∗ (t) = −k 2 (tf − t)x ∗ (tf ) 2

(4.148)

Thus we have derived a simple boundary-value problem for determining the current extremal state from the projected terminal state, x(tf ). An extremal solution to this problem is easily obtained by substituting Eq. (4.148) into Eq. (4.140) and integrating back in time from tf to t: x ∗ (t) − x ∗ (tf ) =



t

(tf − τ )u∗ (τ )dτ = −k 2 x ∗ (tf )

tf



t

(tf − τ )2 dτ

(4.149)

tf

or, x ∗ (tf ) =

x ∗ (t) 1+

k2 3 (tf

− t)3

(4.150)

which substituted into Eq. (4.148) results in the following feedback control law: u∗ (t) = −

k 2 (tf − t) 1+

k2 3 (tf

− t)3

x ∗ (t) ,

(t ≤ tf )

(4.151)

This control law is optimal for any real value of k, hence u(t) ˆ = u∗ (t). The requirement for an exactly zero terminal separation for any given value of tf is satisfied in the limit k → ∞, for which the optimal control law becomes the following: lim u∗ (t) = −

k→∞

3 x ∗ (t) , (tf − t)2

(t ≤ tf )

(4.152)

The simple guidance problem presented in Example 4.2 could be solved in a closed form. However, a general three-dimensional navigation problem is not amenable to a closed-form solution. Even so, certain insights can be derived from this simple example, which cannot be obtained from a numerical solution. These are listed as follows: (a) The optimal guidance problem for interception of a target in a given terminal time, tf , with quadratic cost functions results in a linear feedback control law, which relates the current optimal input, u(t), ˆ to the current state, x(t). ˆ (b) The optimal control law depends upon the remaining time in the control interval, (tf − t), called the time to go. (c) The requirement of precise interception, i.e., for x(t ˆ f ) = 0, translates into a control input which increases in magnitude as the time to-go becomes smaller, reaching an infinite magnitude at t = tf .

4.10 Cross-Product Steering

145

4.10 Cross-Product Steering The Apollo program was perhaps the most complex space mission conducted with rather primitive computational resources. It could be successfully carried out only due to the application of simple guidance techniques. One such method, called cross-product steering, was used to align the velocity vector with an optimal direction at various stages of a mission. This yielded both maneuvering and energy control in a practical way. Its discussion here is along the lines of Battin [8]. The method is based upon the ability to measure the craft’s velocity vector, v(t), and the thrust acceleration vector, a(t), with some precision. Furthermore, a reference (or desired) velocity vector, vR (t), is available from the navigation computer at every instant. Consider vg (t) = vR (t) − v(t) as the velocity-to-be-gained at any given time, t, when the spacecraft’s position vector measured in an inertial reference frame is r(t). For an optimal interception of a final position, rf , in a future time tf , the velocity-to-be-gained must vanish at t = tf , otherwise the fuel is wasted. Hence, vg (tf ) is the objective function to be minimized with respect to a(t), which is applied in the remaining time, tf − t. Normally, such an optimization would require a sophisticated feedback scheme based upon the measurement of position and velocity as well as the integration of the equations of motion. We demonstrate that the same task can be achieved with a simple guidance law. The required acceleration at any time t is given by dvR ∂vR ∂vR dr = + dt ∂t ∂r dt ∂vR ∂vR + v = ∂t ∂r ∂vR ∂vR = + (vR − v) ∂t ∂r

(4.153)

The total acceleration is the sum of the thrust acceleration and that due to gravity: dv = g(r) + a dt

(4.154)

where g(r) denotes the acceleration due to gravity produced by an axisymmetric body. This yields the following time derivative of velocity to be gained: dvg dvR = − g(r) − a dt dt

(4.155)

whose substitution into Eq. (4.153) produces the guidance law, dvR ∂vR = g(r) − vg dt ∂r

(4.156)

146

4 Two-Body Maneuvers with Unbounded Continuous Inputs

or dvg ∂vR =− vg − a dt ∂r

(4.157)

The square matrix, ∂vR /∂r, is pre-computed and stored on board. Equation (4.157)provides the future thrust direction, a(t), which continues to be applied until vg (tf ) = 0 at some final time, tf > t. The velocity-to-be-gained, vg (t), can be driven to zero irrespective of the gravity field under consideration, by selecting a(t) such that the following condition is satisfied: dvg × vg = 0 dt

(4.158)

which requires from Eq. (4.157) that a × vg = −

∂vR vg × vg ∂r

(4.159)

Hence, the velocity-to-be-gained is driven to zero without requiring any integration of the governing differential equations. A post-cross product of Eq. (4.159) with vg (t) results in the following: (a · vg )vg − vg2 a = (p · vg )vg − vg2 p

(4.160)

where p=−

∂vR vg ∂r

(4.161)

and vg =| vg |. From Eq. (4.160) it follows that   vg vg a=p+ − ·p vg vg

(4.162)

where  vg = =a· vg

 a 2 − p2 +

2 vg ·p vg

(4.163)

Equation (4.162) provides the optimal thrust program as a simple algebraic guidance logic, given the measurement of the thrust acceleration magnitude, a, by onboard accelerometers, and the determination of the velocity-to-be-gained from Eq. (4.158).

4.10 Cross-Product Steering

147

Example 4.4 To illustrate the cross-product steering, consider a problem with a constant gravity field, g = const. This model—called the flat-trajectory approximation—is based upon a negligible variation in the radius vector from the vertical direction, and is valid when the craft is very close to the surface of a spherical body (such as during a lunar landing), or is traveling in a nearly circular orbit. The equations of motion for a flat-trajectory model are the following: dr =v dt dv = g+a dt

(4.164)

The solution for the unforced case (a = 0) with the initial conditions, r(0) = r0 , v(0) = v0 , is given by 1 r(t) = r0 + v0 t + gt 2 2 v(t) = v0 + gt

(4.165)

The velocity-to-be-gained at any time t is derived as follows by regarding the current position, r(t), and the reference velocity, vR (t), as the initial conditions, such that the desired final position, rf , is reached at a future time, tf :   1 1 2 vR (t) = rf − r(t) − g(tf − t) tf − t 2

(4.166)

∂vR 1 =− I ∂r tf − t

(4.167)

Thus we have

where I is the identity matrix. The substitution of Eqs. (4.166) and (4.167) into Eq. (4.162) yields the following: dvg 1 = vg − a dt tf − t

(4.168)

A guidance law based upon Eq. (4.168) is necessary for correcting the thrust acceleration, a(t), in the time to go, tf − t, such that vg (tf ) = 0. Since the reference velocity, vR (t), is derived for the unforced case, a successful application of thrust should make vg = 0 at a time either before or at tf . For an optimal thrust program, the engine must cut off at the time tc ≤ tf , exactly when the velocity-to-be-gained becomes zero, otherwise the propellant is wasted. The vehicle travels on a coasting trajectory (i.e., a = 0) in the remaining flight duration, tf − tc .

148

4 Two-Body Maneuvers with Unbounded Continuous Inputs

To derive the optimal thrust profile, take the scalar product of Eq. (4.168) with vg , and obtain 1 d 2 1 (vg ) = v 2 − a · vg 2 dt tf − t g

(4.169)

which is integrated by parts from the present time, t, to the engine cut-off time, tc , with vg (tc ) = 0, as follows:  (tf − t)vg2 (t)

tc

= t

[2(tf − τ )a(τ ) · vg (τ ) − vg2 (τ )]dτ

(4.170)

It is clear from Eq. (4.170) that the powered flight duration (called the remaining burn time), tc − t, which governs the fuel expenditure, is minimized by maximizing the scalar product, a · vg . This is achieved by aligning a(t) with the velocityto-be-gained, vg (t) at every instant, t ≤ tc , thereby making dvg /dt × vg = 0 (cross-product steering). This can be seen by putting a = c(t)vg , with c(t) being an arbitrary function of time, into Eq. (4.168). Hence, the optimal solution is achieved without numerically integrating Eq. (4.168). The velocity-to-be-gained vector can be simply chosen as vg = (1, c1 , c2 )vgx (t), where c1 , c2 are constants to be determined from the initial conditions.

4.11 Energy-Optimal Guidance A practical but simple guidance law for driving the craft from the present initial position, r(t), and velocity, v(t), to a specified final position, rf , and velocity, vf , at a given time, tf , while minimizing the total control power in a known gravity field, g(r, t), can be derived as an energy-optimal problem of minimizing the following objective function: J (t) =

1 2



tf

uT (τ )u(τ )]dτ

(4.171)

t

where u(t) = g(r, t)+a(t) is the net acceleration. The state equations of the problem are given by r˙ = v

(4.172)

v˙ = u with the following Hamiltonian: H =

1 T u u + λTr v + λTv u 2

(4.173)

4.11 Energy-Optimal Guidance

149

and the following costate equations: T  ˙λr = − ∂H =0 ∂r T  ˙λv = − ∂H = −λr ∂v

(4.174)

whose solution can be expressed as λr (t) = c1

(4.175)

λv (t) = −c1 t + c2 where c1 , c2 are constant vectors to be determined from the boundary conditions. The stationary condition for optimality is given by ∂H = u + λv = 0 ∂u

(4.176)

uˆ = −λv = c1 t − c2

(4.177)

or

which also satisfies the sufficient condition for optimality, Huu = I > 0. Thus, the energy-optimal acceleration profile is linear, from which the thrust acceleration, a(t), can be obtained by subtracting the known function, g(r, t). Integration of the state equations beginning from the current position and velocity, r(t), v(t), respectively, for τ = tf − t, yields the following: 1 c1 τ 2 − c2 τ + v(t) 2

(4.178)

1 1 c1 τ 3 − c2 τ 2 + v(t)τ + r(t) 6 2

(4.179)

vˆ (τ ) = and rˆ (τ ) =

from which the constants, c1 , c2 , are derived after applying the terminal boundary condition as follows: vf − v(t) τ c1 − 2 τ

(4.180)

1 τ c1 τ 3 + [vf − v(t)] + v(t)τ 12 2

(4.181)

c2 = rf − r(t) = −

150

4 Two-Body Maneuvers with Unbounded Continuous Inputs

  [vf + v(t)] 12 c1 = 3 r(t) − rf + τ 2 τ

(4.182)

  vf − v(t) [vf + v(t)] 6 c2 = 2 r(t) − rf + τ − 2 τ τ

(4.183)

The substitution of the constants, c1 , c2 , yields the following acceleration profile: ˆ )= u(τ

% % 4$ 6 $ vf − v(t) + 2 r(t) − rf + τ v(t) τ τ

(4.184)

ˆ ˆ = [vf − v(t)]/(tf − t), from which it But we also have u(t) = c1 t − c2 , and u(t) follows that ˆ u(0) = −c2 = −

vf − v(t) τ

(4.185)

Substitution of Eq. (4.185) into Eq. (4.183) produces rf − r(t) =

τ [vf + v(t)] 2

(4.186)

or rf − r(t) − vf τ = r(t) − rf + τ v(t)

(4.187)

whose substitution into Eq. (4.184) yields the following expression for the optimal acceleration profile: ˆ u(t) =

$ % % 4 $ 6 vf − v(t) + rf − r(t) − vf (tf − t) tf − t (tf − t)2

(4.188)

Example 4.5 To consider an example of the energy-optimal guidance, let us return to the flat-trajectory approximation of constant gravity field. In such a case, we have the following optimal thrust acceleration: uˆ =

 % $ % 4 $ 6 vf − v(t) + rf − r(t) + vf (tf − t) − g 2 tf − t (tf − t)

(4.189)

In order to keep u(t) finite in the limit t → tf , the time to go, τ = tf − t, is approximated by a constant, τ tgo , resulting in the following approximation: & uˆ

4 tgo

$ % vf − v(t) +

6 2 (tgo



% $ rf − r(t) + vf tgo − g (t < tf − tgo ) 0

(t ≥ tf − tgo ) (4.190)

4.12 Hill–Clohessy–Wiltshire Model

151

4.12 Hill–Clohessy–Wiltshire Model A guidance method which combines both maneuvering and energy control can be devised by using the approximate equations of relative motion between a spacecraft and a target point moving in a fixed circular orbit. Referred to as the Hill–Clohessy– Wiltshire (or HCW) model [24], these equations of relative motion are valid for small relative separation. Being a linearized system of equations, they have an analytical solution, which is easily used to derive an optimal control strategy. Consider a right-handed reference frame called the HCW frame, (ir , iθ , ih ), attached to a target moving in a circular orbit of radius, C, around a spherical planet, such that ir is pointed radially outward, iθ is tangential to the orbit and pointed in the direction of target’s motion, and ih is normal to the orbital plane, along the direction of the angular momentum. Now consider a spacecraft which can be maneuvered by the application of thrust, located at radius r from the planetary center, and having a relative position ρ from the target. These two position vectors are resolved in the HCW frame as follows: r = Cir + ρ

(4.191)

ρ = δrir + Cδθ iθ + δzih where the coordinates (δr, Cδθ, δz) denote the small radial, in-track, and out-ofplane displacements, respectively, of the maneuvering spacecraft from the . target. Since the HCW frame is rotating with an angular velocity, nih , where n = μ/C 3 , the velocity of the maneuvering spacecraft is given by the following time derivative: r˙ = δ r˙ ir + Cδ θ˙ iθ + δ z˙ ih + nih × r

(4.192)

= (δ r˙ − nCδθ )ir + (Cδ θ˙ + nC + δr)iθ + δ z˙ ih Similarly, the acceleration of the maneuvering spacecraft is resolved in the HCW frame as follows: (4.193) r¨ = (δ r¨ − nCδ θ˙ )ir + (Cδ θ¨ + δ r˙ )iθ + δ z¨ ih + nih × r˙  = (δ r¨ − 2nCδθ + n2 (C + δr) ir + (Cδ θ¨ + 2nδ r˙ − n2 Cδθ )iθ + δ z¨ ih The acceleration due to gravity of the spherical planet on the spacecraft is the following: (C + δr)ir + Cδθ iθ + δzih r = −μ 2 3 r (C + 2Cδr + δr 2 + C 2 δθ 2 + δz2 )3/2 μ μ

−C 3 ir − 3 (−2δrir + Cδθ iθ + δzih ) C C

g = −μ

= n2 (2δr − C)ir − n2 Cδθ iθ − n2 δzih

(4.194)

152

4 Two-Body Maneuvers with Unbounded Continuous Inputs

where the binomial theorem has been applied in the light of the approximation, δr  C, δθ  1, and δz  C. Equations (4.193) and (4.194) substituted into Newton’s second law for the maneuvering spacecraft, r¨ = g + a, produce the following scalar equations of approximate relative motion, commonly referred to as the Hill–Clohessy–Wiltshire (or HCW) equations: δ r¨ − 2nCδθ − 3n2 δr = ar Cδ θ¨ + 2nδ r˙ = aθ

(4.195)

z¨ + n2 z = an where a = ar ir + aθ iθ + an ih is the applied thrust acceleration. It is traditional to abbreviate the deviations such that the in-track displacement is denoted x = Cδθ and the radial displacement as y = δr, with the corresponding thrust accelerations given by ax = aθ , ay = ar , and az = an . Then Eq. (4.195) is expressed as follows: x¨ + 2ny˙ = ax y¨ − 2nx˙ − 3ω2 y = ay

(4.196)

z¨ + n2 z = az Equation (4.196) can be represented by the following state-space model: r˙ = v v˙ = Ar + Bv + a

(4.197)

where r = (x, y, z)T , v = (x, ˙ y, ˙ z˙ )T , a = (ax , ay , az )T , and ⎛

⎞ 0 0 0 A = ⎝ 0 3n2 0 ⎠ , 0 0 −n2

⎞ 0 −2n 0 B = ⎝ 2n 0 0 ⎠ 0 0 0 ⎛

The initial state is specified by r(0) = r0 , and v(0) = v0 . The Hamiltonian for an energy-optimal transfer between two given points is thus the following: H =

1 T a a + λTr v + λTv (Ar + Bv + a) 2

(4.198)

with the costate equations given by   ∂H T ∗ λ˙ r = − = −AT λ∗v ∂r

(4.199)

4.12 Hill–Clohessy–Wiltshire Model

153

  ∂H T ∗ = −λ∗r − BT λ∗v λ˙ v = − ∂v and the boundary conditions on the state and the costate variables being dependent upon the type of transfer. For example, an interception problem is posed as a transfer from the initial state to a final null relative position, r(tf ) = 0 in a given time, tf . The terminal boundary condition, λv (tf ) = 0, is then specified to close the problem. On the other hand, a rendezvous problem would have the terminal boundary conditions as r(tf ) = 0 and v(tf ) = 0 for a given time, tf . For multiple spacecraft involved in a formation flying constellation, the terminal condition on the k th spacecraft is specified by a given separation from the (k − 1) remaining spacecraft, which is specified as a boundary constraint, F[r(tf )] = 0. The stationary condition, Ha = 0, provides the following extremal input profile: a∗ (t) = −λ∗v (t) ,

(0 ≤ t ≤ tf )

(4.200)

The optimality of the extremal history is proved by seeing that it satisfies the sufficient condition, Haa = I > 0. Example 4.6 Consider a rendezvous between a space station in a circular orbit around a spherical Earth of 90 min. period and a spacecraft initially located at x(0) = −100 km, y(0) = 20 km, z(0) = −5 km, relative to the station, having the initial relative velocity components, x(0) ˙ = −1.0 m/s, y(0) ˙ = 2.0 m/s, and z˙ (0) = −0.5 m/s. If the spacecraft is equipped with a low-thrust engine which can provide a continuous acceleration of magnitude 1 m/s2 , select the smallest possible time of transfer, tf , for an energy-optimal rendezvous. Determine the optimal trajectory and input profile. For a rendezvous, the relative separation and velocity should both vanish simultaneously at t = tf , that is r(tf ) = 0, and v(tf ) = 0. The Hamiltonian for an energy-optimal transfer is given by Eq. (4.198), the costate equations by Eq. (4.199), and the extremal input by Eq. (4.200). The associated 2PBVP is numerically solved by a collocation scheme [81], and the results are plotted in Figs. 4.3, 4.4, 4.5, and 4.6. However, the upper limit of input acceleration is not imposed in the solution process, but instead the final time is selected by trial and error such that the maximum input magnitude does not exceed 1 m/s2 . In the next chapter, we will consider how an input acceleration bound can be imposed. Figure 4.4 shows the plots of the trajectory, (x, y, z), as well as the input acceleration magnitude, a, against the traverse angle (converted to degree), nt, for three values of the terminal time of rendezvous, ntf = 55.26◦ , 90◦ and 180◦ . As seen in Fig. 4.4, the smallest possible value of terminal time is for ntf = 55.26◦ (corresponding to tf = 828.9 s), for which the largest required thrust acceleration, a, is just under the maximum allowable value of 1 m/s2 . As the value of tf is increased, the peak value of optimal a decreases. The maximum overshoot of (x, y, z) also become smaller as tf is increased.

4 Two-Body Maneuvers with Unbounded Continuous Inputs 50

40

0

20

y (km)

x (km)

154

−50 −100 −150 0

0 −20

60

120

−40 0

180

ntf = 55.26°

60

ntf = 90°

0

120

180

ntf = 180°

1

a (m/s2 )

z (km)

0.8 −2

−4

0.6 0.4 0.2

−6 0

60

120

nt (deg.)

180

0

0

60

120

180

nt (deg.)

Fig. 4.3 The optimal trajectory for rendezvous, and the corresponding thrust acceleration magnitude plotted for three values of the terminal time, tf

Figure 4.4 shows the velocity components, (x, ˙ y, ˙ z˙ ), for the three values of ntf , whereas Fig. 4.5 plots the thrust acceleration components, (ax , ay , az ). The three-dimensional path taken by the spacecraft in the HCW frame until rendezvous (marked by an “x”) is shown in Fig. 4.6 for the three terminal times, illustrating the case ntf = 55.26◦ results in the shortest possible distance between the initial and final points. Such a shortening of the optimal path (hence the time of transfer) with an increased input magnitude forms the basis of bang–bang profiles, which will be studied in the following chapter when the constraint on the maximum thrust acceleration is enforced. Apart from the HCW model, more accurate models of relative motion can be derived, for which state transition matrices can be analytically evolved, such as the one for elliptical orbit of the target by Yamanaka and Ankersen [86]. These can be applied with greater fidelity to problems requiring accurate relative positioning of spacecraft, as in formation flying applications. Further enhancements are possible by including the atmospheric drag and oblateness effects [44] in a relative motion model, which are crucial in close-proximity formation flying missions in low Earth orbits.

4.12 Hill–Clohessy–Wiltshire Model

155

ntf = 55.26°

ntf = 90°

ntf = 180°

x˙ (m/s)

200 100 0 −100 0

20

40

60

80

100

120

140

160

180

20

40

60

80

100

120

140

160

180

20

40

60

80

100

120

140

160

180

y˙ (m/s)

50 0 −50 −100 0

z˙ (m/s)

10 5 0 −5 0

nt (deg.) Fig. 4.4 The optimal relative velocity components for rendezvous plotted for three values of the terminal time, tf

Exercises 4.1 For the orbital regulation system of a satellite in a circular orbit of radius C around a planet, use the linearized state equations with low-thrust radial acceleration control input, ar (t), to derive the two-point boundary value problem for reducing all initial perturbations, δr(0), δ r˙ (0), δθ (0) to zero in a fixed final time, tf , while minimizing the total control power given by  J = 0

tf

ar2 (t)dt .

4.2 Repeat Exercise 4.1 if the final angular error, δθ (tf ), is unspecified, but both radial displacement and radial velocity errors must finally be made zero at an unspecified time tf , while minimizing the total control power. 4.3 Derive the optimal trajectory for a low-thrust, coplanar orbital transfer using only the input, aθ , between two given positions, (r1 , θ1 ) and (r2 , θ2 ), with a free final time, tf , and an energy-optimal profile. (Hint: Resolve the flight path along radial and tangential directions, and assume that due to the small thrust acceleration, the change in flight path angle is negligible.)

156

4 Two-Body Maneuvers with Unbounded Continuous Inputs ntf = 55.26°

ntf = 90°

ntf = 180°

ax (m/s2 )

1 0 −1 0

20

40

60

80

100

120

140

160

180

20

40

60

80

100

120

140

160

180

20

40

60

80

100

120

140

160

180

ay (m/s2 )

0 −0.5 −1 0

az (m/s2 )

0.05 0 −0.05

0

nt (deg.) Fig. 4.5 The optimal thrust acceleration components for rendezvous plotted for three values of the terminal time, tf t = tf

50

z (km)

0 0

−2 t=0 −4

−50

−6 30

20

−100 10

0

−10

y (km)

x (km) −20

−30

−150

Fig. 4.6 The optimal trajectory for rendezvous plotted in three dimensions for three values of the terminal time, tf , showing the paths taken in the HCW frame

4.4 Consider the low-thrust, energy-optimal rendezvous to a target in a circular orbit of frequency n and radius C, using the approximate equations of relative motion, when only the input aθ is applied. Derive the expression for the energy-

4.12 Hill–Clohessy–Wiltshire Model

157

optimal trajectory for a transfer with tf = π/n with δr(0) = C/1000, δ r˙ (0) = δθ (0) = δ θ˙ (0) = δz(0) = δ z˙ (0) = 0. 4.5 Repeat Exercise 4.4 if only the normal acceleration input, an , is applied, and the initial condition is changed to δz(0) = C/1000, δr(0) = δ r˙ (0) = δθ (0) = δ θ˙ (0) = δ z˙ (0) = 0. 4.6 Derive the terminal boundary conditions to be satisfied by the coefficients of the extremal trajectory in Example 4.1, if the final time, tf , is unspecified. 4.7 A spacecraft is to be launched from the Moon’s surface by applying a known thrust acceleration profile, a(t), at an angle β(t) from the horizontal direction. This must be done such that the following objective is maximized before the engine cuts out in a fixed time, tf : 1 J = gr(tf ) + v 2 (tf ) 2 where g is the constant acceleration due to gravity, assumed to be always vertically downward. Derive a closed-form expression for the optimum thrust direction profile, β(t). 4.8 A lunar lander must make a smooth touchdown on the Moon at a point located (1680, −300, 300)T km in an inertial reference frame. Its final velocity at touchdown should be zero. Assuming that the position and velocity of the spacecraft 500 s before the touchdown are (1750, 300, −300)T km and (0, 1, 1)T km/s, respectively, and the Moon’s acceleration due to gravity is constant at (−1.6, 0, 0) m/s2 , calculate the optimal thrust acceleration profile, and the resulting optimal trajectory. 4.9 Consider two spacecraft traveling in nearly circular orbits of approximately the same radius, C, and in the same plane. It is desired to apply the line-of-sight guidance to one of the spacecraft, such that a constant closure rate v is maintained while the line of sight rotates in the orbital plane at a rate θ˙ , thereby producing a relative lateral acceleration, v θ˙ . A corrective normal acceleration input, u(t), is continuously applied to the maneuvering spacecraft such that an interception takes place at a given terminal time, tf . The projected miss distance is given by 2  z(t) = tf − t v θ˙ Using the HCW model of approximate relative motion, determine an energy-optimal input profile, u(t). 4.10 Derive Edelbaum’s [28] solution for the pitch and yaw angles for a timeoptimal transfer between inclined circular orbits, (a0 , i0 ), and (af , if ), using Gauss’ variational equations, and using only the acceleration components, aθ , an , with , a constant, low-thrust acceleration magnitude, u = aθ2 + an2 , such that the approximately circular shape (e 0, r a) is maintained throughout the transfer.

Chapter 5

Optimal Maneuvers with Bounded Inputs

5.1 Introduction Spacecraft engines can generate only limited thrust magnitudes. This implies that the acceleration inputs in space navigation must necessarily be bounded. The trajectory optimization must therefore be performed taking bounded inputs into account. As discussed in the earlier chapters, the nature of optimal trajectories with bounded acceleration inputs can be classified as either impulsive thrust or continuous thrust maneuvers. A further classification is possible according to the gravity field in which such maneuvers are performed. This chapter presents a general optimization method which can be applied to the various kinds of space navigation problems. While the approach follows the formal development of optimal control presented earlier, the focus is on the so-called primer vector method of Lawden [52] and its further applications.

5.2 Optimal Thrust Direction The equations of motion of a spacecraft in a general gravity field with thrust acceleration input, a(t) = a(t)n(t), where n is a unit vector in the thrust direction, can be expressed as follows: v˙ = g(r, t) + an

(5.1)

r˙ = v

(5.2)

and

© Springer Nature Switzerland AG 2019 A. Tewari, Optimal Space Flight Navigation, Control Engineering, https://doi.org/10.1007/978-3-030-03789-5_5

159

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5 Optimal Maneuvers with Bounded Inputs

with the initial condition, r(t0 ) = r0 , where g(r, t) is the acceleration due to gravity. The thrust acceleration input is bounded in magnitude by the engine’s capacity. The thrust magnitude, T (t), of an ideal rocket engine with an exhaust speed, ve (t), is given by ˙ T (t) = −ve (t)m

(5.3)

where m ˙ is the rate of change of the vehicle’s mass (equal in magnitude and opposite in sign to the mass exhaust rate). A more traditional expression for thrust magnitude is in terms of the specific impulse, Is (t), which is the ratio of the exhaust speed and acceleration due to gravity on Earth’s surface, Is = ve /gs , given by ˙ T (t) = −gs Is (t)m

(5.4)

where gs = 9.81 m/s2 . Note that Eq. (5.4) gives a constant value of Is for a chemical rocket engine, because the exhaust speed ve of such engines is nearly constant. The thrust magnitude of such an engine can still be varied by changing the mass flow rate (by a process called throttling) up to a certain extent: βm ≤ m ˙ ≤0

(5.5)

0 ≤ T (t) ≤ Tm

(5.6)

thereby yielding

where Tm = −gs Is βm is the maximum-thrust magnitude (constant) corresponding to the maximum mass exhaust rate, −βm (also constant). The bounds on the acceleration magnitude provided by the constant specific impulse engine are therefore the following: 0≤a≤

Tm m(t)

(5.7)

This implies that the upper bound of the acceleration magnitude provided by a constant specific impulse engine is not constant, but increases as the vehicle’s mass is consumed. For a constant maximum throttle setting, βm , the mass variation with time is given by m(t) = m0 + βm (t − t0 )

(5.8)

with m0 = m(t0 ), which yields the maximum acceleration limit, 0 ≤ a ≤ am (t), as follows: am (t) =

βm Tm = −gs Is m0 + βm (t − t0 ) m0 + βm (t − t0 )

(5.9)

5.2 Optimal Thrust Direction

161

An electric (or ion) propulsion engine has a constant power output, P = T (t)ve (t) = −mv ˙ e2 (t), instead of a constant exhaust speed, thereby producing a time-variable specific impulse, Is (t). Such an engine also has a variable mass exhaust rate, −m, ˙ inversely proportional to the square of the exhaust speed. The bounds on the mass flow rate are however constant: ˙ ≤0 βm ≤ m

(5.10)

The acceleration magnitude bounds of a constant power engine are thus represented by 0 ≤ a ≤ am (t)

(5.11)

where Tm (t) = am (t) = m(t)

√ −Pβm m(t)

(5.12)

The upper bound, am , can be approximated as being constant, if the propellant consumed, m0 − m(t), is minimized to be a small fraction of the initial mass, m0 , at any time t. It is often useful to define a characteristic speed, c(t), as the integral of the acceleration input magnitude: 

t

c(t) =

a(τ )dτ

(5.13)

t0

This is the maximum possible change in the vehicle’s velocity if the thrust is applied tangentially, and the acceleration due to gravity is negligible in comparison, which is the case for a large (or impulsive) acceleration input. If the vehicle’s current mass, m(t), is to be determined, the state space is enhanced by an additional variable, c(t), where c˙ = a

(5.14)

with the initial condition c(t0 ) = 0. This implies that the characteristic speed, c, is a measure of the total elapsed time, and influences the current mass, m, as well as the upper limit of acceleration, am . The final characteristic speed, c(tf ), gives a measure of the total propellant expenditure, m(t0 ) − m(tf ), and therefore must be minimized for an optimal maneuver. Hence, the performance index to be minimized for the smallest possible propellant requirement, subject to the dynamic equality constraints given by Eqs. (5.1)–(5.2) and (5.14), is the following Mayer type function:  J = c(tf ) =

tf

a(t)dt t0

(5.15)

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5 Optimal Maneuvers with Bounded Inputs

For the special case of a constant specific impulse, the characteristic speed allows an exact integration of Eq. (5.3) for the current mass, m(t), in terms of the initial mass, m0 , and the constant exhaust speed, ve , expressed as follows: c(t)

m(t) = m0 e− ve

(5.16)

and am (t) = −gs Is

βm gc(t)I e ss m0

(5.17)

Irrespective of the type of the engine employed, a minimization of the terminal characteristic speed, J = φ = c(tf ), results in the following Hamiltonian for the optimal control problem: H = λTr v + λTv [g(r, t) + an] + λc a

(5.18)

where λr and λv are the costate vectors corresponding to r and v, respectively, and λc corresponds to the characteristic speed. The costate equations for λr and λv are derived from the necessary conditions of optimality as follows:     ∂H T ∂g(r, t) T λ˙ r = − =− λv ∂r ∂r

(5.19)

  ∂H T λ˙ v = − = −λr ∂v

(5.20)

Here the gravity-gradient matrix, ∂g(r, t)/∂r, depends upon the nature of the gravity field. For example, for a spherical gravity field of the restricted two-body problem, we have a time-invariant acceleration due to gravity: g(r) = −

μ r r3

μ ∂g(r) 3μ = − 3 I3×3 + 5 rrT ∂r r r

(5.21) (5.22)

The general optimal maneuvers with the final time, tf , unspecified can be posed as optimal boundary value problems with the given initial position, r(t0 ) = r0 , and a terminal state lying on a moving hypersurface (see Chap. 2) expressed as follows: F[r(tf ), v(tf ), tf ] = 0

(5.23)

where F : R7×1 → Rk×1 , 1 ≤ k ≤ 5. An example of such maneuvers is the orbital rendezvous with a target moving in a specified orbit, but with the rendezvous

5.2 Optimal Thrust Direction

163

point being unspecified. As covered in Chap. 2, the associated terminal boundary condition from the Euler–Lagrange formulation is the following:  λr (tf ) =  λv (tf ) =

∂F ∂r ∂F ∂v

T d

(5.24)

d

(5.25)

t=tf

T t=tf

where d = (d1 , d2 , . . . , dk )T are the additional adjoint variables used to adjoin the terminal constraint equation, Eq. (5.23), to the performance index J = φ[x(tf ), tf ] + dT F = c(tf ) + dT F

(5.26)

The control variables of this problem are u = (a, nT )T , with n being the unit thrust direction vector. The optimal control problem is singular for a, because of the linearity of the Hamiltonian with respect to a. However, it is possible to minimize ˆ is H relative to the thrust direction, n, such that the optimal thrust direction, n, obtained. Representing the thrust direction unit vector in terms of its direction cosines, n = (1 , 2 , 3 )T , such that 21 + 22 + 23 = 1

(5.27)

we express the Hamiltonian as follows:   H = λTr v + λTv [g(r, t) + an] + λc a + λn 21 + 22 + 23 − 1

(5.28)

where the static constraint, Eq. (5.27), is adjoined with the help of the additional costate parameter, λn . The costate equation corresponding to the static constraint is given by  0=

∂H ∂n

T = aλv + 2λn nˆ

(5.29)

which is a necessary condition for the minimization of H relative to n. On examining Eq. (5.27), it is found that unless both a and λn vanish, the optimal ˆ must be aligned opposite to the direction of the costate vector thrust direction, n, corresponding to velocity, λv , that is nˆ = −

λv λv

(5.30)

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5 Optimal Maneuvers with Bounded Inputs

with λn = 2λa v . This optimal thrust direction is referred to by Lawden [52] as the primer vector, p(t) = −λv (t), and is an important result of the general optimal space navigation problem, irrespective of the bound on the input magnitude, a. Hence we write nˆ =

p p

(5.31)

ˆ becomes indeterminate. If a = 0, then the optimal thrust direction, n, The substitution of Eq. (5.31) into Eqs. (5.19) and (5.20) yields the following differential equations for the costate variables with the optimal thrust direction:  T ∂g λ˙ r = p (5.32) ∂r and p˙ = λr

(5.33)

combining both of which we have the following second-order ordinary differential equation for the primer vector:  T ∂g p¨ = p (5.34) ∂r The boundary conditions for the primer vector and its time derivative are obtained from Eqs. (5.24) and (5.25) as follows:  ∂F T d ∂r t=tf  T ∂F p(tf ) = − d ∂v t=tf 

˙ f) = p(t

(5.35)

(5.36)

Hence, we have the complete set of boundary conditions for integrating Eq. (5.34) backward in time in order to determine the primer vector, p(t). The costate equation corresponding to the characteristic speed is the following: λ˙ c = −



∂H ∂c

T

 = −a

∂H ∂t

T (5.37)

where the Hamiltonian varies with time due to the dependence of the upper limit, am (t), on time. Furthermore, the terminal boundary condition on λc yields the following:   ∂φ =1 (5.38) λc (tf ) = ∂c t=tf

5.2 Optimal Thrust Direction

165

The Hamiltonian can be expressed in terms of the primer vector as follows: H = a(λc − p) + p˙ T v − pT g

(5.39)

Due to the linearity of H relative to a, the optimal control problem is singular in acceleration magnitude. However, the bounds on a given by Eq. (5.7) allow the application of the Pontryagin’s minimum principle (Chap. 2) as follows: a(λ ˆ c − p) + p˙ T v − pT g ≤ a(λc − p) + p˙ T v − pT g

(5.40)

or a(λ ˆ c − p) ≤ a(λc − p)

(5.41)

where aˆ refers to the optimal control, and 0 ≤ a ≤ am (t)

(5.42)

The last two inequalities result in the following switching condition for the optimal control magnitude: & a(t) ˆ =

am (t) , (p − λc ) > 0 0,

(p − λc ) < 0

(5.43)

The sign of the switching function, (p − λc ), is determined by whether the magnitude, p, of the primer vector is either greater than or less than λc . While λc (t) depends solely upon the engine characteristics (Eq. (5.37)), it is the primer vector which gives a control over the optimal control magnitude. The bang–bang profile switches between the “full on” and zero (or “off”) acceleration magnitudes, and thus the optimal trajectory consists of the maximum-thrust (MT) and null-thrust (NT) arcs. Such a feature is common to all singular space navigation problems, such as the fuel minimization mission considered here, as well as the minimum-time orbital transfer missions. It is also possible to have an “intermediate-thrust” (IT) arc satisfying the switching condition, for which the switching function vanishes (p = λc ), hence the control magnitude can fall between the two bounds: 0 < aˆ < am (t) ,

p = λc

(5.44)

Such arcs are typically extremal arcs which satisfy the first-order necessary conditions of optimality. However, due to their being singular arcs, they do not satisfy the Legendre–Clebsch sufficient optimality condition, therefore it is difficult to prove their optimality based upon second-order Hamiltonian variation.

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5 Optimal Maneuvers with Bounded Inputs

5.2.1 Constant Acceleration Bound The problem in which the acceleration upper bound, am (t), varies with time is difficult to analyze and solve. However, if the assumption of a small, minimum propellant mass is made, we write am =

Tm Tm

m(t) m0

(5.45)

Furthermore, if the gravity field is time invariant, it can be shown that the Hamiltonian becomes constant on the optimal trajectory, resulting in the following: λ˙ c = 0

(5.46)

This implies that λc = const., which of course results in λc = λc (tf ) = 1, and yields the following Hamiltonian: H = a(1 − p) + p˙ T v − pT g

(5.47)

The switching condition for the constant acceleration input bound, am , is therefore the following:

a(t) ˆ =

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

am ,

(p − 1) > 0

0,

(p − 1) < 0

0 < aˆ < am ,

p=1

(5.48)

5.2.2 Bounded Exhaust Rate Often it is more convenient to express the acceleration magnitude, a, as a function of the mass exhaust rate, β = −m, ˙ which is always physically bounded by constant limits: 0 ≤ β ≤ βm

(5.49)

These inequalities can alternatively be expressed as the following equation: β(β − βm ) − α 2 = 0

(5.50)

where α is an additional (fictitious) control variable without any physical significance. The equations of motion are recast in terms of β as the input variable instead of a as follows:

5.2 Optimal Thrust Direction

167

r˙ = v v˙ = g + β

ve n m

(5.51)

m ˙ = −β where (β, n) are the input variables. For the constant specific-impulse case (that is, ve = const.), the objective function for minimization, subject to the equality constraints, Eqs. (5.50) and (5.51), is now expressed as follows:  J = c(tf ) =

tf

t0

β(t)ve dt = − m(t)



tf t0

mv ˙ e m0 dt = ve ln m m(tf )

(5.52)

with the initial condition given by r0 = r(t0 ), v0 = v(t0 ), and m0 = m(t0 ), and the terminal state satisfying Eq. (5.23). The minimization of J is thus equivalent to maximizing the final mass, m(tf ), subject to the given constraints, thereby resulting in the minimum propellant consumption. With the optimum thrust direction, n = p/p, the Hamiltonian is then given by H = p˙ T v − pT g −

βve p − λm β + λβ [β(β − βm ) − α 2 ] m

(5.53)

The costate equations, apart from the primer vector differential equation, Eq. (5.34), are the following: λ˙ m = −



∂H ∂m

T =−

βve p m2

0=−

ve ∂H = p + λm − λβ (2β − βm ) ∂β m

0=−

∂H = 2λβ α ∂α

(5.54)

where the last two equations arise from the static equality constraint, Eq. (5.50). Beginning with the last equation of Eq. (5.54), it is seen to be satisfied either by λβ = 0 implying an intermediate-thrust (IT) arc (0 < β < βm ) or by α = 0, which stands for either of the two extremes, that is the maximum-thrust (MT) arc (β = βm ), or the null-thrust (NT) arc (β = 0). Hence, all three possibilities are encompassed by this equation. The terminal boundary condition on λm yields the following:  λm (tf ) =

∂φ ∂m

 =− t=tf

ve m(tf )

(5.55)

This boundary condition and the first equation of Eq. (5.54) imply that λm (t) < 0 for all t0 ≤ t ≤ tf .

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5 Optimal Maneuvers with Bounded Inputs

The switching condition derived from Pontryagin’s minimum principle is the following: v

 v  e p + λm β ≤ p + λm βˆ m m e

which yields λm + vme p > 0 for the NT arc, λm + λm + vme p = 0 for the IT arc.

ve mp

(5.56)

< 0 for the MT arc, and

5.3 Time-Invariant Gravity Field Integration of the primer vector differential equation, Eq. (5.34), is readily made for the null-thrust (NT) and maximum-thrust (MT) arcs in the gravity field of a spherical central body, such as by Lawden [52] and Pines [67]. More generally, if the gravity field is invariant in time, we have g = g(r), and the costate (adjoint) equations expressed in terms of the primer vector are the following: p˙ = λr

(5.57)

and p¨ = λ˙ r =



∂g(r) ∂r

T p

(5.58)

Furthermore, if the acceleration input has a constant upper bound, am , the time derivative of the Hamiltonian H = a(λc − p) + p˙ T v − pT g

(5.59)

vanishes along an optimal arc satisfying Eq. (5.34). Thus, we have H = a(λc − p) + p˙ T v − pT g = const.

(5.60)

on an optimal trajectory. For a null-thrust (NT) arc, (a = 0), we have C = p˙ T v − pT g = const.

(5.61)

Equation (5.61) also holds along an intermediate-thrust (IT) arc (p = λc ). This result is also valid for the case of a bounded exhaust rate, because it also yields a constant Hamiltonian along the optimal trajectory for a time-invariant gravity field. A central gravity field, g(r) = g(r)r/r, has the following gravity gradient:   ∂g(r) g r ∂g g rT = I+ − (5.62) ∂r r r ∂r r r

5.4 Null-Thrust Arc in Central Gravity Field

169

which substituted into Eq. (5.58), yields the following: p¨ =

g r p+ r r



g ∂g − ∂r r



rT p r

(5.63)

For the optimal thrust direction, we have r˙ = v

(5.64)

r p v˙ = g(r) + a r p

(5.65)

and

These, along with Eqs. (5.57) and (5.63), are used to demonstrate the following constant throughout the optimal trajectory: d (p × v − λr × r) = p˙ × v + p × v˙ − λ˙ r × r − λr × r˙ dt =0

(5.66)

D = p × v − λr × r = const.

(5.67)

or

Thus the optimal trajectory in a time-invariant gravity field (such as that of a two-body problem) has important constants of integration, irrespective of the type of input acceleration profile employed.

5.4 Null-Thrust Arc in Central Gravity Field While the primer vector is unnecessary for determining the thrust direction in a nullthrust (NT) arc, its relationship with the radius and velocity vectors can be used to derive important information about the motion. Furthermore, in a trajectory which consists of both NT and powered (IT or MT) arcs, the evolution of primer vector along an NT arc is necessary to determine its value at the beginning of a subsequent powered arc. Consider the flight in a general, central gravity field, g(r) = g(r)r/r. Along an NT arc (a = 0), Eqs. (5.1) and (5.2) yield the following constant angular momentum vector: h = r × v = const.

(5.68)

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5 Optimal Maneuvers with Bounded Inputs

which implies that the flight is confined to a plane normal to h. Other constants for a null-thrust arc in a central gravity field are given by Eqs. (5.61) and (5.67). Yet other integration constants can be derived based upon the specific nature of the central field, such as the inverse-square law gravity field of a spherical body covered in Chap. 3. We collect the various governing equations for the NT arc as follows: r˙ = v v˙ = g(r) p˙ = λr λ˙ r = p¨ =

r r 

∂g(r) ∂r

(5.69)

T p

with constants of the motion given by Eqs. (5.61), (5.67), and (5.68). To resolve the coasting flight in a central gravity field, g(r) = g(r)r/r, consider a rotating, right-handed coordinate frame, (i, j, k), with origin at the center of the field, and oriented such that (i, j) is the plane of the motion containing both r and v (i.e., the plane normal to h = r × v) at all times. Thus k is along the constant angular momentum vector, (k = h/ h). The orientation of the i axis is selected to be always along the radius vector, (i = r/r), and j completes the triad with i × j = k. The various vectors of motion are thus resolved along (i, j, k) as follows: r = ri v = vx i + vy j λr = λx i + λy j + λz k

(5.70)

p = p x i + py j + pz k By substituting Eq. (5.70) into Eq. (5.47) we have the following Hamiltonian along a null-thrust arc: H = p˙ T v − pT g = λx vx + λy vy − px g = const.

(5.71)

which validates Eq. (5.61). The velocity vector can also be expressed in terms of polar coordinates (r, θ ), where θ is the angle made by r with an inertial reference line, I, in the plane of motion: ˙ v = vx i + vy j = r˙ i + r θj

(5.72)

Hence, the constant angular momentum becomes the following: h = r × v = rvy k = r 2 θ˙ k

(5.73)

5.4 Null-Thrust Arc in Central Gravity Field

171

To solve Eq. (5.67) for the primer vector, it is convenient to resolve the vectors of motion in an inertial frame, (I, J, k), where J is normal to both I and k such that I × J = k. This requires the coordinate transformation ⎧ ⎫ ⎛ ⎞⎧ ⎫ cos θ sin θ 0 ⎨ I ⎬ ⎨i⎬ j = ⎝ − sin θ cos θ 0 ⎠ J ⎩ ⎭ ⎩ ⎭ 0 0 1 k k

(5.74)

Consequently, Eq. (5.67) produces the following scalar component equations: − pz vy cos θ − (pz vx − λz r) sin θ = hc1 −pz vy sin θ + (pz vx − λz r) cos θ = hc2

(5.75)

px vy − py vx + λy r = hC3 where the constant vector on the right-hand side of Eq. (5.67) is taken to be ˙ By resolving the velocity components D = h(c1 I + c2 J + c3 k), with h = r 2 θ. in polar coordinates according to Eq. (5.72), the first two equations of Eq. (5.75) are algebraically solved to yield pz = −r (c1 cos θ + c2 sin θ )

(5.76)

Furthermore, scaling the constant Hamiltonian (Eq. (5.71)) by h2 , we write h2 H = λx vx + λy vy − px g

(5.77)

and using the third equation of Eq. (5.75), we solve for λx and λy as follows: 1 2 h H − r 2 θ˙ 2 C3 + px (g + r θ˙ 2 ) − py θ˙ r˙ ˙ 3 − px θ˙ + py r˙ /r λy = r θC

λx =

(5.78)

The time-derivative of the primer vector resolved along the rotating axes can be expressed as follows: p˙ =

δp +ω×p δt

(5.79)

where δp = p˙ x i + p˙ y j + p˙ z k δt

(5.80)

is the partial time derivative derived by treating (i, j, k) to be fixed and ˙ ω = θk

(5.81)

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5 Optimal Maneuvers with Bounded Inputs

is the angular velocity of the orbital motion. A substitution of Eqs. (5.80) and (5.81) into Eq. (5.79) yields p˙ = (p˙ x − py θ˙ )i + (p˙ y + px θ˙ )j + p˙ z k

(5.82)

Similarly, the second-order derivative is obtained to be the following: δ p˙ + ω × p˙ δt   = p¨x − py θ¨ − 2p˙ y θ˙ − px θ˙ 2 i   + p¨y + px θ¨ + 2p˙ x θ˙ − py θ˙ 2 j + p¨z k

p¨ =

(5.83)

5.4.1 Inverse-Square Gravity Field A null-thrust (NT) arc in a gravity field of a spherical body is directly determined from the solution to the two-body problem of Keplerian motion (Chap. 3). As discussed in Chap. 3 for coasting arcs between impulses, the primer vector can be integrated along an NT arc using either spherical coordinates or the classical orbital elements [52]. The result of the integration can also be expressed in terms of Lagrange’s coefficients (Chap. 3), or by an appropriate universal variable (generalized coordinates) [8] to preclude the singularity of the classical orbital elements in near equatorial and circular orbits. A generalized canonical variable approach can also be used to integrate the primer vector along MT and NT arcs in a spherical gravity field. Da Silva [26] extends a canonical variable method via Sundman transformation for analytically determining NT arcs in a closed form. Integrating the primer vector along an NT arc around the inverse-square field of a spherical body, g(r) = −μr/r 3 , requires the expression of the primer-vector differential equation as follows:   ∂g(r) T μ 3μ p¨ = p = − 3 p + 5 (rT p)r (5.84) ∂r r r The right-hand side of Eq. (5.84) is then expressed as follows: −

 μ 3μ μ p + 5 (rT p)r = 3 2px i − py j − pz k 3 r r r

(5.85)

and the following scalar differential equations are derived by the substitution of Eqs. (5.83) and (5.85) into Eq. (5.84): p¨x − py θ¨ − 2p˙ y θ˙ − px θ˙ 2 =

2μ px r3

5.5 Intermediate-Thrust Arc

173

μ py r3 μ p¨z = − 3 pz r

p¨y + px θ¨ + 2p˙ x θ˙ − py θ˙ 2 = −

(5.86)

Eliminating θ from these equations by using θ˙ = h/r 2 and θ¨ = −2h˙r /r 3 yields the following: p¨x + 2py r˙ p¨y − 2px r˙

h h2 h 2μ − 2p˙ y 2 − px 4 = 3 px 3 r r r r h h2 h μ + 2 p ˙ − p = − 3 py x y r3 r2 r4 r μ p¨z = − 3 pz r

(5.87)

For the further development of the integrals of motion, refer to the method presented in Chap. 3.

5.5 Intermediate-Thrust Arc Intermediate-thrust arcs have been of much interest in the literature on optimal space navigation with bounded inputs. Before a rigorous analysis of singular control problems could be undertaken, there was a doubt whether the intermediate-thrust arcs can be regarded as optimal control solutions. In early works [31, 49, 50], it was unclear whether the intermediate-thrust arcs could be solutions of the minimum-fuel, orbital transfer problems. Leitmann [53] first suggested that the intermediate-thrust arcs may be candidate optimal arcs. This was followed by further developments by Lawden [51, 52], wherein it was found that for the minimum-fuel problems in a spherical gravity field with the final time unspecified, the intermediate-thrust arc is a spiral centered on the attracting body. Such an arc became famous in literature as Lawden’s spiral. Many researchers [10, 11, 16, 17, 32, 34, 39, 42, 45, 47, 71, 75, 83], and [3] investigated the optimality of intermediatethrust arcs such as Lawden’s spiral. The primary focus was on the derivation of a generalized form of the Legendre–Clebsch condition (see Chap. 2) involving second- and higher-order variations on control applied on the singular arcs. A survey of these developments can be found in Bell [9] and Bell and Jacobson [12]. Such investigations into the singular optimal control problems led to the maturing of the optimal control theory, which was formalized into texts such as Bryson and Ho [19], and Athans and Falb [4]. We are now ready to determine the optimality of singular, intermediate-thrust arcs of the space navigation problem, such as Lawden’s spiral.

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5 Optimal Maneuvers with Bounded Inputs

5.6 Lawden’s Spiral Consider the flight of a spacecraft in an inverse-square (spherical) gravity field with a constant specific impulse engine having an exhaust speed, ve , and a variable mass exhaust rate, m ˙ = −β. For simplicity, the thrust is assumed to always lie in the plane of flight, thereby producing a two-dimensional motion. The position of the vehicle’s center of mass, o, from the center of the gravity field, O, is given by the radius vector resolved in polar coordinates, (r, θ ). Now consider a moving coordinate frame, (i, k), such that i is along the instantaneous thrust direction, which makes an angle ψ from an inertial reference line, and an angle φ from the normal to the radius vector. The instantaneous velocity vector is resolved along and normal to the thrust direction as follows: v = ui + wk ,

(5.88)

with u denoting the vehicle’s velocity component along the thrust direction, i, and w that along the direction, k, perpendicular to i. The rotational kinematics of the thrust vector gives the following: ˙i = ψk ˙ ,

˙ , k˙ = −ψi

(5.89)

which results in the following expression for acceleration resolved along (i, j): ˙ + (w˙ + uψ)k ˙ v˙ = (u˙ − w ψ)i

(5.90)

The equations of motion are then expressed as follows: r˙ = u sin φ − w cos φ r θ˙ = u cos φ + w sin φ 1 μ βve − 2 sin φ m r μ w˙ = −uψ˙ + 2 cos φ r m ˙ = −β u˙ = w ψ˙ +

(5.91)

where θ =φ+ψ −

π 2

(5.92)

The control variables of this fifth-order system are u = (β, φ), while the initial and final boundary conditions are the following: r(t0 ) = r0

5.6 Lawden’s Spiral

175

θ (t0 ) = θ0 m(t0 ) = m0 r(tf ) = rf θ (tf ) = θf u(t0 ) = r˙ (t0 ) sin φ(t0 ) + r0 θ˙ (t0 ) cos φ(t0 )

(5.93)

w(t0 ) = −˙r (t0 ) cos φ(t0 ) + r0 θ˙ (t0 ) sin φ(t0 ) π ψ(t0 ) + φ(t0 ) = θ0 + 2 u(t0 ) = r˙ (tf ) sin φ(tf ) + rf θ˙ (tf ) cos φ(tf )

(5.94)

w(tf ) = −˙r (tf ) cos φ(tf ) + rf θ˙ (tf ) sin φ(tf ) π ψ(tf ) + φ(tf ) = θf + 2 The objective function for minimization is the final characteristic speed given by  J = c(tf ) =

tf t0

β(t)ve m0 dt = ve ln m(t) m(tf )

(5.95)

which, of course, means a maximization of the final mass, m(tf ). It is clear from the state equations, Eq. (5.92), that the Hamiltonian is linear in β, and hence the optimization problem is singular relative to β:  u w cos φ+ sin φ (5.96) H = λr (u sin φ−w cos φ)+λθ r r      βve μ μ π ˙ ˙ − 2 sin φ + λw −uψ+ + λu w ψ+ cos φ −λm β+ν θ −φ−ψ+ 2 m r 2 r where the mass exhaust rate, β, is physically bounded as follows: 0 ≤ β ≤ βm

(5.97)

with βm being constant. The solution for an extremal intermediate-thrust arc with β falling between the two limits, 0 < β < βm , is obtained from the Euler–Lagrange necessary conditions for optimality. To this end, we switch to the primer vector method introduced earlier in the chapter. Since the mass exhaust rate, β (rather than the acceleration magnitude, a) is now a control variable affecting the thrust magnitude, the application of Pontryagin’s minimum principle results in a switching condition with p = const. along an intermediate-thrust arc (which is equivalent to p = 1 for the constant acceleration bound am ). Therefore, we have p = pi for the optimal thrust direction. The time derivatives of the primer vector are thus the following:

176

5 Optimal Maneuvers with Bounded Inputs

˙ p˙ = p˙i = pψk ¨ − pψ˙ 2 i p¨ = pψk

(5.98)

The last equation is compared with Eq. (5.58),  p¨ =

∂g(r) ∂r

T p

(5.99)

and noting that for a spherical gravity field, g(r) = −(μ/r 3 )r, we have 

∂g(r) ∂r

T p=−

μ 3μ p + 5 rT pr r3 r

(5.100)

Furthermore, we have r = r sin φi − r cos φk

(5.101)

rT p = rp sin φ

(5.102)

and

Hence 

∂g(r) ∂r

T p=p

μ 3μ (3 sin2 φ − 1)i − p 3 sinφ cos φk 3 r r

(5.103)

Substituting Eq. (5.103) into Eq. (5.99), and comparing its components along i, j with those of the second equation of Eq. (5.98), we have the following scalar differential equations: μ (1 − 3 sin2 φ) r3

(5.104)

3μ sinφ cos φ r3

(5.105)

μ (sin φi − cos φk) r2

(5.106)

ψ˙ 2 =

ψ¨ = − By substituting g=−

p = pi

(5.107)

˙ p˙ = p˙i = pψk

(5.108)

and

5.6 Lawden’s Spiral

177

into Eq. (5.61), we have the following constant of the IT arc: p˙ T v − pT g = pw ψ˙ + p

μ sin φ = const. r2

(5.109)

Since p = const. along an optimally directed IT arc, we write w ψ˙ +

μ sin φ = C = const. r2

(5.110)

Another constant of the motion is obtained by using Eqs. (5.104) and (5.105) to yield the following: ψ˙ 2 cos φ − ψ¨ sin φ =

μ cos φ r3

(5.111)

which substituted into the fourth equation of Eq. (5.91), produces w˙ + uψ˙ − r ψ˙ 2 cos φr ψ¨ sin φ = 0

(5.112)

Finally, the radial and transverse velocity components, given by the first two equations of Eq. (5.91), are resolved along i, to yield ˙ cos φ u = r˙ sin φ + r θ˙ cos φ = r˙ sin φ + r(φ˙ + ψ)

(5.113)

which substituted into Eq. (5.112) gives the following result:  d  w + r ψ˙ sin φ = 0 dt

(5.114)

w + r ψ˙ sin φ = A = const.

(5.115)

or,

Equations (5.110) and (5.115) are integrated to obtain Lawden’s spiral. The time can be eliminated from these two differential equations, and instead φ can be used as the independent variable. From Eqs. (5.91) and (5.92), we have w=−

d (r cos φ) + r ψ˙ sin φ dt

(5.116)

which, substituted into Eq. (5.115), results in A − 2w =

d (r cos φ) dt

(5.117)

This equation can be rewritten as follows, where the prime denotes the derivative relative to φ: (A − 2w)t =

d (r cos φ) dφ

(5.118)

178

5 Optimal Maneuvers with Bounded Inputs

which can be integrated to yield φ as a function of time. Similarly, Eq. (5.115) yields ψ =

A−w

t r sin φ

(5.119)

which is integrated to give the shape of the IT arc, with A being a constant shape parameter. Another useful information about the trajectory is derived by eliminating ψ from Eqs. (5.110) and (5.115), while using Eq. (5.104), which yields the following expressions for w and r in terms of the independent variable φ:   μ w(A − w) = Cr − sin φ sin φ r

(5.120)

and w 2 (1 − 3 sin2 φ) =

2 μ r  Cr − sin φ μ r

(5.121)

An integration of the parametric differential equations, Eqs. (5.118) and (5.119), is readily performed for the case where the terminal time, tf , is unspecified. In such a case of open control interval, tf − t0 , we are free to select C = 0 in Eq. (5.110), which results in the following relationships from Eqs. (5.120) and (5.121) [52]: r=

as 6 1 − 3s 2

(5.122)

and 1 w=± 2 s

-

μ a

(5.123)

where s = sin φ and a > 0 is a characteristic length. From Eq. (5.118), it follows that  a 3 s 7 (5s 2 − 3)

t =± (5.124) μ (1 − 3s 2 )2 which is integrated to obtain ±t

c 1 μ 1 31 5 = − c5 + c3 − c − 9 81 9 81 (3c2 − 2) a3 0 √ √ 1 c 3− 2 2 2 ln √ − + const. √ 81 3 c 3+ 2

(5.125)

5.6 Lawden’s Spiral

179

. where c = cos φ. Equations (5.119) and (5.123) yield ψ =

3 −5 s2

(5.126)

which is integrated to obtain ψ = −3 cot φ − 5φ + const.

(5.127)

and from Eq. (5.92) it follows that θ = θ0 − 3 cot φ − 4φ

(5.128)

where θ0 is a constant. Equations (5.122)–(5.128) represent a spiral with a monotonically increasing radius from 0 to ∞ as the angle φ is increased from 0 to √ α = sin−1 (1/ 3). In the same range of φ, the angle θ increases from −∞ to θ0 − 4α − 3/2. The sign ambiguity in Eqs. (5.123) and (5.124) is resolved by always having a positive value of the angular rate, θ˙ , which requires a negative sign when sin φ > 0, and a positive sign when sin φ < 0. To investigate the optimality of Lawden’s spiral, we substitute Eq. (5.110) into the expression for the Hamiltonian:  u w cos φ + sin φ H = λr (u sin φ − w cos φ) + λθ r r     2μ μ βve + λu C − 2 sin φ + + λw −uψ˙ + 2 cos φ m r r   π − λm β + ν θ − φ − ψ + 2

(5.129)

The derivative relative to exhaust rate is then the following: Hβ =

ve ∂H = λu − λm ∂β m

(5.130)

where the costate equations are the following: ∂H ∂r ∂H =− ∂θ ∂H =− ∂u ∂H =− ∂w

2μ (λu sin φ − λw cos φ) r3

λ˙ r = −

=−

λ˙ θ

= −ν

λ˙ u λ˙ w

λθ cos φ r λθ = −λu ψ˙ + λr cos φ − sin φ r

= λw ψ˙ − λr sin φ −

(5.131)

180

5 Optimal Maneuvers with Bounded Inputs

λ˙ m = −

βve ∂H = λu 2 ∂m m

The second-order time derivative of Hβ is now derived as follows: H˙ β =

dHβ ve βve ve = λ˙ u + λu 2 − λ˙ m = λ˙ u dt m m m

(5.132)

and H¨ β =

d2 Hβ ve βve = λ¨ u + λ˙ u 2 m dt 2 m

(5.133)

The necessary condition of optimality is given by Theorem 2.11 to be the following: ve ∂ ¨ Hβ = λ˙ u 2 ≤ 0 ∂β m

(5.134)

for all t ∈ [t0 , tf ]. This condition is violated along Lawden’s spiral, because ve > 0, m > 0 for all t ∈ [t0 , tf ], and λ˙ u > 0 can be chosen to satisfy the boundary conditions given by Eq. (5.93), thereby making the inequality of Eq. (5.134) strictly positive. Hence, an intermediate-thrust arc like Lawden’s spiral does not minimize the propellant consumption, and is therefore not an optimal trajectory. Bell and Jacobson [12] arrive at the same result via a second-order variation of J along Lawden’s spiral.

5.7 Powered Arcs While null-thrust arcs can be derived in a closed form, the maximum-thrust (MT) and intermediate-thrust (IT) arcs require numerical integration of the two-point boundary value problem (2PBVP). Example 5.1 Consider an escape trajectory consisting of null-thrust (NT) and maximum-thrust (MT) arcs, with a constant specific-impulse engine of maximum acceleration am = 1.0 m/s2 . The transfer is made beginning with the geocentric position 8000I km and inertial velocity 1.8269J + 6.8182K km/s, measured in the geocentric, right-handed inertial reference frame, (I, J, K), with K being the polar axis, and I pointed at the initial launch point. Since there is no terminal boundary constraint in this case, the engine is switched on and off at arbitrary intervals, while the primer vector evolves from a given initial condition to maintain a constant thrust direction. The simulation is carried out by a Runge–Kutta solver for 15,000 s of flight, and the resulting trajectory is plotted in Fig. 5.1, with the velocity shown in Fig. 5.2. The abrupt changes in the acceleration caused by switching between NT and MT arcs are evident in the velocity plot.

5.7 Powered Arcs

181 4

x 10 1

Null−thrust arc Max−thrust arc

Z (km)

0.5 0 −0.5 −1 −1.5 2000

3

0

2 1

−2000

4

x 10

0

Y (km)

−4000

−1

X (km)

Fig. 5.1 The escape trajectory consisting of null-thrust and maximum-thrust arcs of Example 5.1 Null−thrust arc Max−thrust arc

10

Z˙ (km/s)

5

0

−5

−10 2 10

0

5 0

−2 −5

Y˙ (km/s)

−4

−10

X˙ (km/s)

Fig. 5.2 The velocity plot for the escape trajectory of Example 5.1

Example 5.2 Consider a variation of the escape trajectory of Example 5.1 wherein ˙ an initial primer vector of p(0) = −10−6 (3, 5, 8)T and p(0) = 0 are specified, signifying a fixed terminal velocity constraint. The optimal trajectory with the same initial state and engine as in Example 5.1 is simulated for 12,000 s, and plotted in Fig. 5.3, with the velocity plot shown in Fig. 5.4. The circular orbit is seen to be preserved during the null-thrust arc, and changes when the thrust is applied. The

182

5 Optimal Maneuvers with Bounded Inputs 4

x 10 1 0.5

Z (km)

0 −0.5 −1 −1.5 4000 2000 1

0

0.5 0

−2000

Y (km)

4

−0.5

−4000

−1

x 10

X (km)

Fig. 5.3 The optimal escape trajectory consisting of a null-thrust, an intermediate-thrust, and a maximum-thrust arc for Example 5.2 10

Z˙ (km/s)

5

0

−5

−10 2 10

0

5 0

−2 −5

Y˙ (km/s)

−4

−10

X˙ (km/s)

Fig. 5.4 The velocity plot for the optimal escape trajectory of Example 5.2

evolution of the primer vector magnitude, p, is seen to be monotonically increasing in Fig. 5.5, thereby yielding a single switching point from NT to MT arc at around 9600 s into the flight. The switching involves an intermediate-thrust (IT) arc with a = am /2, resulting in a much smoother velocity plot when compared to that of Example 5.1. The optimal thrust direction is seen to be constant throughout the

5.7 Powered Arcs

183

2

p

1.5

1

0.5

0

8600

8800

9000

9200

9400

9600

9800

10000

t (s) Fig. 5.5 The monotonically increasing p of the optimal escape trajectory of Example 5.2, indicating the switch from null thrust to powered flight regime at p = 1 40

30

φ = sin −1(px /p) ψ = tan−1(py /pz )

(deg.)

20

10

0

−10

−20 0

2000

4000

6000

8000

10000

12000

t (s) Fig. 5.6 The optimal thrust direction indicated by the angles φ and ψ for the optimal escape trajectory of Example 5.2

trajectory in Fig. 5.6, but is used only in the powered portion of the flight. The radius and speed are plotted in Fig. 5.7, and show the escape nature of the powered trajectory, evident in a steeply increasing radius and a burst in the speed as the engine is switched on at 9600 s point.

184

5 Optimal Maneuvers with Bounded Inputs 12000

r (km)

11000 10000 9000 8000 7000

0

2000

4000

0

2000

4000

6000

8000

10000

12000

6000

8000

10000

12000

8

v (km/s)

7.8 7.6 7.4 7.2 7

t (s) Fig. 5.7 The radius and speed profiles for the optimal escape trajectory of Example 5.2

Example 5.3 In the previous two examples, the terminal constraints were not explicitly given, and instead an initial primer vector was specified. Now consider the planar transfer of a spacecraft around a spherical Earth from an initial position, r(0) = (6000, 4000)T km, and velocity v(0) = (6, 2)T km/s, to a final position of r(tf ) = (7000, 2000)T km by the application of constant specific-impulse engine with the constant acceleration bounds, 0 ≤ a ≤ am . The results of this 2PBVP— computed by a collocation technique [81]—are plotted for am = 0.1, 1.0, and 10 m/s2 , respectively, in Figs. 5.8 and 5.9. Figure 5.8 plots the inertial planar trajectory, r = (x, y)T , for the three chosen values of am . The increasing terminal positional error with the decreasing upper bound of the acceleration magnitude can be clearly observed. Figure 5.9 shows the optimal thrust direction, n = p/p, as the inertial angle made by the primer vector, p = (px , py )T , with the x-axis, tan−1 (py /px ), as a function of time. The trajectory in each case is revealed in this figure to be more than 99% the maximum-thrust (MT) arc, with a = am , with the remaining less than (final) 1% being the null-thrust (NT) arc (p < 1). The value of the terminal time, tf , is seen to increase as the thrust acceleration magnitude is reduced. For the highly underactuated case of am = 0.1 m/s2 , the largest terminal positional error as well as the largest terminal time is observed. The optimal speed profile is plotted in Fig. 5.10. Example 5.4 Let us now consider a three-dimensional transfer for escape from a circular Earth orbit of radius, 42,000 km, inclination 75◦ , when passing the equator with initial velocity, v(0) = 0.7973J + 2.9757K km/s, to the point rf = 51,336I +

5.7 Powered Arcs

185

5000

4500

y (km)

4000

3500

3000

2500

am = 0.1 m/s2 am = 1.0 m/s2 am = 10 m/s2

2000

1500 6000

6500

7000

7500

8000

8500

9000

9500

10000

10500

x (km)

Fig. 5.8 The optimal trajectory of the spacecraft for the interception problem of Example 5.3 60 40

tan−1(py /px )

20 0

am = 0.1 m/s2 am = 1.0 m/s2 am = 10 m/s2

−20 −40 −60 −80 −100 0

500

1000

1500

2000

2500

3000

t (s) Fig. 5.9 The optimal thrust direction of the spacecraft for the interception problem of Example 5.3

8038J + 30,000K km, measured in the geocentric, right-handed inertial reference frame, (I, J, K), with K being the polar axis, and I pointed at the initial launch point. The transfer is made using a constant specific-impulse engine of maximum acceleration am = 1.0 m/s2 . The time of transfer is selected to be 6051 s such

186

5 Optimal Maneuvers with Bounded Inputs 8

am = 0.1 m/s2 am = 1.0 m/s2 am = 10 m/s2

7 6

v (km/s)

5 4 3 2 1 0

0

500

1000

1500

t (s)

2000

2500

3000

Fig. 5.10 The optimal speed of the spacecraft for the interception problem of Example 5.3

that only powered flight is required. The trajectory includes an intermediate-thrust (IT) arc, for which a = am /2 is used. As shown earlier, the IT arc is a suboptimal solution, but it extends the flight time for reducing the terminal positional error, when compared to a null-thrust arc. The performance index for minimization is the following: J = c(tf ) + φ[x(tf ), tf ] ,

φ=

1 [r(tf ) − rf ]2 2

where φ[x(tf ), tf ] =

1 [r(tf ) − rf ]T [r(tf ) − rf ] 2

The boundary conditions are thus the following: r(0) = (42,000, 0, 0)T (km) ,  λr (tf ) =

∂φ ∂r

v(0) = (0, 0.7973, 2.9757)T km/s

T = r(tf ) − rf ,

p(tf ) = 0

t=tf

The solution to the 2PBVP is obtained using a collocation method [81], and the results are plotted in Figs. 5.11, 5.12, 5.13, and 5.14, showing an almost linear

5.7 Powered Arcs

187

4

x 10

Z (km)

3 2 1

0 10000 5.2

8000 5

6000 4.8

4000

4

x 10

4.6 2000

4.4

Y (km)

0

X (km)

4.2

Fig. 5.11 The trajectory of the spacecraft for the three-dimensional transfer problem of Example 5.4 45

φ = sin −1(py /p) 40

ψ = tan−1(pz /p)

35

(deg.)

30 25 20 15 10 5 0 0

1000

2000

3000

t (s)

4000

5000

6000

7000

Fig. 5.12 The optimal thrust direction of the spacecraft for the three-dimensional transfer problem of Example 5.4

increase in the flight speed as the optimal thrust direction (indicated by the angles φ = 45◦ and ψ = 15◦ ) is achieved and maintained throughout the flight. The trajectory culminates with an IT arc covering the final 60 s of flight until the desired position is reached with a hyperbolic escape velocity of 7.802 km/s.

188

5 Optimal Maneuvers with Bounded Inputs 60 58 56

r(×1000 km)

54 52 50 48 46 44 42 0

1000

2000

3000

4000

5000

6000

7000

t (s) Fig. 5.13 The optimal radius of the spacecraft for the interception problem of Example 5.4 8 7.5 7

v (km/s)

6.5 6 5.5 5 4.5 4 3.5 3 0

1000

2000

3000

4000

5000

6000

7000

t (s) Fig. 5.14 The optimal speed of the spacecraft for the interception problem of Example 5.4

The last two examples demonstrate the determination of a fuel-optimal trajectory subject to the specific terminal boundary conditions (also referred to as the transversality conditions). These conditions often cause convergence problems in any numerical scheme used to solve the associated 2PBVP, because of the arbitrary

5.8 Linearization Relative to a Circular Orbit

189

nature of the costate boundary conditions. Since costate variables have no physical meaning, they are generally chosen by trial and error to achieve convergence for a given final time, tf . Resolving such convergence issues in either a shooting or collocation technique is a topic of active research, such as the paper by Minter and Fuller-Rowell [61] who propose setting the initial conditions of the adjoint variables from an analytic solution of a simplified set of differential equations, and that by Pan et al. [65], who propose a list of reduced transversality conditions based upon the classical orbital elements.

5.8 Linearization Relative to a Circular Orbit The optimal transfer relative to a circular orbit in an inverse-square gravity field can be addressed using the linearized Hill–Clohessy–Wiltshire (HCW) model of approximate relative motion (see Chap. 4). Consider the HCW equations of a constant specific impulse powered spacecraft relative . to a point (target) moving in a circular orbit of radius R with orbital frequency, ω = μ/R 3 , expressed as follows: x¨ + 2ωy˙ = anx y¨ − 2ωx˙ − 3ω2 y = any

(5.135)

z¨ + ω2 z = anz Here x = Rδθ is the small in-track displacement, y = δr the small radial displacement, and z = Rδα the small out-of-plane displacement, all measured from the target as the origin of the HCW frame, and n = (nx , ny , ny )T is the thrust direction unit vector, a = an

(5.136)

with the thrust acceleration magnitude, a =| a |, is bounded by the maximum possible acceleration magnitude, am = Tm /m: 0 ≤ a(t) ≤ am

(5.137)

Assuming a negligible change in the vehicle’s mass during the low-thrust maneuver, the upper bound on the acceleration magnitude, am , can be taken to be constant, thereby simplifying the control problem. In a state-space form, Eq. (5.135) can be expressed as follows: r˙ = v v˙ = Ar + Bv + an

(5.138)

190

5 Optimal Maneuvers with Bounded Inputs

where r = (x, y, z)T and v = (x, ˙ y, ˙ z˙ )T , and ⎞ 0 −2ω 0 B = ⎝ 2ω 0 0 ⎠ 0 0 0

⎞ 0 0 0 A = ⎝ 0 3ω2 0 ⎠ , 0 0 −ω2

⎛

⎛

The problem of a fuel-optimal transfer from an initial state, [r(0), v(0)], to a final state [r(tf ), v(tf )], in a specified time, tf , can be stated as the minimization of 

tf

J = c(tf ) =

a(t)dt

(5.139)

0

subject to the constraints of Eqs. (5.137) and (5.138), resulting in the following Hamiltonian: H = a + λTr v + λTv (Ar + Bv + an)

(5.140)

The application of the primer vector method yields the optimal thrust direction to be the following: nˆ =

p , p

(a = 0)

(5.141)

ˆ where p = −λv is the primer vector. If a = 0, then the optimal thrust direction, n, is indeterminate. The Hamiltonian is now expressed in terms of the primer vector as follows: H = a(1 − p) + p˙ T v − pT (Ar + Bv)

(5.142)

Due to the linearity of H relative to a, the optimal control problem is singular in acceleration magnitude. However, the constant bounds on a given by Eq. (5.137) allow the application of the Pontryagin’s minimum principle, resulting in the following switching condition for the optimal control magnitude: & am , (p − 1) > 0 a(t) ˆ = (5.143) 0 , (p − 1) < 0 The costate equations produce the primer vector differential equation as follows: λ˙ r = AT p

(5.144)

p˙ = λr − BT p

(5.145)

p¨ + BT p˙ − AT p = 0

(5.146)

which yields

5.8 Linearization Relative to a Circular Orbit

191

or, considering the symmetric and skew-symmetric nature of A and B, respectively, we have p¨ = Ap + Bp˙

(5.147)

which is the same as the second equation of Eq. (5.138) with a = 0. Hence both the primer vector and the HCW model have the same state transition matrix. The boundary conditions for the primer vector and its time derivative for a transfer with both the end conditions fixed and the terminal time specified are completely arbitrary. They could be expressed as follows: p(0) = p0 ,

˙ p(0) = p˙ 0

(5.148)

and must be chosen for integrating Eq. (5.146) in order to determine the primer vector, p(t), 0 ≤ t ≤ tf , thus the optimal thrust direction and magnitude. The arbitrary nature of the primer vector indicates that infinitely many possible fuel-optimal solutions exist for transferring the spacecraft from an initial state, [r(0), v(0)], to a final state [r(tf ), v(tf )], in a specified time, tf . Such a degeneracy of the singular problem with hard end-point constraints has been explored by Carter [21] using the linearized HCW model, and by Marec [57] using the linearized orbital elements formulation. The degeneracy of the optimal solutions is confined to the linearized problem with specific position and velocity boundary conditions, but the original (nonlinear) transfer problem discussed earlier in this chapter has a unique optimal solution. In order to derive a unique solution, the orbital transfer problem relative to a circular orbit can be posed as either an interception or a rendezvous problem with soft end-point constraints. For this purpose, a terminal cost function is added to the performance index as follows: J = c(tf ) + φ[x(tf ), tf ]

(5.149)

where 1 1 Qr [r(tf ) − rf ]T [r(tf ) − rf ] + Qv [v(tf ) − vf ]T [v(tf ) − vf ] 2 2 (5.150) where rf and vf are the target point’s position and velocity, respectively, at a specific terminal time tf , and the weights Qr > 0, Qv > 0 can be chosen as desired in comparison with the fuel cost, c(tf ). For example, an interception problem can have Qv = 0, while a rendezvous would require both the weights to be selected appropriately. The terminal conditions for the costate vectors can then be obtained as follows: φ[x(tf ), tf ] =

 λr (tf ) =

∂φ ∂r

T = Qr [r(tf ) − rf ] t=tf

192

5 Optimal Maneuvers with Bounded Inputs

 p(tf ) = −

∂φ ∂v

T = −Qv [v(tf ) − vf ]

(5.151)

t=tf

Since the primer-vector differential equation, Eq. (5.146), is linear, it has a closed-form solution, and can be expressed in terms of the individual elements of the vector, p = (px , py , pz )T . We begin by writing Eq. (5.146) in terms of its scalar components as follows: p¨x + 2ωp˙ y = 0 p¨y − 2ωp˙ x − 3ω2 py = 0

(5.152)

p¨z + ω2 pz = 0 which have the same form as the unforced HCW equations, Eq. (5.146), and whose general solution is the following: px (t) = 2A1 sin(ωt + ϕ) + 3c1 t + c2 py (t) = A1 cos(ωt + ϕ) + 2c1 /ω

(5.153)

pz (t) = A2 sin(ωt + ϕ) + A3 cos(ωt + ϕ) The coefficients, A1, A2 , A3 , c1 , c2 , as well as the phase angle, ϕ, are determined from the boundary conditions, Eq. (5.151), which also require the state solution at the terminal time, r(tf ), v(tf ). Thus the iterative nature of the 2PBVP solution is revealed. Another way of expressing the primer vector solution is as follows: ˙ p(t) = rr (t)p(0) + rv (t)p(0)

(5.154)

where rr , rv are the following transition matrices of the unforced HCW problem: ⎤ 1 6(sin ωt − ωt) 0 rr (t) = ⎣ 0 4 − 3 cos ωt 0 ⎦ 0 0 cos ωt ⎡

⎡ rv (t) =

4 2 ω sin ωt − 3t ω (cos ωt − 1) 1 ⎣ − 2 (cos ωt − 1) ω ω sin ωt

0

0

1 ω

⎤ 0 ⎦ 0 sin ωt

(5.155)

(5.156)

˙ The initial vectors, p(0), p(0) are to be determined from the terminal boundary conditions, Eq. (5.151), as well as the state solution at t = tf .

5.8 Linearization Relative to a Circular Orbit

193

Recall that both the primer vector dynamics and the HCW model have the same state transition matrix, expressed by  (t) =

rr (t) rv (t) ˙ rv (t) ˙ rr (t)  

 (5.157)

where ⎤ 0 ω(6 cos ωt − 1) 0 ˙ rr (t) = ⎣ 0 ⎦  3ω sin ωt 0 0 0 −ω sin ωt ⎡

⎡

4 cos ωt − 3 −2 sin ωt cos ωt ˙ rv (t) = ⎢  ⎣ 2 sin ωt

0 0

0

cos ωt

0

(5.158)

⎤ ⎥ ⎦

(5.159)

the respective solutions can be written as follows: 



r(t) v(t)



p(t) ˙ p(t)



 = (t − tf )

p(tf ) ˙ f) p(t

 ,

(0 ≤ t ≤ tf )

& /   t 0 r(0) a(τ ˆ )(t − τ ) p(τ ) dτ , = (t) + v(0) 0 p(τ )

(5.160)



(0 ≤ t ≤ tf ) (5.161)

or 

r(t) v(t)



 = (t)

    t  (t − τ ) p(τ ) r(0) dτ , a(τ ˆ ) ˙ rv + rv (t − τ ) p(τ ) v(0) 0

(0 ≤ t ≤ tf )

(5.162) Equations (5.160) and (5.162) are to be solved simultaneously, given the bound˙ f ), as well as the switching condition, ary conditions, r(0), v(0), p(tf ), and p(t Eq. (5.143). Since the HCW model has decoupled out-of-plane (z) and coplanar (x, y) dynamics, it is possible to solve the two problems independently of each other. However, a three-dimensional rendezvous would require that both the out-of-plane and the coplanar displacement and velocity components should vanish simultaneously (that is, for the same value of tf ).

5.8.1 Out-of-Plane Rendezvous In order to illustrate the procedure of solving the two-point boundary value problem (2PBVP) associated with the linearized HCW model, consider a rendezvous

194

5 Optimal Maneuvers with Bounded Inputs

problem in which the spacecraft has the same circular orbit as the target, but is displaced in plane by an initial displacement, z(0), and velocity, z˙ (0). The rendezvous must be made with the minimum fuel in time tf , using a constant specific-impulse engine of maximum (constant) acceleration, am . The linearized scalar equation of out-of-plane motion is simply the following: z¨ + ω2 z = anz

(5.163)

with nˆ z = 1 being the optimal thrust direction. The switching condition, Eq. (5.143), is governed by the primer vector magnitude, p(t), obeying the following differential equation: p¨ + ω2 p = 0

(5.164)

p(t) = c1 cos ωt + c2 sin ωt

(5.165)

whose solution is given by

where the coefficients c1 , c2 are determined from the terminal boundary condition: p(t ˙ f ) = Qr z(tf ) p(tf ) = −Qv z˙ (tf )

(5.166)

Since p(t) is oscillatory, it can produce infinitely many switching points, p(ts ) = 1. Hence, the rendezvous problem (i.e., z(tf ) = z˙ (tf ) = 0) can have infinitely many solutions, depending upon the acceleration bound, am . To see this fact, the solution for the out-of-plane displacement is expressed as follows: 

z(t) z˙ (t)





  1 sin ωt cos ωt z(0) ω = −ω sin ωt cos ωt z˙ (0)     t cos ω(t − τ ) ω1 sin ω(t − τ ) 0 a(τ ˆ ) dτ (5.167) + 1 −ω sin ω(t − τ ) cos ω(t − τ ) 0

where 0 ≤ t ≤ tf . To apply the terminal boundary condition, we must have 

p(tf ) p(t ˙ f)



 =

0 −Qv Qr 0



z(tf ) z˙ (tf )

 (5.168)

Suppose it were possible to make the rendezvous with only one switching. Then the trajectory must begin with an MT arc, aˆ = am , for which p(0) = c1 > 1, and the switching point, t = ts ≤ tf , should satisfy c1 cos ωts + c2 sin ωts = 1

(5.169)

5.8 Linearization Relative to a Circular Orbit

195

The displacement and velocity at the switching point are then obtained from Eq. (5.167) as follows: 

z(ts ) z˙ (ts )



 =

1 ω

sin ωts cos ωts

cos ωts −ω sin ωts



   ts  1 z(0) ω sin ω(t − τ ) dτ + am z˙ (0) cos ω(t − τ ) 0 (5.170)

or 

z(ts ) z˙ (ts )

&

 =

am z(0) cos ωts + z˙ (0) ω sin ωts + ω2 (1 − cos ωts ) −ωz(0) sin ωts + z˙ (0) cos ωts + aωm sin ωts

/ (0 ≤ ts ≤ tf )

,

(5.171) This is the initial condition for the NT arc, for which it should be true that 

z(t) z˙ (t)



 =

cos ωt −ω sin ωt

1 ω

sin ωt cos ωt



z(ts ) z˙ (ts )

 ,

(ts ≤ t ≤ tf )

(5.172)

By putting t = tf in Eq. (5.172), it may be verified that a perfect rendezvous (z(tf ) = z˙ (tf ) = 0) requires that both z(ts ) and z˙ (ts ) should be made to vanish for an arbitrary initial condition, z(0), z˙ (0), which implies the rendezvous could take place at both t = ts and at t = tf . If this condition were to be satisfied, it would mean that an NT arc is unnecessary, and an MT arc is itself sufficient to make the rendezvous. Since the relative displacement and velocity both vanish at t = ts as well as at t = tf (that is, at either end of the NT arc), it follows that tf = ts + 2π ω . Therefore, we can replace ts with tf in Eqs. (5.169) and (5.171), resulting in the following: c1 cos ωtf + c2 sin ωtf = 1 

z(tf ) z˙ (tf )

&

 =

am z(0) cos ωtf + z˙ (0) ω sin ωtf + ω2 (1 − cos ωtf ) −ωz(0) sin ωtf + z˙ (0) cos ωtf + aωm sin ωtf

(5.173) / (5.174)

or, sin ωtf =

ω˙z(0)(1 − c1 ) c1 ω2 z(0) + c2 ω˙z(0)

(5.175)

Given z(0), z˙ (0), the coefficients c2 and c1 > 1, and the terminal time, tf , should satisfy Eq. (5.175), as well as the requirement  z(0) cos ωtf + z˙ (0) ω sin ωtf 2 >0 (5.176) am = ω cos ωtf − 1 The last equation implies that the terminal time for exact rendezvous in one MT arc is dependent upon the upper bound of the acceleration input, and can have many solutions. A similar result can be obtained for multiple switching points.

196

5 Optimal Maneuvers with Bounded Inputs

For deriving a unique solution for the terminal time, tf , for rendezvous, the soft terminal condition of Eq. (5.168) should be applied, which requires that 

−c1 ω sin ωtf +c2 ω cos ωtf c1 cos ωtf +c2 sin ωtf





 ωQv sin ωtf −Qv cos ωtf (5.177) Qr Qr cos ωtf ω sin ωtf & / am z(0) cos ωts + z˙ (0) sin ωt + (1 − cos ωt ) s s 2 ω ω × −ωz(0) sin ωts + z˙ (0) cos ωts + aωm sin ωts =

Given tf , Qr , and Qv , Eqs. (5.169) and (5.177) are to be solved for c1 , c2 , and ts . However, such a rendezvous is not exact because the displacement and velocity do not vanish simultaneously for any value of tf .

5.8.2 Coplanar Rendezvous The approach considered above for out-of-plane rendezvous can also be applied to the coplanar problem. The state transition matrix for the coplanar HCW model is the following: ⎛

4 1 6(sin ωt − ωt) ω sin ωt − 3t ⎜ 0 4 − 3 cos ωt − 2 (cos ωt − 1) ω (t) = ⎜ ⎝ 0 ω(6 cos ωt − 1) 4 cos ωt − 3 0 3ω sin ωt 2 sin ωt

⎞ 2 ω (cos ωt − 1) 1 ⎟ ω sin ωt ⎟ −2 sin ωt cos ωt

⎠

(5.178)

Let the optimal trajectory begin with an MT arc which, by definition, extends until the first switching time, t = ts ≤ tf , given by p2 (ts ) = px2 (ts ) + py2 (ts ) = 1

(5.179)

The displacement and velocity at the switching point are then obtained to be the following: 

r(ts ) v(ts )





 r(0) = (ts ) v(0) ⎡4 sin ω(t−τ ) − 3(t − τ )  ts ω 2 ⎢ − [cos ω(t − τ ) − 1] ω ⎢ + am ⎣ 4 cos ω(t − τ ) − 3 0 2 sin ω(t − τ )

(5.180) ⎤ 2 ω [cos ω(t−τ ) − 1] & p (τ ) / x 1 ⎥ p(τ ω sin ω(t−τ ) ⎥ p (τ)) dτ y −2 sin ω(t−τ ) ⎦ p(τ ) cos ω(t−τ )

5.8 Linearization Relative to a Circular Orbit

197

with the primer vector elements evolving as follows: 

p(t) ˙ p(t)



 = (t − tf )

p(tf ) ˙ f) p(t

 ,

(0 ≤ t ≤ tf )

(5.181)

from the terminal boundary condition, ˙ f ) = Qr r(tf ) p(t p(tf ) = −Qv v(tf )

(5.182)

This completes the formulation needed to solve the optimal coplanar rendezvous problem with the linearized HCW model.

Exercises 5.1 Formulate a two-point boundary value problem using only the radial thrust acceleration bounded by | ar |≤ c for an open time transfer from an initial circular orbit of radius, r1 , around a spherical body, to a coplanar circular orbit of radius, r2 . 5.2 Find the time-optimal input profile for a plane change maneuver by angle α0 in a circular orbit of radius C, if | an |≤ 1 m/s2 . 5.3 A solar-sail spacecraft is to be transferred from an initial circular orbit of radius, r(0) = r1 , around the Sun to a coplanar circular orbit of radius r, by applying a solar-radiation acceleration of small magnitude, a = c/r 2 , where c  μ is a constant. The direction of the input acceleration from the radial direction is given by the angle β(t), and can be adjusted by rotating the sail in the limits, −bπ/2 ≤ β(t) ≤ bπ/2, where 0 < b ≤ 1 is constant. Formulate the boundary value problem to be solved such that r 2 (tf ) is maximized in a fixed time, tf , relative to β(t). 5.4 Consider an orbital regulation system of a satellite in a circular orbit of radius C and frequency n around a spherical planet, with low-thrust radial acceleration control input, |ar (t)| ≤ Cn2 , and initial perturbations, δr(0) = 0.01C, δ r˙ (0) = 0, δθ (0) = 0. Derive the time-optimal control history and trajectory for driving the final radial response, δr(tf ), δ r˙ (tf ), to zero at an unspecified time tf . 5.5 A satellite in a circular orbit of radius C and frequency n around a planet is hit by space debris, resulting in an initial out-of-plane velocity of Δ˙z(0) = 0.01nC. For restoring the original orbital plane, a normal acceleration input is applied by the rocket engine, which is limited by |an (t)| ≤ Cn2 Calculate the time-optimal trajectory until the first switching time.

198

5 Optimal Maneuvers with Bounded Inputs

5.6 Solve the optimal coplanar transfer problem using only the input, uθ , between two given positions, (r1 , θ1 ) and (r2 , θ2 ), with a free final time, tf , and an energyoptimal profile, if the magnitude of the input is limited by | uθ |≤ 1 m/s2 . (Hint: Resolve the flight path along radial and tangential directions, and assume that due to the small thrust acceleration, the change in flight path angle is negligible.) 5.7 Formulate the optimal control problem for the constant upper limit of acceleration input magnitude, am , for the minimization of 

tf

J =

a(t)dt 0

subject to the dynamic constraints of Eqs. (5.1)–(5.2), but without taking c(t) as a state variable. Show that the same switching condition is obtained for the thrust direction as given by Eq. (5.43).

Chapter 6

Flight in Non-spherical Gravity Fields

6.1 Introduction Spaceflight involving orbital transfers around irregularly shaped bodies or in the gravity field of several large bodies is fundamentally different from the flight in the gravity field of a single spherical body, which was covered in the previous chapters. The primary reason for this difference is that the spacecraft is no longer in a time-invariant gravity field of the two-body problem, but instead encounters a time-dependent field due to the relative motion of the multiple large bodies with respect to one another, or due to the changing position of the spacecraft relative to a rotating, non-spherical body. A general description of such a flight is by the following governing equation: r¨ = g[r(t), t] + a

(6.1)

where r is the instantaneous position vector of the spacecraft in an inertial frame, g is the time-varying acceleration due to gravity caused either by an irregularly shaped body or by multiple large bodies in mutual relative motion, and a is the acceleration input applied by the engines. The determination of an optimal acceleration input profile, a(t), in a specified control interval while minimizing a given objective forms the crux of the spaceflight navigation problem. For example, an orbital transfer between two given positions minimizing a cost function,  tf J = φ[r(tf ), v(tf ), tf ] + aT adt , (6.2) 0

subject to the dynamics of Eq. (6.1), requires the solution of the unbounded acceleration input profile, a, from the following Euler–Lagrange equations (Chap. 2): ∂H = λTv g + aT = 0 ∂a © Springer Nature Switzerland AG 2019 A. Tewari, Optimal Space Flight Navigation, Control Engineering, https://doi.org/10.1007/978-3-030-03789-5_6

(6.3)

199

200

6 Flight in Non-spherical Gravity Fields

   T ∂H T ∂g =− λv λ˙ r = − ∂r ∂r

(6.4)

T  ˙λv = − ∂H = −λr ∂v

(6.5)

  ∂φ H+ =0 ∂t t=tf

(6.6)

where the Hamiltonian is given by H = λTr v + λTv (g + a) + aT a

(6.7)

and appropriate boundary conditions are specified for r, v, λr , and λv . Modeling a non-spherical gravity field thus requires deriving expressions for the acceleration due to gravity, g, and the gravity-gradient matrix, ∂g/r, both of which are time varying at any given location, r, in the inertial space. Such a time-varying gravity model would be cumbersome to employ in the solution of an optimal control problem. However, a great simplification is possible if the gravity field is rendered time invariant by transforming the equations of orbital motion, Eq. (6.1), to an appropriate frame rotating at a constant angular velocity. The essence of a nonspherical gravity model is to find such a coordinate frame where both an orbital stability analysis and an optimal control solution can be conveniently carried out. A prime example of a non-spherical gravity field is the one created by multiple bodies influencing the flight of a spacecraft. In such a case, the gravitational acceleration, g, experienced by the spacecraft—regarded as a test mass—at any instant depends upon the orbital dynamics of the primary bodies (e.g., the Sun and a planet, or a planet and its Moon). The simplest multi-body gravity model is that of spaceflight in the vicinity of two spherical bodies orbiting each other with a constant angular velocity, and is referred to as the restricted three-body problem (R3B) where the spacecraft’s mass is neglected in comparison to the masses of the two primaries. A special case of the R3B model is that of the circular orbit of the two primaries about a common barycenter, and is called the circular restricted three-body problem (CR3BP). The derivation of a time-invariant CR3BP gravity model is carried out in a rotating frame attached to the two primary bodies, which give rise to equilibrium points of the CR3BP model in the rotating frame. The equilibrium points can be utilized for stability analysis of the restricted three-body orbits. An orbital transfer problem can then be posed as an optimal transfer trajectory between two (or more) equilibrium points of the transformed system. Similarly, an optimal control strategy can be devised for stationing the spacecraft close to an equilibrium point in the rotating coordinate frame. The gravity field of an irregularly shaped body (such as an asteroid) rotating uniformly about a body axis is carried out in a body-fixed frame, which renders the gravity field time invariant. A series approximation for the gravitational potential

6.2 Gravity Field of a Non-spherical Body

201

is then carried out using the transformed spatial coordinates, and coefficients are determined by fitting with the observational data obtained by an orbiting spacecraft equipped with accelerometers. Either a spherical harmonics expansion or a finiteelement model (such as a polyhedron model using cubic elements) can be employed for deriving the gravitational potential in the rotating frame, whose gradients yield time-invariant g and ∂g/r. As in the case of the CR3BP model, the equilibrium points in the rotating frame are used for stability analysis of an orbit in the vicinity of the body. An orbital transfer around the irregularly shaped body can be posed as an optimal transfer trajectory between two (or more) equilibrium points in the rotating frame. Another optimal control strategy can be devised for maintaining desired orbit around the body. This chapter introduces the essential methodology necessary to model the nonspherical gravity fields for application to optimal spaceflight trajectories. Due to the advanced nature of these topics, it would be beyond the present scope to go deeply into their modeling details. Instead, the reader is referred to specialized textbooks and research articles which cover those topics in detail, and focus is maintained on the optimal spaceflight applications.

6.2 Gravity Field of a Non-spherical Body The gravity field produced by any heavenly body is non-spherical due to its irregular shape and rotation. The larger bodies such as the planets and large moons, due to their gravitational mass accretion, display a nearly axisymmetric mass distribution with only a slight longitudinal dependence. On the other hand, smaller bodies such as the asteroids and comets depart from any regularity in their mass distribution, thereby producing a highly uncertain gravity field at any given point. Modeling the gravity of such bodies is thus a challenging task.

6.2.1 Gravity of an Axisymmetric Planet A large heavenly body (referred to here by the generic term planet) can be regarded as rotating about an axis of symmetry, but departs from the perfect symmetry of spherical shape due to various reasons. One such departure—called oblateness—is due to the centrifugal mass displacement caused by rotation, causing a bulge at the equator, and flattening near the poles. The net departure from the spherical shape can be approximated as a linear superposition of all such deforming effects, called spherical harmonics. Being a conservative force, gravity is expressed as the gradient of a scalar potential function, U (r), at a point, r, from the planetary center: gT =

∂U ∂r

(6.8)

202

6 Flight in Non-spherical Gravity Fields

Considering the central body’s mass to be distributed over a large number, (N − 1), of elemental masses, mi (i = 1, . . . , N − 1), the net gravitational acceleration acting on the N th body (spacecraft) of mass m is the vector sum of the gravitational accelerations caused by the individual masses, gT =

N −1 # i=1

∂Ui ∂ri

(6.9)

where ri is the position of the test mass, m, from the i th point mass, mi . This approach also yields the N-body problem where the spacecraft of mass m is under the gravitational influence of (N − 1) large bodies. For the purpose of modeling the motion of the spacecraft in the vicinity of a large planet, the test mass, m, is considered to be negligible in comparison with the sum of remaining (N − 1) masses, which are densely clustered together, away from the test mass, the partial derivative can be moved outside the summation sign in Eq. (6.9), resulting in N −1 ∂ # Ui g = ∂ri ri T

(6.10)

i=1

where Ui is the gravitational potential due to the point mass, mi , given by Ui =

Gmi ri

(6.11)

with G being the universal gravitational constant. 4 −1 For all the (N − 1) particles constituting the planetary mass, M = N i=1 mi , we take the limit of an infinitesimal elemental mass, mi → dM and N → ∞, whereby the summation in Eq. (6.10) is replaced by the following integral: gT =

∂ ∂r



G dM s

(6.12)

where s is the distance of the test mass, m, from the elemental mass, dM, and can be expressed as follows: s=

,

r 2 + ρ 2 − 2rρ cos γ

(6.13)

with r and ρ being the position vectors of the test mass, m, and elemental mass, dM, respectively, from the center of mass of the planet, and γ , is the angle between r and ρ Hence, r=s+ρ

(6.14)

6.2 Gravity Field of a Non-spherical Body

203

and ρ is a constant, because the planet is regarded to be a rigid body. Therefore, the gravitational potential of the planetary mass distribution is given by  .

U=

GdM r 2 + ρ 2 − 2rρ cos γ

(6.15)

In order to carry out the integration in Eq. (6.15), it is convenient to expand the integrand in a series (with r > ρ) as follows:  1 ρ ρ 1 1 . 1 + cos γ + ( )2 (3 cos2 γ − 1) = r r 2 r r 2 + ρ 2 − 2rρ cos γ  1 ρ + ( )3 cos γ (5 cos2 γ − 3) + · · · (6.16) 2 r or, 1 ρ ρ 1 . P0 (cos γ ) + P1 (cos γ ) + ( )2 P2 (cos γ ) = r r r r 2 + ρ 2 − 2rρ cos γ  ρ + · · · + ( )n Pn (cos γ ) + . . . (6.17) r where Pn (ν) is the Legendre polynomial of degree n. The first few Legendre polynomials are the following: P0 (ν) = 1 P1 (ν) = ν 1 (3ν 2 − 1) 2 1 P3 (ν) = (5ν 3 − 3ν) 2 1 P4 (ν) = (35ν 4 − 30ν 2 + 3) 8 1 P5 (ν) = (63ν 5 − 70ν 3 + 15ν) 8

P2 (ν) =

(6.18)

Some important properties of Legendre polynomials are as follows: Pn (1) = 1 Pn (−1) = (−1)n Pn (−ν) = (−1)n Pn (ν)

(6.19)

204

6 Flight in Non-spherical Gravity Fields

A Legendre polynomial can be generated from those of lower degree with the help of recurrence formulae, such as Pn (ν) =

(2n − 1)νPn−1 (ν) − (n − 1)Pn−2 (ν) n

(6.20)

Other recursive formulae, called generating functions, such as the following [1]: Pn (ν) =

1 dn 2 (ν − 1) 2n n! dν n

(6.21)

can also be useful in determining the Legendre polynomials. From these properties, it is clear that the Legendre polynomials satisfy the condition | Pn (cos γ ) |< 1, which implies that the series in Eq. (6.17) is convergent. Therefore, one can approximate the integrand of Eq. (6.15) by retaining only a finite number of terms in the series. The exact gravitational potential is expressed as follows in a series of spherical harmonics: ∞

U=

G# r



n=0

ρ ( )n Pn (cos γ )dM r

(6.22)

It is possible to further simplify the gravitational potential, before carrying out the complete integration. The integral arising out of P0 (cos γ ) in Eq. (6.22) yields the mass, M, of the planet; hence, we have  G GM (6.23) P0 (cos γ )dM = r r Furthermore, the integral containing P1 (cos γ ) vanishes, because the axis ρ = 0 is the axis of symmetry of the planet. Therefore, the gravitational potential can be expressed in terms of the Legendre polynomials of the second and higher degrees as ! " ∞  # dM ρ n GM ( ) Pn (cos γ ) U= 1+ (6.24) r r M n=2

A further simplification of the gravitational potential requires the planetary mass distribution. Consider a right-handed triad, (i, j, k), representing an inertial frame with origin at the center of an oblate spheroid of equatorial radius, r0 , such that (i, j) is the equatorial plane, and k locates the north pole. Due to the planetary symmetry about k, the gravitational potential depends only upon the radius, r, from the planetary center and the co-latitude, φ = π/2 − δ, measured from k, as shown in Fig. 6.1. Let the planet’s mass density be given by D(ρ, β, λ) in terms of the spherical coordinates of Fig. 6.1. Hence, the elemental mass can be expressed as dM = D(ρ, β, λ)ρ 2 sin βdρdβdλ

(6.25)

6.2 Gravity Field of a Non-spherical Body

205

k

b

r0

Equatorial Plane

r

•dM

f r

•

ir

m

l

j if

q

i

Fig. 6.1 Geometry for the non-spherical gravity model

where β and λ are the co-latitude and longitude, respectively, of the elemental mass. The assumption of symmetry about the polar axis translates into neglecting any longitudinal dependence of the mass distribution; thus, we have dM = D(ρ, β)ρ 2 sin βdρdβdλ

(6.26)

Using the spherical trigonometry [85] of Fig. 6.1, we can write cos γ = cos β cos φ + sin β sin φ cos(θ − λ)

(6.27)

where φ and θ are the co-latitude and longitude, respectively, of the test mass, m, as shown in Fig. 6.1. Upon neglecting the longitudinal variations in the mass distribution, we have from Eq. (6.27) Pn (cos γ ) = Pn (cos β)Pn (cos φ)

(6.28)

resulting in ∞

μ # An + Pn (cos φ) r r n+1

(6.29)

D(ρ, β)ρ n+2 Pn (cos β) sin βdρdβdλ

(6.30)

U (r, φ) =

n=2

where μ = GM and An = G



A more useful expression for the gravitational potential can be obtained as follows in terms of the non-dimensional distance, rr0 (where r0 is the equatorial radius of the planet):

206

6 Flight in Non-spherical Gravity Fields

/ & ∞   # r0 n μ U (r, φ) = Jn Pn (cos φ) 1− r r

(6.31)

n=2

where Jn = −

An μr0 n

(6.32)

are called Jeffery’s constants, and are unique for a planet. Jeffery’s constants represent the spherical harmonics of the planetary mass distribution, and diminish in magnitude as the order, n, increases. The largest of these constants, J2 , denotes a non-dimensional difference between the moments of inertia about the polar axis, k, and an axis in the equatorial plane (i or j in Fig. 6.1), and is a measure of oblateness of the planet. The higher-order term, J3 , indicates the pear-shaped or triangular harmonic, whereas J4 and J5 are the measures of square and pentagonal shaped harmonics, respectively. For a reasonably large body, it is seldom necessary to include more than the first four Jeffery’s constants. For example, the Earth’s spherical harmonics are given by J2 = 0.00108263, J3 = −0.00000254, and J4 = −0.00000161. The acceleration due to gravity of a non-spherical, axisymmetric planet can be obtained according to Eq. (6.12) by taking the gradient of the gravitational potential, Eq. (6.31), with respect to the position vector, r = rir + rφiφ , as follows: g = −(

∂U T ∂U ∂U ) =− ir − iφ ∂r ∂r r∂φ

(6.33)

or, g = gr ir + g φ iφ

(6.34)

where μ r0 r0 1 − 3J2 ( )2 P2 (cos φ) − 4J3 ( )3 P3 (cos φ) r r r2 r0 4 − 5J4 ( ) P4 (cos φ) r

gr = −

(6.35)

and  3μ r0 2 1 r0 gφ = 2 ( ) sin φ cos φ J2 + J3 ( ) sec φ(5 cos2 φ − 1) 2 r r r  r0 5 + J4 ( )2 (7 cos2 φ − 3) 6 r

(6.36)

6.2 Gravity Field of a Non-spherical Body

207

The unit vectors ir and iφ denote the radial and southward directions in the local horizon frame attached to the test mass (Fig. 6.1). Due to a nonzero transverse gravity component, gφ , the direction of g differs from the radial direction, while its radial component, gr , is smaller in magnitude compared to that predicted by a spherical gravity model.

6.2.2 Gravity Field of an Oblate Planet In many flight applications close to a planet, only the oblateness effect need to be modeled. This greatly simplifies the non-spherical model, as well as the resulting optimal control application. From the series expansion of the gravitational potential, Eq. (6.31), truncated to the largest Jeffery’s constant, J2 , we have U (r, φ) =

   r 2 μ 0 1− J2 P2 (cos φ) r r

(6.37)

This yields the following acceleration due to gravity at the given point, (r, φ):  g=

∂U ∂r

T =−

    3 μ 15z2 2 3z r r − μJ r k + − 2 0 r3 r5 2r 5 2r 7

(6.38)

with z = r cos φ. The gravity-gradient matrix required in the primer vector integration (see Chap. 5) is then evaluated as follows: ! !  "  "   r02 7z2 μ 3μJ2 r02 ∂g 3μ 5z2 ∂r 1− 2 =− 3+ −15μJ2 6 −1 r I+ 5 4 2 ∂r ∂r r 2r r r 2r r     r2 z ∂r ∂z − 3μJ2 05 k (6.39) + 15μJ2 r02 6 k ∂r ∂r r r   z ∂z + 15μJ2 r02 7 r ∂r r where ∂r = rT /r ∂r

(6.40)

∂z = kT ∂r

(6.41)

A substitution of Eqs. (6.40) and (6.41) into Eq. (6.39) yields the following expression for the gravity-gradient matrix:

208

6 Flight in Non-spherical Gravity Fields

!  " 3μJ2 r02 μ 5z2 ∂g 1− 2 =− 3 + I ∂r r 2r 5 r !  " r02 7z2 3μ z − 15μJ2 7 − 1 rrT + 15μJ2 r02 7 krT + r5 2r r2 r   r02 5z T T + 3μJ2 5 rk − kk r r2

(6.42)

6.2.3 Gravity of an Irregular Body Irregular bodies such as asteroids and comets do not have an axisymmetric shape, and thus the longitudinal dependence of the gravity potential in Eq. (6.24) cannot be neglected. The rotation of the body about a principal axis therefore results in a timevarying gravity field at any given point, (r, φ, θ ). Deriving an analytical expression, such as the one given by Eq. (6.33), is generally not feasible by the spherical harmonics expansion method presented above for the axisymmetric planet, because a detailed density distribution, D(ρ, β, λ), is unavailable. Furthermore, the spherical harmonics model, which converges at a point external to the body, may not converge inside a circumscribing sphere centered on the body [62]. The third disadvantage of the spherical harmonics model is that it cannot detect whether a point is located outside or inside the irregularly shaped body. Thus a problem in which a soft landing is to be carried out cannot be studied with a spherical harmonics model. Instead of the spherical harmonics model, a more useful method of deriving a gravity model of an irregularly shaped body is by the polyhedral approximation [84]. This method derives an analytical gravity field by approximating the body as a constant-density polyhedron, which can have peaks, overhangs, cavities, caves, and through holes. Hence there is no penalty associated with detailed surface features, which would require a high-order harmonics model. Furthermore, the model is applicable right up to the surface of the body, and can detect whether a point is located either inside or outside the body. Therefore, the polyhedral gravity model is especially useful for deriving optimal landing strategies on asteroids and comets. The successful orbiting and landing of the comet 67P/Churyumov-Gerasimenko by the ESA Rosetta probe and Philae lander mission in 2014 demonstrated the practical application of a polyhedral gravity model. Yang et al. [87] employ the polyhedral model to devise time-optimal powered descent and landing strategies on an asteroid.

6.3 Circular Restricted Three-Body Problem The three-body problem refers to the motion of three masses under mutual gravitational attraction. It arises when either the two-body orbital motion is perturbed by a distant third body or a much smaller body (like a spacecraft) moves in the

6.3 Circular Restricted Three-Body Problem

209

gravitational field formed by two larger bodies. Examples of the former kind are the orbital motion of a Moon around a planet perturbed by the Sun, and that of a satellite around the Earth perturbed by the Sun or Moon. The latter model is applied to a comet under the combined gravity of the Sun and Jupiter, a spacecraft under mutual gravity of the Earth–Moon system, and an interplanetary probe approaching (or departing) a planet. The three-body problem is a much better approximation of the actual motion of a spacecraft at any given time in the solar system when compared to the two-body problem considered earlier. An important application of the three-body problem is the design of optimal circumlunar and interplanetary trajectories based upon the solutions that take the spacecraft around multiple large bodies on largely natural, fuel-free paths. These will be explored later in this chapter. The three-body problem has attracted attention of mathematicians and physicists over the past three centuries, primarily in studying the motion of the Moon. The first systematic study of the three-body problem was by Lagrange, who derived the particular solution of the coplanar problem, and investigated the stability of the equilibrium solutions. The equations of motion for the three-body problem are expressed as follows: # mj d2 Ri =G rij ; 2 dt rij3 3

(i = 1, 2, 3)

(6.43)

j =i

where G is the universal gravitational constant, Ri denotes the position of the center . of mass of the i th body, and rij = Rj − Ri denotes the relative position of the centers of mass of bodies, i,j (i = j ). Lagrange showed that certain particular solutions of the three-body problem exist when the motion of the three bodies is always confined to a single plane. Each mass in the coplanar three-body problem traces a conic section about the common center of mass (called the barycenter). From the coplanar motion derived by Lagrange, certain stationary solutions can be obtained in terms of fixed locations of the masses in the rotating frame. These solutions represent equilibrium points of the threebody problem, and describe the three bodies moving in concentric, coplanar circles. Such a motion has a constant separation of the three bodies, and the line joining ˙ by angular momentum any two bodies rotates at a constant angular speed, n = θ, conservation. The equations of motion of the coplanar three-body problem are the following: 0

1 n2 m2 m3 m2 − 3 − 3 R1 + 3 R2 + G r12 r13 r12 0 1 n2 m1 m1 m3 − 3 − 3 R2 + R1 + 3 G r12 r12 r23

m3 3 r13

m3 3 r23

R3 = 0 R3 = 0

m1 R1 + m2 R2 + m3 R3 = 0

(6.44)

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6 Flight in Non-spherical Gravity Fields

Particular equilibrium solutions of Eq. (6.44) include the equilateral triangle configurations of the three masses, wherein the masses are at the same constant distance from the common center of mass (r12 = r13 = r23 = R), hence the angular speed is given by n = θ˙ =

G(m1 + m2 + m3 ) R3

(6.45)

Another set of equilibrium points are the collinear solutions, where the three masses share a straight line. Let the axial location of the three masses be given by x1 , x2 , x3 , where we assume x1 < x2 < x3 . Then the equations of motion, written in terms of a distance ratio, α = r23 /r12 , yield the following [8]: x1 = −r12 ω2 =

m2 + (1 + α)m3 m1 + m2 + m3

G(m1 + m2 + m3 ) m2 (1 + α)2 + m3 3 (1 + α)2 m2 + (1 + α)m3 r12

(6.46)

(m1 + m2 )α 5 + (3m1 + 2m2 )α 4 + (3m1 + m2 )α 3 − (m2 + 3m3 )α 2 − (2m2 + 3m3 )α − (m2 + m3 ) = 0 The last of Eq. (6.46), called the quintic equation of Lagrange, has only one positive root, α, which can be seen by examining the signs of the coefficients (they change sign only once). However, for each specific configuration of the three collinear masses, the positive value of α is different. For example, the values of the distance ratio, α, can be numerically determined for the collinear Earth, Moon, spacecraft system by taking the ratio of the Moon’s mass to the Earth’s mass is 1/81.3 and neglecting the spacecraft’s mass in comparison with the masses of heavenly bodies. These are α = 0.1678, 0.1778, or 1.0071, depending upon whether the Moon, the spacecraft, or the Earth, respectively, falls between the other two bodies. When the mass of one of the three bodies, say m3 , is negligible in comparison with that of the other two bodies (called the primaries) which latter are in a circular orbit about the barycenter, a major simplification occurs in the coplanar threebody equations (Eq. (6.44)). In that case, the gravitational pull of m3 on both m1 and m2 is negligible, and the natural motion of m3 relative to the primaries is referred to as the circular restricted three-body problem (or CR3BP). The barycenter is then approximated by the common center of mass of the two primaries, i.e., m1 R1 m2 R2 . For convenience, the CR3BP equations of motion are rendered non-dimensional by dividing each of the masses by the total mass of the primaries, m1 + m2 , and dividing all the distances by the constant separation between the primaries, R12 .

6.3 Circular Restricted Three-Body Problem

. μ= . r1 = . r2 =

211

m2 m1 + m2 r13 r12 r23 r12

(6.47)

Conventionally, we take m1 ≥ m2 , which makes μ ∈ [0, 1/2]. The normalized masses of the primaries, m1 and m2 , are thus (1 − μ) and μ, respectively. The non-dimensional distances of the two primaries from the barycenter are R1 = μ and R2 = (1 − μ), respectively. The value of gravitational constant, G, is nondimensionalized such that the angular velocity, n, of the primaries is unity n = θ˙ =

,

3 = G(m1 + m2 )/r12

.

μ + (1 − μ) = 1

(6.48)

which produces a non-dimensional time, t, such that the orbital period of the primaries is t = 2π . Let the coordinates of the primaries m1 and m2 in an inertial frame with origin at the barycenter be (ξ1 , η1 , 0) and (ξ2 , η2 , 0), respectively. If the position of the negligible mass, m3 , is described by the non-dimensional coordinates, (ξ, η, ζ ), they are governed by the following differential equations: ξ¨ = − η¨ = − ζ¨ = −

(1 − μ)(ξ − ξ1 ) r13 (1 − μ)(η − η1 ) r13 (1 − μ)ζ r13

−

− −

μ(ξ − ξ2 ) r23 μ(η − η2 ) r23

(6.49)

μζ r23

where the overdot represents derivative in non-dimensional time, t, and , r1 = (ξ − ξ1 )2 + (η − η1 )2 + ζ 2 , r2 = (ξ − ξ2 )2 + (η − η2 )2 + ζ 2

(6.50)

Now, consider a frame (i, j, k), with origin at the common center of mass (barycenter) (Fig. 6.2), and rotating with the non-dimensional velocity, k. Such a rotating frame is termed the synodic frame. The position of the mass, m3 , in the rotating frame is given by r = xi + yj + zk. The coordinate transformation between the inertial and the rotating frames is the following: ⎧ ⎫ ⎛ ⎞⎧ ⎫ cos t − sin t 0 ⎨ x ⎬ ⎨ξ ⎬ η = ⎝ sin t cos t 0 ⎠ y ⎩ ⎭ ⎩ ⎭ 0 0 1 ζ z

(6.51)

212

6 Flight in Non-spherical Gravity Fields

j

Orbital plane (i,j) of primaries

3

i 2

1

k Fig. 6.2 Geometry of the circular restricted three-body problem

The substitution of Eq. (6.51) into Eq. (6.49) and elimination of sin t and cos t yields the following: x¨ − 2y˙ − x = − y¨ + 2x˙ − y = − z¨ = −

(1 − μ)(x − x1 ) r13 (1 − μ)(y − y1 ) r13 (1 − μ)z r13

−

− −

μ(x − x2 ) r23 μ(y − y2 ) r23

(6.52)

μz r23

Finally, the direction of the x-axis (i) is chosen to point from the barycenter toward the center of the smaller primary, m2 , and the y-axis (j) being normal to both i and k completes the right-handed triad, i × j = k. Thus we have x1 = −μ, x2 = 1 − μ, y1 = y2 = 0, and the following equations of CR3BP motion in terms of the coordinates, (x, y, z): x¨ − 2y˙ − x = −

(1 − μ)(x + μ) r13

−

μ(x − 1 + μ) r23

6.3 Circular Restricted Three-Body Problem

y¨ + 2x˙ − y = − z¨ = −

213

(1 − μ)y r13 (1 − μ)z r13

−

μy

−

μz

(6.53)

r23 r23

where r1 = r2 =

, ,

(x + μ)2 + y 2 + z2 (x − 1 + μ)2 + y 2 + z2

(6.54)

The set of coupled, nonlinear, ordinary differential equations, Eq. (6.53), representing the CR3BP cannot be solved in a closed form. However, we can obtain certain important analytical insights into the problem without actually solving it. An important aspect of the CR3BP equations is the symmetrical nature of its solutions. If x(t), y(t), z(t) is a solution of the equations, then it is easily verified that x(t), y(t), −z(t) is also a solution, thereby implying a reflection across the (x, y) plane. Similarly, x(−t), −y(−t), z(−t) is also a solution, which implies a reflection across the (x, z) plane with time reversal. The third symmetry is related to the parameter, μ. If x(t), y(t), z(t) satisfies the equations with parameter μ, then −x(t), −y(t), z(t) also satisfies the equations with parameter 1 − μ. This means that one needs to study the CR3BP only for μ ∈ [0, 1/2].

6.3.1 Lagrangian Points and Stability As noted earlier, certain equilibrium solutions are possible to the coplanar threebody problem represented by Eq. (6.44). These equilibrium points are called the Lagrangian points (or libration points), and can be derived for the CR3BP model by equating the time derivatives in Eq. (6.53) to zero, resulting in the following algebraic equations: x= y= 0=

(1 − μ)(x + μ) r13 (1 − μ)y r13 (1 − μ)z r13

+

μy

+

μz

r23

+

μ(x − 1 + μ) r23 (6.55)

r23

From the last of Eq. . (6.55), we have the result z.= 0 for all the Lagrangian points, which implies r1 = (x + μ)2 + y 2 and r2 = (x + μ − 1)2 + y 2 . The coplanar

214

6 Flight in Non-spherical Gravity Fields

equilibrium solutions consist of the three possible collinear positions of m3 , called the collinear Lagrangian points, L1 , L2 , L3 , as well as the two equilateral triangle positions, called triangular Lagrangian points, L4 , L5 , with the two primaries (Fig. 6.2). The points L1 , L2 , L3 are obtained from the solution of the quintic equation of Lagrange (the last of Eq. (6.46)), which for m1 = 1 − μ, m2 = μ, m3 = 0, and α = r23 /r12 is written as follows: α 5 + (3 − μ)α 4 + (3 − 2μ)α 3 − μα 2 − 2μα − μ = 0

(6.56)

As discussed earlier, there is only one real root of Eq. (6.56). For example, with μ = 1/82.3 for the Earth–Moon system, Eq. (6.56) gives α = 0.167833, which yields x = 1.15568 for Lagrangian point L2 . With m1 = μ (Moon), m2 = 1 − μ (Earth), and m3 = 0 (spacecraft), the quintic equation becomes the following: α 5 + (2 + μ)α 4 + (1 + 2μ)α 3 − (1 − μ)α 2 − 2(1 − μ)α − (1 − μ) = 0

(6.57)

whose real root for the Earth–Moon system is α = 0.992912, thereby yielding x = −α − μ = −1.00506 for the Lagrangian point L3 . Finally, the Lagrangian point L1 is obtained by putting m2 = 0 (spacecraft), m1 = μ (Moon), and m3 = 1 − μ (Earth). The quintic equation then becomes μα 5 + 3μα 4 + 3μα 3 − 3(1 − μ)α 2 − 3(1 − μ)α − (1 − μ) = 0

(6.58)

whose real root for the Earth–Moon system is α = r23 /r12 = 5.62538553 with r12 + r23 = 1, thereby yielding x = r23 − μ =

1 − μ = 0.836914719 1 + 1/α

The triangular Lagrangian points, √ L4 , L5 , correspond to r1 = r2 = 1, and are given by the coordinates ( 12 − μ, ± 3/2). In order to investigate the stability of a Lagrangian point, consider infinitesimal displacements, δx, δy, δz, from the equilibrium position, (x0 , y0 , 0). If the displacements remain small, the equilibrium point is said to be stable, otherwise, unstable. Such an analysis easily reveals that the out-of-plane motion represented by δz is unconditionally stable about all the equilibrium points. However, the stability of the coplanar motion described by (δx, δy) crucially depends upon the location (x0 , y0 ) of the given equilibrium point. For the small displacement, coplanar motion about a triangular Lagrangian point, say L4 , the following linearized model can be applied: √ 3 3(μ − 12 ) 3 δy = 0 δ x¨ − 2δ y˙ − δx − 4 2 √ 3 3(μ − 12 ) 9 δ y¨ + 2δ x˙ − δx − δy = 0 2 4

(6.59)

6.3 Circular Restricted Three-Body Problem

215

whose general solution to initial displacement δx0 , δy0 , can be written as δr = δr0 eλt

(6.60)

where δr = (δx, δy)T , δr0 = (δx0 , δy0 )T , and λ is the eigenvalue of the following matrix: ⎛ ⎞ 0 0 1 0 ⎜ 0 0 0 1⎟ ⎜ ⎟ √ ⎟ 1 (6.61) A=⎜ 3(μ− ) 3 3 ⎜ 2 0 2⎟ ⎝ √ 4 1 ⎠ 2 3 3(μ− 2 ) 9 −2 0 2 4 The resulting characteristic equation for λ is thus the following: λ4 + λ2 +

27 μ(1 − μ) = 0 4

(6.62)

whose roots are λ2 =

1 . [± 1 − 27μ(1 − μ) − 1] 2

(6.63)

For stability, the values of λ must be purely imaginary, representing a simple harmonic oscillation about the equilibrium point. If the quantity in the square-root is negative, there is at least one value of λ with a positive real part and represents an unstable system. Therefore, the critical values of μ representing the boundary between stable and unstable behavior of L4 are those that correspond to 1 − 27μ(1 − μ) = 0

(6.64)

Hence, for the stability of the triangular Lagrangian points, we require either μ ≤ 0.0385209 or μ ≥ 0.9614791, which is always satisfied by any two CR3BP primaries in the solar system. Therefore, we expect that the triangular Lagrangian points would provide stable locations for smaller bodies in the solar system. The existence of such bodies for the Sun–Jupiter system has been verified in the form of Trojan asteroids. For the Earth–Moon system, μ = 0.01215 is well within the stable region, hence one could expect the triangular points to be populated. However, the Earth–Moon system is not a good example of the restricted three-body problem, because of an appreciable influence of Sun’s gravity (a fourth body), which renders the triangular points of Earth–Moon system unstable. Hence, L4 , L5 for Earth– Moon system are empty regions in space, where one cannot park a spacecraft. For a given CR3BP system, one can determine the actual small displacement dynamics about the stable triangular Lagrangian points, in terms of the eigenvalues, λ. For example, the Earth–Moon system has non-dimensional natural frequencies 0.2982 and 0.9545, which indicate long-period and short-period modes of the small displacement motion.

216

6 Flight in Non-spherical Gravity Fields

For the stability of the collinear Lagrangian points, we employ y0 = 0, and the resulting equations of small displacement can be written as follows: 

 2(1 − μ) 2μ + + 1 δx (x0 − μ)3 (x0 + 1 − μ)3   μ 1−μ + − 1 δy δ y¨ + 2δ x˙ + (x0 − μ)3 (x0 + 1 − μ)3

δ x¨ − 2δ y˙ −

(6.65)

The eigenvalues of this system are the eigenvalues of the following square matrix: ⎛

0 0 1 ⎜0 0 0 A=⎜ ⎝α 0 0 0 −β −2

⎞ 0 1⎟ ⎟ 2⎠ 0

(6.66)

where α=

2(1 − μ) 2μ + +1 (x0 − μ)3 (x0 + 1 − μ)3

β=

1−μ μ + −1 3 (x0 − μ) (x0 + 1 − μ)3

(6.67)

The resulting characteristic equation for the eigenvalues, λ, is the following: λ4 + (4 + β − α)λ2 − αβ = 0

(6.68)

For the collinear Lagrangian points, α > 0, and β < 0, which implies that the constant term in the characteristic equation is always positive. Therefore, there is always at least one eigenvalue with a positive real part, hence the system is unconditionally unstable. Thus, the collinear Lagrangian points of any restricted three-body system are always unstable. However, as discussed later, a spacecraft can successfully orbit a collinear point with small energy expenditure. Such an orbit around a Lagrangian point is termed a halo orbit , and is useful in practical space missions, such as stationing a probe around the Sun–Earth L1 and L2 points for monitoring the solar and interplanetary zone ahead of and behind the Earth.

6.3.2 Hamiltonian Formulation and Jacobi’s Integral The CR3BP equations of motion can be alternatively derived by the Lagrangian approach using the canonical coordinates, q = (x, y, z)T , as follows: q¨ − 2Kq˙ = ∇Ω

(6.69)

6.3 Circular Restricted Three-Body Problem

217

where ∇ = (∂/∂x, ∂/∂y, ∂/∂z)T is the gradient operator ⎞ 0 1 0 K = ⎝ −1 0 0 ⎠ 0 0 0 ⎛

(6.70)

and (1 − μ) 1 μ (1 − μ) μ 1 + = (x 2 + y 2 ) + + Ω = − qT K2 q + 2 | q − (1 − μ)i | | q + μi | 2 r1 r2 (6.71) is the pseudo-potential function with i = (1, 0, 0)T and r1 , r2 given by Eq. (6.54). The scalar components of Eq. (6.69) are the following: ∂Ω ∂x ∂Ω y¨ + 2x˙ = ∂y x¨ − 2y˙ =

z¨ =

(6.72)

∂Ω ∂z

The Hamiltonian associated with Eq. (6.69) is the following: 1 ˙ = − q˙ T (q˙ − Kq) − L(q, q) ˙ H (q, q) 2

(6.73)

with the Lagrangian given by ˙ = L(q, q)

1 (q˙ − Kq)T (q˙ − Kq) − U (q) 2

(6.74)

and the gravitational potential by 1 (1 − μ) μ U (q) = −Ω − qT K2 q = − − 2 r1 r2

(6.75)

It is to be noted that the terms “Hamiltonian” and “Lagrangian” used here in the dynamical context are different from the same terms occurring in the optimal context (Chap. 2). By defining the generalized momentum vector as follows: p=

∂L = q˙ − Kq ∂ q˙

(6.76)

the Hamiltonian is expressed as H (q, p) =

1 T p p + pT Kq + U (q) 2

(6.77)

218

6 Flight in Non-spherical Gravity Fields

This yields the equations of motion, Eq. (6.69): q˙ =

∂H ∂p

p˙ = −

∂H ∂q

(6.78)

which are compactly expressed in terms of the state vector, X = (qT , pT )T , as follows: ˙ = J∇X H (X) X

(6.79)

where ∇X is the gradient operator relative to X, and  J=

03×3 I3×3 −I3×3 03×3

 (6.80)

An important scalar constant of the restricted three-body motion can be derived ˙ by taking the scalar product of Eq. (6.69) with q: ˙ · q˙ = q¨ · q˙ − 2(Kq)

∂Ω T · q˙ ∂q

(6.81)

or 1 dq˙ 2 dΩ = 2 dt dt

(6.82)

which is an exact differential, and can be integrated to obtain Ω=

1 2 (q˙ + C) 2

(6.83)

Here, C is the constant of integration, called Jacobi’s integral: C = 2Ω − q˙ 2 = (x 2 + y 2 ) − 2

1−μ μ − 2 − (x˙ 2 + y˙ 2 + z˙ 2 ) r1 r2

(6.84)

The Hamiltonian, H , is related to Jacobi’s integral as follows: 1 T p p + pT Kq + U (q) 2 1 1 = q˙ 2 + qT K2 q + U (q) 2 2 1 2 1 2 = q˙ − (x + y 2 ) + U = −C/2 2 2

H =

(6.85)

6.4 Special Three-Body Trajectories

219

Jacobi’s integral can be thought to represent a pseudo-energy of the mass m3 , which is a sum of the relative kinetic energy, gravitational potential energy, and an additional pseudo-potential energy, (x 2 + y 2 ). The value of C at any point is a measure of its relative energy. The most useful interpretation of Jacobi’s integral lies in the contours of zero relative speed, v = 0, which represent the boundary between a motion toward higher potential, or the return trajectory to the lower potential [78]. On such a boundary, the mass must stop and turn back. Hence, the zero relative speed contours demarcate the regions accessible to a spacecraft, and cannot be crossed for the given value of C. The closed zero speed contours around a primary indicate that a flight from one primary to the other is impossible at the given energy level. As the value of C increases, the contours expand in size touch one another for some special values of C, and a further increase of C causes the region around a collinear point to open up, indicating either a flight at nonzero relative velocity between the primaries or an escape from the gravitational influence of the primaries [78]. In such a case, contours enclose the triangular Lagrangian points, indicating that a flight to reach them is impossible from any of the primaries. As the value of C becomes very large, the forbidden regions around L4 , L5 , shrink, but never actually vanish, thereby denoting that the stable triangular Lagrangian points can be approached (but never reached) only with a very high initial energy.

6.4 Special Three-Body Trajectories Although the circular restricted problem of three bodies (CR3BP) appears more amenable to a solution than the general problem due to the availability of Jacobi’s integral, a closed-form solution is still not possible. This fact was first demonstrated by Poincaré using phase-space surfaces. The CR3BP problem is mathematically interesting as it belongs to the group of chaotic systems, wherein a small change in the initial condition results in an arbitrarily large change in the trajectory. Numerical solutions to the CR3BP problem for special cases reveal special trajectories that are possible flight paths for spacecraft in lunar and interplanetary space. Such solutions rarely have a periodic behavior, which requires special integration schemes, such as a higher-order Runge–Kutta technique, to prevent the truncation errors from growing to unacceptable magnitudes within a few orbital time periods of the primaries. Here some typical trajectories are explored which can be derived within one period of primary rotation.

6.4.1 Perturbed Orbits About a Primary When a spacecraft is orbiting one of the primaries, its trajectory is perturbed by the presence of the more distant primary, causing a precession in the plane of the orbit. This leads to both an apsidal rotation and a nodal regression of the two-body orbital plane, thereby producing a quasi-periodic trajectory. Consider, for example,

220

6 Flight in Non-spherical Gravity Fields 0.04 0.02 0

m1

t =0

−0.02

y

−0.04 −0.06 −0.08 −0.1

t =1

−0.12 −0.14 −0.04 −0.02

0

0.02

0.04

0.06

x

0.08

0.1

0.12

0.14

0.16

Fig. 6.3 Trajectory around the Earth perturbed by the Moon simulated for t = 1 (4.8 days)

the apsidal rotation of a spacecraft orbiting the Earth due to the Moon’s gravity, shown in Fig. 6.3, when the three bodies are in the same plane. The value of the nondimensional mass parameter for the Earth–Moon system is μ = 0.01215, which implies the Earth’s location at (−0.01215, 0) marked by the red dot in Fig. 6.3. The trajectory is simulated for t = 1 (recall that a time period of the primaries in the non-dimensional time is t = 2π , or about 30 mean solar days). The initial condition for the given coplanar trajectory is x(0) = 0.15, y(0) = 0, x˙ = 0, and y˙ = 0.45. The elliptical shape of the trajectory is evident in this planar problem. Figure 6.4 shows the apsidal rotation of a spacecraft’s orbital plane around the Moon caused by the Earth. The initial condition for the given coplanar trajectory plotted for t = 1 is x(0) = 1.0, y(0) = 0, x˙ = 0, and y˙ = 0.4. The perturbed orbits shown in Figs. 6.3 and 6.4 are special quasi-periodic solutions of the CR3BP in the vicinity of a collinear Lagrangian point.

6.4.2 Free-Return Trajectories Between the Primaries An interesting periodic solution of the CR3BP is the free-return trajectory, which passes close to both the primaries, and can be used for an essentially fuel-free transfer from one primary to the other. The free-return concept formed the basis of Jules Verne’s famous novel From the Earth to the Moon (1866), and was ultimately realized about a century later by the manned lunar exploration through the Apollo lunar missions, wherein the initial insertion was performed at a given location with the desired velocity, both on the outward and the return journey. Being a natural

6.4 Special Three-Body Trajectories

221

−3

x 10

4

t 2

m2

y

0

−2

−4

−6

−8 0.984

0.986

0.988

0.99

0.992

x

0.994

0.996

0.998

1

Fig. 6.4 Trajectory around the Moon perturbed by the Earth simulated for t = 1 (4.8 days) 1 0.8 0.6 0.4

y

0.2

m1

L1

0

m2 L2

−0.2 −0.4 −0.6 −0.8 −1 −0.4

−0.2

0

0.2

0.4

x

0.6

0.8

1

1.2

Fig. 6.5 Free-return trajectory passing close to the Earth and the Moon in t = 3.5 (16.7 days)

solution, it is a fuel-optimal path of the planar CR3BP. An example of the freereturn Earth–Moon trajectory is plotted in Fig. 6.5 for t = 3.57, which extends from 45,000 km radius near the Earth (green circle) to a point close to the Moon (red circle), and orbits the Lagrangian point, L1 . Another possible free-return trajectory

222

6 Flight in Non-spherical Gravity Fields 0.5 0.4 0.3 0.2

y

0.1

m1

L1

m2

L2

0 −0.1 −0.2 −0.3 −0.4 −0.2

0

0.2

0.4

x

0.6

0.8

1

1.2

Fig. 6.6 Hourglass free-return trajectory passing close to the Earth and the Moon in t = 2.387 (11.4 days)

is the hourglass shaped one (Fig. 6.6) passing between the second primary, m2 , and the Lagrangian point, L2 . The total flight time is reduced significantly in this case to about t = 2.387, because the spacecraft passes around the Moon, between L1 and L2 , and in the process, receives a kinetic energy boost due to the Moon. Such an increase in the Keplerian orbital energy derived by passing close to a third body is called a gravity-assist maneuver (or a swing-by). An extreme case of the lunar swing-by trajectory is depicted in Fig. 6.7, where the spacecraft escapes the Earth–Moon system by passing very close to the Moon. The initial launch speed (from the same initial location) for the escape trajectory of Fig. 6.7 is smaller (v = 6.935, C = 2.3982) than that of the free-return trajectory of Fig. 6.6, which has v = 6.9493 and C = 2.1995. Lunar swing-by trajectories have been employed in cheaply boosting several spacecraft to the Sun–Earth Lagrangian points, such as the ISEE-3, MAP, and ACE missions of NASA [27]. A similar approach of multiple planetary swing-bys has been found useful in reducing the mission cost of spacecraft bound for distant planets. Figure 6.8 shows a trajectory beginning near the Moon, and orbiting all five Lagrangian points of the Earth–Moon system in t = 28.5.

6.5 Optimal Low-Thrust Three-Body Transfer Navigating a spacecraft by continuous thrust in the CR3BP gravity field is a practical problem of space navigation with applications in both lunar and interplanetary flight. Consider a spacecraft with a continuously operating, low-thrust engine

6.5 Optimal Low-Thrust Three-Body Transfer

223

4

3

2

y

1 L3

0

L1

L2

−1

−2

−3 −3

−2

−1

0

x

1

2

3

4

Fig. 6.7 Lunar swing-by escape trajectory simulated for t = 10 (47.76 days) 2 1.5

L4

1

y

0.5

L3

0

L1

L2

−0.5 −1

L5

−1.5 −2 −2

−1.5

−1

−0.5

0

x

0.5

1

1.5

2

Fig. 6.8 Free trajectory orbiting all five Lagrangian points of the Earth–Moon system in t = 28.5 (136 days)

generating an acceleration input, a = a(t)n(t). The non-dimensional state equations representing the spacecraft’s position, r = xi + yj + zk and velocity v = vx i + vy j + vz k in the rotating frame, (i, j, k), with origin at the common center of mass

224

6 Flight in Non-spherical Gravity Fields

(barycenter) (Fig. 6.2) of the circular, restricted three-body problem (CR3BP) can be derived from Eq. (6.53) to be the following: ⎧ ⎫ ⎧ ⎫ ⎨ x˙ ⎬ ⎨ vx ⎬ r˙ = y˙ = vy = v ⎩ ⎭ ⎩ ⎭ z˙ vz ⎧ ⎫ ⎧ ⎫ x + 2vy − (1−μ)(x+μ) − μ(x−1+μ) + ax ⎪ ⎪ 3 3 v ˙ ⎪ ⎪ ⎪ ⎪ x r r ⎪ ⎪ 1 2 ⎨ ⎬ ⎬ ⎪ ⎨ ⎪ (1−μ)y μy y − 2vx − − 3 + ay (6.86) v˙ = v˙y = 3 r1 r2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎪ ⎩ ⎪ ⎩ ⎭ v˙z − (1−μ)z − μz3 + az v˙z 3 r1

r2

where n = x i + y j + z k is the unit vector in the thrust direction, and , r1 = (x + μ)2 + y 2 + z2 , r2 = (x − 1 + μ)2 + y 2 + z2

(6.87)

The thrust direction cosines, (x , y , z ), must satisfy the equality constraint 2x + 2y + 2z = 1

(6.88)

Both the velocity, v, and the thrust acceleration input, a, can alternatively be resolved in spherical coordinates as shown in Fig. 6.9 via the angles, (φ, ψ) and (α, β), respectively, through the following relationships: Fig. 6.9 Resolving the velocity and thrust acceleration in the CR3BP frame using spherical coordinates

a

a

y

v

b

6.5 Optimal Low-Thrust Three-Body Transfer

v = v(cos φ cos ψi + cos φ sin ψj + sin φk)

225

(6.89)

a = a[cos(φ + α) cos(ψ + β)i + cos(φ + α) sin(ψ + β)j + sin(φ + α)k] , which implies φ = tan−1 (vz / (vx2 + vy2 ) and ψ = tan−1 (vy /vx ). Irrespective of whether the Cartesian or spherical coordinates are used to write the state equations, the position, velocity, and acceleration require a transformation from the rotating frame, (i, j, k), to an inertial (sidereal) frame, (I, J, k). Furthermore, the velocity and acceleration must be corrected as follows to yield the corresponding inertial values: v ⇒ v+ω×r v˙ ⇒ v˙ + ω × v

(6.90)

One can use the state equations in the rotating frame to compute the optimal trajectory and control history in the rotating frame, and then transform them to the inertial frame. Thus for the purpose of control derivation, Eq. (6.86) can be regarded as the state equations, and expressed in the form ˙ = f(X, a) X

(6.91)

where X(t) = (rT , vT )T is the state vector, a(t) ∈ R3×1 is the acceleration input vector, and the functional f(.) : R6×1 × R3×1 → R6×1 is given by Eq. (6.86). A maneuver can be represented by the initial state, X(0) = X0

(6.92)

F [X(tf ), tf ] = 0

(6.93)

and a terminal condition

with the terminal time, tf , either specified in advance or not. The optimization problem is of the Mayer type, and can be stated as finding the control history which minimizes the following performance index: J = φ[X(tf ), tf ]

(6.94)

subject to the equality constraints given by Eqs. (6.86)–(6.88). The optimal maneuver is the one that minimizes the propellant consumption, which implies the minimization of the terminal characteristic speed (φ = c(tf )), and is equivalent to the maximization of the final mass (φ = −m(tf )) (see Chap. 5). In either case, an additional scalar state equation must be added, such as c˙ = a

(6.95)

226

6 Flight in Non-spherical Gravity Fields

or m ˙ = −β

(6.96)

with bounds being applied either to the acceleration input magnitude 0 ≤ a ≤ am (t)

(6.97)

0 ≤ β ≤ βm (t)

(6.98)

or to the exhaust rate

depending upon the type of the engine employed. Let us consider the bounded input magnitude formulation, for which we have φ = c(tf ), and the Hamiltonian of the optimal control problem is the following: $ % H = λTr v + λTv g + an + λc a

(6.99)

where λr and λv are the costate vectors corresponding to r and v, respectively, and λc corresponds to the characteristic speed, c(t), obeying Eq. (6.95). Here, the noninertial acceleration due to gravity, g , in the rotating frame is the following: ⎫ ⎧ x + 2vy − (1−μ)(x+μ) − μ(x−1+μ) ⎪ ⎪ ⎪ ⎪ r13 r23 ⎬ ⎨ (1−μ)y μy

y − 2vx − − 3 g = 3 r1 r2 ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ v˙ − (1−μ)z − μz z

r13

(6.100)

r23

The costate equations for λr and λv are derived from the necessary conditions of optimality as follows:    T ∂H T ∂g λ˙ r = − =− λv ∂r ∂r

(6.101)

  ∂H T = −λr λ˙ v = − ∂v

(6.102)

λ˙ c = −



∂H ∂c



 = −a

∂H ∂t

 (6.103)

where the non-inertial gravity-gradient matrix, ∂g /∂r, is given by: ⎛

Ax

0

0

⎞

⎜ ⎟ ∂g

=⎜ 0 Ay 0 ⎟ ⎝ ⎠ ∂r 0 0 Az

(6.104)

6.5 Optimal Low-Thrust Three-Body Transfer

227

with     μ (x − 1 + μ)2 (1 − μ) (x + μ)2 Ax = 1 + −1 + 3 +1 2 2 r13 r2 " !  2 (1 − μ) y μ +1 Ay = 1 − + 3 2 r13 r2 " !   2 (1 − μ) z μ +1 Az = − + 3 2 r13 r2

(6.105)

The boundary conditions for the costate variables are the following:  λr (tf ) =

∂F ∂r

 λv (tf ) =  λc (tf ) =

∂F ∂v

∂φ ∂c

T d

(6.106)

d

(6.107)

=1

(6.108)

t=tf

T t=tf

 t=tf

where d = (d1 , d2 , . . . , dk )T are the additional parameters used to adjoin the terminal constraint equation, Eq. (6.93), to the performance index as follows: Ja = φ[x(tf ), tf ] + dT F = c(tf ) + dT F

(6.109)

The primer vector method (see Chap. 5) produces the following optimal thrust direction in a CR3BP field: nˆ =

p p

(6.110)

where p(t) = −λv (t) is the primer vector governed by the following differential equations: p˙ = λr  p¨ =

∂g ∂r

(6.111)

T p

(6.112)

with the boundary conditions  ˙ f) = p(t

∂F ∂r

T d t=tf

(6.113)

228

6 Flight in Non-spherical Gravity Fields

 p(tf ) = −

∂F ∂v

T d

(6.114)

t=tf

The Hamiltonian expressed in terms of the primer vector is the following: H = a(λc − p) + p˙ T v − pT g

(6.115)

Due to the linearity of H relative to a, the optimal control problem is singular in the acceleration magnitude. However, the bounds on a given by Eq. (6.97) allow the application of the Pontryagin’s minimum principle to yield for the optimal magnitude, a(t): ˆ a(λ ˆ c − p) ≤ a(λc − p)

(6.116)

The inequalities of Eqs. (6.97) and (6.116) result in the following switching condition for the optimal control magnitude:  a(t) ˆ =

am (t) , (p − λc ) > 0 0 , (p − λc ) < 0

(6.117)

If the upper bound am is assumed to be constant, based upon the approximation that the propellant consumed, m0 − m(t), is a small fraction of the initial mass, m0 , at any time t, then the Hamiltonian is time invariant in the non-inertial frame, thereby producing λc = const. = 1. In such a case, the switching condition becomes the following:  a(t) ˆ =

am , p > 1 0, p<1

(6.118)

Example 6.1 Consider a low-thrust interstellar mission which requires sending a spacecraft from a low-Earth orbit to an out-of-ecliptic radial position, while maximizing the terminal speed, vf , as the propellant is exhausted at t = tf . The initial conditions in the CR3BP frame are x(0) = −0.1, y(0) = z(0) = 0, x(0) ˙ = 4, x(0) ˙ = 3, z˙ (0) = 0.01, and an ion-propulsion engine is assumed to produce a maximum acceleration magnitude, am = 0.01 m/s2 , which is transformed to the non-dimensional units of Earth–Moon CR3BP system by multiplying it with 367.1975 s2 /m. A three-dimensional plot of the trajectory (shown in red), simulated with a fourth-order Runge–Kutta integration, is compared with the zero-thrust case (am = 0) (plotted in blue) in Fig. 6.10, showing a more rapidly increasing radius (clearly seen in the planar plot of Fig. 6.11), as well as a much larger out-of-plane motion (Fig. 6.12) achieved with the continuous thrust. The effects of thrusting on the radius and velocity are depicted in Fig. 6.13, clearly showing the variations as the spacecraft initially orbits the Earth, and then departs on the outward journey. Figure 6.14 shows the flight-path angle, φ, and the azimuth angle, ψ, showing the

6.5 Optimal Low-Thrust Three-Body Transfer

229

am = 0 am = 0. 01 m / s2

1 0

z

−1 −2 −3 −4 20 10 0 −10

y

−20 −30

−40

−20

−30

−10

20

10

0

30

x

Fig. 6.10 Optimal low-thrust trajectory (red) for out-of-plane, interstellar mission compared with the free flight case (blue)

am = 0

am = 0. 01 m / s2

20 15 10 5

y

0 −5 −10 −15 −20 −25 −40

−30

−20

−10

x

0

10

20

30

Fig. 6.11 Projection of the optimal low-thrust trajectory (red) in the (x, y) plane compared with the free flight case (blue)

230

6 Flight in Non-spherical Gravity Fields am = 0

am = 0. 01 m /s2

0.5 0 −0.5 −1

z

−1.5 −2 −2.5 −3 −3.5 −4 −40

−30

−20

−10

x

0

10

20

30

Fig. 6.12 Projection of the optimal low-thrust trajectory (red) in the (x, z) plane compared with the free flight case (blue) am = 0

am = 0. 01 m /s2

40

r

30 20 10 0

0

1

2

3

4

5

6

7

8

9

10

0

1

2

3

4

5

6

7

8

9

10

40

v

30 20 10 0

t

Fig. 6.13 Radius and speed of the optimal low-thrust trajectory (red) compared with the free flight case (blue)

6.5 Optimal Low-Thrust Three-Body Transfer

231

am = 0

am = 0. 01 m /s2

φ (deg. )

20 0 −20 −40 −60 0

1

2

3

4

5

6

7

8

9

10

1

2

3

4

5

6

7

8

9

10

90

ψ (deg. )

45 0 −45 −90 0

t

Fig. 6.14 Flight-path and velocity azimuth angles of the optimal low-thrust trajectory (red) compared with the free flight case (blue)

optimal thrusting asymptotically approaches the flight direction, with the azimuth angle varying between ±π/2 on the spiral trajectory. Figure 6.15 is a plot of the optimal thrust direction angles, α and β, resolved in the CR3BP frame (see Fig. 6.9). The maximum thrust acceleration, a = am , is maintained until tf = 10 (approx. 48 days). Zhang and Zhao [88] present a shooting method for solving the minimum-fuel, low-thrust orbital transfer problem in CR3BP field. An interesting application of the optimal low-thrust transfer method is in designing solar-sail missions around heliocentric and Earth centered orbits, such as by Otten and McInnes [64] and Macdonald and McInnes [56]. However, solar-sailing strategies are only locally optimal, and face numerical problems in addressing the transversality conditions, especially for out-of-plane maneuvers. The optimal mission design of a multi-body transfer trajectory involves the solution of a two-point boundary value problem, in which either the fuel or the time of flight is minimized, and the control is applied either impulsively or continuously. The effects of other bodies on either a two-body (Keplerian) or a restricted threebody model (CR3BP or ER3BP) is modeled by patching the numerical solutions at the boundaries of the feasible space for each optimization problem. Such approaches include patching of impulsive two-body trajectories on the spheres of influence [80], or transferring a vehicle to an invariant manifold of a CR3BP model using a

232

6 Flight in Non-spherical Gravity Fields 60

α (deg. )

40 20 0 −20 0

1

2

3

4

5

6

7

8

9

10

1

2

3

4

5

6

7

8

9

10

180

β (deg. )

135 90 45 0 −45 0

t

Fig. 6.15 Optimal thrust direction angles of the low-thrust trajectory (red) compared with the free flight case (blue)

continuous, optimal thrust. The design methods range from the numerical patching of two-body Lambert solutions [20] to analytical searches for particular manifolds of periodic and semi-periodic orbits around collinear Lagrangian points [37, 41]. Such approximate solutions are indispensable, and have served well in practical mission designs, such as the exploration of Jovian and Saturnian systems carried out by Galileo and Cassini missions, respectively.

6.6 Missions Around Collinear Lagrangian Points The collinear Lagrangian points of a planet–Sun system are unique vantage points for continuously observing either the planet or the Sun with a minimum-fuel expenditure. If these equilibrium points were unavailable, a large radius, Keplerian orbit would have to be established around the target body, with the associated high cost of transfer. A similar station near the Lagrangian point, L2 , of the Earth–Moon system can be used to relay communication signals between the far side of the Moon and the Earth. Such an idea was proposed by Farquhar [29] for the Apollo missions, but was first applied to the International Sun–Earth Explorer Satellite (ISEE-3) mission, wherein a spacecraft was stationed near the

6.6 Missions Around Collinear Lagrangian Points

233

Sun–Earth Lagrangian point, L1 , to observe the effects of the solar wind, and its interaction with the Earth’s magnetic field. Further practical examples of stationing a spacecraft near the Sun–Earth L1 or L2 points are the Wind, the Solar and Heliospheric Observatory (SOHO), the Advanced Composition Explorer (ACE), the Wilkinson Microwave Anisotropy Probe (WMAP), the GEOTAIL, the Genesis, ARTEMIS, Herschel, Planck, Chang‘e-5T1, Gaia, DSCOVR, and LISA Pathfinder missions. Although the collinear Lagrangian points are equilibrium points of the CR3BP, and provide excellent vantage points for observational purposes, they are also inherently unstable points. This implies that even a small, arbitrary perturbation will cause a spacecraft stationed at such a point to depart from it, and to approach either one of the two primaries. Such a perturbation occurs naturally due to the neglected dynamics in the CR3BP model, such as the elliptical (rather than circular) orbit of the primaries and the gravitational effect of other large bodies. Due to the inherent instability of a collinear point, a control strategy (called station keeping) is required to maintain the spacecraft in the vicinity of it by the periodic application of thrust. Such a position must be changing periodically in a small amplitude motion around a collinear point (like trying to balance a stick vertically in one’s palm), and is called a halo orbit. It should be noted that halo orbits are not the natural solutions of the CR3BP system to arbitrary initial conditions, but are artificially forced either by choosing special initial conditions that preclude the non-periodic (or secular) part of the motion or by providing external thrust inputs, which requires fuel expenditure. An analytical method of deriving the periodic and semi-periodic, smallamplitude orbits about the collinear Lagrangian points is based upon the local linearization of the CR3BP equations. By neglecting the higher-order terms in a series expansion about a collinear point, and by retaining only those initial conditions which do not excite the unstable response, an approximate linear model of the halo orbits can be obtained [78]. Such special (or particular) solutions of the linearized model, although unrealistic as accurate solutions of the nonlinear, inherently unstable, unforced dynamics, are however useful in providing nominal paths for maintaining the spacecraft in the vicinity of a collinear point, by the application of control. Other small-amplitude orbits of the collinear points are also possible, such as the quasi-halo [36], Lissajous [22], and rectilinear out-of-plane trajectories, which are not exactly periodic, but display a semi-periodic nature. Stable and unstable manifolds around the periodic and semi-periodic orbits can be constructed as feasible paths of transferring the spacecraft to and from, respectively, the collinear points in a practical Lagrangian point mission. The CR3BP equations expanded about the Lagrangian points L2 and L1 in a third-order series were presented by Farquhar and Kamel [30] and Richardson [69], respectively, for deriving the approximate periodic motions by neglecting the non-periodic (secular) part of the response by the classical method of Lindstedt– Poincaré. We will next consider such a model here for illustrative purposes.

234

6 Flight in Non-spherical Gravity Fields

6.6.1 Motion About the Collinear Points Consider a change in the CR3BP coordinates such that the origin is shifted from the barycenter to the collinear Lagrangian point, Li , i = 1, 2, 3. Since the motion around L3 is the same as that around L2 , except for the changed non-dimensional mass from μ to 1 − μ, we need to consider only L1 and L2 , whose locations in the original coordinate system are i = i i, i = 1, 2, where i is given by the root of the quintic equation for the i th Lagrangian point. The vectors from the Lagrangian point to the primaries m1 and m2 in new coordinates are thus the following: d1 = (1 − μ − i )i

(6.119)

d2 = −(μ + i )i and ρ = (ρx , ρy , ρz )T is the location of the third mass, m3 , from the Lagrangian point, normalized by d1 =| d1 |. The Lagrangian function, normalized by d12 , and after dropping the constant and linear terms in ρ˙ (as they do not contribute to the equations of motion) is expressed as follows: ˙ = L(ρ, ρ)

ρx 1 − U (ρ) (ρ˙ − Kρ)T (ρ˙ − Kρ) + i 2 d1

(6.120)

where K is given by Eq. (6.70), and U (ρ) = −

μ d13

(1 − μ) 1 d2 /d1 − 2 |ρ∓i| | ρ + (d2 /d1 )i | d1 d2

(6.121)

is the gravitational potential with ρ =| ρ |. In the denominator of the first term in Eq. (6.121), the upper sign is taken for the Lagrangian point L1 and the lower sign for L2 . A single series expansion of the Lagrangian function with d2 ≥ d1 would give the orbit about L1 for 0 < μ ≤ 1/2, and about L2 for 0 < μ < 1. When the following series expansion by Legendre polynomials, Pn (.), is introduced into Eq. (6.121): ∞

# 1 = ρ n Pn (±ρx /ρ) |ρ∓i|

(6.122)

n=0 ∞

# d2 /d1 = (d1 ρ/d2 )n Pn (−ρx /ρ) | ρ + (d2 /d1 )i |

(6.123)

n=0

we have the following Lagrangian: 1 (ρ˙ − Kρ)T (ρ˙ − Kρ) 2 ∞ # + cn ρ n Pn (ρx /ρ)

˙ = L(ρ, ρ)

n=2

(6.124)

6.6 Missions Around Collinear Lagrangian Points

235

where cn = (±1)n

μ d13

+ (−1)n

(1 − μ)d1n−2

(6.125)

d2n+1

where the upper sign is taken for the Lagrangian point L1 , and the lower sign for L2 . The Lagrangian in the neighborhood of L3 can be derived using the expression for L2 , normalizing ρ by d2 , and replacing the non-dimensional mass μ by 1 − μ. This yields the following coefficient for L3 : cn = (−1)n

1−μ d23

+ (−1)n

μd2n−2

(6.126)

d1n+1

Finally, the Lagrangian results in the following Euler–Lagrange equations of motion in the neighborhood of a Lagrangian point: ρ¨ − 2Kρ˙ + K2 ρ =

∞ #

ncn ρ n−2 Pn (ρx /ρ)ρ

n=2

+

∞ #

cn ρ n−2 Pn (ρx /ρ) (ρi − ρx ρ/ρ)

(6.127)

n=2

where dPn (nu) # = (2n − 4k − 1)Pn−2k−1 (ν) , dν

n−1 ) 2 (6.128) A linearization of Eq. (6.127) in the vicinity of the collinear point requires taking only the quadratic terms in the Lagrangian given by Eq. (6.124). Such an approximation results in the following: Pn (ν) =

¯ ˙ L(ρ, ˙ = L(ρ, ρ) ρ)

(0 ≤ k ≤

1 (ρ˙ − Kρ)T (ρ˙ − Kρ) + c2 ρ 2 P2 (ρx /ρ) 2

(6.129)

where we recall that P2 (ν) = (3ν 2 − 1)/2 and c2 =

μ d13

+

1−μ d23

(6.130)

Substitution of Eqs. (6.129) and (6.130) into Eq. (6.127) with n = 2 yields the following linearized equations of motion: ρ¨ − 2Kρ˙ + (K2 + C)ρ = 0

(6.131)

236

6 Flight in Non-spherical Gravity Fields

where ⎛

⎞ −2c2 0 0 C = ⎝ 0 c2 0 ⎠ 0 0 c2

(6.132)

The characteristic equation of the linearized system is given by det{s 2 I3×3 −2sK+K2 +C} = (s 2 +c2 )[s 4 +s 2 (2−c2 )+1+c2 −2c22 ] = 0

(6.133)

which yields the eigenvalues √ s1,2 = ±j c2 ,

s3,4,...,6

5 , 6 6 c − 2 ± 9c2 − 8c 7 2 2 2 =± 2

√ with j = −1. The coefficient c2 determines the nature of the linearized initial response. Its value can be analyzed as follows: c2 = a + b , a =

μ d13

, b=

1−μ d23

From the location of the collinear point, it follows that i +

μ(1 − μ − i ) (1 − μ)(μ + i ) − =0 | 1 − μ − i |3 | μ + i |3

which implies that a=

i − b(μ + i ) i − 1 + μ

or  c2 = a + b = b 1 −

i + μ i − 1 + μ

 +

i >1 i − 1 + μ

which holds for both L1 and L2 (since | 1 + μ |< 1 and | 2 + μ |> 1). Thus c2 > 1 always holds. Because c2 > 1, the coplanar (x, y) motion involves a conjugate pair of imaginary eigenvalues, ±λj , and a pair of real eigenvalues, ±α, where

λ=

5 , 6 6 2 − c + 9c2 − 8c 7 2 2 2 2

, α=

5 , 6 6 c − 2 + 9c2 − 8c 7 2 2 2 2

6.6 Missions Around Collinear Lagrangian Points

237

and the out-of-plane motion (z) (which is decoupled from the coplanar motion) √ consists of a conjugate pair of imaginary eigenvalues, ± c2 j . Such a configuration of eigenvalues is termed “saddle×center×center” type, and indicates naturally unstable equilibrium. Consequently, the initial response of the unforced linearized system is expressed as follows: x(t) = A1 eαt + A2 e−αt + A3 cos λt + A4 sin λt   y(t) = −k1 A1 eαt − A2 e−αt − k2 (A3 sin λt − A4 cos λt) √ √ z(t) = A5 cos c2 t + A6 sin c2 t

(6.134)

where k1 =

1 + 2c2 − α 2 1 + 2c2 + λ2 , k2 = 2α 2λ

and A1 , A2 , . . . , A6 are real coefficients to be determined from the initial conditions. As pointed out earlier, we are interested only in the periodic and semi-periodic motions around the collinear points. Therefore, we carefully select only those initial conditions which make the secular (or non-periodic) terms vanish, that is A1 = A2 = 0, thereby producing the following initial response of the linear system: x(t) = −Ax cos(λt + φ) y(t) = kAx sin(λt + φ) √ z(t) = Az sin( c2 t + ψ)

(6.135)

where k = k2 , and the amplitudes, Ax , Az , as well as the phase angles, φ, ψ, are to be determined from initial conditions. We note that a purely periodic, three√ dimensional response (i.e., a halo orbit) is possible, if and only if λ/ c2 is an integer. The smallest value of this integer being unity, the smallest energy halo orbit √ would be produced for the resonant condition (λ = c2 in which the coplanar and out-of-plane motions sustain each other). Unfortunately, this is not possible, because √ λ/ c2 is an irrational number. Thus the linear approximation cannot produce a halo orbit, but only gives rise to semi-periodic solutions, such as Lissajous tori. Hence, one must look for higher-order approximations to generate a halo orbit solution.

6.6.2 Lindstedt–Poincaré Method for Halo Orbits In order to devise a higher-order dynamic model for producing periodic orbits— called halo orbits—around the collinear points, the perturbation method of Lindstedt–Poincaré [58] is applied to neglect the secular part of the initial response, while retaining its periodic behavior. The method involves refining a frequency

238

6 Flight in Non-spherical Gravity Fields

scale by successively increasing the order of the series approximation. Consider the periodic motion to have a natural frequency, ω, such as the time scale of motion is changed from t to τ = t/ω. The series expansion of ω in terms of the non-dimensional parameter, 0 <   1, is the following: ω = 1 + ω1 +  2 ω2 + · · ·

(6.136)

where ω1 , ω2 , . . . are the frequency corrections applied to first, second, and higherorder terms. The time derivatives in τ are denoted by primes, and are related to the derivatives in t as follows: d(.) d(.) =ω = ω(.)

dt dτ ( d2 (.) 2 d .) = ω = ω2 (.)

dt 2 dτ 2  = 1 + 2ω1 +  2 (ω12 + 2ω2 ) + · · · (.)

(6.137)

The nonlinear equations of motion in the neighborhood of a collinear point, Eq. (6.127), are expanded up to the third-degree polynomials as follows: 3 c3 (2x 2 − y 2 − z2 ) + 2c4 (2x 2 − 3y 2 − 3z2 )x 2 3 ω2 y

+ 2ωx + (c2 − 1)y = −3c3 xy − c4 (4x 2 − y 2 − z2 )y (6.138) 2 3 ω2 z

+ λ2 z = −3c3 xz − c4 (4x 2 − y 2 − z2 )z + (λ2 − c2 )z 2

ω2 x

− 2ωy − (1 + 2c2 )x =

On expanding the solution in a third-order series of , we have the following result: x(τ ) = x1 (τ ) +  2 x2 (τ ) +  3 x3 (τ ) + · · · y(τ ) = y1 (τ ) +  2 y2 (τ ) +  3 y3 (τ ) + · · ·

(6.139)

z(τ ) = z1 (τ ) +  2 z2 (τ ) +  3 z3 (τ ) + · · · where the assumption of the same order expansion for both the coplanar and out-ofplane motions is made, based upon the approximation that the natural frequencies of the two motions are very close, that is . Δ = λ2 − c2 = O( 2 )

(6.140)

Such a frequency matching (or resonance) of the coplanar and out-of-plane motions is crucial in achieving the periodic motion.

6.6 Missions Around Collinear Lagrangian Points

239

After substituting the linearized equations, Eq. (6.131), the secular terms are made to vanish by taking ω1 = 0, in the second-order response (O()). Next, the third-order equations are derived (O( 2 )) by including the first- and second-order terms, and putting ω1 = 0 in the series expansion. The removal of secular terms from the third-order expansion is not possible simply by choosing an appropriate value for the frequency correction, ω2 . The secular part of the out-of-plane response is due to δ4 sin(2τ1 − τ2 ), which can be adjusted by selecting the phase angles such that φ =ψ +n

π , 2

(n = 0, 1, 2, 3)

(6.141)

Then the secular terms in both of the coplanar equations can be made to vanish by imposing amplitude constraints, as well as selecting a particular third-order frequency correction, ω2 . The final step in the approximation involves removing  from the equations by substituting the amplitudes Ax and Az by Ax / and Az /, respectively, thereby implying the two amplitudes to be of O(). The first- to third-order solutions obtained above are now combined to produce the following result: x(τ ) = a21 A2x + a22 A2z − Ax cos τ1 + (a23 A2x + ηa24 A2z ) cos 2τ1 + (a31 A3x + ηa32 Ax A2z ) cos 3τ1 y(τ ) = (kAx b33 A3x + b34 Ax A2z + ηb35 Ax A2z ) sin τ1 + (b21 A2x + ηb22 A2z ) sin 2τ1 + (b31 A3x + ηb32 Ax A2z ) sin 3τ1 (6.142) z(τ ) = &

(−1)n/2 (Az sin τ1 + d21 Ax Az ) sin 2τ1 + (d31 A3z + d32 A2x Az ) sin 3τ1 ) (n = 0, 2) 2 3 z cos τ1 + d21 Ax Az )(cos 2τ1 − 3) + (d32 Ax Az − d31 Az ) cos 3τ1 ) (n = 1, 3)

(−1)(n−1)/2 (A

where the coefficients a21 − d32 are listed in Thurman and Worfolk [82], and η = (−1)n , (n = 0, 1, 2, 3). The qualitative change in the periodic behavior due to the switching of the out-of-plane phase angle, ψ, between odd and even multiples of π/2 is termed bifurcation. Hence the two distinct classes of periodic solutions are indicated by n = (0, 2) and n = (1, 3), respectively. Example 6.2 Consider a halo orbit around the Sun–Earth Lagrangian point, L1 , representing the ISEE-3 mission parameters [70] with λ = 2.08645, and Δ = 0.292214, and simulated for Az = 125,000 km in Figs. 6.16, 6.17, 6.18, and 6.19. The period of the orbit is 177.704 mean solar days, and the (x, y) symmetry as well as the phase change in the out-of-plane (z) motion from n = 1 (blue) and n = 3 (red) are clearly seen. The coplanar part of the motion is unaffected by the change in ψ (Fig. 6.17), whereas both (y, z) (Fig. 6.18) and (x, z) (Fig. 6.19) projections show a sign change of the z motion from ψ = π/2 (blue) to ψ = 3π/2 (red).

240

6 Flight in Non-spherical Gravity Fields −3

x 10 1 0.5

z

0 −0.5 −1 0.5

0

y

−0.5

−0.15

−0.1

−0.05

x

0

0.05

0.2

0.15

0.1

Fig. 6.16 Three-dimensional halo orbit around the Sun–Earth collinear point L1 with Az = 125,000 km and period 177.704 days, showing out-of-plane phase change from ψ = π/2 (blue) to ψ = 3π/2 (red)

0.5 0.4 0.3 0.2

y

0.1 0 −0.1 −0.2 −0.3 −0.4 −0.5 −0.15

−0.1

−0.05

0

x

0.05

0.1

0.15

0.2

Fig. 6.17 The (x, y) projection of the halo orbit around the Sun–Earth collinear point L1 with Az = 125,000 km and period 177.704 days, unchanged by the out-of-plane phase change from ψ = π/2 (blue) to ψ = 3π/2 (red)

6.6 Missions Around Collinear Lagrangian Points

241

−3

1

x 10

0.8 0.6 0.4 0.2

z

0 −0.2 −0.4 −0.6 −0.8 −1 −0.5

−0.4

−0.3

−0.2

−0.1

0

y

0.1

0.2

0.3

0.4

0.5

Fig. 6.18 The (y, z) projection of the halo orbit around the Sun–Earth collinear point L1 with Az = 125,000 km and period 177.704 days, showing a sign change in the z response due to the out-of-plane phase change from ψ = π/2 (blue) to ψ = 3π/2 (red) −3

1

x 10

0.8 0.6 0.4

z

0.2 0 −0.2 −0.4 −0.6 −0.8 −1 −0.15

−0.1

−0.05

0

x

0.05

0.1

0.15

0.2

Fig. 6.19 The (x, z) projection of the halo orbit around the Sun–Earth collinear point L1 with Az = 125,000 km and period 177.704 days, showing a sign change in the z response due to the out-of-plane phase change from ψ = π/2 (blue) to ψ = 3π/2 (red)

242

6 Flight in Non-spherical Gravity Fields

6.7 Numerical Computation of Halo Orbits and Manifolds The CR3BP equations can be expressed as follows in terms of the canonical coordinate vector, X = (qT , pT )T , in the Hamiltonian form given by Eq. (6.79), repeated here for convenience as follows: ˙ = J∇X H = f(X) X

(6.143)

where ∇X is the gradient operator relative to X,  J=

03×3 I3×3 −I3×3 03×3

 (6.144)

the Hamiltonian is given by H =

1 T p p + pT Kq + U (q) 2

(6.145)

the generalized momentum vector by p = q˙ − Kq ,

(6.146)

⎞ 0 1 0 K = ⎝ −1 0 0 ⎠ 0 0 0

(6.147)

⎛

the gravitational potential by 1 (1 − μ) μ U (q) = −Ω − qT K2 q = − − 2 r1 r2

(6.148)

with (1 − μ) 1 μ (1 − μ) μ 1 + = (x 2 + y 2 ) + + Ω = − qT K2 q + 2 | q − (1 − μ)i | | q + μi | 2 r1 r2 (6.149) and i = (1, 0, 0)T and r1 , r2 given by Eq. (6.54). When expressed in the functional form, Eq. (6.143) is the following: ˙ = f(X) = X



Kq + p −∇q (U ) − KT p

 (6.150)

6.7 Numerical Computation of Halo Orbits and Manifolds

243

where ∇q (.) is the gradient operator relative to q.1

6.7.1 Monodromy Matrix Due to its nonlinear nature, a closed-form solution subject to a given initial condition, X(0) = X0 , is unavailable. However, an approximate numerical solution can be attempted for a periodic orbit where the motion repeats itself after a fixed time period, T , that is X(t + T , X0 ) = X(t, X0 ). Consider the change in the solution due to a small, arbitrary change in the initial condition, (Δt, ΔX0 ), expressed in the following Taylor series:  X(t + Δt, X0 ) = X(t, X0 ) +  +

∂X ∂t



∂X ∂X0

 ΔX0 (t,X0 )

Δt + O(Δ2 t, | ΔX0 |2 )

(6.151)

(t,X0 )

By truncating the series at the linear terms, the solution is treated as if it were that of a linear, time-varying system with the following state transition matrix:  (t, X0 ) =

∂X ∂X0

 (6.152) (t,X0 )

which satisfies the initial condition (0, X0 ) = I6×6

(6.153)

Taking the partial time derivative of , we have ∂ ∂ ∂X = ∂t ∂X0 ∂t

1A

different state-space representation is obtained by replacing the generalized momentum vector, ˙ p, with the velocity, q:     q˙ q˙ = f(X) = KT Kq − ∇q (U ) + 2Kq˙ q¨ whose Jacobian is given by f (X) =



0 Ωqq

I 2K



244

6 Flight in Non-spherical Gravity Fields

 =

∂f(X) ∂X



 X0

∂X ∂X0

 (6.154) (t,X0 )

= f (X) where f (X) =



∂f(X) ∂X



 = X0

K I −Uqq −KT

 (6.155)

is the Jacobian matrix and  Uqq =

∂ 2U ∂q∂q

 q0

For a periodic orbit, the state transition matrix, , computed over one time period is termed the monodromy matrix, denoted by M = (T , X0 ). Since it is evaluated at the initial condition, X(0) = X0 , the monodromy matrix repeats itself after each time period, T , because X(T ) = X(0). It has been noted in both Chaps. 2 and 3 that the state transition matrix of a conservative Hamiltonian system is symplectic [60], hence the same is true for (t, X0 ) as well as the monodromy matrix, that is, they both obey M−1 = −JMT J, where J is given by Eq. (6.144). The symplectic nature (see Chap. 2) of the monodromy matrix, as well as the time-reversal symmetry of the CR3BP solutions across the (x, z) plane (noted above), enables a direct computation of a halo orbit. Computation of the state transition matrix for a fixed initial state, X0 , is carried out by solving the differential equation governing its time evolution, given by ˙ = f (X) , (0) = I  Such an iteration utilizes the symmetry of the CR3BP solutions discussed earlier. For example, consider Newton’s method consisting of the following steps: (i) Assume an initial guess of the solution on the (x, z) plane, X(0) = (x0 , 0, z0 , 0, py0 , 0)T . Then the first return of the symmetric, periodic orbit to the (x, z) plane is given by X(T /2, X0 ) = (x, ¯ 0, z¯ , p¯ x , p¯ y , p¯ z )T , where T is the time period. (ii) To search for the first half of the orbit, the initial conditions (x0 , z0 , py0 ) are to be found which produce v¯x = p¯ x − x¯ = 0 and v¯z = p¯ z = 0 on the (x, z) plane. For this purpose, a correction, ΔX0 = (Δx0 , 0, Δz0 , 0, Δpy0 , 0)T , is applied to the initial state such that the state evolves to cancel the nonzero terms at t = T /2, v¯x , v¯z , by driving them to −v¯x , −v¯z at a time step t = T /2 + δt: X(T /2 + Δt, X0 + ΔX0 ) X(T /2, X0 ) + (T /2, X0 )ΔX0 + f[X(T /2, X0 )]Δt

(6.156)

6.7 Numerical Computation of Halo Orbits and Manifolds

245

which yields ⎫ ⎛⎧ ⎞ ⎪ ⎪ ··· ⎪ ⎪ ⎪ ⎪ ⎪ ⎜⎪ ⎟ ⎪ 0 ⎪ ⎪ ⎜⎪ ⎟ ⎪ ⎨ ⎬ ⎜⎪ ⎟ ··· ⎜ ⎟ −1 − f[X(T /2, X0 )]Δt ⎟ ΔX0 =  (T /2, X0 ) ⎜ ⎜⎪ ⎟ −(p¯ x − x) ¯ ⎪ ⎪ ⎪ ⎜⎪ ⎟ ⎪ ⎪ ⎪ ⎪ ⎝⎪ ⎠ · · · ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ −p¯ z

(6.157)

The three initial variable corrections, (Δx0 , Δz0 , Δpy0 ), and the time step, Δt, are determined from the three equations given by Eq. (6.157). Since there are four unknowns and only three equations, the z0 variation is set to zero, (Δz0 = 0), by selecting an amplitude, z0 , of the out-of-plane motion. This can be done by taking a derivative of Jacobi’s integral at the initial time. (iii) The state transition matrix computed for half orbit, (T /2, X0 + ΔX0 ), is corrected for the new initial condition, and step (ii) is repeated after replacing T /2 → T /2 + Δt. This process is continued until v¯x , v¯z are driven to within a small tolerance. The iterative procedure given above converges to an initial condition, X0 , and a semi-period, T /2, from which the monodromy matrix, M = (T , X0 ), can be computed by using the symmetry (reflection with time reversal) of the orbit about the (x, z) plane as follows [40]: M = (T , X0 ) = A−1 (T /2, X0 )A(T /2, X0 )

(6.158)

⎞ 1 0 0 ,  = ⎝ 0 −1 0 ⎠ 0 0 1

(6.159)

where  A=

 0 0 −



⎛

In the above given calculations, the symplectic nature of  allows its following inversion: −1 (T /2, X0 ) = V−1 S−1 V−T T (T /2, X0 )VT SV

(6.160)

where  V=

I 0 −K I



 , S=

0 I −I 0

 (6.161)

and K is given by Eq. (6.147). The final expression for M is thus the following:     −2K I 0 −I (6.162) A(T /2, X0 ) M=A T (T /2, X0 ) −I 0 I −2K

246

6 Flight in Non-spherical Gravity Fields

The eigenvalues of the monodromy matrix for a nearly planar orbit [40] can be classified as two reciprocal pair of real eigenvalues, (λ1 , λ1 ), (λ2 , λ2 ), respectively, and a complex conjugate pair, λ3 , λ∗3 . Since monodromy matrix is periodic (with a unit determinant due to its symplectic character), a stable motion should have all eigenvalues lying either inside or on the unit circle in the complex plane. However, due to the instability of the collinear points, there is a real eigenvalue greater than unity, λ1 > 1, its reciprocal eigenvalue less than unity, while λ2 = 1 and | λ3 |=| λ∗3 |= 1 are on the unit circle. The two reciprocal eigenvalues, λ1 , λ1 , not falling on the unit circle are termed the unstable and stable eigenvalues, respectively, as they represent either growing or decaying periodic motion.

6.7.2 Stable and Unstable Manifolds A manifold is a set of trajectories in the CR3BP space, x(t), y(t), z(t), which satisfy a particular condition, F[x(t), y(t), z(t)] = 0. The zero-speed contours of constant Jacobi’s integral are an example of a CR3BP manifold. Manifolds associated with a halo orbit around a collinear point are especially useful as a set of transfer trajectories either to or from the collinear point, and are utilized in practical missions for stationing spacecraft around such a point. Manifolds structured around periodic orbits are also termed invariant manifolds, because a trajectory starting on such a manifold always remains on it. Hence manifolds can be used to design essentially fuel-free paths in optimal low-thrust transfer missions, as discussed later in the chapter. A stable manifold begins from a halo orbit and evolves in forward time, t > 0, along the eigenvector, vk− , associated with the stable eigenvalues, λk− < 1, of the monodromy matrix, M = (T , X0 ) [38]. Thus the stable manifold is tangential to both the halo orbit and a stable eigenvector, and evolves away from the halo orbit. An unstable manifold terminates at a halo orbit, and evolves in backward time, t < 0, along the eigenvector, vk+ , associated with an unstable eigenvalue, λk+ > 1, of the monodromy matrix. Thus an unstable manifold serves as an envelope of possible trajectories evolving toward the halo orbit in forward time. For a nearly planar halo orbit, there is only one stable eigenvalue, and also only one unstable eigenvalue. Thus the stable and unstable manifolds of a planar orbit can be constructed by evaluating the corresponding stable and unstable eigenvectors at each point on the orbit, and propagating the point tangentially along the respective eigenvectors using the state transition matrix, (t, X0 ). With a linearized system dynamics, the construction of manifolds becomes much easier, which has been proposed by Thurman and Worfolk [82] and further studied by Bando and Scheeres [6] using the linear optimal control formulation. Howell et al. [41] carry out a similar construction of manifolds around Lissajous (semi-periodic) orbits for the Genesis mission. For the system linearized in the neighborhood of a collinear Lagrangian point, the Hamiltonian is in the following quadratic form: H =

1 1 T p p + pT Kq + qT Cq 2 2

(6.163)

6.7 Numerical Computation of Halo Orbits and Manifolds

247

where ⎛

⎞ −2c2 0 0 C = ⎝ 0 c2 0 ⎠ 0 0 c2

(6.164)

with c2 > 1 being the coefficient associated with the second-degree Legendre polynomial (Eq. (6.130)). Consequently, the linearized canonical equations are time invariant, and expressed in a state-space form as follows: ˙ = X



K I −C −KT

 X = AX

(6.165)

which is the same system as Eq. (6.131), with the eigenvalues ±λj and ±α for the √ coplanar motion and ± c2 j , with

λ=

5 , 6 6 2 − c + 9c2 − 8c 7 2 2 2 2

, α=

5 , 6 6 c − 2 + 9c2 − 8c 7 2 2 2 2

The state transformation from canonical state space, X = (qT , pT )T , to the ¯ = VX, with eigenspace, X ˙¯ = A ¯X ¯ X

(6.166)

such that ⎞ α 0 0 0 0 0 ⎜0 0 0 0 λ 0 ⎟ ⎜ √ ⎟ ⎟ ⎜ 0 0 0 c2 ⎟ ⎜0 0 =⎜ ⎟ ⎜0 0 0 −α 0 0 ⎟ ⎟ ⎜ ⎝ 0 −λ 0 0 0 0 ⎠ √ 0 0 − c2 0 0 0 ⎛

¯ = VAV−1 A

(6.167)

is given by the transformation matrix 0 V=

1 1 1 1 1 1/4 v+ , vR , 1/4 e3 , v− , vI , c2 e3 w1 w2 w w 1 2 c2

1−1 (6.168)

where v+ is the eigenvector of A associated with the unstable eigenvalue, α, v− is that corresponding to the stable eigenvalue, −α, vR ± j vI the eigenvectors corresponding to the imaginary coplanar eigenvalues, ±j λ, e3 = (0, 0, 1, 0, 0, 0)T , e6 = (0, 0, 0, 0, 0, 1)T , and

248

w1 =

6 Flight in Non-spherical Gravity Fields

,

, 2α[(4 + 3c2 )α 2 + 4 + 5c2 − 6c22 ] , w2 = λ[(4 + 3c2 )λ2 − 4 − 5c2 + 6c22 ]

are the normalizing factors associated with α and λ, respectively. Example 6.3 For the halo orbit around the Sun–Earth Lagrangian point, L1 , considered in Example 6.2, we have c2 = 4.06107, which produces the following linearized state dynamics matrix: ⎛

0 1 0 ⎜ −1 0 0 ⎜ ⎜ 0 0 0 ⎜ A=⎜ ⎜ 8.12214 0 0 ⎜ ⎝ 0 −4.06107 0 0 0 −4.06107

1 0 0 0 −1 0

0 1 0 1 0 0

⎞ 0 0⎟ ⎟ ⎟ 1⎟ ⎟ 0⎟ ⎟ 0⎠ 0

has the state transformation matrix ⎞ ⎛ −1.7527 −0.2022 0 −0.5714 −0.3055 0 ⎜ 0.4441 0 0 0 1.2550 0 ⎟ ⎟ ⎜ ⎟ ⎜ 0 0 1.4196 0 0 0 ⎟ ⎜ V=⎜ ⎟ ⎜ 1.7527 −0.2022 0 −0.5714 0.3055 0 ⎟ ⎜ ⎟ ⎝ −0.0000 −2.2299 0 −0.3886 0 0 ⎠ 0 0 0 0 0 0.7044 and the following transformed dynamics matrix in the eigenspace: ⎞ 2.5327 0 0 0 0 0 ⎜ 0 0 0 0 2.0865 0 ⎟ ⎟ ⎜ ⎟ ⎜ 0 0 0 0 0 2.0152 ⎟ ⎜ ¯ =⎜ A ⎟ ⎜ 0 0 0 −2.5327 0 0 ⎟ ⎟ ⎜ ⎝ 0 −2.0865 0 0 0 0 ⎠ 0 0 −2.0152 0 0 0 ⎛

For transferring a spacecraft to (or from) a collinear point orbit, the optimal strategy consists of transferring the spacecraft to an invariant manifold associated with the given collinear point. The remaining journey (or a large part of it) is then carried out on the manifold without fuel expenditure. Anderson and Lo [2] study the practicality of invariant manifolds in designing a low-thrust trajectory. Martin and Conway [25] apply a Gauss–Lobatto collocation method for solving the 2PBVP associated with an optimal transfer on an invariant manifold, and consider optimal transfers to Earth–Moon L1 halo orbit. Such nonlinear programming methods are valuable in determining either the shortest (time optimal) or the most efficient (fuel optimal) low-thrust transfer paths in CR3BP and ER3BP gravity models.

6.8 Optimal Station Keeping around Collinear Points

249

6.8 Optimal Station Keeping around Collinear Points A halo orbit either constructed analytically by the method of the previous section or computed by a numerical scheme can be utilized as the nominal trajectory to station a spacecraft around a collinear Lagrangian point. The design of an orbital regulation system can be carried out using a suitable optimal feedback control law (see Chap. 2) to track the halo orbit in the presence of disturbances. Such disturbances (or noise) are problematic, because as mentioned earlier, the collinear points are naturally unstable equilibrium points of the CR3BP system, and therefore necessitate an automatic regulation system. It is crucial for such a controller to be optimal for a prolonged mission, because frequent, arbitrary corrections can quickly deplete the fuel, thereby shortening the life of the spacecraft. We shall briefly consider some of the optimal station keeping methods that either have been applied in actual missions or are being explored for future applications.

6.8.1 Impulsive Control Impulsive control methods are traditionally applied in most missions, as they are powered by chemical rockets. If N impulses are applied per orbit, 4 the optimization consists of minimizing the net velocity impulse magnitude, N i | Δvi |, per orbit. This invariably requires that each impulse be applied in an optimal direction. The simplest possible optimal impulsive strategy is to determine each velocity impulse, Δvi = (Δx, ˙ Δy, ˙ Δ˙z)T , such that the coefficient of the unstable term, A1 , in the linearized response, Eq. (6.134), is driven to zero as follows: Δvi = −A1

∇A1 | ∇A1 |2

(6.169)

where ∇ = (∂/∂Δx, ˙ ∂/∂Δy, ˙ ∂/∂Δ˙z)T denotes the velocity gradient. Such a problem can be solved either by analytically or by dynamic programming. An alternative approach is to minimize 1 J = T



t0 +T t0

A21 (t)dt

(6.170)

where t0 is the initial time and T the time interval between two impulses. Another method is to determine the velocity impulses for maximizing the time between corrections. Shirobokov et al. [73] survey such impulsive methods, mostly from the Russian archive, and suggest improvements in the optimal strategy. A more sophisticated approach is to apply each impulse opposite to the unstable manifold direction, determined by the eigenvector of the monodromy matrix corresponding to the unstable eigenvalue of the linearized system (see the previous

250

6 Flight in Non-spherical Gravity Fields

section). This would attempt to cancel the instability in the interval between the impulses. However, as the errors in the linearized dynamics increase with time, and the period T between two impulses must be reasonably large for fuel minimization, it is necessary to compute the state transition matrix, Φ(t0 , t), of the first variational equations more accurately. Gomez et al. [35] apply the Floquet mode approach to evolve the state transition matrix by Monte Carlo simulation for stationing a spacecraft near Earth–Moon L2 point. A nonlinear sliding mode application [74] for station keeping using the discrete linear, quadratic regulator (LQR) formulation has also been studied [55], and applied to Lissajous trajectories around Earth–Moon L2 point. Other impulsive methods applied to the station keeping problem include the targeting methods [66] and Chebyshev–Picard iteration methods [5].

6.8.2 Continuous Thrust Control Continuous, low-thrust inputs can be applied optimally to maintain a spacecraft in the vicinity of a collinear point by tracking a given halo orbit in the synodic (CR3BP) frame. Similarly, tracking a manifold of a halo orbit can optimally transfer the spacecraft either toward or away from the collinear point. Consider the following equations of motion linearized about such a nominal trajectory (either the halo or its manifold), qn (t) = {xn (t), yn (t), zn (t)}T : ˙ X(t) = A(t)X(t) + B(t)u(t)

X(0) = X0

(6.171)

where X = (δqT , δpT )T is the state vector comprising first-order deviations from the nominal trajectory, Xn = (qTn , pTn )T : q(t) qn (t) + δq(t) p(t) pn (t) + δp(t)

(6.172)

u the control acceleration input, and 



A = f (X) =

∂f(X) ∂X



 =

Xn

K I −Uqq −KT



  0 ; B= I

(6.173)

are the Jacobian matrices, with f(X) the same as in Eq. (6.143) for the equations of motion referred to the synodic frame with the origin at the barycenter, and the potential Hessian, Uqq , evaluated on the nominal trajectory as follows:  Uqq =

∂ 2U ∂q∂q

 qn

⎞ ⎛ 1−μ 0 0 1−μ⎝ = 3/2 0 1 0⎠ r1 0 0 1

6.8 Optimal Station Keeping around Collinear Points

251

⎧ ⎫ ⎞ ⎛ μ−1 0 0 xn − μ ⎬ ⎨ 3(1 − μ) μ ⎝ − {xn − μ, yn , zn } + 3/2 0 1 0⎠ yn 5/2 ⎩ ⎭ r1 r 2 0 0 1 zn ⎧ ⎫ x + μ − 1⎬ 3μ ⎨ n − 5/2 (6.174) {xn + μ − 1, yn , zn } yn ⎭ r2 ⎩ zn where , (xn − μ)2 + yn2 + zn2 , r2 = (xn + μ − 1)2 + yn2 + zn2 r1 =

(6.175)

An optimal control approach can be applied, such as the linear, quadratic regulator (LQR) (see Chap. 2) for minimizing the following performance index for driving the trajectory variations to zero: 1 J (t) = 2



∞

XT (τ )Q(τ )X(τ ) + uT (τ )R(τ )u(τ ) dτ

(6.176)

t

subject to the state constraint equation, Eq. (6.171), resulting in the linear, full-state feedback control law: u(t) = −K∞ (t)x(t)

(6.177)

Since the plant is slowly time varying, the stabilizing steady-state solution for the feedback gain matrix at each time instant, K∞ (t), is given by K∞ (t) = R−1 (t)BT (t)P(t)

(6.178)

where P∞ (t) is the symmetric, positive semi-definite, quasi-steady solution of the following algebraic Riccati equation: P∞ (t)A(t) + AT (t)P∞ (t) − P∞ (t)B(t)R−1 (t)BT (t)P∞ (t) + Q(t) = 0

(6.179)

For output feedback applications, a Kalman filter can be added in an LQG/LTR formulation [79] to minimize the effects of the measurement noise inputs. Breakwell et al. [18] apply an LQR-based approach to the ISEE-3 mission for station keeping around the Sun–Earth L2 point. Ghorbani and Assadian [33] study the Earth–Moon Lagrangian point missions by the LQR method, and also include the elliptical restricted three-body (ER3BP) model, as well as the gravitational perturbations of Sun and other planets. Jones and Bishop [43] apply the H2 robust control methodology (a variation of the LQG/LTR method using the LQR at its core) [79]

252

6 Flight in Non-spherical Gravity Fields

for tracking Lissajous trajectories around Earth–Moon L2 point, whereas Kulkarni et al. [48] apply a similar H∞ approach for halo orbit station keeping around Sun– Earth L1 point. Other methods applied to collinear point station keeping include the backstepping [63] and model-predictive [54] control. Thus the collinear point station keeping problem has been explored from nearly all possible angles in terms of control theory, and continues to be an active research topic.

Answers to Selected Exercises

Exercises of Chap. 2 2.1 

e5 x (t) = e + 5 e − e−5 ∗



t

 −t  e − et , u∗ (t) = −2



e5 e5 − e−5



e−t , (0 ≤ t ≤ 5)

∗ = 1 > 0. Optimal, because Huu

2.2 y ∗ (t) = −20t 3 + 30t 2 , u∗ (t) = −120t + 60, (0 ≤ t ≤ 1) ∗ = 1 > 0. Optimal, because Huu

2.3

y ∗ (t) =

10k12

   1+k22 1 3 2 −3t + t 1+2k22   , 1+k22 5 1 2 6 + 2 1+2k2

20k12 

∗

u (t) = 5 6

+

1 2

1+k22 1+2k22

!

 −t +

0

1 + k22

1"

1 + 2k22

(0 ≤ t ≤ 1)

k12 = 6.5, k22 = 3000 (Fig. 1).

© Springer Nature Switzerland AG 2019 A. Tewari, Optimal Space Flight Navigation, Control Engineering, https://doi.org/10.1007/978-3-030-03789-5

253

254

Answers to Selected Exercises

y (m)

10

5

0

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0.6

0.7

0.8

0.9

1

t (s) 75

u(m / s2 )

50 25 0 −25 −50 −75 0

0.1

0.2

0.3

0.4

0.5

t (s)

Fig. 1 The optimal trajectory and the corresponding control history for k12 = 6.5, k22 = 3000 (Exercise 2.3)

2.4 tf = 1, y ∗ (t) = −20t 3 + 30t 2 , u∗ (t) = −120t + 60 (0 ≤ t ≤ tf )  2 2.5 No, because it results in the necessary condition, etf + e−tf = −1, which does not have a solution for any real value of tf . 2.7 The optimal displacement and velocity for this case are plotted in Fig. 2: The final value of vertical acceleration input is u(tf ) = g = 1.6 m/s2 , because x˙2 (tf ) = 0. 2.8   %2 ∂φ ∗ 1$ ∗ H+ u (tf ) + λ∗1 (tf )x2∗ (tf ) + λ∗2 (tf )u∗ (tf ) = 0 = ∂t t=tf 2 or −

2 1 c1 tf − c2 + c1 gtf + k 2 tf = 0 2

implying tf =

   ,  1  2 2 2 − c2 c2 c ± c + c g + k c + c g + k c 1 2 1 1 2 1 1 2 c12

Answers to Selected Exercises

255

z*(m)

100 50 0

0

5

10

15

20

25

30

5

10

15

20

25

30

5

10

15

20

25

30

z˙ *(m / s)

5 0 −5 −10 0

u*(m /s2)

2.5 2

1.5

0

t(s) Fig. 2 The optimal trajectory for lunar landing (Exercise 2.7)

A real and positive solution for tf is derived for k = 1/100, for which tf = 57.0248 s, c1 = 0.0249, and c2 = −0.714. The optimal displacement and velocity for this case are plotted in Fig. 3. 2.15  u(t) ˆ =

−1 , λˆ (t) > 0 ˆ 1 , λ(t) <0

H = 1 + λu = 0, x(0) ˆ = 1, uˆ = −1, (λˆ = 1), x(t) ˆ = 1 − t, (0 ≤ t ≤ tf ), x(t ˆ f ) = 0, tf = tˆ = 1. 2.18 (a) No. (b) For a = 1.5g = 2.4 m/s2 , tf =

  , $ % 2 z˙ (0) + z˙ 2 (0) + 5 6gz(0) + z˙ 2 (0) = 15.8712 s 5g tˆ =

z˙ (0) + 12 gtf = 0.5619 s 3g

256

Answers to Selected Exercises 200

z*(m)

150 100 50 0

0

10

20

30

40

50

60

40

50

60

t(s) 10

z˙ *(m/s)

5 0 −5 −10 0

10

20

30

t(s)

Fig. 3 The optimal trajectory for lunar landing (Exercise 2.8)

100

z(m)

50 0 −50 0

2

4

6

8

10

12

14

16

2

4

6

8

10

12

14

16

2

4

6

8

10

12

14

16

z(m ˙ / s)

0 −5 −10 −15 0

u(m / s2 )

5 0

−5 0

t(s) Fig. 4 Time-optimal trajectory for lunar landing (Exercise 2.18)

Answers to Selected Exercises

257

 u(t) =  z(t) =

−1.5g (0 ≤ t ≤ tˆ) 1.5g (tˆ < t ≤ tf )

(0 ≤ t ≤ tˆ) z(0) + z˙ (0)t − 54 gt 2 z(0) + z˙ (0)t + 14 gt 2 + 32 g tˆ2 − 3g tˆt (tˆ < t ≤ tf )

The time-optimal displacement and velocity for this case are plotted in Fig. 4.

Exercises of Chap. 3 3.3 (a) (b) (c) (d) (e) (f)

v = 0.1995 km/s, φ = 53.643◦ . v = 0.1289 km/s, φ = −31.357◦ . 1018.3 s. C is 25.4246◦ behind the radial position of A. z = 0.0024, x = 0.6588, a = 606.591 km, e = 0.3820. Zero.

3.4 (a) v = 0.4164 km/s, φ = 0. (b) v = 0.3875 km/s, φ = 0. 3.5 a = 3875.483 km, e = 0.950225, v1 = 3.32175 km/s, φ1 = 58.221◦ , v2 = 1.855 km/s, φ2 = −28.344◦ . No, because the orbit is Earth intersecting. 3.6 a = 8003.0968 km, e = 0.1754826, i = 63.6658◦ ,  = 98.1301◦ , ω = 59.3285◦ , τ = −373.756 s. 3.7 v1 = −0.193607I − 0.41983J + 3.32604K km/s v2 = 7.28047I + 0.16851J + 6.2595K km/s 3.8 5.5939 km/s. 3.9 lim (v1 + v2 + v3 ) = 14.586 km/s

rt →∞

For rt = 100 A.U., v1 + v2 + v3 = 15.04 km/s.

258

Answers to Selected Exercises

Exercises of Chap. 4 4.1 ar∗ (t) = −λ∗r˙ (t) , ⎧ ⎫ ⎧ ∗ ⎫ c1 cos nt + nc2 sin nt + 2c3 sin nt ⎪ ⎪ λ (t) ⎪ ⎪ ⎪ ⎪ r ⎪ ⎪ ⎪ ⎪ ⎨ c ⎨ ⎬ ⎬ c3 1 ∗ λr˙ (t) = − sin nt + c2 cos nt + 2 (cos nt − 1) (0 ≤ t ≤ tf ) ⎪ ⎪ ⎪ ⎪ n n ⎪ ⎪ ⎪ ⎪ ⎩ ∗ ⎪ ⎪ ⎭ ⎩ ⎭ λθ (t) c 3

where c1 , c2 , c3 are to be solved from the terminal boundary conditions, δr(tf ) = δ r˙ (tf ) = δθ (tf ) = 0 4.2 λ∗θ (tf ) = 0 = c3   ∂φ 1 H+ = 0 = − [λ∗r˙ (tf )]2 ∂t t=tf 2 which implies λ∗r˙ (tf ) = 0. This along with δr(tf ) = δ r˙ (tf ) = 0 yields c1 =

2n2 [nδr(0) cos ntf + δ r˙ (0) sin ntf ] cos ntf ntf − sin ntf cos ntf

c2 =

2n[nδr(0) cos ntf + δ r˙ (0) sin ntf ] sin ntf ntf − sin ntf cos ntf

and tf satisfies the following equation: n2 tf δr(0) sin ntf + δ r˙ (0)(sin3 ntf + cos2 ntf sin ntf − ntf cos ntf ) = 0 4.4 aθ∗ (t) = −λ∗θ˙ (t) ⎫ ⎧ ∗ ⎪ ⎪ ⎪ λr (t) ⎪ ⎪ ⎬ ⎨ λ∗ (t) ⎪ θ ∗ ⎪ ⎪ ⎪ λr˙ (t) ⎪ ⎪ ⎭ ⎩ λ∗ (t) ⎪ θ˙ ⎧ ⎫ ⎪ c1 (4 − 3 cos nt) + 6c2 (nt − sin nt) + 3nc3 sin nt − 6nc4 (1 − cos nt) ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ c2 = (0 ≤ t ≤ tf ) (1−cos nt) sin nt ⎪ ⎪ −c1 n − 2c2 + c3 cos nt + 2c4 sin nt ⎪ ⎪ n ⎪ ⎪ ⎩ ⎭ nt) + c2 (3nt−4nsin nt) − 2c3 sin nt + c4 (4 cos nt − 3) 2c1 (1−cos n

Answers to Selected Exercises

(km)

200

259

δr Cδθ

100 0 −100 0

20

40

60

80

100

120

140

160

(km / s)

0.2

δr˙ Cδ θ˙

0

−0.2 0

180

20

40

60

80

100

120

140

160

180

20

40

60

80

100

120

140

160

180

aθ (m / s2 )

1 0.5 0 −0.5 0

nt (deg. ) Fig. 5 The optimal trajectory for coplanar rendezvous (Exercise 4.4)

where c1 , c2 , c3 , c4 are to be solved from the following terminal boundary conditions:   14 2π 2 3π 2 8 2π − c2 − 2 c3 − 2 c4 = 0.07C c + 1 3 2 2n n n n n n 2 − 3 n

      3π 2 3π 2 π 2π 3π 2 c1 + 3 16 − c2 − 2 c3 +3 16 − c4 = 0.06π C 14 − 2 2 2 n n 8 2π − c1 + c2 − 2π c3 + 12c4 = 0 n n   2π 3π 2 3 12 π c2 − c3 − c4 = 0.12Cn c1 + 2 16 − 2 n n n2 n

The optimal trajectory and control history are plotted in Fig. 5. 4.5 an∗ (t)

=

−λ∗z˙ (t)

 ,

λ∗z (t) λ∗z˙ (t)



 =

c1 cos nt + c2 n sin nt) −c1 sinnnt + c2 cos nt

 (0 ≤ t ≤ tf )

260

Answers to Selected Exercises

where c1 , c2 are solved from the terminal boundary conditions to be the following: c1 = 0.002

n3 C, π

c2 = 0

which give the following extremal trajectory:    n sin nt δz (t) = 0.001C cos nt + − t cos nt (0 ≤ t ≤ tf ) π n ∗

δ z˙ ∗ (t) = 0.001nC

n π

t sin nt − sin nt



(0 ≤ t ≤ tf )

and control history, u∗ (t) = 0.002n2 C sinπnt , (0 ≤ t ≤ tf ). 4.7 tan β ∗ =

vy∗ (tf ) + g(tf − t) vx∗ (tf )

Exercises of Chap. 5 5.2  aˆ n (t) =

−1 , λn > 0 1 , λn < 0

√ with λn = const. With α0 > 0, an = 1, α(t) = Ct/ h, and tf = α0 μ/C. 5.3 ˆ = β(t)



cos−1 (bπ/2) , λ2 > 0 − cos−1 (bπ/2) , λ2 < 0

where λ¨ 2 = −c2

  c 2 4 3 3 λ2 sin β − 3h /r − 2μ/r + 2c cos β/r r2

where c2 is constant and r(t), β(t) are on the optimal trajectory.

Answers to Selected Exercises

261

5.5  aˆ n (t) =

Cn2 , λ2 < 0 −Cn2 , λ2 > 0

where λ2 = −

c2 sin nt + c2 cos nt n

The response until the first switching time, ts = tan−1 (nc2 /c1 ), with c1 > 0, c2 > 0, is the following: δz(t) = C(0.01 sin nt + cos nt − 1) , (0 ≤ t ≤ ts )

References

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Index

A Accessory problem, 64 Algebraic Riccati equation, 135, 137 Angular momentum, 78 Angular speed, 209 Angular velocity, 151 Argument of periapsis, 82 Ascending node, 81 Attitude control, 3 Attitude kinematics, 4 Augmented Lagrangian, 21 Augmented performance index, 21, 68 Autonomous system, 34 Axisymmetric mass distribution, 201, 205 Azimuth, 80

B Backstepping, 252 Bang–bang control, 56 Barker’s equation, 86 Barycenter, 209, 211 Bifurcation, 239 Body axes, 3 Bolza problem, 20 Bounded exhaust rate, 166 Brachistochrone problem, 74

C Celestial frame, 80 Central gravity field, 168, 169 Characteristic speed, 161 Chebyshev–Picard iteration, 250 Chord, 93

Circular restricted three-body problem, 200, 210 Circumferential thrust, 120 Classical orbital elements, 124 Closure rate, 143 Coasting arc, 102 Co-latitude, 205 Collinear Lagrangian points, 210, 214, 232 Collocation method, 71, 186, 248 Constant acceleration bound, 166 Continued fraction, 100, 135 Control-affine system, 55, 61 Control interval, 2, 20 Controllability, 137 Control variables, 2, 8 Coplanar orbital transfer, 122 Coplanar rendezvous, 196 Coplanar three-body problem, 209 Corrective control, 132 Costate equation, 25 Costate variables, 21 Cross-product steering, 145

D Declination, 80 Disturbing acceleration, 132 Docking, 142 Dynamic optimization, 7

E Earth–Moon system, 214, 221 Eccentric anomaly, 83 Eccentricity, 79

© Springer Nature Switzerland AG 2019 A. Tewari, Optimal Space Flight Navigation, Control Engineering, https://doi.org/10.1007/978-3-030-03789-5

267

268 Eccentricity vector, 78 Ecliptic plane, 81 Electric propulsion, 161 Elliptical orbit, 83 Energy control, 138 Energy-optimal guidance, 148, 150 Equality constraint dynamic, 20 static, 11 Equatorial plane, 81 Equatorial radius, 205 Equilateral triangle, 210 Equinoctial elements, 127 Euler–Lagrange equations, 26 Exhaust rate, 160 Exhaust speed, 160 Extremal point, 9 Extremal trajectory, 7

F Feasible region, 19 Feedback controller, 5 Feedforward controller, 5 Flat-trajectory approximation, 147 Flight-path angle, 79 Floquet mode, 250 Focus, 79 Formation flying, 153 Free-return trajectory, 220 Frequency matching, 238

G Gauss’ perturbation equations, 119, 124 Generalized momentum, 217 Generating functions, 204 Global asymptotic stability, 34, 41 Global minimum, 10 Gradient operator, 218 Gravitational potential, 201, 217 Gravity-assist maneuver, 222 Gravity-gradient matrix, 134, 162, 207 Guidance, 138

H Halo orbit, 216, 233, 237, 239 Hamiltonian, 12, 25, 217 Hamilton–Jacobi–Bellman equation, 33 Hessian, 9 Hill–Clohessy–Wiltshire model, 151, 152, 189 Hohmann transfer, 105 Hyperbolic anomaly, 85

Index Hyperbolic orbit, 85 Hypergeometric function, 99

I Impulsive control, 249 Impulsive orbital transfer, 77, 88 Inclination, 81 Inequality constraint, 2, 18, 55 Inertial frame, 3 Initial condition, 20 Initial value problem, 70 Intermediate-thrust arc, 165, 173 Interstellar mission, 228 In-track displacement, 127, 151, 189 Invariant manifold, 246, 248 Inverse-square gravity field, 172 Irregularly shaped body, 199, 208

J Jacobian, 12, 37, 62, 244 Jacobi’s integral, 218 Jacobson’s necessary condition, 68 Jeffery’s constants, 206

K Kalman filter, 133, 251 Kepler’s equation, 83

L Lagrange’s coefficients, 87, 97, 135 Lagrange’s planetary model, 124 Lagrange multipliers, 12 Lagrange problem, 20 Lagrangian, 2, 12, 20, 217, 234 Lagrangian points, 213 Lambert’s problem, 88, 95 Lambert’s theorem, 95 Lawden’s spiral, 173, 177 Legendre–Clebsch generalized necessary condition, 58, 59 necessary condition, 31 sufficient condition, 14, 31, 55 Legendre polynomial, 203, 234 Lindstedt–Poincaré method, 237 Line-of-sight interception, 139 Linear optimal control, 135 quadratic regulator, 38, 64, 135, 250, 251 Line of nodes, 81 Line of sight, 140

Index Lissajous trajectory, 233 Local-horizon frame, 80 Local minimum, 10 Longitude, 205 Low-thrust orbital transfer, 120 Low-thrust three-body transfer, 222 LQG/LTR formulation, 251 Lunar landing problem, 50 Lyapunov function, 34

M Maneuvering, 138 Maneuvering spacecraft, 151 Manifold, 246 Matrix Riccati equation, 40, 64 Maximum-thrust arc, 165 Mayer problem, 20 Mean anomaly, 83 Mean motion, 83 Measurement noise, 132 Minimum energy escape, 86 Minimum energy transfer, 92 Minimum point, 10 Model-predictive control, 252 Monodromy matrix, 244 Monte Carlo simulation, 250

N Navigation, 3 Necessary condition, 7, 8, 13, 23, 25 Neighboring optimal solution, 33 Newton’s method, 48, 84, 100, 244 Nonlinear feedback control, 33 Non-spherical body, 201 Non-spherical gravity field, 199 Normal thrust, 120 Null-thrust arc, 102, 165

O Oblateness, 201, 206, 207 Optimal control, 1, 7 Optimal impulsive transfer, 88, 101 Optimal orbital regulation, 127 Optimal orbital transfer, 123, 199 Optimal return function, 32 Optimal thrust direction, 163, 175 Orbit, 78 Orbital motion, 3, 117 Orbital plane, 78 Orbital tracking, 132 Orbital velocity, 78

269 Orbit equation, 79 Osculating orbit, 118 Out-of-plane displacement, 127, 151, 189 Outer bi-elliptical transfer, 110

P Parabolic orbit, 86 Parameter, 79 Patched two-body trajectory, 231 Performance index, 2, 8, 20 Periapsis, 79 Perifocal frame, 80 Perturbation model, 119 Perturbing acceleration, 124 Plane-change maneuver, 122 Polyhedral model, 208 Pontryagin’s minimum principle, 7, 31, 56, 165, 168, 175 Primaries, 210 Primer vector, 164 Primer vector method, 159 Process noise, 132 Projected miss distance, 141

Q Quadratic cost, 35 Quasi-halo orbit, 233 Quasi-steady approximation, 136 Quintic equation of Lagrange, 210

R Radial displacement, 151, 189 Radial thrust, 120, 128 Recurrence formulae, 204 Remaining burn time, 148 Rendezvous problem, 153, 191, 194 Restricted three-body problem, 200 Right-ascension of the ascending node, 81 Right ascension, 80 Robust control, 135

S Saddle point, 10 Semi-major axis, 79 Shooting method, 70, 231 Singular control, 31, 55, 165 Singular point, 10 Skew-symmetric matrix, 66 Sliding mode, 250 Solar-sail mission, 231

270

Index

Specific energy, 78 Specific impulse, 160 Sphere of influence, 231 Spherical harmonics, 201, 204, 208 Stability augmentation, 137 Stable manifold, 246 State equation, 2, 20 State transition matrix, 42, 87, 134, 193, 243 State variables, 2, 8 Static optimization, 8 Stationary condition, 13, 25 Station keeping, 233, 249 Stumpff function, 97 Sufficient condition, 8, 9, 13, 34 Sun–Earth system, 239 Sun–Jupiter system, 215 Swing-by, 222 Switching condition, 56, 165 Switching function, 56 Symplectic matrix, 42, 88, 134, 244 Synodic frame, 211

Thrust pitch angle, 125 Thrust yaw angle, 125 Time-averaged dynamics, 137 Time of flight, 88 Time-invariant gravity field, 168 Time-invariant system, 34 Time of periapsis, 83 Time to go, 142 Tracking system, 133 Transformational approach, 61 Transversality condition, 25, 188, 231 Triangular Lagrangian points, 214 True anomaly, 79 Two-body problem, 77 Two-point boundary-value problem, 39

T Tangential thrust, 131 Target, 151, 189 Terminal cost, 20 Terminal time, 2 Three-body problem, 77, 208 Throttling, 160 Thrust acceleration, 117, 120, 159 Thrust direction, 159, 163, 189, 224

V Variational model, 119, 124 Velocity-to-be-gained, 145 Vernal equinox, 81

U Universal variable, 97 Unstable equilibrium, 237 Unstable manifold, 246, 249

Z Zermelo’s navigation problem, 74 Zero relative speed contour, 219