Undergraduate Texts in Mathematics
Editors
S. Axler F.W. Gehring K.A. Ribet
Saber Elaydi
An Introduction to Difference Equations
Third Edition
Saber Elaydi
Department of Mathematics Trinity University
San Antonio, Texas 78212 USA
Editorial Board
S. Axler
Mathematics Department San Francisco State
University
San Francisco, CA 94132 USA
F.W. Gehring Mathematics Department East Hall
University of Michigan Ann Arbor, MI 48109 USA
K.A. Ribet Department of
Mathematics University of California
at Berkeley
Berkeley, CA 94720-3840 USA
Mathematics Subject Classification (2000): 12031
Library of Congress Cataloging-in-Publication Data Elaydi, Saber, 1943–
An introduction to difference equations / Saver Elaydi.—3rd ed. p. cm. — (Undergraduate texts in mathematics)
Includes bibliographical references and index. ISBN 0-387-23059-9 (acid-free paper)
1. Difference equations. I. Title. II. Series.
QA431.E43 2005
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ISBN 0-387-23059-9 Printed on acid-free paper.
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Preface to the Third Edition
In contemplating the third edition, I have had multiple objectives to achieve. The first and foremost important objective is to maintain the ac- cessibility and readability of the book to a broad readership with varying mathematical backgrounds and sophistication. More proofs, more graphs, more explanations, and more applications are provided in this edition.
The second objective is to update the contents of the book so that the reader stays abreast of new developments in this vital area of mathematics. Recent results on local and global stability of one-dimensional maps are included in Chapters 1, 4, and Appendices A and C. An extension of the Hartman–Grobman Theorem to noninvertible maps is stated in Appendix D. A whole new section on various notions of the asymptoticity of solutions and a recent extension of Perron’s Second Theorem are added to Chapter 8. In Appendix E a detailed proof of the Levin–May Theorem is presented. In Chapters 4 and 5, the reader will find the latest results on the larval– pupal–adult flour beetle model.
The third and final objective is to better serve the broad readership of this book by including most, but certainly not all, of the research areas in difference equations. As more work is being published in the Journal of Difference Equations and Applications and elsewhere, it became apparent that a whole chapter needed to be dedicated to this enterprise. With the prodding and encouragement of Gerry Ladas, the new Chapter 5 was born. Major revisions of this chapter were made by Fozi Dannan, who diligently and painstakingly rewrote part of the material and caught several errors and typos. His impact on this edition, particularly in Chapters 1, 4, and Chapter 8 is immeasurable and I am greatly indebted to him. My thanks
v
vi Preface to the Third Edition
go to Shandelle Henson, who wrote a thorough review of the book and suggested the inclusion of an extension of the Hartman–Groman Theorem, and to Julio Lopez and his student Alex Sepulveda for their comments and discussions about the second edition.
I am grateful to all the participants of the AbiTuMath Program and to its coordinator Andreas Ruffing for using the second edition as the main reference in their activities and for their valuable comments and dis- cussions. Special thanks go to Sebastian Pancratz of AbiTuMath whose suggestions improved parts of Chapters 1 and 2. I benefited from comments and discussions with Raghib Abu-Saris, Bernd Aulbach, Martin Bohner, Luis Carvahlo, Jim Cushing, Malgorzata Guzowska, Sophia Jang, Klara Janglajew, Nader Kouhestani, Ulrich Krause, Ronald Mickens, Robert Sacker, Hassan Sedaghat, and Abdul-Aziz Yakubu. It is a pleasure to thank Ina Lindemann, the editor at Springer-Verlag for her advice and support during the writing of this edition. Finally, I would like to express my deep appreciation to Denise Wilson who spent many weekends typing various drafts of the manuscript. Not only did she correct many glitches, typos, and awkward sentences, but she even caught some mathematical errors.
I hope you enjoy this edition and if you have any comments or questions, please do not hesitate to contact me at selaydi@trinity.edu.
San Antonio, Texas Saber N. Elaydi April 2004
Suggestions for instructors using this book.
The book may be used for two one-semester courses. A first course may include one of the following options but should include the bulk of the first four chapters:
1. If one is mainly interested in stability theory, then the choice would be Chapters 1–5.
2. One may choose Chapters 1–4, and Chapter 8 if the interest is to get to asymptotic theory.
3. Those interested in oscillation theory may choose Chapters 1, 2, 3, 5, and 7.
4. A course emphasizing control theory may include Chapters 1–3, 6, and 10.
Preface to the Third Edition vii
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Chapter 7
Chapter 6
Chapter 5
Chapter 6
Chapter 9
The diagram above depicts the dependency among the chapters.
Chapter 7
Preface to the Second Edition
The second edition has greatly benefited from a sizable number of com- ments and suggestions I received from users of the first edition. I hope that I have corrected all the errors and misprints in the book. Important revisions were made in Chapters 1 and 4. In Chapter 1, I added two ap- pendices (Global Stability and Periodic Solutions). In Chapter 4, I added a section on applications to mathematical biology. Influenced by a friendly and some not so friendly comments about Chapter 8 (previously Chapter 7: Asymptotic Behavior of Difference Equations), I rewrote the chapter with additional material on Birkhoff’s theory. Also, due to popular demand, a new chapter (Chapter 9) under the title “Applications to Continued Frac- tions and Orthogonal Polynomials” has been added. This chapter gives a rather thorough presentation of continued fractions and orthogonal poly- nomials and their intimate connection to second-order difference equations. Chapter 8 (Oscillation Theory) has now become Chapter 7. Accordingly, the new revised suggestions for using the text are as follows.
The book may be used with considerable flexibility. For a one-semester course, one may choose one of the following options:
(i) If you want a course that emphasizes stability and control, then you may select Chapters 1, 2, and 3, and parts of Chapters 4, 5, and 6. This is perhaps appropriate for a class populated by mathematics, physics, and engineering majors.
(ii) If the focus is on the applications of difference equations to orthogonal polynomials and continued fractions, then you may select Chapters 1, 2, 3, 8, and 9.
ix
x Preface to the Second Edition
I am indebted to K. Janglajew, who used the book several times and caught numerous glitches and typos. I am very grateful to Julio Lopez and his students, who helped me correct some mistakes and improve the exposition in Chapters 7 and 8. I am thankful to Raghib Abu-Saris, who caught some errors and typos in Chapter 4. My thanks go to Gerry Ladas, who assisted in refining Chapter 8, and to Allan Peterson, who graciously used my book and caught some mistakes in Chapter 4. I thank my brother Hatem Elaydi who read thoroughly Chapter 6 and made valuable revisions in the exercises. Many thanks to Fozi Dannan, whose comments improved Chapters 1, 4, and 9. Ronald Mickens was always there for me when I needed support, encouragement, and advice. His impact on this edition is immeasurable. My special thanks to Jenny Wolkowicki of Springer-Verlag.
I apologize in advance to all those whom I did not mention here but who have helped in one way or another to enhance the quality of this edition.
It is my pleasure to thank my former secretary, Constance Garcia, who typed the new and revised material.
San Antonio, Texas Saber N. Elaydi April 1999
Preface to the First Edition
This book grew out of lecture notes I used in a course on difference equa- tions that I have taught at Trinity University for the past five years. The classes were largely populated by juniors and seniors majoring in mathematics, engineering, chemistry, computer science, and physics.
This book is intended to be used as a textbook for a course on difference equations at both the advanced undergraduate and beginning graduate levels. It may also be used as a supplement for engineering courses on discrete systems and control theory.
The main prerequisites for most of the material in this book are calculus and linear algebra. However, some topics in later chapters may require some rudiments of advanced calculus and complex analysis. Since many of the chapters in the book are independent, the instructor has great flexibility in choosing topics for a one-semester course.
This book presents the current state of affairs in many areas such as sta- bility, Z-transform, asymptoticity, oscillations, and control theory. However, this book is by no means encyclopedic and does not contain many impor- tant topics, such as numerical analysis, combinatorics, special functions and orthogonal polynomials, boundary value problems, partial difference equations, chaos theory, and fractals. The nonselection of these topics is dictated not only by the limitations imposed by the elementary nature of this book, but also by the research interest (or lack thereof) of the author.
Great efforts were made to present even the most difficult material in an elementary format and to write in a style that makes the book acces- sible to students with varying backgrounds and interests. One of the main features of the book is the inclusion of a great number of applications in
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xii Preface to the First Edition
economics, social sciences, biology, physics, engineering, neural networks, etc. Moreover, this book contains a very extensive and carefully selected set of exercises at the end of each section. The exercises form an integral part of the text. They range from routine problems designed to build ba- sic skills to more challenging problems that produce deeper understanding and build technique. The asterisked problems are the most challenging, and the instructor may assign them as long-term projects. Another important feature of the book is that it encourages students to make mathematical discoveries through calculator/computer experimentation.
Chapter 1 deals with first-order difference equations, or one-dimensional maps on the real line. It includes a thorough and complete analysis of stability for many popular maps (equations) such as the logistic map, the tent map, and the Baker map. The rudiments of bifurcation and chaos theory are also included in Section 1.6. This section raises more questions and gives few answers. It is intended to arouse the reader’s interest in this exciting field.
In Chapter 2 we give solution methods for linear difference equations of any order. Then we apply the obtained results to investigate the stability and the oscillatory behavior of second-order difference equations. At the end of the chapter we give four applications: the propagation of annual plants, the gambler’s ruin, the national income, and the transmission of information.
Chapter 3 extends the study in Chapter 2 to systems of difference equa- tions. We introduce two methods to evaluate An for any matrix A. In Section 3.1 we introduce the Putzer algorithm, and in Section 3.3 the method of the Jordan form is given. Many applications are then given in Section 3.5, which include Markov chains, trade models, and the heat equation.
Chapter 4 investigates the question of stability for both scalar equations and systems. Stability of nonlinear equations is studied via linearization (Section 4.5) and by the famous method of Liapunov (Section 4.6). Our ex- position here is restricted to autonomous (time-invariant) systems. I believe that the extension of the theory to nonautonomous (time-variant) systems, though technically involved, will not add much more understanding to the subject matter.
Chapter 5 delves deeply into Z-transform theory and techniques (Sections 5.1, 5.2). Then the results are applied to study the stability of Volterra difference scalar equations (Sections 5.3, 5.4) and systems (Sections 5.5, 5.6). For readers familiar with differential equations, Section 5.7 provides a comparison between the Z-transform and the Laplace transform. Most of the results on Volterra difference equations appear here for the first time in a book.
Chapter 6 takes us to the realm of control theory. Here, we cover most of the basic concepts including controllability, observability, observers, and stabilizability by feedback. Again, we restrict the presentation to au-
tonomous (time-invariant) systems, since this is just an introduction to this vast and growing discipline. Moreover, most practitioners deal mainly with time-invariant systems.
In Chapter 7 we give a comprehensive and accessible study of asymp- totic methods for difference equations. Starting from the Poincar ́e Theorem, the chapter covers most of the recent development in the subject. Section 7.4 (asymptotically diagonal systems) presents an extension of Levinson’s Theorem to difference equations, while in Section 7.5 we carry our study to nonlinear difference equations. Several open problems are given that would serve as topics for research projects.
Finally, Chapter 8 presents a brief introduction to oscillation theory. In Section 8.1, the basic results on oscillation for three-term linear difference equations are introduced. Extension of these results to nonlinear differ- ence equations is presented in Section 8.2. Another approach to oscillation theory, for self-adjoint equations, is presented in Section 8.3. Here we also introduce a discrete version of Sturm’s Separation Theorem.
I am indebted to Gerry Ladas, who read many parts of the book and sug- gested many useful improvements, especially within the section on stability of scalar difference equations (Section 4.3). His influence through papers and lectures on Chapter 8 (oscillation theory) is immeasurable. My thanks go to Vlajko Kocic, who thoroughly read and made many helpful comments about Chapter 4 on Stability. Jim McDonald revised the chapters on the Z-transform and control theory (Chapters 5 and 6) and made significant improvements. I am very grateful to him for his contributions to this book. My sincere thanks go to Paul Eloe, who read the entire manuscript and offered valuable suggestions that led to many improvements in the final draft of the book. I am also grateful to Istvan Gyori for his comments on Chapter 8 and to Ronald Mickens for his review of the whole manuscript and for his advice and support. I would like to thank the following math- ematicians who encouraged and helped me in numerous ways during the preparation of the book: Allan Peterson, Donald Bailey, Roberto Hasfura, Haydar Akca, and Shunian Zhang. I am grateful to my students Jeff Bator, Michelle MacArthur, and Nhung Tran, who caught misprints and mistakes in the earlier drafts of this book. My special thanks are due to my student Julie Lundquist, who proofread most of the book and made improvements in the presentation of many topics. My thanks go to Constance Garcia, who skillfully typed the entire manuscript with its many, many revised versions. And finally, it is a pleasure to thank Ina Lindemann and Robert Wexler from Springer-Verlag for their enthusiastic support of this project.
San Antonio, Texas Saber N. Elaydi December 1995
Preface to the First Edition xiii
Contents
Preface to the Third Edition v Preface to the Second Edition ix Preface to the First Edition xi List of Symbols xx
1 Dynamics of First-Order Difference Equations 1
1.1 Introduction …………………….. 1
1.2 Linear First-Order Difference Equations . . . . . . . . . . 2 1.2.1 ImportantSpecialCases …………… 4
1.3 EquilibriumPoints………………….. 9
1.3.1 TheStairStep(Cobweb)Diagrams . . . . . . . . . 13 1.3.2 TheCobwebTheoremofEconomics . . . . . . . . 17 1.4 Numerical Solutions of Differential Equations . . . . . . . 20
1.4.1 Euler’sMethod…………
1.4.2 A Nonstandard Scheme . . . . . . . 1.5 Criterion for the Asymptotic Stability of EquilibriumPoints…………..
……… 20 ……… 24
……… 27
1.6 PeriodicPointsandCycles ……………… 35
1.7 TheLogisticEquationandBifurcation . . . . . . . . . . . 43 1.7.1 EquilibriumPoints ……………… 43 1.7.2 2-Cycles……………………. 45
xv
xvi Contents
1.7.3 22-Cycles …………………… 46
1.7.4 TheBifurcationDiagram…………… 47 1.8 Basin of Attraction and Global Stability (Optional) . . . . 50
2 Linear Difference Equations of Higher Order 57
2.1 DifferenceCalculus………………….. 57 2.1.1 ThePowerShift ……………….. 59
2.1.2 FactorialPolynomials………..
2.1.3 The Antidifference Operator . . . . . . .
2.2 General Theory of Linear Difference Equations
2.3 Linear Homogeneous Equations with Constant
…… 60 …… 61 …… 64
Coefficients ……………………… 75 2.4 Nonhomogeneous Equations: Methods of Undetermind
Coefficeints ……………………… 83 2.4.1 The Method of Variation of Constants
(Parameters)…………………. 89
2.5 LimitingBehaviorofSolutions……………. 91
2.6 Nonlinear Equations Transformable to Linear Equations . 98
2.7 Applications …………………….. 104
2.7.1 PropagationofAnnualPlants . . . . . . . . . . . . 104
2.7.2 Gambler’sRuin ……………….. 107
2.7.3 NationalIncome……………….. 108
2.7.4 The Transmission of Information . . . . . . . . . . 110
3 Systems of Linear Difference Equations 117
3.1 Autonomous (Time-Invariant) Systems . . . . . . . . . . . 117
3.1.1 The Discrete Analogue of the Putzer Algorithm . . 118
3.1.2 The Development of the Algorithm for An . . . . . 119
3.2 TheBasicTheory ………………….. 125
3.3 The Jordan Form: Autonomous (Time-Invariant)
SystemsRevisited ………………….. 135 3.3.1 DiagonalizableMatrices……………. 135 3.3.2 TheJordanForm ………………. 142 3.3.3 Block-DiagonalMatrices …………… 148
3.4 LinearPeriodicSystems……………….. 153
3.5 Applications …………………….. 159 3.5.1 MarkovChains………………… 159 3.5.2 RegularMarkovChains……………. 160 3.5.3 AbsorbingMarkovChains ………….. 163 3.5.4 ATradeModel………………… 165 3.5.5 TheHeatEquation ……………… 167
4 Stability Theory 173
4.1 ANormofaMatrix …………………. 174 4.2 NotionsofStability …………………. 176
4.3 StabilityofLinearSystems ……………… 184 4.3.1 NonautonomousLinearSystems……….. 184 4.3.2 AutonomousLinearSystems…………. 186
4.4 PhaseSpaceAnalysis ………………… 194
4.5 Liapunov’sDirect,orSecond,Method. . . . . . . . . . . . 204
4.6 StabilitybyLinearApproximation. . . . . . . . . . . . . . 219
4.7 Applications …………………….. 229
4.7.1 OneSpecieswithTwoAgeClasses . . . . . . . . . 229
4.7.2 Host–ParasitoidSystems …………… 232
4.7.3 ABusinessCycleModel …………… 233
4.7.4 TheNicholson–BaileyModel…………. 235
4.7.5 TheFlourBeetleCaseStudy . . . . . . . . . . . . 238
5 Higher-Order Scalar Difference Equations 245
5.1 LinearScalarEquations……………….. 246
5.2 SufficientConditionsforStability . . . . . . . . . . . . . . 251
5.3 StabilityviaLinearization………………. 256
5.4 Global Stability of Nonlinear Equations . . . . . . . . . . . 261
5.5 Applications …………………….. 268
5.5.1 FlourBeetles…………………. 268 5.5.2 AMosquitoModel………………. 270
6 The Z-Transform Method and Volterra Difference Equations 273
6.1 DefinitionsandExamples………………. 274 6.1.1 PropertiesoftheZ-Transform . . . . . . . . . . . . 277
6.2 The Inverse Z-Transform and Solutions of Difference Equations………………………. 282
6.2.1 ThePowerSeriesMethod ………….. 282
6.2.2 ThePartialFractionsMethod . . . . . . . . . . . . 283
6.2.3 TheInversionIntegralMethod. . . . . . . . . . . . 287
6.3 Volterra Difference Equations of Convolution Type: The ScalarCase ……………………… 291
6.4 Explicit Criteria for Stability of Volterra Equations . . . . 295
6.5 VolterraSystems…………………… 299
6.6 AVariationofConstantsFormula………….. 305
6.7 The Z-Transform Versus the Laplace Transform . . . . . . 308
7 Oscillation Theory 313
7.1 Three-TermDifferenceEquations ………….. 313
7.2 Self-Adjoint Second-Order Equations . . . . . . . . . . . . 320
7.3 NonlinearDifferenceEquations……………. 327
8 Asymptotic Behavior of Difference Equations 335
8.1 ToolsofApproximation……………….. 335 8.2 Poincar ́e’sTheorem …………………. 340
Contents xvii
xviii
Contents
8.2.1 Infinite Products and Perron’s Example . . . . . . 344
8.3 AsymptoticallyDiagonalSystems . . . . . . . . . . . . . . 351
8.4 High-OrderDifferenceEquations. . . . . . . . . . . . . . . 360
8.5 Second-OrderDifferenceEquations . . . . . . . . . . . . . 369
8.5.1 A Generalization of the Poincar ́e–Perron Theorem . 372
8.6 Birkhoff’sTheorem …………………. 377
8.7 NonlinearDifferenceEquations……………. 382
8.8 Extensions of the Poincar ́e and Perron Theorems . . . . . 387
8.8.1 An Extension of Perron’s Second Theorem . . . . . 387
8.8.2 Poincar ́e’sTheoremRevisited . . . . . . . . . . . . 389
9 Applications to Continued Fractions and Orthogonal Polynomials 397
9.1 Continued Fractions: Fundamental Recurrence Formula
9.2 ConvergenceofContinuedFractions. . . . . . . . . . . .
9.3 Continued Fractions and Infinite Series . . . . . . . . . .
9.4 ClassicalOrthogonalPolynomials . . . . . . . . . . . . .
9.5 The Fundamental Recurrence Formula for Orthogonal
. 397 . 400 . 408 . 413
Polynomials……………………… 417 9.6 Minimal Solutions, Continued Fractions, and Orthogonal
Polynomials……………………… 421
10 Control Theory 429
10.1 Introduction …………………….. 429 10.1.1 Discrete Equivalents for Continuous Systems . . . 431
10.2 Controllability ……………………. 432 10.2.1 Controllability Canonical Forms . . . . . . . . . . . 439
10.3 Observability …………………….. 446 10.3.1 Observability Canonical Forms . . . . . . . . . . . 453
10.4 Stabilization by State Feedback (Design via Pole
Placement) ……………………… 457 10.4.1 Stabilization of Nonlinear Systems by Feedback . . 463
10.5 Observers ………………………. 467 10.5.1 Eigenvalue Separation Theorem . . . . . . . . . . . 468
A Stability of Nonhyperbolic Fixed Points of Maps on the Real Line
477
A.1 Local Stability of Nonoscillatory Nonhyperbolic Maps . . 477
A.2 Local Stability of Oscillatory Nonhyperbolic Maps . . . . 479 A.2.1 Resultswithg(x) ………………. 479
B The Vandermonde Matrix 481
C Stability of Nondifferentiable Maps 483
D Stable Manifold and the Hartman–Grobman–Cushing Theorems 487
D.1 TheStableManifoldTheorem ……………. 487 D.2 The Hartman–Grobman–Cushing Theorem . . . . . . . . . 489
E The Levin–May Theorem 491
F Classical Orthogonal Polynomials 499
G Identities and Formulas 501
Answers and Hints to Selected Problems 503
Maple Programs 517
References 523 Index 531
Contents xix
List of Symbols
B(x0, δ) B(δ)
∆
L K(an/bn) R
R+ Z
Z+
C
Γ
F (a, b; c; z) (ν)n
(α,β)
Pn (x)
Pn(x) Pnν(x) Lαn(x) Hn(x) O(x) ∆n
n−1
i=n0 Sf E fn x(k) ∆−1
Ball centered at x0 with radius δ Ball centered at origin with radius δ The difference operator
Moment functional
Continued fraction
The set of real numbers
The set of nonnegative real numbers The set of integers
The set of nonnegative integers
The set of complex numbers
The gamma function
The hypergeometric function
The Pochhammer symbol
Jacobi polynomials
Legendre polynomials
Gegenbauer polynomials
Laguerre polynomials
Hermite polynomials
The orbit of x
∆n−1 (∆)
Product
The Schwarzian derivative of f Shift operator
The nth iterate of f
Factorial polynomial
The antidifference operator
xxi
xxii List of Symbols
det A W(n) AT
diag ρ(A) ||A||
G Ω(x0) x ̃(z) Z(x(n)) o
O
f ∼ g
The determinant of a matrix A The Casoration
Transpose of a matrix A Diagonal matrix
Spectral radius of A Norm of a matrix A Closure of G
Limit set of x0 z-transform of x(n) z-transform of x(n) Little “oh”
Big “oh”
f is asymptotic to g
1
Dynamics of First-Order Difference Equations
1.1 Introduction
Difference equations usually describe the evolution of certain phenomena over the course of time. For example, if a certain population has discrete generations, the size of the (n+1)st generation x(n+1) is a function of the nth generation x(n). This relation expresses itself in the difference equation
x(n + 1) = f (x(n)). (1.1.1) We may look at this problem from another point of view. Starting from a
point x0, one may generate the sequence
x0, f(x0), f(f(x0)), f(f(f(x0))), . . . .
For convenience we adopt the notation
f2(x0) = f(f(x0)), f3(x0) = f(f(f(x0))), etc.
f(x0) is called the first iterate of x0 under f;f2(x0) is called the second iterate of x0 under f; more generally, fn(x0) is the nth iterate of x0 under f. The set of all (positive) iterates {fn(x0) : n ≥ 0} where f0(x0) = x0 by definition, is called the (positive) orbit of x0 and will be denoted by O(x0). This iterative procedure is an example of a discrete dynamical system. Letting x(n) = fn(x0), we have
x(n + 1) = fn+1(x0) = f[fn(x0)] = f(x(n)),
and hence we recapture (1.1.1). Observe that x(0) = f0(x0) = x0. For
example, let f(x) = x2 and x0 = 0.6. To find the sequence of iterates 1
2 1. Dynamics of First-Order Difference Equations
{fn(x0)}, we key 0.6 into a calculator and then repeatedly depress the x2 button. We obtain the numbers
0.6, 0.36, 0.1296, 0.01679616, . . . .
A few more key strokes on the calculator will be enough to convince the reader that the iterates fn(0.6) tend to 0. The reader is invited to verify that for all x0 ∈ (0,1),fn(x0) tends to 0 as n tends to ∞, and that fn(x0) tends to ∞ if x0 ̸∈ [−1,1]. Obviously, fn(0) = 0,fn(1) = 1 for all positive integers n, and fn(−1) = 1 for n = 1,2,3,… .
After this discussion one may conclude correctly that difference equa- tions and discrete dynamical systems represent two sides of the same coin. For instance, when mathematicians talk about difference equations, they usually refer to the analytic theory of the subject, and when they talk about discrete dynamical systems, they generally refer to its geometrical and topological aspects.
If the function f in (1.1.1) is replaced by a function g of two variables, that is, g : Z+ × R → R, where Z+ is the set of nonnegative integers and R is the set of real numbers, then we have
x(n + 1) = g(n, x(n)). (1.1.2)
Equation (1.1.2) is called nonautonomous or time-variant, whereas (1.1.1) is called autonomous or time-invariant. The study of (1.1.2) is much more complicated and does not lend itself to the discrete dynamical system theory of first-order equations. If an initial condition x(n0) = x0 is given, then for n ≥ n0 there is a unique solution x(n) ≡ x(n,n0,x0) of (1.1.2) such that x(n0, n0, x0) = x0. This may be shown easily by iteration. Now,
x(n0 + 1, n0, x0) = g(n0, x(n0)) = g(n0, x0),
x(n0 +2,n0,x0)=g(n0 +1,x(n0 +1))=g(n0 +1,g(n0,x0)),
x(n0 +3,n0,x0)=g(n0 +2,x(n0 +2))=g[n0 +2,g(n0 +1,g(n0,x0))].
And, inductively, we get x(n, n0, x0) = g[n − 1, x(n − 1, n0, x0)]. 1.2 Linear First-Order Difference Equations
In this section we study the simplest special cases of (1.1.1) and (1.1.2), namely, linear equations. A typical linear homogeneous first-order equation is given by
x(n + 1) = a(n)x(n), x(n0) = x0, n ≥ n0 ≥ 0, (1.2.1) and the associated nonhomogeneous equation is given by
y(n+1)=a(n)y(n)+g(n), y(n0)=y0, n≥n0 ≥0, (1.2.2)
where in both equations it is assumed that a(n) ̸= 0, and a(n) and g(n) are real-valued functions defined for n ≥ n0 ≥ 0.
1.2 Linear First-Order Difference Equations 3
One may obtain the solution of (1.2.1) by a simple iteration:
x(n0 + 1) = a(n0)x(n0) = a(n0)x0,
x(n0 + 2) = a(n0 + 1)x(n0 + 1) = a(n0 + 1)a(n0)x0,
x(n0 + 3) = a(n0 + 2)x(n0 + 2) = a(n0 + 2)a(n0 + 1)a(n0)x0.
And, inductively, it is easy to see that
x(n)=x(n0 +n−n0))
= a(n − 1)a(n − 2) · · · a(n0)x0,
n−1
x(n) =
The unique solution of the nonhomogeneous (1.2.2) may be found as
follows:
y(n0 + 1) = a(n0)y0 + g(n0),
y(n0 +2)=a(n0 +1)y(n0 +1)+g(n0 +1)
= a(n0 + 1)a(n0)y0 + a(n0 + 1)g(n0) + g(n0 + 1). Now we use mathematical induction to show that, for all n ∈ Z+,
n−1 n−1 n−1
i=n0
a(i) x0.
(1.2.3)
y(n) = a(i) y0 + i=n0
a(i) g(r). (1.2.4)
r=n0 i=r+1
To establish this, assume that formula (1.2.4) holds for n = k. Then from (1.2.2), y(k + 1) = a(k)y(k) + g(k), which by formula (1.2.4) yields
y(k+1)=a(k) a(i) y0 + i=n0
=
a(i) y0 + a(i) g(r) i=n0 r=n0 i=r+1
+
k
a(i) g(k) (see footnote 1)
k−1 k−1 k−1
a(i) g(r)+g(k)
a(k) k k−1k
r=n0
i=r+1
i=k+1
k k
=
Hence formula (1.2.4) holds for all n ∈ Z+.
i=n0
a(i) y0 + a(i) g(r). r=n0 i=r+1
1 Notice that ki=k+1 a(i) = 0.
we have adopted the notation ki=k+1 a(i) = 1 and
4 1. Dynamics of First-Order Difference Equations
1.2.1 Important Special Cases
There are two special cases of (1.2.2) that are important in many applications. The first equation is given by
y(n + 1) = ay(n) + g(n), Using formula (1.2.4) one may establish that
The second equation is given by
y(n + 1) = ay(n) + b,
Using formula (1.2.6) we obtain
⎨any0 + b y(n) = ⎪
y(0) = y0.
y(0) = y0.
(1.2.5)
(1.2.6)
(1.2.7)
(1.2.8)
y(n) = a y0 +
a g(k).
⎧⎪
a n − 1 a − 1
if a ̸= 1,
n−1
n n−k−1
⎩y0 + bn
Notice that the solution of the differential equation
dx = ax(t), x(0) = x0, dt
is given by
x(t) = eatx0,
and the solution of the nonhomogeneous differential equation
dy = ay(t) + g(t), y(0) = y0, dt
We now give some examples to practice the above formulas. Example 1.1. Solve the equation
k=0
t 0
ea(t−s)g(s) ds.
Thus the exponential eat in differential equations corresponds to the expo-
y(t) = eaty0 +
nential an and the integral t ea(t−s)g(s) ds corresponds to the summation
0
n−1
an−k−1g(k). k=0
y(n + 1) = (n + 1)y(n) + 2n(n + 1)!, y(0) = 1, n > 0.
if a = 1.
is given by
1.2 Linear First-Order Difference Equations 5 TABLE 1.1. Definite sum.
Number Summation Definite sum
n 1k
2 3 4 5 6 7
y(n)=
k2
k=1
k=1 n
n(n+1) 2
n(n+1)(2n+1) 6
n n(n+1)2
k3 n
ak
2
n(6n4 + 15n3 + 10n2 − 1) 30
(a −1)/(a−1) if a̸=1 n if a = 1
n
n − 1 if a = 1 (a−1)(n+1)an+1 −an+2 +a
n k=1
k=1
k=1
k4 n−1
k=1
n
k=0 n−1
(a −a)/(a−1) if a̸=1
ak kak, a̸=1
Solution
(a−1)2
(i+1) 2k(k+1)!
Solution From (1.2.6), we have
n−1 n−1 n−1
(i+1)+ n−1
i=0 = n! +
k=0 i=k+1 n! 2k
k=0
= 2nn! (from Table 1.1).
Example 1.2. Find a solution for the equation
x(n + 1) = 2x(n) + 3n, x(1) = 0.5.
n−1
1 n−1 n−k−1 k
x(n)=22+2 3 k=1
= 2n−2 + 2n−1
3
n−1 k
= 2n−2 + 2n−1 3 2 − 1
2
k=1
3n−1
23−1 2
= 3n − 5 · 2n−2.
6 1. Dynamics of First-Order Difference Equations
Example 1.3. A drug is administered once every four hours. Let D(n) be the amount of the drug in the blood system at the nth interval. The body eliminates a certain fraction p of the drug during each time interval. If the amount administered is D0, find D(n) and limn→∞ D(n).
Solution We first must create an equation to solve. Since the amount of drug in the patient’s system at time (n + 1) is equal to the amount at time n minus the fraction p that has been eliminated from the body, plus the new dosage D0, we arrive at the following equation:
D(n + 1) = (1 − p)D(n) + D0. Using (1.2.8), we solve the above equation, arriving at
DD D(n)= D0− 0 (1−p)n+ 0.
Hence,
Let D0 = 2 cubic centimeters (cc), p = 0.25. Then our original equation becomes
D(n + 1) = 0.75D(n) + 2, Table 1.2 gives D(n) for 0 ≤ n ≤ 10.
D(0) = 2.
pp
lim D(n) = D0 . n→∞ p
(1.2.9)
It follows from (1.2.9) that limn→∞ D(n) = 8, where D* = 8 cc is the equilibrium amount of drug in the body. We now enter the realm of finance for our next example.
Example 1.4. Amortization
Amortization is the process by which a loan is repaid by a sequence of periodic payments, each of which is part payment of interest and part payment to reduce the outstanding principal.
Let p(n) represent the outstanding principal after the nth payment g(n). Suppose that interest charges compound at the rate r per payment period. The formulation of our model here is based on the fact that the out- standing principal p(n + 1) after the (n + 1)st payment is equal to the outstanding principal p(n) after the nth payment plus the interest rp(n)
incurred during the (n + 1)st period minus the nth payment g(n). Hence p(n + 1) = p(n) + rp(n) − g(n),
TABLE 1.2. Values of D(n).
n
0
1
2
3
4
5
6
7
8
9
10
D(n)
2
3.5
4.62
5.47
6.1
6.58
6.93
7.2
7.4
7.55
7.66
1.2 Linear First-Order Difference Equations 7
or
p(n + 1) = (1 + r)p(n) − g(n), p(0) = p0, where p0 is the initial debt. By (1.2.6) we have
T
p(n) = (1+r)np0 −((1+r)n −1) r . (1.2.12)
If we want to pay off the loan in exactly n payments, what would be the monthly payment T ? Observe first that p(n) = 0. Hence from (1.2.12) we
(1.2.10)
(1.2.11) In practice, the payment g(n) is constant and, say, equal to T. In this case,
n−1
n n−k−1
p(n) = (1+r) p0 − (1+r) g(k). k=0
have
Exercises 1.1 and 1.2
r T =p0 1−(1+r)−n .
1.
2.
3.
Find the solution of each difference equation:
(a) x(n+1)−(n+1)x(n)=0,
(b) x(n+1)−3nx(n)=0, x(0)=c.
(c) x(n+1)−e2nx(n)=0, x(0)=c.
(d) x(n+1)− n x(n)=0, n≥1, x(1)=c.
x(0)=c.
n+1
Find the general solution of each difference equation:
(a) y(n+1)−1y(n)=2, y(0)=c. 2
(b) y(n+1)− n y(n)=4, y(1)=c. n+1
Find the general solution of each difference equation:
(a) y(n+1)−(n+1)y(n)=2n(n+1)!, y(0)=c.
(b) y(n+1)=y(n)+en, y(0)=c.
4. (a)
Write a difference equation that describes the number of regions created by n lines in the plane if it is required that every pair of lines meet and no more than two lines meet at one point.
5.
(b) Find the number of these regions by solving the difference equation in case (a).
The gamma function is defined as Γ(x) = ∞ tx−1 e−t dt, x > 0. 0
(a) Show that Γ(x + 1) = xΓ(x), Γ(1) = 1.
8
1. Dynamics of First-Order Difference Equations
(b) If n is a positive integer, show that Γ(n + 1) = n!.
(c) Showthatx(n) =x(x−1)···(x−n+1)= Γ(x+1) .
Γ(x−n+1)
6. A space (three-dimensional) is divided by n planes, nonparallel, and
no four planes having a point in common.
(a) Write a difference equation that describes the number of regions created.
(b) Find the number of these regions.
7. Verify (1.2.6).
8. Verify (1.2.8).
9. A debt of $12,000 is to be amortized by equal payments of $380 at the end of each month, plus a final partial payment one month after the last $380 is paid. If interest is at an annual rate of 12% com- pounded monthly, construct an amortization schedule to show the required payments.
10. Suppose that a loan of $80,000 is to be amortized by equal monthly payments. If the interest rate is 10% compounded monthly, find the monthly payment required to pay off the loan in 30 years.
11. Suppose the constant sum T is deposited at the end of each fixed period in a bank that pays interest at the rate r per period. Let A(n) be the amount accumulated in the bank after n periods.
(a) Write a difference equation that describes A(n).
(b) Solve the difference equation obtained in (a), when A(0) = 0, T =
$200, and r = 0.008.
12. The temperature of a body is measured as 110◦ F. It is observed that the amount the temperature changes during each period of two hours is −0.3 times the difference between the previous period’s temperature and the room temperature, which is 70◦ F.
(a) Write a difference equation that describes the temperature T(n) of the body at the end of n periods.
(b) Find T(n).
13. Suppose that you can get a 30-year mortgage at 8% interest. How much can you afford to borrow if you can afford to make a monthly payment of $1,000?
14. Radium decreases at the rate of 0.04% per year. What is its half-life? (The half-life of a radioactive material is defined to be the time needed for half of the material to dissipate.)
15. (Carbon Dating). It has been observed that the proportion of carbon- 14 in plants and animals is the same as that in the atmosphere as long as the plant or the animal is alive. When an animal or plant dies, the carbon-14 in its tissue starts decaying at the rate r.
(a) If the half-life of carbon-14 is 5,700 years, find r.
(b) If the amount of carbon-14 present in a bone of an animal is 70%
of the original amount of carbon-14, how old is the bone?
1.3 Equilibrium Points
The notion of equilibrium points (states) is central in the study of the dy- namics of any physical system. In many applications in biology, economics, physics, engineering, etc., it is desirable that all states (solutions) of a given system tend to its equilibrium state (point). This is the subject of study of stability theory, a topic of great importance to scientists and engineers. We now give the formal definition of an equilibrium point.
Definition 1.5. A point x∗ in the domain of f is said to be an equilibrium point of (1.1.1) if it is a fixed point of f, i.e., f(x*) = x*.
In other words, x∗ is a constant solution of (1.1.1), since if x(0) = x∗ is an initial point, then x(1) = f(x*) = x∗, and x(2) = f(x(1)) = f(x*) = x*, and so on.
Graphically, an equilibrium point is the x-coordinate of the point where the graph of f intersects the diagonal line y = x (Figures 1.1 and 1.2). For example, there are three equilibrium points for the equation
x(n + 1) = x3(n)
where f(x) = x3. To find these equilibrium points, we let f(x*) = x∗, or x3 = x, and solve for x. Hence there are three equilibrium points, −1, 0, 1 (Figure 1.1). Figure 1.2 illustrates another example, where f(x) = x2−x+1 and the difference equation is given by
x(n + 1) = x2(n) − x(n) + 1.
Letting x2 − x + 1 = x, we find that 1 is the only equilibrium point. There is a phenomenon that is unique to difference equations and cannot possibly occur in differential equations. It is possible in difference equations that a solution may not be an equilibrium point but may reach one after finitely many iterations. In other words, a nonequilibrium state may go to an equilibrium state in a finite time. This leads to the following definition.
Definition 1.6. Let x be a point in the domain of f. If there exists a positive integer r and an equilibrium point x∗ of (1.1.1) such that fr(x) = x*, fr−1(x) ̸= x*, then x is an eventually equilibrium (fixed) point.
1.3 Equilibrium Points 9
10 1. Dynamics of First-Order Difference Equations
x*=−1 1
x*=0 x*=1 23
FIGURE 1.1. Fixed points of f(x) = x3.
f(x)
y=x
x*=1
FIGURE 1.2. Fixed points of f(x) = x2 − x + 1.
Example 1.7. The Tent Map
Consider the equation (Figure 1.3)
x(n + 1) = T (x(n)),
where
⎧⎪⎨ 2 x
T (x) =
⎪⎩2(1−x)
f o r 0 ≤ x ≤ 1 , 2
for 1
One of the main objectives in the study of a dynamical system is to analyze the behavior of its solutions near an equilibrium point. This study constitutes the stability theory. Next we introduce the basic definitions of stability.
Definition 1.8. (a) The equilibrium point x∗ of (1.1.1) is stable (Figure 1.4) if given ε > 0 there exists δ > 0 such that |x0 −x∗| < δ implies |fn(x0)−x∗| < ε for all n > 0. If x∗ is not stable, then it is called unstable (Figure 1.5).
(b) The point x∗ is said to be attracting if there exists η > 0 such that |x(0) − x∗| < η implies lim x(n) = x∗.
n→∞
If η = ∞, x∗ is called a global attractor or globally attracting.
0 1 2 3 4 5 6 7 8 9 10 n
12 1. Dynamics of First-Order Difference Equations
x(n) x*+ε
x*+δ x0
x* x*- δ x* - ε
FIGURE 1.5. Unstable x*. There exists ε > 0 such that no matter how close x(0) is to x*, there will be an N such that x(N) is at least ε from x*.
x(n)
x* + η x1(0)
x* x2(0)
x-η
n 0 1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10 n
FIGURE 1.6. Asymptotically stable x*. Stable if x(0) is within η of x*; then limn→∞ x(n) = x*.
(c) The point x∗ is an asymptotically stable equilibrium point if it is stable and attracting.
If η = ∞, x∗ is said to be globally asymptotically stable (Figure 1.7).
To determine the stability of an equilibrium point from the above def- initions may prove to be a mission impossible in many cases. This is due to the fact that we may not be able to find the solution in a closed form even for the deceptively simple-looking equation (1.1.1). In this section we present some of the simplest but most powerful tools of the trade to help us understand the behavior of solutions of (1.1.1) in the vicinity of equilib-
x(n) x2(0)
x*
x1(0)
1.3 Equilibrium Points 13
1 2 3 4 5 6 7 8 9 10 n
FIGURE 1.7. Globally asymptotically stable x*. Stable and limn→∞ x(n) = x∗ for all x(0).
rium points, namely, the graphical techniques. A hand-held calculator may fulfill all your graphical needs in this section.
1.3.1 The Stair Step (Cobweb) Diagrams
We now give, in excruciating detail, another important graphical method for analyzing the stability of equilibrium (and periodic) points for (1.1.1). Since x(n+1) = f(x(n)), we may draw a graph of f in the (x(n),x(n+1)) plane. Then, given x(0) = x0, we pinpoint the value x(1) by drawing a vertical line through x0 so that it also intersects the graph of f at (x0 , x(1)). Next, draw a horizontal line from (x0 , x(1)) to meet the diagonal line y = x at the point (x(1), x(1)). A vertical line drawn from the point (x(1), x(1)) will meet the graph of f at the point (x(1), x(2)). Continuing this process, one may find x(n) for all n > 0.
Example 1.9. The Logistic Equation
Let y(n) be the size of a population at time n. If μ is the rate of growth of the population from one generation to another, then we may consider a mathematical model in the form
y(n + 1) = μy(n), μ > 0. (1.3.1) If the initial population is given by y(0) = y0, then by simple iteration we
find that
y(n) = μny0 (1.3.2)
is the solution of (1.3.1). If μ > 1, then y(n) increases indefinitely, and limn→∞y(n)=∞.Ifμ=1,theny(n)=y0 foralln>0,whichmeansthat
14 1. Dynamics of First-Order Difference Equations
the size of the population is constant for the indefinite future. However, for μ < 1, we have limn→∞ y(n) = 0, and the population eventually becomes extinct.
For most biological species, however, none of the above cases is valid as the population increases until it reaches a certain upper limit. Then, due to the limitations of available resources, the creatures will become testy and engage in competition for those limited resources. This competition is proportional to the number of squabbles among them, given by y2(n). A more reasonable model would allow b, the proportionality constant, to be greater than 0,
y(n + 1) = μy(n) − by2(n). (1.3.3) If in (1.3.3), we let x(n) = b y(n), we obtain
μ
x(n + 1) = μx(n)(1 − x(n)) = f (x(n)). (1.3.4)
This equation is the simplest nonlinear first-order difference equation, com- monly referred to as the (discrete) logistic equation. However, a closed-form solution of (1.3.4) is not available (except for certain values of μ). In spite of its simplicity, this equation exhibits rather rich and complicated dynamics. To find the equilibrium points of (1.3.4) we let f(x∗) = μx∗(1 − x∗) = x∗. Thus, we pinpoint two equilibrium points: x∗ = 0 and x∗ = (μ − 1)/μ.
Figure 1.8 gives the stair step diagram of (x(n), x(n + 1)) when μ = 2.5 and x(0) = 0.1. In this case, we also have two equilibrium points. One, x* = 0, is unstable, and the other, x* = 0.6, is asymptotically stable.
Example 1.10. The Cobweb Phenomenon (Economics Application)
Here we study the pricing of a certain commodity. Let S(n) be the number of units supplied in period n, D(n) the number of units demanded in period n, and p(n) the price per unit in period n.
For simplicity, we assume that D(n) depends only linearly on p(n) and is denoted by
D(n) = −mdp(n) + bd, md > 0, bd > 0. (1.3.5)
This equation is referred to as the price–demand curve. The constant md represents the sensitivity of consumers to price. We also assume that the price–supply curve relates the supply in any period to the price one period before, i.e.,
S(n+1)=msp(n)+bs, ms >0, bs >0. (1.3.6)
The constant ms is the sensitivity of suppliers to price. The slope of the demand curve is negative because an increase of one unit in price produces a decrease of md units in demand. Correspondingly, an increase of one unit
1.3 Equilibrium Points 15
x(n+1)
x(n)
in price causes an increase of ms units in supply, creating a positive slope for that curve.
A third assumption we make here is that the market price is the price at which the quantity demanded and the quantity supplied are equal, that is, at which D(n + 1) = S(n + 1).
Thus or
where
−mdp(n+1)+bd =msp(n)+bs, p(n + 1) = Ap(n) + B = f (p(n)),
A = − ms , B = bd − bs . md md
(1.3.7)
(1.3.8)
x0 x*
FIGURE 1.8. Stair step diagram for μ = 2.5.
This equation is a first-order linear difference equation. The equilibrium price p∗ is defined in economics as the price that results in an intersection of the supply S(n + 1) and demand D(n) curves. Also, since p∗ is the unique fixed point of f(p) in (1.3.7), p* = B/(1 − A). (This proof arises later as Exercises 1.3, Problem 6.) Because A is the ratio of the slopes of the supply and demand curves, this ratio determines the behavior of the price sequence. There are three cases to be considered:
(a) −1
(a) Find the positive equilibrium point.
(b) Demonstrate, using the stair step diagram, that the positive equi- librium point is asymptotically stable, taking α = 2 and β = 1.
4. Find the equilibrium points and determine their stability for the equation
5. (a) (b)
x(n+1)=5− 6 . x(n)
Draw a stair step diagram for (1.3.4) for μ = 0.5, 3, and 3.3. What can you conclude from these diagrams?
Determine whether these values for μ give rise to periodic solutions of period 2.
6. (The Cobweb Phenomenon [equation (1.3.7)]). Economists define the equilibrium price p∗ of a commodity as the price at which the demand function D(n) is equal to the supply function S(n + 1). These are defined in (1.3.5) and (1.3.6), respectively.
(a) Show that p* = B , where A and B are defined as in (1.3.8). 1−A
(b) Letms =2,bs =3,md =1,andbd =15.Findtheequilibrium price p*. Then draw a stair step diagram, for p(0) = 2.
7. Continuation of Problem 6:
Economists use a different stair step diagram, as we will explain in the following steps:
(i) Let the x-axis represent the price p(n) and the y-axis represent S(n + 1) or D(n). Draw the supply line and the demand line and find their point of intersection p*.
(ii) Starting with p(0) = 2 we find s(1) by moving vertically to the supply line, then moving horizontally to find D(1) (since D(1) = S(1)), which determines p(1) on the price axis. The supply S(2) is found on the supply line directly above p(1), and then D(2) (= S(2)) is found by moving horizontally to the demand line, etc.
(iii) Is p∗ stable?
1.3 Equilibrium Points 19
8.
9. 10.
Repeat Exercises 6 and 7 for:
(a) ms =md =2,bd =10,andbs =2. (b) ms =1,md =2,bd =14,andbs =2.
Verify that formula (1.3.9) is a solution of (1.3.7). Use formula (1.3.9) to show that:
(a) If−1
Let y(n) =
ha x(n). Then we have 1+ha
y(n + 1) = (1 + ha)y(n)(1 − y(n)) x(t)
x0
1
x0
t
FIGURE 1.13. If a > 0, all solutions with x0 > 0 converge to x∗2 = 1.
24
1. Dynamics of First-Order Difference Equations
x(t)
1
x0
0
x0
t
FIGURE 1.14. If a < 0, all solutions with x0 < 1 converge to x∗1 = 0. or
y(n+1)=μy(n)(1−y(n)), y(0)= ha x(0), and μ=1+ha. 1+ha
The corresponding equilibrium points are y∗ = 0 and y∗ = μ−1 = ha 1 2 μ 1+ha
which correspond to x∗1 = 0 and x∗2 = 1, respectively. Using the Cobweb
diagram, we observe that for 1 < μ < 3 (0 < ha < 2), all solutions
whose initial point y0 in the interval (0, 1) converge to the equilibrium point
y∗ = ha (Figure1.15)andfor0<μ<1(−1
1.4.2 A Nonstandard Scheme
Consider again the logistic differential equation. Now if we replace x2(n) in Euler’s method by x(n)x(n + 1) we obtain
x(n + 1) = x(n) + hax(n) − hax(n)x(n + 1). Simplifying we obtain the rational difference equation
or
x(n + 1) = (1 + ha)x(n) 1 + hax(n)
x(n + 1) = αx(n)
1 + βx(n)
with α = 1 + ha, β = α − 1 = ha.
1.4 Numerical Solutions of Differential Equations 25
x(n+1) 1
0.5
x(n)
x0 0.5 1
FIGURE 1.15. 0 < ha < 2.
x(n+1) 1
0.5
x(n)
x0 0.5 1
FIGURE 1.16. −1 < ha < 0.
This equation has two equilibrium points x∗1 = 0 and x∗2 = 1. From the Cobweb diagram (Figure 1.18) we conclude that lim x(n) = 1 if α > 1.
n→∞
Since h > 0, α > 1 if and only if a > 0. Thus all solutions converge to
the equilibrium point x∗2 = 1 if a > 0 as in the differential equation case regardless of the size of h.
26 1. Dynamics of First-Order Difference Equations
x(n+1) 1
0.5
Exercises 1.4
x(n+1) 1
0.5
x0 0.5 1
FIGURE 1.17. ha > 2.57.
x(n)
x0 0.5 1
FIGURE1.18.α=1+ha, β=α−1=ha.
In Problems 1–5
(a) Find the associated difference equation. (b) Draw an (n, y(n)) diagram.
(c) Find, if possible, the exact solution of the differential equation and draw its graph on the same plot as that drawn in part in (b).
x(n)
1.5 Criterion for the Asymptotic Stability of Equilibrium Points 27
1. y′ =−y2,
2. y′ =−y+ 4,
y
3. y′ =−y+1,
4. y′ =y(1−y),
7. Do Problem 4 using a nonstandard numerical method and compare your results with Euler’s method.
8. Do Problem 5 using a nonstandard numerical method and compare your result with Euler’s method.
9. Use both Euler’s method and a nonstandard method to discretize the differential equation
y′(t)=y2 +t, y(0)=1, 0≤t≤1, h=0.2.
Draw the n − y(n) diagram for both methods. Guess which method
gives a better approximation to the differential equation.
10. (a) Use the Euler method with h = 0.25 on [0, 1] to find the value of y
corresponding to t = 0.5 for the differential equation dy =2t+y, y(0)=1.
dt
(b) Compare the result obtained in (a) with the exact value.
11. Given the differential equation of Problem 10, show that a better approximation is given by the difference equation
y(0)=1, y(0)=1,
0≤t≤1, 0≤t≤1,
h=0.2,0.1. h=0.25. h=0.25.
5. y′ =y2 +2,
y(0)= 1, 4
y(0)=2, y(0)=0.1,
0≤t≤1, 0≤t≤1,
h=0.25. h=0.25.
0≤t≤1,
equation of the differential equation in Problem 1.
6. Use a nonstandard numerical method to find the associated difference
1.5
y(n + 1) = y(n) + 1h(y′(n) + y′(n + 1)). 2
This method is sometimes called the modified Euler method.
Criterion for the Asymptotic Stability of Equilibrium Points
In this section we give a simple but powerful criterion for the asymptotic stability of equilibrium points. The following theorem is our main tool in this section.
Theorem 1.13. Let x∗ be an equilibrium point of the difference equation x(n + 1) = f (x(n)), (1.5.1)
28 1. Dynamics of First-Order Difference Equations
where f is continuously differentiable at x*. The following statements then hold true:
(i) If |f′(x*)| < 1, then x∗ is asymptotically stable. (ii) If |f′(x*)| > 1, then x∗ is unstable.
Proof.
(i) Suppose that |f′(x*)| < M < 1. Then there is an interval J = (x*−γ,
x*+γ) containing x∗ such that |f′(x)| ≤ M < 1 for all x ∈ J. For if
not,thenforeachopenintervalIn =(x∗−1,x∗+1)(forlargen)there nn
isapointxn ∈In suchthat|f′(xn)|>M.Asn→∞,xn →x∗.Since f′ is a continuous function, it follows that
Consequently,
lim f′(xn) = f′(x∗). n→∞
M ≤ lim |f′(xn)| = |f′(x∗)| < M, n→∞
which is a contradiction. This proves our statement. For x(0) ∈ J, we have
|x(1) − x*| = |f (x(0)) − f (x*)|.
By the Mean Value Theorem, there exists ξ between x(0) and x∗ such
that
Thus
Hence
|f(x(0)) − f(x*)| = |f′(ξ)| |x(0) − x*|.
|f(x(0)) − x*| ≤ M|x(0) − x*|.
|x(1) − x*| ≤ M |x(0) − x*|.
Since M < 1, inequality (1.5.2) shows that x(1) is closer to x∗ than
x(0). Consequently, x(1) ∈ J. By induction we conclude that
|x(n) − x*| ≤ Mn|x(0) − x*|. Forε>0weletδ= ε .Thus|x(0)−x*|<δimpliesthat|x(n)−
(1.5.2)
2M
x*| < ε for all n > 0. This conclusion suggests stability. Furthermore,
limn→∞ |x(n) − x*| = 0, and thus limn→∞ x(n) = x*; we conclude asymptotic stability. P
The proof of part (ii) is left as Exercises 1.5, Problem 11.
Remark: In the literature of dynamical systems, the equilibrium point x∗ is said to be hyperbolic if |f′(x*)| ̸= 1.
1.5 Criterion for the Asymptotic Stability of Equilibrium Points 29
g(x)
FIGURE 1.19. Newton’s method.
Example 1.14. The Newton–Raphson Method
x(2) x(1) x0
The Newton–Raphson method is one of the most famous numerical meth- ods for finding the roots of the equation g(x) = 0, where g(x) is continually differentiable (i.e., its derivative exists and is continuous).
Newton’s algorithm for finding a zero x∗ of g(x) is given by the difference equation
(1.5.3) where x(0) = x is your initial guess of the root x*. Here f(x) = x − g(x) .
x
x(n+1)=x(n)− g(x(n)), g′(x(n))
0 g′(x) Note first that the zero x∗ of g(x) is also an equilibrium point of (1.5.3). To determine whether Newton’s algorithm provides a sequence {x(n)} that
converges to x∗ we use Theorem 1.13:
[g′(x*)]2 − g(x*)g′′(x*)
|f′(x*)| = 1 − [g′(x*)]2 = 0,
since g(x*) = 0. By Theorem 1.13, limn→∞ x(n) = x∗ if x(0) = x0 is close
enough to x∗ and g′(x*) ̸= 0.
Observe that Theorem 1.13 does not address the nonhyperbolic case where |f′(x*)| = 1. Further analysis is needed here to determine the sta- bility of the equilibrium point x*. Our first discussion will address the case where f′(x*) = 1.
Theorem 1.15. Suppose that for an equilibrium point x∗ of (1.5.1), f′(x*) = 1. The following statements then hold:
(i) If f′′(x*) ̸= 0, then x∗ is unstable.
(ii) If f′′(x*) = 0 and f′′′(x*) > 0, then x∗ is unstable.
(iii) If f′′(x*) = 0 and f′′′(x*) < 0, then x∗ is asymptotically stable.
30 1. Dynamics of First-Order Difference Equations
x(n+1)
x(n+1)
x(0) x* x(0)
x(n)
FIGURE 1.20. Unstable. f′′(x*) > 0 (semistable from the left).
x(0) x* x0
FIGURE 1.21. Unstable. f′′(x*) < 0 (semistable from the right).
Proof.
(i) If f′′(x*) ̸= 0, then the curve y = f(x) is either concave upward if f′′(x*) > 0 or concave downward if f′′(x*) < 0, as shown in Figures 1.20, 1.21, 1.22, 1.23. If f′′(x*) > 0, then f′(x) > 1 for all x in a small interval I = (x*, x*+ε). Using the same proof as in Theorem 1.13, it is easy to show that x∗ is unstable. On the other hand, if f′′(x*) < 0, then f′(x) > 1 for all x in a small interval I = (x* − ε, x*). Hence x∗ is again unstable. P
x(n)
1.5 Criterion for the Asymptotic Stability of Equilibrium Points 31
x(n+1)
x(0) x* x(0)
FIGURE 1.22. Unstable. f′(x*) = 1,f′′(x*) = 0, and f′′′(x*) > 0.
x(n)
x(n+1)
x(0) x* x(0)
FIGURE 1.23. Asymptotically stable. f′(x*) = 1, f′′(x*) = 0, and f′′′(x*) < 0.
Proofs of parts (ii) and (iii) remain for the student’s pleasure as Exercises 1.5, Problem 14.
We now use the preceding result to investigate the case f′(x*) = −1.
But before doing so, we need to introduce the notion of the Schwarzian derivative of a function f:
Sf(x) = f′′′(x) − 3 f′′(x)2 f′(x) 2 f′(x)
x(n)
32 1. Dynamics of First-Order Difference Equations
Note that if f′(x∗) = −1, then
Theorem 1.16. Suppose that for the equilibrium point x∗ of (1.1.1), f′(x*) = −1. The following statements then hold:
(i) If Sf(x*) < 0, then x∗ is asymptotically stable. (ii) If Sf(x*) > 0, then x∗ is unstable.
Proof. Contemplate the equation
y(n + 1) = g(y(n)), where g(y) = f2(y). (1.5.4)
We will make two observations about (1.5.4). First, the equilibrium point x∗ of (1.1.1) is also an equilibrium point of (1.5.4). Second, if x∗ is asymp- totically stable (unstable) with respect to (1.5.4), then it is so with respect to (1.1.1). (Why?) (Exercises 1.5, Problem 12.) Now,
d g(y) = d f(f(y)) = f′(f(y))f′(y). dy dy
Thus d g(x*) = [f′(x*)]2 = 1. Hence Theorem 1.15 applies to this dy
d2 situation. We need to evaluate dy2 g(x∗) :
Sf(x∗) = −f′′′(x∗) − 3 (f′′(x∗))2 . 2
Hence
d2d2 ′′′ dy2 g(y) = dy2 f(f(y)) = [f (f(y))f (y)]
= [f′(y)]2f′′(f(y)) + f′(f(y))f′′(y). d2
dy2 g(x*) = 0.
Now, Theorem 1.15 [parts (ii) and (iii)] tells us that the asymptotic stability of x∗ is determined by the sign of [g(x*)]′′′. Using the chain rule again, one may show that
[g(x*)]′′′ = −2f′′′(x*) − 3[f′′(x*)]2. (1.5.5)
(The explicit proof with the chain rule remains as Exercises 1.5, Problem 13.) This step rewards us with parts (i) and (ii), and the proof of the theorem is now complete. P
Example 1.17. Consider the difference equation x(n+1) = x2(n)+3x(n). Find the equilibrium points and determine their stability.
Solution The equilibrium points are 0 and −2. Now, f′(x) = 2x + 3. Since f′(0) = 3, it follows from Theorem 1.13 that 0 is unstable. Now, f′(−2) = −1, so Theorem 1.16 applies. Using (1.5.5) we obtain −2f′′′(−2)−
1.5 Criterion for the Asymptotic Stability of Equilibrium Points 33
x0=2.9
x(n+1)
x0=.5
x(n)
FIGURE 1.24. Stair step diagram for x(n + 1) = x2(n) + 3x(n).
3[f′′(−2)]2 = −12 < 0. Theorem 1.16 then declares that the equilibrium point −2 is asymptotically stable. Figure 1.24 illustrates the stair step diagram of the equation.
Remark: One may generalize the result in the preceding example to a gen- eral quadratic map Q(x) = ax2 + bx + c, a ̸= 0. Let x∗ be an equilibrium point of Q(x), i.e., Q(x∗) = x∗. Then the following statements hold true.
(i) If Q′(x∗) = −1, then by Theorem 1.16, the equilibrium point x∗ is asymptotically stable. In fact, there are two equilibrium points for Q(x);
x∗1 = [(1−b)−(b−1)2 −4ac]/2a;
x∗2 = [(1−b)+(b−1)2 −4ac]/2a.
It is easy to see that Q′(x∗1) = −1, if (b−1)2 = 4ac+4 and Q′(x∗2) ̸= −1. Thus x∗1 is asymptotically stable if (b − 1)2 = 4ac + 4 (Exercises 1.5, Problem 8).
(ii) If Q′(x∗) = 1, then by Theorem 1.15, x∗ is unstable. In this case, we have only one equilibrium point x∗ = (1 − b)/2a. Thus, x∗ is unstable if(b−1)2 =4ac.
Remark:
(i) Theorem 1.15 fails if for a fixed point x∗, f′(x∗) = 1, f′′(x∗) = f′′′(x∗) = 0. For example, for the map f(x) = x+(x−1)4 and its fixed point x∗ = 1, f′(x∗) = 1, f′′(x∗) = f′′′(x∗) = 0, and f(4)(x∗) = 24 > 0.
(ii) Theorem 1.16 fails if f′(x∗) = −1, and Sf(x∗) = 0. This may be illustrated by the function f(x) = −x+2×2 −4×3. For the fixed x∗ = 0, f′(x∗) = −1, and Sf(x∗) = 0.
34 1. Dynamics of First-Order Difference Equations
In Appendix A, we present the general theory developed by Dannan, Elaydi, and Ponomarenko in 2003 [30]. The stability of the fixed points in the above examples will be determined.
Exercises 1.5
In Problems 1 through 7, find the equilibrium points and determine their stability using Theorems 1.13, 1.15, and 1.16.
1. x(n+1)= 1[x3(n)+x(n)]. 2
2. x(n+1)=x2(n)+1. 8
3. x(n + 1) = tan−1 x(n).
4. x(n+1)=x2(n).
5. x(n+1)=x3(n)+x(n).
6. x(n+1)= αx(n) , 1 + βx(n)
α>1andβ>0.
7. x(n+1)=−x3(n)−x(n).
8. LetQ(x)=ax2+bx+c,a̸=0,andletx∗ beafixedpointofQ.Prove
the following statements:
(i) If Q′(x∗) = −1, then x∗ is asymptotically stable. Then prove the rest of Remark (i).
(ii) If Q′(x∗) = 1, then x∗ is unstable. Then prove the rest of Remark (ii).
9. Suppose that in (1.5.3), g(x*) = g′(x*) = 0 and g′′(x*) ̸= 0. Prove that x∗ is an equilibrium point of (1.5.3).
10. Prove Theorem 1.13, part (ii).
11. Prove that if x∗ is an equilibrium point of (1.5.1), then it is an equi- librium point of (1.5.1). Show also that the converse is false in general. For what class of maps f(x) does the converse hold?
12. Prove that if an equilibrium point x∗ of (1.5.1) is asymptotically stable with respect to (1.5.4) (or unstable, as the case may be), it is also so with respect to (1.1.1).
13. Verify formula (1.5.5).
14. Prove Theorem 1.15, parts (ii) and (iii).
15. Definition of Semistability. An equilibrium point x∗ of x(n + 1) = f(x(n)) is semistable (from the right) if given ε > 0 there exists δ > 0 such that if x(0) > x*, x(0) − x∗ < δ, then x(n) − x∗ < ε. Semistabil- ity from the left is defined similarly. If in addition, limn→∞ x(n) = x∗
whenever x(0) − x* < η{x* − x(0) < η}, then x∗ is said to be semi- asymptotically stable from the right {or from the left, whatever the case may be}.
Suppose that if f′(x*) = 1, then f′′(x∗) ̸= 0. Prove that x∗ is:
(i) semiasymptotically stable from the right from the right if f′′(x*) < 0;
(ii) semiasymptotically stable from the left from the left if f′′(x*) > 0.
16. Determine whether the equilibrium point x* = 0 is semiasymptotically
stable from the left or from the right. (a) x(n+1)=x3(n)+x2(n)+x(n). (b) x(n+1)=x3(n)−x2(n)+x(n).
1.6 Periodic Points and Cycles
The second most important notion in the study of dynamical systems is the notion of periodicity. For example, the motion of a pendulum is periodic. We have seen in Example 1.10 that if the sensitivity ms of the suppliers to price is equal to the sensitivity of consumers to price, then prices oscillate between two values only.
Definition 1.18. Let b be in the domain of f. Then:
(i) b is called a periodic point of f (or of (1.5.1)) if for some positive integer k, fk(b) = b. Hence a point is k-periodic if it is a fixed point of fk, that is, if it is an equilibrium point of the difference equation
x(n + 1) = g(x(n)), (1.6.1)
where g = fk.
The periodic orbit of b, O(b) = {b,f(b),f2(b),…,fk−1(b)}, is often
called a k-cycle.
(ii) b is called eventually k-periodic if for some positive integer m, fm(b) is
a k-periodic point. In other words, b is eventually k-periodic if fm+k(b) = fm(b).
Graphically, a k-periodic point is the x-coordinate of the point where the graph of fk meets the diagonal line y = x. Figure 1.25 depicts the graph of f2, where f is the logistic map, which shows that there are four fixed points of f2, of which two are fixed points of f as shown in Figure 1.26. Hence the other two fixed points of f2 form a 2-cycle. Notice also that the point x0 = 0.3 (in Figure 1.26) goes into a 2-cycle, and thus it is an eventually
1.6 Periodic Points and Cycles 35
36
1. Dynamics of First-Order Difference Equations
x(n+2)
FIGURE 1.25. Graph of f 2 with four fixed points. f (x) = 3.43x(1 − x).
x(n+1)
x(n)
x(n)
FIGURE 1.26. x0 goes into a 2-cycle. f (x) = 3.43x(1 − x).
2-periodic point. Moreover, the point x* = 0.445 is asymptotically stable relative to f2 (Figure 1.27).
Observe also that if A = −1 in (1.3.7), then f2(p0) = −(−p0 +B)+B = p0. Therefore, every point is 2-periodic (see Figure 1.10). This means that in this case, if the initial price per unit of a certain commodity is p0, then the price oscillates between p0 and B − p0.
Example 1.19. Consider again the difference equation generated by the tent function
T (x) =
⎧⎪⎨ 2 x f o r 0 ≤ x ≤ 1 , 2
⎪⎩2(1−x) for1
Proof. We apply Theorem 1.13 to (1.6.1). Notice that by using the chain rule one may show that
[fk(x(r))]′ =f′(x(0))f′(x(1)),…,f′(x(k−1)).
(See Exercises 1.6, Problem 12.) P
The conclusion of the theorem now follows.
Example 1.22. Consider the map Q(x) = x2 −0.85 defined on the interval [−2, 2]. Find the 2-cycles and determine their stability.
Solution Observe that Q2(x) = (x2 − 0.85)2 − 0.85. The 2-periodic points are obtained by solving the equation
Q2(x)=x, orx4−1.7×2−x−0.1275=0. (1.6.2) This equation has four roots, two of which are fixed points of the map Q(x).
These two fixed points are the roots of the equation
x2 −x−0.85 = 0. (1.6.3)
1.6 Periodic Points and Cycles 39
40 1. Dynamics of First-Order Difference Equations
To eliminate these fixed points of Q(x) from (1.6.2) we divide the left-hand side of (1.6.2) by the left-hand side of (1.6.3) to obtain the second-degree equation
x2 +x+0.15=0. (1.6.4) The 2-periodic points are now obtained by solving (1.6.4). They are given
by
√√ a= −1+ 0.4, b= −1− 0.4.
22
To check the stability of the cycle {a, b} we apply Theorem 1.21. Now,
|Q′(a)Q′(b)| = |(−1 + √0.4)(−1 − √0.4)| = 0.6 < 1. Hence by Theorem 1.21, part (i), the 2-cycle is asymptotically stable.
Exercises 1.6
1. Suppose that the difference equation x(n + 1) = f (x(n)) has a 2-cycle whose orbit is {a, b}. Prove that:
(i) the 2-cycle is asymptotically stable if |f′(a)f′(b)| < 1, (ii) the 2-cycle is unstable if |f′(a)f′(b)| > 1.
2. Let T be the tent map in Example 1.17. Show that 2, 4, 8 is an 999
unstable 3-cycle for T.
3. Let f(x) = −1×2 − x + 1. Show that 1 is an asymptotically stable
2-periodic point of f.
In Problems 4 through 6 find the 2-cycle and then determine its stability.
4. x(n + 1) = 3.5x(n)[1 − x(n)].
5. x(n+1)=1−x2.
6. x(n+1)=5−(6/x(n)).
7. Let f(x) = ax3 −bx+1, where a,b ∈ R. Find the values of a and b for which {0, 1} is an attracting 2-cycle.
Consider Baker’s function defined as follows:
⎧⎪⎨ 2 x f o r 0 ≤ x ≤ 1 ,
2
*8. (Hard). Draw Baker’s function B(x). Then find the number of n-periodic points of B.
22
B(x) =
Problems 8, 9, and 10 are concerned with Baker’s function B(x) on [0, 1].
⎪⎩2x−1 for1
13. Give an example of a decreasing function that has a fixed point and a 2-cycle.
14. (i) Can a decreasing map have a k-cycle for k > 1? (ii) Can an increasing map have a k-cycle for k > 1?
Carvalho’s Lemma. In [18] Carvalho gave a method to find periodic points of a given function. The method is based on the following lemma.
Lemma 1.23. If k is a positive integer and x(n) is a periodic sequence of period k, then the following hold true:
(i) Ifk>1isoddandm=k−1,then 2
for all n≥1.
(ii) Ifkisevenandk=2m,then
1.6 Periodic Points and Cycles 41
for all n≥1.
m 2 j n π 2 j n π
x(n) = c0 +
cj cos k + dj sin k ,
j=1
m−1 n 2jnπ
2jnπ
k ,
x(n)=c0+(−1) cm+ cj cos k j=1
+dj sin
42 1. Dynamics of First-Order Difference Equations
Example 1.24 [23]. Consider the equation
x(n + 1) = x(n) exp(r(1 − x(n)), (1.6.5)
which describes a population with a propensity to simple exponential growth at low densities and a tendency to decrease at high densities. The quantity λ = exp(r(1 − x(n))) could be considered the density- dependent reproductive rate of the population. This model is plausible for a single-species population that is regulated by an epidemic disease at high density.
The nontrivial fixed point of this equation is given by x∗ = 1. Now, f′(1) = 1−r. Hence x∗ = 1 is asymptotically stable if 0 < r ≤ 2 (check r = 2). At r = 2, x∗ = 1 loses its stability and gives rise to an asymptotically stable 2-cycle. Carvalho’s lemma implies
x(n) = a + (−1)nb. Plugging this into equation (1.6.5) yields
a−(−1)nb=(a+(−1)nb) expr(1−a−(−1)nb). The shift n → n + 1 gives
Hence
a+(−1)nb=(a−(−1)nb) expr(1−a+(−1)nb). a2 −b2 =(a2 −b2) exp2r(1−a).
Thus either a2 = b2, which gives the trivial solution 0, or a = 1. Hence a 2-periodic solution has the form x(n) = 1 + (−1)nb. Plugging this again into equation (1.6.5) yields
1 − (−1)nb = (1 + (−1)nb) exp((−1)n+1rb). Let y = (−1)n+1b.Then
1 + y = (1 − y)ery,
1 1+y
r=yln 1−y =g(y).
The function g has its minimum at 0, where g(0) = 2. Thus, for r < 2,g(y) = r has no solution, and we have no periodic points, as predicted earlier. However, each r > 2 determines values ±yr and the corresponding coefficient (−1)nb. Further analysis may show that this map undergoes bifurcation similar to that of the logistic map.
Exercises 1.6 (continued).
In Problems 15 through 20, use Carvalho’s lemma (Lemma 1.23).
1.7 The Logistic Equation and Bifurcation 43
15. Consider Ricker’s equation
x(n + 1) = x(n) exp(r(1 − x(n))).
Find the 2-period solution when r > 2.
16. The population of a certain species is modeled by the difference equa- tion x(n + 1) = μx(n)e−x(n), x(n) ≥ 0, μ > 0. For what values of μ does the equation have a 2-cycle?
17. Use Carvalho’s lemma to find the values of c for which the map Qc(x)=x2 +c, c∈[−2,0],
has a 3-cycle and then determine its stability.
18*.(Term Project). Find the values of μ where the logistic equation x(n+
1) = μx(n)[1 − x(n)] has a 3-periodic solution.
19. Use Carvalho’s lemma to find the values of μ where the logistic
equation x(n + 1) = μx(n)[1 − x(n)] has a 2-periodic solution.
20. Find the 3-periodic solutions of the equation x(n + 1) = ax(n), a ̸= 1.
1.7 The Logistic Equation and Bifurcation
Let us now return to the most important example in this chapter: the logistic difference equation
x(n + 1) = μx(n)[1 − x(n)], (1.7.1) which arises from iterating the function
Fμ(x) = μx(1 − x), x ∈ [0, 1], μ > 0. (1.7.2) 1.7.1 Equilibrium Points
To find the equilibrium points (fixed points of Fμ) of (1.7.1) we solve the equation
Fμ(x*) = x*.
Hence the fixed points are 0, x* = (μ − 1)/μ. Next we investigate the
stability of each equilibrium point separately.
(a) The equilibrium point 0. (See Figures 1.31, 1.32.) Since Fμ′ (0) = μ, it follows from Theorems 1.13 and 1.15 that:
(i) 0 is an asymptotically stable fixed point for 0 < μ < 1, (ii) 0 is an unstable fixed point for μ > 1.
44
1. Dynamics of First-Order Difference Equations
x(n+1)
FIGURE 1.31. 0 < μ < 1 : 0 is an asymptotically stable fixed point.
x0
x(n)
x(n+1)
x0 x*
FIGURE 1.32. μ > 1 : 0 is an unstable fixed point, x∗ is an asymptotically fixed point.
The case where μ = 1 needs special attention, for we have F1′(0) = 1 and F′′(0) = −2 ̸= 0. By applying Theorem 1.15 we may conclude that 0 is unstable. This is certainly true if we consider negative as well as positive initial points in the neighborhood of 0. Since negative initial points are not in the domain of Fμ, we may discard them and consider only positive initial points. Exercises 1.5, Problem 16 tells us that 0 is semiasymptotically stable from the right, i.e., x∗ = 0 is asymptotically stable in the domain [0, 1].
(b) The equilibrium point x* = (μ − 1)/μ, μ ̸= 1. (See Figures 1.32, 1.33.)
In order to have x* ∈ (0, 1] we require that μ > 1. Now, Fμ′ ((μ−1)/μ) = 2− μ. Thus using Theorems 1.13 and 1.16 we obtain the following conclusions:
x(n)
1.7 The Logistic Equation and Bifurcation 45
x(n+1)
x x* 0
x(n)
FIGURE 1.33. μ > 3: x∗ is an unstable fixed point.
(i) x∗ is an asymptotically stable fixed point for 1 < μ ≤ 3 (Figure 1.32).
(ii) x∗ is an unstable fixed point for μ > 3 (Figure 1.33). 1.7.2 2-Cycles
To find the 2-cycles we solve the equation Fμ2(x) = x (or we solve x2 = μx1(1 − x1), x1 = μx2(1 − x2)),
μ2x(1 − x)[1 − μx(1 − x)] − x = 0. (1.7.3) Discarding the equilibrium points 0 and x* = μ−1, one may then divide
μ
(1.7.3) by the factor x(x − (μ − 1)/μ) to obtain the quadratic equation μ2×2 −μ(μ+1)x+μ+1=0.
Solving this equation produces the 2-cycle
x(0) = (1+μ)− (μ−3)(μ+1) 2μ,
x(1) = (1+μ)+ (μ−3)(μ+1) 2μ. (1.7.4) Clearly, there are no periodic points of period 2 for 0 < μ ≤ 3, and there
isa2-cycleforμ>3.Forourreferenceweletμ0 =3.
1.7.2.1 Stability of the 2-Cycle {x(0), x(1)} for μ > 3 From Theorem 1.21, this 2-cycle is asymptotically stable if
or
|Fμ′ (x(0))Fμ′ (x(1))| < 1,
−1 < μ2(1 − 2x(0))(1 − 2x(1)) < 1.
Substituting from (1.7.4) the values of x(0) and x(1) into (1.7.5), we obtain
√ 3<μ<1+ 6≈3.44949.
(1.7.5)
46 1. Dynamics of First-Order Difference Equations
Conclusion This2-cycleisattractingif3<μ<3.44949....
√
Question What happens when μ = 1 + In this case,
6?
Fμ2(x(0))′ = Fμ′ (x(0))Fμ′ (x(1)) = −1. (Verify in Exercises 1.7, Problem 7.)
(1.7.6) Hence we may use Theorem 1.16, part (i), to conclude that the 2-cycle is
√ √1
also attracting. For later reference, let μ = 1 + becomes unstable when μ > μ1 = 1 + 6.
6. Moreover, the 2-cycle
To find the 4-cycles we solve Fμ4(x) = x. The computation now becomes unbearable, and one should resort to a computer to do the work. It turns out that there is a 22-cycle when μ > 1 + √6, which is attracting for 1 + √6 < μ < 3.544090.... This 22-cycle becomes unstable at μ > μ2 = 3.544090….
When μ = μ2, the 22-cycle bifurcates into a 23 cycle. The new 23 cycle is attracting for μ3 < μ ≤ μ4 for some number μ4. This process of double bifurcation continues indefinitely. Thus we have a sequence {μn}∞n=0 where at μn there is a bifurcation from a 2n−1-cycle to a 2n-cycle. (See Figures 1.34, 1.35.) Table 1.4 provides some astonishing patterns.
From Table 1.4 we bring forward the following observations:
(i) The sequence {μn} seems to converge to a number μ∞ = 3.57 . . . .
(ii) The quotient (μn − μn−1)/(μn+1 − μn) seems to tend to a number δ = 4.6692016.... This number is called the Feigenbaum number after its discoverer, the physicist Mitchell Feigenbaum [56]. In fact, Feigenbaum made a much more remarkable discovery: The number δ is universal and is independent of the form of the family of maps fμ. However, the number μ∞ depends on the family of functions under consideration.
1.7.3 22 -Cycles
x
1
μ
1 31+6
FIGURE 1.34. Partial bifurcation diagram for {Fμ}.
1.7 The Logistic Equation and Bifurcation 47
μ
FIGURE 1.35. The bifurcation diagram of Fμ. TABLE 1.4. Feigenbaum table.
— 4.752027 . . . 4.656673 . . . 4.667509 . . . 4.668576 . . . 4.669354 . . .
n μn μn − μn−1 03——
μn − μn−1 μn+1 − μn
1 2 3 4 5 6
1.7.4 The Bifurcation Diagram
Here the horizontal axis represents the μ values, and the vertical axis repre- sents higher iterates Fμn(x). For a fixed x0, the diagram shows the eventual behavior of Fμn(x0). The bifurcation diagram was obtained with the aid of
3.449499 . . . 3.544090 . . . 3.564407 . . . 3.568759 . . . 3.569692 . . . 3.569891 . . .
0.449499 . . . 0.094591 . . . 0.020313 . . . 0.004352 . . . 0.00093219 . . . 0.00019964 . . .
(Feigenbaum [56] (1978)).
ilies of maps (such as Fμ) of an interval into itself, the number δ = 4.6692016 does not in general depend on the family of maps.
Theorem 1.25
For sufficiently smooth fam-
48 1. Dynamics of First-Order Difference Equations
a computer for x0 = 1 , taking increments of 1 for μ ∈ [0, 4] and plotting
n 1 2 5 0 0 μ,Fμ 2 for 200 ≤ n ≤ 500.
all points
Question What happens when μ > μ∞?
Answer From Figure 1.35 we see that for μ∞ < μ ≤ 4 we have a large number of small windows where the attracting set is an asymptotically sta- ble cycle. The largest window appears at approximately μ = 3.828427 . . ., where we have an attracting 3-cycle. Indeed, there are attracting k-cycles for all positive integers k, but their windows are so small that they may not be noticed without sufficient zooming. As in the situation where μ < μ∞, these k-cycles lose stability and then double bifurcate into at- tracting 2nk-cycles. We observe that outside these windows the picture looks chaotic!
Remarks: Our analysis of the logistic map Fμ may be repeated for any quadratic map Q(x) = ax2 + bx + c. Indeed, the iteration of the quadratic map Q (with suitably chosen parameters) is equivalent to the iteration of the logistic map Fμ. In other words, the maps Q and Fμ possess the same type of qualitative behavior. The reader is asked, in Exercises 1.7, Problem 11, to verify that one can transform the difference equation
y(n + 1) = y2(n) + c x(n + 1) = μx(n)[1 − x(n)]
μ μμ2 y(n)=−μx(n)+ 2, c= 2 − 4 .
and μ = 4 corresponds to c = −2. Naturally, we expect to have the same behavior of the iteration of (1.7.7) and (1.7.8) at these corresponding values of μ and c.
Comments: We are still plagued by numerous unanswered questions in connection with periodic orbits (cycles) of the difference equation
x(n + 1) = f (x(n)). (1.7.10)
Question A. Do all points converge to some asymptotically stable periodic orbit of (1.7.8)?
The answer is definitely no.
If f(x) = 1 − 2x2 in (1.7.10), then there are no asymptotically stable (attractor) periodic orbits. Can you verify this statement? If you have some difficulty here, it is not your fault. Obviously, we need some tools to help us in verifying that there are no periodic attractors.
to
by letting
(1.7.7)
(1.7.8)
(1.7.9) Note here that μ = 2 corresponds to c = 0,μ = 3 corresponds to c = −3,
4
1.7 The Logistic Equation and Bifurcation 49
Question B. If there is a periodic attractor of (1.7.10), how many points converge to it?
Once again, we need more machinery to answer this question.
Question C. Can there be several distinct periodic attractors for (1.7.10)?
This question leads us to the Li–Yorke famous result “Period Three Implies Chaos” [92]. To explain this and more general results requires the intro- duction of the so-called Schwarzian derivative of f(x). We will come back to these questions in Chapter 6.
Exercises 1.7
Unless otherwise stated, all the problems here refer to the logistic difference equation (1.7.1).
1. Use the stair step diagram for F4k on [0,1], k = 1, 2, 3, ..., to demon- strate that F4 has at least 2k periodic points of period k (including periodic points of periods that are divisors of k).
2. Find the exact solution of x(n + 1) = 4x(n)[1 − x(n)].
3. Let x* = (μ − 1)/μ be the equilibrium point of (1.7.1). Show that:
(i) For 1 < μ ≤ 3, x∗ is an attracting fixed point. (ii) For μ > 3, x∗ is a repelling fixed point.
4. Provethatlimn→∞F2n(x)=1 if0
*14. (Project). Use a calculator or a computer to develop a bifurcation diagram, as in Figures 1.34, 1.35, for (1.7.6).
*15. (Project). Develop a bifurcation diagram for the quadratic map Qλ(x)=1−λx2 ontheinterval[−1,1], λ∈(0,2].
In Problems 16–19 determine the stability of the fixed points of the difference equation.
16. x(n + 1) = x(n) + 1 sin(2πx(n)). π
17. x(n + 1) = 0.5 sin(πx(n)).
18. x(n + 1) = 2x(n) exp(−x(n)).
19. A population of birds is modeled by the difference equation
3.2x(n) for 0 ≤ x(n) ≤ 1,
x(n+1) =
where x(n) is the number of birds in year n. Find the equilibrium points
0.5x(n) for x(n) > 1, and then determine their stability.
1.8 Basin of Attraction and Global Stability (Optional)
It is customary to call an asymptotically stable fixed point or a cycle an attractor. This name makes sense since in this case all nearby points tend to the attractor. The maximal set that is attracted to an attractor M is called the basin of attraction of M. Our analysis applies to cycles of any period.
Definition 1.26. Let x∗ be a fixed point of map f. Then the basin of attraction (or the stable set) Ws(x∗) of x∗ is defined as
Ws(x∗) = {x : lim fn(x) = x∗}. n→∞
In other words, Ws(x∗) consists of all points that are forward asymptotic to x∗.
Observe that if x∗ is an attracting fixed point, Ws(x∗) contains an open interval around x∗. The maximal interval in Ws(x∗) that contains x∗ is called the immediate basin of attraction and is denoted by Bs(x∗).
Example 1.27. The map f(x) = x2 has one attracting fixed point x∗ = 0. Its basin of attraction Ws(0) = (−1,1). Note that 1 is an unstable fixed point and –1 is an eventually fixed point that goes to 1 after one iteration.
1.8 Basin of Attraction and Global Stability (Optional) 51
5 4 3 2 1
−3 −2 −1 −1
−2 −3
FIGURE 1.36. The basin of attraction W s (0) = (−1, 1) and W s (4) = [−2, −1) ∪ (1, 4]. The immediate basin of attraction B(4) = (1, 4].
Example 1.28. Let us now modify the map f. Consider the map g : [−2, 4] → [−2, 4] defined as
x2 if −2 ≤ x ≤ 1, 3√x−2 if1
To prove the invariance of B(x∗), assume that there exists y ∈ B(x∗) such that fr(y) ∈/ B(x∗) for some r ∈ Z+. Since B(x∗) is an interval, it follows by the Intermediate Value Theorem that fr(B(x∗)) is also an interval. Moreover, this interval fr(B(x∗)) must contain x∗ since fr(x∗) = x∗. Thus fr(B(x∗))∩B(x∗) ̸= 0, and hence B(x∗)∪fr(B(x∗)) is an interval in Ws(x∗), which violates the maximality of B(x∗).
(ii) The proof of this part is analogous to the proof of part (a) and will be left to the reader to verify. P
There are several (popular) maps such as the logistic map and Ricker’s
map in which the basin of attraction, for the attractive fixed point, is the
entire space with the exception of one or two points (fixed or eventually
fixed). For the logistic map Fμ(x) = μx(1 − x) and 1 < μ < 3, the basin of
attraction W s(x∗) = (0, 1) for the fixed point x∗ = μ−1 . And for Ricker’s μ
map Rp(x) = xep−x, 0 < p < 2, the basin of attraction Ws(x∗) = (0,∞), for x∗ = p. Here we will consider only the logistic map and leave it to the reader to prove the statement concerning Ricker’s map.
Notice that |Fμ′ (x)| = |μ−2μx| < 1 if and only if −1 < μ−2μx < 1. This
implies that μ−1 < x < μ+1. Hence |F′(x)| < 1 for all x ∈ μ−1,μ+1 . 2μ2μμ 2μ2μ
Observethatx∗ = μ−1 ∈ μ−1,μ+1 ifandonlyif1<μ<3.Now μ 2μ2μ
1.8 Basin of Attraction and Global Stability (Optional) 53
Fμ μ+1 =Fμ μ−1 = 1 (μ−1)(μ+1) .Noticethatsince1<μ<3, 2μ 2μ 2 2μ
μ−1 < 1 · (μ−1)(μ+1) < μ+1. Hence μ−1, μ+1 ⊂ Ws(x∗). 2μ 22μ2μ 2μ2μ
If z ∈ 0,μ−1 , then F′(z) > 1. By the Mean Value Theorem, 2μ μ
Fμ(z)−Fμ(0) = F′ (γ), for some γ with 0 < γ < z. Hence z−0 μ
Fμ(z) − Fμ(0) = Fμ(z) ≥ βz
forsomeβ>1.Thenforsomer∈Z+,Fr(z)≥βrz> μ−1 andFr−1(z)< μ 2μ μ
μ−1. Moreover, since F is increasing on 0, μ−1 , Fr(z) < F μ−1
2μ
μ−1 μ−1
μ−1 μ+1 μ 4
2μμ μ2μ ∗ s ∗
μ 2μ 1 − 2μ =
= ≤ x . Thus z ∈ W (x ). On the other
hand, Fμ μ+1,1 ⊂ (0,x∗) and hence μ+1,1 ⊂ Ws(x∗). This shows 2μ 2μ
that W s(x∗) = (0, 1). To summarize
Lemma 1.31. For the logistic map Fμ(x) = μx(1 − x), 1 < μ < 3, Ws(x∗) = (0,1) for x∗ = μ−1.
μ
We now turn our attention to periodic points. If x ̄ is a periodic point
of period k under the map f, then its basin of attraction Ws(x ̄) is its
basin of attraction as a fixed point under the map fk. Hence Ws(x ̄) =
{x : lim (fk)n(x) = lim fkn(x) = x ̄}. Let {x ̄1, x ̄2, . . . , x ̄k} be a k-cycle n→∞ n→∞
of a map f. Then clearly for i ̸= j, Ws(x ̄i) ∩ Ws(x ̄j) = ∅. (Why?) More generally, if x is a periodic point of period r and y ̸= x is a periodic point of period s, then W s(x) ∩ W s(y) = ∅ (Exercises 1.8, Problem 6).
Example 1.32. Consider the function f(x) = −x1 . Then x∗ = 0 is the only fixed point. There is a 2-cycle {−1,1} with f(−1) = 1, f2(−1) = −1. The cobweb diagram (Figure 1.37) shows that Ws(1) = (0,∞), Ws(−1) = (−∞, 0).
−1 1
FIGURE 1.37.
3
54 1. Dynamics of First-Order Difference Equations
Exercises 1.8
1. Investigate the basin of attraction of the fixed points of the map
f(x) =
2. Letf(x)=|x−1|.FindWs(1).
x2 if −3 ≤ x ≤ 1, 4√x−3 if1
8. Let x∗ be an attracting fixed point under a continuous map f. If the immediate basin of attraction B(x∗) = (a,b), show that the set {a,b} is invariant. Then conclude that there are only three scenarios in this case: (1) both a and b are fixed points, or (2) a or b is fixed and the other is an eventually fixed point, or (3) {a, b} is a 2-cycle.
9. Show that for Ricker’s map
Rp(x)=xep−x, 0
for 1 ≤ i, k ≤ 7.
13. Use Table 2.1 and formula (2.1.21) to write x3,x4, and x5 in terms of
the factorial polynomials x(k) (e.g., x2 = x(1) + x(2)).
14. Use Problem 13 to find
.
k−1 n−k
ni n−s−1k f(n)= i ∆f(0)+ k−1 ∆f(s).
i=0 s=0
m i=1
si(m)x(i). (2.1.21)
xm =
12. Use (2.1.21) to verify Table 2.1 which gives the Stirling numbers si(k)
64
2. Linear Difference Equations of Higher Order TABLE 2.1. Stirling numbers si(k).
i\k 1 2 3 4 5 6 7 11111111
(i) ∆−1(n3 + 1).
5
(ii) ∆−1
.
2
3
4
5
6 71
1 3 7 15 31 63 1 6 25 90 301 1 10 65 350 1 15 140
121
15. 16. 17.
n(n+3)
Use Problem 13 to solve the difference equation y(n + 1) = y(n) + n3. Use Problem 13 to solve the difference equation y(n + 1) = y(n) − 5n2. Consider the difference equation3
y(n + 1) = a(n)y(n) + g(n). (2.1.22) n−1
(a) Put y(n) = i=0 a(i) u(n) in (2.1.22). Then show that ∆u(n) = g(n)/ ni=0 a(i).
(b) Prove that
n−1 n−1 n−1
y(n) =
(Compare with Section 1.2.)
a(i) g(r),
y0 = y(0).
i=0
r=0 i=r+1
a(i) y0 +
2.2
where pi(n) and g(n) are real-valued functions defined for n ≥ n0 and pk(n) ̸= 0 for all n ≥ n0. If g(n) is identically zero, then (2.2.1) is said to be a homogeneous equation. Equation (2.2.1) may be written in the form
y(n + k) = −p1(n) y(n + k − 1) − · · · − pk(n) y(n) + g(n). (2.2.2)
3This method of solving a nonhomogeneous equation is called the method of variation of constants.
General Theory of Linear Difference Equations
normal form of a kth-order nonhomogeneous linear difference equation y(n + k) + p1(n) y(n + k − 1) + · · · + pk(n) y(n) = g(n), (2.2.1)
The
is given by
2.2 General Theory of Linear Difference Equations 65
By letting n = 0 in (2.2.2), we obtain y(k) in terms of y(k − 1), y(k − 2), · · · , y(0). Explicitly, we have
y(k) = −p1(0)y(k − 1) − p2(0)y(k − 2) − · · · − pk(0)y(0) + g(0). Once y(k) is computed, we can go to the next step and evaluate y(k + 1)
by letting n = 1 in (2.2.2). This yields
y(k + 1) = −p1(1)y(k) − p2(1)y(k − 1) − · · · − pk(1)y(1) + g(1).
By repeating the above process, it is possible to evaluate all y(n) for n ≥ k. Let us now illustrate the above procedure by an example.
Example 2.6. Consider the third-order difference equation
y(n+3)− n y(n+2)+ny(n+1)−3y(n)=n, (2.2.3)
n+1
where y(1) = 0,y(2) = −1, and y(3) = 1. Find the values of y(4), y(5),
y(6), and y(7).
Solution First we rewrite (2.2.3) in the convenient form
y(n+3) = n y(n+2)−ny(n+1)+3y(n)+n. n+1
Letting n = 1 in (2.2.4), we have
y(4) = 1y(3)−y(2)+3y(1)+1 = 5.
For n = 2,
For n = 3,
For n = 4,
(2.2.4)
22
y(5)= 2y(4)−2y(3)+3y(2)+2=−4. 33
y(6)= 3y(5)−3y(4)+3y(3)+3=−3. 42
y(7)= 4y(6)−4y(5)+3y(4)+4=20.9. 5
Now let us go back to (2.2.1) and formally define its solution. A sequence {y(n)}∞n0 or simply y(n) is said to be a solution of (2.2.1) if it satisfies the equation. Observe that if we specify the initial data of the equation, we are led to the corresponding initial value problem
y(k + n) + p1(n)y(n + k − 1) + · · · + pk(n)y(n) = g(n), (2.2.5) y(n0)=a0, y(n0 +1)=a1,…,y(n0 +k−1)=ak−1, (2.2.6)
where the ai’s are real numbers. In view of the above discussion, we conclude with the following result.
66 2. Linear Difference Equations of Higher Order
Theorem 2.7. The initial value problems (2.2.5) and (2.2.6) have a unique solution y(n).
Proof. The proof follows by using (2.2.5) for n = n0,n0 +1,n0 +2,…. Notice that any n ≥ n0 + k may be written in the form n = n0 + k + (n − n0 − k). By uniqueness of the solution y(n) we mean that if there is another solution y ̃(n) of the initial value problems (2.2.5) and (2.2.6), then y ̃(n) must be identical to y(n). This is again easy to see from (2.2.5). P
The question still remains whether we can find a closed-form solution for (2.2.1) or (2.2.5) and (2.2.6). Unlike our amiable first-order equations, obtaining a closed-form solution of (2.2.1) is a formidable task. However, if the coefficients pi in (2.2.1) are constants, then a solution of the equation may be easily obtained, as we see in the next section.
In this section we are going to develop the general theory of kth-order linear homogeneous difference equations of the form
x(n + k) + p1(n)x(n + k − 1) + · · · + pk(n)x(n) = 0. (2.2.7) We start our exposition by introducing three important definitions.
Definition 2.8. The functions f1(n),f2(n),…,fr(n) are said to be lin- early dependent for n ≥ n0 if there are constants a1, a2, . . . , ar, not all zero, such that
a1 f1 (n) + a2 f2 (n) + · · · + ar fr (n) = 0, n ≥ n0 . If aj ̸= 0, then we may divide (2.2.7) by aj to obtain
fj (n) = − a1 f1 (n) − a2 f2 (n) · · · − ar fr (n) aj aj aj
= − ai fi(n). i̸=j aj
(2.2.8)
Equation (2.2.8) simply says that each fj with nonzero coefficient is a linear combination of the other fi’s. Thus two functions f1(n) and f2(n) are linearly dependent if one is a multiple of the other, i.e., f1(n) = af2(n), for some constant a.
The negation of linear dependence is linear independence. Explicitly put, the functions f1(n), f2(n), . . . , fr(n) are said to be linearly independent for n ≥ n0 if whenever
a1 f1 (n) + a2 f2 (n) + · · · + ar fr (n) = 0 foralln≥n0,thenwemusthavea1 =a2 =···=ar =0.
Let us illustrate this new concept by an example.
Example 2.9. Show that the functions 3n , n3n , and n2 3n are linearly independent on n ≥ 1.
2.2 General Theory of Linear Difference Equations 67
Solution Suppose that for constants a1, a2, and a3 we have a13n+a2n3n+a3n23n =0, foralln≥1.
Then by dividing by 3n we get
a1 + a2n + a3n2 = 0, for all n ≥ 1.
This is impossible unless a3 = 0, since a second-degree equation in n pos- sesses at most two solutions n ≥ 1. Hence a1 = a2 = a3 = 0. Similarly, a2 = 0, whence a1 = 0, which establishes the linear independence of our functions.
Definition 2.10. A set of k linearly independent solutions of (2.2.7) is called a fundamental set of solutions.
As you may have noticed from Example 2.9, it is not practical to check the linear independence of a set of solutions using the definition. Fortunately, there is a simple method to check the linear independence of solutions using the so-called Casoratian W (n), which we now define for the eager reader.
Definition 2.11. The Casoratian4 W (n) of the solutions x1 (n), x2 (n), . . . , xr(n) is given by
⎛ x1(n) x2(n) … xr(n) ⎞
⎜ x1(n+1) x2(n+1) … xr(n+1) ⎟ W (n) = det ⎜ . ⎟ .
⎝.⎠ x1(n+r−1) x2(n+r−1) … xr(n+r−1)
(2.2.9)
Example 2.12. Consider the difference equation
x(n + 3) − 7x(n + 1) + 6x(n) = 0.
(a) Show that the sequences 1, (−3)n, and 2n are solutions of the equation.
(b) Find the Casoratian of the sequences in part (a).
Solution
(a) Note that x(n) = 1 is a solution, since 1 − 7 + 6 = 0. Furthermore, x(n) = (−3)n is a solution, since
(−3)n+3 − 7(−3)n+1 + 6(−3)n = (−3)n[−27 + 21 + 6] = 0. Finally, x(n) = 2n is a solution, since
(2)n+3 −7(2)n+1 +6(2)n = 2n[8−14+6] = 0.
4This is the discrete analogue of the Wronskian in differential equations.
68 2. Linear Difference Equations of Higher Order
(b) Now,
⎛1(−3)n 2n⎞ W(n) = det⎜⎝1 (−3)n+1 2n+1⎟⎠
1 (−3)n+2 2n+2
(−3)n+1 =
= (2)n+2(−3)n+1 − (2)n+1(−3)n+2 − (−3)n((2)n+2 − (2)n+1) + (2)n((−3)n+2 − (−3)n+1)
= −12(2)n(−3)n − 18(2)n(−3)n − 4(2)n(−3)n + 2(2)n(−3)n + 9(2)n(−3)n + 3(2)n(−3)n
= −20(2)n(−3)n.
Next we give a formula, called Abel’s formula, to compute the Caso- ratian W(n). The significance of Abel’s formula is its effectiveness in the verification of the linear independence of solutions.
Lemma 2.13 (Abel’s Lemma). Let x1(n),x2(n),…,xk(n) be so- lutions of (2.2.7) and let W(n) be their Casoratian. Then, for n ≥ n0,
pk(i) W(n0). (2.2.10)
(−3)n+2 1
(2)n+1 1 − (−3)n
(2)n+1 (2)n+2
+ (2)n 1
(2)n+2 1 (−3)n+1
(−3)n+2
n−1
W(n) = (−1)
k(n−n0 ) i=n0
Proof. We will prove the lemma for k = 3, since the general case may be established in a similar fashion. So let x1(n), x2(n), and x3(n) be three independent solutions of (2.2.7). Then from formula (2.2.9) we have
⎛⎞
x1(n+1) x2(n+1) x3(n+1) W(n+1)=det⎜⎝x1(n+2) x2(n+2) x3(n+2)⎟⎠.
x1(n+3) x2(n+3) x3(n+3) From (2.2.7) we have, for 1 ≤ i ≤ 3,
(2.2.11)
xi(n + 3) = −p3(n)xi(n) − [p1(n)xi(n + 2) + p2(n)xi(n + 1)] .
(2.2.12)
2.2 General Theory of Linear Difference Equations 69
Now, if we use formula (2.2.12) to substitute for x1(n + 3), x2(n + 3),
and x3(n + 3) in the last row of formula (2.2.11), we obtain
⎛⎞
x1(n+1) x2(n+1) x3(n+1)
⎜ x1(n+2) W (n + 1) = det ⎜ −p3x1(n)
+p1x1(n + 2) +p1x2(n + 2) +p1x3(n + 2) (2.2.13)
Using the properties of determinants, it follows from (2.2.13) that
⎛⎞
x1(n+1) x2(n+1) x3(n+1)
W(n+1) = det⎜⎝ x1(n+2) x2(n+2) x3(n+2) ⎟⎠ (2.2.14)
−p3 (n)x1 (n) −p3 (n)x2 (n) −p3 (n)x3 (n) ⎛⎞
x1(n+1) x2(n+1) x3(n+1) = −p3(n) det ⎜⎝x1(n + 2) x2(n + 2) x3(n2) ⎟⎠
x1 (n) x2 (n) x3 (n) ⎛⎞
x1 (n) x2 (n) x3 (n)
= −p3(n)(−1)2 det ⎜⎝x1(n + 2) x2(n + 2) x3(n + 2)⎟⎠ .
x2(n+2)
x3(n+2) ⎟
−p3x2(n)
−p3x3(n) ⎟ .
⎜⎝−p2x1(n + 1)
−p2x2(n + 1)
−p2x3(n + 1)⎟⎠
x1(n+1) x2(n+1) x3(n+1) W (n + 1) = (−1)3 p3 (n)W (n).
Thus
Using formula (1.2.3), the solution of (2.2.15) is given by
(2.2.15)
n−1
W (n) =
i=n0
3 (−1) p3(i)
n−1 3(n−n0)
i=n0
W (n0) = (−1)
p3(i)W (n0).
This completes the proof of the lemma for k = 3. The general case is left to the reader as Exercises 2.2, Problem 6.
We now examine and treat one of the special cases that arises as we try to apply this Casoratian. For example, if (2.2.7) has constant coefficients p1,p2,…,pk, then we have
(2.2.16)
Formula (2.2.10) has the following important correspondence. Corollary 2.14. Suppose that pk(n) ̸= 0 for all n ≥ n0. Then the
Casoratian W(n) ̸= 0 for all n ≥ n0 if and only if W(n0) ̸= 0.
Proof. This corollary follows immediately from formula (2.2.10) (Exer- cises 2.2, Problem 7). P
P
W(n) = (−1)k(n−n0)p(n−n0)W(n0). k
70 2. Linear Difference Equations of Higher Order
Let us have a close look at Corollary 2.14 and examine what it really says. The main point in the corollary is that either the Casoratian is identically zero (i.e., zero for all n ≥ n0, for some n0) or never zero for any n ≥ n0. Thus to check whether W(n) ̸= 0 for all n ∈ Z+, we need only to check whether W(0) ̸= 0. Note that we can always choose the most suitable n0 and compute W(n0) there.
Next we examine the relationship between the linear independence of solutions and their Casoratian. Basically, we will show that a set of k so- lutions is a fundamental set (i.e., linearly independent) if their Casoratian W(n) is never zero.
To determine the preceding statement we contemplate k solutions x1(n),x2(n),…,xk(n) of (2.2.7). Suppose that for some constants a1, a2,…,ak and n0 ∈ Z+,
a1 x1 (n) + a2 x2 (n) + · · · + ak (n) xk (n) = 0, for all n ≥ n0 . Then we can generate the following k − 1 equations:
a1 x1 (n + 1) + a2 x2 (n + 1) + · · · + ak xk (n + 1) = 0, .
a1 x1 (n + k − 1) + a2 x2 (n + k − 1) + · · · + ak xk (n + k − 1) = 0. This assemblage may be transcribed as
where
X(n)ξ = 0, (2.2.17)
⎛ x1(n) x2(n) … xk(n) ⎞
⎜ x1(n+1) x2(n+1) … xk(n+1) ⎟ X(n) = ⎜ . . . ⎟,
⎝…⎠ x1(n+k−1) x2(n+k−1) … xk(n+k−1)
⎛a1⎞ ⎜a2 ⎟
ξ = ⎜ . ⎟ . ⎝.⎠
ak
Observe that W (n) = det X (n).
Linear algebra tells us that the vector (2.2.17) has only the trivial (or
zero) solution (i.e., a1 = a2 = ··· = ak = 0) if and only if the matrix X(n) is nonsingular (invertible) (i.e., det X(n) = W(n) ̸= 0 for all n ≥ n0). This deduction leads us to the following conclusion.
2.2 General Theory of Linear Difference Equations 71
Theorem 2.15. The set of solutions x1 (n), x2 (n), . . . , xk (n) of (2.2.7) is a fundamental set if and only if for some n0 ∈ Z+, the Casoratian W(n0) ̸= 0.
Proof. Exercises 2.2, Problem 8. P Example 2.16. Verify that {n,2n} is a fundamental set of solutions of
the equation
x(n+2)−3n−2x(n+1)+ 2n x(n)=0. n−1 n−1
Solution We leave it to the reader to verify that n and 2n are solutions of the equation. Now, the Casoratian of the solutions n, 2n is given by
Thus
n 2n W(n)=det n+1 2n+1 .
01
W(0)=det 1 2 =−1̸=0.
Hence by Theorem 2.15, the solutions n, 2n are linearly independent and thus form a fundamental set.
Example 2.17. Consider the third-order difference equation x(n + 3) + 3x(n + 2) − 4x(n + 1) − 12x(n) = 0.
Show that the functions 2n, (−2)n, and (−3)n form a fundamental set of solutions of the equation.
Solution
(i) Let us verify that 2n is a legitimate solution by substituting x(n) = 2n into the equation:
2n+3 + (3)(2n+1) − (4)(2n+1) − (12)(2n) = 2n[8 + 12 − 8 − 12] = 0. We leave it to the reader to verify that (−2)n and (−3)n are solutions
of the equation.
(ii) To affirm the linear independence of these solutions we construct the Casoratian
⎛ 2n
W (n) = det ⎜⎝2n+1 2n+2
(−2)n (−2)n+1 (−2)n+2
(−3)n ⎞ (−3)n+1 ⎟⎠ . (−3)n+2
72 2. Linear Difference Equations of Higher Order
Thus
⎛⎞
111
W(0) = det⎜⎝2 −2 3⎟⎠ = −20 ̸= 0. 449
By Theorem 2.15, the solutions 2n, (−2)n, and 3n are linearly independent, and thus form a fundamental set.
We are now ready to discuss the fundamental theorem of homogeneous linear difference equations.
Theorem 2.18 (The Fundamental Theorem). If pk(n) ̸= 0 for all n ≥ n0, then (2.2.7) has a fundamental set of solutions for n ≥ n0.
Proof. By Theorem 2.7, there are solutions x1 (n), x2 (n), . . . , xk (n) such that xi(n0 +i−1) = 1, xi(n0) = xi(n0 +1) = ··· = xi(n0 +i−2) = xi(n0 + i) = · · · = xi(n0 + k − 1) = 0, 1 ≤ i ≤ k. Hence x1(n0) = 1, x2(n0 + 1) = 1,×3(n0+2)=1,…,xk(n0+k−1)=1.ItfollowsthatW(n0)=detI =1. This implies by Theorem 2.15 that the set {x1(n),x2(n),…,xk(n)} is a fundamental set of solutions of (2.2.7). P
We remark that there are infinitely many fundamental sets of solutions of (2.2.7). The next result presents a method of generating fundamental sets starting from a known set.
Lemma 2.19. Let x1(n) and x2(n) be two solutions of (2.2.7). Then the following statements hold:
(i) x(n) = x1(n) + x2(n) is a solution of (2.2.7).
(ii) x ̃(n) = ax1(n) is a solution of (2.2.7) for any constant a.
Proof. (Exercises 2.2, Problem 9.) P From the preceding lemma we conclude the following principle.
Superposition Principle. If x1(n), x2(n), . . . , xr(n) are solutions of (2.2.7), then
x(n) = a1 x1 (n) + a2 x2 (n) + · · · + ar xr (n)
is also a solution of (2.2.7) (Exercises 2.2, Problem 12).
Now let {x1(n),x2(n),…,xk(n)} be a fundamental set of solutions of
(2.2.7) and let x(n) be any given solution of (2.2.7). Then there are con- stants a1, a2, . . . , ak such that x(n) = ki=1 aixi(n). To show this we use the notation (2.2.17) to write X(n)ξ = xˆ(n), where
⎛ x(n) ⎞
⎜ x(n + 1) ⎟ xˆ ( n ) = ⎜ . . ⎟ .
⎝.⎠ x(n + k − 1)
2.2 General Theory of Linear Difference Equations 73
Since X(n) is invertible (Why?), it follows that ξ = X−1(n)xˆ(n),
and, for n = n0,
The above discussion leads us to define the general solution of (2.2.7).
Definition 2.20. Let {x1(n),x2(n),…,xk(n)} be a fundamental set of solutions of (2.2.7). Then the general solution of (2.2.7) is given by x(n) = ki=1 aixi(n), for arbitrary constants ai.
It is worth noting that any solution of (2.2.7) may be obtained from the general solution by a suitable choice of the constants ai.
The preceding results may be restated using the elegant language of linear algebra as follows: Let S be the set of all solutions of (2.2.7) with the operations +, · defined as follows:
(i) (x+y)(n)=x(n)+y(n), forx,y∈S, n∈Z+, (ii) (ax)(n) = ax(n), for x ∈ S, a constant.
Equipped with linear algebra we now summarize the results of this section in a compact form.
Theorem 2.21. The space (S, +, ·) is a linear (vector) space of dimension k.
Proof. Use Lemma 2.19. To construct a basis of S we can use the fundamental set in Theorem 2.18 (Exercises 2.2, Problem 11). P
Exercises 2.2
1. Find the Casoratian of the following functions and determine whether they are linearly dependent or independent:
(a) 5n,3·5n+2,en.
(b) 5n,n5n,n25n.
(c) (−2)n , 2n , 3.
(d) 0,3n,7n.
2. Find the Casoratian W (n) of the solutions of the difference equations: (a) x(n+3)−10x(n+2)+31x(n+1)−30x(n)=0,ifW(0)=6. (b) x(n+3)−3x(n+2)+4x(n+1)−12x(n)=0,ifW(0)=26.
3. For the following difference equations and their accompanied solutions: (i) determine whether these solutions are linearly independent, and
ξ = X−1(n0)xˆ(n0).
74 2. Linear Difference Equations of Higher Order
(ii) find, if possible, using only the given solutions, the general solution:
(a)x(n+3)−3x(n+2)+3x(n+1)−x(n)=0; 1,n,n2, nπ nπ
(b)x(n+2)+x(n)=0; cos 2 ,sin 2 ,
(c) x(n+3)+x(n+2)−8x(n+1)−12x(n) = 0; 3n, (−2)n, (−2)n+3,
(d) x(n + 4) − 16x(n) = 0; 2n, n2n, n22n.
4. Verify formula (2.2.10) for the general case.
5. Show that the Casoratian W(n) in formula (2.2.9) may be given by the formula
⎛ x1(n) x2(n) … xk(n) ⎞
⎜ ∆x1(n) ∆x2(n) . . . ∆xk(n) ⎟ W (n) = det ⎜ . . . ⎟ .
⎝…⎠ ∆k−1×1(n) ∆k−1×2(n) . . . ∆k−1xk(n)
6. Verify formula (2.2.16).
7. Prove Corollary 2.14.
8. Prove Theorem 2.15.
9. Prove Lemma 2.19.
10. Prove the superposition principle: If x1(n), x2(n), . . . , xr are solutions of (2.2.7), then any linear combination of them is also a solution of (2.2.7).
11. Prove Theorem 2.21.
12. Suppose that for some integer m ≥ n0,pk(m) = 0 in (2.2.1).
(a) What is the value of the Casoratian for n ≥ m?
(b) Does Corollary 2.14 still hold? (Why?)
*13. Show that the equation ∆2y(n) = p(n)y(n + 1) has a fundamental set
of solutions whose Casoratian W (n) = −1.
14. Contemplate the second-order difference equation u(n+2)+p1(n)u(n+ 1) + p2(n)u(n) = 0. If u1(n) and u2(n) are solutions of the equation and W (n) is their Casoratian, prove that
n−1
W (r)/u1(r)u1(r + 1) . (2.2.18)
15. Contemplatethesecond-orderdifferenceequationu(n+2)−(n+3)u(n+ (n+2)
1)+ 2 u(n)=0. (n+2)
u2(n) = u1(n)
r=0
2.3 Linear Homogeneous Equations with Constant Coefficients 75
(a) Verify that u1(n) = 2n is a solution of the equation. n!
(b) Use formula (2.2.18) to find another solution u2(n) of the equation.
16. Show that u(n) = (n+1) is a solution of the equation u(n+2)−u(n+ 1) − 1/(n + 1)u(n) = 0 and then find a second solution of the equation by using the method of Exercises 2.2, Problem 15.
2.3 Linear Homogeneous Equations with Constant Coefficients
Consider the kth-order difference equation x(n+k)+p1x(n+k−1)+p2x(n+k−2)+···+pkx(n)=0, (2.3.1)
where the pi’s are constants and pk ̸= 0. Our objective now is to find a fundamental set of solutions and, consequently, the general solution of (2.3.1). The procedure is rather simple. We suppose that solutions of (2.3.1) are in the form λn, where λ is a complex number. Substituting this value into (2.3.1), we obtain
λk +p1λk−1 +···+pk =0. (2.3.2)
This is called the characteristic equation of (2.3.1), and its roots λ are called the characteristic roots. Notice that since pk ̸= 0, none of the characteristic roots is equal to zero. (Why?) (Exercises 2.3, Problem 19.)
We have two situations to contemplate:
Case (a). Suppose that the characteristic roots λ1, λ2, . . . , λk are distinct. We are now going to show that the set {λn1,λn2,…,λnk} is a fundamental set of solutions. To prove this, by virtue of Theorem 2.15 it suffices to show that W (0) ̸= 0, where W (n) is the Casoratian of the solutions. That is,
⎛11…1⎞
⎜ λ1 λ2 W(0)=det⎜ λ21 λ2
… …
λk ⎟ λ2k ⎟.
(2.3.3)
(2.3.4)
. ⎟ ⎝…⎠
⎜ . .
λk−1 λk−1 . . . λk−1
12k
This determinant is called the Vandermonde determinant. It may be shown by mathematical induction that
W(0)=
The reader will prove this conclusion in Exercises 2.3, Problem 20.
1≤i
(a) Show that F (p) = 5(p−1)/2 mod p. (b) Show that F(p) = ±1 mod p.
19. Show that if pk ̸= 0 in (2.3.1), then none of its characteristic roots is equal to zero.
20. Show that the Vandermonde determinant (2.3.3) is equal to
(λj − λi). 1≤i
2.4 Nonhomogeneous Equations: Methods of Undetermind Coefficeints 89
15. Solve the difference equation
y(n + 2) + y(n) = with y(0) = 0 and y(1) = 1.
1 if 0 ≤ n ≤ 2, −1 ifn>2,
2.4.1 The Method of Variation of Constants (Parameters)
Contemplate the second-order nonhomogeneous difference equation
y(n + 2) + p1(n)y(n + 1) + p2(n)y(n) = g(n) (2.4.17)
and the corresponding homogeneous difference equation
y(n + 2) + p1(n)y(n + 1) + p2(n)y(n) = 0. (2.4.18)
The method of variation of constants is commonly used to find a particular solution yp(n) of (2.4.17) when the coefficients p1(n) and p2(n) are not constants. The method assumes that a particular solution of (2.4.17) may be written in the form
y(n) = u1(n)y1(n) + u2(n)y2(n), (2.4.19)
where y1(n) and y2(n) are two linearly independent solutions of the homo- geneous equation (2.4.18), and u1(n), u2(n) are sequences to be determined later.
16. (a)
(b)
(c)
(d)
Show that
y(n + 1) = u1(n)y1(n + 1) + u2(n)y2(n + 1)
+ ∆u1(n)y1(n + 1) + ∆u2(n)y2(n + 1). The method stipulates that
∆u1(n)y1(n + 1) + ∆u2(n)y2(n + 1) = 0. Use (2.4.20) and (2.4.21) to show that
(2.4.20)
(2.4.21)
y(n + 2) = u1(n)y1(n + 2) + u2(n)y2(n + 2)
+ ∆u1(n)y1(n + 2) + ∆u2(n)y2(n + 2).
By substituting the above expressions for y(n), y(n+1), and y(n+ 2) into (2.4.17), show that
∆u1(n)y1(n + 2) + ∆u2(n)y2(n + 2) = g(n). (2.4.22) Using expressions (2.4.21) and (2.4.22), show that
n−1
−g(n)y2(n + 1) −g(r)y2(r + 1)
∆u1(n) = W(n + 1) , u1(n) =
r=0
W(r + 1) , (2.4.23)
90 2. Linear Difference Equations of Higher Order
g(n)y1(n + 1)
∆u2(n) = W(n + 1) , u2(n) =
r=0 where W(n) is the Casoratian of y1(n) and y2(n).
17. Use formulas (2.4.23) and (2.4.24) to solve the equation
y(n + 2) − 7y(n + 1) + 6y(n) = n.
18. Use the variation of constants method to solve the initial value problem
y(n + 2) − 5y(n + 1) + 6y(n) = 2n, y(1) = y(2) = 0.
19. Use Problem 16(d) to show that the unique solution of (2.4.17) with
n−1
g(r)y1(r + 1)
W(r + 1) , (2.4.24)
y(0) = y(1) = 0 is given by n−1
y(n) = 20. Consider the equation
W (r + 1) . x(n + 1) = ax(n) + f (n).
y1(r + 1)y2(n) − y2(r + 1)y1(n)
r=0
(a) Show that
x(n)=an x(0)+ a + a2 +···+ an
(2.4.25)
(2.4.26)
f(0) f(1) f(n−1) is a solution of (2.4.25).
(b) Show that if |a| < 1 and {f(n)} is a bounded sequence, i.e., |f(n)| ≤ M, for some M > 0, n ∈ Z+, then all solutions of (2.4.25) are bounded.
(c) Suppose that a > 1 and {f(n)} is bounded on Z+. Show that if we choose
f (0) f (1) f (n) ∞ f (i) x(0)=− a + a2 +···+an+1 +··· =− ai+1,
i=0
(2.4.27)
then the solution x(n) given by (2.4.26) is bounded on Z+. Give an explicit expression for x(n) in this case.
(d) Under the assumptions of part (c), show that for any choice of x(0), excepting that value given by (2.4.27), the solution of (2.4.25) is unbounded.
2.5 Limiting Behavior of Solutions 91
2.5 Limiting Behavior of Solutions
To simplify our exposition we restrict our discussion to the second-order difference equation
y(n + 2) + p1y(n + 1) + p2y(n) = 0. (2.5.1) Suppose that λ1 and λ2 are the characteristic roots of the equation. Then
we have the following three cases:
(a) λ1 and λ2 are distinct real roots. Then y1(n) = λn1 and y2(n) = λn2 are two linearly independent solutions of (2.5.1). If |λ1| > |λ2|, then we call y1(n) the dominant solution, and λ1 the dominant characteristic root. Otherwise, y2(n) is the dominant solution, and λ2 is the dominant characteristic root. We will now show that the limiting behavior of the general solution y(n) = a1λn1 + a2λn2 is determined by the behavior of the dominant solution. So assume, without loss of generality, that |λ1| > |λ2|. Then
Since
it follows that
λ n y(n)=λna+a2 .
1 1 2 λ1 λ
λ n 2
λ1
2<1, λ1
→0 asn→∞.
Consequently, limn→∞ y(n) = limn→∞ a1λn1 . There are six different situations that may arise here depending on the value of λ1 (see Figure 2.3).
1. λ1 > 1: The sequence {a1λn1 } diverges to ∞ (unstable system).
2. λ1 = 1: The sequence {a1λn1 } is a constant sequence.
3. 0 < λ1 < 1: The sequence {a1λn1 } is monotonically decreasing to zero (stable system).
4. −1 < λ1 < 0: The sequence {a1λn1 } is oscillating around zero (i.e., alternating in sign) and converging to zero (stable system).
5. λ1 = −1: The sequence {a1λn1 } is oscillating between two values a1 and −a1.
6. λ1 < −1: The sequence {a1λn1} is oscillating but increasing in magnitude (unstable system).
92 2. Linear Difference Equations of Higher Order
FIGURE 2.3. (n, y(n)) diagrams for real roots.
y1(n)
y1(n)
y1(n)
* ****** (6) -1 1
Re z y1(n)
y1(n) n
(2)
(5)
Im z
(4) (3) nn
(1)
n
y1(n)
n
y(n)
y(n)
2.5 Limiting Behavior of Solutions 93
(2)
(1)
n Imz
n
*1 *3
*2
*1
-1
y(n) Re z
*3
FIGURE 2.4. (n, y(n)) diagrams for comples roots.
(b) λ1 =λ2 =λ.
The general solution of (2.5.1) is given by y(n) = (a1 +a2n)λn. Clearly, if |λ| ≥ 1, the solution y(n) diverges either monotonically if λ ≥ 1 or by oscillating if λ ≤ −1. However, if |λ| < 1, then the solution converges to zero, since limn→∞ nλn = 0 (Why?).
(c) Complexroots:λ1 =α+iβandλ2 =α−iβ,whereβ̸=0.
As we have seen in Section 2.3, formula (2.3.12), the solution of (2.5.1) is given by y(n) = arn cos(nθ − ω), where
.
The solution y(n) clearly oscillates, since the cosine function oscillates. However, y(n) oscillates in three different ways depending on the lo- cation of the conjugate characteristic roots, as may be seen in Figure 2.4.
1. r > 1: Here λ1 and λ2 = λ1 are outside the unit circle. Hence y(n) is oscillating but increasing in magnitude (unstable system).
2. r=1:Hereλ1 andλ2 =λ1 lieontheunitcircle.Inthiscasey(n) is oscillating but constant in magnitude.
3. r < 1: Here λ1 and λ2 = λ1 lie inside the unit disk. The solution y(n) oscillates but converges to zero as n → ∞ (stable system).
*2
β
(3)
r = α2 + β2, θ = tan−1
α
n
Finally, we summarize the above discussion in the following theorem.
94 2. Linear Difference Equations of Higher Order
Theorem 2.35. The following statements hold:
(i) All solutions of (2.5.1) oscillate (about zero) if and only if the
characteristic equation has no positive real roots.
(ii) All solutions of (2.5.1) converge to zero (i.e., the zero solution is
asymptotically stable) if and only if max{|λ1|, |λ2|} < 1.
Next we consider nonhomogeneous difference equations in which the
input is constant, that is, equations of the form
y(n + 2) + p1y(n + 1) + p2y(n) = M, (2.5.2)
where M is a nonzero constant input or forcing term. Unlike (2.5.1), the zero sequence y(n) = 0 for all n ∈ Z+ is not a solution of (2.5.2). Instead, we have the equilibrium point or solution y(n) = y*. From (2.5.2) we have
or
y* + p1y* + p2y* = M,
y* = M . (2.5.3)
1+p1 +p2
Thus yp(n) = y* is a particular solution of (2.5.2). Consequently, the
general solution of (2.5.2) is given by
y(n) = y* + yc(n). (2.5.4)
It is clear that y(n) → y* if and only if yc(n) → 0 as n → ∞. Furthermore, y(n) oscillates6 about y* if and only if yc(n) oscillates about zero. These observations are summarized in the following theorem.
Theorem 2.36. The following statements hold:
(i) All solutions of the nonhomogeneous equation (2.5.2) oscillate about the equilibrium solution y* if and only if none of the characteristic roots of the homogeneous equation (2.5.1) is a positive real number.
(ii) All solutions of (2.5.2) converge to y* as n → ∞ if and only if max{|λ1 |, |λ2 |} < 1, where λ1 and λ2 are the characteristic roots of the homogeneous equation (2.5.1).
Theorems 2.35 and 2.36 give necessary and sufficient conditions under which a second-order difference equation is asymptotically stable. In many applications, however, one needs to have explicit criteria for stability based on the values of the coefficients p1 and p2 of (2.5.2) or (2.5.1). The following result provides us with such needed criteria.
6We say y(n) oscillates about y* if y(n)−y* alternates sign, i.e., if y(n) > y*, then y(n + 1) < y*.
have
or
p21 −4p2 <2,
2.5 Limiting Behavior of Solutions 95
Theorem 2.37. The conditions
1+p1 +p2 >0, 1−p1 +p2 >0, 1−p2 >0 (2.5.5)
are necessary and sufficient for the equilibrium point (solution) of equations (2.5.1) and (2.5.2) to be asymptotically stable (i.e., all solutions converge to y*).
Proof. Assume that the equilibrium point of (2.5.1) or (2.5.2) is asymp- totically stable. In virtue of Theorems 2.35 and 2.36, the roots λ1,λ2 of the characteristic equation λ2 + p1λ + p2 = 0 lie inside the unit disk, i.e., |λ1| < 1 and |λ2| < 1. By the quadratic formula, we have
λ1 = −p1 +p21 −4p2 and λ2 = −p1 −p21 −4p2. (2.5.6) 22
Then we have two cases to consider.
Case 1. λ1, λ2 are real roots, i.e., p21 − 4p2 ≥ 0. From formula (2.5.6) we
−2<−p1 +
−2+p1 < p21 −4p2 <2+p1.
(2.5.7)
(2.5.8)
1+p1 +p2 >0.
Similarly, if we square the first inequality in expression (2.5.8) we obtain
1−p1 +p2 >0. (2.5.10) Now from the second inequality of (2.5.7) and the first inequality of (2.5.8)
we obtain
2+p1 >0 and 2−p1 >0 or |p1|<2
since p21 − 4p2 ≥ 0, p2 ≤ p21/4 < 1. This completes the proof of (2.5.5) in
this case.
Case 2. λ1 and λ2 are complex conjugates, i.e., p21 − 4p2 < 0. In this case
Similarly, one obtains
−2+p1 <− p21 −4p2 <2+p1. Squaring the second inequality in expression (2.5.7) yields
we have
−p1 i
λ1,2= 2 ±2 4p2−p21.
(2.5.9)
Moreover, since p21 < 4p2, it follows that −2√p2 < p1 < 2√p2. Now |λ1|2 =
p21 +4p2 −p21 =p2.Since|λ1|<1,itfollowsthat0
√
thatf(0)=1,andf′(x)=1− 1 .Thusx=1isalocalminimumasf(x)
x
decreases for x ∈ (0,1). Hence f(x) > 0 for all x ∈ (0,1).
This completes the proof of the necessary conditions. The converse is left to the reader as Exercises 2.5, Problem 8. P
Example 2.38. Find conditions under which the solutions of the equation y(n + 2) − α(1 + β)y(n + 1) + αβy(n) = 1, α, β > 0,
(a) converge to the equilibrium point y*, and
(b) oscillate about y*.
Solution Let us first find the equilibrium point y*. Be letting y(n) = y* in
the equation, we obtain
y* = 1 , α ̸= 1. 1−α
(a) Applying condition (2.5.5) to our equation yields
α < 1, 1 + α + 2αβ > 0, αβ < 1.
Clearly, the second inequality 1+α+2αβ > 0 is always satisfied, since α, β are both positive numbers.
(b) The solutions are oscillatory about y* if either λ1, λ2 are negative real numbers or complex conjugates. In the first case we have
α2(1+β)2>4αβ, or α> 4β , (1+β)2
and
α(1 + β) < 0,
which is impossible. Thus if α > 4β/(1 + β)2, we have no oscillatory
solutions.
Now, λ1 and λ2 are complex conjugates if
α2(1+β)2<4αβ or α< 4β . (1+β)2
Hence all solutions are oscillatory if
α< 4β .
(1+β)2
For the treatment of the general kth-order scalar difference equations, the reader is referred to Chapter 4, on stability, and Chapter 8, on oscillation.
2.5 Limiting Behavior of Solutions 97
Exercises 2.5.
In Problems 1 through 4:
(a) Determine the stability of the equilibrium point by using Theorem 2.35 or Theorem 2.36.
(b) Determine the oscillatory behavior of the solutions of the equation.
1. y(n+2)−2y(n+1)+2y(n)=0.
2. y(n + 2) + 1 y(n) = 5 . 44
3. y(n+2)+y(n+1)+1y(n)=−5. 2
4. y(n+2)−5y(n+1)+6y(n)=0.
5. Determine the stability of the equilibrium point of the equations
in Problems 1 through 4 by using Theorem 2.37.
6. Show that the stability conditions (2.5.5) for the equation y(n + 2) − αy(n + 1) + βy(n) = 0, where α,β are constants, may be written as
−1 − β < α < 1 + β, β < 1.
7. Contemplate the equation y(n+2)−p1y(n+1)−p2y(n) = 0. Show that if |p1| + |p2| < 1, then all solutions of the equation converge to zero.
8. Prove that conditions (2.5.5) imply that all solutions of (2.5.2) converge to the equilibrium point y*.
9. Determine conditions under which all solutions of the difference equation in Problem 7 oscillate.
10. Determine conditions under which all solutions of the difference equation in Problem 6 oscillate.
11. Suppose that p is a real number. Prove that every solution of the
difference equation y(n + 2) − y(n + 1) + py(n) = 0 oscillates if
and only if p > 1 . 4
*12. Prove that a necessary and sufficient condition for the asymptotic stability of the zero solution of the equation
is
y(n + 2) + p1y(n + 1) + p2y(n) = 0 |p1|<1+p2 <2.
98
2. Linear Difference Equations of Higher Order
13. Determine the limiting behavior of solutions of the equation y(n + 2) = αc + αβ(y(n + 1) − y(n))
if:
(i) αβ = 1,
(ii) αβ = 2,
(iii) αβ = 1, 2
provided that α, β, and c are positive constants.
14. If p1 > 0 and p2 > 0, show that all solutions of the equation
y(n + 2) + p1y(n + 1) + p2y(n) = 0
15. Determine the limiting behavior of solutions of the equation
y(n+2)− βy(n+1)+ βy(n)=0, αα
where α and β are constants, if: (i) β > 4α,
(ii) β < 4α.
Nonlinear Equations Transformable to Linear Equations
are oscillatory.
2.6
In general, most nonlinear difference equations cannot be solved explic- itly. However, a few types of nonlinear equations can be solved, usually by transforming them into linear equations. In this section we discuss some tricks of the trade.
Type I. Equations of Riccati type:
x(n + 1)x(n) + p(n)x(n + 1) + q(n)x(n) = 0.
To solve the Riccati equation, we let z(n)= 1
x(n)
in (2.6.1) to give us
q(n)z(n + 1) + p(n)z(n) + 1 = 0.
The nonhomogeneous equation requires a different transformation
y(n + 1)y(n) + p(n)y(n + 1) + q(n)y(n) = g(n).
(2.6.1)
(2.6.2) (2.6.3)
2.6 Nonlinear Equations Transformable to Linear Equations 99
If we let y(n) = (z(n + 1)/z(n)) − p(n) in (2.6.3) we obtain
z(n + 2) + (q(n) − p(n + 1))z(n + 1) − (g(n) + p(n)q(n))z(n) = 0.
Example 2.39. The Pielou Logistic Equation
The most popular continuous model of the growth of a population is the well-known Verhulst–Pearl equation given by
x′ (t) = x(t)[a − bx(t)], a, b > 0, (2.6.4)
where x(t) is the size of the population at time t; a is the rate of the growth of the population if the resources were unlimited and the individuals did not affect one another, and −bx2(t) represents the negative effect on the growth due to crowdedness and limited resources. The solution of (2.6.4) is given by
Now,
x(t) = a/b . 1 + (e−at/cb)
x(t+1)=
= Dividing by [1 + (e−at/cb)], we obtain
a/b e−a(t+1)/cb
1 +
1+(e−at/cb)+(ea −1)
ea(a/b) . x(t+1)= eax(t) ,
1+ b(ea −1)x(t) a
or
x(n + 1) = whereα=ea andβ=b(ea−1).
a
αx(n) , [1 + βx(n)]
(2.6.5)
This equation is titled the Pielou logistic equation.
Equation (2.6.5) is of Riccati type and may be solved by letting x(n) =
1/z(n). This gives us the equation
z(n + 1) = 1 z(n) + β ,
αα ⎧⎪ β
whose solution is given by
⎨ c−α−1 α−n+(β/(α−1)) ifα̸=1,
z ( n ) = ⎪⎩ c + β n i f α = 1 .
100 2. Linear Difference Equations of Higher Order
x(n+1)
0.5
FIGURE 2.5. Asymptotically stable equilibrium points.
Thus
x(n)=⎩ 1
x(n)
Hence
⎧⎨αn(α−1)/[βαn+c(α−1)−β] ifα̸=1,
c+βn
if α = 1.
lim x(n) = n→∞
(α−1)/β ifα̸=1, 0 if α = 1.
This conclusion shows that the equilibrium point (α − 1)/β is globally asymptotically stable if α ̸= 1. Figure 2.5 illustrates this for α = 3, β = 1, and x(0) = 0.5.
Type II. Equations of general Riccati type:
x(n + 1) = a(n)x(n) + b(n)
c(n)x(n) + d(n)
such that c(n) ̸= 0, a(n)d(n) − b(n)c(n) ̸= 0 for all n ≥ 0.
To solve this equation we let
c(n)x(n) + d(n) = y(n + 1) .
(2.6.6)
(2.6.7)
Then by substituting
y(n)
x(n) = y(n + 1) − d(n) c(n)y(n) c(n)
2.6 Nonlinear Equations Transformable to Linear Equations 101
into (2.6.6) we obtain
y(n + 2) d(n + 1)
y(n+1) d(n) a(n) c(n)y(n) − c(n)
y(n+1) y(n)
y(n + 2) + p1(n)y(n + 1) + p2(n)y(n) = 0, y(0) = 1, y(1) = c(0)x(0) + d(0),
p1(n) = −c(n)d(n + 1) + a(n)c(n + 1), c(n)
p2 (n) = (a(n)d(n) − b(n)c(n)) c(n + 1) . c(n)
+ b(n) .
(2.6.8)
c(n+1)y(n+1) − c(n+1) = This equation simplifies to
where
Example 2.40. Solve the difference equation x(n+1)= 2x(n)+3.
3x(n) + 2
Solution Herea=2,b=3,c=3,andd=2.Hencead−bc̸=0.Usingthe
transformation
3x(n) + 2 = y(n + 1), y(n)
(2.6.9)
y(1) = 3x(0) + 2,
we obtain, as in (2.6.8),
y(n + 2) − 4y(n + 1) − 5y(n) = 0, y(0) = 1,
with characteristic roots λ1 = 5, λ2 = −1. Hence
y(n) = c15n + c2(−1)n. From formula (2.6.9) we have
1 y(n+1) 2 1 c15n+1 +c2(−1)n+1 x(n) = 3 y(n) − 3 = 3 c15n + c2(−1)n
(c15n − c2(−1)n) 5n − c(−1)n = (c15n + c2(−1)n) = 5n + c(−1)n ,
where
c = c1 .
c2
Type III. Homogeneous difference equations of the type
x(n+1)
f x(n) ,n =0.
(2.6.10)
2 − 3
102 2. Linear Difference Equations of Higher Order
Use the transformation z(n) = x(n+1) to convert such an equation to a x(n)
linear equation in z(n), thus allowing it to be solved. Example 2.41. Solve the difference equation
x2(n + 1) − 3x(n + 1)x(n) + 2×2(n) = 0. Solution Dividing by x2(n), equation (2.6.11) becomes
x(n + 1)2 x(n + 1)
x(n) − 3 x(n) + 2 = 0,
which is of Type III.
Letting z(n) = x(n+1) in (2.6.12) creates
x(n)
z2(n)−3z(n)+2 = 0. We can factor this down to
[z(n) − 2][z(n) − 1] = 0, and thus either z(n) = 2 or z(n) = 1.
This leads to
x(n+1)=2x(n) or x(n+1)=x(n). Starting with x(0) = x0, there are infinitely many solutions x(n) of
of the form
x0,…,x0;2×0,…,2×0;22×0,…,22×0;….7
Type IV. Consider the difference equation of the form (y(n + k))r1 (y(n + k − 1))r2 · · · (y(n))rk+1 = g(n).
Let z(n) = ln y(n), and rearrange to obtain
r1z(n + k) + r2z(n + k − 1) + · · · + rk+1z(n) = ln g(n).
Example 2.42. Solve the difference equation x(n+2)= x2(n+1).
(2.6.11)
(2.6.12)
(2.6.11)
(2.6.13) (2.6.14)
(2.6.15)
x2 (n)
Solution Let z(n) = ln x(n) in (2.6.15). Then as in (2.6.12) we obtain
z(n + 2) − 2z(n + 1) + 2z(n) = 0. The characteristic roots are λ1 = 1 + i, λ2 = 1 − i.
Thus,
n/2 nπ nπ z(n) = (2) c1 cos 4 + c2 sin 4 .
7This solution was given by Sebastian Pancratz of the Technical University of Munich.
Therefore,
n/2 nπ nπ (2) c1 cos 4 + c2 sin 4 .
2.6 Nonlinear Equations Transformable to Linear Equations 103
x(n) = exp
Exercises 2.6
1. Find the general solution of the difference equation y2(n + 1) − 2y(n + 1)y(n) − 3y2(n) = 0.
2. Solve the difference equation
y2(n + 1) − (2 + n)y(n + 1)y(n) + 2ny2(n) = 0.
3. Solvey(n+1)y(n)−y(n+1)+y(n)=0.
4. Solvey(n+1)y(n)−2y(n+1)+1y(n)= 5.
3 6 18 5. Solvey(n+1)=5− 6 .
y(n)
6. Solvex(n+1)=x(n)+a,1̸=a>0.
x(n)+1 7. Solve x(n + 1) = x2(n).
8. Solve the logistic difference equation
x(n + 1) = 2x(n)(1 − x(n)).
9. Solve the logistic equation
x(n + 1) = 4x(n)[1 − x(n)].
1a
10. Solvex(n+1)=2 x(n)−x(n) ,a>0.
11. Solve y(n + 2) = y3(n + 1)/y2(n). 12. Solvex(n+1)=2x(n)+4.
x(n)−1 13. Solvey(n+1)= 2−y2(n) .
2(1 − y(n)) 14.Solvex(n+1)= 2x(n).
x(n)+3
15. Solve y(n + 1) = 2y(n)1 − y2(n).
16. The “regular falsi” method for finding the roots of f(x) = 0 is given by
x(n + 1) = x(n − 1)f(x(n)) − x(n)f(x(n − 1)). f(x(n))−f(x(n−1))
104
2. Linear Difference Equations of Higher Order
(a) Show that for f(x) = x2, this difference equation becomes x(n+1) = x(n−1)x(n) .
x(n−1)+x(n)
(b) Let x(1) = 1, x(2) = 1 for the equation in part (a). Show that the solution of the equation is x(n) = 1/F (n), where F (n) is the nth Fibonacci number.
2.7 Applications
2.7.1 Propagation of Annual Plants
The material of this section comes from Edelstein–Keshet [37] of plant propagation. Our objective here is to develop a mathematical model that describes the number of plants in any desired generation. It is known that plants produce seeds at the end of their growth season (say August), after which they die. Furthermore, only a fraction of these seeds survive the winter, and those that survive germinate at the beginning of the season (say May), giving rise to a new generation of plants.
Let
If p(n) denotes the number of plants in generation n, then
γ = number of seeds produced per plant in August,
α = fraction of one-year-old seeds that germinate in May, β = fraction of two-year-old seeds that germinate in May, σ = fraction of seeds that survive a given winter.
plants from
p(n) = +
one-year-old seeds p(n) = αs1(n) + βs2(n),
plants from
,
where s1(n) (respectively, s2(n)) is the number of one-year-old (two-year- old) seeds in April (before germination). Observe that the number of seeds left after germination may be written as
fraction original number seeds left = not germinated × of seeds in April
This gives rise to two equations:
s ̃1(n) = (1 − α)s1(n),
s ̃2(n) = (1 − β)s2(n),
.
two-year-old seeds
(2.7.1)
(2.7.2) (2.7.3)
April-May
Year k=n Year k=n+1 Year k=n+2
August Winter April-May August Winter April-May August
2.7 Applications 105
γ
s0(n)
σ s1(n+1) α
s1(n+1) σ
β
α
γ
s0(n+1)
σ
s0(n+2)
s0(n+2)
s1(n+1)
p(n)
p(n+1)
p(n+2)
FIGURE 2.6. Propogation of annual plants.
where s ̃1(n) (respectively, s ̃2(n)) is the number of one-year (two-year-old) seeds left in May after some have germinated. New seeds s0(n) (0-year-old) are produced in August (Figure 2.6) at the rate of γ per plant,
s0(n) = γp(n). (2.7.4)
After winter, seeds s0(n) that were new in generation n will be one year old in the next generation n+1, and a fraction σs0(n) of them will survive. Hence
s1(n + 1) = σs0(n), or, by using formula (2.7.4), we have
s1(n + 1) = σγp(n).
(2.7.5)
(2.7.6)
Similarly,
which yields, by formula (2.7.2),
s2(n + 1) = σ(1 − α)s1(n),
s2(n + 1) = σ2γ(1 − α)p(n − 1).
s2(n + 1) = σs ̃1(n),
106 2. Linear Difference Equations of Higher Order
Substituting for s1 (n + 1), s2 (n + 1) in expressions (2.7.5) and (2.7.6) into formula (2.7.1) gives
or
p(n + 1) = αγσp(n) + βγσ2(1 − α)p(n − 1),
p(n + 2) = αγσp(n + 1) + βγσ2(1 − α)p(n). The characteristic equation (2.7.7) is given by
λ2 −αγσλ−βγσ2(1−α) = 0 with characteristic roots
λ1=αγσ 1+ 1+4β(1−α) , 2 γα2
λ2=αγσ 1− 1+4β(1−α) . 2 γα2
(2.7.7)
Observe that λ1 and λ2 are real roots, since 1 − α > 0. Furthermore, λ1 > 0 and λ2 < 0. To ensure propagation (i.e., p(n) increases indefinitely asn→∞)weneedtohaveλ1 >1.Wearenotgoingtodothesamewith λ2, since it is negative and leads to undesired fluctuation (oscillation) in the size of the plant population. Hence
αγσ 1+ 1+4β(1−α) >1,
2
αγσ 1+4β(1−α)>1−αγσ. 2 γα2 2
or
γα2
Squaring both sides and simplifying yields
γ> 1 .
(2.7.8)
ασ + βσ2(1 − α)
If β = 0, that is, if no two-year-old seeds germinate in May, then condition
(2.7.8) becomes
γ>1. (2.7.9) ασ
Condition (2.7.9) says that plant propagation occurs if the product of the fraction of seeds produced per plant in August, the fraction of one-year-old seeds that germinate in May, and the fraction of seeds that survive a given winter exceeds 1.
2.7.2 Gambler’s Ruin
A gambler plays a sequence of games against an adversary in which the probability that the gambler wins $1.00 in any given game is a known value q, and the probability of his losing $1.00 is 1 − q, where 0 ≤ q ≤ 1. He quits gambling if he either loses all his money or reaches his goal of acquiring N dollars. If the gambler runs out of money first, we say that the gambler has been ruined. Let p(n) denote the probability that the gambler will be ruined if he possesses n dollars. He may be ruined in two ways. First, winning the next game; the probability of this event is q; then his fortune will be n + 1, and the probability of being ruined will become p(n + 1). Second, losing the next game; the probability of this event is 1 − q, and the probability of being ruined is p(n − 1). Hence applying the theorem of total probabilities, we have
p(n) = qp(n + 1) + (1 − q)p(n − 1). Replacing n by n + 1, we get
p(n + 2) − 1 p(n + 1) + (1 − q) p(n) = 0, n = 0, 1, . . . , N, (2.7.10) qq
with p(0) = 1 and p(N) = 0. The characteristic equation is given by λ2 − 1 λ + 1 − q = 0,
qq
and the characteristic roots are given by
λ1 = 1 + 1 − 2q = 1 − q ,
2.7 Applications 107
2q 2q q λ2 = 1 − 1 − 2q = 1.
2q 2q Hence the general solution may be written as
1−qn 1 p(n) = c1 + c2 q , if q ̸= 2 .
Now using the initial conditions p(0) = 1, P (N ) = 0 we obtain 1−qN
which gives
c1 + c2 = 1, c1 + c2 q = 0, 1−qN
1−q 1−q
−q1
c1 = 1−qN , c2 = 1−qN .
108 2. Linear Difference Equations of Higher Order
Thus
The special case q = 1 must be treated separately, since in this case we 2
have repeated roots λ1 = λ2 = 1. This is certainly the case when we have a fair game. The general solution in this case may be given by
p(n) = a1 + a2n, which with the initial conditions yields
p(n) = 1 − n = N − n . (2.7.12) NN
For example, suppose you start with $4, the probability that you win a dollar is 0.3, and you will quit if you run out of money or have a total of $10. Then n = 4, q = 0.3, and N = 10, and the probability of being ruined is given by
1−qn 1−qN
p(n) = q − q . (2.7.11)
1−qN 1−q
74 710
p(4) = 3 − 3 = 0.994.
710 1−3
On the other hand, if q = 0.5, N = $100.00, and n = 20, then from formula (2.7.12) we have
p(20)=1− 20 =0.8. 100
Observe that if q ≤ 0.5 and N → ∞, p(n) tends to 1 in both formulas (2.7.11) and (2.7.12), and the gambler’s ruin is certain.
The probability that the gambler wins is given by
⎧⎪ 1 − q n
⎪1− q
⎪⎨ 1−qN, ifq̸=0.5,
p ̃(n)=1−p(n)=⎪ 1−
⎪ q
(2.7.13)
⎪⎩ n , N
i f q = 0 . 5 .
2.7.3
National Income
In a capitalist country the national income Y (n) in a given period n may be written as
Y (n) = C(n) + I(n) + G(n), (2.7.14)
where
C(n) = consumer expenditure for purchase of consumer goods,
I(n) = induced private investment for buying capital equipment, and
G(n) = government expenditure,
where n is usually measured in years.
We now make some assumptions that are widely accepted by economists
(see, for example, Samuelson [129]).
(a) Consumer expenditure C(n) is proportional to the national income
Y(n−1) in the preceding year n−1, that is,
C(n) = αY (n − 1), (2.7.15)
where α > 0 is commonly called the marginal propensity to consume.
(b) Induced private investment I(n) is proportional to the increase in
consumption C(n) − C(n − 1), that is,
I(n) = β[C(n) − C(n − 1)], (2.7.16)
where β > 0 is called the relation.
(c) Finally, the government expenditure G(n) is constant over the years,
and we may choose our units such that
G(n) = 1. (2.7.17)
Employing formulas (2.7.15), (2.7.16), and (2.7.17) in formula (2.7.14) produces the second-order difference equation
Y(n+2)−α(1+β)Y(n+1)+αβY(n)=1, n∈Z+. (2.7.18)
Observe that this is the same equation we have already studied, in detail, in Example 2.38. As we have seen there, the equilibrium state of the national income Y * = 1/(1 − α) is asymptotically stable (or just stable in the theory of economics) if and only if the following conditions hold:
α < 1, 1 + α + 2αβ > 0, αβ < 1. (2.7.19) Furthermore, the national income Y (n) fluctuates (oscillates) around
the equilibrium state Y * if and only if
α < 4β . (2.7.20)
2.7 Applications 109
(1+β)2 Nowconsideraconcreteexamplewhereα= 1,β=1.ThenY*=2,i.e.,
2
Y * = twice the government expenditure. Then clearly, conditions (2.7.19) and (2.7.12) are satisfied. Hence the national income Y (n) always converges in an oscillatory fashion to Y * = 2, regardless of what the initial national income Y (0) and Y (1) are. (See Figure 2.7.)
110 2. Linear Difference Equations of Higher Order
Y(n)
3 2 1
n 1 2 3 4 5 6 7 8 9 10
FIGURE 2.7. Solution of Y(n+2)−Y(n+1)+Y(n) = 1,Y(0) = 1,Y(1) = 2. The actual solution may be given by
1n nπ Y(n)=A √2 cos 4 −ω +2.
Figure 2.7 depicts the solution Y (n) if Y (0) = 1 and Y (1) = 2. Here we √
find that A = − 2 and ω = π/4 and, consequently, the solution is 1 n−1 (n+1)
Y(n)=−√2 cos 4π+2.
Finally, Figure 2.8 depicts the parameter diagram (β − α), which shows
regions of stability and regions of instability.
2.7.4 The Transmission of Information
Suppose that a signaling system has two signals s1 and s2 such as dots and dashes in telegraphy. Messages are transmitted by first encoding them into a string, or sequence, of these two signals. Suppose that s1 requires exactly n1 units of time, and s2 exactly n2 units of time, to be transmitted. Let M(n) be the number of possible message sequences of duration n. Now, a signal of duration time n either ends with an s1 signal or with an s2 signal.
α
1 Real roots stable
Real roots unstable
Imaginary roots stable
1
FIGURE 2.8. Parametric diagram (β − α).
Imaginary roots unstable
β
... ; M(n–n1) possible message ... ; M(n–n2) possible message
FIGURE 2.9. Two signals, one ends with s1 and the other with s2.
If the message ends with s1, the last signal must start at n − n1 (since s1 takes n1 units of time). Hence there are M(n−n1) possible messages to which the last s1 may be appended. Hence there are M(n − n1) messages of duration n that end with s1. By a similar argument, one may conclude that there are M(n − n2) messages of duration n that end with s2. (See Figure 2.9.) Consequently, the total number of messages x(n) of duration n may be given by
M(n) = M(n − n1) + M(n − n2).
If n1 ≥ n2, then the above equation may be written in the familiar form
of an n1th-order equation
M(n+n1)−M(n+n1 −n2)−M(n)=0. (2.7.21)
On the other hand, if n1 ≤ n2, then we obtain the n2th-order equation
M(n+n2)−M(n+n2 −n1)−M(n)=0. (2.7.22)
An interesting special case is that in which n1 = 1 and n2 = 2. In this case we have
2.7 Applications 111
or
M(n + 2) − M(n + 1) − M(n) = 0, M(n + 2) = M(n + 1) + M(n),
which is nothing but our Fibonacci sequence {0, 1, 1, 2, 3, 5, 8, . . .}, which we encountered in Example 2.27. The general solution (see formula (2.3.14)) is given by
√n √n
M (n) = a 1 + 5 + a 1 − 5 , n = 0, 1, 2, . . . . (2.7.23)
1222
To find a1 and a2 we need to specify M(0) and M(1). Here a sensible assumption is to let M(0) = 0 and M(1) = 1. Using these initial data in (2.7.23) yields
11 a1 = √ , a2 = −√ ,
55 and the solution of our problem now becomes
√n √n 11+511−5
M(n) = √5 2 − √5 2
.
(2.7.24)
112 2. Linear Difference Equations of Higher Order
In information theory, the capacity C of the channel is defined as
C = lim log2 M(n), (2.7.25)
n→∞ n where log2 denotes the logarithm base 2.
From (2.7.24) we have 1
√ √n
log2√5
n→∞ n n→∞n
1+ 5 2
1− 5 2
1 +lim log2
−
Since 1− 5 ≈0.6<1,itfollowsthat 1− 5 →0asn→∞.
.
(2.7.26)
Observe also that the first term on the right-hand side of (2.7.26) goes to zero as n → ∞.
C=lim √
√ n 22
Thus
Exercises 2.7
√n C=lim1log 1+5 ,
n→∞n2 2 √
C=log 1+ 5 ≈0.7. 22
(2.7.27)
1. The model for annual plants was given by (2.7.7) in terms of the plant population p(n).
(a) Write the model in terms of s1(n).
(b) Letα=β=0.01andσ=1.Howbigshouldγbetoensurethat
the plant population increases in size?
2. An alternative formulation for the annual plant model is that in which we define the beginning of a generation as the time when seeds are produced. Figure 2.10 shows the new method.
Write the difference equation in p(n) that represents this model. Then find conditions on γ under which plant propagation occurs.
3. A planted seed produces a flower with one seed at the end of the first year and a flower with two seeds at the end of two years and each year thereafter. Suppose that each seed is planted as soon as it is produced.
(a) Write the difference equation that describes the number of flowers F (n) at the end of the nth year.
(b) Compute the number of flowers at the end of 3, 4, and 5 years.
4. Suppose that the probability of winning any particular bet is 0.49. If you start with $50 and will quit when you have $100, what is the probability of ruin (i.e., losing all your money):
(i) if you make $1 bets? (ii) if you make $10 bets? (iii) if you make $50 bets?
5. John has m chips and Robert has (N − m) chips. Suppose that John has a probability p of winning each game, where one chip is bet on in each play. If G(m) is the expected value of the number of games that will be played before either John or Robert is ruined:
(a) Show that G(m) satisfies the second-order equation G(m+2)+pG(m+1)+(1−p)G(m) = 0. (2.7.28)
(b) What are the values of G(0) and G(N)?
(c) Solve the difference equation (2.7.28) with the boundary condi-
tions in part (b).
6. Suppose that in a game we have the following situation: On each play, the probability that you will win $2 is 0.1, the probability that you will win $1 is 0.3, and the probability that you will lose $1 is 0.6. Suppose you quit when either you are broke or when you have at least N dollars. Write a third-order difference equation that describes the probability p(n) of eventually going broke if you have n dollars. Then find the solution of the equation.
7. Suppose that Becky plays a roulette wheel that has 37 divisions: 18 are red, 18 are black, and one is green. Becky can bet on either the red or black, and she wins a sum equal to her bet if the outcome is a division of that color; otherwise, she loses the bet. If the bank has one
σ
αβ
2.7 Applications 113
s0(n)
σ
γ
s0(n+1)
p(n)
σ
α
FIGURE 2.10. Annual plant model.
p(n+1)
114 2. Linear Difference Equations of Higher Order
million dollars and she has $5000, what is the probability that Becky can break the bank, assuming that she bets $100 on either red or black for each spin of the wheel?
8. In the national income model (2.7.14), assume that the government expenditure G(n) is proportional to the national income Y (n − 2) two periods past, i.e., G(n) = γY (n − 2), 0 < γ < 1. Derive the difference equation for the national income Y (n). Find the conditions for stability and oscillations of solutions.
9. Determine the behavior (stability, oscillations) of solutions of (2.7.18) for the cases:
(a)α= 4β . (1+β)2
(b)α> 4β . (1+β)2
10. Modify the national income model such that instead of the government having fixed expenditures, it increases its expenditures by 5% each time period, that is, G(n) = (1.05)n.
(a) Write down the second-order difference equation that describes this model.
(b) Find the equilibrium value.
(c) If α = 0.5, β = 1, find the general solution of the equation.
11. Suppose that in the national income we make the following assump- tions:
(i) Y (n) = C(n) + I(n), i.e., there is no government expenditure.
(ii) C(n) = a1Y(n−1)+a2Y(n−2)+K, i.e., consumption in any period is a linear combination of the incomes of the two preceding periods, where a1, a2, and K are constants.
(iii) I(n + 1) = I(n) + h, i.e., investment increases by a fixed amount h > 0 each period.
(a) Write down a third-order difference equation that models the national income Y (n).
(b) Find the general solution if a1 = 1,a2 = 1. 24
(c) Show that Y (n) is asymptotic to the equilibrium Y * = α+βn.
12. (Inventory Analysis). Let S(n) be the number of units of consumer goods produced for sale in period n, and let T(n) be the number of units of consumer goods produced for inventories in period n. Assume that there is a constant noninduced net investment V0 in each period.
Then the total income Y (n) produced in time n is given by Y (n) = T(n) + S(n) + V0.
(a) Develop a difference equation that models the total income Y (n), under the assumptions:
(i) S(n)=βY(n−1),
(ii) T(n)=βY(n−1)−βY(n−2).
(b) Obtain conditions under which:
(i) solutions converge to the equilibrium,
(ii) solutions are oscillatory.
(c) Interpret your results in part (b).
13. Let I(n) denote the level of inventories at the close of period n.
(a) Show that I(n) = I(n − 1) + S(n) + T(n) − βY(n) where
S(n), T (n), Y (n) are as in Problem 12.
(b) Assuming that S(n) = 0 (passive inventory adjustment), show
that
I(n) − I(n − 1) = (1 − β)Y (n) − V0 where V0 is as in Problem 12.
(c) Suppose as in part (b) that s(n) = 0. Show that
I(n + 2) − (β + 1)I(n + 1) + βI(n) = 0.
(d) With β ̸= 1, show that cc
I(n)= I(0)−1−β βn+1−β, where (E − β)I(n) = c.
2.7 Applications 115
14. Consider (2.7.21) with n1 = n2 = 2 (i.e., both signals s1 and s2 take two units of time for transmission).
(a) Solve the obtained difference equation with the initial conditions M(2) = M(3) = 2.
(b) Find the channel capacity c.
15. Consider (2.7.21) with n1 = n2 = 1 (i.e., both signals take one unit of
time for transmission).
(a) Solve the obtained difference equation. (b) Find the channel capacity c.
116 2. Linear Difference Equations of Higher Order
16. (Euler’s method for solving a second-order differential equation.) Re- call from Section 1.4.1 that one may approximate x′(t) by (x(n + 1) − x(n))/h, where h is the step size of the approximation and x(n) = x(t0 + nh).
(a) Show that x′′(t) may be approximated by x(n+2)−2x(n+1)+x(n).
h2
(b) Write down the corresponding difference equation of the differen-
tial equation
x′′(t) = f(x(t),x′(t)).
17. Use Euler’s method described in Problem 16 to write the corresponding
difference equation of
x′′(t) − 4x(t) = 0, x(0) = 0, x′(0) = 1.
Solve both differential and difference equations and compare the results.
18. (The Midpoint Method). The midpoint method stipulates that one may approximate x′ (t) by (x(n + 1) − x(n − 1))/h, where h is the step size of the approximation and t = t0 + nh.
(a) Use the method to write the corresponding difference equation of the differential equation x′(t) = g(t, x(t)).
(b) Use the method to write the corresponding difference equation of x′ (t) = 0.7×2 + 0.7, x(0) = 1, t ∈ [0, 1]. Then solve the obtained difference equation.
(c) Compare your findings in part (b) with the results in Section 1.4.1. Determine which of the two methods, Euler or midpoint, is more accurate.
3
Systems of Linear Difference Equations
In the last chapter we concerned ourselves with linear difference equations, namely, those equations with only one independent and one dependent variable. Since not every situation that we will encounter will be this simple, we must be prepared to deal with systems of more than one dependent variable.
Thus, in this chapter we deal with those equations of two or more depen- dent variables known as first-order difference equations. These equations naturally apply to various fields of scientific endeavor, like biology (the study of competitive species in population dynamics), physics (the study of the motions of interacting bodies), the study of control systems, neurol- ogy, and electricity. Furthermore, we will also transform those high-order linear difference equations that we investigated in Chapter 2 into systems of first-order equations. This transformation will probably prove to be of little practical use in the realm of boundary value problems and oscilla- tions, but will be substantiated as an immensely helpful tool in the study of stability theory later on, see [3], [79], [87].
3.1 Autonomous (Time-Invariant) Systems
In this section we are interested in finding solutions of the following system of k linear equations:
x1(n + 1) = a11x1(n) + a12x2(n) + · · · + a1kxk(n), x2(n + 1) = a21x1(n) + a22x2(n) + · · · + a2kxk(n),
117
118 3. Systems of Linear Difference Equations
. . . .
xk(n + 1) = ak1x1(n) + ak2x2(n) + · · · + akkxk(n). This system may be written in the vector form
x(n + 1) = Ax(n), (3.1.1)
where x(n) = (x1(n), x2(n),…,xk(n))T ∈ Rk, and A = (aij) is a k × k real nonsingular matrix. Here T indicates the transpose of a vector. Sys- tem (3.1.1) is considered autonomous, or time-invariant, since the values of A are all constants. Nonautonomous, or time-variant, systems will be considered later in Section 3.3.
If for some n0 ≥ 0,x(n0) = x0 is specified, then system (3.1.1) is called an initial value problem. Furthermore, by simple iteration (or by direct substitution into the equation), one may show that the solution is given by
(3.1.2)
where A0 = I, the k × k identity matrix. Notice that x(n0, n0, x0) = x0. If n0 = 0, then the solution in formula (3.1.2) may be written as x(n, x0), or simply x(n). We now show that we may assume that n0 = 0 without loss of generality.
Let y(n − n0) = x(n). Then (3.1.1) becomes
y(n + 1) = Ay(n), (3.1.3)
x(n, n0, x0) = An−n0 x0,
with y(0) = x(n0) and
A parallel theory exists for systems of linear differential equations. The
y(n) = Any(0). (3.1.4) solution of the initial value problem
dx = Ax(t), x(t0) = x0, dt
where A is a k × k matrix, x ∈ Rk, is given by x(t) = eA(t−t0)x0.
3.1.1 The Discrete Analogue of the Putzer Algorithm
In differential equations the Putzer algorithm is used to compute eAt. Here, we introduce an analogous algorithm to compute An. First, let us review the rudiments of matrix theory that are vital in the development of this algorithm. In what follows C denotes the set of complex numbers.
Recall that for a real k × k matrix A = (aij), an eigenvalue of A is a real or complex number λ such that Aξ = λξ for some nonzero ξ ∈ Ck. Equivalently, this relation may be written as
(A − λI)ξ = 0. (3.1.5)
3.1 Autonomous (Time-Invariant) Systems 119
Equation (3.1.5) has a nonzero solution if and only if det(A − λI) = 0,
or
λk +a1λk−1 +a2λk−2 +···+ak−1λ+ak =0. (3.1.6)
Equation (3.1.6) is called the characteristic equation of A, whose roots λ are called the eigenvalues of A. If λ1,λ2,…,λk are the eigenvalues of A (some of them may be repeated), then one may write (3.1.6) as
fundamental results of matrix theory.
Theorem 3.1. Every matrix satisfies its characteristic equation. That is,
Ak +a1Ak−1 +a2Ak−2 +···+akI =0. 3.1.2 The Development of the Algorithm for An
k j=1
(λ − λj ). (3.1.7) We are now ready to state the Cayley–Hamilton theorem, one of the
or
p(λ) =
p(A) =
k j=1
(A − λjI) = 0,
(3.1.8)
(3.1.9)
Let A be a k × k real matrix. We look for a representation of An in the form
(3.1.10) where the uj(n)’s are scalar functions to be determined later, and
or
An =
s j=1
uj(n)M(j−1),
M(j) = (A − λjI)M(j − 1), M(0) = I,
(3.1.11)
M(j + 1) = (A − λj+1I)M(j), M(0) = I. By iteration, one may show that
M(n)=(A−λnI)(A−λn−1I)··· (A−λ1I), or, in compact form,
n j=1
M(n) =
(A − λjI).
(3.1.12)
120 3. Systems of Linear Difference Equations
Notice that by the Cayley–Hamilton theorem we have
k
rewrite formula (3.1.10) as
=
Substituting for AM (j − 1) from (3.1.11) yields
k j=1
k j=1
An =
uj(n)M(j−1).
k j=1
(A − λjI) = 0.
M(k) =
Consequently, M(n) = 0 for all n ≥ k. In light of this observation,
we may
(3.1.13)
(3.1.14)
(3.1.15)
j=1
If we let n = 0 in formula (3.1.13) we obtain
A0 =I=u1(0)I+u2(0)M(1)+···+uk(0)M(k−1). Equation (3.1.14) is satisfied if
u1(0)=1 and u2(0)=u3(0)=···=uk(0)=0. From formula (3.1.13) we have
⎡⎤ k k
j=1
uj(n+1)M(j−1)=AAn =A⎣ uj(n)M(j−1)⎦ j=1
k
uj (n)AM (j − 1).
j=1
uj(n + 1)M(j − 1) =
Comparing the coefficients of M(j),1 ≤ j ≤ k, in (3.1.16), and applying
condition (3.1.15), we obtain
u1(n + 1) = λ1u1(n), u1(0) = 1,
uj(n+1)=λjuj(n)+uj−1(n), uj(0)=0, j=2,3,…,k. (3.1.17) The solutions of (3.1.17) are given by
i=0
Equations (3.1.12) and (3.1.18) together constitute an algorithm for com- puting An, which henceforth will be called the Putzer algorithm. For more details and other algorithms, the interested reader may consult the paper by Elaydi and Harris [46].
n−1
n n−1−i
u1(n) = λ1, uj(n) =
λj uj−1(i), j = 2,3,…,k. (3.1.18)
uj(n)[M(j) + λjM(j − 1)].
(3.1.16)
Example 3.2. Find An if
A = ⎜⎝ − 2 3 1 ⎟⎠ .
3 (i2) n−1 i
i=0
3n−1 2
=2 i3 i=0
3.1 Autonomous (Time-Invariant) Systems 121
⎛⎞
TheeigenvaluesofAareλ1 =λ2 =2, λ3 =3, ⎛⎞
−2 1 1 M(0)=I, M(1)=A−2I=⎜⎝−2 1 1⎟⎠,
−3 1 2 ⎛⎞
−1 0 1 M(2)=(A−2I), M(1)=(A−2I)2 =⎜⎝−1 0 1⎟⎠.
Now
u1(n) = 4n,
n−1
(n−1−i) i n−1
u2(n) = 2 ·2=n2, i=0
n−1
(n−1−i) i−1
u3(n) =
3n−1 2 − 1n 2 n − 2 n+1 + 2
= 3 3
2 2 −12
3 = −2n + 3n − n2n−1.
011 −3 1 4
Solution The eigenvalues of A are obtained by solving the characteristic equation
Hence
⎛⎞
−λ 1 1 det(A−λI)=det⎜⎝−2 3−λ 1 ⎟⎠=0.
−3 1 4−λ
p(λ)=λ3 −7λ2 +16λ−12 = (λ − 2)2(λ − 3) = 0.
−2 0 2
3 3 (from Table 1.1)
122
3. Systems of Linear Difference Equations
Thus
Ax(n), where
Now,
n j=1
An =
=⎜⎝ 2n −3n −n2n−1
uj(n)M(j−1)
⎛ 2n−1 −3n −n2n−1
n2n−1 (n+2)2n−1
−2n +3n ⎞
2n+1 −2·3n −n2n−1 ⎛⎞
412
A = ⎜⎝ 0 2 − 4 ⎟⎠ .
016
−2n +3n ⎟⎠. −2n +2·3n
n2n−1
Example 3.3. Find the solution of the difference system x(n + 1) =
Solution The eigenvalues of A may be obtained by solving the characteristic equation det(A − λI) = 0. Now,
⎛⎞
4−λ12
det⎜⎝ 0 2−λ −4 ⎟⎠=(4−λ)(λ−4)2 =0.
016−λ
Hence, the eigenvalues of A are λ1 = λ2 = λ3 = 4. So
⎛⎞
012 M(0)=I, M(1)=A−4I=⎜⎝0 −2 −4⎟⎠,
012
⎛⎞
000 M(2)=(A−4I)M(1)=⎜⎝0 0 0⎟⎠.
u1(n) = 4n, n−1
n−1−i i n−1 (4 )(4 ) = n(4 ),
u2(n) =
u3(n)= 4 (i4 )
i=0 n−1
n−1−i i−1 i=0
n−1 n−2
=4i i=0
= n(n−1) 4n−2. 2
000
3.1 Autonomous (Time-Invariant) Systems 123
Using (3.1.13), we have
⎛⎞⎛⎞
100 012 An = 4n ⎜⎝0 1 0⎟⎠+n4n−1 ⎜⎝0 −2 −4⎟⎠
001 012
⎛⎞
000 +n(n−1) 4n−2⎜⎝0 0 0⎟⎠
2
⎛4n
000
n4n−1 2n4n−1 ⎞
=⎜⎝04n−2n4n−1 −n4n⎟⎠. 0 n4n−1 4n + 2n4n−1
The solution of the difference equation is given by
⎛4nx1(0) + n4n−1×2(0) + 2n4n−1×3(0)⎞ x(n) = Anx(0) = ⎜⎝ (4n − 2n4n−1)x2(0) − n4nx3(0) ⎟⎠ ,
n4n−1×2(0) + (4n + 2n4n−1)x3(0) where x(0) = (x1(0), x2(0), x3(0))T .
Exercises 3.1
In Problems 1 through 4, use the discrete Putzer algorithm to evaluate An.
11 1. A = −2 4 .
−1 2 30
⎡⎤
1 2 −1 3.A=⎢⎣0 1 0⎥⎦.
4 −4 5 ⎡⎤
210
4. A=⎢⎣0 2 1⎥⎦.
002
2. A =
.
5. Solve the system
x1(n + 1) = −x1(n) + x2(n),
x2(n + 1) = 2×2(n), x2(0) = 2.
6. Solve the system
x1(n + 1) = x2(n),
x2(n + 1) = x3(n),
x3(n + 1) = 2×1(n) − x2(n) + x3(n).
x1(0) = 1,
124 3. Systems of Linear Difference Equations
7. Solve the system
⎡⎤⎛⎞
1 −2 −2 1 x(n + 1) = ⎢⎣0 0 −1⎥⎦ x(n), x(0) = ⎜⎝1⎟⎠ .
023 0 8. Solve the system
⎛1 3 0 0⎞ ⎜0 2 1 −1⎟
x ( n + 1 ) = ⎜⎝ 0 0 2 0 ⎟⎠ x ( n ) . 0003
2 −1
9. Verify that the matrix A =
equation (the Cayley Hamilton Theorem).
ρ0 < β.
βn
(a) Showthat|uj(n)|≤(β−ρ0),j=1,2,...,k.
(b) Showthatifρ0 <1,thenuj(n)→0asn→∞.Concludethat
An →0asn→∞.
(c) If α < min{|λ| : λ is an eigenvalue of A}, establish a lower bound
for |uj(n)|.
11. If a k × k matrix A has distinct eigenvalues λ1, λ2, . . . , λk, then one may compute An,n ≥ k, using the following method. Let p(λ) be the characteristic polynomial of A. Divide λn by p(λ) to obtain λn = p(λ)q(λ) + r)(λ), where the remainder r(λ) is a polynomial of degree at most (k − 1). Thus one may write An = p(A)q(A) + r(A).
(a) ShowthatAn =r(A)=a0I+a1A+a2A2+···+ak−1Ak−1. (b) Show that λn1 = r(λ1),λn2 = r(λ2),...,λnk = r(λk).
(c) Use part (b) to find a0,a1,...,ak−1.
12. Extend the method of Problem 11 to the case of repeated roots.
13. Apply the method of Problem 12 to find An for:
11 (i)A= −2 4 .
⎛⎞
1 2 −1 (ii)A=⎜⎝1 0 1⎟⎠.
13
10. Let ρ(A) = max{|λ| : λ is an eigenvalue of A}. Suppose that ρ(A) =
satisfies its characteristic
4 −4 5
y (0,1)
p1 p3
FIGURE 3.1.
14. Apply the method of Problem 12 to find An for ⎛⎞
15.1
412
A = ⎜⎝ 0 2 − 4 ⎟⎠ .
016
Consider the right triangle in Figure 3.1 where p(0) = (0, 0), p(1) =
1, 1, and p(2) = 1,0. For p(n) = (x(n),y(n)) with n ≥ 3, we have 222
3.2 The Basic Theory 125
p0
p2
p4
x
(1,0)
3.2
x(n+3)= 1(x(n)+x(n+1)), 2
y(n+3)= 1(y(n)+y(n+1)). 2
(a) Write each equation as a system z(n + 1) = Az(n). (b) Find limn→∞ p(n).
The Basic Theory
Now contemplate the system
x(n + 1) = A(n)x(n),
(3.2.1)
where A(n) = (aij(n)) is a k × k nonsingular matrix function. This is a homogeneous linear difference system that is nonautonomous, or time- variant.
The corresponding nonhomogeneous system is given by
y(n + 1) = A(n)y(n) + g(n), (3.2.2)
where g(n) ∈ Rk.
We now establish the existence and uniqueness of solutions of (3.2.1).
1Proposed by C.V. Eynden and solved by Trinity University Problem Solving Group (1994).
126 3. Systems of Linear Difference Equations
Theorem 3.4. For each x0 ∈ Rk and n0 ∈ Z+ there exists a unique solution x(n, n0, x0) of (3.2.1) with x(n0, n0, x0) = x0.
Proof. From (3.2.1),
x(n0 + 1, n0, x0) = A(n0)x(n0) = A(n0)x0,
x(n0 + 2, n0, x0) = A(n0 + 1)x(n0 + 1) = A(n0 + 1)A(n0)x0. Inductively, one may conclude that
where
n−1
A(n−1)A(n−2)···A(n0) ifn>n0,
A(i) =
I if n = n0.
x(n, n0, x0) =
n−1
A(i) x0,
(3.2.3)
i=n0
Formula (3.2.3) gives the unique solution with the desired properties. P
We will now develop the notion of a fundamental matrix, a central building block in the theory of linear systems.
Definition 3.5. The solutions x1(n),x2(n),…,xk(n) of (3.2.1) are said to be linearly independent for n ≥ n0 ≥ 0 if whenever c1x1(n) + c2x2(n) + ···+ckxk(n)=0foralln≥n0,thenci =0,1≤i≤k.
Let Φ(n) be a k × k matrix whose columns are solutions of (3.2.1). We write
Now,
Φ(n) = [x1(n), x2(n), . . . , xk(n)].
Φ(n + 1) = [A(n)x1 (n), A(n)x2 (n), . . . , A(n)xk (n)] = A(n)[x1(n), x2(n), . . . , xk(n)]
= A(n)Φ(n).
Hence, Φ(n) satisfies the matrix difference equation
Φ(n + 1) = A(n)Φ(n).
(3.2.4)
Furthermore, the solutions x1(n), x2(n), . . . , xk(n) are linearly indepen- dent for n ≥ n0 if and only if the matrix Φ(n) is nonsingular (det Φ(n) ̸= 0) for all n ≥ n0. (Why?) This actually leads to the next definition.
Definition 3.6. If Φ(n) is a matrix that is nonsingular for all n ≥ n0 and satisfies (3.2.4), then it is said to be a fundamental matrix for system equation (3.2.1).
Note that if Φ(n) is a fundamental matrix and C is any nonsingular matrix, then Φ(n)C is also a fundamental matrix (Exercises 3.2, Problem
i=n0
6). Thus there are infinitely many fundamental matrices for a given system. However, there is one fundamental matrix that we already know, namely,
n−1
i=n0
(Exercises 3.2, Problem 5). In the autonomous case when A is a constant matrix, Φ(n) = An−n0 , and if n0 = 0, then Φ(n) = An. Consequently, it would be much more suitable to use the Putzer algorithm to compute the fundamental matrix for an autonomous system.
Theorem 3.7. There is a unique solution Ψ(n) of the matrix (3.2.4) with Ψ(n0) = I.
Proof. One may think of the matrix difference equation (3.2.4) as a system of k2 first-order difference equations. Thus, to complete the point, we may apply the “existence and uniqueness” Theorem 3.4 to obtain a k2- vector solution ν such that ν(n0) = (1,0,…,1,0,…)T , where 1’s appear at the first, (k + 2)th, (2k + 3)th, . . . slots and 0’s everywhere else. The vector ν is then converted to the k × k matrix Ψ(n) by grouping the components into sets of k elements in which each set will be a column. Clearly, Ψ(n0) = I. P
We may add here that starting with any fundamental matrix Φ(n), the fundamental matrix Φ(n)Φ−1(n0) is such a matrix. This special fundamen- tal matrix is denoted by Φ(n,n0) and is referred to as the state transition matrix.
One may, in general, write Φ(n,m) = Φ(n)Φ−1(m) for any two posi- tive integers n, m with n ≥ m. The fundamental matrix Φ(n, m) has some agreeable properties that we ought to list here. Observe first that Φ(n, m) is a solution of the matrix difference equation Φ(n + 1, m) = A(n)Φ(n, m) (Exercises 3.2, Problem 2). The reader is asked to prove the following statements:
(i) Φ−1 (n, m) = Φ(m, n) (Exercises 3.2, Problem 3).
(ii) Φ(n, m) = Φ(n, r)Φ(r, m) (Exercises 3.2, Problem 3).
(iii) Φ(n, m) = n−1 A(i) (Exercises 3.2, Problem 3). i=m
Corollary 3.8. The unique solution of x(n, n0, x0) of (3.2.1) with x(n, n0, x0) = x0 is given by
(3.2.5)
Checking the linear independence of a fundamental matrix Φ(n) for n ≥ n0 is a formidable task. We will instead show that it suffices to establish linear independence at n0.
Φ(n) =
A(i), with Φ(n0) = I
3.2 The Basic Theory 127
x(n, n0, x0) = Φ(n, n0)x0.
128 3. Systems of Linear Difference Equations
Lemma 3.9 Abel’s Formula. For any n ≥ n0 ≥ 0, n−1
det Φ(n) =
Proof. Taking the determinant of both sides of (3.2.4) we obtain the
scalar difference equation
detΦ(n+1)=det A(n) detΦ(n)
whose solution is given by (3.2.6). P Corollary 3.10. If in (3.2.1) A is a constant matrix, then
(3.2.7) Proof. The proof follows from formula (3.2.6). P
Corollary 3.11. The fundamental matrix Φ(n) is nonsingular for all n ≥ n0 if and only if Φ(n0) is nonsingular.
Proof. This follows from formula (3.2.6), having noted that det A(i) ̸= 0 for i ≥ n0. P
Corollary 3.12. The solutions x1 (n), x2 (n), . . . , xk (n) of (3.2.1) are linearly independent for n ≥ n0 if and only if Φ(n0) is nonsingular.
Proof. This follows immediately from Corollary 3.11. P The following theorem establishes the existence of k linearly independent
solutions of (3.2.1).
Theorem 3.13. There are k linearly independent solutions of system (3.2.1) for n ≥ n0.
Proof. For each i = 1,2,…, k, let ei = (0,0,…, 1,…, 0)T be the standard unit vector in Rk where all the components are zero except the ith component, which is equal to 1. By Theorem 3.4, for each ei, 1 ≤ i ≤ k, there exists a solution x(n, n0, ei) of (3.2.1) with x(n0, n0, ei) = ei. To prove that the set {x(n,n0,ei)|1 ≤ i ≤ k} is linearly independent, according to Corollary 3.11 it suffices to show that Φ(n0) is nonsingular. But this fact is obvious, since Φ(n0) = I. The proof of the theorem is now complete. P
Linearity Principle. An important feature of the solutions of system (3.2.1) is that they are closed under addition and scalar multiplication. That is to say, if x1(n) and x2(n) are solutions of (3.2.1) and c ∈ R, then:
i=n0
[det A(i)] det Φ(n0).
(3.2.6)
det Φ(n) = [det A]n−n0 det Φ(n0).
(1) x1(n) + x2(n) is a solution of (3.2.1),
(2) cx1(n) is a solution of (3.2.1). This is called the linearity principle.
Proof. Statement (1) can be proved as follows. Let x(n) = x1(n)+x2(n). Then
The proof of (2) is similar.
P
x(n + 1) = x1(n + 1) + x2(n + 1) = Ax1(n) + Ax2(n)
= A[x1(n) + x2(n)]
= Ax(n).
An immediate consequence of the linearity principle is that if x1 (n), x2 (n), …,xk(n) are also solutions of system (3.2.1), then so is any linear combination of the form
x(n) = c1x1(n) + c2x2(n) + · · · + ckxk(n). This leads to the following definition.
Definition 3.14. Assuming that {xi(n)|1 ≤ i ≤ k} is any linearly inde- pendent set of solutions of (3.2.1), the general solution of (3.2.1) is defined to be
(3.2.8)
whereci ∈Randatleastoneci ̸=0. Formula (3.2.8) may be written as
x(n) = Φ(n)c, (3.2.9) where Φ(n) = (x1(n),x2(n),…,xk(n)) is a fundamental matrix, and c =
(c1,c2,…,ck)T ∈ Rk.
Remark: The set S of all solutions of system (3.2.1) forms a linear (vector) space under addition and scalar multiplication. Its basis is any fundamental set of solutions and hence its dimension is k. The basis {x1 (n), x2 (n), . . . , xk (n)} spans all solutions of equation (3.2.1). Hence any solution x(n) of equation (3.2.1) can be written in the form (3.2.8) or equivalently (3.2.9). This is why we call x(n) in (3.2.8) a general solution.
Let us now focus our attention on the nonhomogeneous system (3.2.2). We define a particular solution yp(n) of (3.2.2) as any k-vector function that satisfies the nonhomogeneous difference system. The following result gives us a mechanism to find the general solution of system (3.2.2).
3.2 The Basic Theory 129
x(n) =
k i=1
cixi(n),
130 3. Systems of Linear Difference Equations
Theorem 3.15. Any solution y(n) of (3.2.2) can be written as
y(n) = Φ(n)c + yp(n) (3.2.10)
for an appropriate choice of the constant vector c, and a particular solution yp (n).
Proof. Let y(n) be a solution of (3.2.2) and let yp(n) be any particular solution of (3.2.2). If x(n) = y(n) − yp(n), then
x(n + 1) = y(n + 1) − yp(n + 1) = A(n)y(n) − A(n)yp(n)
= A(n)[y(n) − yp(n)] = A(n)x(n).
Thus x(n) is a solution of the homogeneous equation (3.2.1). Hence x(n) = Φ(n)c for some vector constant c. Thus
y(n) − yp(n) = Φ(n)c
which proves (3.2.10). P
We now give a formula to evaluate yp(n).
Lemma 3.16.
with yp(n0) = 0. Proof.
n r=n0
n−1
is given by
y(n + 1) = A(n)y(n) + g(n), y(n0) = y0, (3.2.11)
(3.2.12)
yp(n + 1) = =
Φ(n + 1, r + 1)g(r)
A(n)Φ(n, r + 1)g(r) + Φ(n + 1, n + 1)g(n)
A particular solution of (3.2.2) may be given by n−1
yp(n) =
r=n0
Φ(n, r + 1)g(r)
r=n0
= A(n)yp(n) + g(n).
Hence, yp(n) is a solution of (3.2.2). Furthermore, yp(n0) = 0. P Theorem 3.17 (Variation of Constants Formula). The unique
solution of the initial value problem
y(n, n0, y0) = Φ(n, n0)y0 +
Φ(n, r + 1)g(r),
n−1
r=n0
or, more explicitly, by
n−1 n−1 n−1
Lemma 3.16. P Corollary 3.18. For autonomous systems when A is a constant matrix,
the solution of (3.2.11) is given by
(3.2.14) Example 3.19. Solve the system y(n + 1) = Ay(n) + g(n), where
21n1 A= ,g(n)=,y(0)=.
0210 Solution Using the Putzer algorithm, one may show that
y(n, n0, y0) = A(i) y0 + i=n0
A(i) g(r). (3.2.13) Proof. This theorem follows immediately from Theorem 3.15 and
Hence,
y(n) =
0 2n n n−1
y(n, n0, y0) = An−n0 y0 +
n−1
r=n0
An−r−1g(r).
+
n−1 n−r−1 n−r−2 2 (n−r−1)2 r
2n2 1 0 2n 0
= 0 +
r=0
⎛ n−1 r n−1 r⎞∗
n n−1 An=2n2 .
0
2n−r−1 1 n−r−2
r=0
n n−1 n−r−1
2 r2
+(n−r−1)2 2n−r−1
r=n0 i=r+1
11n−11
n⎜4r2+4 2⎟
3.2 The Basic Theory 131
= 2 +2n⎜ r=1 r=0 ⎟
.
⎜ n−1 2r=0 2
⎟ 0⎝11r⎠
n−1
∗ r a(1−an)−nan+1(1−a)
r=1
ra = (1−a)2
132
3. Systems of Linear Difference Equations
⎛ n n+2 n⎞ n ⎜11−1 −n1 +n−11−1 ⎟ =2+2n⎜2 2 22 2⎟ 0⎝1−1n⎠
2 ⎛ n1n n n1⎞
2n ⎜−4 2 +2−2 2 ⎟ = 0 +2n⎜⎝ 1n ⎟⎠
⎛ 3⎞ 2n n2n−1− n
=+⎝4⎠ 0 2n − 1
1−2
⎛3⎞ = ⎝2n + n2n−1 − 4 n⎠ .
2n − 1
We now revisit scalar equations of order k and demonstrate how to trans- form them into a k-dimensional system of first-order equations. Consider again the equation
y(n + k) + p1(n)y(n + k − 1) + · · · + pk(n)y(n) = g(n). (3.2.15) This relation may be written as a system of first-order equations of
dimension k. We let
z1(n) = y(n),
z2(n) = y(n + 1) = z1(n + 1), z3(n) = y(n + 2) = z2(n + 1),
.
zk(n) = y(n + k − 1) = zk−1(n + 1).
Let z(n) = (z1(n), z2(n), . . . , zk(n)). Hence,
z1(n + 1) = z2(n), z2(n + 1) = z3(n),
.
zk−1(n + 1) = zk(n),
zk(n + 1) = −pk(n)z1(n) − pk−1(n)z2(n), . . . ,
− p1(n)zk(n) + g(n). In vector notation, we transcribe this system as
z(n + 1) = A(n)z(n) + h(n),
(3.2.16)
where
⎛010…0⎞ ⎜0 0 1…0⎟ ⎜0 0 0…0⎟
A(n) = ⎜. . . .⎟ ⎜. . . .⎟
(3.2.17)
⎝0001⎠
−pk(n) −pk−1(n) −pk−2(n) . . . ⎛⎞
0
⎜ 0 ⎟
−p1(n)
and
If g(n) = 0, we arrive at the homogeneous system z(n + 1) = A(n)z(n).
h(n)=⎜ 0 ⎟. ⎜ . ⎟
⎝.⎠ g(n)
The matrix A(n) is called the companion matrix of (3.2.15).
Consider now the kth-order homogeneous equation with constant
coefficients x(n+k)+p1x(n+k−1)+p2x(n+k−2)+···+pkx(n)=0, (3.2.19)
which is equivalent to the system where A is the companion matrix defined in formula (3.2.17) with all pi’s constant,
z(n + 1) = Az(n). (3.2.20)
We first observe that the Casoratian of (3.2.19) is denoted by C(n) = detΦ(n), where Φ(n) is a fundamental matrix of (3.2.20). (Why?) (Ex- ercises 3.2, Problem 14.) The characteristic equation of A is given by
λk + p1λk−1 + p2λk−2 + · · · + pk−1λ + pk = 0 (Exercises 3.2, Problem 13), which correlates with (2.3.2). Hence, the eigenvalues of A are the roots of
the characteristic equation of (2.3.1).
Exercises 3.2
1. Let Φ1(n) and Φ2(n) be two fundamental matrices of system (3.2.1).
Prove that Φ (n)Φ−1(n ) = Φ (n)Φ−1(n ) for any n ≥ 0. 110220 0
3.2 The Basic Theory 133
(3.2.18)
134 3. Systems of Linear Difference Equations
2. Let
(i) Φ(n, m) is a solution of Φ(n + 1, m) = A(n)Φ(n, m).
3. Let
Φ(n, m) be a fundamental matrix of (3.2.1). Show that:
Φ(n, m) be a fundamental matrix of (3.2.1). Show that:
(ii) Φ(n, m) is a solution of Φ(n, m + 1) = Φ(n, m)A−1 (m).
(a) Φ(n, m) = An−m if A(n) ≡ A is a constant matrix.
(b) Φ(n,m)=Φ(n,r)Φ(r,m).
(c) Φ−1 (n, m) = Φ(m, n).
(d) Φ(n, m) = n−1 A(i).
i=m
Φ(n) be a fundamental matrix of (3.2.1). Show that each column
4. Let
of Φ(n) is a solution of (3.2.1).
5. Show that Φ(n) = n−1 A(i) is a fundamental matrix of (3.2.1). i=n0
6. Show that if Φ(n) is a fundamental matrix of (3.2.1) and C is any nonsingular matrix, then Φ(n)C is also a fundamental matrix of (3.2.1).
7. Show that if Φ1(n),Φ2(n) are two fundamental matrices of (3.2.1), then there exists a nonsingular matrix C such that Φ2(n) = Φ1(n)C.
8. Solve the system:
y1(n + 1) = y2(n),
y2(n + 1) = y3(n) + 2,
y3(n + 1) = y1(n) + 2y3(n) + n2.
9. Solve the system:
y1(n + 1) = 2y1(n) + 3y2(n) + 1, y2(n + 1) = y1(n) + 4y2(n), y1(0) = 0, y2(0) = −1.
10. Solve the system y(n + 1) = Ay(n) + g(n) if ⎛⎞⎛⎞
2 2 −2 1
A = ⎜⎝ 0 3 1 ⎟⎠ , g ( n ) = ⎜⎝ n ⎟⎠ .
013 n2
11. For system equation (3.2.18) show that
detA(n) = (−1)kpk(n).
12. If Φ(n) is a fundamental matrix of (3.2.18), prove that
pk(i) detΦ(n0).
detΦ(n) = (−1)
k(n−n0 ) i=n0
n−1
3.3 The Jordan Form: Autonomous (Time-Invariant) Systems Revisited 135
13. Prove by induction that the characteristic equation of
is
−pk −pk−1 −pk−2 . . . −p1
λk +p1λk−1 +p2λk−2 +···+pk−1λ+pk =0.
⎛⎞
010…0 ⎜0 0 1 … 0⎟
A=⎜ . . . ⎟
⎟ ⎝000…1⎠
⎜
14. Let W(n) be the Casoratian of (3.2.15) with g(n) = 0. Prove that there exists a fundamental matrix Φ(n) of (3.2.18) such that W (n) = det Φ(n).
Use the methods of systems to solve the difference equation for Problems 15 through 19.
15. x(n+2)+8x(n+1)+12x(n)=0. 16. x(n+2)−16x(n)=0.
17. y(n+2)−5y(n+1)+4y(n)=4n. 18. ∆2y(n) = 16.
19. ∆2x(n) + ∆x(n) − x(n) = 0.
3.3 The Jordan Form: Autonomous
(Time-Invariant) Systems Revisited
The Jordan form of a matrix is vital for both theoretical and computational purposes in autonomous systems. In this section we will briefly describe the Jordan form and derive a new method for computing fundamental matrices.
3.3.1 Diagonalizable Matrices
We say that the two k × k matrices A and B are similar if there exists a nonsingular matrix P such that P−1AP = B. It may be shown in this case that A and B have the same eigenvalues and, in fact, the eager student will prove this supposition in Exercises 3.3, Problem 15. If a matrix A is similar to a diagonal matrix D = diag(λ1,λ2,…,λk), then A is said to be diagonalizable. Notice here that the diagonal elements of D, namely, λ1, λ2, . . . , λk, are the eigenvalues of A. We remark here that only special types of matrices are diagonalizable. For those particular diagonalizable
136 3. Systems of Linear Difference Equations
matrices, computing An is simple. For if
P−1AP = D = diag[λ1,λ2,…,λk],
then
and, consequently,
Explicitly,
A = PDP−1,
An = (PDP−1)n = PDnP−1.
⎡λn1 0⎤
⎢ λn2 An =P⎢
⎥
.. ⎥P−1.
(3.3.1)
⎣.⎦ 0 λnk
If we are interested in finding another (but simpler) fundamental matrix of the equation
x(n + 1) = Ax(n),
⎡λn1 0⎤
(3.3.2)
(3.3.3)
(3.3.4)
then we let
From formula (3.3.3) we have Φ(0) = P and, consequently, An = Φ(n)Φ−1(0).
⎢ λn2 Φ(n)=AnP =P⎢
⎥ .. ⎥.
⎣.⎦ 0 λnk
Now, formula (3.3.3) is useful only if one can pinpoint the matrix P. Fortunately, this is an easy task. We will now reveal how to compute P .
Let P = (ξ1,ξ2,…,ξk), where ξi is the ith column of P. Since P−1AP = D, then AP = PD. This implies that Aξi = λiξi,i = 1,2,…,k (Exercises 3.3, Problem 15). Thus, ξi, 1 ≤ i ≤ k, is the eigenvector of A corresponding to the eigenvalue λi, and hence the ith column of P is the eigenvector of A corresponding to the ith eigenvalue λi of A. Since P is nonsingular, its columns (and hence the eigenvectors ξ1, ξ2, . . . , ξk of A) are linearly independent. Reversing the above steps, one may show that the converse of the above statement is true. Namely, if there are k linearly independent eigenvectors of a k × k matrix A, then it is diagonalizable. The following theorem summarizes the above discussion.
Theorem 3.20. A k × k matrix is diagonalizable if and only if it has k linearly independent eigenvectors.
3.3 The Jordan Form: Autonomous (Time-Invariant) Systems Revisited 137
Let us revert back to formula (3.3.3), which gives us a computational method to find a fundamental matrix Φ(n). Let λ1, λ2, . . . , λk be the eigen- values of A and let ξ1, ξ2, . . . , ξk be the corresponding linearly independent eigenvectors of A. Then from formula (3.3.3) we have
⎡λn1 0⎤ ⎢ λn2 ⎥
Φ(n) = [ξ1,ξ2,…,ξk]⎢⎣ … ⎥⎦ 0 λnk
= [ λ n1 ξ 1 , λ n2 ξ 2 , . . . , λ nk ξ k ] . ( 3 . 3 . 5 )
Notice that since columns of Φ(n) are solutions of (3.3.2), it follows that for each i,1 ≤ i ≤ k,x(n) = λni ξi is a solution of (3.3.2).
Hence, the general solution of (3.3.2) may be given by
x(n)=c1λn1ξ1 +c2λn2ξ2 +···+ckλnkξk. (3.3.6)
The following example illustrates the above method.
Example 3.21. Find the general solution of x(n + 1) = Ax(n), where
⎛⎞
221
A = ⎜⎝ 1 3 1 ⎟⎠ .
122
Solution The eigenvalues of A may be obtained by solving the characteristic equation
⎛⎞
2−λ21
det(A − λI) = det ⎜⎝ 1 3 − λ 1 ⎟⎠ = 0.
122−λ
This determinant produces (λ − 1)2(λ − 5) = 0. Thus, λ1 = 5, and λ2 = λ3 = 1. To find the corresponding eigenvectors, we solve the equation (A−λI)x=0.Hence,forλ1 =5,
⎛ ⎞⎛⎞⎛⎞
−321×1 0
⎜⎝ 1 −2 1 ⎟⎠⎜⎝x2⎟⎠ = ⎜⎝0⎟⎠.
12−3×3 0 Solving this system gives us the first eigenvector
⎛⎞
1
ξ 1 = ⎜⎝ 1 ⎟⎠ .
1
138 3. Systems of Linear Difference Equations
Forλ2 =λ3 =1,wehave ⎛⎞⎛⎞⎛⎞
121×1 0 ⎜⎝1 2 1⎟⎠ ⎜⎝x2⎟⎠ = ⎜⎝0⎟⎠ .
121×3 0
Consequently, x1 + 2×2 + x3 = 0 is the only equation obtained from this algebraic system. To solve the system, two of the three unknown terms x1,x2,and x3 must be arbitrarily chosen. So if we let x1 = 1 and x2 = 0,
then x3 = −1, and we obtain the eigenvector ⎛⎞
1
ξ 2 = ⎜⎝ 0 ⎟⎠ .
−1
Ontheotherhand,ifweletx1 =0andx2 =1,thenx3 =−2,andwe
obtain the third eigenvector
ξ 3 = ⎜⎝ 1 ⎟⎠ .
or
1 −1 −2 ⎛ c5n+c ⎞
Thus
⎛⎞
Obviously, there are infinitely many choices for ξ2, ξ3. Using formula (3.3.6),
we see that the general solution is
⎛⎞⎛⎞⎛⎞
110 x(n)=c15n⎜⎝1⎟⎠+c2⎜⎝ 0 ⎟⎠+c3⎜⎝ 1 ⎟⎠,
0 −2
12
x ( n ) = ⎜⎝ c 1 5 n + c 3 ⎟⎠ . ( 3 . 3 . 7 )
c15n −c2 −2c3
Suppose that in the above problem we are given an initial value
⎛⎞
0 x(0) = ⎜⎝1⎟⎠
0
and must find the solution x(n) with this initial value. One way of doing this is by letting n = 0 in the solution given by formula (3.3.7) and evaluating the constants c1, c2, and c3.
c1 + c2 = 0,
c1 + c3 = 1, c1 −c2 −2c3 =0.
3.3 The Jordan Form: Autonomous (Time-Invariant) Systems Revisited 139
Solving this system gives c1 = 1,c2 = −1,and c3 = 1, leading us to the 222
solution
15n −
We now introduce yet another method to find the solution. Let
⎛15n − ⎜2
x(n) = ⎜ 1 5n + ⎜⎝2
1⎞ 2⎟
1 ⎟ . 2⎟⎠
1 22
where
and
Thus,
This gives
x(n) = Φ(n)Φ−1(0)x(0), Φ(n) = (λn1 ξ1, λn2 ξ2, λn3 ξ3)
⎛5n 1 0⎞ =⎜⎝5n 0 1⎟⎠
5n −1 −2 ⎛⎞
110
Φ ( 0 ) = ⎜⎝ 1 0 1 ⎟⎠ .
⎛111⎞
⎜ 4 2 4 ⎟
Φ−1(0)=⎜ 3 −1 −1⎟. ⎜⎝4 2 4⎟⎠
−1 1 −1 424
1 −1 −2
⎛5n ⎜ n
1
0
0 ⎞⎜ 4 2 ⎟⎜ 3 1
4 ⎟⎛0⎞ 1⎟⎜ ⎟
−1 −2 ⎝ 1 1
⎛111⎞
x(n)=⎝5 5n
1⎠⎜ 4 −2
−4⎟⎝1⎠
1⎠ 0 −4 2 −4
⎛15n − ⎜2
= ⎜ 1 5n + ⎜⎝2
1⎞ 2⎟
1 ⎟ . 2⎟⎠
15n −
1 22
140 3. Systems of Linear Difference Equations
In the next example we will examine the case where the matrix A has complex eigenvalues. Notice that if A is a real matrix (which we are assum- ing here) and if λ = α+iβ is an eigenvalue of A, then λ = α−iβ is also an eigenvalue of A. Moreover, if ξ is the eigenvector of A corresponding to the eigenvalue λ = α + iβ, then ξ is the eigenvector of A corresponding to the eigenvalue λ = α − iβ. Taking advantage of these observations, one may be able to simplify considerably the computation involved in finding a fundamental matrix of the system of equations (3.3.2).
Suppose that ξ = ξ1 +iξ2. A solution of system (3.3.2) may then be given by x(n) = (α + iβ)n(ξ1 + iξ2). Also, if
r= α2+β2,
β α
x(n)=[rcosθ+isinθ)]n(ξ1 +iξ2)
= rn(cos nθ + i sin nθ)(ξ1 + iξ2)
= rn[(cos nθ)ξ1 − (sin nθ)ξ2] + irn[(cos nθ)ξ2 + (sin nθ)ξ1] = u(n) + i v(n),
where u(n) = rn[(cos nθ)ξ1 − (sin nθ)ξ2] and v(n) = rn[(cos nθ)ξ2 + (sinnθ)ξ1]. One might show (Exercises 3.3, Problem 7) that u(n) and v(n) are linearly independent solutions of system (3.3.2). Hence, we do not need to consider the solution generated by λ and ξ.
Example 3.22. Find a general solution of the system x(n + 1) = Ax(n), where
1 −5 A= 1 −1 .
Solution The eigenvalues of A are λ1 = 2i, λ2 = −2i, and the corresponding eigenvectors are
then
θ = tan−1 This solution may now be written as
.
⎛1 2⎞ ⎛1 2⎞ ξ1 =⎝5−5i⎠, ξ2 =⎝5+5i⎠.
11
Hence,
x(n) = (2i)n ⎝5 − 5i⎠ 1
⎛1 2⎞
3.3 The Jordan Form: Autonomous (Time-Invariant) Systems Revisited 141
is a solution. Since
i = cos π + i sin π , 22
in = cos nπ + i sin nπ , 22
this solution may be written as
⎛1 2⎞
x(n)=2n cos nπ +isin nπ ⎝5−5i⎠ 221
⎛1 nπ 2 nπ⎞ n⎜5cos 2 +5sin2⎟
=2 ⎝ nπ ⎠ cos 2
⎛−2 nπ 1 nπ⎞ n⎜5cos 2 +5sin 2 ⎟
+i2 ⎝ nπ ⎠. sin 2
Thus,
and
n⎜5cos 2 +5sin 2 ⎟ x(n)=c12 ⎝ nπ ⎠
⎛1 nπ 2 nπ⎞ n⎜5cos 2 +5sin 2 ⎟
u(n)=2 ⎝ nπ ⎠ cos 2
⎛−2 nπ 1 nπ⎞ n⎜5cos 2 +5sin 2 ⎟
v(n)=2 ⎝ nπ ⎠ sin 2
are two linearly independent solutions. A general solution may be given as ⎛1 nπ 2 nπ⎞
cos 2
⎛−2 nπ 1 nπ⎞
n⎜5cos 2 +5sin 2 ⎟ +c22 ⎝ nπ ⎠
sin 2
⎡1 2 nπ 2 1 nπ⎤
=2n⎢⎣ 5c1−5c2 cos 2 + 5c1+5c2 sin 2 ⎥⎦. nπ nπ
So far, we have discussed the solution of system (3.3.2) if the matrix A is diagonalizable. We remark here that a sufficient condition for a k×k matrix A to be diagonalizable is that it have k distinct eigenvalues (Exercises 3.3,
c1cos 2 +c2sin 2
142 3. Systems of Linear Difference Equations
Problem 20). If the matrix A has repeated roots, then it is diagonalizable if it is normal, that is to say, if AT A = AAT . (For a proof see [111].) Examples of normal matrices are:
(i) symmetric matrices (AT = A),
(ii) skew symmetric matrices (AT = −A),
(iii) unitary matrices (AT A = AAT = I). 3.3.2 The Jordan Form
We now turn our attention to the general case where the matrix A is not diagonalizable. This happens when A has repeated eigenvalues, and one is not able to generate k linearly independent eigenvectors. For example, the following matrices are not diagonalizable:
⎡2 0 0⎤ ⎡2 0 0 0⎤ 21⎢ ⎥⎢0300⎥
0 2 , ⎣ 0 2 1 ⎦ , ⎢⎣ 0 0 4 1 ⎥⎦ .
002
If a k × k matrix A is not diagonalizable, then it is akin to the so-called
Jordan form, i.e., P−1AP = J, where
J = diag(J1,J2,…,Jr), 1 ≤ r ≤ k,
(3.3.8)
(3.3.9)
and
⎜ Ji = ⎜
.
… . . .
.
⎟ . ⎟ .
⎛λi1 0…0⎞ ⎜0λi 1 0⎟
⎜ 0 0 ⎜⎝ . . . . . .
. ⎟ 1 ⎟⎠
. .
00 λi
The matrix Ji is called a Jordan block.
These remarks are formalized in the following theorem.
Theorem 3.23 (The Jordan Canonical Form). Any k × k matrix A is similar to a Jordan form given by the formula (3.3.8), where each Ji is an si × si matrix of the form (3.3.9), and ri=1 si = k.
The number of Jordan blocks corresponding to one eigenvalue λ is called the geometric multiplicity of λ, and this number, in turn, equals the number of linearly independent eigenvectors corresponding to λ.
0004
. .
3.3 The Jordan Form: Autonomous (Time-Invariant) Systems Revisited 143
The algebraic multiplicity of an eigenvalue λ is the number of times it is repeated. If the algebraic multiplicity of λ is 1 (i.e., λ is not repeated), then we refer to λ as simple. If the geometric multiplicity of λ is equal to its algebraic multiplicity (i.e., only 1 × 1 Jordan blocks correspond to λ), then it is called semisimple. For example, the matrix
⎡⎤
30000 ⎢0 2 0 0 0⎥ ⎢0 0 2 0 0⎥ ⎢⎣0 0 0 5 1⎥⎦ 00005
has one simple eigenvalue 3, one semisimple eigenvalue 2, and one eigenvalue 5, which is neither simple nor semisimple.
To illustrate the theorem, we list below the possible Jordan forms of a 3 × 3 matrix with an eigenvalue λ = 5, of multiplicity 3. In the matrix, different Jordan blocks are indicated by squares.
⎛⎞⎛⎞⎛⎞⎛⎞ ⎜00⎟000
⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎝0 0⎟⎠⎝ 0⎠⎝0 ⎠⎝ ⎠.
0010100
(a) (b) (c)
5
00
(a)
000
(b) (c) (d)
51 05
Recall that si is the order of the ith Jordan block and r is the number of Jordan blocks in a Jordan form. In (a) the matrix is diagonalizable, and we have three Jordan blocks of order 1. Thus, s1 = s2 = s3 = 1,r = 3, and the geometric multiplicity of λ is 3.
In (b) there are two Jordan blocks with s1 = 2,s2 = 1,r = 2, and the geometric multiplicity of λ is 2.
In (c) there are also two Jordan blocks with s1 = 1,s2 = 2,r = 2, and the geometric multiplicity of λ is 2. In (d) there is only one Jordan block with s1 = 3, r = 1, and the geometric multiplicity of λ is 1. The linearly independent eigenvectors corresponding to λ = 5 in (a), (b), (c), (d) are, respectively,
⎛⎞⎛⎞⎛⎞ ⎛⎞⎛⎞ ⎛⎞⎛⎞ 1001010
⎜⎝ 0 ⎟⎠ , ⎜⎝ 1 ⎟⎠ , ⎜⎝ 0 ⎟⎠ ⎜⎝ 0 ⎟⎠ , ⎜⎝ 0 ⎟⎠ ⎜⎝ 0 ⎟⎠ , ⎜⎝ 1 ⎟⎠
⎛⎞
1
⎜⎝ 0 ⎟⎠ .
0 (d)
5
510 051 005
5
51 05
5
5
144 3. Systems of Linear Difference Equations
Note that a matrix of the form
⎛⎞
λ10…0 ⎜0 λ 1 … 0⎟
AP =PJ. Let P = (ξ1, ξ2, . . . , ξk). Equating the first
formula (3.3.10), we obtain
Aξ1 =λ1ξ1,…,Aξi =λ1ξi +ξi−1,
s1 columns i=2,3,…,s1.
(3.3.10) sides in
(3.3.11)
⎜ . . . . ⎟ ⎜ ⎟
⎝0 0 0 … 1⎠ 000…λ
has only one eigenvector, namely, the unit vector e1 = (1, 0, . . . , 0)T . This shows us that the linearly independent eigenvectors of the Jordan form J given by formula (3.3.8) are
e1, es1+1, es1+s2+1, . . . , es1+s2+···+sr−1+1. Now, since P−1AP = J, then
Clearly, ξ1 is the only eigenvector of A in the Jordan chain ξ1, ξ2, . . . , ξs1 . The other vectors ξ2, ξ3, . . . , ξs1 are called generalized eigenvectors of A, and they may be obtained by using the difference equation
(A−λ1I)ξi =ξi−1, i=2,3,…,s1. (3.3.12)
Repeating this process for the remainder of the Jordan blocks, one may find the generalized eigenvectors corresponding to the mth Jordan block using the difference equation
(A−λmI)ξmi =ξmi−1, i=2,3,…,sm. (3.3.13) Now we know that An = (PJP−1)n = PJnP−1, where
0 ⎤ ⎥
Jn =⎢ NoticethatforanyJi,i=1,2,…,r,wehaveJi =λiI+Ni,where
⎛0 1 0 … 0⎞ ⎜0 0 1 0⎟
N i = ⎜⎝ . . . . . . 1 ⎟⎠ 00…0
⎡ J 1n
⎢ J2n
⎥. ⎣.⎦
..
0 Jkn
of both
3.3 The Jordan Form: Autonomous (Time-Invariant) Systems Revisited 145
is an si × si nilpotent matrix (i.e., Nir = 0 for all r ≥ si). Hence,
Jn=(λI+N)n=λnI+ n λn−1N iiii1ii
+ n λn−2N2 +···+ n λn−si+1Nsi−1 2ii si−1i i
⎛⎞ ⎜λn n λn−1 n λn−2 … n λn−si+1⎟
⎜i1i2i si−1i⎟ ⎜ ⎟
⎜ 0 λn n λn−1 … n λn−si+2⎟ ⎜ i 1i si−2i ⎟
=⎜. . . . ⎟.
⎜ . ⎜
⎟ ⎟
⎜ ⎜⎝
nλn−1 ⎟ 1i ⎟⎠
.
00 …λni
The lines inside Jin indicate that the entries in each diagonal are identical. We can now substantiate that the general solution of system (3.3.2) is
or
where
x(n) = Anc = PJnP−1c,
x(n) = PJncˆ, (3.3.15)
cˆ = P − 1 c .
Hence, a fundamental matrix of system (3.3.2) may be given by Φ(n) = PJn. Also, the state transition matrix may be given by Φ(n,n0) = PJn−n0P−1 and thus x(n,n0,x0) = PJn−n0P−1×0.
The following corollary arises directly from an immediate consequence of formula (3.3.14).
Corollary 3.24. Assuming that A is any k×k matrix, then limn→∞ An = 0 if and only if |λ| < 1 for all eigenvalues λ of A.
Proof. (Exercises 3.3, Problem 21.) P
The importance of the preceding corollary lies in the fact that if limn→∞ An = 0, then limn→∞ xn = limn→∞ Anx(0) = 0. This fact re- minds us that if |λ| < 1 for all eigenvalues of A, then all solutions x(n) of (3.3.1) tend toward the zero vector as n → ∞.
..
.
(3.3.14)
146 3. Systems of Linear Difference Equations
Example 3.25. Find the general solution of x(n + 1) = Ax(n) with ⎛⎞
412
A = ⎜⎝ 0 2 − 4 ⎟⎠ .
016
Solution Note that this example uses conclusions from Example 3.3. The eigenvalues are λ1 = λ2 = λ3 = 4. To find the eigenvectors, we solve the equation (A − λI)ξ = 0, or
Hence,
⎛ ⎞⎛⎞⎛⎞
012d1 0 ⎜⎝0 −2 −4⎟⎠ ⎜⎝d2⎟⎠ = ⎜⎝0⎟⎠ .
012d3 0
d2 + 2d3 = 0, −2d2 − 4d3 = 0, d2 + 2d3 = 0.
These equations imply that d2 = −2d3, thus generating two eigenvectors, ⎛⎞ ⎛⎞
01
ξ1 = ⎜⎝−2⎟⎠ and ξ2 = ⎜⎝−2⎟⎠ .
11
We must now find one generalized eigenvector ξ3. Applying formula (3.3.11), let us test (A − 4I)ξ3 = ξ1:
⎛ ⎞⎛⎞⎛⎞
012a1 0
⎜⎝0 −2 −4⎟⎠ ⎜⎝a2⎟⎠ = ⎜⎝−2⎟⎠ .
012a3 1
This system is an inconsistent system that has no solution. The second
attempt will use
or
(A−4I)ξ3 =ξ2,
⎛ ⎞⎛⎞⎛⎞
012a1 1
⎜⎝0 −2 −4⎟⎠ ⎜⎝a2⎟⎠ = ⎜⎝−2⎟⎠ .
012a3 1
Hence, a2 + 2a3 = 1. One may now set ⎛⎞
0
ξ 3 = ⎜⎝ − 1 ⎟⎠ .
1
3.3 The Jordan Form: Autonomous (Time-Invariant) Systems Revisited 147
Thus,
where
1
3/2 1/2 1/2
A = ⎜⎝1/2 5/2 −1/2⎟⎠ .
012
⎛⎞
010
P = ⎜⎝ − 2 − 2 − 1 ⎟⎠ ,
111
⎛⎞
400
J = ⎜⎝ 0 4 1 ⎟⎠ ,
004
⎛4n 0 0 ⎞ J n = ⎜⎝ 0 4 n n 4 n − 1 ⎟⎠ .
0 0 4n
⎛04n n4n−1⎞⎛cˆ⎞
x(n)=PJncˆ=⎜⎝−2·4n −2·4n −4n 4n
and
Hence,
Example 3.26. Solve the system x(n + 1) = Ax(n),
1 −2n4n−1 −4n⎟⎠⎜⎝cˆ2⎟⎠.
⎛⎞
Solution The eigenvalues of A are λ1 = λ2 = λ3 = 2. We have a sole eigenvector,
⎛⎞
1
ξ 1 = ⎜⎝ 0 ⎟⎠ .
1
We now need to compose two generalized eigenvectors, using (3.3.13):
⎛⎞
1
( A − 2 I ) ξ 2 = ξ 1 g i v e s ξ 2 = ⎜⎝ 1 ⎟⎠
n4n−1 + 4n cˆ3 ⎛⎞
1 x(0) = ⎜⎝1⎟⎠ ,
2
148 3. Systems of Linear Difference Equations
and
So
⎛⎞
1 (A−2I)ξ3 =ξ2 givesξ3 =⎜⎝2⎟⎠.
1
⎛⎞⎛⎞
111 210
P = ⎜⎝ 0 1 2 ⎟⎠ , J = ⎜⎝ 0 2 1 ⎟⎠ ,
121 002 ⎛ n(n−1) ⎞
Now,
x(n,x0) = PJnP−1x0
⎛n2 −5n+16
= 2 n − 4 ⎜⎝ 4 n n2 − n
n2 +3n 4 n + 1 6 n2 + 7n
⎞⎛1⎞ ⎟⎠ ⎜⎝ 1 ⎟⎠
2n n2n−1 2n−2 n⎜ 2⎟
J =⎝0 2n n2n−1 0 0 2n
⎠.
−n2 +5n − 4 n
⎛n2 +3n+16⎞ = 2 n − 4 ⎜⎝ 4 n + 1 6 ⎟⎠ .
n2 +7n+16 3.3.3 Block-Diagonal Matrices
−n2 + n + 16 1
In general, the generalized eigenvectors corresponding to an eigenvalue λ of algebraic multiplicity m are the solutions of the equation
(A − λI)mξ = 0. (3.3.16) The first eigenvector ξ1 corresponding to λ is obtained by solving the
equation
(A − λI)ξ = 0.
The second eigenvector or generalized eigenvector ξ2 is obtained by solving
the equation
(A − λI)2ξ = 0.
And so on.
NowifJ istheJordanformofA,thatis,P−1AP =J orA=PJP−1,
then λ is an eigenvalue of A if and only if it is an eigenvalue of J. Moreover, if ξ is an eigenvector of A, then ξ ̃= P−1ξ is an eigenvector of J.
3.3 The Jordan Form: Autonomous (Time-Invariant) Systems Revisited 149
We would like to know the structure of the eigenvectors ξ ̃ of J . For this we appeal to the following simple lemma from Linear Algebra.
A0
Lemma 3.27. Let C = 0 B be a k × k block-diagonal matrix such
that A is an r×r matrix and B is an s×s matrix, with r+s=k. Then the following statements hold true:
(i) If λ is an eigenvalue of A, then it is an eigenvalue of C. Moreover, the eigenvector and the generalized eigenvectors corresponding to λ are of the form ξ = (a1,a2,...,ar,0,...,0)T for some ai ∈ R.
(ii) If λ is an eigenvalue of B, then it is an eigenvalue of C. Moreover, the eigenvector and the generalized eigenvectors corresponding to λ are of the form ξ = (0,...,0,ar+1,ar+2,...,as) for some ar+i ∈ R.
Proof.
(i) Suppose that λ is an eigenvalue of A, and V = (a1,a2,...,ar)T is
the corresponding eigenvector. Define ξ = (a1, . . . , ar, 0, . . . , 0) ∈ Rk.
Then clearly Cξ = λξ, and thus λ is an eigenvalue of C. Let the k × k
identity matrix I be written in the form I = Ir 0 , where Ir and 0 Is
Is are, respectively, the r × r and s × s identity matrices. Let λ be an eigenvalue of A with algebraic multiplicity m. Then
Hence
(C − λI )ξ =
=
⎜0⎟ .
⎜.⎟ ⎝.⎠
has a nontrivial solution
ξ ̃ = ⎜ ⎜ . . ⎟ ⎟ .
A − λIr 0
⎛ξ⎞ 1
⎜.⎟ ⎜ . ⎟
⎜ξr ⎟ ⎜ . ⎟
⎛0⎞ ⎜.⎟
0
B − λIS
⎛⎞⎛⎞
⎜.⎟
ξ1 0 (A−λI )⎜.⎟=⎜.⎟
r ⎝.⎠ ⎝.⎠ ξr 0
⎛⎞
a1
⎝.⎠ ar
⎝.⎠
ξs 0
150
3. Systems of Linear Difference Equations
(ii)
However
has only the trivial solution
⎛⎞
2−1 1
, x(0) = .
042
10
12
⎛⎞⎛⎞
230 0
3. A=⎜⎝4 3 0⎟⎠, x(0)=⎜⎝1⎟⎠.
006 0
⎛⎞
2 −1 0
4. A = ⎜⎝0 4 0⎟⎠.
253
⎛⎞
101
5. A=⎜⎝1 2 3⎟⎠.
003
⎛⎞⎛⎞
110 1
6. A=⎜⎝−1 1 0⎟⎠, x(0)=⎜⎝0⎟⎠.
1. A =
2. A =
.
101 1
⎛⎞⎛⎞
ξr+1 0 (B−λI )⎜ . ⎟=⎜.⎟
s⎝.⎠ ⎝.⎠ ξs 0
0 ⎜.⎟ .
⎝.⎠ 0
Then ξ = (a1,...,ar,0,...,0)T is an eigenvector of C corresponding to λ. The same analysis can be done for generalized eigenvectors by solving (C − λI)iξ = 0, 1 ≤ i ≤ m.
The proof of the second part is analogous and will be omitted. P Exercises 3.3
In Problems 1 through 6, use formula (3.3.6) to find the solution of x(n + 1) = Ax(n), where A is given in the exercise.
3.3 The Jordan Form: Autonomous (Time-Invariant) Systems Revisited 151
7. Suppose that x(n) = u(n) + iv(n) is a solution of (3.3.2), where u(n) and v(n) are real vectors. Prove that u(n) and v(n) are linearly independent solutions of (3.3.2).
8. Utilize Problem 7 to find a fundamental matrix of x(n + 1) = Ax(n) with
⎛1 1 0 1⎞ ⎜−1 1 0 1⎟
A=⎜⎝0 0 2 1⎟⎠. 0 0 −1 2
9. Apply Problem 7 to find a fundamental matrix of x(n + 1) = Ax(n) with
⎛⎞
110
A = ⎜⎝ − 1 1 0 ⎟⎠ .
101
10. Find the eigenvalues and the corresponding eigenvectors and general-
ized eigenvectors for the matrix A.
⎛⎞
210 (b) A = ⎜⎝0 2 1⎟⎠ .
002
⎛2 0 0 0⎞ ⎜0 2 1 0⎟
(a) A =
3 1 . 03
⎛4 23⎞ ⎜ 1 ⎟
(c)A=⎝−2 2 0⎠. (d)A=⎜⎝0 0 2 1⎟⎠.
003
11. Find An for the matrices in Problem 10 using the Jordan form.
12. Use the Jordan form to solve x(n + 1) = Ax(n) with ⎛⎞
321 A=⎜⎝−1 3 2⎟⎠.
1 −3 −2
13. Use the Jordan form to solve x(n + 1) = Ax(n) with
⎛⎞
323
A = ⎜⎝ − 1 / 2 1 0 ⎟⎠ .
002
0002
152 3. Systems of Linear Difference Equations
14. Let A and B be two similar matrices with P−1AP = B.
(i) Show that A and B have the same eigenvalues.
(ii) Show that if ξ is an eigenvector of B, then Pξ is an eigenvector of A.
15. Suppose that P−1AP = D = diag(λ1,λ2,...,λk), where P = [ξ1,ξ2,...,ξk] is a nonsingular k×k matrix. Show that ξ1,ξ2,...,ξk are the eigenvectors of A that correspond to the eigenvalues λ1, λ2, . . . , λk, respectively.
16. Let A be a 4 × 4 matrix with an eigenvalue λ = 3 of multiplicity 4. Write all possible Jordan forms of A.
17. Show that (PJP−1)n = PJnP−1.
18. If λ is an eigenvalue of A, and ξ is the corresponding eigenvector of A,
show that λnξ is a solution of (3.3.2).
19. Let
Then one may write A = λ I + N, where
⎛0 1 ... 0⎞
⎜ .⎟ N=⎜0 0 .⎟.
⎜⎝.. 1⎟⎠ 000
Show that for any α > 0, A is similar to a matrix
⎛λ α … 0⎞
⎜0 λ ⎟ B=λI+αN=⎜. . ⎟.
⎝.. α⎠ 00λ
20. Prove that if a k × k matrix A has k distinct eigenvalues, then: (i) A has k linearly independent eigenvectors.
(ii) A is diagonalizable. (Use mathematical induction.)
21. Prove Corollary 3.24.
⎛⎞
λ1…0 ⎜0 λ … 0⎟
A = ⎜. . .⎟. ⎜ ⎟
⎝0 0 … 1⎠ 00…λ
22.
Consider the companion matrix A of (3.2.17) with the coefficients pi constant. Assume that the eigenvalues of A are real and distinct. Let V denote the Vandermonde matrix
⎛11…1⎞
23.
3.4
12k Show that V −1AV is a diagonal matrix.
Consider the companion matrix A (3.3.16) with pi(n) constants. Sup- pose λ1, λ2, . . . , λr are the distinct eigenvalues of A with multiplicities m1, m2, . . . , mr and ri=1 mi = k. Let V be the generalized Vander- monde matrix (2.3.9). Show that V −1AV = J, where J is in the Jordan form (3.3.8).
Linear Periodic Systems
⎜ λ1 λ2 … λk ⎟ V=⎜ . . . ⎟.
In this section we regard the linear periodic system
x(n + 1) = A(n)x(n), (3.4.1)
where for all n ∈ Z, A( + N) = A(), for some positive integer N .
We now show that the study of the periodic system (3.4.1) simplifies to the study of an associated autonomous system. This inference is the analogue of Floquet theory in differential equations. But before we prove
that analogue, we need the following theorem.
Lemma 3.28. Let B be a k × k nonsingular matrix and let m be any positive integer. Then there exists some k×k matrix C such that Cm = B.
Proof. Let
be the Jordan form of B. Let us write 1
Ji=λi Ii+λNi , i
3.4 Linear Periodic Systems 153
⎝…⎠ λk−1 λk−1 . . . λk−1
⎛J1 ⎞ ⎜ J2 ⎟
P−1BP=J=⎜⎝ …⎟⎠ Jr
154 3. Systems of Linear Difference Equations
where Ii is the si × si identity matrix and ⎛⎞
Observe that
00
Nsi = 0. i
⎜
Ni=⎜ ⎜
0
1 ⎟ … … ⎟.
To motivate our construction, we formally write 1
Hi=exp mlnJi
1 1
=exp m lnλiIi+ln Ii+λNi
1
= exp
Applying formula (3.4.2), we obtain
010
⎟ ⎝ 1⎠
(3.4.2)
i ∞ (−1)s+1 N s
m
ln λiIi +
i s=1 s λi
.
si−1 s
1 (−1)s+1
Hi=exp m lnλiIi+
s=1
Ni
λ .
(3.4.3)
P
s
Hence, Hi is a well-defined matrix. Furthermore, Him = Ji.
Now, if we let
i
H=⎜ .. ⎟, ⎝.⎠
0 Hr
where Hi is defined in formula (3.4.3), then
⎡H1m 0⎤
⎛H1 0⎞ ⎜ H2 ⎟
⎢ H2m Hm =⎢
⎥
.. ⎥=J.
⎣.⎦ 0 Hrm
Define C = PHP−1. Then Cm = PHmP−1 = PJP−1 = B.
Armed with this lemma, we are now prepared to introduce the primary
result for this section.
Lemma 3.29. For system (3.4.1), the following statements hold: (i) If Φ(n) is a fundamental matrix, then so is Φ(n + N ).
(ii) Φ(n + N) = Φ(n)C, for some nonsingular matrix C. (iii) Φ(n+N,N)=Φ(n,0).
Proof.
(i) Let Φ(n) be a fundamental matrix of system (3.4.1). Then Φ(n + 1) = A(n)Φ(n). Now
Φ(n + N + 1) = A(n + N)Φ(n + N) = A(n)Φ(n + N)
Hence Φ(n + N ) is also a fundamental matrix of system (3.4.1).
(ii) Observe that Ψ1(n, n0) = Φ(n + N)Φ−1(n0 + N) and Ψ2(n, n0) = Φ(n)Φ−1(n0) are fundamental matrices of system (3.4.1) with the same initial condition Ψ1(n0,n0) = Ψ2(n0,n0) = I. By the unique-
ness of fundamental matrices (Theorem 3.7) Ψ1(n, n0) = This implies that
Φ(n + N) = Φ(n)Φ−1(n0)Φ(n0 + N) = Φ(n)C
(iii) This is left to the reader as Problem 1.
Ψ2(n, n0).
P
There are many consequences of this lemma, including the following theorem.
Theorem 3.30. For every fundamental matrix Φ(n) of system (3.4.1), there exists a nonsingular periodic matrix P(n) of period N such that
(3.4.4)
Proof. By Lemma 3.28, there exists some matrix B such that BN = C, where C is the matrix specifed in Lemma 3.29(ii). Define P(n) = Φ(n)B−n, where B−n = (Bn)−1. Then
P (n + N ) = Φ(n + N )B−N B−n
= Φ(n)CB−N B−n [using part (ii) of Lemma 3.29]
= Φ(n)B−n = P(n).
We now know that P(n) has period N and is clearly nonsingular. (Why?) From the definition of P(n) it thus follows that Φ(n) = P(n)Bn. P
Remark: If z(n) is a solution of the system
z(n + 1) = Bz(n), (3.4.5)
3.4 Linear Periodic Systems 155
Φ(n) = P(n)Bn.
156 3. Systems of Linear Difference Equations
then
or
x(n) = Φ(n)c = P (n)Bnc,
x(n) = P (n)z(n). (3.4.6)
The value of this remark lies in the fact that the qualitative study of the periodic system of equations (3.4.1) reduces to the study of the autonomous system (3.4.5).
The matrix C = BN , which may be found using Lemma 3.29 part (ii), is referred to as a monodromy matrix of (3.4.1). The eigenvalues λ of B are called the Floquet exponents of (3.4.1); the corresponding eigenvalues λN of BN are called the Floquet multipliers of (3.4.1). The reason we call λN a multiplier is that there exists a solution x(n) of (3.4.1) such that x(n+N) = λN x(n). (See Exercises 3.4, Problem 9.) Notice that the Floquet exponents (multipliers) do not depend upon the monodromy matrix chosen, that is, they do not hinge upon the particular fundamental matrix Φ(n) used to define the monodromy matrix. The following lemma explicitly states this truth.
Lemma 3.31. If Φ(n) and Ψ(n) are two fundamental matrices of (3.4.1) such that
Φ(n + N) = Φ(n)C, Ψ(n + N) = Ψ(n)E,
then C and E are similar (and thus they have the same eigenvalues). Proof. The reader will prove this lemma in Exercises 3.4, Problem
2. P
Lemma 3.32. A complex number λ is a Floquet exponent of (3.4.1) if and only if there is a nontrivial solution of (3.4.1) of the form λnq(n), where q(n) is a vector function with q(n + N) = q(n) for all n.
Proof. First, we assume that λ is a Floquet exponent of (3.4.1). Then, we also know that det(Bn − λnI) = 0. Now choose x0 ∈ Rk, x0 ̸= 0, such that (Bn − λnI)x0 = 0 for all n. (Why?) (See Exercises 3.4, Problem 4.) Hence, we have the equation Bnx0 = λnx0.
Thus, P (n)Bnx0 = λnP (n)x0, where P (n) is the periodic matrix defined in formula (3.4.4). By formula (3.4.4) now,
x(n, n0, y0) = Φ(n, n0)x0 = P (n)Bnx0 = λnP (n)x0 = λnq(n),
and we have the desired periodic solution of (3.4.1), where q(n) = P(n)x0. Conversely, if λnq(n),q(n+N) = q(n) ̸= 0 is a solution of (3.4.1), Theorem 3.30 then implies that
λnq(n) = P (n)Bnx0 (3.4.7)
for some nonzero vector x0. This implies that λn+N q(n) = P (n)Bn+N x0.
But, from (3.4.7),
(3.4.8)
λn+N q(n) = λN P (n)Bnx0.
Equating the right-hand sides of formulas (3.4.8) and (3.4.9), we obtain
P(n)Bn[BN −λNI]x0 =0,
and thus
det[BN −λNI]=0.
This manipulation shows that λ is a Floquet exponent of (3.4.1). P
Using the preceding theorem, one may easily conclude the following results.
Corollary 3.33. The following statements hold:
(i) System (3.4.1) has a periodic solution of period N if and only if it has
a Floquet multiplier equal to 1.
(ii) There is a Floquet multiplier equal to −1 if and only if system (3.4.1)
has a periodic solution of period 2N.
Proof. Use Lemma 3.32 as you prove Corollary 3.33 in Exercises 3.4,
Problem 3. P
Remark: Lemma 3.29, part (ii), gives us a formula to find the monodromy matrix C = BN , whose eigenvalues happen to be the Floquet multipliers of (3.4.1). From Lemma 3.29,
By letting n = 0, we have
C = Φ−1(n)Φ(n + N).
C = Φ−1(0)Φ(N). (3.4.10)
If we take Φ(N) = A(N−1)A(N−2)···A(0), then Φ(0) = I. Thus, formula (3.4.10) becomes
or
C = Φ(N),
C = A(N − 1)A(N − 2) · · · A(0). (3.4.11) We now give an example to illustrate the above results.
3.4 Linear Periodic Systems
157
(3.4.9)
158 3. Systems of Linear Difference Equations
Example 3.34. Consider the planar system
x(n + 1) = A(n)x(n),
0 (−1)n A(n) = (−1)n 0 .
Clearly, A(n + 2) = A(n) for all n ∈ Z. Applying formula (3.4.10),
B2 =C=A(1)A(0)=
−1 0 . 0 −1
Thus the Floquet multipliers are −1, −1. By virtue of Corollary 3.33, the system has a 4-periodic solution. Note that since A(n) has the constant eigenvalues −1, 1, ρ(A(n)) = 1.
The above example may suggest that there is some kind of relationship between the eigenvalues of A(n) and its Floquet multipliers. To dispel any such thoughts we offer the following example.
Example 3.35.
20
This is a system of period 2. The eigenvalues of A are ± 3, and hence
Consider system (3.2.1) with
⎛ 2 + (−1)n ⎞
⎜0 2⎟ A(n) = ⎝ 2 − (−1)n ⎠ .
√ √2
ρ(A)= 3 <1.Now, 2
⎛1 ⎞
B2=C=A(1)A(0)=⎜⎝4 0⎟⎠. 09
4
Thus, the Floquet multipliers are 1 and 9 . Hence, ρ(B) = 3 .
Exercises 3.4
1. Prove Lemma 3.29 (iii).
2. Prove Lemma 3.31.
3. Prove Corollary 3.33.
4. Suppose that (B − λI)x0 = 0 for some x0 ∈ Rk,x0 ̸= 0. Prove that (Bn − λnI)x0 = 0 for all n ∈ Zt.
5. Let a1(n),a2(n) be N-periodic functions and let Ψ1(n),Ψ2(n) be solutions of
442
x(n + 2) + a1(n)x(n + 1) + a2(n)x(n) = 0 (3.4.12)
such that Ψ1(0) = 1, Ψ1(1) = 0, Ψ2(0) = 0, and Ψ2(1) = 1. Show that the Floquet multipliers satisfy the equation λ2 + bλ + c = 0, where
N−1
b = −[Ψ1(N) + Ψ2(N + 1)], c =
6. In Problem 5, let a2(n) ≡ 1. Show that the product of the Floquet
multipliers is equal to 1.
7. In Problem 5, let a2(n) ≡ 1. Show that if b = 2, there is at least one solution of period 2N while for b = −2 there is at least one solution of period N.
8. In Problem 5 it is clear that if λ = 1, then (3.4.12) has a periodic solution of period N. Show that x(n+2)+a1(n)x(n+1)+a2(n)x(n) = 0 has a periodic solution of period 2N if and only if λ = −1.
9. Show that there exists a solution x(n) of (3.4.1) that satisfies x(n + N ) = λx(n) if and only if λ is a Floquet multiplier.
3.5 Applications
3.5.1 Markov Chains
In 1906 the Russian mathematician A.A. Markov developed the concept of Markov chains. We can describe a Markov chain as follows: Suppose that we conduct some experiment with a set of k outcomes, or states, S = {s1, s2, . . . , sk}. The experiment is repeated such that the probability (pij ) of the state si, 1 ≤ i ≤ k, occurring on the (n+1)th repetition depends only on the state sj occurring on the nth repetition of the experiment. In other words, the system has no memory: The future state depends only on the present state. In probability theory language, pij = p(si|sj) is the probability of si occurring on the next repetition, given that sj occurred on the last repetition. Given that sj has occurred in the last repetition, one ofs1,s2,...,sk mustoccurinthenextrepetition.Thus,
p1j +p2j +p3j +···+pkj =1, 1≤j≤k. (3.5.1)
Let pi(n) denote the probability that the state si will occur on the nth repetition of the experiment, 1 ≤ i ≤ k. Since one of the states si must occur on the nth repetition, it follows that
p1 (n) + p2 (n) + · · · + pk (n) = 1. (3.5.2)
To derive a mathematical model of this experiment, we must define pi(n+ 1), 1 ≤ i ≤ k, as the probability that the state si occurs on the (n + 1)th repetition of the experiment. There are k ways that this can happen. The first case is where repetition n gives us s1, and repetition (n + 1) produces
3.5 Applications 159
i=0
a2(i).
160 3. Systems of Linear Difference Equations
si. Since the probability of getting s1 on the nth repetition is p1(n), and the probability of having si after s1 is pi1, it follows (by the multiplication principle) that the probability of the first case occurring is pi1p1(n). The second case is where we get s2 on repetition n and si on repetition (n + 1). The probability of the occurrence of the second case is pi2p2(n). Repeating this for cases 3,4,...,k, and for i = 1,2,...,k, we obtain the k-dimensional system
p1(n + 1) = p11p1(n) + p12p2(n) + · · · + p1kpk(n), p2(n + 1) = p21p1(n) + p22p2(n) + · · · + p2kpk(n),
.
pk(n + 1) = pk1p1(n) + pk2p2(n) + · · · + pkkpk(n),
or, in vector notation,
p(n + 1) = Sp(n), n = 1, 2, 3 . . . , (3.5.3)
where p(n) = (p1 (n), p2 (n), . . . , pk (n))T is the probability vector and S = (pij ) is a k × k transition matrix.
The matrix S belongs to a special class of matrices called Markov matri- ces. A matrix A = (aij) is said to be nonnegative (positive) if aij ≥ 0 (> 0) for all entries aij of A. A nonnegative k × k matrix A is said to be Markov (or stochastic) if ki=1 aij = 1 for all j = 1,2,…,k. It follows from Ta- ble 4.1 that ∥A∥1 = 1, which by inequality (4.1.3) implies that p(A) ≤ 1. Hence |λ| ≤ 1 for all the eigenvalues λ of a Markov matrix. Furthermore, λ = 1 is an eigenvalue of a Markov matrix (Exercises 3.5, Problem 3). Hence p(A) = 1 if A is Markov.
3.5.2 Regular Markov Chains
A regular Markov chain is one in which Sm is positive for some positive integer m. To give a complete analysis of the eigenvalues of such matrices, we need the following theorem, due to O. Perron.
Theorem 3.36 (Perron’s Theorem). Let A be a positive k×k matrix. Then ρ(A) is a simple real eigenvalue (not repeated) of A. If λ is any other eigenvalue of A, then |λ| < ρ(A). Moreover, an eigenvector associated with ρ(A) may be assumed to be positive.
Suppose now that S is the transition matrix of a regular Markov chain with eigenvalues λ1, λ2, . . . , λk. Then ρ(S) = 1. If Sm is positive, then ρ(Sm) = 1. As a matter of fact, the eigenvalues of Sm are λm1 ,λm2 ,...,λmk . By Perron’s theorem, 1 is a simple eigenvalue of Sm. Consequently, S has exactly one simple eigenvalue, say λ1, which equals 1; all other eigenvalues
satisfy |λ | < 1,i = 2,3,...,k. Hence, the Jordan form of S must be of the i
10
form J = 0 J , where the eigenvalues of J∗ are λ2,λ3,...,λk.
∗
By Corollary 3.24, J∗n → 0 as n → ∞, so that Jn → diag(1,0,...,0) as
n → ∞. Therefore, if S = QJQ−1, we have
lim p(n) = lim Snp(0) = lim QJnQ−1p(0) = (ξ1,0,0,...,0)η = aξ1,
n→∞ n→∞ n→∞
(3.5.4) where ξ1 = (ξ11, ξ21, . . . , ξk1)T is the eigenvector of S that corresponds to the eigenvalue λ1 = 1, and a is the first component of η = Q−1p(0). Since finding the matrix Q is not a simple task, we will choose instead to devise
a very easy method to find the constant a. Recall that for p(n) = (p1(n), p2(n), . . . , pk(n))T
we have, from formula (3.5.2), ki=1 pi(n) = 1. Since limn→∞ p(n) = aξ1, it follows that
Therefore,
aξ11 +aξ21 +···+aξk1 =1.
a=1. ξ11 +ξ21 +···+ξk1
3.5 Applications 161
The following example illustrates a regular Markov chain.
Example 3.37. The simplest type of genetic inheritance of traits in ani- mals occurs when a certain trait is determined by a specific pair of genes, each of which may be of two types, say G and g. An individual may have a GG combination, a Gg (which is genetically the same as gG), or a gg combination. An individual with GG genes is said to be dominant; a gg individual is referred to as recessive; a hybrid has Gg genes.
In the mating of two animals, the offspring inherits one gene of the pair from each parent: The basic assumption of genetics is that the selection of these genes is random.
Let us consider a process of continued matings. We begin with an individ- ual of known genetic character (GG) and mate it with a hybrid. Assuming that there is one offspring, we mate that offspring with a hybrid, repeating this process through a number of generations. In each generation there are three possible states, s1 = GG,s2 = Gg, and s3 = gg. Let pi(n) represent the probability that state si occurs in the nth generation and let pij be the probability that si occurs in the (n+1)th generation given that sj occurred in the nth generation.
162 3. Systems of Linear Difference Equations
The difference system that models this Markov chain is denoted by
p1(n + 1) = p11p1(n) + p12p2(n) + p13p3(n), p2(n + 1) = p21p1(n) + p22p2(n) + p23p3(n), p3(n + 1) = p31p1(n) + p32p2(n) + p33p3(n).
Now, p11 is the probability of producing an offspring GG by mating GG
and Gg. Clearly, the offspring receives a G gene from his parent GG with
probability 1 and the other G from his parent Gg with probability 1 . By 112
the multiplication principle, p11 = 1 × 2 = 2 . The probability of creating
an offspring GG from mating a Gg with a Gg is p12. By similar analysis
one may show that p12 = 1 × 1 = 1 . Likewise, p13 is the probability of 224
generating an offspring GG from mating a gg with a Gg. Obviously, p13 = 0. One may show by the same process that
p21 = 1, p22 = 1, p23 = 1, p31 = 0, p32 = 1, p33 = 1. 222 42
Hence, we have
with
p(n + 1) = Sp(n) ⎛⎞
0.5 0.25 0
S = ⎜⎝0.5 0.5 0.5⎟⎠ .
0 0.25 0.5
Notice that all the entries for S2 are positive, and thus this is a regular
Markov chain. The eigenvalues of S are λ1 = 1, λ2 = 1 , and λ3 = 0. Recall 2
from formula (3.5.4) that
Now,
and
implies that
lim p(n) = aξ1. n→∞
⎛⎞
1
ξ 1 = ⎜⎝ 2 ⎟⎠
1
a=1 4
⎛⎞
0.25
lim p(n) = ⎜⎝ 0.5 ⎟⎠. n→∞
0.25
This relation dictates that as the number of repetitions approaches infin- ity, the probability of producing a purely dominant or a purely recessive offspring is 0.25, and the probability of creating a hybrid offspring is 0.5.
3.5.3 Absorbing Markov Chains
A state si in a Markov chain is said to be absorbing if whenever it occurs on the nth repetition of the experiment, it then occurs on every subsequent repetition. In other words, if for some i, pii = 1, then pij = 0 for j ̸= i. A Markov chain is said to be absorbing if it has at least one absorbing state and if from every state it is possible to go to an absorbing state. In an absorbing Markov chain, a state that is not absorbing is called transient.
Example 3.38. Drunkard’s Walk
A man walks along a four-block stretch. He starts at corner x. With prob- ability 1 he walks one block to the right, and with probability 1 he walks
one block to the left. When he comes to the next corner he again randomly chooses his direction. He continues until he reaches corner 5, which is a bar, or corner 1, which is his home. (See Figure 3.2.) If he reaches either home or the bar, he stays there. This is clearly an absorbing Markov chain.
Let us rename the states so that the absorbing states at 1 and 5 are last, and so we refer to them as s4 and s5. The transient states 2, 3, and 4 will be called s1, s2, and s3, respectively. Accordingly, p1(n), p2(n), p3(n), p4(n), and p5(n) will be, respectively, the probabilities of reaching s1,s2,s3,s4, and s5 after n walks. The difference equation that represents this Markov chain is p(n + 1) = Sp(n), where the transition matrix is
⎛1⎞ ⎜020|00⎟
22
3.5 Applications 163
⎜1 1 ⎟
⎜2 ⎜0
0 2 | 0 0⎟ 1 0 | 0 0⎟ T 0
S = ⎜
⎜1 0 0 | 1 0⎟
2 ⎟=QI. ⎜... ... ... ... ... ...⎟
⎜⎝ 2 ⎟⎠ 001|01
2
Let u(n) = (p1(n), p2(n), p3(n))T and v(n) = (p4(n), p5(n))T . Then
u(n+1) T 0 u(n) =,
v(n+1) Q I v(n)
164 3. Systems of Linear Difference Equations
Therefore,
v(n) = v(0) + The eigenvalues of T are
r=0
u(n + 1) = T u(n), v(n + 1) = v(n) + Qu(n).
(3.5.5) (3.5.6)
(3.5.7)
u(n) = Tnu(0).
Substituting from formula (3.5.7) into formula (3.5.6) yields
v(n + 1) = v(n) + QT nu(0). n−1
0, − 1, 1. 22
(3.5.8) By formula (3.2.14), it follows that the solution of (3.5.8) is given by
QT r u(0).
(3.5.9)
Hence, by Corollary 3.24, limn→∞ T n = 0. In this case one may show that ∞ n−1
Now,
r r −1 T = lim T =(I−T)
n→∞
(Exercises 3.5, Problem 5). Using formula (3.5.9), we generate
r=0
lim v(n) = v(0) + Q(I − T )−1u(0). n→∞
⎛3 1 1⎞
⎜2 2⎟ = ⎜⎝ 1 2 1 ⎟⎠ .
113 22
1– 22
s1 s2 s3 s4 s5
s4 s1 s2 s3 s5
FIGURE 3.2. Drunkard’s walk.
r=0
−1 ( I − T )
1–
BAR
HOME
Assume that the man starts midway between home and the bar, that is,
at state s2. Then
and
In this case
⎛⎞
0 u(0) = ⎜⎝1⎟⎠
0
0 0
⎛1
⎞⎛3 1 1⎞⎛0⎞ ⎛1⎞
0⎟⎜2 2⎟⎜ ⎟ ⎜2⎟ ⎠⎜1 2 1⎟⎝1⎠=⎝ ⎠.
v(0) =
.
3.5 Applications 165
⎜2 0 lim v(n)=⎝
n→∞ 001⎝13⎠ 1 2102
22
Thus, the probability that the man ends up at his home is 0.5. The proba-
bility that he ends up at the bar is also 0.5. Common sense could probably have told us this in the first place, but not every situation will be this simple.
3.5.4 A Trade Model
Example 3.39. Consider a model of the trade between two countries, restricted by the following assumptions:
(i) National income = consumption outlays + net investment + exports − imports.
(ii) Domestic consumption outlays = total consumption − imports.
(iii) Timeisdividedintoperiodsofequallength,denotedbyn=0,1,2,....
Let, for country j = 1, 2,
yj(n) = national income in period n, cj(n) = total consumption in period n, ij(n) = net investment in period n,
xj (n) = exports in period n, mj (n) = imports in period n,
dj(n) = consumption of domestic products in period n. For country 1 we then have
y1(n) = c1(n) + i1(n) + x1(n) − m1(n), d1(n) = c1(n) − m1(n),
166 3. Systems of Linear Difference Equations
which, combining those two equations, gives
y1(n) = d1(n) + x1(n) + i1(n). (3.5.10)
Likewise, for country 2, we have
y2(n) = d2(n) + x2(n) + i2(n). (3.5.11)
We now make the following reasonable assumption: The domestic consump- tion dj (n) and the imports mj (n) of each country at period (n + 1) are proportional to the country’s national income yi(n) one time period earlier. Thus,
d1(n + 1) = a11y1(n), m1(n + 1) = a21y1(n), (3.5.12) d2(n + 1) = a22y2(n), m2(n + 1) = a12y2(n). (3.5.13)
The constants aij are called marginal propensities. Furthermore, aij > 0, for i,j = 1,2. Since we are considering a world with only two countries, the exports of one must be equal to the imports of the other, i.e.,
m1(n) = x2(n), m2(n) = x1(n). (3.5.14) Substituting from equations (3.5.12), (3.5.13), and (3.5.14) into (3.5.10)
and (3.5.11) leads to
y1(n + 1) = a11 a12 y1(n) + i1(n + 1) . (3.5.15) y2(n + 1) a21 a22 y2(n) i2(n + 1)
Let us further assume that the net investments i1(n) = i1 and i2(n) = i2
are constants. Then (3.5.15) becomes
y1(n + 1) = a11 a12 y1(n) + i1 . y2(n + 1) a21 a22 y2(n) i2
By the variation of constants formula (3.2.14), we obtain
(3.5.16)
(3.5.17)
n−1
n n−r−1
n
n−1 r
y(n)=A y(0)+
r=0
A
I =A y(0)+
A I,
where I = (i1 , i2 )T . To have a stable economy, common sense dictates that the sum of the domestic consumption dj (n + 1) and the imports mj (n + 1) in period (n + 1) must be less than the national income yj (n) in period n; that is,
or
dj (n + 1) + mj (n + 1) < yj (n), j = 1, 2,
a11 +a21 < 1, a12 +a22 < 1. (3.5.18)
Under conditions (3.5.18), one may show that for all the eigenvalues λ of A, |λ| < 1 (Exercises 3.5, Problem 4).
r=0
This implies from Corollary 3.24 that An → 0 as n → ∞. This fact fur- ther generates the so-called Neumann’s expansion (Exercises 3.5, Problem 4):
n→∞
r=0 r=0
It follows from formula (3.5.17) that
lim y(n) = (I − A)−1i.
n→∞
This equation says that the national incomes of countries 1 and 2 approach equilibrium values independent of the initial values of the national incomes y1(0), y2(0).
However, as we all know, international economics involves many more factors than we can account for here. But in Exercises 3.5, Problem 11, the student will be allowed to create a model for the economic interaction among three countries.
3.5.5 The Heat Equation
Example 3.40. Consider the distribution of heat through a thin bar composed of a homogeneous material. Let x1,x2,...,xk be k equidistant points on the bar. Let Ti(n) be the temperature at time tn = (∆t)n at the point xi, 1 ≤ i ≤ k. Denote the temperatures at the left and the right ends of the bar at time tn by T0(n),Tk+1(n), respectively. (See Figure 3.3.)
Assume that the sides of the bar are sufficiently well insulated that no heat energy is lost through them. The only thing, then, that affects the temperature at the point xi is the temperature of the points next to it, which are xi−1,xi+1. Assume that the left end of the bar is kept at b degrees Celsius and the right end of the bar at c degrees Celsius. These conditions imply that x0(n) = b and xk+1(n) = c, for n ≥ 0.
We assume that the temperature at a given point xi is determined only by the temperature at the nearby points xi−1 and xi+1. Then according to Newton’s law of cooling, the change in temperature Ti(n + 1) − Ti(n) at a point xi from time n to n + 1 is directly proportional to the temperature difference between the point xi and the nearby points xi−1 and xi+1. In
x1 x2 x3 xi xk-2 xk-1 xk
x0 x k+1
FIGURE 3.3. Heat transfer.
n−1 ∞
rr −1
lim A = A =(I−A) . (3.5.19)
3.5 Applications 167
... ...
168 3. Systems of Linear Difference Equations
other words
Ti(n + 1) − Ti(n) = α([Ti−1(n) − Ti(n)] + [Ti+1(n) − Ti(n)])
= α[Ti+1(n) − 2Ti(n) + Ti−1(n)], (3.5.20)
or
Ti(n + 1) = αTi−1(n) + (1 − 2α)Ti(n) + αTi+1(n), i = 2, 3, . . . , k − 1.
Similarly, one may also derive the following two equations: T1(n + 1) = (1 − 2α)T1(n) + αT2(n) + αb,
Tk(n + 1) = αTk−1(n) + (1 − 2α)Tk(n) + αc. This correlation may be written in the compact form
T (n + 1) = AT (n) + g,
⎛(1−2α) α 0 ... 0 ⎞ ⎛αb⎞
where
⎜ . ⎟ ⎜⎟
⎜ α (1−2α) α . ⎟ ⎜0⎟
A= ⎜ 0 α(1−2α)... ⎟ ,g= ⎜ ⎟
⎜0⎟ . ⎜.⎟
⎜⎝ . . . ... α ⎟⎠ 0 0 0 α(1−2α)
⎝.⎠ αc
This is a tridiagonal Toeplitz matrix.2 Its eigenvalues may be found by the formula [111]
nπ
λn =(1−2α)+αcos k+1 , n=1,2,...,k.
Hence |λ| < 1 for all eigenvalues λ of A. Corollary 3.24 then implies that lim An = 0.
n→∞
From the variation of constants formula (3.2.12), it follows that
n−1
T(n) = AnT(0) + Arg. r=0
⎛ a0 a1 a2 ⎜
⎜ a−1 a0 a1 2AisaToeplitzifitisoftheform⎜ a−2 a−1 a0
⎟ ⎝.. a1⎠
⎜. .
a−k+1 ... a−2 a−1 a0
... ak−1⎞ . ⎟
. ⎟ a2 ⎟.
Thus, limn→∞ T(n) = (I − A)−1g. Finally, this equation points out that the temperature at the point xi, 1 ≤ i ≤ k, approaches the ith component of the vector (I − A)−1g, regardless of the initial temperature at the point xi.
Consider the above problem with k = 3, α = 0.4, T0(n) = 10◦ C , T4(n) = 20◦ C.
Then
⎛⎞⎛⎞
lll0.2 0.4 0
A = ⎜⎝ 0 . 4 0 . 2 0 . 4 ⎟⎠ ,
0 0.4 0.2
⎛0.8 −0.4 0⎞−1 ⎜8 4 8⎟
3.5 Applications 169
4
g = ⎜⎝ 0 ⎟⎠ ,
8
⎛15 5 5⎞
= ⎜ 5 5 5⎟. ⎜⎝4 2 4⎟⎠
13 5 15 848
⎛25⎞
⎜ 2 ⎟ ⎟⎝0⎠ = ⎜15⎟.
2
Hence
(I − A)−1 = ⎜⎝−0.4 0.8 −0.4⎟⎠ 0 −0.4 0.8
⎛15 5 5⎞
⎜ 8 ⎜ 5
8 ⎟⎛4⎞
4 5
13 5 15 848
5 ⎟⎜ ⎟
n→∞ ⎜⎝424⎟⎠8 ⎝43⎠
lim T(n) = ⎜
Let ∆x = xi −xi−1 and ∆t = ti−1 −ti. If we assume that the constant of proportionality α depends on both ∆t and ∆x, then we may
Remark:
write
∆t
α = (∆x)2 β, (3.5.21)
where β is a constant that depends on the material of the bar. Formula (3.5.21) simply states that the smaller the value of ∆t, the smaller should be the change in the temperature at a given point. Moreover, the smaller the separation of points, the larger should be their influence on the temperature changes in nearby points. Using formula (3.5.21) in (3.5.20) yields
Ti(n + 1) − Ti(n) Ti+1(n) − 2Ti(n) + Ti−1(n)
∆t =β (∆x)2 . (3.5.22)
Ifwelet∆t→0,∆x→0asn→∞andi→∞,xi =(∆x)i=x,and ti = (∆t)i = t, then (3.5.22) gives the partial differential equation
∂T(x,t) = β∂2T(x,t). (3.5.23) ∂t ∂x2
Equation (3.5.23) is known as the heat equation [137].
170 3. Systems of Linear Difference Equations
Exercises 3.5
1. Consider the difference system
P (n + 1) = RP (n)
where
⎛⎞
0.2 0.1 0.3
R = ⎜⎝0.3 0.5 0.1⎟⎠ .
0.5 0.4 0.6 (a) Show that R is a Markov matrix.
(b) Find limn→∞ P(n).
2. Consider the difference system
where
P (n + 1) = RP (n)
⎛1 0 0.3 0.1⎞
⎜0 1 0.1 0.2⎟ R = ⎜⎝ 0 0 0 . 4 0 . 3 ⎟⎠ .
0 0 0.2 0.4
(a) Show that R is an absorbing Markov matrix.
(b) Find limn→∞ P(n).
3. Show that if A is a k × k Markov matrix, then it has an eigenvalue
equal to 1.
4. LetA=(aij)beak×kpositivematrixsuchthatkj=1aij <1for
i = 1,2,...,k. Show that |λ| < 1 for all eigenvalues λ of A.
5. LetAbeak×kmatrixwith|λ|<1foralleigenvaluesλofA.Show
that:
(i) (I − A) is nonsingular.
(ii) ∞i=0 Ai = (I − A)−1.
6. Modify Example 3.37 by first mating a recessive individual (genes gg) with a dominant individual (genes GG). Then, continuing to mate the offspring with a dominant individual, write down the difference equation that describes the probabilities of producing individuals with genes GG, Gg, and gg. Find limn→∞ p(n) and then interpret your results.
7. In the dark ages, Harvard, Yale, and MIT admitted only male students. Assume that at the time, 80% of the sons of Harvard men went to Harvard and the rest went to MIT, 40% of the sons of MIT men went
to MIT and the rest split evenly between Harvard and Yale; and of the sons of Yale men, 70% went to Yale, 20% to Harvard, and 10% to MIT. Find the transition matrix R of this Markov chain. Find the long-term probabilities that the descendants of Harvard men will go to Yale. (Assume that we start with N men, and each man sends one son to college.)
8. A New York governor tells person A his intention either to run or not to run in the next presidential election. Then A relays the news to B, who in turn relays the message to C, and so forth, always to some new person. Assume that there is a probability α that a person will change the answer from yes to no when transmitting it to the next person and a probability β that he will change it from no to yes. Write down the state transition matrix of this process, then find its limiting state. Note that the initial state is the governor’s choice.
9. A psychologist conducts an experiment in which 20 rats are placed at random in a compartment that has been divided into rooms labeled 1, 2, and 3 as shown in Figure 3.4. Observe that there are four doors in the arrangement. There are three possible states for each rat: It can be in room 1, 2, or 3. Let us assume that the rats move from room
to room. A rat in room 1 has the probabilities p11 = 0, P21 = 1 , and 23
10. In Example 3.38 (drunkard’s walk), assume that the probability of a step to the right is 2 and that of a step to the left is 1. Write down
the transition matrix and determine limn→∞ p(n).
11. In the trade model (Example 3.39) let a11 = 0.4,a21 = 0.5,a12 = 0.3,a22 = 0.6,i1 = 25 billion dollars, and i2 = 20 billion dollars. If y1(n) and y2(n) denote the national incomes of countries 1 and 2 in year n, respectively, and y1(0) = 500 billion dollars and y2(0) = 650 billion dollars, find y1(3) and y2(3). What are the equilibrium national incomes for nations 1 and 2?
3.5 Applications 171
p31 = 3 of moving to the various rooms based on the distribution of doors. Predict the distribution of the rats in the long run. What is the limiting probability that a given marked rat will be in room 2?
33
1
23
FIGURE 3.4. Diagram for Problem 9.
172 3. Systems of Linear Difference Equations
x3 x4 x5 x6
x1 x2
air 50 water 0
50 50 50
000
FIGURE 3.5. Heat flow diagram for Problem 14.
50 0
12. Develop a mathematical model for a foreign trade model among three countries using an argument similar to that used in Example 3.39.
13. In Example 3.40, let k = 4,α = 0.2, and x0(n) = T5(n) = 0◦ C. Compute Ti(n), 1 ≤ i ≤ 4, for n = 1, 2, 3; then find limn→∞ Ti(n).
14. Suppose we have a grid of six points on a bar as shown in Figure 3.5. Part of the bar is in air that is kept at a constant temperature of 50 degrees, and part of the bar is submerged in a liquid that is kept at a constant temperature of 0 degrees. Assume that the temperature at the point xi,1 ≤ i ≤ 6, depends only on the temperature of the four nearest points, that is, the points above, below, to the left, and to the right.
(i) Write a mathematical model that describes the flow of heat in this bar.
(ii) Find the equilibrium temperature at the six points xi.
4
Stability Theory
In Chapter 1 we studied the stability properties of first-order difference equations. In this chapter we will develop the theory for k-dimensional systems of first-order difference equations. As shown in Chapter 3, this study includes difference equations of any order. Here we are interested in the qualitative behavior of solutions without actually computing them. Realizing that most of the problems that arise in practice are nonlinear and mostly unsolvable, this investigation is of vital importance to scientists, engineers, and applied mathematicians.
In this chapter we adapt the differential methods and techniques of Lia- punov [93], Perron [114], and many others, to difference equations. First, we introduce the notion of norms of vectors and matrices in Section 4.1. Next, we give definitions of various notions of stability and some simple examples to illustrate them in Section 4.2. Section 4.3 addresses the question of sta- bility of both autonomous and nonautonomous linear systems and includes the Stable Mainfold Theorem. In Section 4.4 we study the geometrical prop- erties of planar linear systems by means of phase space analysis. Section 4.5 introduces to the reader the basic theory of the direct method of Liapunov, by far the most advanced topic in this chapter. In Section 4.6 we present the stability of nonlinear systems by the method of linear approximation, which is widely used by scientists and engineers. And, finally, in Section 4.7 we investigate mathematical models of population dynamics and a business model. Due to the enormity of the existing literature on Liapunov theory, we have limited our exposition to autonomous equations.
173
174
4. Stability Theory
(i)
(ii)
(iii)
4.1
the l1 norm:
k i=1
x =1 1
x = 1
x = 1 2
∥x∥1 =
the l∞ norm:
|xi| ∥x∥∞ = max |xi|
1≤i≤k
the Euclidean norm l2:
k ∥ x ∥ 2 =
i=1
1/2 x 2i
FIGURE 4.1. A circle in different norms.
A Norm of a Matrix
We start this section by introducing the notion of norms of vectors and matrices.
Definition 4.1. A real-valued function on a vector space V is called a norm, and is denoted by ∥∥, if the following properties hold:
(i) ∥x∥≥0and∥x∥=0onlyifx=0;
(ii) ∥αx∥ = |α|∥x∥ for all x ∈ V and scalars α;
(iii) ∥x+y∥≤∥x∥+∥y∥forallx,y∈V.
The three most commonly used norms on Rk are shown in Figure 4.1.
We remark here that all norms on Rk are equivalent in the sense that if ∥∥, ∥∥′ are any two norms, then there exist constants α, β > 0 such that
α∥x∥ ≤ ∥x∥′ ≤ β∥x∥.
Thus if {xn} is a sequence in Rk, then ∥xn∥ → 0 as n → ∞ if and only if ∥xn∥′ →0asn→∞.
Corresponding to each vector norm ∥∥ on Rk one may define an operator norm ∥∥ on a k × k matrix A as
∥A∥ = max ∥Ax∥. (4.1.1) ∥x∦=0 ∥x∥
It may be shown easily that
(4.1.2)
Using this definition one may easily compute ∥A∥ relative to the above three norms as shown in Table 4.1. (For a proof see [85].)
∥A∥ = max ∥Ax∥ = max ∥Ax∥. ∥x∥≤1 ∥x∥=1
3. Let
⎛⎞
λ10
A = ⎜⎝ 0 λ 1 ⎟⎠ .
TABLE 4.1. Vector and Matrix Norms.
⎡⎤⎡⎤
112 210 (b) ⎢⎣0 2 −1⎥⎦ . (c) ⎢⎣0 2 0⎥⎦ .
(a) 2 1 12
.
2. Give an example of a matrix A such that ρ(A) ̸= ∥A∥∞, ∥A∥1, ∥A∥2.
j=1
4.1 A Norm of a Matrix 175
∥x∥
k
|xi |
i=1
max |xi | 1≤i≤k
1 k 2
x2i i=1
Norm
l1
l∞
l2
∥A∥ m a x k
1≤j≤k
m a x k
|aij |
Sum over columns
Sum over rows
i=1
|aij|
1 ρ(ATA) 2
1≤i≤k
From (4.1.1) we may deduce that for any operator norm on A (Exercises 4.1, Problem 5),
ρ(A) ≤ ∥A∥, (4.1.3) where ρ(A) = max{|λ| : λ is an eigenvalue of A} is the spectral radius of
A.
Exercises 4.1
1. Compute ∥A∥1, ∥A∥∞, ∥A∥2, and ρ(A) for the following matrices:
030 034
00λ
Show that for each ε > 0 there exists a diagonal matrix D such that
∥D−1AD∥ ≤ |λ| + ε for operator norms ∥A∥1 ∥A∥∞.
4. Generalize Problem 3 to any k × k matrix A in the Jordan form
diag(J1, J2, . . . , Jr).
5. Prove that ρ(A) ≤ ∥A∥ for any operator norm ∥∥ on A.
176
4. Stability Theory
6. 7.
4.2
Show that for any two norms ∥∥, ∥∥′ on Rk there are constants α, β > 0 such that α∥x∥ ≤ ∥x∥′ ≤ β∥x∥.
Deduce from Problem 6 that for any sequence {x(n)}, ∥x(n)∥ → 0 as n→∞ifandonlyif∥x(n)∥′ →0.
Notions of Stability
Let us consider the vector difference equation
x(n + 1) = f (n, x(n)), x(n0 ) = x0 , (4.2.1)
where x(n) ∈ Rk,f : Z+ ×Rk → Rk. We assume that f(n,x) is continuous in x. Recall that (4.2.1) is said to be autonomous or time-invariant if the variable n does not appear explicitly in the right-hand side of the equation f(n,x(n)) ≡ f(x(n)). It is said to be periodic if for all n ∈ Z, f(n+N,x) = f(n,x) for some positive integer N.
A point x* in Rk is called an equilibrium point of (4.2.1) if f(n,x∗) = x* for all n ≥ n0. In most of the literature x* is assumed to be the origin 0 and is called the zero solution. The justification for this assumption is as follows: Let y(n) = x(n) − x*. Then (4.2.1) becomes
y(n+1)=f(n,y(n)+x∗)−x∗ =g(n,y(n)). (4.2.2)
Notice that y = 0 corresponds to x = x*. Since in many cases it is not convenient to make this change of coordinates, we will not assume that x∗ = 0 unless it is more convenient to do so.
Recall that in Chapter 3 we dealt with the existence and uniqueness of solutions of linear systems, that is, the case f (n, x(n)) = A(n)x(n), where A(n) is a k × k matrix. The existence and uniqueness of solutions of (4.2.1) may be established in a similar fashion (Exercises 4.2, Problem 9).
We are now ready to introduce the various stability notions of the equilibrium point x* of (4.2.1).
Definition 4.2. The equilibrium point x* of (4.2.1) is said to be:
(i) Stable (S) if given ε > 0 and n0 ≥ 0 there exists δ = δ(ε,n0) such that ∥x0 −x∗∥ < δ implies ∥x(n,n0,x0)−x∗∥ < ε for all n ≥ n0, uniformly stable (US) if δ may be chosen independent of n0, unstable if it is not stable.
(ii) Attracting (A) if there exists μ = μ(n0) such that ∥x0 − x∗∥ < μ implies limn→∞ x(n, n0, x0) = x*, uniformly attracting (UA) if the choice of μ is independent of n0. The condition for uniform attractivity may be paraphrased by saying that there exists μ > 0 such that for every ε and n0 there exists N = N(ε) independent of n0 such that ∥x(n, n0, x0) − x∗∥ < ε for all n ≥ n0 + N whenever ∥x0 − x∗∥ < μ.
x2
x0
x1
4.2 Notions of Stability 177
FIGURE 4.2. Stable equilibrium in phase space.
(iii) Asymptotically stable (AS) if it is stable and attracting, and uniformly asymptotically stable (UAS) if it is uniformly stable and uniformly attracting.
(iv) Exponentially stable (ES) if there exist δ > 0, M > 0, and η ∈ (0,1) such that ∥x(n, n0, x0) − x∗∥ ≤ M∥x0 − x∗∥ηn−n0 , whenever ∥x0 − x∗∥ < δ.
(v) A solution x(n,n0,x0) is bounded if for some positive constant M, ∥x(n,n0,x0)∥ ≤ M for all n ≥ n0, where M may depend on each solution.
If in parts (ii), (iii) μ = ∞ or in part (iv) δ = ∞, the corresponding stability property is said to be global. In Figure 4.2, we suppress the (time) n and show only the movement of a solution that starts inside a ball of radius δ. The figure illustrates that all future states x(n, n0, x0), n ≥ n0, will stay
x 2
n
x1
n
0
FIGURE 4.3. Stable equilibrium.
178
4. Stability Theory
x
2
η
n0
ε
n
x1
FIGURE 4.4. Uniformly asymptotically stable equilibrium.
AS
ES
UA A
UAS
we have the following result.
S US
FIGURE 4.5. Hierarchy of stability notions.
within the ε ball. This diagram is called a phase space portrait and will be used extensively in later sections. In Figure 4.3 the time n is considered part of a three-dimensional coordinate system that provides another perspective on stability. Figure 4.4 depicts the uniform asymptotic stability of the zero solution.
Note that in the above definitions, some of the stability properties auto- matically imply one or more of the others. Figure 4.5 shows the hierarchy of the stability notions.
Important Remark: In general, none of the arrows in Figure 4.5 may be reversed. However, for special classes of equations, these arrows in Figure 4.5 may be reversed. In this section, it will be shown that for linear systems
x(n + 1) = A(n)x(n) (4.2.3)
where A(n) is a k × k matrix defined on Z+, uniform asymptotic stability implies exponential stability (UAS ⇔ ES).
For the autonomous system
x(n + 1) = f (x(n)) (4.2.4)
Theorem 4.3. For the autonomous system (4.2.4), the following state- ments hold for the equilibrium point x∗:
(i) S ↔ US. (ii) AS ↔ UAS.
(iii) A ↔ UA. Proof.
(i) Let x(n, n0, x0) and y(n, m0, x0) be two solutions of (4.2.4), with m0 = n0 +r0, r0 ≥ 0. Notice that x(n−r0, n0, x0) intersects with y(n, m0, x0) at n = m0. By uniqueness of solutions, it follows that y(n,m0,x0) = x(n − r0, n0, x0). This implies that the δ in the definition of stability is independent of the initial time n0 which establishes our result.
The proofs of (ii) and (iii) are similar to the proof of (i). P The following examples serve to illustrate the definitions.
1. The solution of the scalar equation x(n + 1) = x(n) is given by x(n,n0,x0) = x0; hence the zero solution is uniformly stable but not asymptotically stable.
2. The solutions of the scalar equation x(n + 1) = a(n)x(n) are n−1
x(n, n0, x0) = Hence one may conclude the following:
(i) The zero solution is stable if and only if
n − 1
i=n0
a(i) ≤ M(n0) ≡ M,
i=n0
where M is a positive constant that depends on n0 (Exercises 4.2, Problem 2). This condition holds if a(i) = (1 + ηi), where 0 < η < 1.
To show this we write the solution as x(n, n0, x0) = Φ(n)x0, where Φ(n) = n−1 (1 + ηi). Since 1 + ηi < exp(ηi), it follows that
i=n0
n−1 ∞
Φ(n)≤exp
= M(n0) = M.
4.2 Notions of Stability 179
a(i) x0.
(4.2.5)
(4.2.6)
i i ηn0
η ≤exp η ≤exp 1−η i=n0
i=n0
Givenε>0andn0 ≥0,ifweletδ=ε/(2M),then|x0|<δ implies |x(n, n0, x0)| = Φ(n)x0 < ε.
180
4. Stability Theory
(ii) The zero solution is uniformly stable if and only if
n − 1
a(i) ≤ M,
where M is a positive constant independent of n0 (Exercises 4.2,
i=n0
Problem 5). This condition holds if a(i) = sin(i + 1).
(iii) The zero solution is asymptotically stable if and only if
i=n0 (Exercises 4.2, Problem 8).
lim n→∞
a(i) = 0
Now we give two important examples. In the first example we show that the zero solution is stable but not uniformly stable. In the second example the zero solution is attracting but not stable (personal communication by Professor Bernd Aulbach).
Example 4.4. The solution of the equation x(n + 1) = n+1 [x(n)]2 is 2
given by
2 x(n,n0,x0)= n n−1
n − 1
(4.2.8) (Exercises 4.2, Problem 5). This condition clearly holds if a(i) =
i=n0
i+1 . The solution is given by x(n, n0, x0) = (n0 + 1)/(n + 1)x0.
stable (globally), but not uniformly asymptotically stable. (Why?) (See Exercises 4.2, Problem 6.)
(iv) The zero solution is uniformly asymptotically stable (and thus exponentially stable) if and only if
(4.2.7)
i+2
Thus, the zero solution is uniformly stable and asymptotically
n − 1
n−n0
a ( i ) ≤ M η , ( 4 . 2 . 9 ) for some M > 0,0 < η < 1. This may be satisfied if a(i) = 1/i
4 2222
2n−n0−1
n−2
n0 +1
···
(x0)2n−n0,
x(n0) = x0.
If |x0| is sufficiently small, then limn→∞ x(n) = 0. Thus, the zero solution is attracting. However, it is not uniformly attracting. For if δ > 0 is given and n0 is chosen such that (n0 + 1)δ2 ≥ 2, then, for |x0| = δ,
2
Let us now check the stability of the zero solution. Given ε > 0 and n0 ≥ 0, let δ = ε/(n0 +1). If |x0| < δ, then |x(n,n0,x0)| < ε for all n ≥ n0. Since δ
|x(n0 +1,n0,x0)|= 0
|x0|2 ≥1.
n +1
depends on the choice of n0, the zero solution is stable but not uniformly stable.
Example 4.5.
Consider the difference equation (in polar coordinates)
r(n+1)= r(n), r>0,
show this, observe that
r(n) = r2−n , r
= r(0), θ(n) = (2π)(1−2−n) · (θ0)2−n , θ0 = θ(0).
00
Clearly, limn→∞ r(n) = 1 and limn→∞ θ(n) = 2π. Now, if r0 ̸= 0, θ0 = 0, then (r(n), θ(n)) = ((r0)2−n , 0), which converges to the equilibrium point
(1, 0). However, if θ0 = δπ, 0 < δ < 1, then the orbit of (r0, θ0) will spiral around the circle counterclockwise to converge to the equilibrium point (1, 0). Hence the equilibrium point (1, 0) is attracting but not stable. (See Figure 4.6.)
Remark: The situation in Example 4.5 is a higher dimension phenomenon. In 1997, Sedaghat [132] showed that a continuous map on the real line cannot have an attracting unstable fixed point.
To demonstrate this phenomenon, let us contemplate the following example:
Example 4.6. Consider the map
−2x ifx<μ, Gμ(x) =
0 if x ≥ μ,
4.2 Notions of Stability 181
θ(n + 1) = 2πθ(n), 0 ≤ θ ≤ 2π.
We claim that the equilibrium point (1, 0) is attracting but not stable. To
FIGURE 4.6. Attracting but not stable equilibrium.
182 4. Stability Theory
G (x) 2
xx 0
FIGURE 4.7. G2(x).
where μ ∈ R+. Equivalently, we have the difference equation x(n + 1) =
Gμ(x(n)) whose solution is given by
n n−1
x(n)=Gn(x )= (−2) x0 if(−2) x0 <μ, μ 0 0 if (−2)n−1x0 ≥ μ,
where x(0) = x0.
Now,ifx0 ≥μ,thenGnμ(x0)=0foralln≥1.Ontheotherhand,if
x0 < μ, then for some k ∈ Z+, Gkμ(x0) ≥ μ. Hence, Gnμ(x0) = 0 for all n ≥ k.
Hence the fixed point x∗ = 0 is globally attracting. However, x∗ = 0 is unstable, for points x0 that are close to 0 are mapped to points further away from 0 until they exceed μ (see Figure 4.7 for G2).
Theorem 4.7 [132]. A continuous map f on the real line cannot have an attracting unstable fixed point.
To facilitate the proof of the theorem, we first establish a stability result that is of independent interest, since it does not require differentiability of f.
Criterion for asymptotic stability of fixed points of nondifferentiable maps.
Theorem 4.8 [135]. A fixed point x∗ of a continuous map f is asymp- totically stable if and only if there is an open interval (a,b) containing x∗ such that f2(x)>x for a
Notice that by Theorem 4.8, case (a) implies that x∗ is asymptotically stable and must be discarded. It remains to rule out case (b). So assume that f2(x) < x for x < x∗. Now let x0 < x∗. Then by iteration, we have ···
Exercises 4.2
1. Meditate upon the scalar equation x(n + 1) = ax(n). Prove that:
(i) If |a| < 1, the zero solution is uniformly asymptotically stable.
(ii) If |a| = 1, the zero solution is uniformly stable.
(iii) If |a| > 1, the zero solution is not stable.
2. (a)
(b) Show that the zero solution of the equation x(n + 1) = (1 + ηn)x(n), 0 < η < 1, is stable.
3. (a)
4.
5.
6. 7.
(b) Show that the zero solution of the equation x(n + 1) = sin(n + 1)x(n) is uniformly stable.
Show that the zero solution of the equation x(n + 1) = n+1 x(n) is n+2
Prove that the zero solution of the scalar equation x(n + 1) = n−1
a(n)x(n) is stable if and only if i=n0 a(i) ≤ M(n0), where M depends on n0.
Prove that the zero solution of the equation x(n + 1) = a(n)x(n) n−1
is uniformly stable if and only if i=n0 a(i) ≤ M, where M is a positive constant independent of n0.
4.2 Notions of Stability 183
asymptotically stable.
Prove that the zero solution of the equation x(n + 1) = a(n)x(n) is
n−1 asymptotically stable if and only if limn→∞ i=n0 a(i) = 0.
Show that the zero solution of the equation in Problem 4 is not uniformly asymptotically stable.
Prove that the zero solution of the equation x(n) = a(n)x(n) is uni-
n−1 n−n0 formly asymptotically stable if and only if i=n0 a(i) ≤ Mη
some M > 0, 0 < η < 1.
, for
184
4. Stability Theory
8.
9. 10.
Show that the zero solution of the scalar equation x(n + 1) = (1/n)x(n), n ≥ 1, is uniformly asymptotically stable.
Establish the existence and uniqueness of solutions of (4.2.1). Consider the system
x(n + 1) = x(n) + y(n + 1) = y(n) +
x2(n)(y(n) − x(n)) + y5(n) , [x2(n) + y2(n)] + [x2(n) + y2(n)]3
y2(n)(y(n) − 2x(n)) , [x2(n) + y2(n)] + [x2(n) + y2(n)]3
11.
4.3
which can be written as
x(n + 1) = x(n) + g1(x(n), y(n)),
y(n + 1) = y(n) + g2(x(n), y(n)).
Show that the zero solution is globally attracting but unstable.
Define the difference equation on the unit circle as
r(n + 1) = 1, √
θ(n+1)= 2πθ, 0≤θ<2π.
Show that the fixed point (1, 0) is globally attracting but unstable.
Stability of Linear Systems
4.3.1 Nonautonomous Linear Systems
In this subsection we investigate the stability of the linear nonautonomous (time-variant) system given by
x(n + 1) = A(n)x(n), n ≥ n0 ≥ 0. (4.3.1)
It is always assumed that A(n) is nonsingular for all n ≥ n0.
If Φ(n) is any fundamental matrix of system (4.3.1) or (4.3.6), then recall that Φ(n, m) = Φ(n)Φ−1 (m) is the state transition matrix. In the following result we express the conditions for stability in terms of a fundamental
matrix Φ(n) of system (4.3.1).
Theorem 4.9. Consider system (4.3.1). Then its zero solution is
(i) stable if and only if there exists a positive constant M such that
∥Φ(n)∥≤M for n≥n0 ≥0; (4.3.2)
(ii) uniformly stable if and only if there exists a positive constant M such that
∥Φ(n,m)∥≤M for n0 ≤m≤n<∞; (4.3.3)
(iii) asymptotically stable if and only if
lim ∥Φ(n)∥ = 0; (4.3.4)
n→∞
(iv) uniformly asymptotically stable if and only if there exist positive
constants M and η ∈ (0, 1) such that:
∥Φ(n,m)∥ ≤ Mηn−m for n0 ≤ m ≤ n < ∞. (4.3.5)
Proof. Without loss of generality we may assume that Φ(n0) = I, since conditions (4.3.2) through (4.3.5) hold true for every fundamental matrix if they hold for one. Thus x(n, n0, x0) = Φ(n)x0.
(i) Suppose that inequality (4.3.2) holds. Then ∥x(n, n0, x0)∥ ≤ M∥x0∥. So for ε > 0, let δ < ε/M. Then ∥x0∥ < δ implies ∥x(n,n0,x0)∥ < ε and, consequently, the zero solution is stable. Conversely, suppose that ∥x(n,n0,x0)∥ = ∥Φ(n)x0∥ < ε whenever ∥x0∥ ≤ δ. Observe that ∥x0∥ ≤ δ if and only if 1∥x0∥ ≤ 1. Hence
δ
∥Φ(n)∥ = sup ∥Φ(n)ξ∥ = 1 sup ∥Φ(n)x0∥ ≤ ε = M. ∥ξ∥≤1 δ ∥x0∥≤δ δ
Parts (ii) and (iii) remain as Exercises 4.3, Problems 9 and 10.
(iv) Suppose finally that inequality (4.3.5) holds. The zero solution of sys- tem (4.3.1) would then be uniformly stable by part (ii). Furthermore, forε>0,0<ε
∥Φ(n,n0)∥≤∥Φ(n,n0 +mN)∥∥Φ(n0 +mN,n0 +(m−1)N)∥× ···×∥Φ(n0 +N,n0)∥
M 1 (m+1)N
≤Mεm≤ε εN =M ̃η(m+1)N,
4.3 Stability of Linear Systems 185
≤ M ̃ η ( n − n 0 ) ,
formN ≤n−n ≤(m+1)N whereM ̃ = M,η=ε1 .Thisconcludes
0εN
the proof of the theorem. P
The following result arises as an immediate consequence of the above theorem. [See the Important Remark, part (i).]
186 4. Stability Theory
Corollary 4.10. For the linear system (4.3.1) the following statements hold:
(i) The zero solution is stable if and only if all solutions are bounded.
(ii) The zero solution is exponentially stable if and only if it is uniformly
asymptotically stable.
Proof. Statements (i) and (ii) follow immediately from conditions (4.3.3) and (4.3.5), respectively (Exercises 4.3, Problem 6). P
The following is another important consequence of Theorem 4.9: Corollary 4.11. For system (4.3.1), every local stability property of the
zero solution implies the corresponding global stability property.
Proof. Use Theorem 4.9 (Exercises 4.3, Problem 7). P We now give a simple but powerful criterion for uniform stability and
uniform asymptotic stability. Theorem 4.12 [17].
(i) If ki=1 |aij(n)| ≤ 1,1 ≤ j ≤ k,n ≥ n0, then the zero solution of system (3.2.15) is uniformly stable.
(ii) If ki=1 |aij(n)| ≤ 1 − ν for some ν > 0,1 ≤ j ≤ k,n ≥ n0, then the zero solution is uniformly asymptotically stable.
Proof.
(i) From condition (i) in Theorem 4.12, ∥A(n)∥1 ≤ 1 for all n ≥ n0. Thus,
i=m 1
This now implies uniform stability by Theorem 4.9, part (ii).
(ii) The proof of statement (ii) is so similar to the proof of statement (i) that we will omit it here. P
4.3.2 Autonomous Linear Systems
In this subsection we specialize the results of the previous section to autonomous (time-invariant) systems of the form
x(n + 1) = Ax(n). (4.3.6) In the next theorem we summarize the main stability results for the
linear autonomous systems (4.3.6).
n−1
∥Φ(n, m)∥1 = A(i) ≤ ∥A(n − 1)∥1∥A(n − 2)∥1 · · · ∥A(m)∥1 ≤ 1.
Theorem 4.13. The following statements hold:
(i) The zero solution of (4.3.6) is stable if and only if ρ(A) ≤ 1 and the eigenvalues of unit modulus are semisimple.1
(ii) The zero solution of (4.3.6) is asymptotically stable if and only if ρ(A) < 1.
Proof.
(i) Let A = PJP−1, where J = diag(J1,J2,...,Jr) is the Jordan form of
A and
⎛⎞
J i n = ⎜ ⎜ 0 ⎜
⎜⎝
. .
⎟ ⎟ . ⎟
⎟⎠
Ji =⎜ ⎜
... ⎟. ⎟
⎜
λi1 0 λi
⎟
4.3 Stability of Linear Systems 187
... ⎝1⎠
0 λi
From Theorem 4.9 the zero solution of (4.3.6) is stable if and only if ∥An∥ = ∥PJnP−1∥ ≤ M or ∥Jn∥ ≤ M ̃, where M ̃ = M/(∥P∥∥P−1∥). Now, Jn = diag(J1n,J2n,...,Jrn), where
⎛n n ⎞ λn λn−1 ··· λn−si+1 ⎜i 1i si−1i ⎟
⎜ .⎟
λ ni
0 0 ··· λni
Obviously, Jin becomes unbounded if |λi| > 1 or if |λi| = 1 and Ji isnot1×1.If|λi|<1,thenJin →0asn→∞.Toprovethis conclusion it suffices to show that |λi|n nl → 0, as n → ∞ for any positive integer l. This conclusion follows from L’Hoˆpital’s rule, since |λi|n nl = nle(ln |λi|)n (Exercises 4.3, Problem 8).
(ii) The proof of statement (ii) has already been established by the above argument. This completes the proof of the theorem. P
Explicit Criteria for Stability of Two-Dimensional Systems
In many applications one needs explicit criteria on the entries of the matrix for the eigenvalues to lie inside the unit disk. So consider the matrix
a11 a12 a21 a22
1An eigenvalue is said to be semisimple if the corresponding Jordan block is diagonal.
A=
· · ·
n λn−1
1i
188 4. Stability Theory
whose characteristic equation is given by
λ2 − (a11 + a22)λ + (a11a22 − a12a21) = 0
or
λ2 −(trA)λ+detA=0. Comparing (4.3.7) with the equation
(4.3.7)
λ2 + p1λ + p2 = 0,
where p1 = −tr A, p2 = det A, we conclude from Theorem 2.37 that the
eigenvalues of A lie inside the unit disk if and only if 1+trA+detA>0, 1−trA+detA>0, 1−detA>0 (4.3.8)
or, equivalently,
(4.3.9) It follows that under condition (4.3.9), the zero solution of the equation
x(n + 1) = Ax(n)
is asymptotically stable.
We now describe the situation when some eigenvalues of A in (4.3.6) are
inside the unit disk and some eigenvalues are outside the unit disk. The result below is called the Stable Subspace (Manifold) Theorem. The result does not require that A is invertible.
Let λ be an eigenvalue of A of multiplicity m and let ξ1,ξ2,…,ξm be the generalized eigenvectors corresponding to λ. Then for each i, 1 ≤ i ≤ m, either
Aξi = λξi (ξi is an eigenvector of A), or Aξi = λξi + ξi−1.
It follows that the generalized eigenvectors corresponding to λ are the solutions of the equation
(A − λJ)mξ = 0. (4.3.10)
The set of all linear combinations, or the span of the generalized eigenvec- tors corresponding to λ is invariant under A and is called the generalized eigenspace Eλ of the eigenvalue of A. Clearly, if λ1 ̸= λ2, then Eλ1 ∩Eλ2 = {0}. Notice that each eigenspace Eλ includes the zero vector.
Assume that A is hyperbolic, that is, none of the eigenvalues of A lie on theunitcircle.ArrangetheeigenvaluesofAsuchthat∆s ={λ1,λ2,…,λr} arealltheeigenvaluesofAwith|λi|<1,1≤i≤rand∆u = {λr+1,λr+s,...,λk} are all the eigenvalues of A with |λi| > 1, r+1 ≤ i ≤ k. The eigenspace spanned by the eigenvalues in ∆s is denoted by Ws, where W s = ri=1 λi and the eigenspace spanned by the eigenvalues in ∆u is denoted by W u, where W u = ki=r+1 λi.
|tr A| < 1 + det A < 2.
Theorem 4.14 (The Stable Subspace (Manifold) Theorem). If A is hyperbolic, then the following statements hold true:
(i) If x(n) is a solution of (4.3.6) with x(0) ∈ Ws, then for each n, x(n) ∈ Ws. Furthermore,
lim x(n) = 0. n→∞
(ii) If x(n) is a solution of (4.3.6) with x(0) ∈ Wu, then x(n) ∈ Wu for each n. Moreover,
lim x(n) = 0. n→−∞
Proof.
(i) Let x(n) be a solution of (4.3.6) with x(0) ∈ Ws. Since AEλ = Eλ, it follows that AWs = Ws. Hence x(n) ∈ Ws for all n ∈ Z+. To prove the second statement, observe that x(0) = ri=1 ciξi, where 1 ≤ ξi ≤ r are the generalized eigenvectors corresponding to elements in ∆s. Let J = P−1AP be the Jordan form of A. Then J may be written in the form
Js 0
J=
0 Ju
where Js has the eigenvalues in ∆s and Ju has the eigenvalues in
∆u′. By Lemma 3.27 in Chapter 3, the corresponding generalized
eigenvectorsξ ̃,1≤i≤r,ofJ areoftheformξ ̃ =P−1ξ = isii
(ai1,ai2,...,air,0,0,...,0)T . Now x(n) = Anx(0)
n − 1 r =PJ P
ciξi
r Jnξ ̃ =P ci s i .
i=1 00 Thuslimn→∞x(n)=0sinceJsn →0asn→∞.
(ii) The proof of (ii) is analogous to (i) and will be left to the reader as Problem 11. P
Remark:
(i) Part (i) may be obtained without the condition of hyperbolicity of A, and similarly for part (ii).
= P J n
c ξ ̃ i=1
4.3 Stability of Linear Systems 189
r
ii
i=1
190 4. Stability Theory
(ii) TheGeneralStableManifoldTheoremforNonlinearMapswillbegiven in Appendix D.
We now use the above result to investigate the stability of the periodic system
x(n + 1) = A(n)x(n), A(n + N ) = A(n). (4.3.11)
Recall from Chapter 3 that if Φ(n, n0 ) is a fundamental matrix of (4.3.11), then there exist a constant matrix B whose eigenvalues are called the Floquet exponents and a periodic matrix P(n,n0) such that Φ(n,n0) = P(n,n0)Bn−n0, where P(n + N,n0) = P(n,n0). Thus if Bn is bounded, thensoisΦ(n,n0),andifBn →0asn→∞,thenitfollowsthat Φ(n, n0 ) → 0 as n → ∞. This proves the following result.
Theorem 4.15. The zero solution of (4.3.11) is:
(i) stable if and only if the Floquet exponents have modulus less than or
equal to 1; those of modulus of 1 are semisimple;
(ii) asymptotically stable if and only if all the Floquet exponents lie inside
the unit disk.
For practical purposes, the following corollary is of paramount impor- tance.
Corollary 4.16. The zero solution of (4.3.11) is:
(i) stable if and only if each eigenvalue of the matrix C = A(N − 1)A(N − 2)···A(0) has modulus less than or equal to 1; those solutions with modulus of value 1 are semisimple;
(ii)asymptotically stable if and only if each eigenvalue of C = A(N − 1)A(N − 2) · · · A(0) has modulus less than 1.
Let us summarize what we have learned thus far. First, for the au- tonomous (time-invariant) linear system x(n + 1) = Ax(n), the eigenvalues of A determine the stability properties of the system (Theorem 4.13). But for a periodic system x(n + 1) = A(n)x(n), the eigenvalues of A(n) do not play any role in the determination of the stability properties of the sys- tem. Instead, the Floquet multipliers of A(n) determine those properties. The following example should dispel any wrong ideas concerning the role of eigenvalues in a nonautonomous system.
Example 4.17. Let us again consider the periodic system in Example 3.35 where
⎛ 2 + (−1)n ⎞ ⎜0 2⎟
A(n) = ⎝ 2 − (−1)n ⎠ . 20
4.3 Stability of Linear Systems 191
√
Here the eigenvalues of A are ± 3/2, and thus ρ[A(n)] < 1. By applying
Corollary 4.16, one may quickly check the stability of this system. We have ⎛ 3⎞⎛ 1⎞ ⎛9 ⎞
⎜0 2⎟⎜0 2⎟ ⎜4 0⎟ C =A(1)A(0)=⎝1 ⎠⎝3 ⎠=⎝ 1⎠.
202004
Hence, by Corollary 4.16, the zero solution is unstable, since C has an eigenvalue 9/4 which is greater than 1.
For the eager reader, perpetually searching for a challenge, we might determine the stability by explicitly transcribing the fundamental matrix as follows:
⎛21−n −(−2)1−n 3n −−3n⎞
⎜22⎟ Φ(n)=⎜ 2 2 ⎟.
⎝2−n−(−2)−n 3 n− −3 n⎠ 22
22
Hence, these are unbounded solutions. Consequently, the zero solution is unstable. This example demonstrates without any doubt that eigen- values do not generally provide any information about the stability of nonautonomous difference systems.
Exercises 4.3
1. Determine whether the zero solution of the system x(n + 1) = Ax(n) is stable, asymptotically stable, or unstable if the matrix A is:
⎛5 0 1⎞ ⎜12 2⎟
(a) 1 0 . (b)⎜−1 −1 5⎟. −21 ⎜⎝24⎟⎠
100 ⎛3⎞
−1 5 1.5 1 −1 −0.5 2 . (d) ⎜⎝−1.5 −0.5 1.5⎟⎠ .
(c)
2. Give another example (see Example 4.17) of a matrix A(n) such that
0.5 1 0
ρ[A(n)] < 1 and the zero solution of x(n + 1) = A(n)x(n) is unstable.
3. Give an example of a stable matrix A (i.e., ρ(A) < 1) with ∥A∥ > 1, for some matrix norm ∥∥.
4. Consider the autonomous (time-invariant) system (4.3.6). Prove the following statements:
(i) The zero solution is stable if and only if it is uniformly stable.
192 4. Stability Theory
(ii) The zero solution is asymptotically stable if and only if it is uniformly asymptotically stable.
5. Use Corollary 4.16 to determine whether or not the zero solution of x(n + 1) = A(n)x(n) is uniformly stable or uniformly asymptotically stable, where A(n) is the matrix:
⎛1⎞ ⎛n⎞ ⎜−1 4 cos(n)⎟ ⎝n + 1 0⎠
(a)⎝1⎠. (b) . 0 2 sin(n) −1 1
⎛1 1⎞⎛n+2⎞ ⎜n+1 0 2sin(n)⎟ ⎜n+1 0 0⎟
(c)⎜ 1 1 sin(n) 1 cos(n)⎟. (d)⎜ 0 1 0⎟. ⎜⎝ 4 2 4 ⎟⎠ ⎝ 1 ⎠
100n+101 5
6. Prove Corollary 4.10.
7. Prove Corollary 4.11.
8. Show that if |λ| < 1, then limn→∞ |λ|nns = 0 for any given positive integer s.
9. Prove that the zero solution of system (4.3.1) is uniformly stable if and onlyifthereexistsM >0suchthat∥Φ(n,m)∥≤M,forn0 ≤m≤ n < ∞.
10. Prove that the zero solution of system (4.3.1) is asymptotically stable if and only if limn→∞ ∥Φ(n)∥ = 0.
11. Prove Theorem 4.14, part (ii).
Iterative Methods
Consider the system of linear algebraic equations Ax = b,
(4.3.12) Iterative methods are used widely to solve (4.3.12) numerically. We
where A = (aij ) is a k × k matrix.
generate a sequence x(n) using the difference equation
x(n + 1) = Bx(n) + d, (4.3.13)
where the choice of B and d depends on the iterative method used. The iterative method (4.3.13) is consistent with (4.3.12) if a solution x* of (4.3.12) is an equilibrium point of (4.3.13), i.e., if
Bx∗ + d = x∗. (4.3.14)
We now describe one such consistent method, the Jacobi iterative method. Assuming that the diagonal elements aii of A are nonzero, then D =
diag(a11, a22, . . . , akk) is nonsingular. In (4.3.13) define
B = I − D−1A, d = D−1b. (4.3.15)
This method is consistent (Exercises 4.3, Problem 12). If A is nonsingular, then x* is unique. The associated error equation may be derived by letting e(n) = x(n) − x*. Equations (4.3.13) and (4.3.14) then yield the equation
e(n + 1) = Be(n). (4.3.16)
The quantity e(n) represents the error in approximating the solution x* by the nth iterate x(n) of (4.3.13).
12.
13. 14.
(i)Provethatx(n)→x*asn→∞ifandonlyife(n)→0as n → ∞. In other words, the iterative method ((4.3.13) converges to the solution x* of (4.3.12) if and only if the zero solution of (4.3.16) is asymptotically stable).
(ii) Use Theorem 4.9 to show that the iterative method converges if and only if ρ(B) < 1.
Show that the Jacobi iterative method is consistent.
Consider (4.3.12) and (4.3.13) with the assumption that the diagonal elements of A are nonzero. Let L be the lower triangular part of A and let U be the strictly upper triangular part of A (i.e., the main diagonal of U is zero). Then A = L + U . The Gauss–Seidel iterative method defines B = −L−1U and d = L−1b. Show that this method is consistent.
In Problem 15 we consider the k-dimensional system x(n + 1) = A(n)x(n).
*15. (a)
Define H(n) = AT (n)A(n). Prove the Lagrange identity
∥x(n + 1)∥2 = xT (n + 1)x(n + 1) = xT (n)H(n)x(n). (4.3.17)
4.3 Stability of Linear Systems 193
(b) Show that all eigenvalues of H(n) as defined in part (a) are real and nonnegative.
(c) Let the eigenvalues of H(n) be ordered as λ1(n) ≤ λ2(n) ≤ · · · ≤ λk(n). Show that, for all x ∈ Rn,
λ1(n)xT x ≤ xT H(n)x ≤ λk(n)xT x. (4.3.18)
(d) Use the Lagrange identity (4.3.17) in formula (4.3.18) to show
that
n−1
λ1(i) xT (n0)x(n0) ≤ xT (n)x(n) n−1
i=n0
≤
λk(i) xT (n0)x(n0).
i=n0
194
4. Stability Theory
4.4
λ1(i) ≤ ∥Φ(n,n0)∥ ≤ λk(i). (4.3.19) i=n0 i=n0
Phase Space Analysis
or where
x1(n + 1) = a11x1(n) + a12x2(n), x2(n + 1) = a21x1(n) + a22x2(n),
x(n + 1) = Ax(n),
A= a11 a12 . a21 a22
(4.4.1)
(e) Show that
n−1 n−1 2
In this section we will study the stability properties of the second-order linear autonomous (time-invariant) systems
Recall that x* is an equilibrium point of system (4.4.1) if Ax* = x* or (A−I)x* = 0. So if (A−I) is nonsingular, then x* = 0 is the only equilibrium point of system (4.4.1). On the other hand, if (A−I) is singular, then there is a family of equilibrium points, as illustrated in Figure 4.8. In the latter case we let y(n) = x(n) − x* in (4.4.1) to obtain the system y(n + 1) = Ay(n), which is identical to system (4.4.1). Thus the stability properties of any equilibrium point x∗ ̸= 0 are the same as those of the equilibrium point x* = 0. Henceforth, we will assume that x* = 0 is the only equilibrium point of system (4.4.1).
Let J = P−1AP be the Jordan form of A. Then J may have one of the
following canonical forms:
λ1 0 , 0 λ2
(a)
distinct real eigenvalues λ1, λ2
If we let or
λ 1 , 0 λ
(b) repeated real
eigenvalue λ
α β . −β α
(c)
complex conjugate eigenvalues
λ = α ± iβ
(4.4.2)
y(n) = P −1x(n), x(n) = P y(n),
(4.4.3)
FIGURE 4.8. λ1 < λ2 < 1, asymptotically stable node.
then system (4.4.1) becomes
y(n + 1) = Jy(n). (4.4.4)
If x(0) = x0 is an initial condition for system (4.4.1), then y(0) = y0 = P−1x0 will be the corresponding initial condition for system (4.4.4). Notice that the qualitative properties of the equilibrium points of systems (4.4.1) and (4.4.4) are identical.
Our program is to sketch the phase space of system (4.4.4) in cases (a),
(b), and (c). Starting with an initial value
y10 y20
in the y1y2-plane, we trace the movement of the points y(1), y(2), y(3), . . . . Essentially, we draw the orbit {y(n,0,y0)|n ≥ 0}. An arrow on the orbit indicates the direction of motion as time increases.
Case (a). In this case the system becomes
y1(n + 1) = λ1y1(n),
Hence
and thus
y2(n + 1) = λ2y2(n).
y 1 ( n ) = λ n1 y 1 0 , y 2 ( n ) λ n2 y 2 0
y(n) λny 2=2 20.
y0 =
4.4 Phase Space Analysis 195
y 2
y 1
y1(n) λ1 y10
If |λ1| > |λ2|, then limn→∞ y2(n)/y1(n) = 0, and if |λ1| < |λ2|, then
limn→∞ |y2(n)| = ∞ (see Figures 4.8, 4.9, 4.10, 4.11, 4.12). |y1 (n)|
196 4. Stability Theory
y 2
y 1
FIGURE 4.9. λ1 > λ2 > 1, unstable node.
y 1
FIGURE 4.10. 0 < λ1 < 1, λ2 > 1, saddle (unstable).
Case (b). In this case,
n n−1
or
y1(n) =Jn y10 = λ nλ y10
,
y2(n)
y20 0 λn y20
y1(n) = λny10 + nλn−1y20, y2(n) = λny20.
Thus
FIGURE 4.11. 0 < λ1 = λ2 < 1, asymptotically stable node.
y2
y1
FIGURE 4.12. λ1 = 1, λ2 < λ1, degenerate node.
lim y2(n)=0. n→∞ y1(n)
y2
4.4 Phase Space Analysis 197
y1
(See Figures 4.13, 4.14.)
Case (c). In this case, the matrix A has two complex conjugate eigenvalues,
λ1 =α+iβ and λ2 =α−iβ, β̸=0.
198 4. Stability Theory
y2
y1
FIGURE 4.13. λ1 = λ2 < 1, asymptotically stable. y2
y1
FIGURE 4.14. λ1 = λ2 = 1, degenerate case (unstable). All points on the y1-axis are equilibrium points.
1 The eigenvector corresponding to λ1 = α + iβ is given by ξ1 = i
, and
the solution may be given by
1i (α+iβ)n = 1i = |λ1|n
|λ1|n(cosnω+isinnω),
cos nω + i|λ1|n sin nω , −sin nω cos nω
where ω = tan−1(β/α).
A general solution may then be given by
y1(n) = |λ1|n c1 cos nω + c2 sin nω . y2(n) −c1 sin nω + c2 cos nω
Given the initial values y1 (0) = y10 and y2 (0) = y20 , one may obtain c1 = y10 and c2 = y20. The solution is denoted by
y1(n) = |λ1|n(y10 cos nω + y20 sin nω), y2(n) = |λ1|n(−y10 sin nω + y20 cos nω).
Ifweletcosγ=y /r andsinγ=y /r,wherer =y2 +y2 ,we 100 20001020
have y1(n) = |λ1|nr0 cos(nω − γ) and y2(n) = −|λ1|nr0 sin(nω − γ). Using polar coordinates we may now write the solution as
r(n) = r0|λ1|n, θ(n) = −(nω − γ).
If |λ1| < 1, we have an asymptotically stable focus, as illustrated in Figure 4.15. If |λ1| > 1, we find an unstable focus, as shown in Figure 4.16. When |λ1| = 1, we obtain a center where orbits are circles with radii (Figure 4.17)
r0= y120+y20. y2
4.4 Phase Space Analysis 199
y1
FIGURE 4.15. |λ| < 1, asymptotically stable focus. y2
y1
FIGURE 4.16. |λ| > 1, unstable focus.
200 4. Stability Theory
y1
FIGURE 4.17. |λ| = 1, center (stable).
Using (4.4.3) one may sketch the corresponding phase space portraits in the x1x2-plane for the system of equations (4.4.1). The following example illustrates this method.
Example 4.18. Sketch the phase space portrait of the system
x(n+1)=Ax(n), whereA=
Solution The eigenvalues of A are λ1 = 1.5 and λ2 = 1 ; the corresponding
by noticing that 1 in the y1–y2 system corresponds to P1 = 2 in 0 0 0 1
the x1–x2 system, and in the y1-y2 system corresponds to the point 0 2 1 −1
P 1 = −1 in the x1–x2 system. The y1-axis is rotated by θ1 = tan (0.5) to the x1-axis, and the y2-axis is rotated by θ2 = tan−1(−0.5) to the x2-axis.
Furthermore, the initial point
y10 = 1
y20 0
for the canonical system corresponds to the initial point
x10 =P 1 = 2 . x20 0 1
The phase space portrait of our system is shown in Figure 4.19. Basically,
theaxisx1 iscξ1 =2c,c∈R,andtheaxisx2 iscξ2 =2c,c∈R. c −c
11 0.25 1
.
2 2 2 eigenvectors are ξ1 = 1 and ξ2 = −1 , respectively. Thus
P−1AP=J= 1.5 0 , whereP= 2 2 . 0 0.5 1 −1
Figure 4.18 shows the phase space portrait for y(n+1) = Jy(n). To find the
corresponding phase space portrait of our problem, we let x(n) = Py(n).
We define the relationship between the y1–y2 system and the x1–x2 system
y2
4.4 Phase Space Analysis 201
y1
FIGURE 4.18. Canonical saddle.
x2
x1
FIGURE 4.19. Actual saddle.
Example 4.19. Sketch the phase space portrait of the system x(n + 1) = Ax(n) with
13 A= −1 1 .
√√ Solution The eigenvalues of A are λ1 = 1+ 3i and λ2 = 1− 3i. The
eigenvector corresponding to λ1 is
√ √
330 ξ1= i = 0 +i 1 .
202 4. Stability Theory
If we let
then
√
P=, 01
√ −1 13
30
PAP=J=√ , −31
which is in the canonical form (4.4.2) (c). Hence, the solution of y(n + 1) = Jy(n) is
and
where
r(n) = r0|λ1|n = y120 + y120(2)n θ(n) = α − nω,
y 20 ,
√ π 3 = .
α = tan−1
Figure 4.20 depicts the orbit of −1,0. The solution is given by r(n) =
ω = tan−1
3
√ √ 3 0 −1/16 3/16
x0= 0 1 0 = 0
and is depicted in Figure 4.21. Notice that no axis rotation has occurred here.
y10
1 n n−4 16
16 (2) = 2 , θ(n) = π − (nπ)/3. The corresponding orbit in the original system has an initial point
FIGURE 4.20. Canonical unstable focus.
Exercises 4.4
FIGURE 4.21. Actual unstable focus.
1. Sketch the phase space diagram and determine the stability of the
equation x(n + 1) = Ax(n), where A is given by
0.5 0
(a) .
0.5 0 (b) .
0 2
2 1 −0.5 1
(c) 0 2 . (d) 0 −0.5 .
2. Sketch the phase space diagram and determine the stability of the
system x(n + 1) = Ax(n), where A is given by
0 2 0.6 −0.5
(a) −20. (b) 0.5 0.6 .
1 0.5 0.6 0.8 (c) −0.5 1 . (d) −0.8 0.6 .
In Problems 3 through 6, sketch the phase space diagram and determine
the stability of the system x(n + 1) = Ax(n).
11 3.A= −1 3 .
−2 1 4.A= −1 3 .
−2 1 5.A= −7 3 .
12 6.A= −1 −1 .
0 0.5
4.4 Phase Space Analysis 203
204 4. Stability Theory
7. If the eigenvalues of a real 2×2 matrix A are α+iβ, α−iβ, show
that the Jordan canonical form of A is
αβ −β α .
4.5 Liapunov’s Direct, or Second, Method
In his famous memoir, published in 1892, the Russian mathematician A.M. Liapunov introduced a new method for investigating the stability of non- linear differential equations. This method, known as Liapunov’s direct method, allows one to investigate the qualitative nature of solutions with- out actually determining the solutions themselves. Therefore, we regard it as one of the major tools in stability theory. The method hinges upon finding certain real-valued functions, which are named after Liapunov. The major drawback in the direct method, however, lies in determining the appropriate Liapunov function for a given equation.
In this section we adapt Liapunov’s direct method to difference equations. We begin our study with the autonomous difference equation
x(n + 1) = f (x(n)), (4.5.1)
where f : G → Rk,G ⊂ Rk, is continuous. We assume that x* is an equilibrium point of (4.5.1), that is, f(x∗) = x*.
Let V : Rk → R be defined as a real-valued function. The variation of V relative to (4.5.1) would then be defined as
∆V (x) = V (f(x)) − V (x)
∆V (x(n)) = V (f(x(n))) − V (x(n)) = V (x(n + 1)) − V (x(n)).
Notice that if ∆V (x) ≤ 0, then V is nonincreasing along solutions of (4.5.1). The function V is said to be a Liapunov function on a subset H of Rk if:
(i) V is continuous on H, and
(ii) ∆V (x) ≤ 0, whenever x and f(x) belong to H.
Let B(x,γ) denote the open ball in Rk of radius γ and center x defined by B(x,γ) = {y ∈ Rk|∥y − x∥ < γ}. For the sake of brevity, B(0,γ) will henceforth be denoted by B(γ). We say that the real-valued function V is positive definite at x* if:
(i) V(x∗)=0,and
(ii) V(x)>0forallx∈Bx∗,γ),x̸=x∗,forsomeγ>0.
and
x1
4.5 Liapunov’s Direct, or Second, Method 205
V (x1, x2)
FIGURE 4.22. A quadratic Liapunov function.
We now present to the reader an informal geometric discussion of the first Liapunov stability theorem. For simplicity, we will assume that our system is planar with x* = 0 as the equilibrium point. Suppose that (4.5.1) has a positive definite Liapunov function V defined on B(η). Figure 4.22 then illustrates the graph of V in a three-dimensional coordinate system, while Figure 4.23 gives the level curves V (x1, x2) = c of V in the plane. If we now let ε > 0, B(ε) then contains one of the level curves of V , say V (x) = c ̃2. The level curve V (x) = c ̃2 contains the ball B(δ) for some δ with 0 < δ ≤ ε. If a solution x(n, 0, x0) starts at x0 ∈ B(δ), then V (x0) ≤ c ̃2. Since ∆V ≤ 0, V is a monotonic nonincreasing function along solutions of (4.5.1). Hence, V (x(n)) ≤ V (x0) ≤ c ̃2 for all n ≥ 0. Thus, the solution x(n, 0, x0) will stay forever in the ball B(ε). Consequently, the zero solution is stable. The above argument contains the essence of the proof of the first Liapunov stability theorem.
Theorem 4.20 (Liapunov Stability Theorem). If V is a Liapunov function for (4.5.1) in a neighborhood H of the equilibrium point x∗, and V is positive definite with respect to x∗, then x∗ is stable. If, in addition, ∆V(x) < 0 whenever x,f(x) ∈ H and x ̸= x*, then x∗ is asymptotically stable. Moreover, if G = H = Rk and
V(x)→∞ as ∥x∥→∞, (4.5.2) then x∗ is globally asymptotically stable.
Proof. Choose α1 > 0 such that B(x∗, α1) ⊂ G ∩ H. Since f is continu- ous, there is α2 > 0 such that if x ∈ Bx∗,α2), then f(x) ∈ B(x∗,α1). Let 0<ε≤α2 begiven.Defineψ(ε)=min{V(x)|ε≤∥x−x∗∥≤α1}.Bythe Intermediate Value Theorem, there exists 0 < δ < ε such that V (x) < ψ(ε) whenever ∥x − x∗∥ < δ.
x2
206 4. Stability Theory
x2
V=c3 V=c2 V=c1
FIGURE 4.23. Level curves.
B( )
B( )
x1
Realize now that if x0 ∈ B(x∗,δ), then x(n) ∈ B(x∗,ε) for all n ≥ 0. This claim is true because, if not, there exist x0 ∈ B(x∗, δ) and a positive integer m such that x(r) ∈ B(x∗, ε) for 1 ≤ r ≤ m and x(m+1) ̸∈ B(x∗, ε). Since x(m) ∈ B(x∗, ε) ⊂ B(x∗, α2), it follows that x(m + 1) ∈ B(x∗, α1). Consequently, V (x(m+1)) ≥ ψ(ε). However, V (x(m+1)) ≤ · · · ≤ V (x0) < ψ(ε), and we thus have a contradiction. This establishes stability.
To prove asymptotic stability, assume that x0 ∈ B(x∗,δ). Then x(n) ∈ B(x∗,ε) holds true for all n ≥ 0. If {x(n)} does not converge to x*, then it has a subsequence {x(ni)} that converges to y ∈ Rk. Let E ⊂ B(x∗,α1) be an open neighborhood of y with x∗ ̸∈ E. Having already defined on E the function h(x) = V (f (x))/V (x), we may consider h as well-defined and continuous, and h(x) < 1 for all x ∈ E. Now, if η ∈ (h(y),1), then there exists α > 0 such that x ∈ B(y, α) implies h(x) ≤ η. Thus, for sufficiently large ni,
V (f(x(ni)))≤ηV (x(ni −1))≤η2V (x(ni −2))≤···≤ηniV(x0). Hence,
lim V (x(ni)) = 0. ni →∞
But since limni→∞ V (x(ni)) = V (y), this statement implies that V (y) = 0 and, consequently, y = x*.
To prove the global asymptotic stability, it suffices to show that all solu- tions are bounded and then repeat the above argument. Begin by assuming that there exists an unbounded solution x(n), and then some subsequence {x(ni)} → ∞ as ni → ∞. By condition (4.5.2), this assumption im- plies that V (x(ni)) → ∞ as ni → ∞, which is a contradiction, since V (x0) > V (x(ni)) for all i. This concludes the proof. P
4.5 Liapunov’s Direct, or Second, Method 207
The result on boundedness has its own independent importance, so we give it its due respect by stating it here as a separate theorem.
Theorem 4.21. If V is a Liapunov function on the set {x ∈ Rk|∥x∥ > α} for some α > 0, and if condition (4.5.2) holds, then all solutions of (4.5.1) are bounded.
Proof. (Exercises 4.5, Problem 7.) P Example 4.22. Consider the following second-order difference equation:
x(n+1)= αx(n−1), β>0. 1 + βx2(n)
This equation is often called an equation with delay. There are three equilibrium points, namely, x* = 0 and
x(n − 1) and y2(n) = x(n). Then we obtain the system y1(n + 1) = y2(n),
y2(n + 1) = αy1(n) . 1 + βy2(n)
Consider the stability of the equilibrium point (0, 0). Our first choice of a Liapunov function will be V (y1, y2) = y12 + y2. This is clearly continuous and positive definite on R2:
∆V (y1(n), y2(n)) = y12(n + 1) + y2(n + 1) − y12(n) − y2(n).
x∗ = ±
if α > 1. Let us first change the equation into a system by letting y1(n) =
(α−1) β
Thus,
∆V (y1(n), y2(n)) = [1 + βy2(n)]2 − 1 y12(n) ≤ (α2 − 1)y12(n). (4.5.3)
Ifα2 ≤1,then∆V ≤0.Inthiscasex∗ =0wouldbetheonlyequilib- rium point, and by Theorem 4.20, the origin is stable (Figure 4.24). Since lim∥x∥→∞ V (x) = ∞, Theorem 4.21 implies that all solutions are bounded. Since ∆V = 0 for all points on the y2-axis, Theorem 4.20 fails to determine asymptotic stability for this equation.
This situation is typical in most of the problems encountered in applica- tions in science and engineering. Therefore, a finer and more precise analysis is required. This need leads us to LaSalle’s invariance principle, which will be presented shortly.
To prepare for the introduction of our major theorem, we ought to familiarize ourselves with some vital terminology:
(i) For a subset G ⊂ Rk, x is a limit point of G if there exists a sequence {xi}inGwithxi →xasi→∞.
α2
208 4. Stability Theory
(ii) The closure G of G is defined to be the union of G and all of its limit points.
(iii) After considering (4.5.1), the positive orbit O+(x0) is defined as O+(x0) = {x(n,0,x0)|n ∈ Z+}. Since we will only deal with posi- tive orbits, O+(x) will be denoted by O(x). We will denote O+(x0) by O(x0).
(iv) The limit set Ω(x0), also referred to as the positive limit set, of x0 is the set of all positive limit points of x0. Thus, Ω(x0) = {y ∈ Rk|x(ni) → y as ni → ∞ for some subsequence {ni} of Z+}.
(v) A set A is positively invariant if O(x0) ⊂ A for every x0 ∈ A. One may easily show that both O(x0) and Ω(x0) are (positively) invariant.
The nagging question still persists as to whether or not Ω(x0) is nonempty for a given x0 ∈ Rk. The next lemma satisfies that question.
Theorem 4.23. Let x0 ∈ Rk and let Ω(x0) be its limit set in (4.5.1). Then the following statements hold true:
∞ ∞ n
(i) Ω(x0) = {f (x0)} =
∞ ∞ i=0 n=i
{x(n)} .
i=0 n=i
(ii) If fj(x0) = y0, j ∈ Z+, then Ω(y0) = Ω(x0) .
(iii) Ω(x0) is closed and invariant.
(iv) If the orbit O(x0) is bounded, then Ω(x0) is nonempty and bounded. Proof.
(i) Lety∈Ω(x0).Thenfni(x0)→yasni →∞.Nowforeachi,there
∞ i=0
exists a positive integer Ni such that fnj (x0) ∈
{fn(x0)} for all
nj ≥ Ni. Thus y ∈ ∞ ∞ n
∞ n n=i
∞ n
{f (x0)}. Thus for each i
n=i
{f (x0)} for every N and, consequently, y ∈
i=0 n=i
∞ ∞ n
{f (x0)}. This proves one inclusion and, conversely, let y ∈
{f (x0)}. Then for each i, y ∈
there exists fni(x0) ∈ By(x0), with n1 < n2 < n3 < ··· and ni → ∞
i=0 n=i
asi→∞.Clearly,fni(x0)→yasnN →∞andhencey∈Ω(x0). (ii) The proof of (ii) is left to the reader as Problem 5.
(iii) Since the closure of a set is closed,
∞ n
{f (x0)} is closed. Now that
n=i
Ω(x0) is closed follows from the fact that the intersection of closed sets
is closed.
4.5 Liapunov’s Direct, or Second, Method 209
To show that Ω(x0) is invariant, let y ∈ Ω(x0). Then fni(x0) → y as ni → ∞. Since f is continuous, it follows that fni+1(x0) = f (fni (x0) → f(y)). Hence f(y) ∈ Ω(x0) and Ω(x0) is thus invariant.
(iv) This is left to the reader as Problem 6. P Let V be a positive Liapunov function on a subset G of Rk. Define
E = {x ∈ G|∆V (x) = 0}.
Let M be the maximal invariant subset of E, that is, define M as the union
of all invariant subsets of E.
Theorem 4.24 (LaSalle’s Invariance Principle) [88]. Suppose that V is a positive definite Liapunov function for (4.5.1) in G ⊂ Rk. Then for each bounded solution x(n) of (4.5.1) that remains in G for all n ∈ Z+, there exists a number c such that x(n) → M ∩ V −1(c) as n → ∞.
Proof. Let x(n) be a bounded solution of (4.5.1) with x(0) = x0 and such that x(n) is bounded and remains in G. Then, by Theorem 4.23, φ̸=Ω(x0)⊂G.Thus,ify∈Ω(x0),thenx(ni)→y asni →∞for some subsequence ni ∈ Z+. Since V (x(n)) is nonincreasing and bounded below, limn→∞ V (x(n)) = c for some number c. By the continuity of V, it follows that V(x(ni)) → V(y) as ni → ∞, and thus V(y) = c. This implies that V (Ω(x0)) = c and, consequently, Ω(x0) ⊂ V −1(c). Moreover, ∆V (y) = 0 for every y ∈ Ω(x0). This implies that Ω(x0) ⊂ E. But, since Ω(x0) is (positively) invariant, Ω(x0) ⊂ M. Therefore, x(n) → Ω(x0) ⊂ M ∩ V −1(c) as n → ∞. P
Example 4.22 revisited. Let us reexamine Example 4.22 in light of LaSalle’s invariance principle. We will consider three cases:
Case 1. α2 = 1. The set E consists of all the points on the x- and y- axes. We have two subcases to consider. Subcase (i): α = 1. If y1(0) = a and y2(0) = 0, then y1(1) = 0 and y2(1) = a, and y1(2) = a, y2(2) = 0. Therefore, any solution starting on either axis is of period 2, and M = E. Subcase (ii): α = −1. Then 0+ (a, 0) = {(a, 0), (0, −a), (−a, 0), (0, a)}. Thus any solution starting on either axis is of period 4, and M = E again. Hence all solutions converge to (a, 0), (−a, 0), (0, a), or (0, −a). Clearly, the zero solution is not asymptotically stable.
Case 2. Because α2 < 1, E is equal to the y-axis and M = {(0, 0)}. Thus, all solutions converge to the origin. Hence the origin is globally asymptoti- cally stable. Figure 4.24 depicts the phase portrait for α = 0.5. Notice the difference in the way solutions in quadrants I and III begin, compared to the way the solutions in quadrants II and IV commence.
Case 3. α2 > 1. In this case, LaSalle’s invariance principle does not aid us in determining the stability of the solution. In other words, the stability is indeterminable.
210 4. Stability Theory
(-0.9 , 0.8)
y 2
(0.9 , 0.8)
y 1
(0.9 , -0.8)
(-0.9 , -0.8)
FIGURE 4.24. A globally asymptotically stable equilibrium.
Sometimes, we may simplify the difference equation by applying a simple basic transformation to the system. For instance, one might translate the system into polar coordinates (r, θ), where x1 = r cos θ, x2 = r sin θ. The following example demonstrates the effectiveness of this method.
Example 4.25. Consider the difference system x1(n + 1) = x21(n) − x2(n),
x2(n + 1) = 2×1(n)x2(n). Let x1(n) = r(n) cos θ(n) and x2(n) = r(n) sin θ(n).
Then
r(n + 1) cos θ(n + 1) = r2(n) cos2 θ(n) − r2(n) sin2 θ(n)
and
= r2(n) cos 2θ(n),
r(n + 1) sin θ(n + 1) = 2r2 sin θ(n) cos θ(n)
= r2(n) sin 2θ(n).
(4.5.4)
(4.5.5)
Dividing (4.5.4) by (4.5.5), we get
θ(n + 1) = 2θ(n).
Substituting this into (4.5.4), we obtain
r(n + 1) = r2(n).
We may write this solution as r(n) = [r(0)]2n and θ(n) = 2nθ(0). The equilibrium points are (0, 0) and (1, 0).
4.5 Liapunov’s Direct, or Second, Method 211
1
FIGURE 4.25. Unstable limit cycle. Three initial values (0.6,0.8), (0.6,0.81), (0.6, 0.79).
For r(0) < 1, limn→∞ r(n) = 0. Thus solutions starting inside the unit disk spiral toward the origin. Consequently, the origin is asymptotically stable (not globally), as shown in Figure 4.25.
For r(0) > 1, we have limn→∞ r(n) = ∞, and hence solutions that start outside the unit disk spiral away from the unit circle to ∞. This occurrence makes the equilibrium point (1, 0) unstable.
For r(0) = 1,r(n) = 1, for all n ≥ 0. Therefore, the circle is an
invariant set, with very complicated dynamics. For instance, the solu-
tion starting at 1,π will reach the equilibrium point (1,0) in three π 4π
iterations: 1,
2π 4 2
, 1, (1, π), (1, 0). However, the solution that starts at
1, 3 is a 2-cycle. In general, (1,θ) is periodic, with period m, if and
only if 2mθ = θ+2kπ for some integer k, i.e., if and only if θ =
(2kπ)/2m − 1, k = 0, 1, 2, . . . , 2m. For m = 3, θ = 2π , 4π , 6π , 8π , 10π , 12π . 2π 4π 2π 8π 2π 4π 14π 16π 7 7 7 7 7 7
Form=4,θ= 15,15, 5 ,15, 3 , 5 , 15 , 15 ,….
Notice here that θ is essentially the (2m − 1)th root of 1. Hence, the set
of periodic points (1, θ) densely fills the unit circle (Exercises 4.5, Problem 8). Furthermore, for every m = 1,2,…, there is a periodic point on the unit circle of that period m.
Now, if θ = απ, α irrational, then obviously, θ ̸= (2kπ)/2m − 1 for any m, and thus any solution starting at (1, απ) cannot be periodic. However, its orbit is dense within the unit circle, that is, O(x) is the unit circle (Exercises 4.5, Problem 8).
Sometimes, some simple intuitive observations make it much easier to show that an equilibrium point is not asymptotically stable. The following example illustrates this remark.
212 4. Stability Theory
Example 4.26. Consider the planar systems
x1(n + 1) = 2×2(n) − 2×2(n)x21(n),
x2(n + 1) = 1×1(n) + x1(n)x2(n). 2
We find three equilibrium points:
111 1 (0,0), √ ,√ , −√ ,−√ .
2222
Let us consider the stability of (0, 0). If V (x1, x2) = x21 + 4×2, then
∆V (x1(n), x2(n)) = 4×2(n) − 8×2(n)x21(n) + 4×2(n)x41(n) + x21(n) + 4×21(n)x2(n) + 4×21(n)x42(n) − x21(n) − 4×2(n)
= 4×21(n)x2(n)[x21(n) + x2(n) − 1].
If x21 + x2 ≤ 1, then ∆V (x1, x2) ≤ 0. a
For any real number a, the solution with an initial value of x0 = a 0 0
is periodic with period 2 and with orbit 0 , a/2 , and a solution with an initial value of x0 = 0 is also periodic with period 2. Hence, the zero
a
solution cannot be asymptotically stable. However, it is stable according to Theorem 4.20. (Figure 4.26 depicts the phase space portrait near the origin.)
We now turn our attention to the question of instability. We are inter- ested in finding conditions on Liapunov functions under which the zero solution is unstable. Here is a widely used theorem in this area.
(-0.4 , 0.3)
(0.4 , 0.3)
x2(n)
x (n) 1
(-0.4 , -0.3)
(0.4 , -0.3)
FIGURE 4.26. Stable equilibrium.
2 L e t V ( x 1 , x 2 ) = x 21 + 1 6 x 2 2 . T h e n
4.5 Liapunov’s Direct, or Second, Method 213
Theorem 4.27. If ∆V is positive definite in a neighborhood of the origin and there exists a sequence ai → 0 with V (ai) > 0, then the zero solution of (4.5.1) is unstable.
Proof. Let ∆V(x) > 0 for x ∈ B(η),x ̸= 0,V(0) = 0. We will prove Theorem 4.27 by contradiction, first assuming that the zero solution is stable, in which case, for ε < η, there would exist δ < ε such that ∥x0∥ < δ implies ∥x(n, 0, x0)∥ < ε, n ∈ Z+.
Sinceai →0,pickx0 =aj forsomejwith∆V(x0)>0and∥x0∥<δ. Hence 0(x0) ⊂ B(ε) ⊂ B(η) is closed and bounded (compact). Since its domain is compact, V (x(n)) is also compact, and therefore bounded above. Since V (x(n)) is also increasing, it follows that V (x(n)) → c. Fol- lowing the proof of LaSalle’s invariance principle, it is easy to see that limn→∞ x(n) = 0. Therefore, we would be led to believe that 0 < V (x0) < limn→∞ V (x(n)) = 0. This statement is infeasible—so the zero solution cannot be stable, as we first assumed. The zero solution of (4.5.1) is thus unstable.
The conclusion of the theorem also holds if ∆V is negative definite and
V (ai) < 0.
Example 4.28. Consider the systems
P
if x1(n) ̸= 0.
(4.5.6) (4.5.7)
x1(n + 1) = 4x2(n) − 2x2(n)x21(n), x2(n + 1) = 1x1(n) + x1(n)x2(n).
∆V (x1(n), x2(n)) = 3x21(n) + 16x21(n)x42(n) + 4x41x2 > 0 Hence, by Theorem 4.27 the zero solution is unstable. Example 4.29. First, contemplate the systems
x1(n + 1) = x1(n) + x2(n) + x21(n), x2(n + 1) = x2(n).
Notice that (0, 0) is an equilibrium of the systems. Its linear component is denoted by x(n + 1) = Ax(n), where
10 A=,
01
and thus ρ(A) = 1. Let V (x) = x1 + x2 be a Liapunov function. Then
∆V[x(n)]=x21+x2 >0, if(x1,x2)̸=(0,0).
Theorem 4.27 implies that the zero solution of system (4.5.6) is unstable.
214 4. Stability Theory
Let us now ponder system (4.5.7), with the same linear component as system (4.5.6):
x1(n + 1) = x1(n) − x31(n)x2(n),
x2(n + 1) = x2(n). (4.5.8)
We let V (x) = x21 + x2 be a Liapunov function for system (4.5.7). Then ∆V [x(n)] = x41(n)x2(n) −2 + x21(n)x2(n) .
Hence, ∆V (x) ≤ 0 if x21x2 ≤ 2. It follows from Theorem 4.27 that the zero solution of system (4.5.7) is stable.
We conclude from this discussion that if ρ(A) = 1, then the zero solution of the nonlinear equation may be either stable or unstable, thus proving part (i) of Theorem 4.38.
We conclude this section with a brief discussion of Liapunov functions for linear autonomous systems. In Section 4.3, we noticed that the condition for asymptotic stability of the difference equation (4.3.6) is that ρ(A) < 1. This condition requires the computation of the eigenvalues of A. Using the second method of Liapunov, such computation is unnecessary. Before introducing Liapunov’s method, however, we need to recall the definition of a positive definite matrix. Consider the quadratic form V (x) for a k × k real symmetric matrix B = (bij ):
k k i=1 j=1
A matrix B is said to be positive definite if V(x) is positive definite. Sylvester’s criterion is the simplest test for positive definiteness of a matrix. It merely notes that a real symmetric matrix B is positive definite if and only if the determinants of its leading principal minors are positive, i.e., if and only if
bbb b11 b12 11 12 13
b11 >0, b b >0, b21 b22 b23>0,…,detB>0. 12 22 b31 b32 b33
The leading principal minors of matrix B are B itself and the minors
obtained by removing successively the last row and the last column. For
instance, the leading principal minors of
are
0 −1 1
32 25
V (x) = xT Bx =
bijxixj.
⎛⎞
320 B=⎜⎝2 5 −1⎟⎠
(3),
, B,
4.5 Liapunov’s Direct, or Second, Method 215
all of which have positive determinants. Hence, B is positive definite. Notice that, for x = (x1, x2, x3)T ,
V(x)=xTBx=3×21 +5×2 +x23 +4x1x2 −2x2x3 >0 for all x ̸= 0, and V (0) = 0.
On the other hand, given
V(x)=ax21 +bx2 +cx23 +dx1x2 +ex1x3 +fx2x3,
one may write
where
V (x) = xT Bx, ⎛⎞
a d/2 e/2
B = ⎜⎝ d / 2 b f / 2 ⎟⎠ .
e/2 f/2 c
Hence V is positive definite if and only if B is. We now make a useful observation. Note that if B is a positive definite symmetric matrix, then all eigenvalues of B are positive (Exercises 4.5, Problem 14). Furthermore, if λ1,λ2,…,λk are the eigenvalues of B with
then
λmin = min{|λi||1 ≤ i ≤ k},
λmax = ρ(A) = max{|λi||1 ≤ i ≤ k},
λmin∥x∥2 ≤ V (x) ≤ λmax∥x∥2, (4.5.9)
for all x ∈ Rk, where V(x) = xTBx, and ∥.∥ is the Euclidean norm (Exercises 4.5, Problem 15).
If B is a positive definite matrix, we let V (x) = xT Bx be a Liapunov function of (4.3.6). Then, relative to (4.3.6),
∆V (x(n)) = xT (n)AT BAx(n) − xT (n)B(n)
= xT (AT BA − B)x. (4.5.10)
Thus∆V <0ifandonlyif
AT BA − B = −C (4.5.11)
for some positive definite matrix C. Equation (4.5.11) is labeled the Li- apunov equation of the system of equations (4.3.6). The above argument establishes a sufficient condition for the asymptotic stability of the zero solution of (4.3.6). It is also a necessary and vital condition, as may be seen by the following result.
216 4. Stability Theory
Theorem 4.30. The zero solution of (4.3.6) is asymptotically stable if and only if for every positive definite symmetric matrix C, (4.5.11) has a unique solution B that is also symmetric and positive definite.
Proof. Assume that the zero solution of (4.3.6) is asymptotically stable. Let C be a positive definite symmetric matrix. We will show that the Liapunov equation (4.5.11) has a unique solution B. Multiply (4.5.11) from the left by (AT )r and from the right by Ar to obtain
Hence
n→∞
n n
and
lim
(AT )r+1BAr+1 − (AT )rBAr ∞
= − lim n→∞
(AT )rCAr
(4.5.12)
r=0
lim
r=0
n→∞
B − (AT )n+1BAn+1 =
(AT )rCAr.
(AT )r+1BAr+1 − (AT )rBAr = −(AT )rCAr.
Using Theorem 4.13, part (ii), we conclude that ρ(A) < 1 and, consequently, ρ(AT ) < 1. This implies that limn→∞(AT )n+1BAn+1 = 0. Thus formula (4.5.12) yields
∞ r=0
It is straightforward to prove that formula (4.5.13) gives a solution of (4.5.11) (Exercises 4.5, Problem 16). But since there is a norm such that ∥AT∥ < 1 and ∥A∥ < 1, it may be shown that the series in formula (4.5.13) converges (Exercises 4.5, Problem 16). It is easy to verify that B is symmetric and positive definite (Exercises 4.5, Problem 16). P
Remark: Note that from the proof preceding the statement of Theorem 4.30, the zero solution of (4.3.6) is asymptotically stable if (4.5.11) has a unique, symmetric, and positive definite matrix B for some (not all) positive definite matrices C. Indeed, one may allow C to be the identity matrix I. In this case a solution of (4.5.11) is given by
∞ r=0
Corollary 4.31. If ρ(A) > 1, then there exists a real symmetric matrix B that is not positive semidefinite such that (4.5.11) holds for some positive definite matrix C.
Proof. This follows from Theorem 4.30 and is left to the reader as Problem 17 of Exercises 4.5. P
B =
(AT )rCAr. (4.5.13)
B =
(AT )rAr. (4.5.14)
r=0
4.5 Liapunov’s Direct, or Second, Method 217
Exercises 4.5
1. Consider the planar system
x1(n + 1) = x2(n)/[1 + x21(n)], x2(n + 1) = x1(n)/[1 + x2(n)].
Find the equilibrium points and determine their stability.
2. Consider the planar system
x1(n + 1) = g1(x1(n), x2(n)),
x2(n + 1) = g2(x1(n), x2(n)),
with g1(0, 0) = g2(0, 0) = 0 and g1(x1, x2)g2(x1, x2) > x1x2, for every point x = (x1, x2) in a neighborhood of the origin. Show that the origin is unstable.
*3. Consider the system
x1(n + 1) = ax2(n)/[1 + x21(n)], x2(n + 1) = bx1(n)/[1 + x2(n)].
(i) Find conditions on a and b under which:
(a) the zero solution is stable, and
(b) the zero solution is asymptotically stable.
(ii) Find the attractor when a2 = b2 = 1.
4. Prove that the zero solution of
x1(n + 1) = x2(n) − x2(n)[x21(n) + x2(n)], x2(n + 1) = x1(n) − x1(n)[x21(n) + x2(n)],
is asymptotically stable.
5. Prove Theorem 4.23, part (ii).
6. Prove Theorem 4.23, part (iv).
7. Prove Theorem 4.21.
8. In Example 4.25:
(a) Show that the orbit starting at the point (1, απ), where α is any irrational number, is dense on the unit circle.
(b) Show that the set of periodic points (1,θ) is dense on the unit circle.
218
4. Stability Theory
*9.
Suppose that:
(i) V is a Liapunov function of system equation (4.5.1) on Rk,
(ii) Gλ = {x|V (x) < λ} is bounded for each λ, and
(iii) M is closed and bounded (where M is the maximal invariant set
in E).
(a) Prove that M is a global attractor, i.e., Ω(x0) ⊂ M for all
x0 ∈Rk.
(b) Suppose that M = {0}. Verify that the origin is globally
asymptotically stable.
Show that the sets Gλ defined in the preceding problem are bounded
if V (x) → ∞ as ∥x∥ → ∞.
(Project). Suppose that V : Rk → R is a continuous function with ∆2V(x(n)) > 0 for x(n) ̸= 0, where x(n) is a solution of (4.5.1). Prove that for any x0 ∈ Rk, either x(n,x0) is unbounded or limn→∞ x(n, x0) = 0.
(Project). Wade through Problem 11 again, after replacing the condition ∆2V (x(n)) > 0 by ∆2V (x(n)) < 0.
Contemplate the planar system
x(n + 1) = y(n),
y(n + 1) = x(n) + f(x(n)).
If ∆[y(n)f(x(n))] > 0 for all n ∈ Z+, prove that the solutions are
either unbounded or tend to the origin.
Prove that if B is a positive definite symmetric matrix, then all its eigenvalues are positive.
Let B be a positive definite symmetric matrix with eigenvalues λ1 ≤ λ2 ≤···≤λk.ForV(x)=xTBx,showthatλ1∥x∥2 ≤V(x)≤λ2∥x∥2 for all x ∈ Rk.
10. *11.
*12. 13.
14. 15.
16. (a) Show that the matrix B = ∞r=0(AT)rCAr is symmetric and positive definite if ∥A∥ < 1 and C is a positive definite symmetric matrix.
(b) Show that the matrix B in formula (4.5.13) is a solution of (4.5.11).
17.
Prove Corollary 4.31.
4.6 Stability by Linear Approximation 219
4.6 Stability by Linear Approximation
The linearization method is the oldest method in stability theory. Scien- tists and engineers frequently use this method in the design and analysis of control systems and feedback devices. The mathematicians Liapunov and Perron originated the linearization method, each with his own unique approach, in their work with the stability theory of differential equations. In this section we adapt Perron’s approach to our study of the nonlinear systems of difference equations
y(n + 1) = A(n)y(n) + g(n, y(n)) (4.6.1) using their linear component
z(n + 1) = A(n)z(n), (4.6.2)
whereA(n)isak×kmatrixforalln∈Z+ andg:Z+×G→Rk,G⊂Rk,is a continuous function. One may perceive system (4.6.1) as a perturbation of system (4.6.2). The function g(n, y(n)) represents the perturbation due to noise, inaccuracy in measurements, or other outside disturbances. System (4.6.1) may arise from the linearization of nonlinear equations of the form
x(n + 1) = f (n, x(n)), (4.6.3) where f : Z+ ×G → Rk , G ⊂ Rk , is continuously differentiable at an equilib-
rium point y* (i.e., ∂f |y* exists and is continuous on an open neighborhood ∂yi
of y* for 1 ≤ i ≤ k). We now describe the linearization method applied to system (4.6.3). Let us write f = (f1 , f2 , . . . , fk )T . The
⎛∂f1(n,0) ∂f1(n,0) ··· ∂f1(n,0)⎞
⎜ ∂y1 ∂y2 ∂yk ⎟ ∂f(n,y) ∂f(n,0) ⎜∂f2(n,0) ∂f2(n,0) ··· ∂f2(n,0)⎟
∂y = ∂y =⎜ ∂y1 y=0 ⎜ .
⎜⎝ ∂ fk (n, 0) ∂y1
∂y2 .
∂ fn (n, 0) ∂y2
· · ·
∂yk ⎟. . ⎟
∂ fk (n, 0) ⎟⎠ ∂yk
(4.6.4)
For simplicity, ∂f(n,x∗) is denoted by Df(n,x∗). Letting ∂x
in (4.6.3) yields
y(n) = x(n) − x∗
y(n + 1) = f(n, y(n) + x∗) − x∗
= ∂f (n, x∗)y(n) + g(n, y(n)) ∂x
where g(n, y(n)) = f(n, y(n) + x∗) − x∗ − ∂f (n, x∗)y(n). ∂x
220 4. Stability Theory
If we let A(n) = ∂f (n, x∗), then we obtain (4.6.1). From the assumptions ∂x
on f, we conclude that g(n,y) = o(∥y∥) as ∥y∥ → 0. This means, given ε > 0, there exists δ > 0 such that ∥g(n, y)∥ ≤ ε∥y∥ whenever ∥y∥ < δ, for all n ∈ Z+.
Notice that when x∗ = 0, we have
g(n, y(n)) = f (n, y(n)) − Df (n, 0)y(n)
= f (n, y(n)) − A(n)y(n).
An important special case of system (4.6.3) is the autonomous system
y(n + 1) = f (y(n)), (4.6.5)
which may be written as
y(n + 1) = Ay(n) + g(y(n)), (4.6.6)
where A = f′(0) is the Jacobian matrix of f at 0, and g(y) = f(y) − Ay. Since f is differentiable at 0, it follows that g(y) = o(y) as ∥y∥ → 0. Equivalently,
lim ∥g(y)∥ = 0. ∥y∥→0 ∥y∥
Remarks:
(a) Observe that whether the linearization is about a nontrivial equilib- rium point x∗ ̸= 0 or a trivial equilibrium x∗ = 0, g(n, 0) = 0 (g(0) = 0) for all n ∈ Z+. Hence the zero solution of (4.6.1) corresponds to the equilibrium point x∗ that we linearize about.
(b) If one wishes to study a nontrivial equilibrium point x∗ ̸= 0, then by virtue of (a), we have two options. The first option is to linearize about x∗. The second option is to make the change of variable y(n) = x(n) − x∗ as in (4.6.4). In the new system, y∗ = 0 corresponds to x∗. Then we linearize the new system about y∗ = 0. The latter option is simple in computation as it is usually used if x∗ is given explicitly. The former option is used normally if x∗ is given implicitly or we have multiequilibria.
Before commencing our stability analysis we must consider a simple but important lemma. This lemma is the discrete analogue of the so-called Gronwall inequality, which is used, along with its variations, extensively in differential equations.
Lemma 4.32 (Discrete Gronwall Inequality). Let z(n) and h(n) be two sequences of real numbers, n ≥ n0 ≥ 0 and h(n) ≥ 0. If
⎡⎤
z(n) ≤ M z(n0) +
h(j)z(j)
n−1 ⎣⎦
j =n0
for some M > 0, then
z(n) ≤ z(n0)
z(n) ≤ z(n0) exp ⎡⎤
n−1 ⎣⎦
j =n0
Proof. Let u(n) = M
u(n0) +
h(j)u(j)
Thus
y(n, n0, y0) = Φ(n, n0)y0 +
n−1
j =n0
j =n0
∥y(n)∥ ≤ Mη
(n−n0)
∥y0∥ + Mη
n−1
−1 (n−j)
4.6 Stability by Linear Approximation
221
n−1
j =n0
n ≥ n0, n ≥ n0.
, u(n0) = z(n0).
j =n0
[(1 + Mh(j)], ⎡⎤
(4.6.7)
(4.6.8)
(4.6.9)
n−1 ⎣⎦
j =n0
Since h(j) ≥ 0 for all j ≥ n0, it follows that z(n) ≤ u(n) for all n ≥ n0. From (4.6.9) we have u(n+1)−u(n) = Mh(n)u(n), or u(n+1) = [1+ Mh(n)]u(n). By formula (1.2.3) we obtain
n−1
u(n) =
This proves formula (4.6.7). The conclusion of formula (4.6.8) follows by
noting that 1 + Mh(j) ≤ exp(Mh(j)). P
Theorem 4.33. Assume that g(n,y) = o(∥y∥) uniformly as ∥y∥ → 0. If the zero solution of the linear system (4.6.2) is uniformly asymptotically stable, then the zero solution of the nonlinear system (4.6.1) is exponentially stable.
Proof. From (4.3.5) it follows that ∥Φ(n,m)∥ ≤ Mη(n−m),n ≥ m ≥ n0, for some M ≥ 1 and η ∈ (0,1). By the variation of constants formula (3.2.12), the solution of (4.6.6) is given by
For a given ε > 0 there is δ > 0 such that ∥g(j,y)∥ < ε∥y∥ whenever ∥y∥ < δ. So as long as ∥y(j)∥ < δ, (4.6.10) becomes
η−n∥y(n)∥ ≤ M η−n0 ∥y0∥ +
εη−j−1∥y(j)∥ .
(4.6.11)
Mh(j) ,
[1 + Mh(j)]u(n0).
⎡⎤
n−1 ⎣⎦
j =n0
Φ(n, j + 1)g(j, y(j)).
η ∥g(j, y(j)∥.
(4.6.10)
222 4. Stability Theory
Letting z(n) = η−n∥y(n)∥ and then applying the Gronwall inequality (4.6.7), one obtains
Thus,
η−n∥y(n)∥ ≤ η−n0 ∥y0∥
n−1
j =n0
[1 + εη−1M].
∥y(n)∥ ≤ ∥y0∥(η + εM)(n−n0).
(4.6.12)
Choose ε < (1 − η)/M. Then η + εM < 1. Thus ∥y(n)∥ ≤ ∥y0∥ < δ for all n ≥ n0 ≥ 0. Therefore, formula (4.6.11) holds and, consequently, by virtue of formula (4.6.12), we obtain exponential stability. P
Corollary 4.34. If ρ(A) < 1, then the zero solution of (4.6.6) is exponentially stable.
Proof. Using Theorem 4.13, the corollary follows immediately from Theorem 4.33. P
Corollary 4.35. If ∥f′(0)∥ < 1, then the zero solution of (4.6.5) is exponentially stable.
Proof. Since ρ(f′(0)) ≤ ∥f′(0)∥, the proof follows from Corollary 4.34. P
A Remark about Corollaries 4.34 and 4.35
It is possible that ∥A∥ ≥ 1 but ρ(A) < 1. For example,
A = 0.5 1 , ∥A∥2 = ρ(AT A) = 0.75 + (√2/2) > 1, 0 0.5
∥A∥∞ = 3, ∥A∥1 = 3. 22
However, ρ(A) = 1. With the above matrix A, the zero solution of the 2
system x(n + 1) = Ax(n) + g(x(n)) is exponentially stable, provided that g(x) = o(x) as ∥x∥ → 0. Obviously, Corollary 4.35 fails to help us in deter- mining the stability of the system. However, even with all its shortcomings, Corollary 4.35 is surprisingly popular among scientists and engineers.
It is also worthwhile to mention that if ρ(A) < 1, there exists a nonsin- gular matrix Q such that ∥Q−1AQ∥ < 1 [85]. One may define a new norm on A,∥A∥ = ∥Q−1AQ∥, and then apply Corollary 4.35 in a more useful way.
Let us return to our example where
0.5 1 A=.
0 0.5
Let
Then
and
10 Q=.
0α
Q−1AQ= 0.5 α . 0 0.5
4.6 Stability by Linear Approximation 223
1⎟⎠
In this case, we let Q = diag(1,α,α2,...,αk−1), where k is the order of A.
⎜ . ∥Q−1AQ∥ = ⎜. ⎜⎝ . . .
λ
.⎟, α⎟⎠
10 0 1/α
Q−1 =
We have ∥Q−1AQ∥1 = α + 0.5. If we choose α < 0.5, then ∥Q−1AQ∥1 < 1. The above procedure may be generalized to any Jordan block
⎛λ 1 ··· 0⎞
⎜ . A=⎜. λ
. ⎟ .⎟.
⎜⎝. . 00λ
Hence,
and ∥Q−1AQ∥1 = |λ| + |α| (see Exercises 4.1, Problems 3 and 4). Con- sequently, if |λ| < 1, one may choose an α such that |λ| + |α| < 1, so that under the matrix norm ∥A∥ = ∥Q−1 AQ∥1 , ∥A∥ < 1. We now give two examples to illustrate the preceding conclusions.
Example 4.36. Investigate the stability of the zero solution of the planar system
⎛λ α
··· 0⎞ . ⎟
. . . 00λ
y1(n + 1) = ay2(n)/[1 + y12(n)], y2(n + 1) = by1(n)/[1 + y2(n)].
(4.6.13)
224 4. Stability Theory
Solution Let f = (f1, f2)T , where f1 = ay2(n)/[1 + y12(n)] and f2 = by1(n)/[1 + y2(n)]. Then the Jacobian matrix is given by
⎛∂f1(0,0) ∂f1(0,0)⎞ ∂f(0,0)=⎜ ∂y1 ∂y2 ⎟= 0 a .
∂y ⎝∂f2(0,0) ∂f2(0,0)⎠ b 0 ∂y1 ∂y2
Hence system (4.6.13) may be written as
22 y1(n + 1) = 0 a y1(n) + −ay2(n)y1(n)/[1 + y1(n)] ,
y2(n + 1) b 0 y2(n) −by2(n)y1(n)/[1 + y2(n)] or as
y(n + 1) = Ay(n) + g(y(n)).
√√
The eigenvalues of A are λ1 = ab, λ2 = − ab. Hence, if |ab| < 1, the zero
solution of the linear part x(n + 1) = Ax(n) is asymptotically stable. Since g(y) is continuously differentiable at (0, 0), g(y) is o(y). Corollary 4.34 then implies that the zero solution of (4.6.13) is exponentially stable.
Example 4.37. Pielou Logistic Delay Equation [119] In Example 2.39 we investigated the Pielou logistic equation
x(n+1)= αx(n) . 1 + βx(n)
If we now assume that there is a delay of time period 1 in the response of the growth rate per individual to density change, then we obtain the difference delay equation
y(n + 1) = αy(n) , α > 1, β > 0. (4.6.14) 1+βy(n−1)
An example of a population that can be modeled by (4.6.14) is the blowfly
(Lucilia cuprina) (see [107]). Find conditions on α, β for which the positive
equilibrium point y* = α−1 is asymptotically stable. β
Solution
Method (1): Let y(n) = y(n) − (α − 1)/β. Then (4.6.14) becomes
y(n + 1) = αy(n) − (α − 1)y(n − 1). α+βy(n−1)
(4.6.15) The equilibrium point y*(n) = 0 of (4.6.15) corresponds to y* = (α −
1)/β. To change (4.6.15) to a planar system, we let x1(n) = y(n − 1) and x2(n) = y(n).
Then
⎛ x2(n) ⎞ x1(n+1) =⎝αx2(n)−(α−1)x1(n)⎠.
x2(n + 1) α + βx1(n)
By linearizing (4.6.16) around (0, 0) we give it the new form
(4.6.16)
4.6 Stability by Linear Approximation 225
where
and
x(n + 1) = Ax(n) + g(x(n)), ⎛⎞
01 A = ⎝1 − α ⎠
α1 ⎛0⎞
g(x)=⎝β(α−1)x21 −αβx1x2⎠. α(α + βx1)
The characteristic equation of A is λ2 − λ + α−1 = 0. Thus by condition α
(4.3.3) the eigenvalues of A are inside the unit disk if and only if 1 < α−1 +1<2,or0<α−1 <1,whichisalwaysvalid,sinceα>1.
Therefore, ρ(A) < 1 for all α > 1. Since g(x) is continuously differen- tiable at (0,0), the zero solution of (4.6.16) is uniformly asymptotically stable. Consequently, the equilibrium point x* = (α − 1)/β of (4.6.14) is asymptotically stable.
Method (2):Lettingy(n)=(α−1)/βexp(x(n))in(4.6.14),weobtainthe new equation
αα
exp(x(n + 1)) = exp(x(n)) . {1 + (α − 1) exp(x(n − 1))}/α
Taking the logarithm of both sides, we get x(n+1)−x(n)+ α−1f[x(n−1)]=0,
or
where
α
x(n+2)−x(n+1)+ α−1f[x(n)]=0, α
α (α−1)ex +1 f(x) = α − 1 ln α .
(4.6.17)
The Taylor expansion of f around 0 is given by f(x) = x+g(x), where g(x) is a polynomial in x that contains terms of degree higher than or equal to 2. Thus g(x) = o(x). The linearized equation of (4.6.17) is denoted by
x(n+2)−x(n+1)+ α−1x(n)=0. (4.6.18) α
226 4. Stability Theory
Since the characteristic roots of (4.6.18) are the same as the eigenvalues of A, it follows that the zero solution (4.6.18) is asymptotically stable. Corol- lary 4.34 then implies that the zero solution of (4.6.17) is asymptotically stable. Since the equilibrium point y* = (α − 1)/β corresponds to the zero solution of (4.6.17), it then follows that y* = (α−1)/β is an asymptotically stable equilibrium point of (4.6.15).
Our final result deals with the cases ρ(A) = 1 and ρ(A) > 1. Theorem 4.38. The following statements hold:
(i) If ρ(A) = 1, then the zero solution of (4.6.6) may be stable or unstable.
(ii) If ρ(A) > 1 and g(x) is o(x) as ∥x∥ → 0, then the zero solution of
(4.6.6) is unstable.
Proof.
(i) See Example 4.29.
(ii) Assume that ρ(A) > 1. Then by Corollary 4.31, there exists a real symmetric matrix B that is not positive semidefinite for which BT AB − B = −C is negative definite. Thus the Liapunov function V (x) = xT Bx is negative at points arbitrarily close to the origin. Fur- thermore, ∆V (x) = −xT Cx + 2xT AT B g(x) + V (g(x)). Now, (4.5.6) allowsustopickγ>0suchthatxTCx≥4γ∥x∥2 forallx∈Rk. There exists δ > 0 such that if ∥x∥ < δ, then ∥Bg(x)∥ ≤ γ∥x∥ and V (g(x)) ≤ γ∥x∥. Hence ∆V (x(n)) ≤ −γ∥x(n)∥2. Hence by Theorem 4.27, the zero solution is unstable. P
Example 4.39. Let S(n) and I(n) denote the number of susceptibles and infectives, respectively, of a population at time n. Let d > 0 be the per capita natural death rate of the population and α ≥ 0 be the disease related death rate. In the following model, suggested by Elaydi and Jang [47], a simple mass action βSI is used to model disease transmission, where β > 0 and a fraction γ ≥ 0 of these infectives recover. Hence we have the following system
S(n+1)= S(n)+A+γI(n), 1+βhI(n)+dh
I(n + 1) = I(n) + βS(n)I(n), 1 + (d + γ + α)
S(0), I(0) ≥ 0. We make the assumption that
(4.6.19)
ω = βA − d(d + γ + α) > 0
(4.6.20)
4.6 Stability by Linear Approximation 227
under assumption (4.6.20) equation (4.6.19) has two equilibria
A d+γ+α βA−d(d+γ+α) X1∗=d,0 andX2∗= β , (d+α)β .
The linearization of (4.6.19) about X2∗ = (S∗,J∗) yields the Jacobian
matrix
⎛ 1 J=⎜ 1+βI∗+d
γ + dγ − S∗β − AB ⎞ (1+βI∗+d)2 ⎟.
⎝ βI∗
1+d+γ+α 1
⎠
Notice that detJ =
and
One may show that
1 − βI∗(γ+dγ−S∗β−Aβ) >0 1+βI∗ +d (1+d+γ+α)(1+βI∗ +d)2
tr J = 1 + 1 > 0. 1+βI∗ +d
tr J < 1 + det J < 2.
Hence by Theorem 4.33 and equation (4.3.9), the equilibrium point X2∗ is asymptotically stable.
We now turn our attention to the equilibrium point X∗ = A , 0 . The 1d
linearization of (4.6.19) about X1∗ yields the Jacobian matrix ⎛ 1 γ + dγ − A β − AB ⎞
⎜d⎟ J = ⎜ 1 + d (1 + d)2 ⎟ .
⎝1+βA⎠ 0d
1+d+γ+α
The eigenvalues of J are given by
1+d 1+d+γ+α
By virtue of assumption (4.6.20), λ2 > 1 and hence by Theorem 4.38(ii),
the equilibrium point X1∗ is unstable. Exercises 4.6
1. Determine the stability of the zero solution of the equation
x(n + 2) − 1 x(n + 1) + 2x(n + 1)x(n) + 13 x(n) = 0. 2 16
2. Judge the stability of the zero solution of the equation x(n + 3) − x(n + 1) + 2×2(n) + 3x(n) = 0.
1 1+βA λ1= andλ2= d .
228
4. Stability Theory
3.
4.
5.
6.
7.
8.
9.
Consider Example 4.25. Determine the stability and asymptotic
1111 stability for the equilibrium points √ , √ , −√ ,−√ .
2222
(a) Hunt down the equilibrium points of the system: x1(n + 1) = x1(n) − x2(n)(1 − x2(n)),
x2(n + 1) = x1(n),
x3(n+1)= 1×3(n). 2
(b) Determine the stability of all the equilibrium points in part (a). Investigate the stability of the zero solution of the system:
x1(n + 1) = 1×1(n) − x2(n) + x3(n), 2
x2(n + 1) = x1(n) − x2(n) + x3(n), x3(n + 1) = x1(n) − x2(n) + 1×3(n).
2
Linearize the equation
x1(n + 1) = sin(x2) − 0.5×1(n), x2(n + 1) = x2/(0.6 + x1(n)),
around the origin and then determine whether the zero solution is stable.
(a) Find the equilibrium points of the system: x1(n + 1) = cos x1(n) − x2(n),
x2(n + 1) = −x1(n).
(b) Is the point (π/2, −π/2) asymptotically stable?
Determine conditions for the asymptotic stability of the zero solution
of the system
x1(n + 1) = ax1(n)/[1 + x2(n)],
x2(n + 1) = [bx2(n) − x1(n)][1 + x1(n)].
The following model of combat was proposed by Epstein [52], Sedaghat [133].
u(n + 1) = u(n) + 1(a − u(n))[a − u(n)(1 − v(n))], a
v(n + 1) = v(n) + 1 − v(n)[u(n)(1 − v(n)) − d)], 1−d
where d < a, a > 0.
Investigate the stability of the positive equilibrium point.
10. The following system represents a discrete epidemic model. The pop- ulation is divided into three groups: susceptibles S(n), infectives I(n), and immune or removed individuals R(n), n ∈ Z+. If we assume the total population size equals N for all time, then S(n)+I(n)+R(n) = N and we then can eliminate one of the variables, say R(n). The model is given by
S(n+1)=S(n)− αI(n)S(n)+β(N−S(n)), N
I(n+1)=I(n)(1−γ−β)+ αI(n)S(n), N
with 0 < β + γ < 1 and 0 < α < 1. This model is called an SIR epidemic model.
(a) Find all the equilibrium points.
(b) Determine the stability of the equilibrium points.
11. Consider system (4.6.19) under the assumption that
σ = βA − d(d + γ + α) < 0.
(i) Show that there is only one equilibrium point X∗ = A,0.
(ii) Show that X1∗ is asymptotically stable. (iii)∗Show that X1∗ is globally asymptotically stable.
12. Show that if the zero solution of (4.6.2) is uniformly stable (uni- formly asymptotically stable), then the zero solution of (4.6.1) is also uniformly stable (uniformly asymptotically stable), provided that
∞ n=0
13. Suppose that the zero solution of x(n + 1) = Ax(n) is asymptotically stable. Prove that the zero solution of y(n + 1) = [A + B(n)]y(n) is asymptotically stable if ∞n=0 ∥B(n)∥ < ∞.
4.7 Applications
4.7.1 One Species with Two Age Classes
Consider a single-species, two-age-class system, with X(n) being the number of young and Y (n) that of adults, in the nth time interval:
X(n + 1) = bY (n),
Y (n + 1) = cX(n) + s Y (n) − D Y 2(n). (4.7.1)
4.7 Applications 229
1d
∥g(n, y(n))∥ ≤ an∥y(n)∥, where an > 0 and
an < ∞.
230 4. Stability Theory
A proportion c of the young become adult, and the rest will die before reaching adulthood. The adults have a fecundity rate b and a density- dependent survival rate sY (n) − DY 2(n). Equation (4.7.1) may be written in a more convenient form by letting X ̃(n) = DX(n)/b and Y ̃(n) = DX(n). Hence we have
X ̃ ( n + 1 ) = Y ̃ ( n + 1 ) ,
Y ̃ ( n + 1 ) = a X ̃ ( n ) + s Y ( n ) − Y 2 ( n ) , ( 4 . 7 . 2 )
with a = cb > 0.
The nontrivial fixed point is (X ̃∗,Y ̃∗), with X ̃∗ = Y ̃∗ and Y ̃∗ = a+s−1.
Note that the equilibria X ̃ ∗ and Y ̃ ∗ must be positive in order for the model to make sense biologically. This implies that a+s−1 > 0. Since it is easier to do stability analysis on the zero equilibrium point, we let x(n) = X ̃(n)−X ̃∗ and y(n) = Y ̃ (n) − Y ̃ ∗. This yields the system
x(n + 1) = y(n),
y(n + 1) = ax(n) + ry(n) − y2(n), r = 2 − 2a − s. (4.7.3)
The fixed point (0,0) corresponds to the fixed point (X ̃∗,Y ̃∗). Local stability can now be obtained by examining the linearized system
x(n + 1) = y(n),
y(n + 1) = ax(n) + ry(n),
whose eigenvalues are the roots of the characteristic equation λ2 − rλ − a = 0.
By criteria (4.3.9), the trivial solution is asymptotically stable if and only if:
(i) 1−r−a>0 or a+s>1,and (ii) 1+r−a>0 or 3a+s<3.
Hence the range of values of a and s for which the trivial solution is asymp- totically stable is bounded by the region a = 1, s = 1, a+s = 1, and 3a + s = 3, as shown in Figure 4.27.
The shaded region represents the range of parameters a, s for which the trivial solution is asymptotically stable.
To find the region of stability (or the basin of attraction) of the trivial solution we resort to the methods of Liapunov functions. Let
V (x, y) = a2x2 + 2arxy + y2. 1−a
Recall from calculus that Ax2+2Bxy+Cy2 = D is an ellipse if AC−B2 > 0,ora2− a2r2 >0,ora−1
(1−a)2
and s < 3 − 3a, which is the shaded region in Figure 4.27. By rotating the
s
s3−3a a1
4.7 Applications 231
1
s1
1
FIGURE 4.27.
a
axes, one may eliminate the mixed term xy to obtain A′x2 + C′y2 = D, with A′ + C′ = a2 + 1 > 0. Moreover, A′C′ > 0. Hence both A′ and C′ are positive and, consequently, D is positive. Thus in the shaded region in Figure 4.27, V (x, y) is positive definite.
After some computation we obtain ∆V (x, y) = y2W (x, y),
where
W (x, y) = (y − r)2 − 2ax − 2ar(r − y) + a2 − 1. 1−a
Hence ∆V (x, y) ≤ 0 if W (x, y) < 0, that is, if (x, y) is in the region
2ar(r − y) G= (x,y):(y−r)2 −2ax− 1−a +a2 −1<0
.
The region G is bounded by the parabola W (x, y) = 0. Now, in the region G, ∆V (x, y) = 0 on the x-axis y = 0. Hence E is the x-axis. But since (c, 0) is mapped to (0, ac), the largest invariant set M in E is the origin. Hence by Theorem 4.24 every bounded solution that remains in G will converge to the origin.
We now give a crude estimate of the basin of attraction, that is, the set of all points in G that converges to the origin. Define
Vmin = min{V (x0, y0) : (x0, y0) ∈ ∂G}, Jm ={X ̃,Y ̃}:X ̃ =x0 +X ̃∗,Y ̃ =y0 +Y ̃∗, V(x(m),y(m))
(iii) 1−a−b > 0.
Taking into account that a > 0 and 0 < b < 1, the region of stability S is
given by
S={(b,a)|0
P (n + 1) = cH (n)(1 − exp(−aP (n))). The equilibrium points are solutions of
(4.7.21)
H∗
1=expr 1−K −aP∗ , P∗=cH∗(1−exp(−aP∗)).
Hence ∗ ∗
P∗ = r 1− H = r(1−q), H∗ = P aKa (1−e ̄)
.
(4.7.22)
(4.7.23)
Thus
r(1 − H∗ )
K =1−exp −r 1−
H∗ K
ap∗
acH∗
.
Clearly, H1∗ = K, P1∗ = 0 is an equilibrium state. The other equilibrium point may be obtained by plotting the left- and right-hand sides of (4.7.19) against H∗. From Figure 4.30 we see that there is another equilibrium point with 0 < H2∗ < K. Then we may find P2∗ from (4.7.18).
To perform the stability analysis of the equilibrium point (H2∗,P2∗), we putH(n)=x(n)+H2∗, P(n)=y(n)+P2∗.Henceweobtain
x(n+1)=−H2∗+(x(n)+H2∗)exp r 1−
y(n + 1) = −P2∗ + c(x(n) + H2∗)[1 − exp(−a(y(n) + P2∗))].
By linearizing around (0, 0) we obtain the linear system
with
where q = H2∗ and K
4.7 Applications 237
x(n)+H∗
2 −a(y(n)+P2∗) ,
x(n + 1) y(n + 1)
x(n) y(n)
(4.7.24)
(4.7.25)
(4.7.26)
= A
1 − rq −arq
A= c(1−exp(−r(1−q)) φ−r(1−q) ,
K
φ = r(1 − q) . 1 − exp(−r(1 − q))
The details of obtaining the matrix A will be left to the reader. Observe that the value of q = H2∗ is a measure of the extent to which the predator
K
can depress the prey below the carrying capacity.
The characteristic equation of A is given by
λ2 −λ(1−r+φ)+(1−rq)φ+r2q(1−q)=0. (4.7.27)
By criterion (4.3.9), the eigenvalues of A lie inside the unit disk if and only if
|1 − r + φ| < 1 + (1 − rq)φ + r2q(1 − q) < 2. q
1.1 1
.9 .8 .7 .6 .5 .4 .3 .2 .1
12345
FIGURE 4.31. The origin is asymptotically stable within the shaded area.
238 4. Stability Theory
Hence
(1 − rq)φ + r2q(1 − q) < 1, (4.7.28)
1 + (1 − rq)φ + r2q(1 − q) > |1 − r + φ|. (4.7.29)
Plotting (4.7.24) and (4.7.25) gives the region of asymptotic stability indicated by the shaded area in Figure 4.31.
The origin is asymptotically stable within the shaded area. Note that the area of stability narrows as r increases.
Beddington et al. [7] conducted a numerical simulation for the specific value q = 0.4. As r grows past a certain value, the equilibrium point becomes unstable and a hierarchy of stable limit cycles of increasing, non- integral period, ultimately breaking down to cycles of period 5, appears. These are followed by cycles of period 2×5, 22 ×5,…,2n ×5,….
4.7.5 The Flour Beetle Case Study
The team of R.F. Costantino, J.M. Cushing, B. Dennis, R.A. Deshar- nais, and S.M. Henson [27] have studied the flour beetles extensively. They conducted both theoretical studies as well as experimental studies in the laboratory. To describe their model, we will give a brief background of the life-cycle of the flour beetles. The life-cycle consists of larval and pupal stages each lasting approximately two weeks, followed by an adult stage (see Figure 4.32).
As is shown in Figure 4.32, cannibalism occurs among the various groups. Adults eat pupae and eggs, larvae eat eggs. Neither larvae nor adults eat mature adults. Moreover, larvae do not feed on larvae. The cannibalism of larvae by adults and of pupae by larvae is assumed negligible since it typically occurs at much reduced rates.
Let L(n) be the larvae population at time period n, let P (n) be the pupal population at time period n, and let A(n) be the adult population at time
CC
EA
PA
Pupae
Eggs
Adults
Larvae
C EL
FIGURE 4.32. The arrows show the cannibalistic interaction between difference life-cycle stages.
period n. Then the larval–pupal–adult (LPA) model is given by
L(n + 1) = bA(n) exp(−cEAA(n) − cELL(n)),
P (n + 1) = (1 − μL)L(n), (4.7.30) A(n + 1) = P (n) exp(−cP AA(n)) + (1 − μA)A(n),
where L(0) ≥ 0, P (0) ≥ 0, A(0) ≥ 0.
The constants μL, μA are the larval and adult probability of dying from
causes other than cannibalism, respectively. Thus 0 ≤ μL ≤ 1 and 0 ≤ μA ≤ 1. The term exp(−cEAA(n)) represents the probability that an egg is not eaten in the presence of A(n) adults, exp(−cELL(n)) represents the probability that an egg is not eaten in the presence of L(n) larvae, and exp(−cPAA(n)) is the survival probability of a pupa in the presence of A(n) adults. The constants cEA ≥ 0, cEL ≥ 0, cPA ≥ 0 are called the cannibalism coefficients. We assume that adult cannibalism is the only significant cause of pupal mortality.
There are two equilibrium points (0, 0, 0)T and (L∗, P ∗, A∗) ∈ R3+, L∗ > 0, P∗ > 0, A∗ > 0. The positive equilibrium point may be obtained by solving the three equations
Eliminating P yields
L exp(cELL) = bA exp(−cEAA), P = (1 − μL)L,
μA exp(cPAA) = P.
(1 − μL)L = μAA exp(cP AA), L exp(cELL) = bA exp(−cEAA).
(4.7.31)
Dividing the second equation by the first yields exp(cELL) = b(1 − μL) exp[(−cEA − cP A)A].
(4.7.32)
4.7 Applications 239
The number
μA
N = b(1−μL)
μA
is called the inherent net reproductive number. This number will play a significant role in our stability analysis. Observe that if N < 1, equa- tion (4.7.32) has no solution and we have no positive equilibrium point. However, if N > 1, then equation (4.7.32) has a solution which is the inter- section of the curve (1 − μL )L = μA exp(cP A A) and the straight line from (0, ln N/cEL) to (ln N/(cEA + cP A), 0) in the (A, L)-plane represented by equation (4.7.32). To investigate the local stability of the equilibrium point
240 4. Stability Theory
(L,P,A) of equation (4.7.30), we compute the Jacobian J, ⎛−cELbAe(−cEAA−cELL) 0 be(−cELL−cEAA)(1−cEAA)⎞
J=⎜⎝ 1−μL 0 0 ⎟⎠. 0 e(−cPAA) 1−μA −cPAPe(−cPAA)
At the equilibrium point (0, 0, 0)T we have ⎛⎞
00b
J 1 = J | ( 0 , 0 , 0 ) T = ⎜⎝ 1 − μ L 0 0 ⎟⎠ .
0 11−μA The characteristic polynomial of J1 is given by
P(λ)=λ3 −(1−μA)λ2 −b(1−μL)=0 which is of the form
P(λ)=λ3 +p1λ2 +p2λ+p3 =0,
(4.7.33)
(4.7.34)
withp1 =−(1−μA), p2 =0, p3 =−b(1−μL).
According to (5.1.17), the eigenvalues of J1 are inside the unit circle if
and only if
|p3 +p1|<1+p2 and |p2 −p3p1|<1−p23. Applying the first condition yields
or
|−b(1 − μL) − (1 − μA)| < 1, b(1 − μL) + (1 − μA) < 1,
N = b(1 − μL) < 1. μA
(4.7.35)
The second condition gives
|−(1 − μA)(1 − μL)b| < 1 − b2(1 − μL)2,
b2(1−μL)2 +(1−μA)(1−μL)b<1.
But this inequality is satisfied if we assume (4.7.35). For if N < 1 we have b2(1−μL)2 +(1−μA)(1−μL)b<μ2A +μA(1−μA)=μA ≤1.
We conclude that the trivial equilibrium is asymptotically stable if and only if N < 1, and thus attracts all orbits in the nonnegative cone. As N increases past 1, a “bifurcation” occurs which results in the instability of the trivial equilibrium and the creation of the positive equilibrium. In fact, for N > 1 there exists one and only one positive equilibrium. The Jacobian
at the positive equilibrium (L∗, P ∗, A∗) satisfying (4.7.31) is given by ⎛ L∗ ⎞
− cEAL∗ ⎝1−μL 0 0 ⎠
−cELL∗ 0
0 A∗exp(cPA) 1−μA−A∗μAcPA
⎟.
λ3 + (cELL∗ + μAcP AA∗ − (1 − μA))λ2 − cELL∗(1 − μA)λ− L∗
A∗ −cEAL∗ (1−μL)exp(−cPAA∗)=0. As of writing this edition of the book, a condition for the stability of the
4.7 Applications 241
J2 = J|(L∗,P∗,A∗) = ⎜
The characteristic equation is given by
A∗
positive equilibrium is known only in special cases.
Case (i) If cEL = 0, the positive equilibrium is globally attracting if
1 < N < e min{(1, (cEA/cP A)((1 − μA)/μA)} [83].
Case (ii) In several long term experiments reported in [27], the adult death rate was manipulated to equal 96% and hence μA = 0.96. Motivated by this data, Cushing [25] assumed that μA = 1. In this case we have N = b(1−μL) and equation (4.7.30) becomes
L(n + 1) = N A(n) exp(−cELL(n) − cEAA(n)), 1−μL
P (n + 1) = (1 − μL)L(n), (4.7.36) A(n + 1) = P (n) exp(−cP A A(n)).
Theorem 4.40 [25]. For N > 1, the trivial equilibrium of equation (4.7.36) is unstable and there exists a unique positive equilibrium. This positive equilibrium, which bifurcates from the trivial equilibrium at N = 1, is unstable for N = 1 + δ, where δ is sufficiently small.
A subcase of Case (ii) is the case of synchronous orbits. A triple
(L(n),P(n),A(n)) is said to be synchronous at time n if one com-
ponent equals zero and at least one component is nonzero. One can
see immediately from equation (4.7.36), that an orbit that is syn-
chronous at time n0 is synchronous for all n ≥ n0. Notice that a
point (L0, P0, 0)T in the L, P -plane is mapped to the point (0, (1 −
μL)L0,P0)T in the P,A-plane, which in turn is mapped to the point
N P0 exp(−cEAP0), 0, (1 − μL)L0 exp(−cP AP0)T in the L, A-plane. 1−μL
Hence points are mapped from one nonnegative quadrant of the co- ordinate planes to the next in sequential order. A synchronous triplet (L(n), P (n), A(n))T is said to be fully synchronous at time n if it has two zero components. This is the case for points on the positive coordinate axes. An orbit is fully synchronous if and only if its initial point is fully
242 4. Stability Theory
A
(0,(1− μ
P
(L , P ,0)T 00
T L00
) L , P )
((N/(1− μ
) P exp(−c P ), 0,(1− μ
L0 EA0 L0 PA0
L
) T )
exp(−c P
FIGURE 4.33.
synchronous. This notion is derived from the fact that the three life-cycle stages are synchronized temporarily in such a way that they never overlap.
Denote the map (4.7.36) by F, ⎛⎞⎛⎞
L(n + 1) L(n)
⎜⎝P (n + 1)⎟⎠ = F ⎜⎝P (n)⎟⎠ . (4.7.37)
A(n + 1) A(n)
Then F3 maps the nonnegative quadrant of a coordinate plane to itself.
A fixed point of F3 corresponds to a 3-cycle of F and so on. The map F3
is defined by the equations
x(n+1)=Nx(n)exp −cPAy(n)exp(−cPAz(n))
− cEA(1 − μL)x(n) exp(−cP Ay(n) exp(−cP Az(n)))
− c N y(n) exp(−c y(n) − c y(n) exp(−c EL 1 − μ P A EA P A
L
) L
z(n))) −c N z(n)exp−c N z(n)exp(−c z(n)
EL 1 − μ EA 1 − μ EA LL
− cELx(n)) ,
y(n+1)=Ny(n)exp −cPAz(n)−cEAy(n)exp(−cPAz(n)) − c N z(n) exp(−c z(n) − c x(n)),
(4.7.38)
(4.7.39)
If(x0,0,z0)T isapointinthex,z-plane,thenitsorbitisdescribedbythe two-dimensional system
x(n + 1) = N x(n) exp(−cx(n)), (4.7.40) z(n + 1) = [N exp(−αx(n))]z(n) exp(−βz(n)), (4.7.41)
EA 1 − μ EA L
EL
z(n+1)=Nz(n)exp −cEAz(n)−cELx(n) −cPA(1−μL)x(n)exp(−cPAy(n)exp(−cPAz(n))) .
where c = cEA(1 − μL), α = cEL + cP A(1 − μL), β = cEA.
The first equation (4.7.40) is the well–known Ricker’s map, where
limn→∞ x(n) = 0 and the convergence is exponential. Hence equation
(4.7.41) may be looked at as a perturbation of (4.7.40). Hence by Corollary
8.27, limn→∞ z(n) = 0 which is consistent with what we had earlier. For
N > 1, Ricker’s map has a unique positive equilibrium x∗ = ln N/c. Con-
sequently, there exists a fully synchronous 3-cycle of equation (4.7.36). As
N increases, Ricker’s map undergoes a period-doubling bifurcation route
to chaos. If 1 < N < e2, then (x∗,z∗)T = (1 lnN,0)T is an asymptotically c
stable equilibrium point of equations (4.7.40) and (4.7.41) and globally at- tracts all positive initial conditions in the x, z-plane. This fixed point of F 3 corresponds to the fully synchronous 3-cycle of the LPA model (4.7.36)
4.7 Applications 243
⎛ lnN ⎞
⎛0⎞⎛0⎞
⎜cEA(1−μL)⎟→⎜ 1 lnN⎟→⎜ 0 ⎟. ⎝ 0 ⎠⎝cEA ⎠⎝1 ⎠
0 0 clnN EA
(4.7.42)
Thus we have the following result. Theorem 4.40 [25] for 1 < N < e2, the LPA model (4.7.36) has a unique, nontrivial fully synchronous 3-cycle given by (4.7.42). This 3-cycle attracts all fully synchronous orbits or equation (4.7.36). For N > e2, the system has a period-doubling cascade of fully synchronous (3×2n)-cycle attractors and, for sufficiently large N, has “fully synchronous chaotic” attractors (with respect only to fully synchronous orbits).
This is the first proof of the presence of chaos in a population model.
Details about synchronous but not fully synchronous orbits may be found in Cushing [25]. There are still many open problems that need to be tackled. We invite the reader to solve them.
Open Problem 1. Investigate the LPA model for the general case μA ̸= 1. For a starting point, try the case with μA = 0.96.
Open Problem 2. Investigate the behavior of orbits that are not syn- chronous provided that μA = 1.
5
Higher-Order Scalar Difference Equations
In Chapter 4 we investigated the qualitative behavior of systems of dif- ference equations, both linear and nonlinear. In this chapter we turn our attention to linear and nonlinear higher-order scalar difference equations. Although one may be able to convert a scalar difference equation to a system, it is often advantageous to tackle the scalar difference equation directly. Moreover, since a system of difference equations may not be con- vertible to a scalar difference equation, results on the latter may not extend to the former. Every section in this chapter was written with this statement in mind. Section 5.1 gives explicit necessary and sufficient conditions for the stability of the zero solution of a kth-order scalar difference equation. This task is accomplished either via the Schur–Cohn criterion or by using spe- cial techniques that were developed by Levin and May [90], Kuruklis [86], Dannan [28], and Dannan and Elaydi [29]. Section 5.2 provides easy com- putable sufficient conditions for asymptotic stability using Gerschgorin’s Theorem which provides a rough estimate of the location of eigenvalues of matrices.
In the first and second editions of this book I have used Rouch ́e’s The- orem from complex analysis to obtain the results in Section 5.2. However, the new approach is not only more accessible to readers with no background in complex analysis but, more importantly, it is much more intuitive. Sec- tion 5.3 treats nonlinear equations via linearization and follows closely the exposition in Section 4.4 for systems. Section 5.4 collects the main results in global stability of nonlinear scalar difference equations. It remains an open question of whether or not these results extend to nonlinear systems of difference equations. Finally, Section 5.5 presents the larval–pupal–adult
245
246 5. Higher-Order Scalar Difference Equations
(LPA) model of flour beetles with no larval cannibalism on eggs, and a mosquito model.
5.1 Linear Scalar Equations
Consider the kth-order difference equation x(n+k)+p1x(n+k−1)+p2x(n+k−2)+···+pkx(n)=0 (5.1.1)
where the pi’s are real numbers.
It follows from Corollary 2.24 that the zero solution of (5.1.1) is asymp-
totically stable if and only if |λ| < 1 for all characteristic roots λ of (5.1.1), that is, for every zero λ of the characteristic polynomial
p(λ)=λk +p1λk−1 +···+pk. (5.1.2)
Furthermore, the zero solution of (5.1.1) is stable if and only if |λ| ≤ 1 for all characteristic roots of (5.1.1) and those characteristic roots λ with |λ| = 1 are simple (not repeated). On the other hand, if there is a repeated characteristic root λ with |λ| = 1, then according to Corollary 2.24 the zero solution of (5.1.1) is unstable.
One of the main tools that provides necessary and sufficient conditions for the zeros of a kth-degree polynomial, such as (5.1.2), to lie inside the unit disk is the Schur–Cohn criterion. This is useful for studying the stability of the zero solution of (5.1.1). Moreover, one may utilize the Schur–Cohn criterion to investigate the stability of a k-dimensional system of the form
x(n + 1) = Ax(n) (5.1.3)
where p(λ) in (5.1.2) is the characteristic polynomial of the matrix A. But before presenting the Schur–Cohn criterion we introduce a few
preliminaries.
First let us define the inners of a matrix B = (bij ). The inners of a matrix
are the matrix itself and all the matrices obtained by omitting successively the first and last rows and the first and last columns. For example, the inners for the following matrices are highlighted:
A 3 × 3 matrix ⎛⎞
⎜ b11 b12 b13 ⎟ ⎜ ⎜⎝ b 2 1 b 2 3 ⎟ ⎟⎠ ,
b31 b32 b33
A 4 × 4 matrix ⎛⎞
⎛ ⎜
⎜ b21
⎞ ⎟
⎜ b11 b12 b13 b14 ⎟ ⎜ b21 b24 ⎟
b11
b12 b13 b14 b15 b25 ⎟
⎜ ⎟, ⎜⎝ b 3 1 b 3 4 ⎟⎠
⎜ b31 ⎜
b35 ⎟ . ⎟
b41 b42 b43 b44
⎜⎝ b 4 1 b51
b45 ⎟⎠ b52 b53 b54 b55
A 5 × 5 matrix
b22 b23 b24 b32 b34 b42 b43 b44
b22 b23 b32 b33
b22
b33
A matrix B is said to be positive innerwise if the determinants of all of its inners are positive.
Theorem 5.1 (Schur–Cohn Criterion) [74]. The zeros of the charac- teristic polynomial (5.1.2) lie inside the unit disk if and only if the following hold:
(i) p(1) > 0,
(ii) (−1)kp(−1) > 0,
(iii) the (k − 1) × (k − 1) matrices ⎛⎞⎛⎞
⎜⎝ p k − 3 ⎟⎠ ⎜⎝ 0 p k p 3 ⎟⎠ pk−2 pk−3 … p1 1 pk pk−1 … p3 p2
are positive innerwise.
Using the Schur–Cohn criterion (Theorem 5.1), one may obtain necessary and sufficient conditions on the coefficients pi’s such that the zero solution of (5.1.1) is asymptotically stable. Neat and compact, necessary and suffi- cient conditions for the zero solution of (5.1.1) to be asymptotically stable are available for lower-order difference equations. We will present these conditions for second- and third-order difference equations.
For the second-order difference equation x(n+2)+p1x(n+1)+p2x(n) = 0
the characteristic polynomial is
p(λ)=λ2 +p1λ+p2.
The characteristic roots are inside the unit disk if and only if
p(1)=1+p1 +p2 >0, p(−1) = 1−p1 +p2 > 0,
B 1± = 1 ± p 2 > 0 .
(5.1.4)
(5.1.5)
(5.1.6) (5.1.7) (5.1.8)
It follows from (5.1.6) and (5.1.7) that 1 + p2 > |p1| and 1 + p2 > 0. Now (5.1.8) reduces to 1 − p2 > 0 or p2 < 1. Hence the zero solution of (5.1.4) is asymptotically stable if and only if
(5.1.9)
For the third-order difference equation x(n+3)+p1x(n+2)+p2x(n+1)+p3x(n) = 0 (5.1.10)
5.1 Linear Scalar Equations 247
1 0 ... 0 0 0 ... 0 pk ⎜p1 1 ... 0⎟ ⎜0 0 ... pk pk−1⎟
±⎜. .⎟⎜.. .⎟ Bk−1=⎜ . .⎟±⎜. . . ⎟
|p1|<1+p2 <2.
248 5. Higher-Order Scalar Difference Equations
the characteristic polynomial is
λ3 +p1λ2 +p2λ+p3 =0.
The Schur–Cohn criterion are
1+p1 +p2 +p3 >0,
(5.1.11)
(5.1.12) (5.1.13)
> 0 .
> 0 .
Thus
and
Hence
1 + p 2 − p 1 p 3 − p 23 > 0
(5.1.14)
(5.1.15)
(5.1.16)
(−1)3[−1+p1 −p2 +p3]=1−p1 +p2 −p3 >0
| B 2+ | = 1 0 p1 1
+ 0 p 3 = 1 p 3
p3 p2 p1+p3 1+p2
| B 2− | =
− p 3 p1 1 p3 p2 p1−p3 1−p2
1 0 − 0 p 3 = 1
1 − p 2 + p 3 p 1 − p 23 > 0 .
Using (5.1.12), (5.1.13), (5.1.14), and (5.1.16), we conclude that a nec- essary and sufficient condition for the zero solution of (5.1.10) to be asymptotically stable is
(5.1.17)
It is now abundantly clear that the higher the order of the equation, the more difficult the computation involved in applying the Schur–Cohn criterion becomes. However, Levin and May [90], using a very different technique, were able to obtain a simple criterion for the asymptotic stability of the following special equation
x(n + 1) − x(n) + qx(n − k) = 0, (5.1.18) x(n + k + 1) − x(n + k) + qx(n) = 0. (5.1.19)
|p1 +p3|<1+p2 and |p2 −p1p3|<1−p23.
or, equivalently,
Theorem 5.2. The zero solution of (5.1.18) is asymptotically stable if
and only if
kπ
0
are parts of |b| = |a2 + 1 − 2|a| cos φ| 1 , where φ is the solution in (0, π(k + 1)) of sin(kθ)/sin(k + 1)θ = 1/|a|.
Contemplate the equation
x(n+1)−ax(n)+bx(n−k)=0, n∈Z+. (5.1.21)
Theorem 5.3. Let a be a nonnegative real number, b an arbitrary real number, and k a positive integer. The zero solution of (5.1.21) is asymptotically stable if and only if |a| < (k + 1)/k, and:
(i) |a|−1
2
0 < b < (2 − 2 cos φ) 1 . (5.1.25)
250 5. Higher-Order Scalar Difference Equations
Also note that
(2−2cosφ)1 =[2(1−cosφ)]1 =[4sin2(φ/2)]1 =2sin(φ/2)
222
and thus (5.1.25) can be written as
0 < b < 2 sin(φ/2).
Furthermore, (5.1.24) yields sin(kφ) = sin[(k + 1)φ] and so either kφ + (k + 1)φ = (2n + 1)π
or
kφ = (k + 1)φ + 2nπ,
where n is an integer. Since (5.1.28) cannot be valid for 0 < φ < π/(k + 1) we have that (5.1.27) holds. In fact, 0 < φ < π/(k+1) forces n = 0 and so φ = π/2 and thus condition (5.1.25) may be written as 0 < b < 2 cos[kπ/(2k + 1)], which is the condition of Theorem 5.3. P
Dannan [28] considered the following more general equation x(n+k)+ax(n)+bx(n−l)=0, n∈Z+, (5.1.29)
where k ≥ 1 and l ≥ 1 are integers.
Theorem 5.4 [28]. Let l ≥ 1 and k > 1 be relatively prime odd integers. Then the zero solution of (5.1.29) is asymptotically stable if and only if |a| < 1 and
2 |a|−1
Remark: If l and k in Theorems 5.4, 5.5, and 5.6 are not relatively prime, then l = s ̃l and k = sk ̃ for some positive integers s, ̃l, and k ̃, where ̃l and k ̃ are relatively prime. The asymptotic stability of (5.1.29) is equivalent to the asymptotic stability of
x(n + k ̃) + ax(n) + bx(n − ̃l) = 0. (5.1.37) (Why?) The reader is asked to prove this in Exercises 5.1, 5.2, Problem 5.
Example 5.7. Consider the difference equation x(n+25)+ax(n)+bx(n−15)=0, n=0,1,2,….
The corresponding characteristic equation is λ40 +aλ25 +b = 0,
and in the reduced form is
Here we have ̃l = 5 and k ̃ = 3. Therefore, Theorem 5.5 is applicable and
λ8 +aλ5 +b=0.
the given equation is asymptotically stable if and only if |a| < 1 and
2 |a| − 1 < b min(1 + a2 − 2|a| cos 3θ) 1 ,
θ∈S
where S is the solution set of 1 = sin5θ on the interval (0,π). If we let
|a| sin 8θ
a = 0.6, then θ = 2.007548968 and the given equation is asymptotically
stable if and only if −0.4 < b < 0.4477703541.
5.2 Sufficient Conditions for Stability
Clark [21] considered the equation
x(n + k) + px(n + k − 1) + qx(n) = 0 (5.2.1)
where p, q ∈ R. When p = −1, we revert back to the Levin and May equa- tion (5.1.18). He showed that the zero solution of (5.2.1) is asymptotically stable if
|p| + |q| < 1. (5.2.2)
252 5. Higher-Order Scalar Difference Equations
Moreover, the zero solution is unstable if
|p| − |q| > 1. (5.2.3)
Here we will extend Clark’s Theorem to the general equation (5.1.1). The novelty of Clark’s proof is the use of Rouch ́e’s Theorem from Complex Analysis to locate the characteristic roots of the equation. In the first two editions of this book, I followed Clark’s proof. In this edition, I am going to deviate from this popular method, and give instead a simpler proof based on Gerschgorin’s Theorem [111] which we now state.
Theorem 5.8. Let A be a k × k real or complex matrix. Let Si be the disk
k j=1
j ̸=i
in the complex plane with center at aii and radius ri =
Since Av = λv, equating the ith row in both sides yields Hence
|λ − arr| = |(λ − arr)vr| ≤
|aij|. Then all
Then
so that λ is in the disk Sr.
The following example illustrates the above theorem.
P
k i=1
the eigenvalues of A lie in S =
Proof. Let λ be an eigenvalue of A with a corresponding eigenvector
Si.
v = (v1, v2, . . . , vk)T such that ||v||∞ = max{|vi|} = 1. (Why?).
k j ̸=i
(λ−aii)vi =
Since ||v||∞ = 1, there exists r, 1 ≤ r ≤ k, such that ||v||∞ = |vr| = 1.
Example 5.9. Consider the difference equation
x(n + 3) + 1 x(n + 2) − 1 x(n + 1) + 1 x(n) = 0. (5.2.4)
i
aijvj, i=1,2,…,k.
k j ̸=r
|arj||vj| ≤ rk
k j=1
aijvj = λvi.
245 This equation can be converted to the system
x1(n + 1) = x2(n), x2(n + 1) = x3(n),
x3 (n + 1) = − 1 x1 (n) + 1 x2 (n), − 1 x3 (n) 542
where x1(n) = x(n), x2(n) = x(n + 1), x3(n) = x(n + 2).
5.2 Sufficient Conditions for Stability 253
Imaginary axis
Real axis
−1 −1 1 2
FIGURE 5.2. Gerschgorin disks.
This can be written in the compact form
y(n + 1) = Ay(n),
y(n) = (x1(n), x2(n), x3(n))T ,
⎛010⎞ A=⎜⎝0 0 1⎟⎠.
−1 1 −1 542
The eigenvalues of A are the characteristic roots of (5.2.4). By Ger-
schgorin’s Theorem, all the eigenvalues of A lie in the union of the disks S1
and S2, where S1 is the disk centered at the origin with radius 1 and S2 is
centered at −1 and with radius 1 + 1 = 9 = 0.45 (Figure 5.2). Thus 2 4510
the spectral radius of A, ρ(A) ≤ 1. In fact we can do better if we realize
that an eigenvalue λ0 of A is also a characteristic root of (5.2.4). Hence
λ3+1λ2−1λ +1 =0orλ3 =−1λ2+1λ −1 =0.Nowif|λ|=1, 020405 0 20405 0
1 = |λ3| ≤ |− 1λ |+|1λ |+|− 1| = 19, a contradiction. Hence ρ(A) < 1. 0 2040 520
Thus by Corollary 2.24, the zero solution of (5.2.4) is asymptotically stable. Theorem 5.10. The zero solution of (5.1.1) is asymptotically stable if
(5.2.5)
Proof. We first convert (5.1.1) into a system of first-order difference equations
k
|pi| < 1.
i=1
x(n + 1) = Ax(n) (5.2.6)
254 5. Higher-Order Scalar Difference Equations
where
⎛010···0⎞ ⎜0 0 1 ··· 0⎟
A=⎜ . . . . . ⎟. (5.2.7) ⎜ ⎟
⎝000···1⎠ −pk −pk−1 −pk−2 · · · −p1
By Gerschgorin’s Theorem all the eigenvalues lie in S1 ∪ S2, where S1 is
the unit disk, and S2 is the disk with center −p1 and radius r = ki=2 |pi|.
By assumption (5.2.5), |λ| ≤ 1. To eliminate the possibility that |λ| = 1,
we assume that A has an eigenvalue λ0 such that |λ0| = 1. Now λ0 is also a
characteristic root of (5.1.1) and thus from (5.1.2), p(λ ) = λk + p λk−1 + 0010
···+pk =0.Thisimpliesthat
1=|λk|≤|−p λk−1|+|−p λk−2|+···+|−p |
01020k k
|pi|
which leads to a contradiction. It follows by Corollary 2.24 that the zero solution (5.1.1) is asymptotically stable. P
A partial converse of this theorem may be obtained by a more refined statement of Gerschgorin’s Theorem.
Theorem 5.11. Let Si, i = 1,2,...,k, be the Gerschgorin disks of a k×k matrix A. Suppose that for some 1 ≤ m ≤ k, (mi=1 Si) ∩ (ki=m+1 Si) = ∅. Then there are exactly m eigenvalues (counting multiplicities) of A that lie in mi=1 Si and k − m eigenvalues that lie in ki=m+1 Si.
Proof. The proof may be found in [111]. P We are now ready to provide the promised partial converse to Theorem
5.10.
Theorem 5.12. The zero solution of (5.1.1) is unstable if
(5.2.8)
Proof. We first convert (5.1.1) into system (5.2.6). Then by Ger- schgorin’s Theorem all the eigenvalues of A lie in the disks S1 and S2 where S1 is the unit disk and S2 is the disk centered at −p1 and with ra- dius r = ki=2 |pi|. Condition (5.2.8) implies that S1 ∩S2 = ∅. By Theorem 5.11, S2 must contain an eigenvalue λj of A. Moreover, |λj | > 1. Hence the zero solution of (5.1.1) is unstable. P
=
i=1 <1
|p1| −
k i=2
|pi| > 1.
5.2 Sufficient Conditions for Stability 255
Exercises 5.1 and 5.2
1. Show that the zero solution of x(n+4)+p1x(n+3)+p2x(n+2)+ p3x(n + 1) + p4x(n) = 0 is asymptotically stable if and only if |p4| < 1,|p3 +p1| < 1+p2 +p4, and |p2(1−p4)+p4(1−p24)+p1(p4p1 −p3)| < p4p2(1 − p4) + (1 − p24) + p3(p4p1 − p3).
2. Extend the result in Problem 1 to the fifth-order equation
x(n+5)+p1x(n+4)+p2x(n+3)+p3x(n+2)+p4x(n+1)+p5x(n) = 0.
3. For what values of α is the zero solution of x(n+3)−x(n+2)+ α−1 x(n) = 0 asymptotically stable?
α
4. Considertheequationx(n+k)−x(n+k−1)+α−1x(n)=0,k≥2. α
(i) Show that the zero solution is asymptotically stable if and only if (k−1)π (k−1)π
(L) 1 < α < 1+ 2 cos 2k − 1 / 1 − 2 cos 2k − 1 . (ii) Show that in (L) as k increases to ∞, α decreases monotonically
to 1.
5. Prove that the zero solution of (5.1.37) is asymptotically stable if and only if the zero solution of (5.1.29) is asymptotically stable.
6. Apply Theorem 5.1 to show that the zero solution of the difference equation ∆x(n) = −qx(n − 1), q > 0, is asymptotically stable if and only if q < 1.
*7. (Hard). Prove that the zero solution of the difference equation ∆x(n) = −qx(n − k), q > 0, k > 1, is asymptotically stable if qk < 1.
*8. Consider the linear difference equation
m i=1
pix(n − ki) = 0, n ∈ Z+, p1,p2,...,pm ∈ (0,∞) and k1,k2,...,km are positive integers. Show
x(n + 1) − x(n) +
that the zero solution is asymptotically stable if mi=1 kipi < 1. 9. Consider the difference equation
x(n+1)−x(n)+px(n−k)−qx(n−m)=0, n∈Z+,
where k and m are nonnegative integers, p ∈ (0, ∞), and q ∈ [0, ∞).
Show that the zero solution is (globally) asymptotically stable if kp<1 and q
Then f(z) and f(z) + g(z) have the same number of zeros, counting multiplicities, inside γ.
Use Rouch ́e’s Theorem to prove Theorem 5.10. Use Rouch ́e’s Theorem to prove Theorem 5.12.
Stability via Linearization
LetIbeasubsetofthereallineR.ThenIm =I×I×···×I⊂Rm is the product of m copies of I equipped with any of the norms l1,l2, or l∞ as discussed in Chapter 3.
Now a function f : Im → I is continuous at x = (x1,x2,…,xm) if given ε > 0, there exists δ > 0 such that if y = (y1,y2,…,ym) ∈ Im and ||x−y|| < δ, then |f(x)−f(y)| < ε. Notice that the l1-norm gives ||x−y||1 = mi=1 |xi − yi|. For the l2-norm, we have ||x − y||2 = mi=1(x2i − yi2), and finally the l∞-norm gives ||x − y||∞ = max |xi − yi|.
1≤i≤m
Following the work of Ladas and his collaborators [85] we will use the
l1-norm, unless otherwise noted.
Consider the following difference equation of order k + 1,
x(n + 1) = f (x(n), x(n − 1), . . . , x(n − k)) (5.3.1)
where f : Ik+1 → I is a continuous function.
Given a set of (k + 1) initial conditions x−k, x−k+1, . . . , x0 ∈ I, there ex-
ists a unique solution {x(n)}∞n=−k of (5.3.1) such that x(−k) = xk , x(−k + 1) = x−k+1,...,x(0) = x0. Of course one may convert (5.3.1) to a
system of first-order difference equations of order k + 1 as follows. Let y1(n) = x(n − k),y2(n) = (x − k + 1),...,yk+1(n) = x(n). Then (5.3.1) may be converted to the system
y(n + 1) = F (y(n)), (5.3.2)
where
F(y(n)) = (y2(n),y3(n),...,yk+1(n),f(yk+1(n),yk(n),...,y1(n))T .
We may write F = (F1,F2,...,Fk+1)T, where F1(y1) = y2,F2(y2) = y3,...,Fk+1(yk+1) = f(yk+1,...,y1).
A point x∗ ∈ I is an equilibrium point of (5.3.1) if f(x∗,x∗,...,x∗) = x∗. This corresponds to the equilibrium point (x∗, x∗, . . . , x∗) ∈ Rk+1 for system (5.3.2). Notions of stability of equilibrium points and periodic points of (5.3.1) may be stated via (5.3.2) and the use of proper interpretations of the notions in regards to (5.3.1). Here is a sample.
Definition 5.14. An equilibrium point x∗ of (5.3.1) is stable (S) if, given ε > 0, there exists δ > 0 such that if {x(n)}∞n=−k is a solution of (5.3.1) with
then
(|x(−k) − x∗| + |x(−k + 1) − x∗| + · · · + |x(0) − x∗|) < δ, |x(n)−x∗|<ε forall n≥−k.
y(n) = (y1(n), y2(n), . . . , yk+1(n))T ,
Analogous definitions can be given for the remaining notions of stability as defined in Chapter 4.
If f is continuously differentiable in some open neighborhood of X∗ = (x∗, x∗, . . . , x∗), then one can linearize (5.3.1) around X∗. One way to do this is to revert to system (5.3.2) to obtain the linear system
z(n + 1) = Az(n), (5.3.3)
where A is the Jacobian of F at X∗, A = DF(X∗). Then convert (5.3.3) to a scalar equation. However, one may also linearize (5.3.1) directly using the chain rule. Thus the linearized equation around x∗ is given by
u(n + 1) = p0u(n) + p1u(n − 1) + · · · + pku(n − k), pi = ∂f (x ̄,x ̄,...,x ̄),
∂ui
with f(u0,u1,...,uk).
The characteristic equation of (5.3.4) is given by
(5.3.4)
where
5.3 Stability via Linearization 257
λk+1 −p0λk −p1λk−1 −···−pk =0.
(5.3.5)
258 5. Higher-Order Scalar Difference Equations
Using the stability results in Chapter 4 for system (5.3.2), one may easily establish the following fundamental stability result.
Theorem 5.15 (The Linearized Stability Result). Suppose that f is continuously differentiable on an open neighborhood G ⊂ Rk+1 of (x∗, x∗, . . . , x∗), where x∗ is a fixed point of (5.3.1). Then the following statements hold true:
(i) If all the characteristic roots of (5.3.5) lie inside the unit disk in the complex plane, then the equilibrium point x∗ of (5.3.1) is (locally) asymptotically stable.
(ii) If at least one characteristic root of (5.3.5) is outside the unit disk in the complex plane, the equilibrium point x∗ is unstable.
(iii) If one characteristic root of (5.3.5) is on the unit disk and all the other characteristic roots are either inside or on the unit disk, then the equilibrium point x∗ may be stable, unstable, or asymptotically stable.
Proof. The proofs of (i) and (ii) follow from Corollary 4.34 and Theorem 4.38.
(iii) This part may be proved by the following examples. First consider the logistic equation x(n + 1) = x(n)(1 − x(n)) = f (x(n)). The linearized equation around the equilibrium point x∗ = 0 is given by u(n + 1) = u(n) with the characteristic roots λ = 1. But we know from Section 1.6 that x∗ = 0 is unstable.
Now we give an example that produces a different conclusion from the above example. Consider the equation
x(n + 1) = x(n) − x3(n) = f(x(n)). The linearized equation is given by
u(n + 1) = u(n)
with the characteristic root λ = 1. Now for the equilibrium point x∗ = 0, we have f′(0) = 1, f′′(0) = 0, f′′′(0) = −6 < 0. This implies by Theorem 1.15 that x∗ = 0 is asymptotically stable. P
Example 5.16 [15]. Consider the difference equation
x(n + 1) = ax(n) + F(x(n − k)) (5.3.6)
which models whale populations. Here x(n) represents the adult breeding population, a, 0 ≤ a ≤ 1, the survival coefficient, and F (x(n − k)) the recruitment to the adult stage with a delay of k years. The equilibrium point x∗ of (5.3.6) is given by the equation
x∗ =ax∗ +F(x∗),
x∗ =F(x∗)/(1−a). (5.3.7)
Since F(x∗) = (1 − a)x∗, (1 − a) is the annual mortality rate of the whale population. The linearized equation associated with (5.3.6) is given by
u(n + 1) = au(n) + bu(n − k), (5.3.8)
where b = F ′(x∗).
Equation (5.3.8) may be written in the form
u(n + 1) − au(n) − bu(n − k) = 0. (5.3.9) By Theorem 5.10, a sufficient condition for the zero solution of (5.3.9)
to be asymptotically stable is
|a| + |b| < 1,
a + |b| < 1. (5.3.10)
Condition (5.3.10) is a sufficient condition for the asymptotic stability of the equilibrium point x∗ given by (5.3.7).
Exercises 5.3
1. Consider the delayed recruitment model
x(n+1) = 1x(n)+F(x(n−k)).
2
Let x∗ be the equilibrium point and let b = F′(x∗). Assume that F is continuously differentiable in an open neighborhood of x∗. Find sufficient conditions for x∗ to be asymptotically stable if:
(i) k=2, (ii) k = 3.
2. Consider the single species, age-structured population model x(n + 2) = x(n) exp(r − ax(n + 1) − x(n)),
wherexn ≥0foralln∈Z+,a,r>0.
(i) Show that all solutions are bounded.
(ii) Find conditions on r and α under which the positive equilibrium is asymptotically stable.
In Subsection 4.7.5 we studied in detail the larval–pupal–adult (LPA) of the flour beetle.
L(n + 1) = bA(n) exp(−cEAA(n) − cELL(n)),
P (n + 1) = (1 − μL)L(n), (5.3.11) A(n + 1) = P (n) exp(−cP AA(n)) + (1 − μA)A(n).
Kuang and Cushing [83] considered the simplified case when larval cannibalism of eggs is not present, i.e., cEL = 0. Problems 3 though 5 refer to this simplified model.
5.3 Stability via Linearization 259
260 5. Higher-Order Scalar Difference Equations
3. Prove that (5.3.11) reduces to x(n+1)−αx(n)−βx(n−2)exp(−c1x(n−2)−c2x(n))=0, (5.3.12)
where α = 1−μA, β = b(1−μL), c1 = cEA, c2 = cPA, x(n) = A(n+2), n ≥ −2.
Then show that if α + β ≤ 1, equation (5.3.12) has only the trivial equilibrium x∗1 = 0. Furthermore, if α + β > 1, then (5.3.12) has two equilibria, x∗1 = 0 and x∗2 > 0, with x∗2 = (1/c1 + c2) ln(β/(1 − α)).
4. Show that the linearized equation of (5.3.12) around an equilibrium point x∗ is given by
y(n + 1) − [α − βc2x∗ exp{−(c1 + c2)x∗}]y(n)
− β(1 − c1x∗) exp{−(c1 + c2)x∗}y(n − 2) = 0.
5. Prove that:
(i) The trivial solution x∗1 = 0 is asymptotically stable
b(1 − μL) < 1. μA
(ii) The positive equilibrium x∗2 = (1/c1 + c2)ln(β/(1 − α)) is asymptotically stable if and only if
|A+B|<1, |A−3B|<3, and B(B−A)<1, (5.3.13)
where
c2(1 − α) β
A= c +c ln1−α −α,
12
cβ B=(1−α) 1 ln −1.
c1 + c2 1 − α
6. Consider the difference equation N(n+1)−N(n) = N(n)[a+bN(n−
k) − cN2(n − k)], n ∈ Z+, where a, c ∈ [0, ∞) and b ∈ R, k ∈ Z+.
(a) Prove the equation has a unique positive equilibrium N∗.
(b) Show that N∗ is (locally) asymptotically stable if ∗ kπ
N b2 +4ac<2cos 2k+1 .
7. Consider the rational difference equation
k x(n + 1) = a +
i=0
k
aix(n − i) b +
bix(n − i) ,
where k is a nonnegative integer, a0,a1,...,ak,b0,b1,...,bk ∈ [0,∞), a,b ∈ (0,∞), ki=0ai = 1 and B = ki=0bi > 0. Find the pos-
i=0
9.
*10.
5.4
Show that the positive equilibrium x∗ of equation (5.3.14) is asymp- totically stable if either k = 1 and a > 0 or k ≥ 2 and A > b.
Show that for any positive solution x(n) of equation (5.3.14) there exists positive constants C and D such that C ≤ x(n) ≤ D, n ∈ Z+ provided that b > 1.
Global Stability of Nonlinear Equations
a = 0 and
kπ b>A>b 1−2cos2k+1 .
5.4 Global Stability of Nonlinear Equations 261
itive equilibrium point x∗ of the equation. Then show that x∗ is asymptotically stable if b > 1.
Consider the rational difference equation
x(n + 1) = [a + bx(n)]/[A + x(n − k)], (5.3.14)
n ∈ Z+, a, b ∈ [0, ∞), with a + b > 0, A ∈ (0, ∞), and k is a positive integer. Problems 8, 9, and 10 refer to equation (5.3.14) with the above assumptions.
8. (i) Showthatifeithera>0ora=0andb>A,equation(5.3.14) has a positive equilibrium x∗. Then find x∗.
(ii) Show that x∗ is (locally) asymptotically stable if either b = 0 or
Results on global asymptotic stability are scarce and far from complete. In Chapter 4 we have seen how Liapunov functions can be used to establish both local and global asymptotic stability. In this section we will utilize the special nature of scalar equations to present a few results that can deal with several types of equations. Roughly speaking, we will be mainly concerned with “monotone” equations. A more general investigation on monotone discrete dynamical systems is beyond the scope of this book and the interested reader is referred to the work of Hal Smith [138]. More detailed expositions may also be found in Sedaghat [133], Kocic and Ladas [80], and Kulenovic and Ladas [85].
Consider the following difference equation of order (k + 1),
x(n + 1) = f (x(n), x(n − 1), . . . , x(n − k)), (5.4.1)
n ∈ Z+ and k ≥ 1 is a positive integer. The main result in this section is due to Hautus and Bolis [65]. (See also Kocic and Ladas [80] and Kulenovic and Ladas [84].)
262 5. Higher-Order Scalar Difference Equations
Theorem 5.17. Consider (5.4.1) with f ∈ C(Ik+1, R), where I is an open interval in R and x∗ ∈ I is an equilibrium point. Suppose that f satisfies the following assumptions:
(i) f is nonincreasing in each of its arguments, i.e., if a ≤ b, then f(·,…,a,…,·) ≤ (f(·,…,b,…,·).
(ii) (u−x∗)[f(u,u,…,u)−u]<0 for all u∈I/{x∗}.
Then with initial values (x(0), x(−1), . . . , x(−k)) ∈ I, we have x(n) ∈ I
for n ≥ 0 and limn→∞ x(n) = x∗.
Proof. Condition (ii) ensures that x∗ is the only equilibrium point in I. Forify∗ ∈I isanotherequilibriumpoint,then(y∗−x∗)[f(y∗,y∗,...,y∗)− y∗] = 0 which violates condition (ii). Let x(n) be a solution of (5.4.1) with x(0),x(−1),...,x(−k) ∈ I. Set m = min{x∗,x(0),...,x(−k)}, M = max{x∗, x(0), . . . , x(−k)}. By condition (ii) and since m ≤ x∗, we have m ≤ f(m,m,...,m). Moreover, by condition (i) we obtain f(m,m,...,m) ≤ f(x(0),x(−1),...,x(−k)) = x(1). Thus m ≤ f(m,m,...,m) ≤ x(1). Sim- ilarly, one may show that x(1) ≤ f(M,M,...,M) ≤ M. By induction onn,itiseasytoshowthatm≤x(n)≤M foralln≥−k.Inpartic- ular, since [m,M] ⊂ I, it follows that x(n) ∈ I, for all n ≥ −k. Since x(n) is bounded, both limn→∞ inf x(n) = L1 and limn→∞ sup x(n) = L2 exist. Furthermore, m ≤ L1 ≤ L2 ≤ M. Let ε > 0 be sufficiently small such that [m + ε, M + ε] ⊂ I. There exists a positive integer N such that L1 −ε < x(n−k) for all n ≥ N. This implies by condition (i) that f(L1 −ε,L1 −ε,...,L1 −ε) ≤ f(x(n),x(n−1),...,x(n−k)) = x(n+1), for all n ≥ N. Consequently, f(L1 −ε,L1 −ε,...,L1 −ε) ≤ L1. Since f is continuous, and ε is arbitrary, it follows that f(L1,L1,...,L1) ≤ L1. This implies by condition (ii) that x∗ ≤ L1. By a similar argument (Problem 12). one may show that L2 ≤ x∗ and, consequently, L2 ≤ x∗ ≤ L1. Hence x∗ =L1 =L2,andlimn→∞x(n)=x∗. P
Example 5.18. The Beverton–Holt model [10]
x(n + 1) = rKx(n) , K > 0, r > 0, (5.4.2)
K + (r − 1)x(n)
has been used to model populations of bottom-feeding fish, including the North Atlantic plaice and haddock. These species have very high fertility rates and very low survivorship to adulthood. Furthermore, recruitment is essentially unaffected by fishing.
Local stability analysis (Theorem 1.13) reveals that:
(i) if 0 < r < 1, then the zero solution x∗1 = 0 is asymptotically stable,
(ii) if r > 1, then the equilibrium x∗2 = K is asymptotically stable.
Using Theorem 5.17 one can say more about x∗2. Since f(x) = rKx/K + (r − 1)x is monotonically increasing for r > 1, condition (i) in Theorem
5.4 Global Stability of Nonlinear Equations 263
5.17 holds. Now for any u ∈ (0, ∞),
(u − K)(f(u) − u) = (u − K) K + (r − 1)u
rKu − u(K + (r − 1)u) (K − u)2
=−u(r−1)K+(r−1)u <0
and condition (ii) in Theorem 5.17 is satisfied. It follows that x∗2 = K is globally asymptotically stable.
Cushing and Henson [26] used more elementary methods to prove the above results.
Now consider the second-order (modified) Beverton–Holt model
x(n + 1) = rK(αx(n) + βx(n − 1)) (5.4.3)
K + (r − 1)x(n − 1)
where r > 0, K > 0, and α + β = 1.
In this model, the future generation x(n + 1) depends not only on the
present generation x(n) but also on the previous generation x(n − 1). This model has two equilibrium points as before x∗1 = 0 and x∗2 = K.
We first investigate the local stability of these two equilibria. Using Theorem 5.15, we have the following conclusions:
(a) The zero solution x∗1 = 0 of (5.4.3) is locally asymptotically stable if and only if 0 < r < 1.
(b) The equilibrium x∗2 = K is locally asymptotically stable if and only if r > 1.
In fact, one can say more about x∗2 = K. Following the same analysis used for (5.4.2), one may conclude by employing Theorem 5.17 that x∗2 = K is in fact globally asymptotically stable. A higher-order Beverton–Holt equation has been investigated in [81].
The second result that we will present is of a different flavor. It is much more flexible than Theorem 5.17 since it allows f to be either nondecreas- ing or nonincreasing in its arguments. This leads to the notion of weak monotonicity.
Definition 5.19. The function f(u1,u2,…,uk+1) is said to be weakly monotonic if f is nondecreasing or nonincreasing in each of its arguments, i.e., for a given integer j, 1 ≤ j ≤ k + 1, if a ≤ b, then either
f(·,…,a,…,·) ≤ f(·,…,b,…,·) or f(·,…,a,…,·) ≥ f(·,…,b,…,·),
where a and b are in the jth slot, and all the other slots are filled with fixed numbers z1,z2,…,zj−1,zj,…,zk+1.
Theorem 5.20 [60]. Suppose that f in (5.4.1) is continuous and weakly monotonic. Assume, in addition, that whenever (m, M ) is a solution of the
264 5. Higher-Order Scalar Difference Equations
system
m = f(m1,m2,…,mk+1) and M = f(M1,M2,…,Mk+1),
where, for each i=1,2,…,k+1,
and
m if f is nondecreasing in zi,
mi =
M if f is nonincreasing in zi,
M if f is nondecreasing in zi,
Mi =
m if f is nonincreasing in zi,
then m = M.
Then (5.4.1) has a unique equilibrium point x∗ = m which is globally
attracting.
Proof. Letm0 =a,M0 =b,andforeachi=1,2,… set Mi = f(Mi−1,Mi−1,…,Mi−1)
and
where
m0i =
and
Mi0 =
mi = f(mi−1,mi−1,…,mi−1 ), 1 2 k+1
a b
b a
For r > 0,
mri= i
if f is nondecreasing in the ith slot, if f is nonincreasing in the ith slot,
if f is nondecreasing in the ith slot, if f is nonincreasing in the ith slot.
if f is nondecreasing in the ith slot, if f is nonincreasing in the ith slot,
if f is nondecreasing in the ith slot,
and
Mir= i
mr−1 Mr−1
i
Mr−1 mr−1
1 2 k+1
if f is nonincreasing in the ith slot. It follows from the assumptions on f that, for i ≥ 0,
i
m0 ≤m1 ≤···
p+y(n−1)
This equation has only two fixed points y1∗ = 0 and y2∗ = 1. Local stability
analysis shows that y2∗ is locally asymptotically stable. Now assume that
y(n) is a solution of (5.4.4), such that y(n) ≥ 1 for all n ∈ Z+. Then from
(5.4.4) we have y(n + 1) − y(n) = (1 − y(n)) y(n−1) ≤ 0. This implies p+y(n−1)
that y(n) is nonincreasing and thus has a limit in [1,∞). But this leads to a contradiction since 0 < y2∗ < 1. Hence for some positive integer N, y(N ) ∈ (0, 1). Writing (5.4.4) in the form
y(n + 1) − 1 = [y(n) − 1] p p+y(n−1)
we conclude that y(N + r) ∈ (0, 1) for all r ∈ Z+. Now in the interval (0,1),thefunctionf(u,v)=pv+u isincreasinginbothargumentsand,by
p+u
Theorem 5.20, y2∗ is globally asymptotically stable.
The following corollary of Theorem 5.20 is easy to apply to establish the global asymptotic stability of the zero solution.
An independent proof of this conclusion may be found in Grove et al. [61].
Corollary 5.22. Contemplate the difference equation k
x(n+1) =
x(n−i)fi(x(n),x(n−1),...,x(n−k)), n ∈ Z+, (5.4.5)
i=0
with initial values x(0), x(−1), . . . , x(−k) ∈ [0, ∞) such that
(i) k ∈ Z+;
(ii) f0,f1,...,fk ∈ C [0,∞)k+1,[0,1);
(iii)f0,f1,...,fk arenonincreasingineachargument;
(iv) ki=0 fi(u0,u1,...,uk) < 1 for all (u0,u1,...,uk) ∈ (0,∞)k+1;
(v) f0(u,u,...,u) > 0 for all u ≥ 0.
Then the trivial solution x∗ = 0 of (5.4.5) is globally asymptotically stable. Proof. This will be left as Exercises 5.4, Problem 13. P
266 5. Higher-Order Scalar Difference Equations
Example 5.23. Consider again (5.4.4). If p + 1 < q, then y1∗ = 0 is the only equilibrium point in [0,∞). By Theorem 5.15, y1∗ = 0 is locally asymptotically stable. Now we write (5.4.4) in the form
y(n + 1) = y(n) · p + y(n − 1) · 1 . q + y(n − 1) q + y(n − 1)
Hence
f0(u,v)= p , f1(y,v)= 1 . q+v q+v
It is easy to show that f0 and f1 satisfy all the conditions (i) through (iv) in Corollary 5.22. Hence by Corollary 5.22, y1∗ = 0 is globally asymptotically stable.
Exercise 5.4
1. Consider a modified Beverton–Holt equation x(n+1)=rK(αx(n)+βx(n−1)), α+β=1, α,β>0.
K + (r − 1)x(n − 1)
Show that the zero solution is globally asymptotically stable if 0 < r <
1.
2. Show that the zero solution of the equation x(n + 1) = ax(n − k)exp[−b(x2(n) + ··· + x2(n − m))], |a| ≤ 1, b > 0, is globally asymptotically stable.
3. Consider the LPA model of the flour beetle (5.3.11) with no larval cannibalism on eggs:
x(n + 1) = αx(n) + βx(n − 2) exp[−c1x(n − 2) − c2x(n)]
where α = 1−μA, β = b(1−μL), c1 = cEA, c2 = cPA, x(n) = A(n+2). Show that the zero solution is globally asymptotically stable if α+β ≤ 1 and β > 0.
4. The following equation describes the growth of a mosquito population: x(n + 1) = (ax(n) + bx(n − 1)e−x(n−1))e−x(n), n ∈ Z+,
where a ∈ (0, 1), b ∈ [0, ∞). Prove that the zero solution is globally asymptotically stable if a + b ≤ 1.
5. A variation of the mosquito model in Problem 4 is given by the equation
x(n + 1) = (ax(n) + bx(n − 1))e−x(n), n ∈ Z+, (5.4.6)
where a ∈ [0, 1), b ∈ (0, ∞). Prove that the zero solution is globally asymptotically stable if a + b ≤ 1 and b < 1.
5.4 Global Stability of Nonlinear Equations 267
6.1 Show that the positive equilibrium of the equation x(n+1)=p+qx(n−1), p,q>0,
1+x(n) is globally asymptotically stable if q < 1.
7. Show that the positive equilibrium of the equation x(n+1)= p+x(n−1) , p,q>0,
qx(n)+x(n−1)
is globally asymptotically stable if q ≤ 1 + 4p.
8. Consider equation (5.4.4). Show that if y(−1)+y(0) > 0 and p+1 > q, then the positive equilibrium y∗ = p + 1 − q is globally asymptotically stable.
9. Consider the equation
x(n + 1) = x(n) + p , p, q > 0, n ∈ Z+.
x(n)+qx(n−1)
Show that the positive equilibrium point of the equation is globally
asymptotically stable if q ≤ 1 + 4p.
10. Show that the positive equilibrium point of the equation
x(n+1)= p+qx(n) , p,q>0, n∈Z+, 1+x(n−1)
is globally asymptotically stable if one of the following two conditions holds:
(i) q<1,
(ii) q≥1andeitherp≤qorq
qx(n)+x(n−1)
is globally asymptotically stable if q < pq + 1 + 3p and p < q.
12. Complete the proof of the Theorem 5.17 by showing that L2 ≤ x∗.
13. Prove Corollary 5.22.
*14. (Term pro ject). Consider equation (5.4.6) with the assumption a + b >
1 and b < e−a (where ex is the exponential function). Show that the e+1
positive equilibrium is globally asymptotically stable, with basin of attraction [0, ∞) × [0, ∞)/{(0, 0)}.
1Problems 6–11 are from Kulenovic and Ladas [85].
268 5. Higher-Order Scalar Difference Equations
*15. Conjecture [85]. Prove that every positive solution of the equation x(n+1)= α+βx(n)+γx(n−1),
cx(n − 1)
where n ∈ Z+, αc > 0, converges to the positive equilibrium of the
equation.
*16. Conjecture [85]. Consider the equation
x(n + 1) = α + γx(n − 1) , A+Bx(n)+Cx(n−1)
where α, γ, A, B, C > 0. Prove that if the equation has no solutions of prime period 2, then the positive equilibrium is globally asymptotically stable.
(Open Problems: Kulenovic and Ladas.) Assume that p,q,r ∈ [0,∞), k ≥ 2 is a positive integer. Investigate the global stability of the following equations.
*17.y(n+1) = *18.y(n+1) = *19.y(n+1) =
p+qy(n) . 1+y(n)+ry(n−k)
p+qy(n−k) . 1+y(n)+ry(n−k)
py(n)+y(n−k) . r + qy(n) + y(n − k)
5.5 Applications
5.5.1 Flour Beetles
In this section we consider again the LPA model [83] of the flour beetle (5.3.12) with no larval cannibalism on eggs:
x(n + 1) = αx(n) + βx(n − 2) exp[−c1x(n − 2) − c2x(n)]
= f (x(n), x(n − 1), x(n − 2)) (5.5.1)
whereα=1−μA,β=b(1−μL),c1 =cEA,c2 =cPA,x(n)=A(n+2). Using the Schur–Cohn Criterion one may show that the positive equilib-
rium x∗ = 1 ln β is (locally) asymptotically stable if and only if 2 c1+c2 1−α
the following conditions hold:
|A+B|<1, |A−3B|<3, B(B−A)<1, (5.5.2)
where
1 − α
c β
1 ln −1 . c1 + c2 1 − α
c(1−α) β
A= 2 ln c1 + c2
−α, B=(1−α)
In the sequel, we will go one step further and prove that the equilib- rium point x∗2 is in fact globally asymptotically stable under the following assumptions:
A1 :α+β>1 and β
The following two lemmas from Kuang and Cushing [83] will facilitate the proof of the main result and make it more transparent.
Lemma5.24. Ifα+β>1,then lim sup x(n) ≤
n→∞ Proof. Consider the function h(x)
critical point xc, where
β . c1e(1 − α)
= c1 exe−c1 x . This
function
(5.5.3) has a
h′(xc) = c1e1−c1xc − c21xce1−c1xc = 0.
Hencexc = 1 ,andh 1 =1isthemaximumvalueofh.Itfollowsthat c1 c1
5.5 Applications 269
c1exe−c1x ≤ 1 and, consequently,
xe−c1 x ≤ 1 .
c1e Going back to equation (5.5.1) we obtain
(5.5.4)
(5.5.5)
P
x(n + 1) ≤ αx(n) + βx(n − 2) exp(−c1x(n − 2)) ≤ αx(n) + β (by (5.5.4)).
c1e
By (1.2.8), the solution of (5.5.5) is given by
x(n) ≤ αnx(0) + β c1e(1 − α)
and since α ∈ (0, 1), it follows that
lim sup x(n) ≤ β
(1 − αn)
.
n→∞ c1e(1 − α)
The next lemma shows that (5.5.1) satisfies condition (ii) in Theorem
5.17.
Lemma 5.25. For any u > 0, u ̸= x∗2,
(u − x∗2 )[f (u, u, u) − u] < 0. (5.5.6) Proof. Letg(u)=f(u,u,ur)−u.Then
g(u) = u[α + β exp(−(c1 + c2)u) − 1].
Clearly g(u) = 0 if and only if u = 0 or u = x∗2. We now have two cases to consider.
270 5. Higher-Order Scalar Difference Equations
Case(a):If0
Case (b): If u > x∗2, then
α+βexp[−(c1 +c2)u]−1<α+βexp[−(c1 +c2)x∗2]−1=0,
and (5.5.6) holds true. P We are now ready to prove the following theorem.
Theorem 5.26. If conditions A1 and A2 hold, then lim x(n) = x∗2.
n→∞
Proof. By virtue of Lemma 5.25 it remains to show that condition (ii)
of Theorem 5.17 holds. To accomplish this task we need to show that
∂f , ∂ x(n)
∂f , ∂f ≥ 0 on a region D. Simple computations show that ∂ x(n−1) ∂ x(n−2)
∂f = α − c2βx(n − 2) exp[−c1x(n − 2) − c2x(n)], ∂x(n)
∂f = 0, ∂x(n − 1)
∂f = β(1 − c1x(n − 2)) exp[−c1x(n − 2) − c2x(n)]. ∂x(n − 2)
Now by Lemma 5.24, lim supn→∞ x(n) ≤ β[c1e(1 − α)]−1. Since β < e(1 − α), this implies that lim sup ≤ 1 . Hence, for n > N , for some integer
n→∞ c1 Ngreaterthan2,x(n−2)< 1.LetI= 0,1 andD=I3.Thenx∗∈I
c1 c 2
and ∂f ≥ 0 for i = 1,2. Furthermore, ∂f ≥ α−c2βexp(−c1x(n−
∂ x(n−i) ∂ x(n)
2)) ≥ α − c2β ≥ 0. This shows that f is nondecreasing in each of its
c1e
arguments restricted to D. Since max{x(−2), x(−1), x(0)} > 0, there is n0
suchthatforn≥n0,0
Moreover, the positive equilibrium point is given by
√2 x ∗2 = l n a + a + 4 b .
2
(5.5.8)
Theorem 5.27. Suppose that
1−a
Our main objective in this subsection is to prove the following result.
a+1
Then the positive equilibrium x∗2 of (5.5.7) is globally asymptotically stable
whose basin of attraction is (0, ∞).
Proof. The proof is divided into two parts. The first part establishes lo- cal asymptotic stability and the second part establishes global attractivity. Here we denote x∗2 by x∗.
Part I: Local stability analysis.
By criterion (5.1.9), it follows that the equilibrium point x∗2 is locally
asymptotically stable if and only if the following three inequalities hold:
b − bx∗ − e2x∗ < aex∗ − ax∗ex∗ − bx∗, aex∗ − ax∗ex∗ − bx∗ < e2x∗ − b + bx∗,
bx∗ − b < e2x∗ . Observe that x∗ satisfies the equation
x∗ = (ax∗ + bx∗e−x∗ )e−x∗ or, equivalently, if x∗2 ̸= 0,
e2x∗ = aex∗ + b.
Inequality (5.5.11) is satisfied since by using (5.5.13) we obtain
(5.5.10) (5.5.11) (5.5.12)
(5.5.13)
e2x∗ −b+2bx∗ +ax∗ex∗ −aex∗ =2bx∗ +ax∗ex∗ >0.
272 5. Higher-Order Scalar Difference Equations
Inequalities (5.5.10) and (5.5.12) both hold using (5.5.13) if 0 < x∗ < 2. But this is true in general for any positive solution x(n) since from (5.5.7)
it follows that
x(n + 1) < ax(n)e−x(n) + bx(n − 1)e−x(n−1) ≤ a + b = a + b .
eee
(Why?) (The function x attains its maximum at x = 1.) But by assump-
ex
tion(5.5.9),b≤ a(e−a)
j →∞ x(j )z
R=lim j =|a|. j→∞ a
n ∞aj 1 z
for|z|>|a|.
Z(a )=
z =1−(a/z)=z−a
(6.1.2)
j=0
x ̃(z) = Z(x(n)) =
∞ j=0
x(j)z−j ,
Im z
6.1 Definitions and Examples 275
Region of convergence
R
Region of divergence
FIGURE 6.1. Regions of convergence and divergence for x ̃(z).
Im z
Re z
FIGURE 6.2. Regions of convergence and divergence for Z(an).
A special case of the above result is that of a = 1. In this case we have
Z(1) = z for |z| > 1. z−1
Example 6.2. Find the Z-transform of the sequences {nan} and {n2an}. Solution Recall that an infinite series Z(an) may be differentiated, term by
term, any number of times in its region of convergence [20]. Now,
for|z|>|a|.
for |z| > |a|.
Re z
Regio
n of
Region of convergence
a
divergence
∞ z
ajz−j =(z−a) Taking the derivative of both sides yields
j=0
∞ −a
j=0
−jajz−j−1 = (z − a)2
276 6. The Z-Transform Method and Volterra Difference Equations Hence
Therefore,
Z(nan)=
∞ j=0
∞
jajz−j =−z −jajz−j−1.
j=0
Z(nan) = az for |z| > 1. (z − a)2
(6.1.3)
Again taking the derivative of both sides of the identity
∞ az
jajz−j =(z−a)2 Z(n2an) = az(z + a)
for|z|>|a|
yields
j=0
for |z| > |a|.
Example 6.3. The unit impulse sequence, or the Kronecker delta sequence,
is defined by
The Z-transform of this function is ∞
Z(δk(n)) =
δk(j)z−j = z−k.
(z − a)3
(6.1.4)
1 ifn=k,
δk(n) =
0 ifn̸=k.
j=0
If k = 0, we have the important special case Z(δ0(n)) = 1.
Notice that the radius of convergence of Z(δk(n)) is R = 0. Example 6.4. Find the Z-transform of the sequence {sin(ωn)}.
(6.1.5)
Solution Recall that the Euler identity gives eiθ = cos θ + i sin θ for any real number θ. Hence e−iθ = cos θ − i sin θ. Both identities yield
Thus
eiθ + e−iθ eiθ + e−iθ cos θ = 2 and sin θ = 2i .
Z(sinωn)= 1[Z(eiωn)−Z(e−iωn)]. 2i
Using formula (6.1.2) we obtain 1zz
or
Z(sinωn)=2i z−eiω −z−e−iω
= zsinω
(z − eiω)(z − e−iω)
= zsinω , z2 −(eiω +e−iω)z+1
Z(sin ωn) = z sin ω
z2 −2zcosω+1
for|z|>1
for |z| > 1. (6.1.6)
6.1 Definitions and Examples 277
6.1.1 Properties of the Z-Transform
We now establish some useful properties of the Z-transform that will be
needed in the sequel.
(i) Linearity.
Let x ̃(z) be the Z-transform of x(n) with radius of convergence R1, and let y ̃(z) be the Z-transform of y(n) with radius of convergence R2. Then for any complex numbers α, β we have
Z[αx(n)+βy(n)]=αx ̃(z)+βy ̃(z) for |z|>max(R1,R2). (6.1.7) The proof of property (6.1.7) is left to the reader as Exercises 6.1,
Problem 18.
(ii) Shifting. Let R be the radius of convergence of x ̃(z).
(a) Right-shifting: If x(−i) = 0 for i = 1,2,…,k, then
(b) Left-shifting:
(6.1.8)
(6.1.9)
Z[x(n+k)]=z x ̃(z)−
x(r)z for|z|>R.
The proofs are left as Exercises 6.1, Problem 16. The most commonly used cases of formula (6.1.9) are
Z[x(n + 1)] = zx ̃(z) − zx(0) for |z| > R,
Z[x(n + 2)] = z2x ̃(z) − z2x(0) − zx(1) for |z| > R. (iii) Initial and final value.
(a) Initial value theorem:
Z[x(n − k)] = z−kx ̃(z) for |z| > R.
k−1
k k−r
r=0
lim x ̃(z) = x(0). |z|→∞
(6.1.10)
278
6. The Z-Transform Method and Volterra Difference Equations (b) Final value theorem:
(6.1.11) The proof of formula (6.1.10) follows immediately from the definition
of x ̃(z). To prove formula (6.1.11) we first observe that
∞ j=0
∞ j=0
to
Thus
(z − 1)x ̃(z) = zx(0) +
[x(j + 1) − x(j)]z−j .
x(∞) = lim x(n) = lim (z − 1)x ̃(z). n→∞ z→1
lim(z − 1)x ̃(z) = x(0) + z→1
[x(j + 1) − x(j)] = lim x(n). n→∞
(iv) Convolution.
The convolution∗ of two sequences x(n), y(n) is defined by
n x(n) ∗ y(n) =
j=0
x(n − j)y(j) =
n j=0
x(n)y(n − j).
Now,
Z[x(n) ∗ y(n)] = Interchanging the summation signs yields
Z[x(n) ∗ y(n)] =
y(j) And if we put m − i = s, we obtain
Z[x(n) ∗ y(n)] =
j=0
,
⎝
−j⎠ −s y(j)z x(s)z
∞ j=0
[x(j + 1) − x(j)]z−j.
Z[x(n + 1) − x(n)] =
Using formula (6.1.9) on the left-hand side of the above identity leads
⎡⎤ ∞ m
⎣ x(m − j)y(j)⎦ z−m. m=0 j=0
∞ j=0
∞ m=j
x(m − j)z−m. ⎛∞⎞∞
s=0
Z[x(n) ∗ y(n)] = x ̃(z)y ̃(z).
(6.1.12)
∞ j=0
(v) Multiplication by an property.
S u p p o s e t h a t x ̃ ( z ) i s t h e Z – t r a n s f o r m o f x ( n ) w i t h convergence R. Then
to the reader as Exercises 6.1, Problem 19. Example 6.5. Determine the Z-transform of
g(n) = an sinωn, n = 0,1,2,…. Using Example 6.4 and formula (6.1.13) we have
g ̃(z) = Z(an sin ωn) = (z/a) sin ω
(z/a)2 − 2(z/a) cos ω + 1
= az sin ω , for |z| > |a|. (6.1.14) z2 −2azcosω+a2
x(n) ∗ y(n) =
x(n − j)y(j).
n z Z[a x(n)] = x ̃ a
6.1 Definitions and Examples 279
It is interesting to know that one may obtain formula (6.1.12) if the convolution is defined as
r a d i u s
o f
, for |z| > |a|R.
The proof of (6.1.13) follows easily from the definition and will be left
(6.1.13)
(vi) Multiplication by nk.
In Example 6.2 it was shown that Z(nan) = written in the form
Z(nan) = −z d Z(an). dz
az , which may be (z−a)2
Similarly, formula (6.1.4) may be written in the form
dd Z(n2an) = −z dz −z dz Z(an) .
This may be written in the compact form
d2 Z(n2an) = −z dz Z(an).
Generally speaking, we write
dk d d d −zdz x ̃(z)= −zdz −zdz ··· −zdzx ̃(z) ··· .
It may be shown (Exercises 6.1, Problem 7) that
dk
Z[nkx(n)] = −zdz Z(x(n)). (6.1.15)
280 6. The Z-Transform Method and Volterra Difference Equations Exercises 6.1
1. Find the Z-transform and its region of convergence of the given sequence {x(n)}.
(a) cos ωn. (b)n sin 2n. (c)n.
2. Find the Z-transform and its region of convergence of the sequence
1 for n = 1,3,5, f(n) =
0 for all other values of n.
3. Find the Z-transform and its region of convergence of the sequence
⎧
⎪⎨0 for n = 0,−1,−2,…,
f(n)=⎪−1 forn=1,
⎩an for n = 2,3,4,….
4. Let x(n) be a periodic sequence of period N , i.e., x(n + N ) = x(n) for
all n ∈ Z+. Prove that x ̃(z) = [zn/(zn − 1)]x ̃1(z) for |z| > 1, where
x ̃1(z) = N−1 x(j)z−j(x ̃1)(z) is called the Z-transform of the first j=0
period.
5. Determine the Z-transform of the periodic sequence shown in Figure 6.3.
6. Use Problem 4 to find the Z-transform and its radius of convergence for the periodic sequence of period 4
1 for n = 0, 1, f(n) =
−1 forn=2,3.
7. Let R be the radius of convergence of x ̃(z). Show that
Z[nkx(n)] =
2
1
dk −z dz
x ̃(z) for |z| > R.
1
n 0 1 2 3 4 5 6 7 8 9 10 11
FIGURE 6.3. A periodic sequence.
8. Prove that the Z-transform of the sequence
(n−1)an−2, n=0,1,2,…, x(n) = .
0 n = −1,−2,…, is x ̃(z) = 1 for |z| > |a|.
6.1 Definitions and Examples 281
z−a2
9. Find the Z-transform and its region of convergence of the sequence
defined by
n = −1,−2,….
The first backward difference for a sequence x(n) is defined by ∇x(n) =
x(n)−x(n−1).
10. Find Z[∇x(n)], Z[∇2x(n)].
11. Generalize the results of Problem 10 and show that Z[∇kx(n)] = z−1k x ̃(z).
z
⎧⎨(n−1)(n−2)an−3, n=0,1,2,…,
x(n) = ⎩ 2 0
12. Find Z[∆x(n)], Z[∆2x(n)].
13. Show that Z[∆kx(n)] = (z − 1)kx ̃(z) − z k−1(z − 1)k−j−1∆jx(0).
j=0
14. Let y(n) = n x(i),n ∈ Z+. Show that y ̃(z) = z x ̃(z) for |z| >
i=1 z−1 max{1, R}, where R is the radius of convergence of x ̃(z).
15. Let y(n) = n ix(i). Prove that y ̃(z) = −z2 d x ̃(z).
i=0
16. Prove formulas (6.1.8) and (6.1.9).
17. Find the Z-transform of:
(a) x(n) = nr=0 an−r sin(ωr). (b) nr=0 cos ω(n − r).
18. Prove expression (6.1.7).
19. ShowthatZ[anx(n)]=x ̃z for|z|>|a|R,whereRistheradiusof
a
convergence of x ̃(z).
20. Find the Z-transform and its radius of convergence of the sequence
g(n) = an cos(ωn).
21. Use the initial value theorem to determine x(0) for the sequence {x(n)}
whose Z-transform is given by: (a) 2 , for|z|>a.
z−a
(b)3z, for|z|>3. z−6
z−1 dz
282 6. The Z-Transform Method and Volterra Difference Equations 22. Extend the initial value theorem to finding x(1), x(2) by proving:
6.2
(i) x(1) = lim [z(x ̃(z) − x(0))], |z|→∞
(ii) x(2) = lim [z(x ̃(z) − zx(0) − x(1))]. |z|→∞
The Inverse Z-Transform and Solutions of Difference Equations
As we have mentioned in the introduction to this chapter, the Z-transform transforms a difference equation of an unknown sequence x(n) into an algebraic equation in its Z-transform x ̃(z). The sequence x(n) is then ob- tained from x ̃(z) by a process called the inverse Z-transform. This process is symbolically denoted by
Z−1[x ̃(z)] = x(n). (6.2.1)
The uniqueness of the inverse Z-transform may be established as follows: Suppose that there are two sequences x(n), y(n) with the same Z-transform, that is,
Then
∞ i=0
x(i)z−i =
∞ i=0
y(i)z−i,
for |z| > R.
∞ i=0
[x(i) − y(i)]z−i = 0,
It follows from Laurent’s theorem [20] that x(n) ≡ y(n). The most
commonly used methods for obtaining the inverse Z-transform are: 1. power series method;
2. partial fractions method;
3. inversion integral method.
It is imperative to remind the reader that when finding the inverse Z- transform, it is always assumed that for any sequence x(n), x(k) = 0 for k = −1,−2,….
6.2.1 The Power Series Method
In this method we obtain the inverse Z-transform by simply expanding
x ̃(z) into an infinite power series in z−1 in its region of convergence:
x ̃(z) = ∞ aiz−i for |z| > R. Then by comparing this with Z[x(n)] = ∞ i=0
i=0 x(i)z−i for |z| > R, one concludes that x(n) = an, n = 0,1,2,….
for |z| > R.
6.2 The Inverse Z-Transform and Solutions of Difference Equations 283
If x ̃(z) is given in the form of a rational function x ̃(z) = g(z)/h(z), where g(z) and h(z) are polynomials in z, then we simply divide g(z) by h(z) to obtain a power series expansion x ̃(z) in z−1. The only possible drawback of this method is that it does not provide us with a closed-form expression of x(n).
Example 6.6. Obtain the inverse Z-transform of x ̃ ( z ) = z ( z + 1 ) .
(z − 1)2
Solution We first write x(z) as a ratio of two polynomials in z−1: 1+z−1
x ̃(z)= 1−2z−1 +z−2. Dividing the numerator by the denominator, we have
x ̃ ( z ) = 1 − 3 z − 1 + 5 z − 2 + 7 z − 3 + 9 z − 4 + 1 1 z − 5 + · · · . x(0)=1, x(2)=3, x(3)=5, x(4)=7,…, x(n)=2n+1.
6.2.2 The Partial Fractions Method
This method is used when the Z-transform x ̃(z) is a rational function in z, analytic at ∞, such as
x ̃(z)= b0zm +b1zm−1 +···+bm−1z+bm, m≤n. (6.2.2) zn +a1zn−1 +···+an−1z+bn
If x ̃(z) in expression (6.2.2) is expressed by a partial fraction expression, x ̃ ( z ) = x ̃ 1 ( z ) + x ̃ 2 ( z ) + x ̃ 3 ( z ) + · · · ,
then by the linearity of the inverse Z-transform one obtains x(n) = Z−1[x ̃1(z)] + Z−1[x ̃2(z)] + Z−1[x ̃3(z)] + · · · .
Then a Z-transform table (Table 6.1; see the end of this chapter) is used to find Z−1[x ̃i(z)], i = 1,2,3,….
Before giving some examples to illustrate this method we remind the reader that the zeros of the numerator of expression (6.2.2) are called zeros of x ̃(z), and zeros of the denominator of expression (6.2.2) are called poles of x ̃(z).
Remark: Since x ̃(z) is often an improper fraction, it is more convenient to expand x ̃(z)/z rather than x ̃(z) into sums of partial fractions.
Thus
284 6. The Z-Transform Method and Volterra Difference Equations
Example 6.7. Simple Poles
Solve the difference equation
x(n+2)+3x(n+1)+2x(n)=0, x(0)=1, x(1)=−4.
Solution Taking the Z-transform of both sides of the equation, we get x ̃(z) = z(z − 1) .
(z + 1)(z + 2) We expand x ̃(z)/z into partial fractions as follows:
x ̃(z)/z = (z − 1) = a1 + (z + 1)(z + 2) z + 1
a2 . z + 2
Clearing fractions, we obtain
z − 1 = a1(z + 2) + a2(z + 1).
This reduces to
z−1=(a1 +a2)z+(2a1 +a2). Comparing coefficients of like powers of z, we get
a1 + a2 = 1, 2a1 + a2 = −1.
Hence a1 = −2, a2 = 3. Consequently,
x ̃ ( z ) = − 2 z + 3 z .
Thus
z+1 z+2 x(n) = −2(−1)n + 3(−2)n.
Remark: If x ̃(z) has a large number of poles, a computer may be needed to determine the constants a1, a2, . . . .
Example 6.8. Repeated Poles
Solve the difference equation
x(n + 4) + 9x(n + 3) + 30x(n + 2) + 44x(n + 1) + 24x(n) = 0,
x(0)=0, x(1)=0, x(2)=1, x(3)=10. Solution Taking the Z-transform, we get
x ̃(z) = z(z − 1) . (z + 2)3(z + 3)
It is convenient here to expand x ̃(z)/z into partial fractions as follows:
x ̃(z)/z= z−1 =b+a1 +a2 +a3.(6.2.3) (z + 2)3(z + 3) z + 3 (z + 2)3 (z + 2)2 z + 2
6.2 The Inverse Z-Transform and Solutions of Difference Equations 285 This time we use a smarter method to find a1,a2,a3, and a4. To find b we
multiply (6.2.3) by (z + 3) and then evaluate at z = −3. This gives b= (z−1) =4.
(z + 2)3 z=−3
To find a1 we multiply (6.2.3) by (z + 2)3 to get
z − 1 2 (z + 2)3 z+3 =a3(z+2) +a2(z+2)+a1 +4 (z+3)
and evaluate at z = −2. This gives
a 1 = z − 1 = − 3 .
(6.2.4)
(6.2.5)
z−3 z=−2
To find a2 we differentiate (6.2.4) with respect to z to get
4
(z+3)2 =2a3(z+2)+a2 +
and again evaluate at z = −2. This gives a 2 = d z − 1
r(2z + 7)(z + 2)2 (z+3)2
= 4 . Finally, to find a3 we differentiate (6.2.5) to obtain
dz z+3 z=−2
−8 d2 (z+2)3
(z+3)3 =2a3+4dz2 (z+3), and if we let z = −2, then we have
Hence
a = 1 d 2 z − 1 = − 4 . 3 2 dz2 z + 3 z=−2
x ̃(z)=−4z+ 4z − 3z +4z. z + 2 (z + 2)2 (z + 2)3 z + 3
The corresponding sequence is (Table 6.1, at the end of this chapter) x(n) = −4(−2)n − 2n(−2)n + 3 n(n − 1)(−2)n + 4(−3)n
3 11 4
= 4n2 − 4 n−4 (−2)n +4(−3)n.
Remark: The procedure used to obtain a1,a2, and a3 in the preceding example can be generalized. If x ̃(z)/z has a pole of multiplicity m at z = z0, then the corresponding terms in the partial fraction expansion can be written
···+ a1 +···+ am +···, (z − z0)m z − z0
286 6. The Z-Transform Method and Volterra Difference Equations and a1, a2, . . . , am can be found using the formula
1 d r − 1 x ̃ ( z ) ar= r−1z−z0)m .
(r − 1)! dz z z=z0 Example 6.9. Complex Poles
Solve the difference equation
x(n+3)−x(n+2)+2x(n)=0, x(0)=1, x(1)=1.
Solution Taking the Z-transform of the equation, we get z3
x ̃(z)= (z2 −2z+2)(z+1).
Next we expand x ̃(z)/z as a sum of the partial fraction in the form
z2 a1 a2 a3 x ̃(z)/z = (z2 − 2z + 2)(z + 1) = [z − (1 + i)] + [z − (1 − i)] + (z + 1).
Using the method of the preceding example we obtain a= z2 =1,
3 z2 − 2z + 2z=−1 5 a1= z2
= 1 =2−1i, [z−(1−i)](z+1) z=1+i 2+i 5 5
a 2 = a ̄ 1 = 2 + 1 i . 55
Hence
5
n ̄n n √nnπ
1 z a 1 z a ̄ 1 z x ̃(z)=5 + + ̄,
where λ = 1 + i. Thus
x(n) = 1(−1)n + a1λn + a ̄1λ ̄n.
z+1 z−λ z−λ
But
a1λ +a ̄1λ =2Re(a1λ )=2|a ̄1|( 2) cos 4 +arga1 ,
where |a1| = 1√5 and arga1 = tan−1(1) = 0.46 radians. Thus 52
1 n 2√√n nπ x(n)=5(−1) +5 5( 2) cos 4 +0.46 .
6.2 The Inverse Z-Transform and Solutions of Difference Equations 287 6.2.3 The Inversion Integral Method1
From the definition of the Z-transform, we have
∞
x(i)z−i.
Multiplying both sides of the above equation by zn−1, we get
∞
i=0
1
x(n) = 2πi
and by the residue theorem [20] we obtain
x ̃(z) =
i=0
x(i)zn−i−1
= x(0)zn−1 + x(1)zn−2 + · · · + x(n)z−1 + x(n + 1)z−2 + · · · .
(6.2.6)
Equation (6.2.6) gives the Laurent series expansion of x ̃(z)zn−1 around z = 0.
Consider a circle C, centered at the origin of the z-plane, that encloses all poles of x ̃(z)zn−1. Since x(n) is the coefficient of z−1, it follows by the Cauchy integral formula [20] that
x ̃(z)zn−1 =
Suppose that
c
x(n) = sum of residues ofx ̃(z)zn−1. x ̃(z)zn−1 = h(z).
x ̃(z)zn−1 dz,
(6.2.7)
(6.2.8)
g(z)
In evaluating the residues of x ̃(z)zn−1, there are two cases to consider:
(i) g(z) has simple zeros (i.e., x ̃(z)zn−1 has simple poles) (see Figure 6.4). In this case the residue Ki at a pole zi is given by
h(z)
Ki = lim (z−zi) . (6.2.9)
z→zi g(z)
(ii) g(z) has multiple zeros (i.e., x ̃(z)zn−1 has multiple poles). If g(z) has
a multiple zero zi of order r, then the residue Ki at zi is given by
1 dr−1
r−1 (z−zi)r 1Requires some knowledge of residues in complex analysis [20].
Ki = lim (r−1)!z→zi dz
h(z) g(z)
.
288 6. The Z-Transform Method and Volterra Difference Equations Im z
FIGURE 6.4. Poles of x ̃(z).
Example 6.10. Obtain the inverse Z-transform of
Solution Notice that
x ̃(z) =
n−1
z(z − 1) . (z − 2)2(z + 3)
(z − 1)zn
= (z − 2)2(z + 3).
x ̃(z)z
.z1
.z2 .z3
Re z
Thus x ̃(z)zn−1 has a simple pole at z1 = −3 and a double pole at z2 = 2. Thus from formula (6.2.8), we get x(n) = K1 + K2, where K1, K2 are the residues of x(z)zn−1 at z1, z2, respectively. Now,
K1 = lim z→−3
(−3) ,
(z + 3)(z − 1)zn
2 =
−4 n
(z−2) (z+3)
K2 = 1 lim d (z−2)2(z−1)zn
25
Thus
(2−1)!z→2 dz (z−2)2(z+3)
= lim zn−1[(z+3)(z+nz−n)−z(z−1)]
z→2 (z + 3)2 = (8 + 5n)(2)n−1.
25
x(n)=−4(−3)n+(8+5n)(2)n−1, n=0,1,2,…. 25 25
Example 6.11. Electric Circuits or a Ladder Network
Consider the electric network shown in Figure 6.5. Here i(n) is the current in the nth loop; R is the resistance, which is assumed to be constant in every loop; and V is the voltage. By Ohm’s law, the voltage (or electric potential) between the ends of a resistor R may be expressed as V = iR.
6.2 The Inverse Z-Transform and Solutions of Difference Equations 289 RRRRRR
R
Now, Kirchhoff’s2 second law states that “in a closed circuit the impressed voltage is equal to the sum of the voltage drops in the rest of the circuit.” By applying Kirchhoff’s law to the loop corresponding to i(n+1) we obtain
V
RRRRRRR
i(0) i(1) i(n) i(n+1) i(n+2) i(k)
FIGURE 6.5. A ladder network.
R[i(n + 1) − i(n + 2)] + R[i(n + 1) − i(n)] + Ri(n + 2) = 0,
or
i(n + 2) − 3i(n + 1) + i(n) = 0. For the first loop on the left we have
(6.2.10)
(6.2.11)
or
V = Ri(0) + R(i(0) − i(1)),
i(1) = 2i(0) − V . R
Taking the Z-transform of (6.2.10) with the data (6.2.11) yields the equation
⎡⎤
z2−1+V z
̃ı(z) = z[zi(0) − 3i(0) + i(1)] = ⎣ Ri(0) ⎦ i(0).
(6.2.12) becomes
i(0) V 2 zsinhω + 2 +R √5 z2−2zcoshω+1 .
2 Gustav Kirchhoff, a German physicist (1824–1887), is famous for his contributions to electricity and spectroscopy.
(6.2.12) Let ω > 0 be such that coshω = 3. Then sinhω = 5. Then expression
z2 −3z+1 z2 −3z+1
√
z2−zcoshω ̃i(z)=i(0) z2 −2zcoshω+1
22
290 6. The Z-Transform Method and Volterra Difference Equations Taking the inverse Z-transform (Table 6.1, at the end of this chapter), we
obtain
i(0) V 2 i(n) = i(0) cosh(ωn) + 2 + R √5
sinh(ωn).
Exercises 6.2
1. Use the partial fractions method to find the inverse Z-transform of: (a) z .
(b)
2. Use (a)
(b)
3. Use (a)
(b)
4. Use find
(z− 1 )(z+1) 2
z(z+1) . (z+2)2 (z−1)
the power series method to find the inverse Z-transform of: z−2 .
(z−1)(z+3) e−az .
(z−e−a)2
the inversion integral method to find the inverse Z-transform of:
z(z−1). (z+2)3
z(z+2) . (z− 1 )(z+i)(z−i)
2
the partial fractions method and the inversion integral method to the inverse Z-transform of:
(a) z(z+1). (z−2)2
(b) z2+z+1 . (z−1)(z2 −z+1)
In Problems 5 through 7, use the Z-transform method to solve the given difference equation.
5. (The Fibonacci Sequence). x(n + 2) = x(n + 1) + x(n), 0, x(1) = 1.
6. x(n+2)−3x(n+1)+2x(n)=δ0(n), x(0)=x(1)=0.
7. (n+1)x(n+1)−nx(n)=n+1, x(0)=0.
x(0) =
8. Consider the continued fraction
aa a K=Kn=0+1
bn b0 b1 + a2a3 b2+b3+…
= a0 a1 a2 …. b0+ b1+ b2+
Letai =bi =1foralli∈Z+,andx(n)= a1 a2 … an . b1+ b2+ bn+
(a) Showthatx(n+1)=1+ 1 .Findx(n). x(n)
6.3 Volterra Difference Equations of Convolution Type: The Scalar Case 291
(b) Find K = 1 + limn→∞ x(n).
9. Prove that the convolution product is commutative and associative
(i.e., x ∗ y = y ∗ x; x ∗ (y ∗ f ) = (x ∗ y) ∗ f ).
10. Solve, using convolution, the equation x(n+1) = 2+4 nr=0(n−r)x(r).
11. Solve the equation x(n) = 1 − n−1 en−r−1x(r). r=0
6.3 Volterra Difference Equations of Convolution Type: The Scalar Case3
Volterra difference equations of convolution type are of the form
n j=0
where A ∈ R and B : Z+ → R is a discrete function. This equa- tion may be considered as the discrete analogue of the famous Volterra integrodifferential equation
t
0
Equation (6.3.2) has been widely used as a mathematical model in popu- lation dynamics. Both (6.3.1) and (6.3.2) represent systems in which the future state x(n+1) does not depend only on the present state x(n) but also on all past states x(n − 1), x(n − 2), . . . , x(0). These systems are sometimes called hereditary. Given the initial condition x(0) = x0, one can easily gen- erate the solution x(n, x0) of (6.3.1). If y(n) is any other solution of (6.3.1) with y(0) = x0, then it is easy to show that y(n) = x(n) for all n ∈ Z+ (Exercises 6.3, Problem 8).
One of the most effective methods of dealing with (6.3.1) is the Z-transform method. Let us rewrite (6.3.1) in the convolution form
x(n + 1) = Ax(n) +
B(n − j)x(j), (6.3.1)
x′(t) = Ax(t) +
B(t − s)x(s) ds. (6.3.2)
x(n + 1) = Ax(n) + B ∗ x.
Taking formally the Z-transform of both sides of (6.3.3), we get
(6.3.3)
(6.3.4)
which gives or
zx ̃(z) − zx(0) = Ax ̃(z) + B ̃(z)x ̃(z),
[z − A − B ̃(z)]x ̃(z) = zx(0),
x ̃(z) = zx(0)/[z − A − B ̃(z)].
3This section requires some rudiments of complex analysis [20].
292 6. The Z-Transform Method and Volterra Difference Equations Let
g(z) = z − A − B ̃(z). (6.3.5)
The complex function g(z) will play an important role in the stability analysis of (6.3.1). Before embarking on our investigation of g(z) we need to present a few definitions and preliminary results.
Definition 6.12. Let E be the space of all infinite sequences of com- plex numbers (or real numbers) x = (x(0), x(1), x(2), . . .). There are three commonly used norms that may be defined on subsets of E. These are
(i) the l1 norm: ∥x∥1 = ∞i=0 |x(i)|;
(ii) the l2, or, Euclidean norm: ∥x∥2 = ∞i=0 |x(i)|21/2 ; (iii) the l∞ norm: ∥x∥∞ = supi≥0 |x(i)|.
The corresponding normed spaces are called l1,l2, and l∞, respectively. One may show easily that (Exercises 6.3, Problem 6)
l1 ⊂l2 ⊂l∞.
Definition 6.13. A complex function g(z) is said to be analytic in a region in the complex plane if it is differentiable there. The next result establishes an important property of l1 sequences.
Theorem 6.14. If x(n) ∈ l1, then:
(i) x ̃(z) is an analytic function for |z| ≥ 1;
(ii) |x ̃(z)| ≥ ∥x∥ for |z| ≥ 1. Proof.
(i) Since x(n) ∈ l1, the radius of convergence of x ̃(z) = ∞n=0 x(n)z−n is R = 1. Hence x ̃(z) can be differentiated term by term in its region of convergence |z| > 1. Thus x ̃(z) is analytic on |z| > 1. Furthermore, since x(n) ∈ l1, x ̃(z) is analytic for |z| = 1.
(ii) This is left as Exercises 6.3, Problem 9. P
We now turn our attention to the function g(z) = z−A−B ̃(z) in formula (6.3.5). This function plays the role of the characteristic polynomial of linear difference equations. (See Chapter 2.) In contrast to polynomials, the function g(z) may have infinitely many zeros in the complex plane. The following lemma sheds some light on the location of the zeros of g(z).
Lemma 6.15 [39]. The zeros of
g ( z ) = z − A − B ̃ ( z )
all lie in the region |z| < c, for some real positive constant c. Moreover, g(z) has finitely many zeros z with |z| ≥ 1.
6.3 Volterra Difference Equations of Convolution Type: The Scalar Case 293
Proof. Suppose that all the zeros of g(z) do not lie in any region |z| < c for any positive real number c. Then there exists a sequence {zi} of zeros of g(z) with |zi| → ∞ as i → ∞. Now,
̃ ∞ |zi − A| = |B(zi)| ≤
n=0
Notice that the right-hand side of inequality (6.3.6) goes to B(0) as i → ∞, while the left-hand side goes to ∞ as i → ∞, which is a contradiction. This proves the first part of the lemma. P
To prove the second part of the lemma, we first observe from the first part of the lemma that all zeros z of g(z) with |z| ≥ 1 lie in the annulus 1 ≤ |z| ≤ c for some real number c. From Theorem 6.14 we may conclude that g(z) is analytic in this annulus (1 ≤ |z| ≤ c). Therefore, g(z) has only finitely many zeros in the region |z| ≥ 1 [39].
Next we embark on a program that will reveal the qualitative behavior of solutions of (6.3.1). In this program we utilize (6.3.4), which may be written as
x ̃(z) = x(0)zg−1(z). (6.3.7) Let γ be a circle that includes all the zeros of g(z). The circle γ is guaranteed
to exist by virtue of Lemma 6.15. By formula (6.2.7) we obtain
1
x(n) = 2πi
and by formula (6.2.8) we get
x(n) = sum of residues of [x(0)zng−1(z)].
This suggests that
z =[zr−(zr−z)] =
i zr (z−zr).
x(n) =
γ
pr(n)zrn,
where the sum is taken over all the zeros of g(z) and where pr(n) is a polynomial in n of degree less than k − 1 if zr is a multiple root of order k. To show the validity of formula (6.3.10), let zr be a zero of g(z) of order k. We write the following Laurent’s series expansion [20]:
∞
g−1(z) = gn(z − zr)n, for some constants gn,
n=−k
n nnnn−ii
i=0
|B(n)||zi|−n. (6.3.6)
x(0)zng−1(z) dz,
(6.3.8)
(6.3.9)
(6.3.10)
294 6. The Z-Transform Method and Volterra Difference Equations
The residue of x(0)zng−1 at zr is x(0) times the coefficient of (z − zr)−1 in g−1(z)zn. The coefficient of (z − zr)−1 in g−1(z)zn is given by
g n zn−k+1 +g n zn−k+2 +· · ·+g n zn. (6.3.11) −k k−1 r −k+1 k−2 r −1 0 r
It follows from formula (6.3.9) that x(n) may be given by formula (6.3.10). Formula (6.3.10) has the following important consequences.
Theorem 6.16 [39]. The zero solution of (6.3.1) is uniformly stable if and only if:
(a) z−A−B ̃(z)̸=0 for all |z|>1, and
(b) if zr is a zero of g(z) with |zr| = 1, then the residue of zng−1(z) at zr
is bounded as n → ∞.
Proof. Suppose that conditions (a) and (b) hold. If zr is a zero of g(z) with |zr| < 1, then from formula (6.3.10) its contribution to the solution x(n) is bounded. On the other hand, if zr is a zero of g(z) with |zr| = 1 at which the residue of x(0)zng−1(z) is bounded as n → ∞, then from formula (6.3.9) its contribution to the solution x(n) is also bounded. This shows that |x(n)| ≤ L|x(0)| for some L > 0, and thus we have uniform stability. The converse is left to the reader as Exercises 6.3, Problem 10. P
We observe here that a necessary and sufficient condition for condition (b) is that each zero z of g(z) with |z| = 1 must be simple (Exercises 6.3, Problem 11).
The next result addresses the question of asymptotic stability. Theorem 6.17 [39]. The zero solution of (6.3.1) is uniformly asymptot-
ically stable if and only if
(6.3.12) Proof. The proof follows easily from formula (6.3.10) and is left to the
reader as Exercises 6.3, Problem 12. P Exercises 6.3
1. Solve the Volterra difference equation x(n+1) = 2x(n)+nr=0 2n−rx(r), and then determine the stability of its zero solution.
z−A−B ̃(z)̸=0, forall|z|≥1.
2. Solve the Volterra difference equation x(n + 1) = −1x(n) + n r−n 2
r=0 3 x(r), and then determine the stability of its zero solution.
3. Use Theorems 6.16 and 6.17 to determine the stability of the zero solutions of the difference equations in Problems 1 and 2.
6.4 Explicit Criteria for Stability of Volterra Equations 295
4. Without finding the solution of the equation
1 n1r−n
x(n + 1) = −4x(n) + determine the stability of its zero solution.
2 x(r),
5. Determine the stability of the zero solution of x(n + 1) = 2x(n) −
12 nr=0(n − r)x(r), using Theorem 6.16 or 6.17.
6. Prove that l1 ⊂ l2 ⊂ l∞.
7. Let x = {xn} and y = {yn} be two l1 sequences. Prove that x∗y ∈ l1 by following these steps:
(i)If∞ x(i)=a,∞ y(i)=b, and c(n)=n x(n− i=0 i=0 i=1
i)y(i), show that ∞i=0 c(i) = ab.
∞ ∞ ∞
(ii) Prove that n=0 |c(n)| ≤ ( i=0 |x(i)|) j=0 |y(j)| .
8. Prove the uniqueness of solutions of (6.3.1), that is, if x(n) and y(n) are solutions of (6.3.1) with x(0) = y(0), then x(n) = y(n) for all n ∈ Z+.
9. If x(n) ∈ l1 show that |x ̃(z)| ≤ ∥x∥1 for |z| ≥ 1.
*10. Suppose that the zero solution of (6.3.1) is uniformly stable. Prove
that:
(a) g(z)=z−A−Bˆ(z)̸=0forall|z|>1.
(b) If zr is a zero of g(z) with |zr| = 1, then the residue of zng−1(z) at zr is bounded.
11. Prove that a necessary and sufficient condition for condition (b) in Theorem 6.16 is that zr be a simple root of g(z).
*12. Prove Theorem 6.17.
6.4 Explicit Criteria for Stability of Volterra Equations
The stability results in Section 6.3 are not very practical, since locating the zeros of g(z) is more or less impossible in most problems. In this section we provide explicit conditions for the stability of (6.3.1). The main tools in this study are Theorems 6.17 and Rouch ́e’s Theorem (Theorem 5.13).
r=0
296 6. The Z-Transform Method and Volterra Difference Equations Theorem 6.18 [39]. Suppose that B(n) does not change sign for n ∈ Z+.
Then the zero solution of (6.3.1) is asymptotically stable if ∞
|A| +
Proof. Let β = ∞n=0 B(n) and D(n) = β−1B(n). Then ∞n=0 D(n) = 1. Furthermore, D ̃(1) = 1 and |D ̃(z)| ≤ 1 for all |z| ≥ 1. Let us write g(z) in the form
g(z) = z − A − βD ̃(z). (6.4.2)
To prove uniform asymptotic stability of the zero solution of (6.3.1), it suffices to show that g(z) has no zero z with |z| ≥ 1. So assume that there exists a zero zr of g(z) with |zr| ≥ 1. Then by (6.4.2) we obtain |zr − A| = |βD ̃(z)| ≤ |β|. Using condition (6.4.1) one concludes that |zr| ≤ |A| + |β| < 1, which is a contradiction. This concludes the proof of the theorem. P
Unfortunately, we are not able to show that condition (6.4.1) is a neces- sary condition for asymptotic stability. However, in the next result we give a partial converse to the above theorems.
Theorem 6.19 [39]. Suppose that B(n) does not change sign for n ∈ Z+. Then the zero solution of (6.3.1) is not asymptotically stable if any one of the following conditions holds:
(i) A+∞n=0B(n)≥1.
(ii) A+∞n=0B(n)≤−1andB(n)>0forsomen∈Z+.
(iii) A+∞n=0 B(n) < −1 and B(n) < 0 for some n ∈ Z+, and ∞n=0 B(n) is sufficiently small.
Proof. Let β and D(n) be as defined in the proof of Theorem 6.18.
(i) Assume condition (i). If A+β = 1, then clearly z = 1 is a root of g(z) defined in (6.4.2). Hence by Theorem 5.17 the zero solution of (6.3.1) is not asymptotically stable. If A+β > 1, say A+β = 1+δ, then there are two areas to consider.
(a) If β < 0, then we let γ be the circle in the complex plane with center at A and radius equal to |β| + 1 δ. Then on γ (Figure 6.6)
2
we have |z| > 1 and thus
|βD ̃ (z)| ≤ |β| < |z − A|. (6.4.3)
Let h(z) = −βD ̃(z),f(z) = z − A. Then from inequality (6.4.3) |h(z)| < |f(z)| on γ. Hence by Rouch ́e’s Theorem (Theorem 5.13), g(z) = f(z)+h(z) and f(z) have the same number of zeros inside γ. Since A is the only zero of f(z) inside γ, then g(z) has exactly
n=0
B(n) < 1. (6.4.1)
6.4 Explicit Criteria for Stability of Volterra Equations 297
Im z
11+δA
B +1δ 2
FIGURE 6.6. A circle with center A and radius |β| + δ . 2
one zero z0 inside γ with |z0| > 1. Again by using Theorem 6.16, the zero solution of (6.3.1) is not asymptotically stable.
(b) Suppose that β > 0. Since A+β > 1, it follows that g(z) = 1−A−β < 0. Moreover, |D ̃(A+β)| ≤ 1. Thus g(A+β) = β[1 − D ̃ (A + β)] ≥ 0. Therefore, g(z) has a zero between 1 and A + β and, consequently, by Theorem 6.17, the zero solution of (6.3.1) is not asymptotically stable. This completes the proof of condition (i).
Parts (ii) and (iii) are left to the reader as Exercises 6.4, Problems 7 and 8. P
The above techniques are not expendable to uniform stability. This is mainly due to the lack of easily verifiable criteria for condition (b) of The- orem 6.16. Therefore, new techniques are needed to tackle the problem of uniform stability. These techniques involve the use of Liapunov functionals (functions), which we have encountered in Chapter 4.
Let E by the space of all infinite sequences of complex numbers as defined in Definition 6.12. Then a function V : E → R is said to a Liapunov functional if, for x = {x(n)} ∈ E,
(i) V (x) is positive definite (Chapter 4), (ii) ∆V (x) ≤ 0,
where ∆V (x) = V (xˆ) − V (x) and xˆ(n) = x(n + 1) for all n ∈ Z+.
The next result illustrates the use of Liapunov functionals in stability
theory.
Theorem 6.20 [39]. The zero solution of (6.3.1) is uniformly stable if
n j=0
Re z
|A| +
|B(j)| ≤ 1 for all n ∈ Z+. (6.4.4)
298 6. The Z-Transform Method and Volterra Difference Equations
Proof.
For x ∈ E, let
V (x) = |x(n)| +
Then
n n∞
∆V (x) = Ax(n) + B(n − j)x(j) +
|B(s − r)||x(r)| (6.4.6)
(6.4.7)
(6.4.8)
j=0
n−1 ∞
r=0s=n+1 |B(s − r)||x(r)|
⎞
|B(j)| − 1⎠ |x(n)|.
By assumption (6.4.4) we thus have
∆V (x) ≤ 0.
⎛ r=0 s=n
− |x(n)| −
≤ ⎝|A| +
∞
j=0
n−1 ∞
r=0 s=n
|B(s − r)||x(r)|.
(6.4.5)
From (6.4.5) we obtain |x(n)| ≤ V(x). Using inequality (6.4.8) and expression (6.4.5) again we obtain
|x(n)| ≤ V (x) ≤ |x(0)|.
Consequently, the zero solution is uniformly stable (Chapter 4). P
Exercises 6.4
Use Theorem 6.19 to determine the stability and instability of the zero solution of the equations in Problems 1, 2, and 3.
1. x(n+1)=−1x(n)+n 1n+1−rx(r). 4 r=0 3
2. x(n+1)= 1x(n)+n (n−r)x(r). 2 r=0
3. x(n+1)= 1x(n)+n er−nx(r). 3 r=0
4. Find the values of a for which the zero solution of the equation x(n) = n−1(n − r − 1)an−r−1x(r) is:
r=0
(i) uniformly stable,
(ii) asymptotically stable, (iii) not asymptotically stable.
5. Determine the values of a for which the zero solution of the equation ∆x(n)=−2x(n)+n (n−r)2an−rx(r)isasymptoticallystable.
3 r=0
6. Prove Theorem 6.18 using the method of Liapunov functionals used in the proof of Theorem 6.20.
7. Prove part (ii) of Theorem 6.19.
8. Prove part (iii) of Theorem 6.19.
9. Provide details of how inequality (6.4.7) is obtained from inequality (6.4.6).
10. (Open problem). Discuss the stability of the zero solution of (3.5.1) under the condition A + ∞n=0 B(n) = −1 and ∞n=0 B(n) < 0.
11. (Open problem). Can we omit the assumption that ∞n=0 B(n) is sufficiently small in Theorem 6.19, part (iii)?
12. (Open problem). Develop a necessary and sufficient condition for the asymptotic stability of the zero solution of (6.3.1).
6.5 Volterra Systems
In this section we are mainly interested in the following Volterra system of convolution type:
n j=0
where A = (aij) is a k×k real matrix and B(n) is a k×k real matrix defined on Z+. It is always assumed that B(n) ∈ l1, i.e., ∞j=0 |B(j)| < ∞. The Z-transform for sequences in Rk and matrices Rk×k is defined in the natural way, that is,
Z[x(n)] = Z(x1(n)), Z(x2(n)), . . . , Z(xk(n))T , Z[B(n)] = (Z(bij(n)).
Thus all the rules and formulas for the Z-transform of scalar sequences hold for vector sequences and matrices.
Taking the Z-transform of both sides of (6.5.1), one obtains zx ̃(z) − zx(0) = Ax ̃(z) + B ̃(z)x ̃(z), |z| > R,
which yields
x ̃(z) = [zI − A − B ̃(z)]−1zx(0), |z| > R. (6.5.2) Theorem 6.17 for scalar equations has the following counterpart for
systems.
Theorem 6.21. A necessary and sufficient condition for uniform asymptotic stability is
(6.5.3)
x(n + 1) = Ax(n) +
B(n − j)x(j), (6.5.1)
6.5 Volterra Systems 299
det(zI−A−B ̃(z))̸=0, forall|z|≥1.
300 6. The Z-Transform Method and Volterra Difference Equations Proof. See [39]. P
An application of the preceding theorem will be introduced next. This will provide explicit criteria for asymptotic stability. But before introducing our result we need the following lemma concerning eigenvalues of matrices.
Lemma 6.22 [14]. Let G = (gij) be a k×k matrix. If z0 is an eigenvalue of G, then:
(i) |z0−gii||z0−gjj|≤′r|gir|′r|gjr|,forsomei,j,i̸=j,and (ii) |z0 − gtt||z0 − gss| ≤ ′r |grt| ′r |grs|, for some t, s, t ̸= s,
′ k where r gir means r=1 gir − gii.
Using the above lemma we can prove the next result. Let
ically stable if either one of the following conditions holds: (i) kj=1(|aij|+βij)<1,foreachi,1≤i≤k,or
(ii) ki=1(|aij|+βij)<1,foreachj,1≤j≤k.
Proof. (i) To prove uniform asymptotic stability under condition (i) we
need to show that condition (6.5.3) holds. So assume the contrary, that is,
det(z0I − A − B ̃(z0)) = 0 for some z0 with |z0| ≥ 1.
Then z0 is an eigenvalue of the matrix A + B ̃(z0). Hence by condition (i) in Lemma 6.22, we have
Similarly,
∞ n=0
|bij(n)|, 1 ≤ i,j ≤ k.
Theorem 6.23 [39]. The zero solution of (6.5.1) is uniformly asymptot-
|z0 −aii − ̃bii(z0)||z0 −ajj − ̃bjj(z0)| ≤ But
′ ′ |air + ̃bir(z0)| rr
|ajr + ̃bjr(z0)|. (6.5.4)
βij =
|z0 − aii − ̃bii(z0)| ≥ |z0| − |aii| − | ̃bii(z0)| ≥ 1 − |aii| − | ̃bii(z0)| ′
(|air | + |βir )|
̃ ′ |z0 − ajj − bjj(z0)| >
>
(by condition (i)).
(|ajr| + βjr).
r
r
Combining both inequalities, we get
̃ ̃ ′ ′
|z0 − aii − bii(z0)||z0 − ajj − bjj(z0)| > (|air| + βir) (|ajr| + βjr).
rr
It is clear that this contradicts inequality (6.5.4) if one notes that for any 1 ≤ s, m ≤ k,
|ast| + βst ≥ |ast| + | ̃bst(z0)| ≥ |ast + ̃bst(z0)|. P
As in the scalar case, the above method may be extended to provide criteria for uniform stability. Again, the method of Liapunov functionals will come to the rescue.
Theorem 6.24 [39]. The zero solution of (6.5.1) is uniformly stable if (6.5.5)
for all j = 1,2,…,k.
Proof. Define the Liapunov functional
Then
V (x) =
|xi(n)| + |bij(s − r)||xj(r)| . j=1 r=0 s=n
⎡⎤
k k n−1 ∞ ⎣⎦
i=1
∆V(6.5.1) (x) ≤
k k i=1 j=1
6.5 Volterra Systems 301
+
A crucial but simple step is now in order. Observe that
(6.5.6)
and
k k
|aij||xj(n)| =
i=1 j=1
k k ∞
|bij (s − n)||xj (n)| =
i=1 j=1 s=n (Exercises 6.5, Problem 1).
k
|aij|+βij ≤1
i=1
|aij ||xj (n)| − |xi (n)| k ∞
j=1 s=n
k k |aji||xi(n)|,
i=1 j=1
k k ∞
|bij (s − n)||xi (n)|
i=1 j=1 s=n
|bij(s−n)||xj(n)| .
302 6. The Z-Transform Method and Volterra Difference Equations
Hence inequality (6.5.6) now becomes
This implies that
which proves uniform stability.
P
|x(n)| ≤ V (x) ≤
k i=1
|xi(0)| = ∥x(0)∥,
Example 6.25. An Epidemic Model [89]
⎡⎤ k k
⎣ |aji| + bji − 1⎦ |xi(n)| i=1 j=1
∆V(6.5.1)(x) ≤
≤ 0 (by condition (6.5.5)).
Let x(n) denote the fraction of susceptible individuals in a certain popula-
tion during the nth day of an epidemic, and let a(k) > 0 be the measure of how infectious the infected individuals are during the kth day. Then the spread of an epidemic may be modeled by the equation
1 n
(1+ε−x(n−j))a(j), where ε is a small positive number, n ∈ Z+.
Then we obtain
n j=0
Since x(n) ∈ [0,1], we have y(n) ≥ 0 for all solutions of (6.5.8). Observe
that during the early stages of the epidemic x(n) is close to 1 and, con-
(6.5.9)
(6.5.10) If a(n) has a simple form, one may be able to compute y(n). For example,
ln x(n+1) =
To transform (6.5.7) into a Volterra-type equation, we put x(n) = e ̄ .
y(n+1) =
y(j)
a(n−j)(1+ε−e ̄ ). (6.5.8)
j=0
(6.5.7)
y(n)
sequently y(n) is close to zero. Hence it is reasonable to linearize (6.5.8) y(j)
around zero. So if we replace e ̄ by 1 − y(j), (6.5.8) becomes
zy ̃(z) = a ̃(z) εz + a ̃(z)y ̃(z), z−1
y(n + 1) =
n j=0
a(n − j)(ε + y(j)), y(0) = 0. Taking the Z-transform of both sides of the equation yields
y ̃(z) = ifa(n)=can,thena ̃= c z .Hence
εz a ̃(z)
(z + 1)(z − a ̃(z))
z−a
εcz εc 1 a+c
y ̃(z)=(z−1)(z−(a+c))=1−a−c z−1−z−a−c .
Thus
Nowif0
rr
Assumethatνij = ∞n=0bij(n)<∞for1≤i,j≤k.Supposethat
the following two conditions hold: (i) aii +νii >1, 1≤i≤k.
(ii) (aii+νii−1)(ajj+νjj−1)>′r|air+νir|for1≤i, j≤k,i̸=j. Prove that:
(a) If k is odd, then the zero solution of (6.5.1) is not asymptotically stable.
(b) If k is even, then the zero solution of (6.5.1) may or may not be asymptotically stable.
*7. (Open problem). Discuss the stability of the zero solution of (6.5.1) under the condition aii + νii ≤ 1, 1 ≤ i ≤ k. Consider the Volterra system with infinite delay
stability of the zero solution of (6.5.7).
9. Mimic the proof of Theorem 6.24 to find criteria for the uniform stability of the zero solution of (6.5.7).
10. Prove Theorem 6.23 using the method of Liapunov functionals (as in Theorem 6.24).
6.6 A Variation of Constants Formula
Associated with the homogeneous system (6.5.1) we contemplate the following nonhomogeneous system:
n j=0
where g(n) ∈ Rk.
The existence and uniqueness of solutions of system (6.5.1) may be es-
tablished by a straightforward argument (Exercises 6.6, Problem 13). Let ei = (0,…,1,…,0)T be the standard ith unit vector in Rk, 1 ≤ i ≤ k. Then there are k vector solutions x1(n),x2(n),x3(n),…,xk(n) of system (6.5.1) with xi(n) = ei,1 ≤ i ≤ k. The solutions are linearly independent on Z+. For if there is a nontrivial linear relation c1x1(n) + c2x2(n) + · · · + ckxk(n) = 0 on Z+, then at n = 0 we have c1e1 +c2e2 +···+ckek = 0.
B(n − j)x(j). (6.5.14) 8. Mimic the proof of Theorem 6.23 to find criteria for the asymptotic
x(n + 1) = Ax(n) +
y(n + 1) = Ay(n) +
B(n − j)y(j) + g(n), (6.6.1)
∞ j=0
306 6. The Z-Transform Method and Volterra Difference Equations
This proves that c1 = c2 = ··· = ck = 0, which is a contradiction. The k × k matrix X(n) whose ith column is xi(n), is called the fundamental matrix of system (6.5.1). Notice that X(n) is a nonsingular matrix with X (0) = I . Moreover, x(n) = X (n)x0 is a solution of system equations (6.5.1) with x(0) = x0 (Exercises 6.6, Problem 1). Furthermore, the funda- mental matrix X(n) satisfies the matrix equation (Exercises 6.6, Problem 2)
n j=0
It should be pointed out that the fundamental matrix X(n) enjoys all the properties possessed by its counterpart in ordinary difference equations (Chapter 3).
Next we give the variation of constants formula.
Theorem 6.27. Suppose that the Z-transforms of B(n) and g(n) exist.
Then the solution y(n) of system (6.6.1) with y(n0) = y0 is given by (6.6.3)
Proof. We first observe that (Why?) n
X(n + 1) = AX(n) +
Taking the Z-transform of both sides of (6.6.4), we obtain, for some R > 0,
This yields
r=0
zX ̃(z) − zX(0) = AX ̃(z) + B ̃(z)X ̃(z), |z| > R.
[zI − A − B ̃(z)]X ̃(z) = zI, |z| > R. (6.6.5)
X(n + 1) = AX(n) +
B(n − j)X(j). (6.6.2)
y(n, 0, y0) = X(n)y0 +
n−1
r=0
X(n − r − 1)g(r).
Since the right-hand side of (6.6.5) is nonsingular, it follows that the matrix zI − A − B ̃(z) is also nonsingular. (Why?) This implies that
X ̃(z) = z[zI − A − B ̃(z)]−1, |z| > R. (6.6.6) In the next step of the proof we take the Z-transform of both sides of
system (6.6.1) to obtain
y ̃(z) = [zI − A − B ̃(z)]−1[zy0 + g ̃(z)], |z| > R1,
for some R1 ≥ R, and by using formula (6.6.6) this gives
y ̃(z) = X ̃(z)y0 + 1X ̃(z)g ̃(z), |z| > R1. 2
B(n − r)X(r). (6.6.4)
Thus
1
y ( n ) = Z − 1 [ X ̃ ( z ) y 0 ] + Z − 1 n−1
= X(n)y0 + (using formulas (6.1.4) and (6.1.8)).
Exercises 6.6
6.6 A Variation of Constants Formula 307
r=0
1. Let X(n) be the fundamental matrix of system equation (6.5.1). Prove that x(n) = X(n)x0 is a solution of (6.5.1) for any vector x0 ∈ Rk.
2. Prove that the fundamental matrix X(n) satisfies (6.6.2).
3. Prove that the zero solution of (6.5.1) is uniformly stable if and only
if |x(n,n0,x0)| ≤ M|x0| for some M > 0.
4. Prove that the zero solution of (6.5.1) is uniformly asymptotically sta-
ble if and only if there exist M > 0, ν ∈ (0,1) such that |x(n,n0,x0)| ≤ Mνn−n0.
5. Solve the equation
x(n+1) = −2 3x(n)+
2n−r 32(n−r+1)
√ n 1
x(r)+2n(3n/2), x(0) = 0 :
(n−r)x(r)+n:
X ̃ ( z ) g ̃ ( z ) X(n − r − 1)g(r)
2
P
r=0
(a) by the Z-transform method,
(b) by using Theorem 6.27.
6. Solvetheequationx(n+1)=1x(n)+n
2 r=0 (a) by the Z-transform method,
(b) by using Theorem 6.27.
7. Consider the planar system x(n + 1) = Ax(n) + nj=0 B(n − j)x(j) + g(n), x(0) = 0, where
√ −n/2 −2020
A= 0 −√6 , B(n)= 0 6−n/2 .
(a) Find the fundamental matrix X(n) of the homogeneous equation.
n
(b) Use Theorem 6.27 to solve the equation when g(n) =
8. Consider the system ∆x(n) = nj=0 B(n − j)x(j) + g(n), where
10 B(n)= 0 2n .
0
.
308
6. The Z-Transform Method and Volterra Difference Equations
(a) Solve the homogeneous part when g(n) = 0.
(b) Use Theorem 6.27 to find the solution of the nonhomogeneous
equation when g(n) =
a
0
, where a is a constant.
9. Consider the system
y(n + 1) = Ay(n) + g(n), y(0) = y0. (6.6.7)
Use the Z-transform to show that:
(a) An = Z−1[z(zI − A)−1].
(b) n−1 An−r−1g(r) = Z−1[(zI − A)−1g ̃(z)]. r=0
(c) Conclude that the solution of the given equation is given by y(n) = Z−1[z(zI − A)−1]y0 + Z−1[(zI − A)−1g ̃(z)].
10. Use (6.6.5) to show that for some R > 0, det(zI − A − B ̃(z)) ̸= 0 for |z| > R.
Apply the method of Problem 9 to solve equation (6.6.7) in Problem 9
if A and g(n) are given as follows:
⎛⎞
11. A= 3 −2 , g(n)=⎜⎝3⎟⎠, y(0)=0.
10
6.7 The Z-Transform Versus the Laplace Transform5
The Laplace transform plays the same role in differential equations as does the Z-transform in difference equations. For a continuous function f(t), the
Laplace transform is defined by ˆ
0.5 1 0 0.5
0 , g(n)=0.
n
12.A=
13. Prove the existence and uniqueness of the solutions of (6.6.1).
f(s) = L (f(t)) =
If we discretize this integral we get ∞
f(t) dt.
e−snf(n). If further we let
. Hence
5This section may be skipped by readers who are not familiar with the Laplace transform.
s n=0
z = e , we get the Z-transform of f(n), namely n=0 f(n)z
∞ 0
−st e
∞ −n
6.7 The Z-Transform Versus the Laplace Transform 309 given s = α + iβ in the s-plane (commonly called the frequency domain in
engineering), then
z=eα+iβ =eα ·eiβ =eα ·ei(β+2nπ), n∈Z.
Hence a point in the z-plane corresponds to infinitely many points in the s- plane. Observe that the left half of the s-plane corresponds to the interior of the unit disk |z| < 1 in the z-plane. Thus asymptotic stability of a differential equation is obtained if all the roots of its characteristic equation have negative real parts. In difference equations this corresponds to the condition that all the roots of the characteristic equation lie inside the unit disk.
There is another method that enables us to carry the stability analysis from the s-plane to the z-plane, i.e., from differential equations to difference equations. Suppose that the characteristic equation of a difference equation is given by
P(z)=a0zn +a1zn−1 +···+an =0. The bilinear transformation defined by
z=s+1 s−1
maps the interior of the unit disk to the left half-plane in the complex plane (Figure 6.7). To show this we let s = α + iβ. Then
α + i β + 1
|z| = α + iβ − 1 < 1,
Im z
Im s
z-plane
s-plane
Re z
Re s
FIGURE 6.7. A bilinear transformation.
310 6. The Z-Transform Method and Volterra Difference Equations or
(α+1)2 +β2 (α−1)2+β2 <1,
which gives α < 0.
Now substituting z = s+1 into P(z) we obtain
or
s−1
s + 1n s + 1n−1
a0 s−1 +a1 s−1 +···+an =0, Q(s)=b0sn +b1sn−1 +···+bn =0.
We now can apply the Routh stability criterion [102] to Q(s) to check whether all the zeros of Q(s) are in the left half-plane. If this is the case, then we know for sure that the zeros of P(z) all lie inside the unit disk. We are not going to pursue this approach here, since the computation involved is horrendous.
6.7 The Z-Transform Versus the Laplace Transform 311
TABLE 6.1. Z-transform pairs.
No.
x(n) for n = 0,1,2,3,... x(n) = 0 for n = −1,−2,−3,...
x ̃(z) = ∞n=0 x(n)z−n
1.
1
z/z − 1
2.
an
z/z − a
3.
an−1
1
z−a
4.
n
z/(z − 1)2
5.
n2
z(z + 1)/(z − 1)3
6.
n3
z(z2 + 4z + 1)/(z − 1)4
(−1)kDk z ;D=z d z−1 dz
7.
nk
8.
nan
az/(z − a)2
9.
n2 an
az(z + a)/(z − a)3
10.
n3 an
az(z2 + 4az + a2)/(z − a)4
(−1)kDk z ;D=z d z−a dz
11.
nk an
12.
sin nω
z sin ω/
(z2 −2z cos ω +1)
13.
cos nω
z(z− cos ω)/ (z2 −2z cos ω + 1)
14.
an sin nω
az sin nω/
(z2 −2az cos ω + a2)
15.
an cos nω
z(z−a cos ω)/
(z2 −2az cos ω + a2)
16.
δ0 (n)
1
17.
δm (n)
z−m
18.
an /n!
ea/z
19.
cosh nω
z(z− cosh ω)/ (z2 −2z cosh ω + 1)
20.
sinh nω
z sinh ω/
(z2 −2z cosh ω + 1)
21.
1, n>0 n
ln (z/z−1)
22.
e−ωnx(n)
x ̃ ( e ω z )
23.
n(2) = n(n − 1)
2z/(z − 1)3
24.
n(3) = n(n−1)(n−2)
3!z/(z − 1)4
25.
n(k) =n(n−1)···(n−k+1)
k!z/(z − 1)k+1
26.
x(n−k)
z − k x ̃ ( z )
zk x ̃(z) − k−1 x(r)zk−r r=0
27.
x(n+k)
7
Oscillation Theory
In previous chapters we were mainly interested in the asymptotic behav- ior of solutions of difference equations both scalar and nonscalar. In this chapter we will go beyond the question of stability and asymptoticity. Of particular interest is to know whether a solution x(n) oscillates around an equilibrium point x∗, regardless of its asymptotic behavior. Since we may assume without loss of generality that x∗ = 0, the question that we will address here is whether solutions oscillate around zero or whether solutions are eventually positive or eventually negative.
Sections 7.1 and 7.3 follow closely the paper of Erbe and Zhang [53] and the book of Gyori and Ladas [63]. In Section 7.2 we follow the approach in the paper of Hooker and Patula [67]. For more advanced treatment of oscillation theory the reader is referred to [3], [63], [64], [79].
7.1 Three-Term Difference Equations
In this section we consider the three-term difference equation (of order k + 1)
x(n+1)−x(n)+p(n)x(n−k)=0, n∈Z+, (7.1.1)
where k is a positive integer and p(n) is a sequence defined for n ∈ Z+.
A nontrivial solution x(n) is said to be oscillatory (around zero) if for every positive integer N there exists n ≥ N such that x(n)x(n + 1) ≤ 0. Otherwise, the solution is said to be nonoscillatory. In other words, a solution x(n) is oscillatory if it is neither eventually positive nor eventually
313
314 7. Oscillation Theory
negative. The solution x(n) is said to be oscillatory around an equilibrium point x∗ if x(n) − x∗ is oscillatory around zero. The special case, where k = 1 and p(n) = p is a constant real number, has been treated previously in Section 2.5. In this case (7.1.1) may be written in the more convenient form
x(n + 2) − x(n + 1) + px(n) = 0. (7.1.2) The characteristic roots of (7.1.2) are given by
λ1,2 = 1 ± 11−4p. 22
Recall from Section 2.5 that all solutions of (7.1.2) oscillate if and only if
λ1 and λ2 are not positive real numbers. Hence every solution of (7.1.2)
oscillates if and only if p > 1 . 4
Let us now turn our attention back to (7.1.1). This equation is the discrete analogue of the delay differential equation
x′(t) + p(t)x(t − k) = 0. (7.1.3)
The oscillatory behavior of (7.1.3) is remarkably similar to that of its dis- crete analogue (7.1.1), with one exception, when k = 0. In this case, the equation
has the solution
x′(t) + p(t)x(t) = 0 t
x(t)=x(t0) exp −
which is never oscillatory. However, the discrete analogue
x(n + 1) = (1 − p(n))x(n) n−1
has the solution x(n) = j=n0 (1 − p(j)) x(n0), which oscillates if 1 −
p(j) < 0 for all j ≥ n0.
To prepare for the study of the oscillatory behavior of (7.1.1) we first
investigate the solutions of the following associated difference inequalities: x(n + 1) − x(n) + p(n)x(n − k) ≤ 0, (7.1.4)
x(n + 1) − x(n) + p(n)x(n − k) ≥ 0. (7.1.5) In the sequel we make use of the notions of the limit superior and the
limit inferior of a sequence {a(n)}, denoted by lim sup a(n) and lim inf a(n),
respectively.
Definition 7.1. Let {a(n)} be a sequence of real numbers. Let β(n) be the least upper bound of the set {a(n), a(n + 1), a(n + 2), . . .}. Then either
t0
p(s)ds ,
n→∞
n→∞
7.1 Three-Term Difference Equations 315
β(n) = ±∞ for every n, or the sequence {β(n)} is a monotonically decreas- ing sequence of real numbers, and thus limn→∞ β(n). Similarly, let α(n) be the greatest lower bound of the set {a(n), a(n + 1), a(n + 2), . . .}. Then:
(i) lim sup a(n) = limn→∞ β(n). n→∞
(ii) lim inf a(n) = limn→∞ α(n). n→∞
Note that limn→∞ a(n) exists if and only if lim sup a(n) = lim inf a(n) = n→∞ n→∞
lim a(n). n→∞
Example 7.2. Find the limit superior and the limit inferior for the following sequences:
S1 :0,1,0,1,....
S2 :1,−2,3,−4,...,(−1)n+1n,....
S3 :3,−1, 4,−1, 5,−1, 6,−1,.... 22334455
Solution
Theorem 7.3 [53]. Suppose that
limsupS1 = 1, n→∞
limsupS2 = ∞, n→∞
limsupS3 = 1, n→∞
liminf S2 = 0, n→∞
liminf S2 = −∞, n→∞
liminf S3 = 0. n→∞
kk
liminf p(n) = p > k+1 . n→∞ (k + 1)
Then the following statements hold:
(i) Inequality (7.1.4) has no eventually positive solution.
(ii) Inequality (7.1.5) has no eventually negative solution.
Proof. (i) To prove statement (i), assume the contrary, that is, there exists a solution x(n) of inequality (7.1.4) that is eventually positive. Hence there exists a positive integer N1 such that x(n) > 0 for all n ≥ N1. Dividing inequality (7.1.4) by x(n), we get, for n ≥ N1,
x(n + 1) ≤ 1 − p(n)x(n − k). (7.1.7) x(n) x(n)
(7.1.6)
316 7. Oscillation Theory
If we let z(n) = x(n) , then x(n+1)
x(n−k) = x(n)
x(n−k) x(n−k+1),…,x(n−1) x(n − k + 1) x(n − k + 2) x(n)
= z(n − k)z(n − k + 1), . . . , z(n − 1). Substituting into inequality (7.1.7) yields
1 ≤1−p(n)z(n−k)z(n−k+1)···z(n−1), n≥N1 +k. (7.1.8) z(n)
Now, condition (7.1.6) implies that there exists a positive integer N2 such
that p(n) > 0 for all n ≥ N2. Put N = max{N2,N1 +k}. Then for
n ≥ N, x(n + 1) − x(n) ≤ −p(n)x(n − k) ≤ 0. Consequently, x(n) is non-
increasing, and thus z(n) ≥ 1. Let lim inf z(n) = q. Then from inequality n→∞
(7.1.8) we have
limsup 1 = 1 =1/q
n→∞ z(n)
which yields
lim infz(n) n→∞
≤ 1 − lim inf[p(n)z(n − k)z(n − k − 1), . . . , z(n − 1)], n→∞
1 ≤ 1 − pqk, q
p≤q−1. (7.1.9) qk+1
or
Let h(q) = (q − 1)/qk+1. Then h(q) attains its maximum at q = (k + 1)/k. Hence maxq≥1 h(q) = (kk)/(k + 1)k+1. Hence from inequality (7.1.9) we have p ≤ (kk)/(k+1)k+1, a contradiction. This completes the proof of part (i) of the theorem. The proof of part (ii) is left to the reader as Exercises 7.1, Problem 6. P
Corollary 7.4. If condition (7.1.6) holds, then every solution of (7.1.1) oscillates.
Proof. Suppose the contrary and let x(n) be an eventually positive solu- tion of (7.1.1). Then inequality (7.1.4) has an eventually positive solution, which contradicts Theorem 7.3. On the other hand, if (7.1.1) has an eventu- ally negative solution, then so does inequality (7.1.5), which again violates Theorem 7.3. P
The above corollary is sharp, as may be evidenced by the following example, where we let
kk
p(n) = (k + 1)k+1 .
Example 7.5. Consider the difference equation
x(n + 1) − x(n) + (kk/(k + 1)k+1)x(n − k) = 0.
Then x(n) =
, n > 1, is a nonoscillatory solution of the equation.
(7.1.10)
k
n−1
7.1 Three-Term Difference Equations 317
k+1
Next we give a partial converse of Corollary 7.4.
Theorem 7.6 [53]. Suppose that p(n) ≥ 0 and kk
sup p(n) < k + 1k+1 . Then (7.1.1) has a nonoscillatory solution.
Proof. As in the proof of Theorem 7.3, we let z(n) = x(n)/x(n + 1) in (7.1.1) to obtain
1/z(n) = 1 − p(n)z(n − k)z(n − k + 1) · · · z(n − 1). (7.1.11) To complete the proof, it suffices to show that (7.1.11) has a positive
solution. To construct such a solution we define z(1−k)=z(2−k)=···=z(0)=a= k+1 >1 (7.1.12)
k
and
z(1) = [1 − p(1)z(1 − k)z(2 − k) · · · z(0)]−1. (7.1.13) Then z(1) > 1 also. We claim that z(1) < a. To show this, we have
z(1) = 1
a a[1−p(1)z(1−k)···z(0)]
≤ k =1. (k+1) 1− kk · k+1 k
(k+1)k+1 k
Hence by induction, we may show that 1 < z(n) < a, with n = 1, 2, 3,.... Moreover, z(n) is a solution of (7.1.11). Now let x(1) = 1, x(2) = x(1)/z(1), x(3) = x(2)/z(2), and so on. Then x(n) is a nonoscillatory solution of (7.1.1). P
For the special case where p(n) = p is a constant real number we have the following stronger result.
Theorem 7.7. Consider the equation
x(n + 1) − x(n) + px(n − k) = 0, (7.1.14)
where k is a positive integer and p is a nonnegative real number. Then every solution of (7.1.14) oscillates if and only if p > kk/(k + 1)k+1.
Proof. Combining the results of Corollary 7.4, Example 7.5, and Theorem 7.6 yields the proof. P
318 7. Oscillation Theory
Remark: Gyori and Ladas [63] showed that every solution of the kth-order equation
x(n + k) + p1x(n + k − 1) + · · · + pkx(n) = 0 (7.1.15)
oscillates if and only if its characteristic equation has no positive roots. Based on this theorem (Exercises 7.1, Problem 8), they were able to show that every solution of the three-term equation (7.1.14), where k ∈ Z – {−1, 0}, oscillates if and only if p > kk/(k + 1)k+1.
Exercises 7.1
1. Find the limit superior and limit inferior of the following sequences: (a) S1: 2,1,3,1,4,1,5,1,….
33445566 (b) S2 : (−1)n+1.
(c) S3 : αn/(1 + βn). (d) S4 : 1 + (−1)n+1.
2. Prove the following statements:
(a) limsup(1/a(n))=1/liminfa(n).
n→∞
(b) If a(n) > 0, then lim sup(−a(n)) = − lim inf a(n).
n→∞
n→∞ (c) liminfa(n)≤limsupa(n).
n→∞
n→∞
3. Show that the difference equation
n→∞
∆2x(n−1)+ 1x(n)=0, n≥1, n
is oscillatory on [0, ∞).
4. (a)
Show that every solution of the equation
x(n + 1) − x(n) + px(n) = 0
oscillates if and only if p > 1, where p ∈ R. (b) Show that every solution of the equation
x(n + 1) − x(n) + px(n − 1) = 0 oscillates if and only if p > 1 .
4
5. Consider the difference equation
∆2x(n) + p(n)x(n + 1) = 0,
where p(n) > a > 0 for n ∈ Z+. Show that every nontrivial solution of the equation is oscillatory.
6. Prove part (ii) of Theorem 7.3.
*9. [53] Assume that
and
lim inf p(n) = q > 0, n→∞
lim sup p(n) > 1 − q. n→∞
x(n + 1) − x(n) + where kj are positive integers.
10. Suppose that pi(n) ≥ 0 and m
m j=1
pj(n)x(n − kj) = 0,
> 1.
(7.1.16)
7.1 Three-Term Difference Equations 319
7. The characteristic equation of (7.1.14) is given by λk+1 −λk +p=0, where p≥0.
Show that the characteristic equation has no positive roots if and only if p > kk/(k + 1)k+1. Then give a proof of Theorem 7.7.
8. Show that every solution of (7.1.15) oscillates if and only if its characteristic equation has no positive real roots.
Prove that all conclusions of Theorem 7.3 hold.
In Problems 10 through 12 consider the equation with several delays
(ki +1)ki+1 lim inf pi(n) ki
i=1 n→∞ (ki) Show that every solution of (7.1.16) oscillates.
11. Suppose that pi(n) ≥ 0 and ⎛⎞
(7.1.16) oscillates. *12. Suppose that
and
m
n→∞
i=1
m
m lim inf ⎝
(k ̄)k ̄
̄ ,
pi(n)⎠ >
where k ̄ = min{k1,k2,…,km} ≥ 1. Show that every solution of
n→∞ j=1
(k ̄ + 1)k+1
pi(n) = c > 0
liminf
pi(n) = 1 − c. Prove that every solution of (7.1.16) oscillates.
lim sup
n→∞ i=1
320 7. Oscillation Theory
7.2 Self-Adjoint Second-Order Equations
In this section we consider second-order difference equations of the form
∆[p(n − 1)∆x(n − 1)] + q(n)x(n) = 0, (7.2.1)
where p(n) > 0,n ∈ Z+. Equation (7.2.1) is called self-adjoint, a name borrowed from its continuous analogue
[p(t)x′(t)]′ + q(t)x(t) = 0. Equation (7.2.1) may be written in the more familiar form
p(n)x(n + 1) + p(n − 1)x(n − 1) = b(n)x(n),
where
b(n) = p(n − 1) + p(n) − q(n). As a matter of fact, any equation of the form
p0(n)x(n + 1) + p1(n)x(n) + p2(n)x(n − 1) = 0,
(7.2.2)
(7.2.3)
(7.2.4)
with p0(n) > 0, and p2(n) > 0, can be written in the self-adjoint form (7.2.1) or (7.2.2). To find p(n) and q(n) from p0(n),p1(n), and p2(n), multiply both sides of (7.2.4) by a positive sequence h(n). This yields
p0(n)h(n)x(n + 1) + p1(n)h(n)x(n) + p2(n)h(n)x(n − 1) = 0. Comparing (7.2.5) with (7.2.2), we obtain
(7.2.5)
Thus or
Hence
p(n) = p0(n)h(n), p(n − 1) = p2(n)h(n).
p2(n + 1)h(n + 1) = p0(n)h(n), h(n + 1) = p0(n) h(n).
p2(n + 1)
(7.2.6)
h(n) =
is a solution of (7.2.6). This gives us
p(n) = p0(n)
Also from (7.2.3) we obtain
q(n) = p1(n)h(n) + p(n) + p(n − 1).
n−1
p0(j) p2(j + 1)
j =n0
n−1
p0(j) p2(j + 1).
j =n0
or
p(r − 1)∆2x(r − 1) + ∆x(r)∆p(r − 1) = 0, p(r)x(r + 1) = −p(r − 1)x(r − 1).
7.2 Self-Adjoint Second-Order Equations 321
In [64] Hartman introduced the notion of generalized zeros in order to obtain a discrete analogue of Sturm’s separation theorem in differential equations. Next we give Hartman’s definition.
Definition 7.8. A solution x(n), n ≥ n0 ≥ 0, of (7.2.1) has a generalized zero at r > n0 if either x(r) = 0 or x(r−1)x(r) < 0. In other words, a generalized zero of a solution is either an actual zero or where the solution changes its sign.
Theorem 7.9 (Sturm Separation Theorem). Let x1(n) and x2(n) be two linearly independent solutions of (7.2.1). Then the following statements hold:
(i) x1(n) and x2(n) cannot have a common zero, that is, if x1(r) = 0, then x2(r) ̸= 0.
(ii) If x1(n) has a zero at n1 and a generalized zero at n2 > n1, then x2(n) must have a generalized zero in (n1,n2].
(iii) If x1(n) has generalized zeros at n1 and n2 > n1, then x2(n) must have a generalized zero in [n1, n2].
Proof.
(i) Assume that x1(r) = x2(r) = 0. Then the Casoratian
x2(r) = 0.
(ii) Assumethatx1(n1)=0,x1(n2−1)x(n2)<0(orx1(n2)=0).Wemay assume that n2 is the first generalized zero greater than n1. Suppose thatx1(n)>0forn1
W(r) = x1(r)
x2(r + 1)
x1(r + 1)
It follows from Corollary 2.14 that x1(n) and x2(n) are linearly
dependent, a contradiction.
∆[p(r − 1)∆x(r − 1)] + q(r)x(r) = 0. Since x(r) = 0, we have
(7.2.7)
322 7. Oscillation Theory
Since x(r + 1) ̸= 0, x(r − 1) ̸= 0, and p(n) > 0, equation (7.2.7) implies that x(r − 1)x(r + 1) < 0, which is a contradiction. This completes the proof of part (ii). The proof of part (iii) is left to the reader as Exercises 7.2, Problem 6. P
Remark: Based on the notion of generalized zeros, we can give an alterna- tive definition of oscillation. A solution of a difference equation is oscillatory on [n2,∞) if it has infinitely many generalized zeros on [n0,∞). An im- mediate consequence of the Sturm separation theorem (Theorem 7.9) is that if (7.2.1) has an oscillatory solution, then all its solutions are oscil- latory. We caution the reader that the above conclusion does not hold in general for non-self-adjoint second-order difference equations. For example, the difference equation x(n + 1) − x(n − 1) = 0 has a nonoscillatory so- lution x1(n) = 1 and an oscillatory solution x2(n) = (−1)n. Observe that this equation is not self-adjoint. We are now ready to give some simple criteria for oscillation.
Lemma 7.10. If there exists a subsequence b(nk) ≤ 0, with nk → ∞ as k → ∞, then every solution of (7.2.2) oscillates.
Proof. Assume the contrary, that there exists a nonoscillatory solution x(n) of (7.2.2). Without loss of generality, suppose that x(n) > 0 for n ≥ N . Then
p(nk)x(nk + 1) + p(nk − 1)x(nk − 1) − b(nk)x(nk) > 0, for nk > N, which is a contradiction.
P
One of the most useful techniques in oscillation theory is the use of the so- called Riccati transformations. We will introduce only one transformation that is needed in the development of our results. Two more transformations will appear in the exercises. In (7.2.2) let
Then z(n) satisfies the equation c(n)z(n) +
where
1 = 1, z(n−1)
z(n) = b(n + 1)x(n + 1) . p(n)x(n)
(7.2.8)
(7.2.9)
(7.2.10)
c(n) =
Next we give a crucial result that relates (7.2.2) with (7.2.9).
p2(n) b(n)b(n + 1)
.
Lemma 7.11. Suppose that b(n) > 0 for n ∈ Z+. Then every solution x(n) of (7.2.2) is nonoscillatory if and only if every solution z(n) of (7.2.9) is positive for n≥N, for some N >0.
7.2 Self-Adjoint Second-Order Equations 323
Proof. Suppose that x(n) is a nonoscillatory solution of (7.2.2). Then x(n)x(n + 1) > 0 for n ≥ N . Equation (7.2.8) then implies that z(n) > 0. Conversely, assume that z(n) is a positive solution of (7.2.9). Using this solution we construct inductively a nonoscillatory solution x(n) of (7.2.2) as follows: Let x(N ) = 1, x(n + 1) = (p(n)/b(n + 1))z(n)x(n), with n > N . Then one may verify that x(n), with n ≥ N, is indeed a solution of (7.2.2) that is nonoscillatory. By the Sturm separation theorem, every solution of (7.2.2) is thus nonoscillatory. We need a comparison result concerning (7.2.9) that will be needed to establish the main result of this section. P
Lemma7.12. Ifc(n)≥a(n)>0foralln>0andz(n)>0isasolution of the equation
c(n)z(n) + 1 z(n−1)
a(n)y(n) + 1 y(n−1)
Proof. Since c(n) > 0 and z(n) > 0, it follows from (7.2.1) that 1/(z(n− 1)) < 1. This implies that z(n−1) > 1 for all n ≥ 1. We now define inductively a solution y(n) of (7.2.12). Choose y(0) ≥ z(0) and let y(n) satisfy (7.2.11). Now, from (7.2.12) and (7.2.12), we have
a(n)y(n) + 1 = c(n)z(n) + 1 y(n−1) z(n−1)
a(1)y(1) + 1 = c(1)z(1) + 1 . y(0) z(0)
.
= 1,
(7.2.11)
(7.2.12)
then the equation
= 1 has a solution y(n)≥z(n)>1 for all n∈Z+.
So
Since y(0) ≥ z(0), we have 1/y(0) ≤ 1/z(0), and hence a(1)y(1) ≥ c(1)z(1), or
y(1)≥ c(1)z(1)≥z(1)>1. a(1)
Inductively, one may show that
y(n) ≥ z(n) > 1. P
Theorem 7.13. If b(n)b(n + 1) ≤ (4 − ε)p2(n) for some ε > 0 and for all n ≥ N, then every solution of (7.2.2) is oscillatory.
Proof. If b(n)b(n−1) ≤ (4−ε)p2(n) for some ε ≥ 4, then b(n)b(n−1) ≤ 0. The conclusion of the theorem then follows from Lemma 7.10. Hence we may assume that 0 < ε < 4. Now assume that (7.2.2) has a nonoscillatory
324 7. Oscillation Theory
solution. Then by Lemma 7.11, (7.2.9) has a positive solution z(n) for n ≥ N. Using the assumption of the theorem in formula (7.2.10) yields
c(n) = p2(n) ≥ p2(n) = 1 . b(n)b(n+1) (4−ε)p2(n) 4−ε
Then it follows from Lemma 7.12 that the equation
1 y(n) + 1 4−ε y(n−1)
= 1
(7.2.13)
has a solution y(n),n ≥ N, such that y(n) ≥ z(n) > 1 for all n ≥ N. Define a positive sequence x(n) inductively as follows: x(N) = 1,x(n+1) = (1/√4 − ε)y(n)x(n) for n ≥ N. Then
√ x(n+1) y(n) = 4 − ε x(n)
. (7.2.14) Substituting for y(n) in (7.2.14) into (7.2.13) yields x(n+1)−√4 − εx(n)+
x(n − 1) = 0, n ≥ N , whose characteristic roots are √4−ε √ε
λ1,2 = 2 ±i 2 .
Thus its solutions are oscillatory, which gives a contradiction. The proof of
the theorem is now complete.
It is now time to give some examples.
Example 7.14. Consider the difference equation 1
P
y(n+1)+y(n−1) = 2+ 2(−1)n Here p(n) = 1 and b(n) = 2 + 1 (−1)n:
y(n). 2
b(n)b(n+1)= 2+1(−1)n 2+1(−1)n+1 22
= 33. 4
Thus b(n) b(n + 1) ≤ 4 − 1 p2(n). By Theorem 7.13, we conclude that 5
every solution is oscillatory.
The following example will show the sharpness of Theorem 7.13 in the sense that if ε is allowed to be a sequence tending to zero, then the theorem fails.
Example 7.15 [67]. Consider the equation x(n+1)+x(n)=b(n)x(n−1), n=1,2,3,…,
where
√n+1+√n−1 b(n) = √n .
7.2 Self-Adjoint Second-Order Equations 325
Now,
(n+1)(n+2)+ (n−1)(n+2)+ n(n+1)+ n(n−1) =.
n(n+1)
But limn→∞ b(n)b(n + 1) = 4. Hence if one takes εn = 4 − b(n)b(n + 1), then εn → 0 as n → ∞. However, Theorem 7.13 fails, since x(n) = √n, n ≥ 1, is a nonoscillatory solution of the equation.
A partial converse of Theorem 7.13 now follows.
Theorem 7.16. If b(n)b(n + 1) ≥ 4p2(n) for n ≥ N, then every solution
of (7.2.2) is nonoscillatory.
Proof. From formula (7.2.10) and the assumption we obtain c(n) ≤ 1 .
4
We now construct inductively a solution z(n) of (7.2.9) as follows: Put z(N) = 2, and
b(n)b(n + 1)
11 z(n)=c(n) 1−z(n−1) , n>N.
Observe that
z(N + 1) = 1
1 z(N)
1 − 1 2
≥ 4
we conclude that every solution of (7.2.2) is nonoscillatory. P
Example 7.17. Consider the difference equation ∆(n∆x(n−1)− 1x(n)=0.
1 −
Similarly, one may show that z(n) ≥ 2 for n ≥ N. Hence by Lemma 7.11,
c(N + 1)
= 2.
n
Here p(n) = n + 1 and q(n) = 1 . Using formula (7.2.3) we obtain
Now,
n
b(n)=2n+1+ 1. n
11 b(n)b(n+1)= 2n+1+n 2n+3+n+1
=4n2 +8n+7+ 2n+4 n(n+1)
≥ 4p2(n) for all n ≥ 1.
Hence by Theorem 7.16, every solution is nonoscillatory.
Exercises 7.2
In Problems 1 through 5 determine the oscillatory behavior of all solutions.
1. ∆[(n−1)x(n−1)]+ 1x(n)=0. n
326 7. Oscillation Theory
2. x(n+1)+x(n−1)=2− 1x(n). n
3. x(n+1)+x(n−1)=2+ 1x(n). n
4. ∆2[x(n−1)]+ 1 x(n)=0, n>1. n ln(n)
5. ∆[(n−1)x(n−1)]+x(n)=0.
6. Prove part (iii) of Theorem 7.9.
7. [112] Show that if b(n) ≤ min{p(n), p(n − 1)} for n ≥ N , for some positive integer N, then every solution of (7.2.2) is oscillatory.
8. Show that if b(n) ≤ p(n) and p(n) is eventually nonincreasing, then every solution of (7.2.2) is oscillatory.
9. Show that if b(n) ≤ p(n − 1) and p(n) is eventually nondecreasing, then every solution of (7.2.2) is oscillatory.
10. (A second Riccati transformation). Let z(n) = x(n+1)/x(n) in (7.2.2).
(i) Show that z(n) satisfies the equation
p(n)z(n) + p(n − 1) = b(n). (7.2.15) z(n−1)
(ii) Assuming p(n) > 0, show that every solution of (7.2.2) is nonoscil- latory if and only if (7.2.15) has a positive solution z(n),n ≥ N, for some N > 0.
*11. Use the second Riccati transformation in Problem 10 to show that if b(n) ≤ p(n − 1) and lim sup(p(n))/p(n − 1) > 1 , then every solution
n→∞ 2
12. [67] Show that if b(n) ≥ max{p(n − 1), 4p(n)}, for all n ≥ N , for some
of (7.2.2) oscillates.
N > 0, then every solution of (7.2.2) is nonoscillatory.
13. Show that if p(nk) ≥ b(nk)b(nk + 1) for a sequence nk → ∞, then every solution of (7.2.2) is oscillatory.
14. As in formula (7.2.10), let
c(n) = p2(n) , n ≥ 0.
b(n)b(n + 1)
Show that either one of the following implies that every solution of
(7.2.2) oscillates:
(i) limsupc(n)>1.
n→∞
(ii) lim sup 1 n c(j) > 1. n→∞ n j=1
15.
16.
Show that if p(n) is bounded above on [a, ∞) and (7.2.1) is nonoscilla-
7.3
tory on [a, ∞), then either to −∞.
∞ n=a+1
q(n) exists and is finite or it is equal
(i) p(n) is bounded on [a, ∞) and (ii) p(n) is bounded on [a, ∞) and
n −∞ ≤ liminf
n→∞ s=a+1
Nonlinear Difference Equations
7.3 Nonlinear Difference Equations 327
Use Problem 15 to prove that (7.2.1) is oscillatory if either one of the following conditions hold:
∞ n=a+1
q(s) ≤ limsup
In this section we will investigate the oscillatory behavior of the nonlinear difference equation
x(n + 1) − x(n) + p(n)f(x(n − k)) = 0, (7.3.1) where k ∈ Z+ and N ∈ Z+. The first theorem is due to Erbe and Zhang
[53].
Theorem 7.18. Suppose that f is continuous on R and satisfies the following assumptions:
(i) xf(x)>0, x̸=0,
(ii) liminf f(x) =L, 0
(k+1)k+1
0.
Then every solution of (7.3.1) oscillates.
n→∞
Proof. Assume the contrary and let x(n) be a nonoscillatory solution of (7.3.1). Suppose that x(n) > 0 for n ≥ N. This implies by assumption (i) that f(x(n)) > 0. Hence x(n+1)−x(n) = −p(n)f(x(n−k)) < 0, and thus x(n) is decreasing. Hence limn→∞ x(n) = c ≥ 0.
Taking the limit of both sides of (7.3.1) yields f(c) = 0, which by as- sumption (i) gives c = 0. Hence limn→∞ x(n) = 0. Dividing (7.3.1) by x(n) and letting z(n) = x(n)/x(n + 1) ≥ 1 yields
1 =1−p(n)z(n−1)···z(n−k)f(x(n−k)). (7.3.2) z(n) x(n − k)
328 7. Oscillation Theory
Let lim inf z(n) = r. By taking the limit superior in (7.3.2) we obtain n→∞
or
1 ≤ 1 − pLrk, r
pL≤ r−1. rk+1
(7.3.3)
It is easy to see that the function h(r) = (r − 1)/rk+1 attains its maximum at r = (k + 1)/k, and its maximum value is thus kk/(k + 1)k+1. Hence inequality (7.3.3) becomes
which contradicts assumption (iii). P Remark: If we let lim inf f (x)/x = 1, then the linearized equation associated
n→∞
with (7.3.1), where p(n) is equal to a constant real number p, is given by
y(n + 1) − y(n) + py(n − k) = 0, (7.3.4)
which has been studied in Section 7.1. We may now rephrase Theorem 7.18 as follows: Suppose that assumptions (i) and (ii) hold with L = 1 and that p(n) is constant. If every solution of (7.3.4) oscillates, then so does every solution of (7.3.1). Gyori and Ladas [63] considered the more general equation with several delays
with 1 ≤ i ≤ m. They obtained the following result. Theorem 7.19. Suppose that the following hold:
(i) pi >0,ki ∈Z+,andmi=1(pi+ki)̸=1,1≤i≤m,
(ii) f is continuous on R, and xfi(x)>0, for x̸=0, 1≤i≤m,
(iii) liminf fi(x) ≥1, 1≤i≤m, x→0 x
(iv) m p (ki+1)ki+1 > 1. i=1 i kki
i
Then every solution of (7.3.5) oscillates.
To facilitate the proof of this theorem we present the following lemma.
Lemma 7.20 [63]. Suppose that condition (i) in Theorem 7.19 holds and let {qi(n) : 1 ≤ i ≤ m} be a set of sequences of real numbers such that
liminf qi(n) ≥ pi, 1 ≤ i ≤ m. (7.3.6) n→∞
kk pL≤ (k+1)k+1,
m i=1
x(n + 1) − x(n) +
where pi > 0, ki is a positive integer, and fi is a continuous function on R,
pifi(x(n − ki)) = 0, (7.3.5)
If the linear difference inequality
y(n+1)−y(n)+ also has an eventually positive solution.
m i=1
qi(n)x(n − ki) ≤ 0, n ∈ Z+, (7.3.7)
7.3 Nonlinear Difference Equations 329
x(n + 1) − x(n) +
has an eventually positive solution x(n), then the corresponding limiting
equation
m i=1
piy(n−ki) = 0
(7.3.8)
Proof. There are two distinct cases to consider
Case (a): Assume that ki = 0, 1 ≤ i ≤ m. Then (7.3.7) and (7.3.8)
simplify to
m
1 − qi(n) x(n),
i=1 m
1 −
follows that, for sufficiently large n, m
qi(n) < 1. (7.3.11) i=1
Now, from assumption (7.3.6) for any ε > 0 there exists N > 0 such that
x(n + 1) ≤ y(n + 1) =
(7.3.9) (7.3.10)
pi y(n).
Let x(n) be an eventually positive solution of (7.3.9). Then from (7.3.9) it
This implies that 0<
i=1
pi ≤ Since ε was arbitrarily chosen,
qi(n)+ε<1+ε forn≥N.
m
m
m
0
⎛⎞
u − 1 +
piu−ki ≤ 0. (7.3.14)
m ki−1 ⎝1⎠
u(n + 1) ≤ 1 − (1 − ε) pi u(n − j) for n ≥ Nε. i=1 j=1
Choose N ̃ε such that N ̃ε ≥ Nε + k and u(n)≤(1−ε)u forn≥N ̃ε−k.
Then, for n ≥ N ̃ε − k,
u(n + 1) ≤ 1 − (1 − ε)
Consequently,
m
u ≤ 1 − (1 − ε)
Since ε was arbitrarily chosen, it follows that
m i=1
u ≤ 1 −
piu−ki ,
i=1
which proves claim (7.3.14).
To complete the proof of the lemma we consider the characteristic
polynomial of (7.3.8), p(λ) = λ − 1 + mi=1 piλ−ki .
Observe that p(0+) = ∞ and p(u) ≤ 0. This implies by the intermediate
value theorem that p(λ) has a positive root and, consequently, (7.3.8) has a positive solution. P
m i=1
piu−ki (1 + ε)−ki .
piu−ki (1 + ε)−ki .
7.3 Nonlinear Difference Equations 331
Proof of Theorem 7.19. Assume that (7.3.5) has a nonoscillatory
solution x(n). Without loss of generality, assume that x(n) is eventu-
ally positive. Then it follows from assumption (iii) that x(n + 1) ≤
x(n) − mi=1 pix(n − ki). As in the proof of Theorem 7.18, one may show
that lim x(n) = 0. We now need to put (7.3.5) in the form of (7.3.7). This n→∞
can be accomplished by setting
qi(n) = pi f(x(n − ki)).
x(n − ki) Thus from assumption (iii) we obtain
lim inf qi(n) ≥ pi. n→∞
By Lemma 7.20, it follows that the limiting equation (7.3.8) also has an eventually positive solution, which is a contradiction. This completes the proof of the theorem. P
In fact, under additional conditions it was proved ([63, , Corollary 7.4.1]) that every solution of the nonlinear equation oscillates if and only if every solution of the corresponding linearized equation oscillates.
We now apply the obtained results to study the oscillatory behavior of the Pielou logistic delay equation. The stability of this equation has been determined previously in Example 4.37.
Example 7.21 [63]. Consider the Pielou logistic delay equation
equilibrium point y∗ = (α − 1)/β if
α−1 kk
y(n+1) = αy(n) 1+βy(n−k)
, α > 1,β > 0, k a positive integer. (7.3.15) Show that every positive solution of (7.3.15) oscillates about its positive
α > (k + 1)k+1 . (7.3.16) Solution We follow Method 2 in Example 4.37 by letting y(n) = ((α −
1)/β)ex(n) in (7.3.15). We obtain the equation x(n+1)−x(n)+ α−1f(x(n−k))=0,
(7.3.17)
where
α
α (α−1)ex +1
f(x) = α − 1 ln α .
It may be shown that the function f satisfies conditions (i) and (ii) in Theorem 7.18 with L = 1. Hence by Theorem 7.18 every solution of (7.3.17) oscillates about 0. This implies that every solution of (7.3.15) oscillates about the equilibrium point y∗ = (α − 1)/β.
332 7. Oscillation Theory
Exercises 7.3
1. Consider the difference equation
∆x(n) + ex(n−1) − 1 = 0.
Determine the oscillatory behavior of all solutions.
2. Consider the difference equation
x(n)
x(n+1)=x(n)exp r 1− α , r>0, α>0, x(0)>0.
(a) Show that x∗ = α is the only positive equilibrium point. (b) Show that every solution oscillates about α if r > 1.
(c) Show that if r = 1, every solution converges monotonically to α.
3. Consider the difference equation
x(n−1)
x(n+1)=x(n)exp r 1− α , r>0, α>0, x(0)>0.
Show that every solution oscillates about x∗ = α if r > 1 . 4
4. Consider the difference equation
x(n−1) x(n−2)
x(n+1)=x(n)exp r 1− α − β , r>0, α>0, β>0, x(0)>0.
Show that every solution oscillates about x∗ = (αβ)/(α + β) if r > 4(α + β)/(27α + 16β).
5. Consider the difference equation ∆x(n)+p(1+x(n))x(n)=0, p>0, 1+x(n)>0.
Show that every solution oscillates if p > 1.
6. Consider the difference equation
∆x(n)+p(1+x(n))x(n−1)=0, p>0, 1+x(n)>0. Show that every solution oscillates if p > 1 .
4
7. [63] Consider the difference equation ∆x(n)+p(n)[1+x(n)]x(n−k)=0, p(n)>0,
for n ≥ 1, x(n)+1 > 0 for n ≥ −k. Prove that every solution oscillates
if lim inf p(n) = c > kk/((k + 1)k+1). n→∞
7.3 Nonlinear Difference Equations 333
8. Consider the difference equation
x(n + 1) = αx(n)
withα>1, β>0, γ>0, k∈Z+.Findconditionsunderwhichall solutions oscillate.
1+βx(n−k)+γx(n−1)
8
Asymptotic Behavior of Difference Equations
In Chapters 4 and 5 we were mainly interested in stability questions. In other words, we wanted to know whether solutions of a difference equa- tion converge to zero or to an equilibrium point. In asymptotic theory, we are concerned rather with obtaining asymptotic formulas for the manner in which solutions tend to zero or a constant. We begin this chapter by introducing the reader to the tools of the trade.
8.1 Tools of Approximation
The symbols ∼, o, and O are the main tools of approximating functions and are widely used in all branches of science. For the benefit of our readers, we shall give our definitions for functions defined on the real or complex numbers. Hence sequences will be treated as a special case of the general theory.
We start with the symbol O (big oh).
Definition 8.1. Let f(t) and g(t) be two functions defined on R or C. Then we say that f(t) = O(g(t)), t → ∞, if there is a positive constant M such that
f ( t )
Equivalently, f(t) = O(g(t)) if g(t) is bounded for t ≥ t0. In other
words, f = O(g) if f is of order not exceeding g.
|f(t)| ≤ M|g(t)| for all t ≥ t0.
335
336 8. Asymptotic Behavior of Difference Equations
Example 8.2.
(a) Show that
n n 1
t2 + n2 = O tn , n → ∞, for n ∈ Z+.
Solution Without loss of generality we assume t > 1. We have t2 +n2 = (t − n)2 + 2nt ≥ 2nt. Hence
nn 1 111
≤tn,
11
sin nπ+n =O n , n→∞.
+
t2+n2 ≤(2t)n =2n tn
forn∈Z ,≈>.
It follows that
n n
1 =O tn
t2+n2
with the constant M = 1 being independent of n.
(b) Show that
sinnπ + 1 sin 1 n=n. 1/n 1/n
Solution Recall that sin nπ + 1 = (−1)n sin 1 . Thus nn
sin 1 Ifweletu= 1,thenlimn→∞ n =limu→0 sinu =1.
n 1/n u
Hence we conclude that sin 1 /(1/n) is bounded, which gives the
n
required result.
(c) Showthatt2logt+t3 =O(t3),t→∞.
t2 logt+t3 logt Solution t3 =1+ t .
Using the first derivative test one may show that the function y = log t/t attains its maximum value 1 as t = e. Hence |log t/t| ≤ 1 < 1, and thus
2 33 e e (t log t + t )/t ≤ 2. This proves the required result.
Remark: We would like to point out here that the relation defined by O is not symmetric, i.e., if f = O(g), then it is not necessarily true that g = O(f). To illustrate this point we cite some simple examples such as x=O(x2),x→∞,butx2 ̸=O(x),x→∞,ore−x =O(1),x→∞,but 1̸=O(e−x),x→∞,since1/e−x →∞,x→∞.
However, it is true that the relation O is transitive, that is to say if f = O(g) and g = O(h), then f = O(h) (Exercises 8.1, Problem 1). In this case we say that f = O(h) is a better approximation of f than f = O(g).
Next we give the definition of the symbol o (little oh). Definition 8.3. If limt→∞ f(t) = 0, then we say that
Example 8.4.
g(t)
f(t) = o(g(t)), t → ∞.
(a) Showthatt2logt+t3 =o(t4),t→∞.
t2 log t + t3 log t 1
Solution limt→∞ t4 = limt→∞ t2 + limt→∞ t . Using L’Hoˆpital’s rule we have
8.1 Tools of Approximation 337
limlogt=lim 1 =0. t→∞ t2 t→∞2t2
Hence
t2 logt+t3 lim 4
= 0,
t→∞ t and the required conclusion follows.
(b) Show that o(g(t)) = g(t)o(1), t → ∞. Solution Let f(t) = o(g(t)),t → ∞. Then
lim f(t) = 0, t→∞ g(t)
which implies that f(t) = o(1),t → ∞. Consequently, f(t) = g(t)o(1),
g(t)
t → ∞.
The reader may sense correctly that the symbol o plays a much less
important role than the symbol O.
Finally, we introduce the asymptotic equivalence relation ∼.
Definition 8.5. If limt→∞ f(t) = 1, then we say that f is asymptotic to g(t)
g, t → ∞, and we write f ∼ g, t → ∞. Notice that if f ∼ g as t → ∞, then
This implies from Definition 8.3 that f(t) − g(t) = o(g(t)) = g(t)o(1) (Example 8.4). Hence we have
f (t) = g(t)[1 + o(1)].
Thus, it appears that the symbol ∼ is superfluous, since, as has been demonstrated above, f ∼ g can be conveniently written as f = g(1 + o(1)).
lim f(t)−g(t)=0. t→∞ g(t)
338 8. Asymptotic Behavior of Difference Equations
Example 8.6.
(a) Show that sinht ∼ 1et,t → ∞.
sinht t→∞ 1et
= lim 2 = 1. t→∞ 1et
2
1(et −e−t)
Solution lim
(b) Showthatt2logt+t3 ∼t3,t→∞.
Solution
22
t2 logt+t3 logt lim 3 = 1 + lim
t→∞t t→∞t
= 1 + 0 (using L’Hoˆpital’s rule)
= 1.
Notice that from Examples 8.2(c) and 8.6(b) we have t3 ∼ t2 log t + t3 and t2 logt + t3 = O(t3). It is also true that t2 logt + 2t3 = O(t3), but t2 log t + 2t3 is not asymptotic to t3, since
t2 logt+2t3
lim 3 = 2.
Before ending this section we will entertain the curious reader by intro- ducing the prime number theorem, well known in the discipline of number theory. It says that the number of primes π(t) that are less than the real number t is asymptotic to t/(log t), t → ∞, that is,
π(t)∼ t , t→∞. log t
For a proof of this result the reader may consult [144]. Another interesting asymptotic result is Stirling’s formula
n! ∼ nn√2πn e−n, n → ∞. A proof of this formula may be found in [127].
Exercises 8.1
t2
1. Showthat 1+t3 +log(1+t2)=O(logt), t→∞.
2. Show that sinht = O(et), t → ∞.
3. Show that O(g(t)) = g(t)O(1), t → ∞.
t→∞ t
4. Show that:
1 1 1 1
(i)t−1=t 1+t+O t2 , t→∞,
1 1 1 1 (ii)t−1=t1+t+ot, t→∞.
8.1 Tools of Approximation 339
1
5. Showthatsinh t =o(1), t→∞.
6. Show that:
(i) [O(t)]2 = O(t2) = o(t3),
(ii) t + o(t) = O(t).
7. Show that:
(i) sin(O(t−1)) = O(t−1),
(ii) cos(t + α + o(1)) = cos(t + α) + o(1), for any real number α.
8. Prove that ∼ is an equivalence relation.
9. Prove that both relations o and O are transitive.
10. Suppose that f(t) = O(t),t → ∞, and g(t) = O(t2),t → ∞. Show that
for any nonzero constants a, b, af (t) + bg(t) = O(g(t)), t → ∞.
11. If f = O(g),t → ∞, show that:
(i) O(o(f)) = o(O(f)) = o(g), (ii) O(f)o(g) = o(f)o(g) = o(fg).
12. Let f be a positive nonincreasing function of t, and let f(t) ∼ g(t), t → ∞. Prove that sups>t f(s) ∼ g(t), t → ∞.
13. Suppose that the functions f and g are continuous and have convergent integrals on [1, ∞). If f (t) ∼ g(t), t → ∞, prove that
∞ ∞
f(s) ds ∼ g(s) ds, t → ∞.
tt
14. Consider the exponential integral En(x) defined by
∞ e−xt
En(x) =
(a) Show that En(x) satisfies the difference equation
tn dt, (x > 0), where n is a positive integer. En+1(x) = 1 e−x − xEn(x) .
1
(b) Use integration by parts to show that e−x 1
En(x)= x 1+0 x , x→∞.
(c) Show that
En(x)=n−1 1+O n−2 , n→∞.
n
e−x 1
340 8. Asymptotic Behavior of Difference Equations
15. Show that
16. Show that
∞e−1 1 1 1 0x+t=x1−x+Ox2 ,x→∞.
n 1
kk =nn 1+O n , n→∞. k=1
8.2 Poincar ́e’s Theorem
In this section we introduce to the reader the theorems of Poincar ́e and Perron. Both theorems deal with the asymptotic behavior of linear differ- ence equations with nonconstant coefficients. It is widely accepted among researchers in difference equations that the theorem of Poincar ́e [123] marks the beginning of research in the qualitative theory of linear differ- ence equations. Thirty-six years later, Perron [117] made some significant improvements to Poincar ́e’s theorem.
To motivate our study we will take the reader on a short excursion to the much simpler linear equations with constant coefficients of the form
x(n + k) + p1x(n + k − 1) + · · · + pkx(n) = 0, (8.2.1) where the pi’s are real or complex numbers. The characteristic equation of
(8.2.1) is given by
λk +p1λk−1 +···+pk =0. (8.2.2) Let λ1, λ2, . . . , λk be the characteristic roots of (8.2.2). Then there are two
main cases to consider.
Case 1. Suppose that distinct characteristic roots have distinct moduli, i.e., ifλi ̸=λj,then|λi|̸=|λj|forall1≤i,j≤k.
For the convenience of the reader we will divide Case 1 into two subcases.
Subcase (a) Assume that all characteristic roots are distinct. So, by relabeling them, one may write the characteristic roots in descending order
|λ1| > |λ2| > ··· > |λk|. Then the general solution of (8.2.1) is given by
x(n)=c1λn1 +c2λn2 +···+ckλnk. (8.2.3)
Hence if c1 ̸= 0, we have
lim
n→∞
x(n+1) c λn+1 +c λn+1 +···+c λn+1 = lim 1 1 2 2 k k
⎡
⎢c1 +c2
n+1⎤ λk ⎥ λ1
n+1 λ2
λ1
=λ1, since i<1, i=2,...,k.
8.2 Poincar ́e’s Theorem 341
x(n) n→∞ c1λn1 +c2λn2 +···+ckλnk
=limλ1⎣
n→∞ c1+c2 λ2
n ⎦ λ1
λ1 n
+···+ck +···+ck λk
λ λ1
Similarly, if c1 = 0, c2 ̸= 0, we obtain
lim x(n+1)=λ2.
n→∞ x(n) And,ingeneral,ifc1 =c2 =···=ci−1 =0,ci ̸=0,then
lim x(n+1)=λi. n→∞ x(n)
Subcase (b) Now suppose that there are some repeated characteristic roots. For simplicity assume that λ1 is of multiplicity r, so λ1 = λ2 = · · · = λr,|λ1| = |λ2| = ··· = |λr| > |λr+1| > ··· > |λk|. Then the general solution of (8.2.1) is given by
x(n)=(c1 +c2n+···+crnr−1)λn1 +cr+1λnr+1 +···+ckλnk.
Then one may show easily that this case is similar to Subcase (a) (Exercises
8.2, Problem 1).
Case 2. There exist two distinct characteristic roots λr,λj with |λr| = |λj|.
This may occur if λr and λj are conjugates, i.e., λr = α+iβ, λj = α−iβ for
some real numbers α and β. For simplicity, let us assume that r = 1, j = 2,
soλr ≡λ1 andλj ≡λ2.Wewriteλ1 =α+iβ=reiθ,λ2 =α−iβ=re−iθ,
where r = (α2 +β2)1/2, θ = tan−1 β . Then the general solution of (8.2.1) α
is given by
x(n) = c1rneinθ + c2rne−inθ + c3λn3 + · · · + ckλnk . lim x(n+1)
n→∞ = lim
n→∞
Since einθ = cosnθ + isinnθ,e−inθ = cosnθ − isinnθ do not tend to definite limits as n → ∞, we conclude that the limit (8.2.4) does not exist. For particular solutions the limit may exist. For example, if |λ1| = |λ2| > |λ3| > ··· > |λk|, and
Hence
x(n)
rn+1(c ei(n+1)θ +c e−i(n+1)θ)+c λn+1 +···+c λn+1
1 2 3 3 k k . (8.2.4) rn(c1einθ +c2e−inθ)+c3λn3 +···+ckλnk
342 8. Asymptotic Behavior of Difference Equations
(a)c1̸=0,c2=0,then limx(n+1)=reiθ=λ1, n→∞ x(n)
(b)c1=0,c2̸=0,then limx(n+1)=re−iθ=λ2. n→∞ x(n)
Case 2 may also occur if λi = −λj. It is left to the reader as Exercises 8.2, Problem 2, to verify that in this case, too, limn→∞ x(n + 1)/x(n) does not exist.
We now summarize the above discussion in the following theorem. Theorem 8.7. Let x(n) be any nonzero solution of (8.2.1). Then
(8.2.5)
for some characteristic root λm, provided that distinct characteristic roots have distinct moduli. Moreover, if there are two or more distinct roots λr , λj with the same modulus (|λr| = |λj|), the limit (8.2.5) may not exist in general, but particular solutions can always be found for which the limit (8.2.5) exists and is equal to a given characteristic root λm.
Example 8.8. Consider the difference equation x(n + 2) + μx(n) = 0.
(a) If μ = β2, then the characteristic equation is given by λ2 + β2 = 0.
Hence the characteristic roots are λ1 = βi = βeiπ/2 and λ2 = −βi = βe−iπ/2. The general solution is given by
So
x(n) = c1βneinπ/2 + c2βne−inπ/2.
c1ei(n+1)π/2 + c2e−i(n+1)π/2 inπ/2 −inπ/2 ,
then
x(n + 1) n→∞ x(n)
lim x(n+1) =λm n→∞ x(n)
lim
which does not exist. However, if we pick the particular solution
= β
c1e +c2e x(n) = c1βneinπ/2,
lim x(n + 1) = βeinπ/2 = βi. n→∞ x(n)
Similarly, for the solution xˆ(n) = c2βne−inπ/2,
l i m xˆ ( n + 1 ) = − β i . n→∞ xˆ(n)
(b) If μ = −β2, then the characteristic roots are λ1 = β,λ2 = −β. The general solution is given by x(n) = c1βn + c2(−β)n.
Hence
The limit (8.2.6) does not exist, since x(n + 1)/x(n) oscillates between β(c1 + c2)/(c1 − c2) and β(c1 − c2)/(c1 + c2). Notice that for the solution x(n) = c1βn,
lim x(n+1)=β, n→∞ x(n)
8.2 Poincar ́e’s Theorem 343
lim x(n + 1) = lim c1βn+1 + c2(−β)n+1 n→∞ x(n) n→∞ c1βn + c2(−β)n
c1 + c2(−1)n+1
=βlim n. (8.2.6)
n→∞ c1 + c2(−1)
and for the solution x ̃(n) = c2(−β)n,
l i m x ̃ ( n + 1 ) = − β .
n→∞ x ̃(n)
In 1885 the French mathematician Henri Poincar ́e [123] extended the
above observations to equations with nonconstant coefficients of the form x(n + k) + p1(n)x(n + k − 1) + · · · + pk(n)x(n) = 0 (8.2.7)
such that there are real numbers pi, 1 ≤ i ≤ k, with
lim pi(n)=pi, 1≤i≤k. (8.2.8)
n→∞
We shall call an equation of the form (8.2.7), (8.2.8) a difference equation
of Poincar ́e type. The characteristic equation associated with (8.2.7) is
λk +p1λk−1 +···+pk =0. (8.2.9)
The underlying idea behind Poincar ́e’s theorem is that since the coefficients of a difference equation of Poincar ́e type are nearly constant for large n, one would expect solutions of (8.2.7) to exhibit some of the properties of the solutions of the corresponding constant coefficient difference equation (8.2.1) as stated in Theorem 8.7.
An important observation which carries over from autonomous to nonau- tonomous systems is the following. If lim x(n+1) = α, then α must be a
n→∞ x(n) characteristic root, i.e., a root of (8.2.9).
Theorem 8.9 (Poincar ́e’s Theorem). Suppose that condition (8.2.8) holds and the characteristic roots λ1, λ2, . . . , λk of (8.2.9) have distinct moduli. If x(n) is a solution of (8.2.7), then either x(n) = 0 for all large n or
344 8. Asymptotic Behavior of Difference Equations
lim x(n+1) =λi n→∞ x(n)
(8.2.10)
for some i, 1≤i≤k.
Proof. The proof will be given in Section 8.8. P
Note that Poincar ́e’s Theorem does not tell us whether or not each char- acteristic root λi can be written in the form (8.2.10). In 1921, Oscar Perron [117] gave an affirmative answer to this question.
Theorem 8.10 (Perron’s First Theorem). Assume that pk(n) ̸= 0 for all n ∈ Z+ and the assumptions of Theorem 8.9 hold. Then (8.2.7) has a fundamental set of solutions {x1(n),x2(n),…,xk(n)} with the property
(8.2.11)
Proof. A proof of this theorem may be found in Meschkowski [99, p. 10]. Theorem 8.10 is commonly referred to as the Poincar ́e–Perron Theorem. Perron [117] later formulated and proved a result of a different nature that does not suffer from the restriction on the characteristic roots. P
Theorem 8.11 (Perron’s Second Theorem). Suppose that pk(n) ̸= 0 for all n ∈ Z+. Then (8.2.7) has a fundamental set of solutions {x1(n), x2(n), . . . , xk(n)} such that
lim sup n |xi(n)| = |λi|. (8.2.12) n→∞
It is questionable whether Poincar ́e–Perron Theorem remains valid if (8.2.7) has characteristic roots with equal moduli. Perron himself addressed this question and gave the following example, which shows that Poincar ́e’s theorem may fail in this case.
But in order to understand this example we need to make a detour to infinite products.
lim xi(n+1) =λi, 1≤i≤k. n→∞ xi (n)
8.2.1 Infinite Products and Perron’s Example
An expression of the form
infinite product (8.2.13) is said to converge if lim
∞ n=1
(1 + a(n)), a(n) ̸= −1 for all n ∈ Z+, is called an infinite product. The partial products are
(8.2.13)
(1 + a(j)). The n (1 + a(j)) is
n→∞ j=1 finite and nonzero. Otherwise, it is said to be divergent.
n j=1
Theorem 8.12.
Consider the infinite series
Then the following statements hold:
∞ n=1
∞ n=1
a(n),
a2 (n).
(8.2.14)
(8.2.15)
(i) The convergence of any two of (8.2.13), (8.2.14), (8.2.15) implies that of the third.
(ii) If ∞n=1 |a(n)| converges, then both (8.2.13) and (8.2.15) converge.
(iii) If (8.2.14) converges conditionally, then:
(a) (8.2.13) converges if (8.2.15) converges,
(b) (8.2.13) diverges to zero if (8.2.15) diverges.
Proof. See [109]. P
Example 8.13. Consider the difference equation (−1)n
x(n+2)− 1+ n+1 x(n)=0, n≥0. (8.2.16)
Then the associated characteristic equation of (8.2.16) is λ2 − 1 = 0. Hence
the characteristic roots are λ1 = 1 and λ2 = −1, with |λ1| = |λ2| = 1. We
8.2 Poincar ́e’s Theorem 345
now have two cases to consider:
(i) Forn=2k, x(2k+2)= 1+ 1 x(2k),andhence 2k+1
k 1
1 + 2j − 1 x(0).
(ii) Forn=2k−1, x(2k+1)=1+ 1 x(2k−1),andhence
(8.2.17)
x(2k) =
j=1
2k
k−1 1
1 − 2j ⎡⎤
x(2k − 1) =
x(1). (8.2.18)
Hence
lim
k→∞x(2k−1) k→∞ j=1
j=1
k k−1
x(2k) ⎣ 1 1 ⎦
= lim 1+
2j−1
x(0)
1− x(1) . 2j
j=1
(8.2.19)
346 8. Asymptotic Behavior of Difference Equations
In the sequel we will show that this limit does not exist. To accomplish this task, we need to evaluate the infinite products
∞ 1
1 + 2j − 1 ∞ 1
(8.2.20)
and
j=1
1 − 2j .
∞ 1 converges, it follows by Theorem 8.12(i) that if (8.2.20) con-
(8.2.21) Let us now apply Theorem 8.12(i) to the infinite product (8.2.20). Since
j=1
j=1 (2j−1)2 ∞ 1
verges, then so does j=1 2j−1, which is false. Thus the infinite product
(8.2.20) diverges to ∞, since each term 1 + 1 is greater than 1. 2j −1
Next we consider the infinite product (8.2.21). By a similar argument,
we show that it diverges to zero, since each term 1 − 1 is less than 1. 2j
It follows that the limit (8.2.21) does not exist. Example 8.14. Consider the difference equation
x(n+2)− n x(n+1)+ 1x(n)=0. n+1 n
The associated characteristic equation is given by λ2 − λ = 0
with characteristic roots λ1 = 1, λ2 = 0. Hence by Perron’s theorem there exist solutions x1(n), x2(n) such that
lim x1(n+1) =1 and lim x2(n+1) =0. n→∞ x1 (n) n→∞ x2 (n)
What can we conclude about the solutions x1(n) and x2(n)? The solution x1(n) may be equal to a constant c, a polynomial in n such as
aknk +ak−1nk−1 +···+a0,
or a function such as 1 , log n, among others. The solution x2(n) may be
The reader may correctly conclude from the preceding examples that Poincar ́e’s or Perron’s theorem provides only partial results about the asymptotic behavior of solutions of linear difference equations. The ques- tion remains whether we can use Perron’s theorem to write an asymptotic expression of solutions of equations of Poincar ́e type. Using null sequences, Wimp [145] devised an elegant and simple method to address the above question. Recall that ν(n) is called a null sequence if limn→∞ ν(n) = 0.
n equal to 0, e−2n , e−n2 , etc.
Lemma 8.15. Suppose that limn→∞ x(n+1) = λ. x(n)
(a) If λ ̸= 0, then
for some null sequence ν(n). (b) If λ=0, then
for some positive null sequence μ(n). Proof.
(8.2.22)
(8.2.23)
Hence
lim
x(n)
x(n) = ±λnenν(n)
8.2 Poincar ́e’s Theorem 347
(a) Let
Then
x(n) y(n)=λn .
|x(n)| = e−n/μ(n)
y ( n + 1 ) y(n)
1 x ( n + 1 )
= lim n→∞ λ
=1. lim log
= 0.
Hence for a given ε > 0 there exists a positive integer N such that
|z(n+1)−z(n)|<ε/2 foralln≥N. Moreover, for n ≥ N, we obtain
n ε
n<21−n+n < ε + ε = ε,
n→∞
If we let z(n) = log y(n), then we have
lim z(n+1)−z(n) = n→∞
=log lim y(n+1) n→∞ y(n)
y(n + 1) n→∞ y(n)
|z(n)−z(N)|≤
|z(r)−z(r−1)|< 2(n−N). z(n) ε N z(N)
r=N+1
22
for sufficiently large n. It follows that limn→∞ z(n) = 0 or z(n) = nν(n)
n
for some null sequence ν(n).
(b) This is left to the reader as Exercises 8.2, Problem 6. P
348 8. Asymptotic Behavior of Difference Equations
Example 8.16. Use Lemma 8.15 and the Poincar ́e–Perron Theorem to find asymptotic estimates of a fundamental set of solutions of the difference equation
y(n+2)+n+1y(n+1)− 2n y(n)=0. n+2 n+2
Solution The associated characteristic equation is given by λ2 + λ − 2 = 0
with roots λ1 = 1, λ2 = −2. By Perron’s Theorem, there is a fundamental set of solutions y1(n), y2(n) with
lim y1(n+1) =1, n→∞ y1 (n)
Thus by Lemma 8.15 we obtain y1(n) = enν(n),
lim y2(n+1) =−2. n→∞ y2 (n)
y2(n) = (−2)nenμ(n),
for some null sequences ν(n) and μ(n).
For the curious reader we note that an exact fundamental set of solutions
is given by
Exercises 8.2
1 (−2)n y1(n) = n, y2(n) = 2 .
1. Prove that each nontrivial solution x(n) of the second-order difference equation
x(n + 2) + p1x(n + 1) + p2x(n) = 0
with double characteristic roots λ1 = λ2 = λ satisfies limn→∞(x(n +
1))/x(n) = λ.
2. Suppose that the characteristic roots λ1, λ2 of
x(n + 2) + p1x(n + 1) + p2x(n) = 0
are such that λ1 = −λ2. Prove that limn→∞(x(n + 1))/x(n) does not
exist for some solution x(n).
3. Consider the difference equation
x(n+3)−(α+β+γ)x(n+2)+(αβ+βγ+γα)x(n+1)−αβγu(x) = 0, where α, β, γ are constants.
(a) Show that the characteristic roots are λ1 = α, λ2 = β, and λ3 = γ.
(b) If |α| > |β| > |γ|, find a fundamental set of solutions x1(n), x2(n), and x3(n) with
lim x1(n+1) =α, lim x2(n+1) =β, n→∞ x1 (n) n→∞ x2 (n)
lim x3(n+1)=γ. n→∞ x3 (n)
(c) If |α| = |β|,α ̸= β,|α| > |γ|, find a fundamental set of solu- tions x1(n), x2(n), and x3(n) such that limn→∞ x1(n+1)/x1(n) = α, limn→∞ x2(n + 1)/x2(n) = β, limn→∞ x3(n + 1)/x3(n) = γ.
8.2 Poincar ́e’s Theorem 349
4. Consider the difference equation
x(n+2)+ 1 x(n+1)−n+1x(n)=0.
n+4 n+4
Use iteration to show that limn→∞ x(n + 1)/x(n) does not exist for
every solution x(n).
5. Consider the equation
x(n + 2) − ((n + 2) + 2(−1)n)/(n + 2)3(n + 3)x(n) = 0.
Use iteration to show that limn→∞(x(n + 1))/x(n) does not exist for
any solution x(n).
6. Prove part (b) of Lemma 8.15.
7. Show that the difference equation
x(n + 1) − n + 7 x(n) − n2 + 1 x(n − 1) = 0 n + 5 n2 + 4
has an oscillatory solution and a nonoscillatory solution.
8. Consider the difference equation
2n−1 2n−1 x(n+2)− 3+n2−2n−1 x(n+1)+2 1+n2−2n−1 x(n)=0.
(a) Use Lemma 8.15 and Perron’s theorem to find asymptotic estimates of a fundamental set of solutions of the equation.
(b) Verify that x1(n) = 2n and x2(n) = n2 constitute a fundamental set of solutions.
9. Let x(n) be a nontrivial solution of (8.2.7) such that limn→∞ x(n + 1)/x(n) = α. Show that α is a characteristic root, i.e., a root of (8.2.9).
10. Let α be a number whose modulus is greater than all of the charac- teristic roots of a difference equation of Poincar ́e type (8.2.7). Prove
350 8. Asymptotic Behavior of Difference Equations
that
lim x(n) = 0 n→∞ αn
for any solution x(n) of the equation.
11. Suppose that limn→∞ x(n + 1)/x(n) = λ > 0. Prove that for any
δ ∈ (0, λ):
(i) |x(n)| = O(λ + δ)n, and
(ii) (λ + δ)n = O(x(n)).
12. Consider the equation x(n + 2) − (n + 1)x(n + 1) − 2n2x(n) = 0.
(a)
(b)
Transform the equation into an equation of Poincar ́e type by letting x(n) = (n − 1)! y(n).
Use part (a) to get an asymptotic estimate of a fundamental set of solutions.
13. Use
fundamental set of solutions of the equation
the scheme of Problem 11 to find an asymptotic set of a
x(n + 2) + 4nx(n + 1) + 4n(n − 1)x(n) = 0.
14. Prove Theorem 8.12.
15. Consider the equation
(n + 2)x(n + 2) − (n + 3)x(n + 1) + 2x(n) = 0. (8.2.24)
(a) Show that 1, 0 are the characteristic roots of the equation.
(b) Put
x(n + 1) = 1 + μ(n) (8.2.25) x(n)
in (8.2.23), where μ(n) is a null sequence, and show that the equation becomes
(c) Show that
(n + 2)μ(n + 1) = 1 − 2 . 1+μ(n)
2 = 2 + O(μ(n)). 1+μ(n)
(8.2.26)
(8.2.27)
(d) Use part (c) to show that (8.2.27) is equivalent to
1 1 μ(n+1)=−n+1+O n2 .
n 1 (f) Provethatx(n)∼ c,n→∞.
8.3 Asymptotically Diagonal Systems 351
(e) Show that
x(n+1) = n+1 1+O n2 x(n). (8.2.28)
n
16. Show that (8.2.24) has another solution x ∼ c 2n , n → ∞. n!
17. Use the scheme of Problem 15 to find asymptotic estimates of a fundamental set of solutions of the equation
(n + 1)x(n + 2) − (n + 4)x(n + 1) + x(n) = 0.
18. Showthattheequationx(n+2)−(n+1)x(n+1)+(n+1)x(n)=0
has solutions x1(n), x2(n) with asymptotic estimates x1(n) ∼ c(n − 2)!, x2(n) = an, n → ∞.
*19. (Hard). Consider the equation of Poincar ́e type
x(n + 2) − (2 + p1(n))x(n + 1) + (1 + p2(n))x(n) = 0,
where p1(n) ≥ p2(n) for all n ∈ Z+. Show that if x(n) is a solution that is not constantly zero for large values of n, then limn→∞(x(n + 1))/x(n) = 1.
*20. (Hard). Consider the equation
x(n + 2) + P1(n)x(n + 1) + P2(n)x(n) = 0
with limn→∞ P1(n) = p1, limn→∞ P2(n) = p2. Let η be a positive constant such that |x(n + 1)/x(n)|2 > |p2| + η for sufficiently large n. Suppose that the characteristic roots λ1, λ2 of the associated equation are such that |λ1| ≥ |λ2|.
Prove that limn→∞ x(n + 1)/x(n) = λ1.
8.3 Asymptotically Diagonal Systems
In this section we derive conditions under which solutions of a perturbed diagonal system are asymptotic to solutions of the unperturbed diagonal system. As a byproduct we obtain asymptotic results for nonautonomous kth-order scalar difference equations.
We begin our study by considering the perturbed diagonal system
y(n + 1) = (D(n) + B(n))y(n) (8.3.1)
and the unperturbed diagonal system
x(n + 1) = D(n)x(n), (8.3.2)
352 8. Asymptotic Behavior of Difference Equations
where D(n) = diag(λ1(n),λ2(n),…,λk(n)),λi(n) ̸= 0, for all n ≥ n0 ≥ 0,1≤i≤k,andB(n)isak×kmatrixdefinedforn≥n0 ≥0.The fundamental matrix of system (8.3.2) is given by
by letting
Φ1 (n) = diag(μ1 (n), μ2 (n), . . . , μk (n)) ⎧ n−1
Φ(n) = diag λ1(r), r=n0
λ2(r), . . . , λk(r) . r=n0
(8.3.3)
(8.3.4)
⎩0, Define Φ2(n) = Φ(n) − Φ1(n).
n−1 n−1 n−1
r=n0
Let S be a subset of the set {1,2,3,…,k}. Define
⎪ ⎪⎨ μi(n) = ⎪r=n0
λi(r),
if i ∈ S, otherwise.
We are now ready for the definition of the important notion of dichotomy. Definition 8.17. System (8.3.2) is said to possess an ordinary dichotomy
if there exists a constant M such that:
(i) ∥Φ1(n)Φ−1(m)∥ ≤ M, for n ≥ m ≥ n0, (ii) ∥Φ2(n)Φ−1(m)∥ ≤ M, for m ≥ n ≥ n0.
Notice that if D(n) is constant, then system (8.3.2) always possesses an ordinary dichotomy.
After wading through the complicated notation above, here is an example.
Example 8.18. Consider the difference system x(n + 1) = D(n)x(n) with ⎛1⎞
⎜1+n+10 0 0⎟
⎜ 0 0.5 0 0 ⎟ D(n)=⎜0 0n+10⎟.
⎜⎝
0001 n+2
⎟⎠
Then a fundamental matrix of the system may be given by
⎛ ⎞
n−1 n−1 n−1
⎝ 1 n 1⎠
Φ(n)=diag 1+j+1 ,(0.5) , (j+1), j+2 j=0 j=0 j=0
1 =diag n+1,(0.5)n,n!,(n+1)! .
From this we deduce that
Φ1(n) = diag 0, (0.5)n, 0, (n + 1)!
and
Hence
1 Φ2(n) = diag(n + 1, 0, n!, 0).
8.3 Asymptotically Diagonal Systems
353
Finally,
Φ1(n)Φ−1(m) = diag 0, (0.5)n−m, 0, (n + 1)(n) · · · (m + 2)
1 ∥Φ1(n)Φ−1(m)∥ ≤ 1, for n ≥ m ≥ 0.
.
Similarly,
Φ2(n)Φ−1(m)=diag m+1,0,m!,0 , form≥n≥n0.
Hence
n+1 n! ∥Φ2(n)Φ−1(m)∥ ≤ 1, for m ≥ n ≥ n0.
We are now ready to establish a new variation of constants formula that is very useful in asymptotic theory.
Theorem 8.19 (Variation of Constants Formula). Suppose that system (8.3.2) possesses an ordinary dichotomy and the following condition holds:
∞
∥B(n)∥ < ∞. (8.3.5)
n=n0
Then for each bounded solution x(n) of (8.3.2) there corresponds a bounded
solution y(n) of (8.3.1) given by
(8.3.6)
The converse also holds; for each bounded solution y(n) of (8.3.1) there corresponds a bounded solution x(n) of (8.3.2).
Proof. Let x(n) be a bounded solution of (8.3.2). By using the method of successive approximation, we will produce a corresponding bounded so- lution y(n) of (8.3.1). We define a sequence {yi(n)} (i = 1, 2, . . .) by letting
y(n) = x(n) +
n−1
j =n0
Φ1(n)Φ−1(j + 1)B(j)y(j)
−
∞ −1
Φ2(n)Φ (j + 1)B(j)y(j).
j=n
354 8. Asymptotic Behavior of Difference Equations
y1(n) = x(n) and
yi+1(n) = x(n) +
−
n−1
−1
j=n
on the discrete interval [n0, ∞).
Using (8.3.7) we have, for i = 1,2,...,
⎡ ⎤i
∞ j =n0
|yi+2(n) − yi+1(n)| ≤ M
Hence by induction on i (Exercises 8.3, Problem 8)
|yi+1(n) − yi(n)| ≤ ⎣M We choose n0 sufficiently large such that
∞ j =n0
∥B(j)∥⎦ c1.
(8.3.8)
(8.3.9)
M
∥B(j)∥ = η < 1.
Φ1(n))Φ (j + 1)B(j)yi(j) Φ2(n)Φ−1(j + 1)B(j)yi(j).
j =n0 ∞
First we prove that yi(n) is bounded on the discrete interval [n0,∞). This task will be accomplished by induction on i. From our assumption we have |y1(n)| = |x(n)| ≤ c1, for some constant c1. Now assume that |yi(n)| ≤ ci, for some constant ci. Then by Definition 8.17 we have
∞ j =n0
Hence yi(n) is bounded for each i.
In the next step we show that the sequence {yi(n)} converges uniformly
|yi+1(n)| ≤ c1 + Mci
∥B(j)∥ = ci+1.
∞ j =n0
Thus |yi+1(n) − yi(n)| ≤ c1ηi and, consequently, ∞i=1{yi+1(n) − yi(n)} converges uniformly on n ≥ n0 (by the Weierstrass M-test).1
We define
∞ i=1
1Weierstrass M-test: Let un(x),n = 1,2,..., be defined on a set A with range in R. Suppose that |un(x)| ≤ Mn for all n and for all x ∈ A. If the series of constants ∞n=1 Mn converges, then ∞n=1 un(x) and ∞n=1 |un(x)| converge uniformly on A.
y(n) = y1(n) +
{yi+1(n) − yi(n)} = lim yi(n). i→∞
∥B(j)∥|yi+1(j) − yi(j)|.
(8.3.7)
y(n) =
λi(r)z(n),
for a specific i,
(8.3.13)
r=n0 Then (8.3.1) becomes
n−1
1 ≤ i ≤ k.
8.3 Asymptotically Diagonal Systems 355
Hence |y(n)| ≤ L, for some constant L. Letting i → ∞ in (8.3.7), we obtain (8.3.6). The second part of the proof of the theorem is left to the reader as Exercises 8.3, Problem 10. P
If the condition of ordinary dichotomy is strengthened, then we obtain the following important result in asymptotic theory.
Theorem 8.20. Suppose that the following assumption holds: ⎧
⎨(i) Systems (8.3.2) posses an ordinary dichotomy; Condition (H) ⎩(ii) lim Φ1(n) = 0.
n→∞
If, in addition, condition (8.3.5) holds, then for each bounded solution x(n)
of (8.3.2) there corresponds a bounded solution y(n) of (8.3.1) such that
y(n) = x(n) + o(1). (8.3.10)
Proof. Let x(n) be a bounded solution of (8.3.2). Then by using formula (8.3.6) we obtain, for a suitable choice of m (to be determined later),
y(n) = x(n) + Φ1(n)
m−1
−1
j =n0
(j + 1)B(j)y(j) + Ψ(n),
(8.3.11)
Φ
where
Ψ(n) = Φ1(n) Φ (j + 1)B(j)y(j) − Φ2(n)Φ
j=m j=n
Now recall that from Theorem 8.20, ∥y∥ ≤ L, for some L > 0. Hence from
formula (8.3.12) it follows that |Ψ(n)| ≤ ML
Thus for ε > 0, there exists a sufficiently large m such that |Ψ(n)| < ε/2. Since Φ1(n) → 0 as n → ∞, it follows from formula (8.3.11) that |y(n) − x(n)| < ε, for sufficiently large n. Therefore, y(n) = x(n) + o(1).
Our next objective is to apply the preceding theorem to produce a dis- crete analogue of Levinson’s theorem [91], [36]. We start our analysis by making the change of variables
n−1
−1 −1
z(n + 1) = (Di(n) + Bi(n))z(n),
(8.3.14)
∞ j=m
∥B(j)∥.
∞
(j + 1)B(j)y(j).
(8.3.12)
356 8. Asymptotic Behavior of Difference Equations
where
λ1(n) λk(n) Di(n) = diag λi(n),...,1,..., λi(n) ,
Bi(n) = 1 B(n). λi (n)
Associated with (8.3.14) is the unperturbed diagonal system x(n + 1) = Di(n)x(n).
(8.3.15P) To make the proof of our main theorem more transparent we introduce
the following lemma.
Lemma 8.21 [9]. Assumption (H) hold for every equation (8.3.15), 1 ≤ i ≤ k, if the following conditions hold.
There exist constants μ > 0 and K > 0 such that for each pair λi, λj,
as n → ∞,
for all 0 ≤ n1 ≤ n2,
for all 0≤n1 ≤n2.
The proof is omitted and left to the reader to do as Exercises 8.3, Problem
7.
Example 8.22. Consider the diagonal matrix
i ̸= j, either
⎪ λj (r)
⎧⎪n λi(r) → +∞,
⎪n ⎨ 2
Condition (L)⎪and
⎪ r=n1
λ i ( r )
λj(r) ≥ μ > 0,
⎪r=0
where
Notice that: (i)
λ1(n) = 2 + sin λ2(n) = 2 − sin
λ3(n) = 2.
2n+1
2 π,
2n+1
2 π,
⎪n ⎪ ⎪ ⎪ ⎪ 2 λ i ( r )
⎪⎩or λj(r)≤K, r=n1
D(n) = diag(λ1(n), λ2(n), λ3(n))
n2
λ1(r) = 1
λ2(r) ⎪ r=n1 ⎪⎩1
if both n1 and n2 are odd,
⎧ ⎪⎨3
if both n1 and n2 are even,
if n1 is odd and n2 is even or vice versa,
3
and
n λ1(r) → ∞ as n → ∞.
λ3(r)
⎧⎪1 if both n and n are even,
n2⎪⎨ 1 2
8.3 Asymptotically Diagonal Systems 357
(ii)
λ2(r) 3
r=0
and
(iii)
n λ2(r) → ∞
as n → ∞.
r=n1
λ1(r) = ⎪1 if n1 is odd and n2 is even or vice versa, ⎪⎩3 if both n1 and n2 are odd,
λ3(r)
n λ3(r) → ∞
r=0
as n → ∞,
but no subsequence of the product converges to zero, and
λ1(r)
n λ3(r) → ∞ as n → ∞,
λ2(r)
but no subsequence of the product converges to zero.
r=0
r=0
Thus the system x(n + 1) = D(n)x(n) satisfies Condition (L) and, consequently, it satisfies Condition (H).
Next we give the fundamental theorem in the asymptotic theory of difference equations; the discrete analogue of Levinson’s theorem [91].
Theorem 8.23. Suppose that Condition (L) holds and for each i, 1 ≤ i ≤ k,
∞ 1 ∥ B ( n ) ∥ < ∞ . ( 8 . 3 . 1 6 ) n=n0 |λi(n)|
Then system (8.3.1) has a fundamental set of k solutions yi(n) such that
(8.3.17)
where ei is the standard unit vector in Rk where its components are all zero, except that the ith component is 1.
Proof. Notice that under Condition (L) it follows from Lemma 8.21 that (8.3.15) satisfies Condition (H). Moreover, from assumption (8.3.16), Bi(n)
yi(n) = (ei + o(1))
n−1
r=n0
λi(r),
358 8. Asymptotic Behavior of Difference Equations
satisfies condition (8.3.5). Thus we can apply Theorem 8.20 to (8.3.14) and (8.3.15). Observe that since the ith diagonal element in Di(n) is 1, it follows that x(n) = ei is a bounded solution of (8.3.15). By Theorem 8.20, there corresponds a solution z(n) of (8.3.14) such that z(n) = ei + o(1). Now conclusion (8.3.17) follows immediately by substituting for z(n) from formula (8.3.13). P
Theorem 8.23 will be referred to as the Benzaid–Lutz theorem. Example 8.24. Consider the difference system y(n+1) = A(n)y(n), where
⎛n2+2 0 1⎞
⎜ 2n2 n3⎟ A(n) = ⎜ 0 1 0 ⎟ .
⎝1⎠ 2n 0n
To apply Theorem 8.23 we need to write A(n) in the form D(n) + B(n) with D(n) a diagonal matrix and B(n) satisfying condition (8.3.16). To achieve this we let
⎛1⎞ ⎛101⎞ ⎜2 0 0⎟ ⎜n2 n3⎟
D ( n ) = ⎝ 0 1 0 ⎠ , B ( n ) = ⎜⎝ 0 0 0 ⎟⎠ . 00n 100
2n
Hence λ1 = 1,λ2 = 1, and λ3 = n. Thus for n0 = 2, our system satisfies
2
the hypotheses of Theorem 8.23. Consequently, there are three solutions:
⎛⎞
1
⎜⎝ 0 ⎟⎠ ,
0
⎛⎞
0 y2(n) ∼ ⎜⎝1⎟⎠ ,
0
y1(n) ∼
1n 2
y3(n) ∼
⎝ ⎠⎜ ⎟
j ⎝0⎠=(n−1)!⎝0⎠.
⎛ ⎞⎛0⎞ n−1
⎛0⎞ ⎜ ⎟
j=1
11
Remark: Before ending this section we make one further comment on the conditions in Theorem 8.23. This comment concerns the necessity for some condition on B(n) such as (8.3.16). Certainly, condition (8.3.16) holds when B(n) = O(n−α),n → ∞, for some α > 1, i.e., nα∥B(n)∥ ≤ L for all n ≥ n0. On the other hand, the condition B(n) = O(n−1) is not sufficient for formula (8.3.17) to hold, and a simple example illustrates this point.
Let us take k = 1,D(n) = 1, and B(n) = 1. Then (8.3.1) takes the n
8.3 Asymptotically Diagonal Systems 359
formy(n+1)=n+1y(n)=1+ 1y(n),whichhasthegeneralsolution nn
y(n) = cn, for some constant c. Hence no solution satisfies formula (8.3.17).
Exercises 8.3
In Problems 1 through 5 find asymptotic estimates (using Theorem 8.23) for a fundamental set of solutions of the given system.
1. y(n + 1) = (D(n) + B(n))y(n), where ⎛3⎞⎛13⎞
D(n)=⎝n+2 0 ⎠, B(n)=⎜⎝n2 n3 ⎟⎠. 0n+1 05
n3/2
⎛ ⎞ ⎛sinnn ⎞
2. y(n + 1) = (D(n) + B(n))y(n), where
cosπn00 ⎜n3en 0⎟
D(n) = ⎜ 0 n 0⎟ , B(n) = ⎜ 0 0 n ⎟ . ⎝n+1⎠ ⎜⎝ 3n⎟⎠
003 10n 2n n3 + 5
3. y(n + 1) = A(n)y(n), where
⎛11⎞
⎜1 + n 0 n(n + 1) ⎟ A(n) = ⎜ 1 ⎟ .
⎝0n0⎠ 0 0 1 + (−1)n cos nπ
4. y(n + 1) = A(n)y(n), where
⎛ne−n 0⎞
A ( n ) = ⎜⎝ 0 3 − e − 2 n 0 ⎟⎠ . 2−n 0 1+n
5. Give an example of a two-dimensional difference system where Theorem 8.20 does not hold.
6. Define a diagonal matrix P = diag(a1, a2, . . . , ak), where
ai =
0 ifi̸∈S, 1 ifi∈S,
where S is a subset of the set {1,2,…,k}.
Prove the following statements:
(a) P 2 = P (a projection matrix).
(b) Φ(n)P = Φ1(n) as defined in (8.3.4).
360 8. Asymptotic Behavior of Difference Equations
(c) Φ(n)(I − P ) = Φ2(n), where Φ2(n) = I − Φ1(n). (d) Φ1(n)P = Φ1(n), Φ2(n)(I − P ) = Φ2(n).
7. Prove Lemma 8.21.
8. Prove formula (8.3.8) using mathematical induction on i.
9. Prove that the solution y(n) of (8.3.1) defined by (8.3.6) is bounded forn≥n0 ≥0.
10. Prove that under the assumption of Theorem 8.19, for each bounded solution y(n) of (8.3.1), there exists a bounded solution x(n) of (8.3.2).
*11. (Open Problem). Improve Theorem 8.19 by relaxing condition (8.3.5), requiring only conditional convergence of ∞n=n0 B(n).
*12. (Hard). Extend Theorem 8.19 to the case where D(n) is a constant matrix in a one-block Jordan form, then extend it to the case when D(n) is a constant matrix in the general Jordan form.
*13. (Hard). Extend Theorem 8.19 to the case where D(n) has an eigenvalue equal to zero.
*14. (Open Problem). Suppose that there are r distinct eigenvalues λ1(n), λ2(n),. . . , λr(n) with distinct moduli. Prove that with the conditions of Theorem 8.19 holding for 1 ≤ i ≤ r, there are solutions yi(n), 1 ≤ i ≤ r, of system equation (8.3.1) that satisfy formula (8.3.12).
8.4 High-Order Difference Equations
In this section we turn our attention to the kth-order scalar equations of the form
y(n+k)+(a1 +p1(n))y(n+k−1)+···+(ak +pk(n))y(n)=0, (8.4.1)
where ai ∈ R and pi(n),1 ≤ i ≤ k, are real sequences. As we have seen in Chapter 3, (8.4.1) may be put in the form of a k-dimensional system of first-order difference equations that is asymptotically constant. Thus we are led to the study of a special case of (8.4.1), namely, the asymptotically constant system
y(n + 1) = [A + B(n)]y(n), (8.4.2)
where A is a k × k constant matrix that is not necessarily diagonal. This system is, obviously, more general than the system induced by (8.4.1). The first asymptotic result concerning system equation (8.4.2) is a consequence of Theorem 8.23.
Theorem 8.25 [9]. Suppose that the matrix A has k linearly independent eigenvectors ξ1, ξ2, . . . , ξk and k corresponding eigenvalues λ1, λ2, . . . , λk.
Then we obtain
or
Tz(n + 1) = [A + B(n)]Tz(n),
z(n + 1) = [D + B ̃(n)]z(n),
8.4 High-Order Difference Equations 361
If condition (8.3.16) holds for B(n), then system equation (8.4.2) has solutions yi(n), 1 ≤ i ≤ k, such that
(8.4.3) Proof. In order to be able to apply Theorem 8.23 we need to diagonalize
the matrix A. This may be accomplished by letting y = Tz
in (8.4.2), where
T = (ξ1,ξ2,…,ξk), that is, the ith column of T is ξi.
(8.4.4)
(8.4.5)
(8.4.6)
yi(n) = [ξi + o(1)]λni .
where D = T−1AT = diag(λ1,λ2,…,λk) and B ̃(n) = T−1B(n)T. It is easy to see that B ̃(n) satisfies condition (8.3.16). Now formula (8.4.3) follows by applying Theorem 8.23. P
Example 8.26. Find an asymptotic estimate of a fundamental set of solutions of
where
y(n + 1) = [A + B(n)]y(n), ⎛⎞
221
A = ⎜⎝ 1 3 1 ⎟⎠ ,
(8.4.7)
1 2 1, ⎛1/n2+1 0
(0.5)n ⎞ B(n)=⎜⎝ 0 (0.2)n 0 ⎟⎠.
e−n 0 logn/n2
Solution The eigenvalues of A are λ = 5,λ = 1, and λ = 1, and the cor- 123
⎛⎞⎛⎞⎛⎞
111 responding eigenvectors are ξi = ⎜⎝1⎟⎠,ξ2 = ⎜⎝ 0 ⎟⎠, and ξ3 = ⎜⎝ 0 ⎟⎠.
1 −1 −2 Furthermore, B(n) satisfies condition (8.3.16). Thus by Theorem 8.25,
362 8. Asymptotic Behavior of Difference Equations
equation (8.4.7) has the solutions
11
y1(n) = (1 + o(1))(5n) ⎜⎝1⎟⎠ ∼ ⎜⎝1⎟⎠ (5n),
11
⎛⎞⎛⎞
11 y2(n)=(1+o(1))⎜⎝ 0 ⎟⎠∼⎜⎝ 0 ⎟⎠,
−1 −1 ⎛⎞⎛⎞ 11
y3(n)=(1+o(1))⎜⎝ 0 ⎟⎠∼⎜⎝ 0 ⎟⎠. −2 −2
Next, we apply Theorem 8.23 to establish the following asymptotic result for (8.4.1).
Corollary 8.27. Suppose that the polynomial p(λ)=λk +a1λk−1 +···+ak
hasdistinctrootsλ1,λ2,…,λk andthat
∞ n=1
Then (8.4.1) has k solutions y1(n),y2(n),…,yk(n) with yi(n) = [1 + o(1)]λni .
(8.4.8)
(8.4.9)
|pi(n)|<∞, for 1≤i≤k.
(8.4.10) Proof. First we put (8.4.1) into the form of a k-dimensional system
where
z(n + 1) = [A + B(n)]z(n),
⎛01...0⎞
(8.4.11)
⎜0 0 1 0⎟ A=⎜ . . ⎟,
⎛⎞ ⎛⎞
⎝. .⎠
⎛−ak −ak−1 ... −a1 ⎞
00...0 B(n)=⎜⎝0 0...0⎟⎠,
−pk(n) −pk−1(n) . . . −p1(n) z(n) = (y(n), y(n + 1), . . . , y(n + k − 1))T .
Notice that polynomial (8.4.8) is the characteristic polynomial of the matrix
A. Furthermore, for each eigenvalue λi there corresponds the eigenvector
ξi = (1, λi, λ2i , . . . , λk−1)T . In addition, the matrix B(n) satisfies condition i
zi(n)=⎜ ⎜
⎜ 2 ⎟ yi(n+2) ⎟=(1+o(1))λni ⎜ λi ⎟.
⎛
⎜ ⎜
yi(n) ⎞
yi(n+1) ⎟ ⎟
⎛1⎞ ⎜ λi ⎟
8.4 High-Order Difference Equations 363
(8.3.16). Hence one may apply Theorem 8.25 to conclude that there are k solutions z1(n), z2 (n), . . . , zk(n) of (8.4.11) such that, for 1 ≤ i ≤ k,
. ⎟ ⎜ . ⎟ ⎝.⎠⎝.⎠
yi(n + k − 1) λk−1 i
Hence yi(n) = [1 + o(1)]λni .
Example 8.28. Find asymptotic estimates of fundamental solutions to
the difference equation
−n−2 1
y(n+3)− 2+e y(n+2)− 1+n2+1 y(n+1)+2y(n)=0.
P
Solution The characteristic equation is given by λ3 − 2λ2 − λ + 2 = 0 with
roots λ1 = 2,λ2 = 1, λ3 = −1. Notice that p1(n) = −e−n−2,p2(n) =
− 1 , and p (n) = 0 all satisfy condition (8.4.8). Hence Corollary 8.27 n2+1 3
applies to produce solutions y1(n),y2(n), and y3(n) defined as follows: y1(n) = [1 + o(1)]2n, y2(n) = 1 + o(1), y3(n) = [1 + o(1)](−1)n.
Corollary 8.27 is due to Evgrafov. It says that for each characteristic root of polynomial (8.4.8), at least one solution behaves as in formula (8.4.10), provided that the rate of convergence of the coefficients is not too slow.
What happens if all the roots of the characteristic equation (8.4.8) are equal? This same question was addressed by Coffman [22], where he obtained the following result.
Theorem 8.29. Suppose that the polynomial (8.4.8) has a k-fold root of 1 and that
(8.4.12)
Then (8.4.1) has k solutions y1(n),y2(n),...,yk(n) with
yi(n) = ni−1(1 + o(1)), n → ∞. (8.4.13)
We remark here that the actual result of Coffman is stronger than the statement of Theorem 8.29. Indeed, he proved that
∞
nk−1|pi(n)|<∞, for 1≤i≤k.
n=1
⎧
⎪ n + o i − m
for 1 ≤ m ≤ i,
for i ≤ m ≤ k − 1.
⎪⎨i−m n ∆myi(n)=⎪ i−m
⎪⎩o
n
364 8. Asymptotic Behavior of Difference Equations
The curious reader might wonder whether Coffman’s theorem (Theorem 8.29) applies if the polynomial (8.4.8) has a k-fold root not equal to 1. Luckily, by a very simple trick, one is able to do exactly that. Assume that the characteristic equation (8.4.8) has a k-fold root μ ̸= 1. Then polynomial (8.4.8) may be written as
(λ − μ)k = 0. (8.4.14) Letting y(n) = μnx(n) in (8.4.1), we obtain
μn+kx(n+k)+μn+k−1(a1+p1(n))x(n+k−1)+···+μn(ak+pk(n))x(n) = 0, or
x(n+k)−1(a1+p1(n))x(n+k−1)+···+ 1 (ak+pk(n))x(n) = 0. (8.4.15) μ μk
The characteristic equation (8.4.15) is given by λk+a1λk−1+a2λk−2+···+ak =0,
which has a k-fold root λ = 1. Moreover, if pi(n),1 ≤ i ≤ k, satisfies condition (8.4.1), then so does (1/μi)pi(n). Hence Theorem 8.29 applies to (8.4.15) to yield solutions x1(n),x2(n),...,xk(n) with
xi(n) = ni−1(1 + o(1)), n → ∞.
Consequently, there are solutions y1 (n), y2 (n), . . . , yk (n) of (8.4.1) such that yi(n) = ni−1(1 + o(1))μn.
We now summarize the above observations in the following corollary.
Corollary 8.30. Suppose that the polynomial (8.4.8) has a k-fold root μ and that condition (8.4.12) holds. Then (8.4.1) has k solutions y1(n), y2(n), . . . , yk(n) such that
(8.4.16) Example 8.31. Investigate the asymptotic behavior of solutions of the
μμ2 μk
yi(n) = ni−1(1 + o(1))μn.
difference equation
y(n+3)−(6+e−n−2)y(n+2)+ 12−(n+1)4 y(n+1)−8y(n)=0.
Solution The characteristic equation is given by λ3 − 6λ2 + 12λ − 8 = 0 with roots λ1 = λ2 = λ3 = 2. Also, p1(n) = −e−n−2, p2(n) = −1/(n + 1)4, and p3(n) = 0 all satisfy condition (8.4.12). Hence, by Corollary 8.30 there are three solutions y1(n) = (1 + o(1))2n, y2(n) = n(1 + o(1))2n, and y3(n) = n2(1 + o(1))2n.
Example 8.32. Consider the difference equation
x(n + 2) + p1(n)x(n + 1) + p2(n)x(n) = 0, (8.4.17)
1
where
and where
p1(n)̸=0, n≥n0 ≥0, lim 4p2(n) =p
x±(n) ∼ where ν = √1 − p.
Solution Let
−2
x(n) =
p1(j) 1 ± ν ∓ 2ν ⎛⎞
,
j =n0
8.4 High-Order Difference Equations 365
n→∞ p1(n)p1(n − 1) exists. Let α(n) be defined by
α(n) = 4p2(n) p1(n)p1(n − 1)
Assume that p ̸= 0, p < 1, and ∞
|α(j)| < ∞. Show that (8.4.17) has two solutions,
− p.
(8.4.18)
(8.4.19)
(8.4.20)
(8.4.21)
(8.4.22)
(8.4.23)
nn−1 1 α(j)
j =n0
−2
p1(j) y(n).
Then (8.4.17) is transformed to
y(n + 2) − 2y(n + 1) + (p + α(n))y(n) = 0.
n n−2 1⎝⎠
Letz(n)=(z1(n),z2(n))T =(y(n),y(n+1))T.Then(8.4.23)maybeput
into a system of the form
z1(n + 1) = 0 1 z1(n) . (8.4.24) z2(n+1) ν2 −1−α(n) 2 z2(n)
Again we let
z1(n) = 1 1 u1(n) . z2(n) −(ν − 1) ν + 1 u2(n)
Then (8.4.24) becomes
u1(n + 1) = (1 − ν + (α(n)/2ν)) α(n)/2ν u1(n) . u2(n + 1) −α(n)/2ν (1 + ν − (α(n)/2ν)) u2(n)
j =n0
(8.4.25)
366 8. Asymptotic Behavior of Difference Equations
If we let u(n) = (u1 (n), u2 (n))T , then we may write (8.4.25) in the form
where
Hence
j =n0
D(n)= B(n) =
u(n + 1) = (D(n) + B(n))u(n),
(1 − ν + α(n)/2ν) 0
0 (1+ν−(α(n)/2ν)) ,
0 α(n)/2ν −α(n)/2ν 0 .
(8.4.26)
By Theorem 8.23, there are two solutions of (8.4.26) given by
u+(n) ∼ u−(n) ∼
These two solutions produce two solutions of (8.4.24),
y+ (n) z+(n) =
y+(n + 1)
= −(ν−1) ν+1
y− (n) z−(n) =
j =n0
(1−ν+(α(j)/2ν)) 0 ,
Using (8.4.22) we obtain
x+(n) ∼
n−1 1
y−(n + 1)
⎡ ⎤
n−1
⎣ ⎦1
(1 − ν + (α(j)/2ν)) 0 , ⎡ ⎤
j =n0 n−1
⎣ ⎦0
j =n0
(1 + ν − (α(j)/2ν)) 1
.
⎛⎞
n−1 11⎝⎠1
⎛⎞
n−1 11⎝⎠1
= −(ν−1) ν+1 n−1
y+(n) ∼ y−(n) ∼
(1+ν−(α(j)/2ν)) 0 .
j =n0 n−1
(1 − ν + (α(j)/2ν)), (1 + ν − (α(j)/2ν)).
j =n0
n n−2
−2 p1(j) (1 − ν + (α(j)/2ν)). (8.4.27)
j =n0 j =n0
x−(n) ∼
(See Exercises 8.4, Problem 11.)
−2
j =n0
8.4 High-Order Difference Equations 367
Similarly, one may show that
n n−2 n−1 1
Exercises 8.4
In Problems 1 through 4 find an asymptotic estimate of a fundamental set of solutions of the given equation y(n + 1) = [A + B(n)]y(n).
p1(j) (1 + ν − (α(j)/2ν)). (8.4.28) j =n0
⎛e−n 0⎞ 1.A= 2 0 , B(n)=⎝ 1 n⎠.
0 3 (n+1)2 (0.1)
0 e−n−1 16
2.A=52,B(n)=2−n n. en
0 e (n+1)3 ⎛5 4 2⎞ ⎛ 0 (0.2)n 0 ⎞
⎛3−n 02−n⎞
⎛⎞
−1 0 0 ⎜ sinn ⎟
3. A=⎜⎝ 0 1 0⎟⎠, B(n)=⎜(n+1)2 0 0 ⎟. 004 ⎝ −n 1⎠
4. A = ⎜⎝4 5 2⎟⎠, B(n) = ⎜⎝(0.1)n 0 e−n2⎟⎠. 222010
of the given equation.
1
5. y(n+2)−(5+e−n)y(n+1)+ 6−(n+1)2 y(n)=0.
6. y(n+2)−(4+ne−n)y(n)=0.
7. y(n+2)+(4+ne−n)y(n)=0.
8. y(n+3)−6y(n+2)+(11+(sinn)e−n)y(n+1)−6y(n)=0. 9. y(n+3)−(3+2−n)y(n+2)+3y(n+1)−y(n)=0.
10. y(n+3)−15y(n+2)+75y(n+1)−(125+(0.1)n)y(n)=0.
11. Complete the proof of Example 8.32 by verifying formula (8.4.28).
n2 + 1
In Problems 5 through 10 investigate the asymptotic behavior of solutions
368 8. Asymptotic Behavior of Difference Equations
*12. Consider the second-order difference equation
x(n + 2) + p1(n)x(n + 1) + p2(n)x(n) = 0.
Assume that p1(n) ̸= 0 for n ≥ n0 and that: (i) limn→∞ 4p2(n)/(p1(n)p1(n − 1)) = p,
(ii) ∞n=n0 |α(n)| < ∞, where
α(n) = [4p2(n)/(p1(n)p1(n − 1)] − p).
x±(n)∼ −2 where ν = √1 − p.
(8.4.29)
(8.4.30)
If p is neither 0 nor 1, show that (8.4.29) has two solutions ⎛⎞
n n−2 1⎝⎠n
j =n0
p1(j) (1±ν) , (n→∞),
13. In Problem 12, suppose that p = 1 and that all the assumptions there hold except that the condition ∞n=n0 |α(n)| < ∞ is replaced by ∞n=n0 n|α(n)| < ∞.
Show that there are two solutions x1(n) ∼ −1n n−2 p1(j) and 1nn−2 2 j=n0
x2(n)∼n −2 j=n0 p1(j),n→∞.
14. Consider the difference equation (8.4.29) such that p1(n) ̸= 0 for n ≥ n0. Assume that limn→∞(p2(n))/(p1(n)p1(n − 1)) = 0 and α(n) = (p2(n))/(p1(n)p1(n − 1)).
(a) Use the transformation x(n) = −1n n−2 p1(j)z(n) to trans- 2 j=n0
form (8.4.29) to z(n + 2) − 2z(n + 1) + α(n)z(n) = 0.
(b) Show that (8.4.29) has two solutions x1 (n) ∼ (−1)n n−2 p1 (j )
and x2(n) = o(νn|x1(n)|) for any ν with 0 < ν < 1.
*15. Consider the difference (8.4.17) with conditions (8.4.17) and (8.4.18) satisfied. If p is real and p > 1, show that formula (8.4.21) remains valid if we assume that
∞
j|α(j)| < ∞. (8.4.31)
j =n0
*16. Show that formula (8.4.21) remains valid if one replaces hypothesis
(8.4.20) by
∞
|α(j)|σ < ∞ (8.4.32)
j =n0
for some real number σ with 1 ≤ σ ≤ 2.
j =n0
8.5
∞
|α(j)|σjτ−1
j =n0
for some real numbers σ and τ such that 1 ≤ σ ≤ 2 and τ > σ.
Second-Order Difference Equations
−2 Then (8.5.1) is transformed to
p1(j)
y(n).
y(n) = 0.
(8.5.2)
(8.5.3)
(8.5.4)
8.5 Second-Order Difference Equations 369
*17. Show that the conclusions of Problem 16 remain valid if condition (8.4.32) is replaced by
The asymptotics of second-order difference equations play a central role in many branches of pure and applied mathematics such as continued frac- tions, special functions, orthogonal polynomials, and combinatorics. In this section we will utilize the special characteristics of second-order equations to obtain a deeper understanding of the asymptotics of their solutions. Consider the difference equation
x(n + 2) + p1(n)x(n + 1) + p2(n)x(n) = 0. (8.5.1) One of the most effective techniques to study (8.5.1) is to make the
change of variables
x(n) =
⎛⎞ n−1 n−2
1⎝⎠
y(n + 2) − 2y(n + 1) +
q = lim 4p2(n) , n→∞ p1(n)p1(n − 1)
j =n0
4p2(n) p1(n)p1(n − 1)
Put
α(n) = 4p2(n) − q. p1(n)p1(n − 1)
Then the characteristic roots associated with (8.5.3) are λ1 = 1 − √1 − q andλ2 =1+√1−q.
Here there are several cases to consider:
Case I. If −∞ < q < 1, then λ1 and λ2 are real distinct roots with
|λ1| ̸= |λ2|. Case I may be divided into subcases.
(a) If α(n) → 0, then by invoking the Poincar ́e–Perron theorem we obtain
two linearly independent solutions y1(n) and y2(n) such that
lim y1(n+1) =λ1, lim y2(n+1) =λ2. (8.5.5)
Although this does not provide us with explicit representations of the solutions y1(n) and y2(n) it does guarantee the existence of a special
n→∞ y1 (n) n→∞ y2 (n)
370
8. Asymptotic Behavior of Difference Equations
solution, called a minimal solution. As we will see later, minimal solu- tions play a central role in the convergence of continued fractions and the asymptotics of orthogonal polynomials.
Definition 8.33. A solution φ(n) of (8.5.1) is said to be minimal (subdominant, recessive) if
for any solution x(n) of (8.5.1) that is not a multiple of φ(n). A non- minimal solution is called dominant. One may show that a minimal solution is unique up to multiplicity (Exercises 8.5, Problem 1).
Returning to (8.5.3), let us assume that |λ1| < |λ2|. Then there exist μ1,μ2 such that |λ1| < μ1 < μ2 < |λ2|. By (8.5.5) it follows that, for sufficiently large n,
(b)
|α(n)| < ∞, then by Corollary 8.27, (8.5.3) has a fundamental set of solutions y1(n) and y2(n) such that
which implies that
lim 1
Hence
y1(n) = λn1 (1 + o(1)), y2(n) = λn2 (1 + o(1)). ⎛ ⎞
lim φ(n) = 0 n→∞ x(n)
|y1(n+1)| ≤μ1 and |y2(n+1)| ≥μ2.
Hence
|y1 (n)|
|y1(n)| ≤ μn1 |y1(0)|,
= lim
|y2 (n)|
|y2(n)| ≥ μn2 |y2(0)|,
μ n |y (0)|
1 1 =0. (8.5.6)
|y (n)| n→∞ |y2(n)|
μ2 |y2(0)|
n→∞
Therefore, y1(n) is a minimal solution of (8.5.3) (Why?) (Exercises
8.5, Problem 1).
If α(n) ∈ l1(Z+), that is,
∞ n0
n−1 n−2 1⎝⎠n
x1(n)= −2
⎛ ⎞
x2(n)= −2
p1(j) λ1(1+o(1)), 1⎝⎠n
j =n0
j =n0 n−1 n−2
p1(j) λ2(1+o(1)).
Notice that we have treated this case thoroughly in Example 8.32, where we obtained formulas (8.4.27) and (8.4.28).
x1(n) =
x2(n) =
−2 1⎝⎠
1⎝⎠
8.5 Second-Order Difference Equations 371
(c) Suppose that α(n) ∈ l2(Z+), that is, ∞n0 α2(n) < ∞. Then using the scheme of Example 8.32, Elaydi [38] showed that (8.5.1) has two linearly independent solutions x1(n), x2(n) obeying formulas (8.4.27) and (8.4.28). In other words,
⎛⎞
n n−2 n−1
(1 − ν + (α(j)/2ν))(1 + o(1)), (8.5.7)
j =n0 j =n0
(8.5.8)
Moreover, x1(n) is a minimal solution, and x2(n) is a dominant solution.
We remark here that the above results may be extended to systems as well as to higher-order difference equations [38].
CaseII.Ifq=1,thenλ1 =λ2 =1.InthiscaseweuseCoffman’sresult (Theorem 8.29) to produce two solutions of (8.5.3),
y1(n) ∼ 1 and y2(n) ∼ n,
n|α(n)| < ∞.
Case III. If q > 1, then λ1 and λ2 are complex conjugates λ ̄1 = λ2,
|λ1| = |λ2|. In this case we use another result from [43].
Theorem 8.34 [43]. Suppose that q > 1 and the following condition
p1(j) ⎛⎞
j =n0
provided that
∞ n0
holds.
j =n0 n n−2
n−1
−2 p1(j) (1 + ν − (α(j)/2ν))(1 + o(1)).
∞
|α(n + 1) − α(n)| < ∞.
n=n0
Then (8.5.3) has two solutions y1(n), y2(n) with
where
yi(n) = (1 + o(1))
n−1
m=n0
βi(m), i = 1, 2, . . . ,
1 − q + α(n) is semidefinite for a fixed branch of the square root (0 ≤ arg √z < π).
provided that Re
(8.5.9)
β1(n)=1− 1−q+α(n), β2(n)=1+ 1−q+α(n),
372 8. Asymptotic Behavior of Difference Equations
8.5.1 A Generalization of the Poincar ́e–Perron Theorem
[57] In many applications related to (8.5.1) the coefficients p1(n) and p2(n) are of the form
p1(n)∼anα, p2(n)∼bnβ, ab̸=0, α,βreal; n→∞.
The asymptotics of the solutions of (8.5.1) can be determined by means of the Newton–Puiseux diagram formed with the points P0(0, 0), P1(1, α), P2(2, β) (Figure 8.1)
Theorem 8.35 [114], [82].
(a) If the point P1 is above the line P0P2 (i.e., α > β/2), then (8.5.1) has
a fundamental set of solutions x1(n) and x2(n) such that
Moreover, x2(n) is a minimal solution.
(8.5.10)
lim x1(n+1)=−anα, n→∞ x1 (n)
lim x2(n + 1) = −bn(β−α). n→∞ x2(n) a
(b) Suppose that the points P0, P1, P2 are collinear (i.e., α = β/2). Let λ1, λ2 be the roots of the equation λ2+aλ+b = 0, such that |λ1| ≥ |λ2|. Then (8.5.1) has a fundamental set of solutions x1(n) and x2(n) with
(8.5.11) provided that |λ1| ̸= |λ2|. Moreover, x2(n) is a minimal solution.
lim x1(n+1)=λ1nα, n→∞ x1 (n)
lim x2(n+1)=λ2nα, n→∞ x2 (n)
If |λ1| = |λ2|, then
for all nontrivial solutions x(n) of (8.5.1).
(c) If the point P1 lies below the line segment P0P2, then
(8.5.12)
(8.5.13)
|x(n)| 1/n
lim sup α =|λ1|
n→∞ (n!)
|x(n)| 1/n lim sup β/2 = |b|
n→∞ (n!)
for all nontrivial solutions of (8.5.1).
Proof. Let p1(n) = anα + ν(n), P2(n) = bnβ + μ(n), where ν(n) and
μ(n) are null sequences. Then we may write (8.5.1) as x(n+2)+(anα+ν(n))x(n+1)+(bnβ+μ(n))x(n)=0. (8.5.14)
8.5 Second-Order Difference Equations 373
α
β /2 β
P 0012
FIGURE 8.1. Newton–Puiseux diagram for (8.5.14).
Making the change of variable x(n) = (n!)αy(n) in (8.5.14) yields n α ν(n)
y(n+2)+ a n+2 +(n+2)α y(n+1)
bnβ μ(n)
P 1
P 1
+ (n+1)α(n+2)α +(n+1)α(n+2)α y(n)=0. (8.5.15)
(a) If 2α > β, the characteristic equation of (8.5.14) is λ2 + aλ = 0. The first solution x1(n) in (8.5.1) corresponds to λ1 = −a in the Poincar ́e– Perron theorem. The second solution x2(n) may be obtained by using formula (2.2.18) and is left to the reader as Exercises 8.5, Problem 2.
The proofs of parts (b) and (c) are left to the reader as Problem 2. P Remark: The above theorem is valid for kth-order scalar difference
equations. The interested reader may consult [113], [82], [146].
Exercises 8.5
1. Consider a kth-order scalar difference equation of Poincar ́e type (8.2.7) such that its characteristic roots have distinct moduli.
(a) Show that the equation has a minimal solution.
(b) Show that the minimal solution is unique up to multiplicity.
2. Complete the proofs of parts (a), (b), (c) in Theorem 8.35.
3. Investigate the asymptotic behavior of the equation 1
y(n+2)−2y(n+1)+ 1−n3 y(n)=0.
4. Investigate the asymptotic behavior of solutions of the equation
2 (−1)n
∆ y(n)= nα+1 y(n+1)
where α > 1.
5. Investigate the asymptotic behavior of solutions of
∆2y(n)= p(n)y(n+1), nα+1
374 8. Asymptotic Behavior of Difference Equations
where α > 1, and
n
p(j) ≤ M < ∞ j=1
for all n > 1.
6. Show that the difference equation
∆2x(n) = p(n)x(n + 1)
has two linearly independent solutions x1(n) and x2(n) such that
det x1(n) x2(n) = −1. ∆x1 (n) ∆x2 (n)
7. (Multiple Summation). Show that for any sequence f(n), n ∈ Z+, (8.5.16)
n−1 r−1 n−1
r=n0 j=n0
f(j) = (n − j)f(j). j=n0
8. Consider the second-order difference equation [34], [35] ∆2y(n) + p(n)y(n) = 0
such that
∞ j=1
Show that (8.5.17) has two solutions y1(n) ∼ 1 and y2(n) ∼ n as n → ∞, without using Coffman’s theorem (Theorem 8.29). You may use the following steps:
r
n−1
j=r
j|p(j)| < ∞.
(8.5.17)
(8.5.18)
nj
0
0
n−1
n
FIGURE 8.2.
(a) Use ∆2y(n) = −p(n)y(n) to show that n−1 r−1
y(n) = c1 + c2(n) − (b) Use formula (8.5.16) to show that
8.5 Second-Order Difference Equations 375
n−1 |y(n)|
|y(j)| n ≤|c1|+|c2|+ j|p(j)| j .
*9. (Generalized Gronwall’s Inequality). Suppose that n−1
c(j)uγ (j),
where 1 ̸= γ > 0, a ≥ 0, b > 0, c(j) > 0, and u(j) > 0, for j ≥ n0.
Prove that
u(n) ≤
⎤1/(1−γ ) c(j) ,
a + b(1 − γ) provided that, for γ > 1,
n−1 j =n0
(8.5.19)
u(n) ≤ a + b
⎡
⎣1−γ ⎦
n−1 1−γ
a + b(1 − γ)
*10. Generalize the result of Problem 8 to the so-called Emden–Fowler
j =n0
∆2y(n)+p(n)|y(n)|γ sgny(n)=0, where γ ̸= 1 is a positive real number, and
equation
sgn y(n) =
1 if y(n) > 0, −1 if y(n) < 0.
j =n0
r=1 j=1
c(j) > 0 for n ≥ n0.
p(j)y(j).
j=1
(c) Use the discrete Gronwall’s inequality (Lemma 4.32) to show that
|y(n)| ≤ c3n. n−1
p(j)y(j)
lim ∆y(n) = c2 − M. n→∞
(d) Substitute back into
∆y(n) = c1 −
to obtain
j=1
(8.5.20)
376
8. Asymptotic Behavior of Difference Equations
Show that if
∞
jγp(j) = M < ∞,
j =n0
then each solution y(n) with the initial condition y(n0) with
∆y(n0)+ n
is such that y(n) ∼ n as n → ∞. You may use the following steps.
nA(n) + B(n).
(b) Show that
∆A(n)=p(n)[|nA(n)+B(n)|]γ sgny(n), ∆B(n)=(n+1)p(n)[|nA(n)+B(n)|]γ sgny(n).
(c) Use the antidifference operator ∆−1 to obtain A(n) and B(n) and then use the generalized Gronwall’s inequality.
(d) Suppose that x1(n) and x2(n) are two linearly independent solutions of the equation
∆2x(n) = p(n)x(n + 1).
In addition, assume that for a sequence q(n) we have
∞
|q(j)|u(j) = M < ∞,
j =n0 where for a specific m ∈ Z+,
u(n) = max |x1(n + 1)||x1(n)|2m+1, |x1(n + 1)||x2(n)|2m+1, |x2(n + 1)||x1(n)|2m+1, |x2(n + 1)||x2(n)|2m+1 .
Show that there exist solutions y(n) of the equation ∆2y(n) = p(n)y(n + 1) + q(n)y2m+1(n)
y ( n 0 ) 1 − γ
−∆y(n0) +2(1−γ)M >0
(a) Let A(n) = ∆y(n), B(n) = y(n) − n∆y(n). Show that y(n) =
such that
with
y(n) = α(n)x1(n) + β(n)x2(n), lim α(n) = a, lim β(n) = b,
n→∞ n→∞ for some constants a, b.
0
8.6 Birkhoff’s Theorem
Consider again the second-order difference equation
x(n + 2) + p1(n)x(n + 1) + p2(n)x(n) = 0,
where p1(n) and p2(n) have asymptotic expansions of the form
(8.6.1)
( 8 . 6 . 2 )
with b0 ̸= 0.
The characteristic equation associated with (8.6.1) is λ2 + a0λ + b0 = 0
p 1 ( n ) ∼ ∞ a j ,
p 2 ( n ) ∼ ∞ b j ,
8.6 Birkhoff’s Theorem
377
with roots
λ1,λ2=−1a0± 1a20−b0. (8.6.3) 24
nj j=0
nj j=0
Extensive work by Birkhoff [11], [12] Birkhoff and Trjitzinsky [13], and Adams [2] has been done concerning the asymptotics of equations of type (8.6.1) with expansions (8.6.2). Due to the limitations imposed by the intro- ductory nature of this book, we will restrict our exposition to second-order difference equations. Our presentation here follows closely the excellent papers by Wong and Li [147], [148].
Theorem 8.36 (Birkhoff–Adams).
(a) If λ1 ̸= λ2, i.e., a20 ̸= 4b0, then equation (8.6.1) has two linearly inde- pendent solutions x1(n),x2(n), which will be called normal solutions, of the form
ci(j) = 0, ci(0) = 1. In particular, we obtain
∞ c ( r )
xi(n)∼λninαi i ,i=1,2, r=0 nr
αi=a1λi+b1, i=1,2, a0λi + 2b0
(8.6.4)
(8.6.5)
(8.6.6)
s−1⎡s ⎤
⎣2s−j αi−j
αi−j ⎦ +λi r−j as−r +bs−j
r=j
λi2 s−j j=0
ci(1)= −2λ2iαi(αi −1)−λi(a2 +λia1 +αi(αi −1)a0/2)−b2. 2λ2i(αi −1)+λi(a1 +(λi −1)a0)+b1
378 8. Asymptotic Behavior of Difference Equations
(b) Ifλ1 =λ2 =λbutλ=−1a0 isnotarootoftheequationa1λ+b1 =0 2
(i.e., 2b1 ̸= a0a1), then equation (8.6.1) has two linearly independent solutions, x1(n), x2(n), which will be called subnormal solutions, of the form
√ ∞c(j) xi(n) ∼ λneγi nnα i
, i = 1,2,
where
α= 1+ b1 , 4 2b0
24b20 γi
(The general recursive formula for ci(n) is too complicated to be
included here. The interested reader is referred to Wong and Li [147].) (c) If λ1 = λ2 = λ and 2b1 = a0a1, then we consider the equation
α(α−1)λ2 +(a1α+a2)λ+b2 =0.
Let α1, α2 (Reα2 ≥ Reα1) be the roots of this equation. Then there
are three subcases to contemplate.
(c1) If α2 − α1 ̸= 0, 1, 2, . . . , then equation (8.6.1) has two linearly independent solutions of the form
(8.6.10)
(c2) If α2 − α1 = 1, 2, . . . , then equation (8.6.1) has two solutions, x1(n) given by (8.6.10) and x2(n) = z(n)+c(ln n)x1(n), where c is a constant that may be zero, and
(8.6.11) (c3 ) If α2 = α1, then equation (8.6.1) has two solutions: x1(n) given by
a0a1 −2b1, γ2 =−2 2b0
(8.6.7)
a0a1 −2b1, (8.6.8) 2b0
γ1 =2 c0 = 1, ci(1) = 1
(a20a21 − 24a0a1b0 + 8a0a1b1
− 24a0a2b0 − 9b20 − 32b21 + 24b0b1 + 48b0b2).
(8.6.9)
xi(n) ∼ λnnαi i . j=0 nj
(8.6.10), and x2(n) = z(n) + c(ln n)x1(n), c ̸= 0,
where r is an integer ≥ 3.
j=0 nj/2
z(n) ∼ λnnα2 s . s=0 ns
∞ c ( j )
∞ d
z(n) ∼ λnnα1−r+2 ∞ ds , s=0 ns
(8.6.12)
Proof. The main idea of the proof is to substitute the formal solu- tions (8.6.4) and (8.6.7) back into (8.6.1) and then compare coefficients of powers of n in the resulting expression. Details of the proof will not be included here, and we refer the interested reader to the paper of Wong and Li [147]. P
Example 8.37. The Ap ́ery Sequence
The sequence [141]
n2 2 n n+k
u(n) =
k=0k k
satisfies the second-order difference equation
(n+2)3u(n+2)−(34n3 +153n2 +231n+117)u(n+1)+(n+1)3u(n) = 0. (8.6.13)
Writing (8.6.13) in the form (8.6.1), we have
p1(n) = −(34n3 + 153n2 + 231n + 117)
8.6 Birkhoff’s Theorem 379
(n+2)3 = a0 + a1 + a2 + · · · ,
(8.6.14) (8.6.15)
To find a0 we just take the limit of both sides of (8.6.14) to obtain a0 = −34.
Subtracting a0 from both sides, multiplying by n, and then taking the limit
as n → ∞ yields a1 = 51. Repeating this process, we obtain a2 = −129.
Similarly, one may obtain b = 1, b = −3, b = 9. Hence by formula (8.6.3) 012√
the characteristic roots are λ1, λ2 = 17 ± 12 2. From formula (8.6.5) we
have
nn2
n+13 b1b2
p2(n)= n+2 =b0+n+n2+···.
√
51(17+12 2)−3 −3
α1 = (−34)(17+12√2)+2 = 2 , √
51(17−12 2)−3 −3 α2 = (−34)(17−12√2)+2 = 2 .
Hence we have two solutions u1(n) and u2(n) such that
√ c(1)c(2) u1(n)∼(17+12 2)nn−3/2 1+ 1 + 1 +··· ,
with c1(1) ≈ −15.43155325, and
√ c(1)c(2) u2(n)∼(17−12 2)nn−3/2 1+ 2 + 2 +··· ,
with c2(1) ≈ −1.068446129. Since u2(n) → 0, it follows that u(n) = cu1(n) for some constant c.
n n2
n n2
380 8. Asymptotic Behavior of Difference Equations
Example 8.38. Laguerre Polynomials [147]
Laguerre polynomials Lβn(x) are defined for β > −1,0 < x < ∞, by the
following formula, called Rodrigues’ formula:
n nm Lβn(x)= 1exx−β d e−x xn+α= (−1)m n+β x .
n! dxn
It can be shown (see Appendix F) that Lβn(x) satisfies a second-order
difference equation of the form
ny(n) + (x − 2n − β + 1)y(n − 1) + (n + β − 1)y(n − 2) = 0.
(8.6.16)
Writing this equation in the form (8.6.1) yields
y(n+2)+ x−2n−β−3y(n+1)+ n+β+1y(n)=0.
m=0
n−m m!
n+2 n+2
Following the procedure in the preceding example, we obtain
an =−2+x−β+1−2(x−β+1)+···, n n2
bn =1+β−1−2(β−1)+··· . n n2
The characteristic equation is λ2 − 2λ + 1 = 0, which has a multiple root
λ1 = λ2 = 1. This root does not satisfy (x−α+1)λ+α−1 = 0, and hence
we have two subnormal solutions of the form (8.6.7). Using formula (8.6.8)
weobtainα= 1,β−1,γ1 =2√xi,γ2 =−2√xi.Henceitfollowsfrom 24
formula (8.6.7) that we have two solutions yr(n) = e(−1)r+12√nxinβ/2−1/4 ∞ cr(j),
j=0 nj/2
with c1(0) = c2(0) = 1,
(−1)ri 2 2
√ 1 1 Lβn(x) = π−1/2ex/2x−β/2−1/4nβ/2−1/4 cos 2 nx − 2βπ − 4
+ O(nβ/2−3/4).
r = 1,2,
(8.6.17)
cr(1)= 48√x 4x −12β −24xβ−24x+3 , r=1,2. Thus y2(n) = y1(n). But we know from [98, p. 245] that
Thus
β 1 −1/2 x/2 −β/2−1/4 (βπ/2+π/4)i (βπ/2+π/4)i
Ln(x)=2π e x e ̄ y1(n)+e y2(n)
√ 1 1 ∞ A ( x )
8.6 Birkhoff’s Theorem 381
= π1/2ex/2x−β/2−1/4nβ/2−1/4 cos 2 nx − βπ − π s
2 4s=0ns/2
∞ +sin 2√nx−1βπ−1π Bs(x) ,
2. (a)
Verify that the sequence u(n) = n n4 satisfies the equation k=0 k
3 7 (n+2)3u(n+2)−12 n+ 2 n2 +3n+ 3 u(n+1)
3 5
−64 n+4 (n+1) n+4 u(n)=0.
2 4s=0ns/2 B1(x) = 48√x(4x − 12β − 24xβ − 24x + 3).
where A0(x) = 1, A1(x) = 0, B0(x) = 0, and 122
Remark: In [148] the authors extended their analysis to equations of the form
x(n + 2) + nrp1(n)x(n + 1) + nsp2(n)x(n) = 0, with r and s integers and p1(n), p2(n) of the form (8.6.2).
Exercises 8.6
1. (Binomial Sums)
(a) Verify that the sequence u(n) = n n3 satisfies the equation
k=0 k
(n+2)2u(n+2)−(7n2 +21n+16)u(n+1)−8(n+1)2u(n) = 0.
(b) Find an asymptotic representation of u(n).
(b) Find an asymptotic representation of u(n).
3. Find asymptotic representations for the solutions of the equation
(n+2)5u(n+2)−((n+2)5 +(n+1)5)u(n+1)+(n+1)5u(n)=0.
4. Find asymptotic representations for the solutions of the difference equation
u(n + 2) − u(n + 1) − (n + 1)u(n) = 0.
382
8. Asymptotic Behavior of Difference Equations
5.
6.
8.7
Find asymptotic representations for the solutions of the difference equation
(n+1)(n+2)x(n+2)−(n+1)[(2n+b+c+1)+z] ×x(n+1)+n+b)(n+c)x(n)=0, z̸=0.
Find asymptotic representations for the solutions of the second-order difference equation
(n+1)(n+2)y(n+2)−(n+1)(2n+2b+1)y(n+1)+(n+b)2y(n) = 0.
Nonlinear Difference Equations
In this section we consider the nonlinearly perturbed system
y(n + 1) = A(n)y(n) + f (n, y(n)) (8.7.1)
along with the associated unperturbed system
x(n + 1) = A(n)x(n), (8.7.2)
where A(n) is an invertible k × k matrix function on Z+ and f (n, y) is a function from Z+ × Rk → Rk that is continuous in y. Let Φ(n) be the fundamental matrix of system (8.7.2). The first step in our analysis is to extend the variation of constants formula (Theorem 8.19) to system (8.7.1). Since A(n) is not assumed here to be a diagonal matrix, we need to replace Definition 8.17 by a more general definition of dichotomy.
Definition 8.39. System (8.7.2) is said to possess an ordinary dichotomy if there exists a projection matrix P and a positive constant M such that
|Φ(n)PΦ−1(m)| ≤ M, for n0 ≤ m ≤ n,
|Φ(n)(I − P )Φ−1(m)| ≤ M, for n0 ≤ n ≤ m. (8.7.3)
Notice that if A(n) = diag(λ1 (n), . . . , λk (n)), then this definition reduces to Definition 8.17 if we let Φ1(n) = Φ(n)P and Φ2(n) = Φ(n)(I − P).
Theorem 8.40 [44], [121], [131]. Suppose that system (8.7.2) possesses an ordinary dichotomy. If, in addition,
and
∞ j =n0
|f (j, 0)| < ∞
(8.7.4)
(8.7.5)
|f(n, x) − f(n, y)| ≤ γ(n)|x − y|,
where γ(n) ∈ l1([n0,∞)) i.e., ∞j=n0 γ(j) < ∞, then for each bounded solution x(n) of (8.7.2) there corresponds a bounded solution y(n) of (8.7.1)
and vice versa. Furthermore, y(n) is given by the formula n−1
Proof.
y(n) = x(n) + −
−1
Φ(n)PΦ (j + 1)f(j,y(j))
j =n0
∞ j=n
8.7 Nonlinear Difference Equations 383
Φ(n)(I −P)Φ−1(j +1)f(j,y(j)). (8.7.6)
The proof mimics that of Theorem 8.19 with some obvious mod- ifications. Let x(n) be a bounded solution of (8.7.2). Define a sequence {yi (n)} (i = 1, 2, . . .) successively by letting y1 (n) = x(n) and
yi+1(n) = x(n) + −
n−1
−1
j =n0 ∞
j=n
(8.7.7)
Φ(n)PΦ (j + 1)f(j,yi(j)) Φ(n)(I −P)Φ−1(j +1)f(j,yi(j)).
We use mathematical induction to show that yi(n) is bounded on [n0,∞) for each i. First we notice that by assumption, |y1(n)| ≤ c1. Now suppose that |yi(n)| ≤ ci. Then it follows from (8.7.4), (8.7.5), and (8.7.7) that
where
n−1 ∞
−1 −1
∞ j =n0
|yi+1(n)|≤c1 +M
≤c1 +M⎣ ciγ(j)+M ̃⎦=ci+1,
[γ(j)|yi(j)|+|f(j,0)|] ⎡⎤
∞
|f (j, 0)| = M ̃ .
j =n0
∞ j =n0
Hence yi(n) is bounded for each i.
As in the proof of Theorem 8.19, one may show that the sequence {yi(n)}
converges uniformly on [n0, ∞) to a bounded solution y(n) of (8.7.1). Con- versely, let y(n) be a bounded solution of (8.7.1). Then one may verify easily that
y ̃(n) = Φ(n)PΦ (j+1)f(j,y ̃(j))− Φ(n)(I−P)Φ (j+1)f(j,y ̃(j)) j=n0 j=n
is another bounded solution of (8.7.1). Hence x(n) = y(n) − y ̃(n) is a bounded solution of (8.7.2). P
The preceding result does not provide enough information about the asymptotic behavior of solutions of system equation (8.7.1). In order to obtain such results we need one more assumption on (8.7.2).
384 8. Asymptotic Behavior of Difference Equations
Theorem 8.41 [44]. Let all the assumptions of Theorem 8.40 hold. If Φ(n)P → 0 as n → ∞, then for each bounded solution x(n) of (8.7.2) there corresponds a bounded solution y(n) of (8.7.1) such that
y(n) = x(n) + o(1), (8.7.8)
or
y(n) ∼ x(n).
Proof. The proof is similar to the proof of Theorem 8.20 and is left to
the reader as Exercises 8.7, Problem 7. P
Example 8.42. Consider the equation 2
y1(n+1) = 3 0 y1(n) + siny1(n)/n . (8.7.9) y2(n + 1) 0 1/2 y2(n) (1 − cos y2(n))/n2
Here
2 A(n) = 3 0 , f(n,y) = siny1/n .
0 1/2 (1 − cos y2)/n2
Using the Euclidean norm we obtain ∞j=1 |f(j,0)| = 0. Moreover, for
we have |f(n,x)−f(n,y)|= 1 sinx1 −siny1
x= x1 , y= y1 , x2 y2
n2 cosx2 −cosy2
= 1 (sin x1 − sin y1)2 + (cos x2 − cos y2)2. (8.7.10) n2
By the Mean Value Theorem,
|sin x1 − sin y1| = |cos c|, |x1 −y1|
for some c between x1 and y1,
and
≤ 1,
|cosx2 −cosy2| ≤1.
|x2 −y2| Hence substituting into (8.7.10), we obtain
|f(n,x)−f(n,y)|≤ 1|x−y|. n2
The associated homogeneous equation
x(n + 1) = A(n)x(n)
8.7 Nonlinear Difference Equations 385
has a fundamental matrix Φ(n) = 3n 0 and two solutions; one 0 (1/2)n
bounded, x1(n) = 0 (1)n, and one unbounded, x2(n) = 1 3n. If we let 120
the projection matrix be
then
00 P=,
01
00
0 (1/2)n
→ 0
Φ(n)P =
Hence all the conditions of Theorem 8.41 hold. Thus corresponding to
as n → ∞.
the bounded solution x1(n) = 0(1)n there corresponds a solution y(n) of
(8.7.9) such that
12
01n y(n)∼ 1 2 .
Next we specialize Theorem 8.41 to the following kth-order nonlinear equation of Poincar ́e type:
y(n+k)+(a1 +p1(n))y(n+k−1)+···+(ak +pk(n))y(n)=f(n,y(n)). (8.7.11)
Corollary 8.43. Suppose that the characteristic equation λk + a1λk−1 + ···+ak =0hasdistinctrootsλi,1≤i≤k,and∞n=1|pj(n)|<∞,1≤ j ≤ k. Assume further that conditions (8.7.4) and (8.7.5) hold. Then for each λj with |λj | ≤ 1 there corresponds a solution yj of (8.7.11) such that y j ( n ) ∼ λ nj .
Proof. By Corollary 8.27, the homogeneous part of (8.7.11), x(n+k)+(a1 +p1(n))x(n+k−1)+···+(ak +pk(n))x(n)=0,
has solutions x1(n),x2(n),...,xk(n) with xj(n) ∼ λnj . If |λj| ≤ 1, then xj(n) is bounded. Corresponding to this bounded solution xj(n) there is a solution yj (n) of (8.7.11) with yj (n) ∼ xj (n). Thus yj (n) ∼ λnj . P
Example 8.44. Investigate the asymptotic behavior of solutions of the equation
3 1 e−n y(n + 2) − 2y(n + 1) + 2y(n) = 1 + y2(n).
Solution The characteristic equation is given by λ2 − 3 λ + 1 = 0
(8.7.12)
with distinct roots
22
λ1=1, λ2=1. 2
386 8. Asymptotic Behavior of Difference Equations
Now,
∞ ∞ f(n,0)=
n=0 n=0
e−n <∞.
1
Moreover,
|f(n,x)−f(n,y)|=e−n1+x2 −1+y2
1
= e−n |x + y| · |x − y|
(1+x2 +y2 +x2y2) ≤ |x − y|.
Hence all the assumptions of Corollary 8.43 are satisfied. Consequently, (8.7.12) has two solutions y1(n) ∼ 1 and y2(n) ∼ (1)n.
2
Exercises 8.7
In Problems 1 through 3 investigate the asymptotic behavior of solutions of the equation y(n + 1) = A(n)y(n) + f (n, y(n)).
⎛1 ⎞ 1. A(n) = ⎜⎝2 0 ⎟⎠, f(n,y) = e−n cosy1 .
010 n+2
⎛1⎞ 2. A(n)= 2 3 , f(n,y)=⎝n3⎠.
3−1 0 −n
3. A(n)= 3 2 , f(n,y)= y1ne . 2 1 y2e−n
4. Study the asymptotic behavior of solutions of 3 1
siny(n) n2 .
y(n + 2) − 4y(n + 1) + 3y(n) = 1 , n ≥ 1. n2 + y2
6. Study the asymptotic behavior of solutions of y(n+2)+(1+e−n)y(n) = e−n .
7. Prove Theorem 8.41.
In Problems 8 through 10 investigate the asymptotic behavior of the difference equation.
y(n + 2) + 2 + n2 y(n + 1) − (1 + e−n)y(n) = 5. Study the asymptotic behavior of solutions of
n2
10. ∆2y(n) + 1 (y(n))1/3 = 0.
8.8 Extensions of the Poincar ́e and Perron Theorems 387
8. ∆2y(n) + e−ny3(n) = 0.
9. ∆2y(n) − 1 (|y(n)|)1/2 = 0.
n2
In Problems 11 and 12 we consider the nonlinear equation x(n + 1) = f (x(n))
with f(0) = 0, 0 < |f′(0)| < 1, and f′ continuous at 0. 11. Showthatthereexistδ>0and0<α<1suchthat
|x(n)| ≤ αn|x(0)|, n ∈ Z+, for all solutions x(n) of (8.7.13) with |x(0)| < δ.
(8.7.13)
12. Suppose that f′′ is bounded near 0, and |f′(x0)| < 1 for |x0| < δ. Prove that for any solution x(n) of (8.7.11) with |x(0)| < δ we have x(n) ∼ cx(0)(f′(0))n as n → ∞, where c depends on x(0).
13. Use Problem 12 to find an asymptotic representation of solutions of the equation
x(n + 1) = x(n)/(1 + x(n)), x(0) = 0.1.
14. Find an asymptotic representation of solutions of the equation
u(n+1)=u(n)+ 1 . u(n)
8.8 Extensions of the Poincar ́e and Perron Theorems
8.8.1 An Extension of Perron’s Second Theorem
Coffman [22] considers the nonlinear system
y(n + 1) = Cy(n) + f(n, y(n))
(8.8.1)
where C is a k×k matrix and f : Z+ ×Rk → Rk is continuous. The following result extends Perron’s Second Theorem to nonlinear systems.
Theorem 8.45 [22]. Suppose that ∥f (n,y)∥ → 0 as (n, y) → (∞, 0). If ∥y∥
y(n) is a solution of (8.8.1) such that y(n) ̸= 0 for all large n and y(n) → 0 as n→∞, then
lim n ∥y(n)∥ = |λi| (8.8.2) n→∞
for some eigenvalue λi of A. Moreover, |λi| ≤ 1.
388 8. Asymptotic Behavior of Difference Equations
Using the above theorem, Pituk [122] was able to improve Perron’s Sec- ond Theorem such that conclusion (8.2.12) is now valid for all nontrivial solutions. As a bonus we get a system version of this new result.
Consider again the perturbed linear system
y(n + 1) = [A + B(n)]y(n)
such that A is a k × k constant matrix and lim ∥B(n)∥ = 0.
n→∞
(8.8.3)
(8.8.4) Theorem 8.46 [122]. Under condition (8.8.4), for any solution y(n) of
(8.8.3), either y(n) = 0 for all large n or
lim n ∥y(n)∥ = |λi| (8.8.5) n→∞
for some eigenvalue λi of A.
Proof. Let y(n) be a solution of (8.8.3). Clearly if y(N) = 0 for some N, then y(n) = 0 for all n ≥ N. Hence we assume without loss of generality that y(n) ̸= 0 for n ≥ n0. Let λ1,λ2,...,λk be the eigenvalues of A. Let μ > max |λi| and let
1≤i≤k
z(n) = x(n)/μn, n ≥ n0. 11
(8.8.6) z(n) or
(8.8.7) where C = 1A, E(n) = 1B(n). Notice that the eigenvalues of C are
Substituting in (8.8.3) yields z(n + 1) = μ A + μ B(n) z(n + 1) = Cz(n) + E(n)z(n)
μμ
1 λ1, 1 λ2,…, 1 λk, where λi, 1 ≤ i ≤ k, are the eigenvalues of A. Moreover μμμ
1λi < 1, for 1 ≤ i ≤ k. By virtue of (8.8.4), ∥f(n,z)∥ ≤ μ−1∥B(n)∥ →
μ ∥z∥
0 as n → ∞. Hence by Corollary 4.34, the zero solution of (8.8.7) is (glob-
ally) exponentially stable. Thus z(n) → 0 as n → ∞ for every solution z(n)
of (8.8.7). By Coffman’s Theorem 8.45 we have lim n ∥z(n)∥ = 1 |λ |, n→∞ μi
for some 1 ≤ i ≤ k. This implies that limn→∞ n ∥y(n)∥ = |λi|, for some 1 ≤ i ≤ k. P
Now we specialize the preceding result to the scalar difference equation of Poincar ́e type
x(n + k) + p1(n)x(n + k − 1) + · · · + pk(n)x(n) = 0. (8.8.8)
Using the l∞-norm ∥y(n)∥∞ = max{|xi(n)| | 1 ≤ i ≤ k}, we obtain the following extension of Perron’s Second Theorem.
8.8 Extensions of the Poincar ́e and Perron Theorems 389
Theorem 8.47. Consider the difference equation of Poincar ́e type (8.8.8). If x(n) is a solution, then either x(n) = 0 for all large n or
lim sup n |x(n)| = |λi|, (8.8.9) n→∞
for some characteristic root λi of (8.8.8). 2 3
Proof. We first convert (8.8.8) to a system of the form (8.8.3), where y(n) = (x(n),x(n + 1),...,x(n + k − 1))T. Notice that ∥y(n)∥∞ = max{|x(n)|, |x(n + 1)|, . . . , |x(n + k − 1)|}. Conclusion (8.8.9) follows from Theorem 8.46. P
Using the l1-norm ∥y(n)∥1 we obtain the following interesting result.
Theorem 8.48. If x(n) is a solution of the difference equation of Poincar ́e type (8.8.8), then either x(n) = 0 for all large n or
lim n |x(n)|+|x(n+1)|+···+|x(n+k−1)|. n→∞
Remarks:
(i) The conclusion (8.8.9) cannot be improved, the limsup cannot be replaced by lim as shown by considering the equation
x(n + 2) − x(n) = 0.
This equation has the solution x(n) = 1 + (−1)n, where
lim supn→∞ n |x(n)| = 1 = |λ1 | = |λ2 |. However, limn→∞ n |x(n)| does not exist.
(ii) For a direct proof of Theorem 8.46 without the use of Coffman’s Theorem, the reader may consult the paper by Pituk [122].
8.8.2 Poincar ́e’s Theorem Revisited
The main objective of this subsection is to extend Poincar ́e’s Theorem to systems of the form (8.8.3). So as a by-product we prove Poincar ́e’s Theorem for scalar difference equations. The exposition here is based on a recent paper by Abu-Saris, Elaydi, and Jang [1]. The following definitions were developed in a seminar at Trinity University led by Ulrich Krause of the University of Bremen and the author.
2This was conjectured by U. Krause and S. Elaydi in a seminar at Trinity University.
3Mih ́aly Pituk, a Professor of Mathematics at the University of Veszpr ́em, received the best paper award (2002) from the International Society of Difference Equations for proving Theorem 8.47 and other related results.
390 8. Asymptotic Behavior of Difference Equations
Definition 8.49. Let y(n) be a solution of (8.8.3). Then y(n) is said to be of:
(1) Weak Poincar ́e type (WP) if
n ∥y(n)∥ = |λ|
lim
n→∞ for some eigenvalue λ of A.
(2) Poincar ́e type (P) if
for some eigenvalue λ of A.
(3) Strong Poincar ́e type (SP) if
lim ∥y(n+1)∥=|λ| n→∞ ∥y(n)∥
lim y(n) = C n→∞ λn
for some eigenvalue λ of A and a nonzero vector C. (4) Ergodic Poincar ́e type (EP) if
for some eigenvector v of A.
The following examples [1] illustrate the interrelationship among the
above concepts.
Example 8.50.
Then
is a solution. Notice that
n ∥y(n)∥ = 1
lim y(n) =v n→∞ ∥y(n)∥
Consider the system
12
y(n+1)= 0 1 y(n), n≥0.
y(n)=α(−1)n 1 +β 1 = β+α(−1)n 01β
but
lim
n→∞
lim ∥y(n+1)∥= n→∞ ∥y(n)∥
if n is even, (β+α)/β ifnisodd.
β/(β + α)
Hence y(n) is weak Poincar ́e but not Poincar ́e.
8.8 Extensions of the Poincar ́e and Perron Theorems 391
Example 8.51. Consider the system
⎛n+1⎞
y(n+1)=⎝− 2n 0⎠y(n), n≥1, y(1)= 10 . 01
The solution is given by
y(n)= (−1)n−1n 1
, n≥1.
Notice that limn→∞ ∥y(n+1)∥ = 1, where −1 is an eigenvalue. How-
2n−1 0
∥y(n)∥ 22
ever, limn→∞ y(n) = limn→∞(−1)n−1 1 does not exist. Thus y(n) is ∥y(n)∥ 0
Poincar ́e but not ergodic Poincar ́e. Example 8.52. Contemplate the system
⎛n+1 ⎞ y(n+1)=⎝ n 0⎠y(n), n≥1, y(1)= 10 .
01 The solution is given by
Notice that
y(n) = ∥y(n)∥
1 , where 0
1 is an eigenvector that corresponds 0
1
y(n)=n 0 , n≥1.
y(n) diverges. Hence y(n) is ergodic n→∞ 1n
To facilitate the proof of the main result we present a definition and two lemmas.
Definition 8.53. A solution y(n) = (y1(n), y2(n), . . . , yi(n))T of (8.8.3) is said to have the index for maximum property (IMP) if there exists an index r ∈ {1,2,...,k} such that, for sufficiently large n,
Observe that solutions in Examples 8.51 and 8.52 possess the IMP, while the solution in Example 8.50 does not possess the IMP.
Lemma 8.54. Suppose that A = diag(λ1, λ2, . . . , λk) such that 0 < |λ1| < |λ2| < ··· < |λk| and (8.8.4) holds. Then every solution of (8.8.3) possesses the IMP.
to the eigenvalue 1. However lim Poincar ́e but not strong Poincar ́e.
∥y(n)∥ = max |yi(n)| = |yr(n)|. 1≤i≤k
392 8. Asymptotic Behavior of Difference Equations
Proof. Since limn→∞ ∥B(n)∥ = 0, for any ε > 0, there exists N1 > 0 suchthat∥B(n)∥=max k |b (n)|<εforn≥N.Wechoose
ε > 0 such that
|λi|+ε <1 for1≤i
s = sup |yj(n)|. n |yr (n)|
1≤i≤k j=1 ij 1
8.8 Extensions of the Poincar ́e and Perron Theorems 393
Then there exists a subsequence ni such that lim |yj(ni)| = s.
ni→∞ |yr(ni)|
|yj(ni + 1)| ≥ |λj| |yj(ni)| − ε|yr(ni)| = |λj| |yj(ni)|/|yr(ni)| − ε
Observe that |yr(ni + 1)|
(|λr| + ε)|yr(ni)|
s ≥ |λj |s − ε
|λj|+ε s≤ε
|λr| + ε
for ni > N. Therefore,
and, consequently,
|yj(n + 1)| |yr(n + 1)| ≤
for n > N. Thus
|λr |−ε
limsup |yj(n)| ≤ ε
n→∞ |yr(n)| |λr| − |λj| − ε
|λj|−|λr|−ε
for all sufficiently small ε. This implies that s = 0 and the assertion is shown.
On the other hand, if j < r, then
|λj| |yj(n)| ε |λr| − ε |yr(n)| + |λr| − ε
|λj| |yj(n)| + ε|yr(n)| (|λr| − ε)|yr(n)| =
⎡
|λj| n−N |yj(N)| ⎢1− |λr|−ε
|yr(n)|≤ |λr|−ε |yr(N)|+⎣ 1− |λj|
|yj(n)| and as a result
⎥ ε ⎦|λr|−ε
|λj| n−N⎤
for all sufficiently small ε. This implies that lim sup |yj (n)| = 0
n→∞ |yr (n)|
By using Lemmas 8.54 and 8.55 we present a sufficient condition for
P
Theorem 8.56. Suppose that the eigenvalues of A have distinct moduli and limn→∞ ∥B(n)∥ = 0. Then (8.8.3) possesses the Poincar ́e property P.
Proof. We may assume, without loss of generality, that A is in diagonal form, i.e., A = diag(λ1,λ2,...,λk), where 0 < |λ1| < ··· < |λk|. Let y(n) be a nontrivial solution of (8.8.3). It follows from Lemma 8.54 that
and complete the proof.
which (8.8.3) has the Poincar ́e property.
∥y(n)∥ = |yr(n)|
394 8. Asymptotic Behavior of Difference Equations
for all large n, for some 1 ≤ r ≤ k. Moreover, Lemma 8.55 implies lim |yi(n)| = 0
n→∞ |yr(n)|
for 1 ≤ i ≤ k such that i ̸= r. Therefore, if i ̸= r, then ⎡⎤
yi(n+1)
lim = lim
⎣ yi(n) k λi +
yi(n)⎦ bij(n) =0
n→∞ |yr (n)| n→∞ and if i = 1, we have
|yr (n)|
j =1
|yr (n)|
lim
n→∞ Consequently,
Since
it follows that
yr(n + 1) − λryr(n) |yr(n)|
k n→∞ j=1
yj(n) |yr(n)|
=0.
= lim
bij(n)
lim ∥y(n+1)−λry(n)∥=0. n→∞ ∥y(n)∥
∥y(n + 1)∥ − |λr| ≤ ∥y(n + 1) − λr∥, ∥y(n)∥ ∥y(n)∥
lim ∥y(n+1)∥=|λr|. n→∞ ∥y(n)∥
The proof is now complete.
P
As an immediate consequence of the above result, we obtain the original Poincar ́e’ Theorem.
Proof of Theorem 8.9. Write equation (8.2.7) as a system of the form (8.8.3). Then a solution y(n) of (8.8.3) is of the form y(n) = (x(n), x(n + 1),...,x(n+k−1))T. By Theorem 8.56 we have
lim ∥y(n+1)∥=|λ| (8.8.10) n→∞ ∥y(n)∥
for some eigenvalue λ. By Lemma 8.54, there exists r ∈ {1, 2, . . . , k} such that
∥y(n)∥ = |yr(n)| = |x(n + r − 1)|. Substituting (8.8.10) yields
lim |x(n+r)| = lim |x(n+1)| =|λ| n→∞ |x(n + r − 1)| n→∞ |x(n)|
8.8 Extensions of the Poincar ́e and Perron Theorems 395
x(n+1) x(n+1) where λ is a characteristic root of (8.2.7). Since x(n) ≤ x(n) ,
limn→∞ x(n+1) exists. Hence limn→∞ x(n+1) = λ. (See Exercises 8.2, x(n) x(n)
Problem 9.)
Term Projects 8.8
P
1. Find the relationships among the notions of WP, P, SP, and EP for both scalar equations and systems of Poincar ́e type.
2. Find sufficient conditions for EP and SP. 3. Extend Perron’s First Theorem to systems.
9
Applications to Continued Fractions and Orthogonal Polynomials
9.1 Continued Fractions: Fundamental Recurrence Formula
Continued fractions are intimately connected with second-order difference equations. Every continued fraction may be associated with a second-order difference equation; and conversely, every homogeneous second-order dif- ference equation may be derived from some continued fraction. The first point of view is useful for computing continued fractions, the second for computing the minimal solutions.
Let {an}∞n=1 and {bn}∞n=0 be two sequences of real or complex numbers. A continued fraction is of the form
or, in compact form,
or or
b+ a1
0 a2
b1 + .. an b2+.+ ..
bn+ .
b0+ a1 a2 a3 ... (9.1.1)
b1 + b2 + b3 + b0 + K(an/bn)
b +K∞ (a /b ). 0 n=1nn
397
398 9. Applications to Continued Fractions and Orthogonal Polynomials
The nth approximant of a continued fraction is defined as C(n)= A(n) =b +Kn (a /b )
B(n) 0 j=1 j j
=b0+ a1 a2 ... an. (9.1.2)
b1 + b2 + bn
The sequences A(n) and B(n) are called the nth partial numerator and
the nth partial denominator of the continued fraction, respectively. It is
always assumed that A(n) is in reduced form, that is, A(n) and B(n) are B(n)
coprime (i.e., they have no common factors). An alternative way of defining a continued fraction is through the use of a “Mo ̈bius” transformation, which is defined as
t0(u)=b0 +u, tn(u)= an , n=1, 2, 3, .... (9.1.3) bn + u
Then it is easy to see that the nth approximant is given by
C(n)=(t0 ◦ t1 ◦···◦tn)(0). (9.1.4)
Definition 9.1. The continued fraction (9.1.1) is said to converge to a finite limit L if limn→∞ C(n) = L, and it is said to diverge otherwise.
Next we show that both the nth partial numerator A(n) and the nth partial denominator B(n) of the continued fraction (9.1.1) satisfy a second-order difference equation commonly known in the literature as the fundamental recurrence formula for continued fractions. The explicit statement of this important result now follows.
Theorem 9.2. Consider the continued fraction b +K∞ (a /b ) with nth 0 n=1 n n
approximant C(n) = A(n)/B(n). Then A(n) and B(n) satisfy, respectively, the difference equations
(9.1.5) Proof. The proof of (9.1.5) will be established using mathematical
induction on n.
Observe that from (9.1.5) we obtain
A(1) = b1A(0) + a1A(−1) = b1b0 + a, B(1) = b1B(0) + a1B(−1) = b1.
Hence, (9.1.5) is valid for n = 1. Next, we assume that it is true for n = m, that is,
A(m)=bmA(m−1)+amA(m−2), A(−1)=1, B(m)=bmB(m−1)+amB(m−2), B(−1)=0.
A(n)=bnA(n−1)+anA(n−2), A(−1)=1, A(0)=b0, B(n)=bnB(n−1)+anB(n−2), B(−1)=0, B(0)=1.
Similarly,
Hence
9.1 Continued Fractions: Fundamental Recurrence Formula 399
Now A(m + 1)/B(m + 1) is obtained from A(m)/B(m) by replacing bm by bm + am+1 . Then we can write A(m + 1)/B(m + 1) = A∗(m)/B∗(m),
a
A∗(m)= bm + m+1 A(m−1)+amA(m−2)
bm+1 where
B∗(m) = b−1 m+1
B(m − 1)) .
bm+1
= b−1 (b m+1
B(m + 1) bm+1B(m) + am+1B(m − 1)
which establishes (9.1.5), and the proof of the theorem is now com-
plete. P
The converse of the preceding theorem is also true. In other words, every homogeneous second-order difference equation gives rise to an associated continued fraction.
Suppose now we are given the difference equation x(n)−bnx(n−1)−anx(n−2)=0, an ̸=0 forn∈Z+. (9.1.6)
m+1
A(m) + a A(m − 1)) . m+1
(b
C(m+1) = A(m+1) = bm+1A(m)+am+1A(m−1),
m+1
B(m) + a m+1
Dividing (9.1.6) by x(n − 1) and then setting y(n) = x(n)
bn=an . y(n−1)
Hence,
yields y(n) −
y(n − 1) =
Applying this formula repeatedly, with n successively increasing, we obtain
x(n−1)
an . −bn + y(n)
y(n − 1) = an an+1 an+2 . −bn+ −bn+1+ −bn+2 + . . .
In particular, when n = 1, we have
y(0) = x(0) = a1 a2 a3
.
(9.1.7)
x(−1) −b1+ −b2+ −b3+
Remark 9.3. We would like to make the following important observations
concerning (9.1.7):
(a) Formula (9.1.7) is formal in the sense that it does not tell us whether
the continued fraction K(an/−bn) converges or diverges.
(b) Even if the continued fraction K(an/−bn) converges, formula (9.1.7) does not show us how to pick the particular solution x(n) for which x(0)/x(−1) is the limit of the continued fraction.
400 9. Applications to Continued Fractions and Orthogonal Polynomials
(c) It is not difficult to show that if K(an/−bn) converges to x(0)/x(−1), then K(an/bn) converges to −x(0)/x(−1).
A Formula for C(n)
We end this section by providing a formula for computing the nth
approximant C(n) = A(n) of the continued fraction b0 + K(an/bn). B(n)
To find the formula, we multiply the first equation in (9.1.5) by B(n−1) and the second by A(n − 1) and then subtract one from the other. This yields
A(n)B(n−1)−B(n)A(n−1) = −an[A(n−1)B(n−2)−B(n−1)A(n−2)], which is equivalent to
u(n) = −anu(n − 1), u(0) = −1, where u(n) = A(n)B(n − 1) − B(n)A(n − 1). Hence,
u(n)=A(n)B(n−1)−B(n)A(n−1)=(−1)n+1a1a2···an, n≥1. (9.1.8)
Dividing both sides by B(n)B(n − 1) yields A(n−1) (−1)n+1a1a2 ···an
formula (2.1.16))
A(n − 1) A(0) C(n−1)=B(n−1)=B(0)+
This produces the desired formula
∆ B(n−1) = B(n−1)B(n) . (9.1.9) Taking the antidifference ∆−1 of both sides of (9.1.9), we obtain (see
n−1
(−1)k+1a1a2 · · · ak
k=1
B(k−1)B(k) .
9.2 Convergence of Continued Fractions
Two continued fractions K(an/bn) and K(a∗n/b∗n) are said to be equiva- lent, denoted by the symbol K(an/bn) ≈ K(a∗n/b∗n), if they have the same sequence of approximants.
Let {dn}∞n=1 be any sequence of nonzero complex numbers. Then the Mo ̈bius transformation tn(u) = an can be represented as tn(u) =
(9.1.10)
bn +u
dnan ; which may be repeated as a composition of two transformations
dnbn+dnu
tn = sn ◦ rn, where
C(n)=b0 +
n (−1)k+1a1a2 · · · ak
k=1
B(k−1)B(k) .
sn(u) = dnan and rn(u) = dnu. dnbn +u
Hence we have
t1 ◦t2 ◦···◦tn =s1 ◦r1 ◦s2 ◦r2 ◦···◦sn ◦rn
9.2 Convergence of Continued Fractions 401
=s1 ◦(r1 ◦s2)◦(r2 ◦s3)◦···◦(rn−1 ◦sn)◦rn. Definet∗(u):=r ◦s (u)=dn−1 dn an.Thenifd :=1,
nn−1ndnbn+u 0 C(n)=(t1 ◦t2 ◦···◦tn)(0)=(t∗1 ◦t∗2 ◦···◦t∗n)(0).
This yields the important equivalence relation
d da K(an/bn)≈K n−1 n n ,
dn bn
(9.2.1)
which holds for any arbitrary sequence of nonzero complex numbers d0 = 1, d1, d2,....
Observe that if we choose the sequence {dn} such that dnbn = 1, then (9.2.1) becomes
Similarly, one can show that
where d1 = 1 , dn = 1 . Hence a1 an dn−1
b ba K(an/bn) ≈ K n−1 n n
.
(9.2.2)
(9.2.3)
(9.2.4)
1 K(an/bn) ≈ K(1/bndn),
d2n=a1a3···a2n−1, d2n+1= a2a4···a2n , a2a4 ···a2n a1a3 ···a2n+1
(Exercises 9.1 and 9.2, Problem 8).
We are now ready to give the first convergence theorem.
Theorem 9.4. Let bn > 0, n = 1,2,3,…. Then the continued fraction K(1/bn) is convergent if and only if the infinite series ∞n=1 bn is divergent.
Proof. From (9.1.10) we have
∞ (−1)r+1
B(r − 1)B(r). (9.2.5)
Hence K(1/bn) converges if and only if the alternating series on the right- hand side of (9.2.5) converges. Now the fundamental recurrence formula for K(1/bn) is
B(n) = B(n − 2) + bnB(n − 1), B(0) = 1, B(1) = b1. (9.2.6)
K(1/bn) =
r=1
This implies that B(n+1) > B(n−1), n = 1,2,3,…, and, consequently, (−1)n+1
B(n)B(n + 1) > B(n − 1)B(n), n = 1,2,3,…. Thus B(n−1)B(n) is monotonically decreasing. Hence the series (9.2.5) converges if and only if
lim B(n − 1)B(n) = ∞. (9.2.7) n→∞
402 9. Applications to Continued Fractions and Orthogonal Polynomials
Again from (9.2.6) we have B(n) ≥ γ = min(1,b1), n = 1,2,3,…, and, consequently,
B(n − 1) B(n) = B(n − 2)B(n − 1) + bnB2(n − 1) ≥ B(n − 2)B(n − 1) + bnγ2
≥(b1 +b2 +···+bn)γ2.
Thus if ∞i=1 bi diverges, (9.2.7) holds and K(1/bn) converges. On the other
hand, we have, from (9.2.6),
B(n−1)+B(n) = B(n−2)+(1+bn)B(n−1) ≤ (1+bn)[B(n−1)+B(n−2)]. It follows by induction that
B(n − 1) + B(n) ≤ (1 + b1)(1 + b2) · · · (1 + bn) < eb1+b2+···+bn . ∞
Thus if
n=1
bn converges to L, then B(n − 1) + B(n) ≤ eL. Therefore, B(n − 1)B(n) ≤ 1(B(n − 1) + B(n))2 ≤ 1e2L.
44
Consequently, (9.2.7) does not hold, and hence the continued fraction
diverges. P
A more general criterion for convergence was given by Pincherle in his fundamental work [120] on continued fractions. Consider again the difference equation
x(n)−bnx(n−1)−anx(n−2)=0, an ̸=0 forn∈Z+. (9.2.8) Theorem 9.5 (Pincherle). The continued fraction
a1 a2 a3 ... (9.2.9) b1+ b2+ b3+
converges if and only if (9.2.8) has a minimal solution φ(n), with φ(0) ̸= 0. In case of convergence, moreover, one has
−φ(n−1) = an an+1 an+2 ..., n=1,2,3,.... (9.2.10) φ(n − 2) bn+ bn+1+ bn+2+
Proof.
(a) Assume that the continued fraction (9.2.9) converges. Hence if A(n) and B(n) are the the nth partial numerator and nth partial denominator of (9.2.9), respectively, then
lim A(n) = L. n→∞ B(n)
It follows from Theorem 9.2 that A(n) and B(n) are solutions of (9.2.8) with A(−1) = 1, A(0) = 0, and B(−1) = 0, B(0) = 1.
9.2 Convergence of Continued Fractions 403
We claim that
φ(n) = A(n) − LB(n) (9.2.11) is a minimal solution of (9.2.8).
To prove the claim, let y(n) = αA(n) + βB(n), for some scalars α and β, be any other solution of (9.2.8). Then
lim φ(n) = lim A(n) − LB(n) = lim (A(n)/B(n)) − L = 0. n→∞ y(n) n→∞ αA(n) + βB(n) n→∞ α (A(n)/B(n)) + β
This establishes the claim. Furthermore, φ(−1) = 1 ̸= 0.
(b) Conversely, assume that (9.2.8) possesses a minimal solution φ(n),
with φ(−1) ̸= 0. From (9.1.7), the associated continued fraction to
(9.2.8) is K(a /−b ) with the nth approximant C∗(n) = A∗(n) . Since n n B∗(n)
A∗(n) and B∗(n) are two linearly independent solutions of (9.2.8) with A∗(−1) = 1, A∗(0), B∗(−1) = 0, B∗(0) = 1 (Theorem 9.2), it follows that
φ(n) = A∗(n) − LB∗(n), n ≥ 0. 0= lim φ(n) = lim A∗(n)−L.
Observe that
Hence
n→∞ B∗(n) φ(−1) From Remark 9.3(c), we conclude that
lim C(n) = lim A(n) = − φ(0) . n→∞ n→∞ B(n) φ(−1)
n→∞ B∗(n) n→∞ B∗(n) lim A∗(n) = φ(0) .
(9.2.12)
This proves the first part of the theorem as well as (9.2.10) for n = 1. The proof of (9.2.10) for n > 1 is left to the reader as Exercises 9.1 and 9.2 Problem 5. P
The following example illustrates Theorem 9.5. Example 9.6. Contemplate the continued fraction
a a a …, (9.2.13) 1+ 1+ 1+
where a is any complex number. Find conditions on a under which the continued fraction converges.
404 9. Applications to Continued Fractions and Orthogonal Polynomials
Solution
Method 1: Let A(n) and B(n) be the nth partial numerator and denominator, respectively. Then from (9.1.5) we have
A(n)−A(n−1)−aA(n−2)=0, A(−1)=1, A(0)=0, B(n)−B(n−1)−aB(n−2)=0, B(−1)=1, B(0)=1.
The characteristic equation of either equation is given by λ2 − λ − a = 0,
whose roots are
λ1,2 = 1±√1+4a. 2
Now, if |λ1| ̸= |λ2|, then the difference equation x(n)−x(n−1)−ax(n−2) = 0 (9.2.14)
has a minimal solution and, consequently by Pincherle’s theorem the continued fraction (9.2.13) converges.
Suppose that |λ2| < |λ1|. Then φ(n) = λn2 is a minimal solution of
(9.2.14). Hence by (9.2.12) the continued fraction (9.2.13) converges to
−λ2=−1+1√1+4a.
22111n Ontheotherhand,ifa=− ,thenλ1 =λ2 = .ThusA(n)=c1 +
1n 422
c2n 2 . Using the initial conditions A(−1) = 1, A(0) = 0, we get c1 = 0,
c2 = −1. Hence A(n) = −n1n+1. Similarly, we obtain B(n) = (n + 1n 2 2
1) 2 . Thus
K(a/1) = lim A(n) = −1 = −λ2. n→∞ B(n) 2
124
Conclusion. If a is complex, K(a/1) converges to −1 + onlyifa∈/ x∈R:x<−4 .
1 +a if and
Method 2: Let
Then
Hence
x=aaa.... 1+ 1+ 1+
x= a , x2+x−a=0. 1+x
x1=−1+ 1+a, x2=−1− 1+a, 24 24
are two solutions. If a is real we require that a ≥ −1 in order for x to be 4
real. By inspection, we conclude that the continued function converges to x1.
Exercises 9.1 and 9.2
1. Show that
9.2 Convergence of Continued Fractions 405
1−aa... 1− 1−
converges to (1 + √1 − 4a)/2 if 0 < 4a < 1.
2. Prove that the continued fraction
111... b1+ b2+ b3+
converges if bi ≥ 1 for i = 1,2,3,....
3. Discuss the convergence of the continued fraction
a a a..., b+ b+ b+
where a, b are complex numbers, a ̸= 0.
4. Show that the continued fraction
is equivalent to
where α0(x) =
λ1 λ2 λ3 ... (x − c1)− (x − c2)− (x − c3)−
α0(x) α1(x) α2(x), 1− 1− 1−
λ1 ,αn(x) = λn+1 ,n = 1,2,3,.... x−c1 (cn −x)(cn+1 −x)
5. Prove (9.2.10) for n > 1.
6. [19]Provethatifb +K∞ (a /b )=L̸=0,then 0 n=1nn
K∞ (a/b)=a0. n=0 n n L
7. Consider the continued fraction b0 +K(an/bn) and let A(n) and B(n) be the nth partial numerator and denominator, respectively.
406
9. Applications to Continued Fractions and Orthogonal Polynomials
Show that
0
,
1 b n
0
.
1 b n
tn(u) = an , an ̸= 0, n = 0,1,2,…. bn + u
Let T0(u) = t0(u), Tn(u) = Tn−1(tn(u)). Show that Tn(u)= A(n)+A(n−1)u
B(n)+B(n−1)u
and A(n)B(n−1)−A(n−1)B(n) ̸= 0, n = 0,1,2,…, where A(n) and B(n) are the nth partial numerator and denominator of b0 +K(an/bn), respectively.
10. Let {A(n)} and {B(n)} be sequences of complex numbers such that A(−1)=1, A(0)=b0, B(−1)=0, B(0)=1,
and
A(n)B(n−1)−A(n−1)B(n)̸=0, n=0,1,2,….
(a) Show that there exists a uniquely determined continued fraction b0 + K (an/bn) with nth partial numerator A(n) and nth partial denominator B(n).
8. Prove (9.2.4).
A(n) =
0
b 1 1
B(n) =
… … …
b 0 1
a1 b1 1
0a2b21
a n − 1
1
a3 b3 1
a2 b2
0
9. Let {tn} be a sequence of M ̈obius transformations defined as
… … … a n − 1 b n − 1
b n − 1 a n
a n
is
1− 1− 1− B(n)=(1−a1)(1−a2)···(1−an−1), n≥1.
9.2 Convergence of Continued Fractions 407
(b) Show that
b0 = A(0), b1 = B(1), a1 = A(1) − A(0)B(1),
an = A(n−1)B(n)−A(n)B(n−1) , A(n−1)B(n−2)−A(n−2)B(n−1)
bn = A(n)B(n−2)−A(n−2)B(n) . A(n−1)B(n−2)−A(n−2)B(n−1)
11. Show that the nth partial denominator of the continued fraction 1− 1 a1 (1−a1)a2 …
12. [19]Letαn =(1−an−1)an,a0 =0,0
bnbn+1 =∞. n=1 an+1
Show that the continued fraction
a1 a2 a3 … b1+ b2+ b3+
∞ a1a2···an
L =
13. Letβn =(1−bn−1)bn,0≤b0 <1,0
*16. (Term Project, [31]). Consider the continued fraction τ2(a)=2− a a ….
2− 2−
Let t1(a) = 2, t2(a) = 2−a/2, t3(a) = 2−a/(2−a/2),… be the
approximant sequence.
(a) Showthattn+1(a)=2− a ,t1 =2,n=1,2,3,….
tn (a)
(b) Show that if a ≤ 1, the continued fraction converges to 1+√1 − a.
(c) A number a is said to be periodic (of period n) if tn+k(a) = tk for k = 1,2,3,….
Show that if a is of period n, then tn−1(a) = 0.
⌊ n−1 ⌋ 2 k n k
(d) Let Pn(x) =
greatest integer function. Show that
k=0
(−1) 2k+1 x , where ⌊ ⌋ denotes the
Our main objective in this section is to show that every infinite series can be represented by a continued fraction and vice versa. Let {cn} be a sequence of complex numbers, with cn ̸= 0, n = 1, 2, 3, …, and let
n ρ1ρ2ρ3ρn ρ0+ ρ1ρ2···ρk=ρ0+1−(1+ρ)−(1+ρ)−···1+ρ .
n k=0
un =
Let ρ0 = c0, ρ1 = c1, ρn = cn/cn−1. Then c0 = ρ0, cn = ρ1ρ2···ρn.
Moreover,
ck, n=0,1,2,….
k=1 23n
9.3 Continued Fractions and Infinite Series
409
Hence
ck = b0 + K(an/bn), whereb0 =c0,a1 =c1,b1 =1,and
an =− cn , bn =1+ cn , n=2,3,4,…. cn−1 cn−1
To illustrate the above method observe that
∞ k=0
(9.3.1)
∞ czcz/ccz/c ckzk=c0+ 1 2 1 3 2 ….
1− 1 + c2z − 1 + c3z k=0 c1 c2
Here is a more interesting example.
Example 9.7. Consider the Riemann zeta function, defined by
n−1k n−1k an=− n ,andbn=1+ n .
∞ r=1
r−k =1+2−k +3−k +···, k=2,3,4,…. Thenb0 =0,a1 =1,b1 =1,
Thus
u(n + 2) − 1 + n + 2 u(n + 1) + n + 2 u(n) = 0.
ζ(k)=
ζ(k) = K(an/bn).
If we let u(n) = Kn (a /b ), then it follows from (9.1.6) that
j=1 j j
n+1k n+1k
(9.3.2)
An equivalent representation of ζ(k) may be obtained using (9.3.2): ζ(k) = K(1/bndn),
(2/3)k (4/5)k · · · (2n − 2/(2n − 1))k d1 =1, d2n = (1/2)k(3/4)k···(2n−1/(2n))k ,
(1/2)k (3/4)k · · · (2n − 1/(2n))k d2n+1 = (2/3)k (4/5)k · · · (2n/(2n + 1))k .
Example 9.8. (Regular continued fractions) [73].
A regular continued fraction b0 + K(1/bn) of a positive real number x is defined by letting
bn = ⌊xn⌋, n = 0,1,2,…,
410 9. Applications to Continued Fractions and Orthogonal Polynomials
where
x0 =x, xn = 1 , n=1,2,3,…, Frac(xn−1 )
where ⌊ ⌋ denotes the greatest integer function, and Frac(xn) denotes the fractional part of xn. If Frac(xn−1) = 0, the regular continued fraction expansion terminates with bn−1. Suppose now that x = k/l is a rational number.
Set x0 = x and r1 = l. Then by the Euclidean algorithm, x0 =b0+r2/r1 =b0+1/x1, withx1 =r1/r2, r2
determined for r = 0, 1, . . . , n−1. This establishes the first part. The second part is left to the reader as Exercises 9.5, Problem 1. P
We are now ready to present some of the main classical orthogonal polynomials.
(α,β)
1. Jacobi polynomials Pn (x), α > −1, β > −1.
These polynomials are orthogonal on (−1,1) with the weight func-
tion w(x) = (1−x)α(1+x)β, andann = 2−n 2n+α+β . An n
(α,β) explicit expression for the Jacobi polynomials Pn
Rodrigues’ formula
(x) are given by
P(α,β)(x)=(−1)n(1−x)−α(1+x)−β dn (1−x)n+α(1+x)n+β. n 2nn! dxn
(9.4.3) To write Pn (x) more explicitly, we need to utilize Leibniz’s formula
(9.4.4)
(See Appendix G.)
(α,β)
d n n n d n − k u d k v
dxn (uv) = k dxn−k dxk . k=0
Hence,
dn
dxn (1 − x)n+α(1 + x)n+β
Therefore, P(α,β)(x) = 2−n
9.4 Classical Orthogonal Polynomials 415
n
nk Dn−k(1 − x)n+αDk(1 + x)n+β n
× n+α n+β (x−1)k(x+1)n−k. k=0 n−k k
n
n + α n + β (x − 1)k(x + 1)n−k (9.4.5)
=
= (−1)n(1 − x)α(1 + x)βn!
nn−kk k=0
k=0
with the leading coefficient
n
ann =2−n n+α n+β =2−n 2n+α+β . (9.4.6)
k=0n−k k n
(See G.)
To verify (9.4.3), let Qn(x) denote the right-hand side of the equation and let g(x) be another polynomial. Then successive application of integration by parts yields
1
(1 − x)α(1 − x)βQn(x)g(x) dx
−1
11
(1 − x)n+α(1 + x)n+βg(n)(x) dx, where g(n)(x) denotes the nth derivative of g(x).
(9.4.7)
= 2nn!
−1
Observe that if g(x) is a polynomial of degree less than n, then g(n)(x) = 0. Hence, Qn(x) satisfies (9.4.2). Furthermore, the coefficient
of xn in Qn(x) is
2−n 2n+α+β . n
(α,β) Hence, by uniqueness (Theorem 9.5), Qn(x) = Pn (x).
2. Legendre polynomials Pn(x): These are special Jacobi polynomials obtained by letting α = β = 0. Hence (9.4.3) is reduced to
(9.4.8)
(0,0) (−1)n dn 2n Pn(x)=Pn (x)= 2nn! dxn{(1−x ) }.
416
9. Applications to Continued Fractions and Orthogonal Polynomials
The Legendre polynomials are orthogonal on (−1,1) with the weight function w(x) = 1. Moreover, using Leibniz’s formula yields
n
Pn(x) = 2−n n n (x − 1)k(x + 1)n−k, (9.4.9)
k=0 n−k k
with leading coefficient ann = 2−n(2n)!/(n!)2.
3. Gegenbauer (or ultraspherical) polynomials Pnν(x): These are special Jacobi polynomials obtained by setting α = β and α = ν − 1 .
2
The Gegenbauer polynomials are orthogonal on (−1, 1) with the weight function w(x) = (1 − x2)ν−1/2, ν > −1, and ann = 2nν+n−1. By
Rodrigues’ formula we have
ν (−1)n 2 1/2−ν dn
2 ν+n−1/2 Pn(x)= n! (1−x ) dxn(1−x )
2
ν−1
(9.4.10)
−1
= 2ν − 1 n + 2ν − 1 P (ν−1/2,ν−1/2)(x).
ν ν − 1/2 n Using Leibniz’s formula yields
−1 P nν ( x ) = 2 − n 2 ν − 1 n + 2 ν − 1
ν ν − 1/2
n
× n+ν −1/2 n+ν −1/2 (x−1)k(x+1)k. k=0 n−k k
(9.4.11) 4. LaguerrepolynomialsLαn(x),α>−1,areorthogonalon(0,∞)with
the weight function w(x) = e−xxα, and ann = (−1)n/n!. Moreover, (9.4.12)
By Leibniz’s formula we can show that
n k
Lαn(x)= n+α (−x) . (9.4.13)
5. Hermite polynomials Hn(x) are orthogonal on (−∞,∞) with the
weight function w(x) = e−x2 , and ann = 2n. These are given by Rodrigues’ formula
(9.4.14)
α exx−α dn −x n+α Ln(x)= n! dxn(e x ).
k=0 n−k k!
nx2dn −x2 Hn(x) = (−1) e dxn (e ).
9.5 The Fundamental Recurrence Formula for Orthogonal Polynomials 417
By Taylor’s theorem,
2xw−w2 ∞ wn e = Hn(x) n! .
n=0
Expanding e2xw and e−w2 as power series in w and taking the Cauchy
product1 of the result gives
2xw−w2 e
=
∞ ⌊n/2⌋
(−1)k(2x)n−2kwn
9.5
The Fundamental Recurrence Formula for Orthogonal Polynomials
Hence,
Hn(x)=n!
⌊n/2⌋
(−1)k(2x)n−2k
n=0 k=0
(n − 2k)! k! .
k=0
(n−2k)!k! .
We now show why difference equations, particularly those of second order, are of paramount importance in the study of orthogonal polynomials. The following is the main result.
Theorem 9.11. Any sequence of monic orthogonal polynomials {Pn(x)} with Pn(x) = nk=0 akxk must satisfy a second-order difference equation of the form
with
(9.5.1)
An = an+1,n+1 . (9.5.2) an,n
Pn+1(x) − (Anx + Bn)Pn(x) + CnPn−1(x) = 0
(9.4.15)
Proof. Choose An such that Pn+1(x) − AnxPn(x) does not possess any term in xn+1. Hence, we may write
n r=0
Multiplying both sides of (9.5.3) by w(x)Ps(x) and integrating from a to b yields
Pn+1(x) − AnxPn(x) =
dnrPr(x). (9.5.3)
b b dns w(x){Ps(x)}2 dx = −An
aa
1Given two series ∞ an and ∞ ∞ n=0 n=0
xw(x)Ps(x)Pn(x) dx. bn, we put cn = n
k=0
(9.5.4)
akbn−k, n = 0, 1, 2, . . . . Then n=0 cn is called the Cauchy product of the two series.
418 9. Applications to Continued Fractions and Orthogonal Polynomials
Since xPs(x) is of degree s+1, and Pn(x) is orthogonal to all polynomials of degree less than n, it follows that the right-hand side of (9.5.4) vanishes except possibly when s = n − 1 and s = n. Hence, dnn and dn,n−1 are possibly not zero. Therefore, from (9.5.3) we have
Pn+1(x) − (Anx + dnn)Pn(x) − dn,n−1Pn−1(x) = 0,
which is (9.5.1) with Bn = dnn, Cn = −dn,n−1. P Remark:
(i) A monic sequence of orthogonal polynomials {Pˆn(x)} satisfies the difference equation
where
Pˆn+1(x) − (x − βn)Pˆn(x) + γnPˆn−1(x) = 0, βn = −Bnann , γn = Cnan+1,n+1 .
(9.5.5)
(9.5.6)
an+1,n+1 an−1,n−1
This may be shown easily if one writes Pˆ(x) = a−1P (x).
(ii) If Pn(−x) = (−1)nPn(x), then {Pn(x)} is often called symmetric. In this case, one may show that Bn = βn = 0. To show this, let Qn(x) = (−1)nPn(−x). Then
Qn+1(x) + (Bn − Anx)Qn(x) + CnQn−1(x) = 0. (9.5.7) If Qn(x) = Pn(x), then subtracting (9.5.1) from (9.5.7) yields Bn = 0.
(iii) The converse of Theorem 9.11 also holds and is commonly referred to as Favard’s theorem. Basically, this theorem states that any polynomial sequence that satisfies a difference equation of the form of (9.5.1) must be an orthogonal polynomial sequence.
Let us illustrate the preceding theorem by an example.
Example 9.12. Find the difference equation that is satisfied by the
Legendre polynomials Pn(x).
Solution From (9.4.8) the coefficients of xn, xn−1, xn−2 are, respectively,
ann = (2n)! , an,n−1 =0, an,n−2 = (2n−2)! . 2n(n!)2 2n(n − 2)! (n − 1)!
nn n
Furthermore, {Pn(x)} is symmetric, since Pn(x) = (−1)nPn(−x). Thus, from Remark (ii) above, we have Bn = 0. From (9.5.2), we have An = 2n+1 .
n+1 It remains to find Cn. For this purpose, we compare the coefficients of xn−1
in (9.5.1). This yields
an+1,n−1 − Anan,n−2 + Cnan−1,n−1 = 0.
9.5 The Fundamental Recurrence Formula for Orthogonal Polynomials 419
Thus,
Cn = Anan,n−2 − an+1,n−1 = −n . an−1,n−1 n + 1
Hence the Legendre polynomials satisfy the difference equation
(n + 1)Pn+1(x) − (2n + 1)xPn(x) + nPn−1(x) = 0. (9.5.8)
Example 9.13. Find the difference equation that is satisfied by the Jacobi (α,β)
polynomials Pn (x).
Solution This time we will use a special trick! Notice that from (9.4.5) we
obtain
P (α,β)(−x) = (−1)nP (β,α)(x), nn
(9.5.9) (9.5.10)
(9.5.11)
(9.5.12)
(9.5.13)
P(α,β)(1)= n+α , nn
P(α,β)(−1)= n+β . nn
From (9.4.6) and (9.5.2) we get
An = (2n+2+α+β)(2n+1+α+β).
2(n+1)(n+1+α+β)
Using (9.5.10) and setting x = 1 in (9.5.1) yields
n+α+1 −(An+Bn) n+α + n+α−1 =0. n+1 n n−1
Similarly, setting x = 1 in (9.5.1) and using (9.5.11) yields
(−1)n+1
n+β+1 −(−An +Bn)(−1)n n+β n+1 n
+Cn(−1)n−1 n+β−1 =0. n−1
(9.5.14) Multiplying (9.5.13) by n+β and (9.5.14) by n+α and adding, produces
nn
an equation in An and Cn which, by substitution for An from (9.5.12),
gives
Cn = (n+α)(n+β)(2n+2+α+β) . (n + 1)(n + α + β + 1)(2n + α + β)
(9.5.15)
(9.5.16)
Substituting for Cn in (9.5.13) yields
Bn = (2n+1+α+β)(α2 −β2) . 2(n + 1)(n + α + β + 1)(2n + α + β)
420 9. Applications to Continued Fractions and Orthogonal Polynomials
Exercises 9.4 and 9.5
1. Let {Pn(x)} be a sequence of orthogonal polynomials on the inter- val (a,b) relative to the weight function w(x). Prove that Pn(x) is orthogonal to any polynomial of lower degree.
2. Verify formula (9.4.9).
3. Verify formula (9.4.12).
4. Verify formula (9.4.14).
5. Find the difference equation that represents the Gegenbauer polyno- mials {Pnν(x)}.
6. Find the difference equation that represents the Laguerre polynomials {Lαn (x)}.
7. Find the difference equation that represents the Hermite polynomials Hn (x).
8. (Charlier polynomials). Let C(a)(x) denote the monic Charlier poly- n
nomials defined by
C(a)(x) = n! n
n n−k x (−a) .
(9.5.17)
k (n−k)! (x)} satisfies the difference equation
k=0
(a) Show that {Cn
C(a) (x)=(x−n−a)C(a)(x)−anC(a) (x), n≥0. n+1 n n−1
9. (The Bessel function). Let n ∈ Z, z ∈ C. The Bessel function Jn(z) is defined by
Find the corresponding difference equation.
*10. (Christoffel–Darboux identity). Let {Pn(x)} satisfy
Pn(x)=(x−cn)Pn−1(x)−λPn−2(x), n=1,2,3,…,
Jn(z)=(z/2)
j!(n+j)! , n=0,1,2,….
n ∞ (−1)j(z2/4)j
j=0
P−1(x) = 0, P0(x) = 1, λ ̸= 0. Prove that
n Pk(x)(Pk(u) = (λ1λ2 ···λn+1)−1 k=0 λ1λ2 ···λk+1
× Pn+1(x)Pn(u) − Pn(x)Pn+1(u). x−u
(9.5.18)
(9.5.19)
9.6 Minimal Solutions, Continued Fractions, and Orthogonal Polynomials 421
11. (Confluent form of (9.5.19)). Show that
n P2(x) P′ (x)P′ (x)−Pn(x)Pn+1(x)
k=n+1n . k=0 λ1λ2 ···λk+1 λ1λ2 ···λn+1
12. Consider the sequence {Pn(x)} satisfying (9.5.18) and let Qn(x) = a−nPn(ax + b), a ̸= 0.
(a) Show that Qn(x) = (x − cn − b)Qn−1(x) − λn Qn−2(x). a a2
(b) If {Pn(x)} is an orthogonal polynomial sequence with respect to the moments μn, show that {Qn(x)} is an orthogonal polynomial sequence with respect to the moments
n
k=0
nk (−b)n−kμk.
13. Suppose that {Qn(x)} satisfies (9.5.18), but with the initial conditions
νn=a−n Q−1(x) = −1 and Q0(x) = 0.
(a) Show that Qn(x) is a polynomial of degree n − 1. (1) −1
(b) Put Pn (x) = λ1 Qn+1(x) and write the difference equation (1)
corresponding to {Pn (x)}.
14. Let {Pn(x)} be a sequence of orthogonal polynomials on the interval
(a,b). Show that the zeros of Pn(x) are real, distinct, and lie in (a,b). 15. In the following justify that y(x) satisfies the given differential
equations:
(a) y′′ −2xy′ +2ny=0; y(x)=Hn(x).
(b) xy′′ +(α+1−x)y′ +ny=0; y(x)=Lαn(x).
(c) (1−x2)y′′ +{(β−α)−(α+β+2)x}y′ +n(n+α+β+1)y = (α,β)
9.6 Minimal Solutions, Continued Fractions, and Orthogonal Polynomials
The intimate connection between continued fractions and orthogonal poly- nomials is now apparent in light of the fundamental recurrence formulas for continued fractions and orthogonal polynomials. If {Pn(x)} is a monic orthogonal polynomial sequence on the interval (a, b), with P−1(x) = 0 and P0(x) = 1, then it must satisfy the difference equation
0; y(x) = Pn (x).
Pn+1 − (x − βn)Pn(x) + γnPn−1(x) = 0, n ∈ Z+. (9.6.1)
422 9. Applications to Continued Fractions and Orthogonal Polynomials
To find the continued fraction that corresponds to (9.6.1) we take bn = −(x − βn ), an = γn in (9.1.6). We then have the continued fraction
K= γ0 γ1 γ2 …. (9.6.2) (x−β0)− (x−β1)− (x−β2)−
Moreover, Pn(x) is the nth partial denominator of the continued fraction (9.6.2).
Next we focus our attention on finding a minimal solution of (9.6.1). Re- call from Pincherle’s theorem that (9.6.1) has a minimal solution if and only if the continued fraction K converges. To accomplish our task we need to find another polynomial solution Qn(x), called the associated polynomials, that forms with Pn(x) a fundamental set of solutions of (9.6.1).
Define
(9.6.3) Lemma 9.14. The set {Pn(x),Qn(x)} is a fundamental set of solutions
of (9.6.1).
Proof. From (9.6.1), we have
Pn+1(x) − Pn+1(t) = (x − t)Pn(t) + (x − βn)[Pn(x) − Pn(t)] − γn[Pn−1(x) − Pn−1(t)].
Qn(x) =
b (Pn(x) − Pn(t))
Dividing by x − t and integrating yields b
Qn+1(x) = with
a
n = 0, 1, 2, . . . ,
0
μ0 ifn=0.
Pn(t)w(t) dt+(x−βn)Qn(x)−γnQn−1(x), Q−1(x) = 0, Q0(x) = 0.
Notice that by the orthogonality of {Pn(x)} we have
b b Pn(t)w(t) dt =
a a
if n > 0,
a
x − t w(t) dt.
Pn(t)P0(t)w(t) dt =
Hence, we obtain
Qn+1(x) − (x − βn)Qn(x) + γnQn−1(x) = 0,
Q0(x) = 0, Q1(x) = μ0. (9.6.4)
Since P0(x) = 1, P1(x) = x − β0, the Casoratian W (0) of Pn and Qn at n = 0 is equal to μ0 ̸= 0, which implies that {Pn(x)} and {Qn(x)} are two linearly independent solutions of (9.6.1).
Observe that the polynomial sequence Qn(x) is the nth partial numerator of the continued fraction K. Hence if
lim Qn(x)=F(x) n→∞ Pn(x)
9.6 Minimal Solutions, Continued Fractions, and Orthogonal Polynomials 423
exists, then by Pincherle’s theorem, the minimal solution of (9.6.1) exists and is defined by
Sn(x) = F(x)Pn(x) − Qn(x). (9.6.5)
Furthermore, if we let γ0 = μ0 and Q−1(x) = −1, then Sn(x) satisfies (9.6.1) not only for n ≥ 1, but also for n = 0, with S−1(x) = −1.
To find an explicit formula for the minimal solution Sn(x) we need to utilize complex analysis. Henceforth, we replace the variable x by z in all the considered functions. From a result of [113], F(z) has the integral representation
b w(t)
z − t d t , z ∈/ [ a , b ] . ( 9 . 6 . 6 ) Combining this formula with (9.6.3) produces the following integral
Remark:
Sn(z) is always guaranteed (see [113]).
(ii) If (a, b) is a half-infinite interval of the form (a, ∞) or (−∞, b), then a sufficient condition for the existence of the minimal solution is
∞
μ−1/2n = ∞
n n=1
(see [66]).
(iii) If (a, b) = (−∞, ∞), then a sufficient condition for the existence of the
minimal solution Sn(z) is
representation of Sn(z):
Sn(z) =
b Pn(t)
F ( z ) =
a
(z − t) w(t) dt.
(i) If (a, b) is a finite interval, then the existence of the minimal solution
∞
n=1 n=1
a
(9.6.7)
P
∞
2n n
μ−1/2n = ∞, or
Proofs of these remarks are beyond the scope of this book and will be
omitted. For details the reader may consult [66], [113]. Example 9.15 [145]. Consider the difference equation
with
Pn+1(z) − zPn(z) + γnPn−1(z) = 0, n ≥ 0, (9.6.8)
γn = n(n + 2ν − 1) , ν > 1. 4(n+ν)(n+ν −1)
γ−1/2 = ∞.
424 9. Applications to Continued Fractions and Orthogonal Polynomials
Notice that Pn(z) is related to the Gegenbauer polynomials Pnν(z) by the relation
where
Pn(z) = n! Pnν(z), 2n(ν)n
(ν)n =ν(ν+1)(ν+2)···(ν+n−1)= Γ(ν+n) Γ(ν )
(9.6.9)
(9.6.10)
denotes the Pochhammer symbol. Hence by formula (9.6.7), a minimal solution of (9.6.8) is given by
Using the value of the integral found in Erd ́elyi et al. ([54, p. 281]) yields Sn(z) = (n + ν)n n! √π eiπ(ν−1/2)(z2 − 1)(2ν−1)/4Q1/2−λ (z), (9.6.11)
2n+ν −3/2 n+ν −1/2 where Qβα(z) is a Legendre function defined as
β α (z−1)(β/2)−α−1 2 Qα(z)=2 Γ(α+1) (z+1)β/2 F α+1,α−β+1;2α+2;1−z ,
with
∞ ( a ) s ( b ) s z s
Γ(c + s)s! (|z| < 1). (9.6.12) The function F (a, b; c; z) is called a hypergeometric function [110].
Quite often it is possible to find an asymptotic representation of the minimal solution using the methods of Chapter 8. The following example demonstrates this point.
Example 9.16. (Perturbations of Chebyshev Polynomials [71]). Consider again a monic orthogonal polynomial sequence {Pn(x)} repre-
sented by the second-order difference equation
Qn+1(z) − (z − an)Qn(z) + bnQn−1(z) = 0 (9.6.13)
with Q−1(z) = 0, Q0(z) = 1.
The (complex) Nevai class M(a,b) [106] consists of all those orthogonal
polynomial sequences such that limn→∞ an = a, limn→∞ bn = b. Without loss of generality we take a = 0 and b = 1. (Why?) Then the limiting
4
Pn+1(z) − zPn(z) + 1Pn−1(z) = 0. (9.6.14) 4
n! 1 Pnν(t)(1 − t2)ν−1/2
Sn(z)=2n(ν)n −1 z−t dt, z∈/[−1,1].
F (a, b; c; z) =
equation associated with (9.6.1) is given by
s=0
9.6 Minimal Solutions, Continued Fractions, and Orthogonal Polynomials 425
Observe that {Pn(z)} are the monic Chebyshev polynomials, which can be obtained (Appendix F, No. 4) by setting Pn(z) = 21−n Tn(z) (since 2n−1 is the leading coefficient in Tn(z)), where {Tn(x)} are the Chebyshev polynomials of the first kind. Equation (9.6.14) has two solutions,
Pn+(z) = λn+(z), and Pn−(z) = λn−(z), z ̸= 1, −1, (9.6.15)
where
λ+(z)=(z+ z2 −1)/2 and λ−(z)=(z− z2 −1)/2.
Observe that if z ∈ C\[−1, 1], we can choose the square root branch such
that
λ+(z) < 1. λ−(z)
Hence Pn−(z) is a minimal solution on z ∈ C\[−1,1], and Pn+(z) is a dominant solution on z ∈ C\[−1, 1].
Now, the boundary values of the minimal solution on the cut Pn−(x + i0) = lim Pn−(x + iε) = (λ−(x))n,
ε→0+
Pn−(x − i0) = lim Pn−(x − iε) = (λ+(x))n,
ε→0+
with
λ−(x)=(x− 1−x2)/2, λ+(x)=(x+i 1−x2)/2, x∈(−1,1),
yields a ratio of solutions that oscillates as n → ∞. Thus there is no minimal solution for z = x ∈ (−1, 1).
Next we turn our attention to (9.6.13). Since |λ+(z)| ̸= |λ−(z)|, by virtue of the Poincar ́e–Perron theorem there are two linearly independent solutions Q+n (z), Q−n (z) of (9.6.13) such that
lim Q+n+1(z) = λ+(z), lim Q−n+1(z) = λ−(z). n→∞ Q+n (z) n→∞ Q−n (z)
Furthermore, Q+n (z) is a dominant solution and Q−n (z) is a minimal solution of (9.6.1) for z ∈ C\[−1, 1].
Moreover, if
∞ 1
|an| + bn − 4 < ∞,
(9.6.16)
n=0 then by Corollary 8.30, we have
Q−n (z) = λn−(z)(1 + o(1)), z ∈ C\[−1, 1],
Q+n (z) = λn+(z)(1 + o(1)), z ∈ C\[−1, 1],
where Q−n and Q+n are, respectively, minimal and dominant solutions of (9.6.13). Furthermore, for z = x ∈ (−1,1), there are two linearly
426 9. Applications to Continued Fractions and Orthogonal Polynomials
independent solutions
Qn(x + i0) = (λ−(x))n(i + o(1)), Qn(x − i0) = (λ+(x))n(i − o(1)),
where
λ−(x)=x−i 1−x2, λ+(x)=x+i 1−x2.
For relaxing condition (9.6.16) and more generalizations, the reader is referred to [38].
Exercises 9.6
1. Show that
2. Show that
H
(x) = (−1)n22nn! L(−1/2)(x2), 2n n
H
2n+1
(x) = (−1)n22n+1n! xL(−1/2)(x2). n
22n (n!)2 Tn(x) = (2n)!
(−1/2,−1/2)
Pn (x),
U (x) = 22nn!(n + 1)!P(1/2,1/2)(x). n (2n+1)! n
In Problems 3 through 6 investigate the existence of a minimal solution for the given polynomial. If a minimal solution exists, find an explicit representation for it.
3. Legendre polynomials {Pn(x)} (See Appendix F, No. 3).
4. Hermite polynomials {Hn(x)} (See Appendix F, No. 6).
5. Laguerre polynomials {Lαn(x)} (See Appendix F, No. 7). (a)
6. Charlier polynomials {Cn } (See Appendix F, No. 8).
*7. Use Rodrigues’ formula for the Legendre polynomial Pn(x) and the Cauchy integral formula2 for the nth derivative to derive the formula (Sch ̈afli’s integral)
1 (t2−1)n
Pn(x) = 2n+1πi γ (t − x)n+1 dt, (9.6.17)
where γ is any positively directed simple closed curve enclosing the point x (x may be real or complex).
2f(n)(z0) = n! f(z) dz, n = 0, 1, 2, ..., where γ is any positively 2πi γ (z−z0 )n+1
directed closed curve enclosing z0.
9.6 Minimal Solutions, Continued Fractions, and Orthogonal Polynomials 427
Generating functions [110]. Suppose that G(x, u) is a polynomial with the Maclaurin expansion
∞ n=0
Then G(x,u) is called a generating function for {Qn(x)}. *8. (a) Show that
1 (t2−1)h−1 dt ∞
2πi 1− 2(t−x) t−x = Pn(x)hn =G(x,h).
γ n=0
(b) Deduce from (a) that the generating function of Pn(x) is given by
G(x, u) = (1 − 2xu + u2)−1/2.
*9. Consider the Chebyshev polynomials of the first kind {Tn(x)}.
(a) Show that T (x) = (−1)nn! √1−x2 (1−z2)n−1/2 dz. n 1·3···(2n−1) 2πi γ (z−x)n+1
(b) Then use (a) to verify that the generating function of Tn(x) is given by
1−u2 1 G(x, u) = 2(1 − 2xu + u2 ) − 2 .
*10. Consider the Gegenbauer polynomials Pnν(x). (a) Show that
(−1)n(2ν + n − 1)!(ν − 1 )! Pν(x) = 2
G(x, u) =
Qn(x)un.
(1 − z2)ν+n−1/2 (z−x)n+1
dz. (b) Show that the generating function of Pnν (x) is G(x, u) = (1 −
(a) Show that
n (2ν−1)!(ν+n−1)!(1−x2) γ 2
2xu+u2)−ν.
*11. Consider the Hermite polynomials {Hn(x)}.
(a) Show that
(b) Show that
Hn(x) = (−1)nn! exp(x2 − z2)dz. nπi γ (z − x)n+1
exp(2ux−u)=
n! u.
2 ∞Hn(x)n
n=0 12. Consider the Laguerre polynomials {Lαn(x)}.
Lαn(x)=x−α zn+λexp(x−z)dz. 2πi γ (z − x)n+1
428 9. Applications to Continued Fractions and Orthogonal Polynomials
(b) Show that the generating function of {Lαn(x)} is
−xu G(x,u)=(1−u)−α−1exp 1−x .
10
Control Theory
10.1 Introduction
In the last three decades, control theory has gained importance as a dis- cipline for engineers, mathematicians, scientists, and other researchers. Examples of control problems include landing a vehicle on the moon, con- trolling the economy of a nation, manufacturing robots, and controlling the spread of an epidemic. Though a plethora of other books discuss contin- uous control theory [6], [75], [96], we will present here an introduction to discrete control theory.
We may represent a physical system that we intend to control by the homogeneous difference system
x(n + 1) = Ax(n), (10.1.1)
where A is a (k × k) matrix. We extensively studied this equation in Chapters 3 and 4; here we will refer to it as an uncontrolled system.
To control this system, or to induce it to behave in a predetermined fashion, we introduce into the system a forcing term, or a control, u(n). Thus, the controlled system is the nonhomogeneous system
x(n + 1) = Ax(n) + u(n). (10.1.2)
In realizing system (10.1.2), it is assumed that the control can be applied to affect directly each of the state variables x1(n),x2(n),...,xk(n) of the system. In most applications, however, this assumption is unrealistic. For
429
430 10. Control Theory
example, in controlling an epidemic, we cannot expect to be able to affect directly all of the state variables of the system.
We find another example in the realm of economics. Economists, and certain politicians even, would pay dearly to know how the rate of inflation can be controlled, especially by altering some or all of the following vari- ables: taxes, the money supply, bank lending rates. There probably is no equation like (10.1.2) that accurately describes the rate of inflation. Thus, a more reasonable model for the controlled system may be developed: We denote it by
x(n + 1) = Ax(n) + Bu(n), (10.1.3)
where B is a (k × m) matrix sometimes called the input matrix, and u(n) is an m×1 vector. In this system, we have m control variables, or components, u1(n), u2(n), . . . , um(n), where m ≤ k.
In engineering design and implementation, the system is often repre- sented by a block diagram, as in Figures 10.1, 10.2.
The delay is represented traditionally by z−1, since 1 Z[x(n + 1)] = z
Z[x(n)]. (See Figure 10.3.)
Delay x(n+1)
x(n)
A
-1 z
u(n)
FIGURE 10.1. Uncontrolled system.
Delay x(n+1)
+
+
FIGURE 10.2. Controlled system.
x(n)
B
z-1
A
x(n+1) x(n)
~~
z x(z) x(z)
FIGURE 10.3. Representation of system delay.
10.1 Introduction 431
HT
u(t) x(t)
Continuous System Σc
ST
Discrete system Σd
u(k)
x(k)
FIGURE 10.4. A continuous system with ideal sampler and zero-order hold.
10.1.1 Discrete Equivalents for Continuous Systems
One of the main areas of application for the discrete control methods de- veloped in this chapter is the control of continuous systems, i.e., those modeled by differential and not difference equations. The reason for this is that while most physical systems are modeled by differential equations, control laws are often implemented on a digital computer, whose inputs and outputs are sequences. A common approach to control design in this case is to obtain an equivalent difference equation model for the continuous system to be controlled.
The block diagram of Figure 10.4 shows a common method of interfacing
a continuous system to a computer for control. The system c has state vector x(t) and input u(t) and is modeled by the differential equation
x ̇(t) = Aˆ(t)x(t) + Bˆu(t). (10.1.4) ThesystemST isanidealsamplerthatproduces,givenacontinuoussignal
x(t), a sequence x(k) defined by
x(k) = x(kT ). (10.1.5)
The system HT is a zero-order hold that produces, given a sequence u(k), a piecewise-constant continuous signal uc(t) defined by
u(t)=u(k), t∈[kT,(k+1)T). (10.1.6)
It is not hard to check that the solution to (10.1.4) for t ∈ [kT,(k + 1)T) is given by
(10.1.7)
x(t) = eAˆtx(kT) +
t kT
eAˆ(t−τ)Bˆu(τ) dτ.
-1 z
432 10. Control Theory
Thus a difference equation model for the overall system d (indicated by the dotted box in Figure 10.4) can be obtained by evaluating formula (10.1.7) at t = (k + 1)T and using (10.1.5) and (10.1.6):
x(k + 1) = Ax(k) + Bu(k), (10.1.8)
where
A = eAˆT and B = TeAˆT Bˆ. (10.1.9) Example 10.1. A current-controlled DC motor can be modeled by the
differential equation
x ̇ ( t ) = − 1 x ( t ) + K u ( t ) , ττ
where x is the motor’s angular velocity, u is the applied armature current, and K and τ are constants. A difference equation model suitable for the design of a discrete control system for this motor can be found using (10.1.8) and (10.1.9):
where
x(k + 1) = Ax(k) + Bu(k),
A=eAˆT =e−T/τ and B=TeAˆTBˆ=KT e−T/τ. τ
10.2 Controllability
In this section we are mainly interested in the problem of whether it is possible to steer a system from a given initial state to any arbitrary state in a finite time period. In other words, we would like to determine whether a desired objective can be achieved by manipulating the chosen control variables. Until 1960, transform methods were the main tools in the analysis and design of controlled systems. Such methods are referred to now as classical control theory. In 1960, the Swiss mathematician/engineer R.E. Kalman [77] laid down the foundation of modern control theory by introducing state space methods. Consequently, matrices have gradually re- placed transforms (e.g., Z-transform, Laplace transform), as the principal mathematical machinery in modern control theory [88], [108], [142].
Definition 10.2. System (10.1.3) is said to be completely controllable (or simply controllable) if for any n0 ∈ Z+, any initial state x(n0) = x0, and any given final state (the desired state) xf , there exists a finite time N > n0 and a control u(n),n0 < n ≤ N, such that x(N) = xf.1
1In some books such a system may be referred to as completely reachable.
Remark: Since system (10.1.3) is completely determined by the matrices A and B, we may speak of the controllability of the pair {A, B}.
In other words, there exists a sequence of inputs u(0), u(1), . . . , u(N − 1) such that this input sequence, applied to system (10.1.3), yields x(N ) = xf .
Example 10.3.
Here
Consider the system governed by the equations x1(n + 1) = a11x1(n) + a12x2(n) + bu(n),
x2(n + 1) = a22x2(n).
A= a11 a12 , B= b . 0 a22 0
It will not take much time before we realize that this system is not completely controllable, since u(n) has no influence on x2(n). Moreover, x2(n) is entirely determined by the second equation and is given by x2(n) = an22x2(0).
The above example was easy enough that we were able to determine controllability by inspection. For more complicated systems, we are going to develop some simple criteria for controllability.
The controllability matrix W of system (10.1.3) is defined as the k × km matrix
(10.2.1)
The controllability matrix plays a major role in control theory, as may be seen in the following important basic result.
Theorem 10.4. System (10.1.3) is completely controllable if and only if rank W = k.
Before proving the theorem, we make a few observations about it and then prove a preliminary result.
First, consider the simple case where the system has only a single input, and thus the input matrix B reduces to an m × 1 vector b. Hence the controllability matrix becomes the k × k matrix
The controllability condition that W has rank k means that the matrix W is nonsingular or its columns are linearly independent. For the general case, the controllability condition is that from among the km columns there are k linearly independent columns. Let us now illustrate the theorem by an example.
10.2 Controllability 433
W = [B,AB,A2B,...,Ak−1B].
W = [b,Ab,...,Ak−1b].
434 10. Control Theory
Example 10.5. Contemplate the system
y1(n + 1) = ay1(n) + by2(n),
y2(n + 1) = cy1(n) + dy2(n) + u(n),
where ad − bc ̸= 0. Here A=,B=,
and u(n) is a scalar control. Now,
W =(B,AB)=
has rank 2 if b ̸= 0. Thus the system is completely controllable by Theorem 10.4 if and only if b ̸= 0.
Lemma 10.6. For any N ≥ k, the rank of the matrix [B,AB,A2B,...,AN−1B]
is equal to the rank of the controllability matrix W.
Proof. (I) Consider the matrix W (n) = [B, AB, . . . , An−1B], n = 1, 2, 3, . . .. As n increases by 1, either the rank of W (n) remains constant or increases by at least 1. Suppose that for some r > 1, rank W(r + 1) = rank W(r). Then every column in the matrix ArB is linearly dependent on the columns of W (r) = [B, AB, . . . , Ar−1B]. Hence
ArB = BM0 + ABM1 + · · · + Ar−1BMr−1, (10.2.2) where each Mi is an m×m matrix. By premultiplying both sides of (10.2.2)
by A, we obtain
Ar+1B = ABM0 + A2BM1 + · · · + ArBMr−1.
Thus the columns of Ar+1B are linearly dependent on the columns of W (r+ 1). This implies that rank W(r + 2) = rank W(r + 1) = rank W(r). By repeating this process, one may conclude that
rank W(n) = rank W(r) for all n > r.
We conclude from the above argument that rank W(n) increases by at least 1 as n increases by 1 until it attains its maximum k. Hence the rank maximum of W(n) is attained in at most k steps. There- fore, the maximum rank is attained at n ≤ k and, consequently, rank W (≡ rankW(k))= rankW(N)forallN≥k. P
Proof. (II) In the second proof we apply the Cayley–Hamilton theorem (Chapter 3). So if p(λ) = λk + p1λk−1 + · · · + pk is the characteristic polynomial of A, then p(A) = 0, i.e.,
ab0 cd1
Ak +p1Ak−1 +···+pkI =0,
0b 1d
or
or
where
x(k) − Akx(0) = Wu ̄(k), ⎛u(k − 1)⎞
⎜u(k − 2)⎟ u ̄ ( k ) = ⎜ . . ⎟ .
⎝.⎠ u(0)
(10.2.5)
AkB =
qiAk−iB. (10.2.4)
x(k) − A x(0) =
A Bu(r),
k
k−1
k−r−1
k
qiAk−1, (10.2.3)
Ak =
where qi = −pi. Multiplying expression (10.2.3) by B, we obtain
k i=1
Thus the columns of AkB are linearly dependent on the columns of W(k) ≡ W . Therefore, rank W (k + 1) = rank W . By multiplying expression (10.2.4) by A we have
Ak+1B = q1Ak + q2Ak−1 + · · · + qkA.
Consequently, rank W (k + 2) = rank W (k + 1) = rank W . By repeating
the process, one concludes that rank W(N) = rank W for all N ≥ k. P We are now ready to prove the theorem.
Proof of Theorem 10.4.
Sufficiency Suppose that rank W = k. Let x0 and xf be two arbitrary vectors in Rk. Recall that by the variation of constants formula (3.2.14) we have
i=1
r=0
SincerankW =k,rangeW =Rk.Henceifweletx(0)=x0 andx(k)=xf, then xf −Akxo ∈ range W. Thus xf −Akx0 = Wu ̄ for some vector u ̄ ∈ Rk. Consequently, system (10.1.3) is completely controllable.
Necessity Assume that system (10.1.3) is completely controllable and rank W < k. From the proof of Lemma 10.6 (Proof I) we conclude that there exists r ∈ Z+ such that
10.2 Controllability 435
436 10. Control Theory
rank W(1) < rank W(2) < ··· < rank W(r) = rankW(r+1)=···= rankW.
Moreover, rank W(n) = rank W for all n > k. Furthermore, since W(j + 1) = (W (j), Aj B), it follows that
range W(1) ⊂ range W(2) ⊂ ··· ⊂ range W(r)
= rangeW(r+1)=···=rangeW =···= rangeW(n)
for any n > k.
Since rank W < k, range W ̸= Rk. Thus there exists ξ ̸∈ range W.
Thisimpliesthatξ̸∈rangeW(n)foralln∈Z+.Ifweletx0 =0in formula (10.2.5) with k replaced by n, we have x(n) = W(n)u ̄(n). Hence for ξ to be equal to x(n) for some n, ξ must be in the range of W(n). But ξ ̸∈ range W(n) for all n ∈ Z+ implies that ξ may not be reached at any time from the origin, which is a contradiction. Therefore, rank W = k. P
Remark 10.7. There is another definition of complete controllability in the literature that I will call here “controllability to the origin.” A system is controllable to the origin if, for any n0 ∈ Z+ and x0 ∈ Rk, there exists a finitetimeN>n0 andacontrolu(n),n0
u(n)
+
x(n+1) x(n)
y(n)
B
-1
z
A
C
A
FIGURE 10.7. Input–output system: y(n) = Cx(n).
u(n)
x1(n+1) +
x2(n+1)
x1(n)
x2(n)
y(n)
10.3 Observability 447
b1
b2
+
FIGURE 10.8. A nonobservable system.
Example 10.12. Consider the system (Figure 10.8)
x1(n + 1) = a1x1(n) + b1u(n), x2(n + 1) = a2x2(n) + b2u(n),
y(n) = x1(n).
This system is not observable, since the first equation shows that x1(n) = y(n) is completely determined by u(n) and x1(0) and that there is no way to determine x2(0) from the output y(n).
In discussing observability, one may assume that the control u(n) is iden- tically zero. This obviously simplifies our exposition. To explain why we can do this without loss of generality, we write y(n) using the variation of constants formula (3.2.14) for x(n):
y(n) = Cx(n),
or
Since C, A, B, and u are all known, the second term on the right-hand side of this last equation is known. Thus it may be subtracted from the observed value y(n). Hence, for investigating a necessary and sufficient condition for complete observability it suffices to consider the case where u(n) ≡ 0.
We now present a criterion for complete observability that is analogous to that of complete controllability.
z-1
1
a1
z-1
a2
y(n) = CAn−n0 x0 +
n−1
j =n0
CAn−j−1Bu(j).
448 10. Control Theory
Theorem 10.13. System (10.3.1) is completely observable if and only if the rk × k observability matrix
Let
yˆ(n) = y(n) −
n−1
n−r−1 CA
⎡C⎤ ⎢ C A ⎥
V= ⎢ CA2 ⎥ ⎢ . ⎥
(10.3.2)
has rank k.
Proof. By applying the variation of constants formula (3.2.14) to
(10.3.1) we obtain y(n)=Cx(n)=C A x0 +
⎣.⎦ CAk−1
n−1
n n−r−1 A
Bu(r) .
(10.3.3)
(10.3.4)
(10.3.5)
(10.3.6)
r=0
Using formula (10.3.3), equation (10.3.4) may now be written as
yˆ(n) = CAnx0. Putting n = 0, 1, 2, . . . , k − 1 in (10.3.5) yields
⎡ yˆ(0) ⎤ ⎡ C ⎤ ⎢ yˆ(1) ⎥ ⎢ CA ⎥
⎢ . ⎥ = ⎢ . ⎥ x0. ⎣.⎦⎣.⎦
yˆ(k − 1) CAk−1
Suppose that rank V = k. Then range V = Rk. Now, if y(n),u(n) are given for 0 ≤ n ≤ k − 1, then it follows from (10.3.4) that yˆ(n),0 ≤ n ≤ k − 1, is also known. Hence there exists x0 ∈ Rk such that (10.3.6) holds. Hence system (10.3.1) is completely observable. Conversely, suppose system (10.3.1) is completely observable. Let us write
⎡C⎤ ⎢ C A ⎥
V(N)=⎢ CA2 ⎥=CT,ATCT,(AT)2CT,…,(AT)N−1CTT . ⎢ . ⎥
⎣.⎦ CAN−1
r=0
Bu(r).
(10.3.7)
Then from Theorem 10.13, V (N ) is of rank k if and only if the observability matrix V ≡ V (k) has rank k. Therefore, if x0 can be uniquely determined fromN observationsy(0),y(1), …,y(N−1),itcanbesodeterminedfrom y(0),y(1), …,y(k−1).ThusrankV =k. P
Notice that the matrix B does not play any role in determining observ- ability. This confirms our earlier remark that in studying observability, one may assume that u(n) ≡ 0. Henceforth, we may speak of the observability of the pair {A, C}.
Example 10.3 revisited. Consider again Example 10.3. The system may bewrittenas
x1(n+1) = a1 0 x1(n) + b1 u(n), x2(n+1) 0 a2 x2(n) b2
y(n) = 1 0 x1(n) .
x2 (n)
a1 0 and C = 1 0. It follows that the observability
Thus A =
matrixisgivenbyV = 1 a1 0 0 .SincerankV =1<2,thesystem
is not completely observable by virtue of Theorem 10.13. Finally, we give an example to illustrate the above results.
Example 10.14. Consider the input–output system (Figure 10.9)
x1(n + 1) = x2(n),
x2(n + 1) = −x1(n) + 2x2(n) + u(n),
y(n) = c1x1(n) + c2x2(n).
010
Then A = −1 2 ,B = 1 , and C = (c1,c2). The observability
matrix is given by
V=C=c1 c2 . CA −c2 c1 +2c2
0 a2
By adding the first column to the second column in V we obtain the matrix
Vˆ= c1 c1+c2 . −c2 c1 + c2
ObservethatrankVˆ =2ifandonlyifc1+c2 ̸=0.SincerankV =rankVˆ, it follows that the system is completely observable if and only if c1 + c2 ̸= 0 (or c2 ̸= −c1).
We may also note that the system is completely controllable.
10.3 Observability 449
450 10. Control Theory
u(n)
+
x2(n+1) x2(n) x2(n)
+ y(n)
FIGURE 10.9. A completely observable and controllable system.
Example 10.15. Example 10.10 looked at the controllability of a cart attached to a fixed wall via a flexible linkage using an applied force u. A dual question can be posed: If the force on the cart is a constant, can its magnitude be observed by measuring the cart’s position? In order to answer this question, the state equation (10.2.8) must be augmented with one additional equation, representing the assumption that the applied force is constant:
⎡⎤⎡ ⎤⎡⎤
x ̇ 0 1 0 x ⎢⎣v ̇ ⎥⎦ = ⎢⎣−k/m −b/m 1/m⎥⎦ ⎢⎣v⎥⎦ ,
u ̇ 0 0 0 u ⎡⎤
x
y = 1 0 0 ⎢⎣ v ⎥⎦ .
u
Using the same values m = 1, k = 2, and b = 3 as in Example 6.10,
can be written as
where
z-1
⎡⎤
010 Aˆ = ⎢⎣−2 −3 1⎥⎦
000
Aˆ = P Λ P − 1 , ⎡⎤⎡⎤
−1 0 0 1 1 1
Λ = ⎢⎣ 0 − 2 0 ⎥⎦ , P = ⎢⎣ − 1 − 2 0 ⎥⎦ .
c2
z-1
2
-1
000 002
c1
Thus
A=eAˆT =PeΛTP−1
⎡11⎤ 2e−T −e−2T e−T −e−2T +e−T + e−2T
⎢ 22⎥ =⎣−2e−T +2e−2T −e−T +2e−2T −e−T −e−2T ⎦
001 To check observability, we must compute
⎡⎤
C
V = ⎢⎣ C A ⎥⎦
CA2 ⎡100⎤
10.3 Observability 451
⎢2e−T −e−2T =⎢⎣
2e−2T − e−4T and its determinant
e−T −e−2T e−2T − e−4T
1 +e−T + 1e−2T ⎥
2 2 ⎥⎦ 1 + 2e−T + e−2T + 1e−4T 22
detV =e−T +2e−2T −4e−3T −2e−4T +3e−5T = e−T (1 + e−T )(1 − e−T )2(1 + 3e−T ).
The factored form above shows that since T is real, det V = 0 only if T = 0. The system is therefore observable for all nonzero T .
Theorem 10.13 establishes a duality between the notions of controllability and observability. The following definition formalizes the notion of duality.
Definition 10.16. The dual system of (10.3.1) is given by x(n + 1) = AT x(n) + CT u(n),
y(n) = BT x(n). (10.3.8) Notice that the controllability matrix W ̄ of system (10.3.8) may be given
by
W ̄ =CT,ATCT,(AT)2CT, ...,(AT)k−1CT. Furthermore, the observability matrix V of system (10.3.1) is the transpose
of W ̄ , i.e.,
But since rank W ̄ = rank W ̄ T = rank V , we have the following conclusion.
Theorem 10.17 (Duality Principle). System (10.3.1) is completely controllable if and only if its dual system (10.3.8) is completely observable.
V = W ̄ T .
452 10. Control Theory
complete controllability
controllability to the origin
dual
det A ≠ 0
det A ≠ 0
FIGURE 10.10.
dual
complete observability
constructibility
Remark: In Remark 10.7 we introduced a weaker notion of controllability, namely, controllability to the origin. In this section we have established a duality between complete controllability and complete observability. To complete our analysis we need to find a dual notion for controllability to the origin. Fortunately, such a notion does exist, and it is called constructibility. System (10.3.1) is said to be constructible if there exists a positive integer N such that for given u(0),u(1), ...,u(N −1) and y(0),y(1), ...,y(N − 1) it is possible to find the state vector x(N) of the system. Since the knowledge of x(0) yields x(N) by the variation of constants formula, it follows that complete observability implies constructibility. The two notions are in fact equivalent if the matrix A is nonsingular. Figure 10.10 illustrates the relations among various notions of controllability and observability.
Finally, we give an example to demonstrate that constructibility does not imply complete observability.
Example 10.18. Contemplate a dual of the system in Example 10.8:
x1(n+1) = 0 0 x1(n) + 1 u(n), x2(n+1) 1 0 x2(n) 0
y(n) = 1 0 x1(n) .
x2 (n) The observability matrix is given by
10 V=,
00
whose rank is 1. It follows from Theorem 10.13 that the system is not completely observable. However, if we know u(0), u(1) and y(0), y(1), then from the second equation we find that x1(1) = y(1). The first equation
yields x1(2) = u(1) and x2(2) = x1(1). Thus
x1 (2)
x(2) =
x2 (2)
is now obtained and, consequently, the system is constructible.
10.3.1 Observability Canonical Forms
Consider again the completely observable system x(n + 1) = Ax(n) + bu(n),
y(n) = Cx(n),
(10.3.9)
where b is a k×1 vector and C is a 1×k vector. Recall that in Section 10.2 we constructed two controllability canonical forms of system (10.3.1). By exactly parallel procedures we can obtain two observability canonical forms corresponding to system (10.3.9). Both procedures are based on the nonsingularity of the observability matrix
⎛C⎞ ⎜ C A ⎟
V=⎜ . ⎟. ⎝.⎠
CAk−1
If we let z(n) = Vx(n) in (10.3.11), we obtain the first observability
canonical form (Exercises 10.3, Problem 10)
z ( n + 1 ) = A ̄ z ( n ) + ̄b u ( n ) ,
where
y(n) = c ̄ z(n),
⎛⎞
010...0 ⎜0 0 1 ... 0⎟
(10.3.10)
A ̄=⎜. . . ... .⎟, ⎜ ⎟
⎝000...1⎠ −pk −pk−1 −pk−2 . . . −p1
c ̄=(1 0 0 ··· 0), ̄b = V b.
(10.3.11)
10.3 Observability 453
454 10. Control Theory
In Exercises 10.3, Problem 10, the reader is asked to find a change of variable that yields the other observability canonical form {A ̃, c ̃}, with
⎛0 0 ... 0 −p ⎞ k
⎜1 0 ... 0 −pk−1⎟ A ̃=⎜0 1 ... 0 −pk−2⎟, c ̃= 0 0 ··· 1 .
(10.3.12)
⎜. . . ⎟ ⎝.. . ⎠
0 0 ... 1 −p1
Exercises 10.3
1. Consider the input–output system
x(n + 1) = Ax(n) + Bu(n),
01 whereA= 2 −1 .
y(n) = Cx(n),
(a) If C = (0,2), show that the pair{A,C} is observable. Then find x(0) if y(0) = a and y(1) = b.
(b) If C = (2, 1), show that the pair {A, C } is unobservable. 2. Determine the observability of the pair {A,C}, where
⎛0 1 0⎞
A = ⎜ ⎜⎝ 0 0 1 ⎟ ⎟⎠ , C = 2 − 3 − 2 .
−1 1 1 2 3 1 44
3. Consider the system defined by
x1(n+1) = a b x1(n) + 1 u(n), x2(n+1) c d x2(n) 1
y(n) = 1 0 x1(n) .
x2 (n)
Determine the conditions on a,b,c, and d for complete state control- lability and complete observability. In Problems 4 and 5, determine the observability of the pair {A, C}.
4.
5.
⎛⎞
21000 ⎜02100⎟
A=⎜0 0 2 0 0⎟, C= 1 ⎜⎝ 0 0 0 − 3 1 ⎟⎠ 0
000 0−3 ⎛⎞
1 1 0 1 . 1 1 1 0
21000 ⎜02100⎟
A=⎜0 0 2 0 0⎟, C= 1 ⎜⎝ 0 0 0 − 3 1 ⎟⎠ 0
000 0−3
1 1 0 1 . 1 1 0 0
A21 : m × r; C1 : p × r;
6. Show that the pair {A, C }, where
A11 0 , C=(C1 0), A11:r×r; A21 A22 A22 : m × m;
is not completely observable for any submatrices A1, A21, A22, and C1.
7. Prove that system (10.3.2) is completely observable if and only if rank[C,CA,...,CAm−r]T =k,wheremisthedegreeoftheminimal polynomial of A, and r = rank C.
8. Prove that system (10.3.2) is completely observable if and only if the matrix V T V is positive definite, where V is the observability matrix of {A,C}.
9. Show that the kth-order scalar equation
z(n + k) + p1z(n + k − 1) + · · · + pkz(n) = u(n),
y(n) = c z(n), is completely observable if c ̸= 0.
10. Verify that the change of variable z(n) = Vx(n) produces the observability canonical pair {A ̄, c ̄} defined in expression (6.3.13).
11. Consider system (10.3.2), where P−1AP is a diagonal matrix. Show that a necessary and sufficient condition for complete observability is that none of the columns of the r × k matrix CP consist of all zero elements.
12. Consider system (10.3.2), where P−1AP is in the Jordan form J. Show that necessary and sufficient conditions for complete observability of the system are:
10.3 Observability 455
456 10. Control Theory
(i) no two Jordan blocks in J correspond to the same eigenvalue of A,
(ii) none of the columns of CP that correspond to the first row of each Jordan block consists of all zero elements, and
(iii) no columns of CP that correspond to distinct eigenvalues consist of all zero elements.
P be a nonsingular matrix. Show that if the pair {A,C} is 14. Show that if the matrix A in system equation (10.3.2) is nonsingular,
13. Let
completely observable, then so is the pair {P−1AP,CP}.
then complete observability and constructibility are equivalent. 15. Consider the completely observable system
x(n + 1) = Ax(n) + Bu(n), y(n) = Cx(n),
whereaisak×kmatrixandCisa1×kvector.LetM = (CT ,AT CT ,...,(AT )k−1CT ).
⎛⎞
010...0 ⎜0 0 1 ... 0⎟
(a) ShowthatMTA(MT)−1 =⎜ . . . . ⎟, ⎜ ⎟
−ak −ak−1 −ak−2 . . . −a1 where ai,1 ≤ i ≤ k, are the coefficients of the characteristic
polynomial
|λI−A|=λk +a1λk−1 +···+ak.
(b) Write down the corresponding canonical controllable system by letting x ̃(n) = M T x(n). Then deduce a necessary condition on C for complete observability of the original system.
16. Consider the system
x(n + 1) = Ax(n), y(n) = Cx(n),
where A is a k×k matrix and C is a 1×k vector. Prove that the system is completely observable if and only if the matrix
⎝000...1⎠
G = (C, CA−1, CA−2, . . . , CA−k+1) is nonsingular.
10.4 Stabilization by State Feedback (Design via Pole Placement) 457
u(n)
x(n+1)
x(n)
B
+
+
z-1
or
x(n + 1) = Ax(n) − BKx(n), x(n + 1) = (A − BK)x(n).
x(n+1)
+
+
FIGURE 10.12. Closed-loop system.
u(n)
(10.4.2)
x(n)
FIGURE 10.11. Open-loop system.
10.4 Stabilization by State Feedback (Design via Pole Placement)
Feedback controls are used in many aspects of our lives, from the braking system of a car to central air conditioning. The method has been used by engineers for many years. However, the systematic study of stabilization by state feedback control is of more recent origin (see [3], [4]) and dates to the 1960s. The idea of state feedback is simple: It is assumed that the state vector x(n) can be directly measured, and the control u(n) is adjusted based on this information. Consider the (open-loop) time-invariant control system shown in Figure 10.11, whose equation is
x(n + 1) = Ax(n) + Bu(n), (10.4.1)
where, as before, A is a k×k matrix and B a k×m matrix.
Suppose we apply linear feedback u(n) = −Kx(n), where K is a real m × k matrix called the state feedback or gain state matrix. The resulting (closed-loop) system (Figure 10.12) obtained by substituting u = −Kx into
(10.4.1) is
A
B
z-1
A
-K
458 10. Control Theory
The objective of feedback control is to choose K in such a way such that the resulting system (10.4.2) behaves in a prespecified manner. For example, if one wishes to stabilize system (10.4.1), that is, to make its zero solution asymptotically stable, K must be chosen so that all the eigenvalues of A − BK lie inside the unit disk.
We now give the main result in this section.
Theorem 10.19. Let Λ = {μ1, μ2, . . . , μk} be an arbitrary set of k com- plex numbers such that Λ ̄ = {μ ̄1,μ ̄2,...,μ ̄k} = Λ. Then the pair {A,B} is completely controllable if and only if there exists a matrix K such that the eigenvalues of A − BK are the set Λ.
Since the proof of the theorem is rather lengthy, we first present the proof for the case m = 1, i.e., when B is a k×1 vector and u(n) is a scalar. We start the proof by writing the characteristic polynomial of A, |A − λI| = λk + a1λk−1 + a2λk−2 + · · · + ak. Suppose also that
k
(λ−μi)=λk +b1λk−1 +b2λk−2 +···+bk.
i=1
Define T = WM, where W is the controllability matrix of rank k defined
in (10.2.1) as
and
W = (B,AB,...,Ak−1B)
⎛⎞
ak−1 ak−2 ... a1 1
⎜ak−2
M=⎜ . ⎜
ak−3 .
. . .
1 0⎟
. .⎟. ⎟
⎝a1 1...00⎠ 1000
Then (Exercises 10.4, Problem 12)
⎛⎞
010...0 ⎜0 0 1 ... 0⎟
⎝000...1⎠ −ak −ak−1 −ak−2 . . . −a1
and
B ̄=T−1B=0 0 ··· 0 1T.
Letting x(n) = T x ̄(n) in system (10.4.2), we get the equivalent system
A ̄ = T−1AT = ⎜ . . . ⎟ (10.4.3) ⎜ ⎟
x ̄ ( n + 1 ) = ( A ̄ − B ̄ K ̄ ) x ̄ ( n ) , ( 1 0 . 4 . 4 )
with
1−3 1 A=,B=.
10.4 Stabilization by State Feedback (Design via Pole Placement) 459
where Choose Then
Observe that A − BK is similar to A ̄ − B ̄K ̄, since A ̄ − B ̄K ̄ = T−1AT −
K ̄ = K T . ( 1 0 . 4 . 5 ) K ̄ =(bk −ak,bk−1 −ak−1,...,b1 −a1). (10.4.6)
⎛⎞
00...0 ⎜0 0...0⎟
B ̄ K ̄ = ⎜ ⎜ . . . . . . . . . ⎟ ⎟ . ⎜ ⎟
T−1BKT =T−1(A−BK)T.Thus
λ 1 ... 0 0 λ ... 0
⎝00...0⎠ bk −ak bk−1 −ak−1 b1 −a1
. . 0 0 ... 1
−bk −bk−1 ... λ−b1 =λk +b1 +λk−1 +···+bk,
which has Λ as its set of eigenvalues. Hence the required feedback (gain) matrix is given by
K=K ̄T−1=(bk−ak,bk−1−ak−1, ...,b1−a1)T−1. Example 10.20. Consider the control system x(n + 1) = Ax(n) + Bu(n)
|λI−A+BK|=|λI−A ̄+B ̄K ̄|= .
421
Find a state feedback gain matrix K such that the eigenvalues of the closed
loop system are 1 and 1 . 24
Solution Method 1
So
|A−λI|=1−λ
−3 =λ2 −3λ +14.
4 2−λ
a1 =−3, a2 =14.
460 10. Control Theory
Also
So
Now,
Hence
and
Therefore,
or
Method 2 In this method we substitute K = (k1k2) into the characteristic polynomial |A − BK − λI| and then match the coefficients of powers in λ with the desired characteristic polynomial (10.4.7).
1−k1 −λ
1 1 3 1
λ − 2 λ − 4 = λ2 − 4 λ + 8 (10.4.7)
b1=−3 and b2=1. 48
1 −2 −3 1 W=,M=.
16 10
1−2−31 −51 T = WM = =
161031
T−1=−1 1 −1 . 8 −3 −5
K = (b2 − a2 , b1 − a1 )T −1 ,
K= −137 21 · −1 1 −1 = 165 −21 . 84 8−3−5 6464
−3−k2 =λ2 −λ(3−k1 −k2)+14−5k1 +3k2. (10.4.8)
4−k1
Comparing the coefficients of powers in λ in (10.4.7) and (10.4.8), we obtain
2−k2−λ
3−k1 −k2 = 3, 4
14−5k1 +3k2 = 1. 8
Thisgivesusk1=165 andk2=−21. 64 64
Hence
K= 165 −21 . 64 64
10.4 Stabilization by State Feedback (Design via Pole Placement) 461
To prove the general case m > 1 of Theorem 10.19 we need the following preliminary result.
Lemma 10.21. If the pair {A,B} is completely controllable and the columns of B, assumed nonzero, are b1, b2, . . . , bm, then there exist ma- trices Ki, 1 ≤ i ≤ m, such that the pairs {A − BKi, bi} are completely controllable.
Proof. Let us consider the case i = 1. Since the controllability matrix W has rank k (full rank), one may select a basis of Rk consisting of k columns of W . One such selection would be the k × k matrix
M = b1,Ab1,…,Ar1−1b1,b2,Ab2,…,ar2−1b2,…,
where ri is the smallest integer such that Aribi is linearly dependent on all the preceding vectors. Define an m × k matrix L having its r1th column equal to e2 = (0,1,…,0)T , its (r1 + r2)th column equal to e3 =(0,0,1,…,0)T,andsoon,allitsothercolumnsbeingzero.Weclaim that the desired matrix K1 is given by K1 = LM−1. To verify the claim, we compare the corresponding columns on both sides of K1M = L. It follows immediately that
K1b1 = 0, K1Ab1 = 0,…,K1Ar1−1b1 = e2, K1b2 = 0, K1Ab2 = 0,…,K1Ar2−1b2 = e3, K1b3=0, etc.
Thus we have
b ,(A−BK )b ,(A−BK )2b ,…,(A−BK )k−1b =W(k),
1112121
which has rank k by assumption. This proves our claim. We are now ready to give the proof of the general case m > 1 in Theorem 10.19. P
Proof of Theorem 10.19. Let K1 be the matrix in Lemma 10.21. Then by Lemma 10.21, it follows that the pair {A−BK1,b1} is completely controllable. And by the proof of Lemma 10.21 for m = 1, there exists a 1×k vector ξ such that the eigenvalues of A+BK1 +b1ξ are the set Λ. Let K ̄ be the m × k matrix whose first row is ξ and all other rows are zero. Then the desired feedback (gain) matrix is given by K = K1 + K ̄ . Since u = −Kx, this gives
x(n+1)=(A−BK)x(n)=(A−BK1 −b1ξ)x(n).
To prove the converse, select K0 such that (A − BK0)n → 0 as n →
∞, that is, the spectral radius ρ(A − BKo) is less than 1, and select K1
such that ρ(A−BK1) = exp2πn:n=0,1,…,k−1, the kth roots k
of unity. Then clearly, (A − BK1)k = 1. Suppose that for some vector
462 10. Control Theory
ξ ∈ Rk, ξT AnB = 0 for all n ∈ Z+. Then for any matrix K, ξT(A−BK)n =ξT(A−BK)(A−BK)n−1
= (ξT A − ξT BK)(A − BK)n−1
= ξT A(A − BK)n−1 (since ξT B = 0)
= ξT A(A − BK)(A − BK)n−2
= ξT A2(A − BK)n−2 (since ξT AB = 0).
Continuing this procedure we obtain
ξT(A−BK)n =ξTAn, foralln∈Z+.
This implies that
ξT[(A−BKo)n−(A−BK1)n]=0, foralln∈Z+,
or
Letting r → ∞, we have (A − BK0)kr → 0 and, consequently, ξT = 0.
ξT[(A−BKo)kr −1]=0, forallr∈Z+.
This implies that the pair {A, B} is completely controllable. P
An immediate consequence of Theorem 10.19 is a simple sufficient condi- tion for stabilizability. A system x(n + 1) = Ax(n) + Bu(n) is stabilizable if one can find a feedback control u(n) = −Kx(n) such that the zero solution of the resulting closed-loop system x(n + 1) = (A − BK)x(n) is asymp- totically stable. In other words, the pair {A,B} is stabilizable if for some matrix K, A − BK is a stable matrix (i.e., all its eigenvalues lie inside the unit disk).
Corollary 10.22. System (10.4.1) is stabilizable if it is completely controllable.
The question still remains whether or not we can stabilize an uncontrol- lable system. The answer is yes and no, as may be seen by the following example.
Example 10.23. Consider the control system x(n + 1) = Ax(n) + Bu(n),
where
Let us write
⎛⎞⎛⎞
0 a b 1 β1
A = ⎜⎝ 1 d e ⎟⎠ , B = ⎜⎝ 0 β 2 ⎟⎠ .
00h 00
A= A11 A12 , B= B1 , 0 A22 0
If x =
, then our system may be written as
y(n + 1) = A11y(n) + A12z(n) + B1u(n),
z(n + 1) = A22z(n).
10.4 Stabilization by State Feedback (Design via Pole Placement)
463
where
A11= 0 a , A12= b , A22=(h), B1= 1 β1
1de 0β2
y z
.
It is easy to verify that the pair {A11 , B1 } is completely controllable. Hence by Theorem 10.19, there is a 2 × 2 gain matrix K ̄ such that A11 + B1K ̄ is a stable matrix. Letting K = (K ̄ )(0), then
̄ A−BK= A11−B1K ∗ .
0h
Hence the matrix A − BK is stable if and only if |h| < 1.
In the general case, a system is stabilizable if and only if the uncon- trollable part is asymptotically stable (Exercises 10.4, Problem 8). In this instance, from the columns of the controllability matrix W we select a basis for the controllable part of the system and extend it to a basis S for Rk. The change of variables x = P y, where P is the matrix whose columns are
the elements of S, transforms our system to
y(n + 1) = A ̄y(n) + B ̄u,
where
A ̄= A11 A12 , B ̄= B1 . 0 A22 0
Here the pair {A11,B1} is controllable. Hence the system is stabilizable if and only if the matrix A22 is stable.
10.4.1 Stabilization of Nonlinear Systems by Feedback
Before ending this section, let us turn our attention to the problem of stabilizing a nonlinear system
x(n + 1) = f (x(n), u(n)), (10.4.9) where f : Rk × Rm → Rk. The objective is to find a feedback control
u(n) = h(x(n)) (10.4.10) in such a way that the equilibrium point x∗ = 0 of the closed-loop system x(n + 1) = f (x(n)), h(x(n)), (10.4.11)
464 10. Control Theory
is asymptotically stable (locally!). We make the following assumptions: (i) f(0,0) = 0, and
(ii) f is continuously differentiable, A = ∂f (0, 0) is a k × k matrix, B =
with
x(n + 1) = g(x(n))
∂ g = A − B K . ∂x x=0
(10.4.12)
(10.4.13)
∂f (0, 0) is a k × m matrix. ∂u
∂x
Under the above conditions, we have the following surprising result.
Theorem 10.24. If the pair {A,B} is controllable, then the nonlinear system (10.4.9) is stabilizable. Moreover, if K is the gain matrix for the pair {A,B}, then the control u(n) = −Kx(n) may be used to stabilize system (10.4.9).
Proof. Since the pair {A,B} is controllable, there exists a feedback control u(n) = −Kx(n) that stabilizes the linear part of the system, namely,
y(n + 1) = Ay(n) + Bv(n).
We are going to use the same control on the nonlinear system. So let g: Rk → Rk be a function defined by g(x) = f(x,−Kx). Then system equation (10.4.9) becomes
Since by assumption the zero solution of the linearized system y(n + 1) = (A − BK)y(n)
is asymptotically stable, it follows by Theorem 4.20 that the zero solution of system (10.4.12) is also asymptotically stable. This completes the proof of the theorem. P
Example 10.25. Consider the nonlinear difference system x1(n + 1) = 2 sin(x1(n)) + x2 + u1(n),
x2(n + 1) = x21(n) − x2(n) − u2(n). Find a control that stabilizes the system.
Solution One may check easily the controllability of the linearized system
{A, B}, where
211 A= 0 −1 , B= −1 ,
after some computation. A gain matrix for the linearized system is K =
(2.015, 0.975), where the eigenvalues of A−BK are 1 and 0.1. As implied by 2
10.4 Stabilization by State Feedback (Design via Pole Placement) 465
Theorem 10.24, the control u(n) = −Kx(n) would stabilize the nonlinear system, where K = (2.015, 0.975).
Exercises 10.4
In Problems 1 through 3 determine the gain matrix K that stabilizes the system {A, B}.
010 1.A= −0.16 −1 , B= 1 .
⎛⎞⎛⎞
211 01
2. A = ⎜⎝−2 1 0⎟⎠ , B = ⎜⎝1 0 ⎟⎠ .
−2 −1 0 0 −2 ⎛⎞⎛⎞
010 0
3. A = ⎜⎝ 0 0 1⎟⎠ , B = ⎜⎝0⎟⎠ .
−2 1 3
4. Determine the matrices B
1
⎛1 −1 2⎞
for which
the system
{A, B}, A =
⎜0 1 1⎟, is (a) controllable and (b) stabilizable. ⎝2⎠
22
5. Consider the second-order equation
1 −1 1
x(n + 2) + a1x(n + 1) + a2x(n) = u(n).
Determine a gain control u(n) = c1x(n) + c2x(n + 1) that stabilizes
the equation.
6. Describe an algorithm for decomposing the system x(n+1) = Ax(n)+ Bu(n) into its controllable and uncontrollable parts when A is a 3 × 3 matrix and B is a 3×2 matrix.
7. Generalize the result of Problem 6 to the case where A is a k×k matrix and B is a k×r matrix.
*8. Show that the pair {A,B} is stabilizable if and only if the uncontrol- lable part of the system is asymptotically stable.
9. Deadbeat Response. If the eigenvalues of the matrix A − BK are all zero, then the solutions of the system x(n + 1) = (A − BK)x(n) will read 0 in finite time. It is then said that the gain matrix K produces a deadbeat response. Suppose that A is a 3×3 matrix and B a 3×1 vector.
466 10. Control Theory
(a) Show that the desired feedback matrix K for the deadbeat response is given by
where
x1(n + 1) = 3x1(n) + x2(n) − sat(2x2(n) + u(n)), x2(n + 1) = sin x1(n) − x2(n) + u(n),
where
K=1 0 0ξ ξ ξ−1, 123
ξ1 = A−1B, ξ2 = (A−1)2B, ξ3 = (A−1)3B.
(b) Show that the vectors ξ1,ξ2, and ξ3 are generalized eigenvectors of the matrix A−BK [i.e., (A−BK)ξ1 = 0,(A−BK)ξ2 = ξ1,(A−BK)ξ3 =ξ2].
10. Ackermann’s Formula:
Let Λ = {μ1, μ2, . . . , μk} be the desired eigenvalues for the completely controllable pair {A,B}, with Λ = Λ ̄. Show that the feedback (gain) matrix K can be given by
K = 0 0 ··· 0B AB ··· Ak−1B−1 p(A), where
k i=1
11. Let Λ = {μ1,μ2,...,μk} be a set of complex numbers with Λ = Λ ̄. Show that if the pair {A, C } is completely observable, then there exists a matrix L such that the eigenvalues of A − LC are the set Λ.
12. Verify formula (10.4.3).
13. Find a stabilizing control for the system
p(λ)=
(λ−μi)=λk +α1λk−1 +···+αk.
sat y =
y if |y| ≤ 1,
signy if|y|>1.
14. Find a stabilizing control for the system
x1(n + 1) = 2×1(n) + x2(n) + x3(n) + u1(n) + 2u2(n), x2(n + 1) = x21(n) + sin x2(n) + x2(n) + u21(n) + u2(n),
x3(n + 1) = x41(n) + x32(n) + 1×3(n) + u1(n). 2
15. (Research problem). Find sufficient conditions for the stabilizability of a time-variant system
x(n + 1) = A(n)x(n) + B(n)u(n).
16. (Research problem). Extend the result of Problem 15 to nonlinear time-variant systems.
10.5 Observers
Theorem 10.19 provides a method of finding a control u(n) = −Kx(n) that stabilizes a given system. This method clearly requires the knowledge of all state variables x(n). Unfortunately, in many systems of practical importance, the entire state vector is not available for measurement. Faced with this difficulty, we are led to construct an estimate of the full state vector based on the available measurements. Let us consider again the system
x(n + 1) = Ax(n) + Bu(n),
y(n) = Cx(n). (10.5.1)
To estimate the state vector x(n) we construct the k-dimensional observer (Figure 10.13)
z(n + 1) = Az(n) + E[y(n) − Cz(n)] + Bu(n), (10.5.2)
where E is a k × r matrix to be determined later. Notice that unlike x(n), the state observer z(n) can be obtained from available data. To see this, let us write the observer (10.5.2) in the form
z(n + 1) = (A − EC)z(n) + Ey(n) + Bu(n). (10.5.3)
We observe here that the inputs to the observer involve y(n) and u(n), which are available to us.
The question remains whether the observer state z(n) is a good estimate of the original state x(n). One way to check the goodness of this estimator is to ensure that the error e(n) = z(n)−x(n) goes to zero as n → ∞. To achieve this objective we write the error equation in e(n) by subtracting (10.5.2) from (10.5.1) and using y(n) = Cx(n). Hence
z(n + 1) − x(n + 1) = [A − EC][z(n) − x(n)],
u(n)
+ x(n+1) x(n) y(n) +
+ z(n+1) z(n) Cz(n)
10.5 Observers 467
B
B
z-1
A
C
_
FIGURE 10.13. Observer.
E
z-1
A
C
468 10. Control Theory
or
e(n + 1) = [A − EC]e(n). (10.5.4)
Clearly, if the zero solution of system (10.5.4) is asymptotically stable (i.e., the matrix A − EC is stable), then the error vector e(n) tends to zero. Thus the problem reduces to finding a matrix E such that the matrix A − EC has all its eigenvalues inside the unit disk. The following result gives a condition under which this can be done.
Theorem 10.26. If system (10.5.1) is completely observable, then an observer (10.5.2) can be constructed such that the eigenvalues of the matrix A − EC are arbitrarily chosen. In particular, one can choose a matrix E such that the error e(n) = z(n) − x(n) in the estimate of the state x(n) by the state observer z(n) tends to zero.
Proof. Since the pair {A,C} is completely observable, it follows from Section 4.3 that the pair {AT , C T } is completely controllable. Hence by Theorem 10.19 the matrix E can be chosen such that AT − CT ET has an arbitrary set of eigenvalues, which is the same as the set of eigenvalues of the matrix A − EC.
Moreover, if we choose the matrix E such that all the eigenvalues of the matrix A − EC are inside the unit disk, then e(n) → 0 (see Corollary 3.24). P
10.5.1 Eigenvalue Separation Theorem
Suppose that the system
x(n + 1) = Ax(n) + Bu(n),
y(n) = Cx(n),
is both completely observable and completely controllable. Assuming that the state vector x(n) is available, we can use Theorem 10.24 to find a feedback control u(n) = −Kx(n) such that in the closed-loop system
x(n + 1) = (A − BK)x(n)
the eigenvalues of A − BK can be chosen arbitrarily. Next we use Theorem 10.26 to choose a state observer z(n) to estimate the state x(n) in such a way that the eigenvalues of A − EC in the observer
z(n + 1) = (A − EC)z(n) + Ey(n) + Bu(n)
can also be chosen arbitrarily.
In practice, a feedback control may be obtained using the state ob-
server z(n) instead of the original state x(n) (whose components are not all available for measurement). In other words, we use the feedback control
u(n) = −Kz(n). (10.5.5)
The resulting composite system is given by x(n + 1) = Ax(n) − BKz(n),
z(n + 1) = (A − EC)z(n) + ECx(n) − BKz(n). It follows that
e(n + 1) = z(n + 1) − x(n + 1) = (A − EC)e(n). Hence we have the following composite system:
x(n + 1) = (A − BK)x(n) + BKe(n), e(n + 1) = (A − EC)e(n).
The system matrix is given by
A−BK BK
A= 0 A−EC ,
whose characteristic polynomial is the product of the characteristic poly- nomials of (A − BK) and (A − EC). Hence the eigenvalues of A are either eigenvalues of A − BK or eigenvalues of A − EC, which we can choose arbitrarily. Thus we have proved the following result.
Theorem 10.27 (Eigenvalue Separation Theorem).
Consider the
system
with the observer
x(n + 1) = Ax(n) + Bu(n), y(n) = Cx(n),
z(n + 1) = (A − EC)z(n) + Ey(n) + Bu(n) and the feedback control
u(n) = −Kz(n).
Then the characteristic polynomial of this composite system is the product of the characteristic polynomials of A − BK and A − EC. Furthermore, the eigenvalues of the composite system can be chosen arbitrarily.
Example 10.28. Consider the system
x(n + 1) = Ax(n) + Bu(n),
where
y(n) = Cx(n), ⎛ 1⎞
A=⎝0 −4⎠, B= 0 , C=0 1.
10.5 Observers 469
1 −1
1
470 10. Control Theory
Design a state observer so that the eigenvalues of the observer matrix A − EC are 1 + 1i and 1 − 1i.
Solution The observability matrix is given by
C01 CA = 1 −1 ,
which has full rank 2. Thus the system is completely observable, and the desired observer feedback gain matrix E may be now determined. The characteristic equation of the observer is given by det(A − EC − λI) = 0.
If
E= E1 , E2
4
11 11
22 22
thenwehave⎛ ⎝ 0
1⎞ − 4 ⎠ − E 1
1 −1 E2
0 = 0 , By assumption the desired characteristic equation is given by
or
0
λ2 + (1 + E2)λ + E1 + 1 = 0.
1 − λ
0 λ
which reduces to
λ−2−2i λ−2+2i =0, λ2 − λ + 1 = 0.
(10.5.6)
(10.5.7)
2 Comparing (10.5.6) and (10.5.7) yields
⎛1⎞ Thus E = ⎝ 4 ⎠ .
−2
E1=1, E2=−2. 4
Example 10.29. Figure 10.14 shows a metallic sphere of mass m sus- pended in a magnetic field generated by an electromagnet. The equation of motion for this system is
mx ̈t = mg − k u2t , (10.5.8) xt
where xt is the distance of the sphere from the magnet, ut is the cur- rent driving the electromagnet, g is the acceleration of gravity, and k is a constant determined by the properties of the magnet.
ut
10.5 Observers 471
xt
m
FIGURE 10.14. A metallic sphere suspended in a magnetic field.
It is easy to check that (10.5.8) has an equilibrium at
xt =x0 =1,
ut=u0= mg/k.
Linearizing (10.5.8) about this equilibrium gives the following approximate
modelintermsofthedeviationsx=xt−x0 andu=ut−u0: x ̈ − g x = − 2 k g / m u ,
k
or, in state variable form,
0 1
Thus
x ̇ x 0
v ̇ = g 0 v + −2kg/m u.
k
01 0 Aˆ = g 0 , Bˆ = − 2 k g / m .
k
The matrix Aˆ can be written in the form
where
Aˆ = P Λ P − 1 ,
⎡g ⎤ ⎡ ⎤
⎢0⎥11
⎢k ⎥ 1⎣ ⎦ Λ=⎣ g⎦, P=√2 g − g .
0−kkk
472 10. Control Theory
So
⎡⎤
⎢ cosh gT A=eAˆT =⎢ k
ksinh gT⎥
⎣g g k sinh kT
g
cosh
g/m cosh g T k
k ⎥, g ⎦
kT g ⎤ ˆ ⎢ k/msinh kT⎥ B = T eAT Bˆ = −2T ⎢⎣ ⎥⎦
⎡
The discrete equivalent system is thus controllable, since detW=B AB
⎡g gg⎤
⎢ s i n h k T 2 s i n h k T c o s h k T ⎥
=−2Tk/m⎢⎣ ⎥⎦
gsinh gT 2 gsinh gTcosh gT kkkkk
=ce gTsinh gT, kk
wherec=0onlyifT =0.
If the position deviation x of the ball from equilibrium can be measured,
then the system is also observable, since then we have the measurement
equation
and hence
x
y=10
v
C=10. Observability is easily verified by computing
⎡ 1
0 ⎤ C ⎢ ⎥
detV = CA = ⎣cosh gT k sinh gT⎦ kgk
= ksinh gT, gk
which is zero only if T = 0. Before continuing, fix m = k = 0.1,g = 10,
and T = 0.01. Thus
A = 1.0017 1.0050 , B = −0.2010 .
1.0050 0.0100 −0.0020
Note that A is unstable, with eigenvalues λ1 = 1.1052 and λ2 = 0.9048. Controllability of {A,B} implies that a stabilizing state feedback gain K = [k1, k2] can be found. Moreover, the eigenvalues of the resulting system
matrix A − BK can be assigned arbitrarily. In our example,
A − BK = 1.0050 + 0.0020k1 0.0100 + 0.0020k2 , 1.00017 + 0.2010k1 1.0050 + 0.2010k2
so that
|λI − A + BK| = λ2 − (2.0100 + 0.002k1 + 0.201k2)λ + 0.2000k2 + 1,
and eigenvalues λ1 = 1 and λ2 = −1 (both inside the unit circle) can be 22
obtained by choosing
K = [k1, k2] = [−376.2492 − 6.2500].
Observability of {A,C} implies that an asymptotic observer can be constructed to produce an estimate of the system state vector from mea- surements of x. The observer gain L = [l1,l2]T can be chosen not only to ensure that the state estimate converges, but to place the observer
eigenvalues arbitrarily. In our example,
A−LC = 1.0050−l1 0.0100 , 1.0017 − l2 1.0050
so that
|λI−A+LC|=λ2 +(l1 −2.0100)λ−1.0050l1 +0.0100l2 +1,
and eigenvalues λ1 = 1 and λ2 = −1 can be obtained by choosing 44
L= l1 2.0100 . l2 95.5973
The eigenvalue separation theorem ensures that combining this observer
with the state feedback controller designed above will produce a stable
closed-loop system with eigenvalues ±1 and ±1. 24
Exercises 10.5
In Problems 1 through 4 design an observer so that the eigenvalues of the
matrix A − EC are as given.
11 1. A = 0 −1 , C = 1 1 ,
10.5 Observers 473
λ1=1, λ2=−1. 24
01
, C = 0 1 ,
λ1=1−1i, λ2=1+1i. 24 24
2. A =
10
474 10. Control Theory
⎛⎞ 100
3.A=⎜⎝0 0 1⎟⎠, C=0 1 0, 010
λ1=1, λ2=1−1 i, λ3=1+1 i. 24444
4.A=⎜⎝0 −1 1⎟⎠, C= 0 1 0 ,
⎛⎞
010 0 0 −1
101 λ1 = 1, λ2 =−1, λ3 =−1.
242
5. (Reduced-Order Observers):
Consider the completely observable system
x(n + 1) = Ax(n) + Bu(n), y(n) = Cx(n),
(10.5.9)
where it is assumed that the r × k matrix C has rank r (i.e., the r measurements are linearly independent). Let H be a (k−r)×k matrix
such that the matrix
is nonsingular. Let
Then x ̄ may be written as
H
P=
C
x ̄(n) = P x(n).
w(n) x ̄= ,
(10.5.10)
y(n)
where w(n) is (k − r)-dimensional and y(n) is the r-dimensional vector
of outputs.
(a) Use (10.5.10) to show that system equation (10.5.9) may be put
in the form
w(n + 1) = A11 A12 w(n) + B1 u(n). (10.5.11) y(n + 1) A21 A22 y(n) B2
(b) Multiply the bottom part of (10.5.11) by any (k − r) × r matrix E to show that
W(n + 1) − Ey(n + 1) = (A11 − EA21)[W(n) − Ey(n)]
+ [A11E − EA21E + A12 − EA22]y(n)
+ (B1 − EB2)u(n).
(c) If v(n) = w(n) − F y(n), show that
v(n + 1) = (A11 − EA21)v(n)
+ [A11E − EA21E + A12 − EA22]y(n)
+(B1 −FB2)u(n).
(d) Explain why we can take an observer of system equation (10.5.9)
as the (k − r)-dimensional system
z(n + 1) = (A11 − EA21)z(n)
+ [A11E − EA21E + A12 − EA22]y(n)
+(B1 −FB2)u(n). (10.5.12) (e) Let e(n) = z(n) − v(n). Show that
e(n + 1) = (A11 − EA21)e(n). (10.5.13)
6. Prove that if the system equation (10.5.9) is completely observable,
then the pair {A11,A21} in (10.5.11) is completely observable.
7. Prove the eigenvalue separation theorem, Theorem 10.27, for reduced-
order observers.
8. Consider the system
x1(n + 1) = x2(n),
x2(n + 1) = −x1(n) + 2×2(n) + u(n),
y(n) = x1(n).
Construct a one-dimensional observer with a zero eigenvalue.
10.5 Observers 475
Appendix A
Stability of Nonhyperbolic Fixed Points of Maps on the Real Line
A.1 Local Stability of Nonoscillatory Nonhyperbolic Maps
Our aim in this appendix is to extend Theorems 1.15 and 1.16 to cover all the remaining unresolved cases. The exposition is based on the recent paper by Dannan, Elaydi and Ponomarenko [30]. The main tools used here are the Intermediate Value Theorem and Taylor’s Theorem which we are going to state.
Theorem A.1 (The Intermediate Value Theorem). Let f be a con- tinuous function on an interval I = [a,b] such that f(a) ̸= f(b). If c is between f(a) and f(b), then there exists x0 ∈ (a,b) such that f(x0) = c. In particular, if f(a) and f(b) are of opposite sign, then since 0 is between f(a) and f(b), there exists x0 between a and b such that f(x0) = 0.
Theorem A.2 (Taylor’s Theorem). Suppose that the (n+1)th deriva- tive of the function f exists on an interval containing the points a and b. Then
f(b) = f(a) + f′(a)(b − a) + f′′(a)(b − a)2 + f(3)(a)(b − a)3 2! 3!
+···+ f(n)(a)(b−a)n + f(n+1)(z)(b−a)n+1 (A.1.1) n! (n + 1)!
for some number z between a and b.
477
478 A. Stability of Nonhyperbolic Fixed Points of Maps on the Real Line
Notation:Thenotationoff ∈Cr meansthatthederivativesf′,f′′,…,f(r) exist and are continuous.
Theorem A.3. Let x∗ be a fixed point of f, then the following statements hold true:
(i) Suppose that f ∈ C2k. If f′(x∗) = 1, and f′′(x∗) = ··· = f(2k−1)(x∗) = 0 but f(2k)(x∗) ̸= 0, then x∗ is semi-asymptotically stable:
(a) from the left if f(2k)(x∗) > 0, and
(b) from the right if f(2k)(x∗) < 0.
(ii) Suppose that f ∈ C(2k+1). If f′(x∗) = 1, and f′′(x∗) = ··· =
f(2k)(x∗) = 0 but f(2k+1)(x∗) ̸= 0, then:
(a) x∗ is asymptotically stable if f(2k+1)(x∗) < 0, and
(b) x∗ is unstable if f(2k+1)(x∗) > 0.
Proof.
(i) Assume that f′(x∗) = 1, f′′(x∗) = ··· = f(2k−1)(x∗) = 0 but f(2k)(x∗) ̸= 0.
(a) If f(2k)(x∗) > 0, then by Taylor’s Theorem, for a sufficiently small number δ > 0, we have
f(x∗ +δ)=f(x∗)+f′(x∗)δ+… f(2k−1)(x∗)δ(2k−1) f(2k)(ξ)δ2k
+ (2k − 1)! + (2k)! (A.1.2) for some ξ ∈ (x∗, x∗ +δ). If δ is sufficiently small, we may conclude
that f(2k)(ξ) > 0. Substituting in (A.1.2) yields ∗ ∗ f(2k)(ξ)δ2k
f(x +δ)=x +δ+ (2k)! . Similarly one may show that
∗ ∗ f(2k)(ξ)δ2k f(x −δ)=x −δ+ (2k)! .
(A.1.3)
(A.1.4)
Hence from (A.1.3), it follows that f (x∗ + δ) > x∗ + δ. And from (A.1.4) we have x∗ − δ < f(x∗ − δ) < x∗. This proves semi- asymptotic stability from the left.
(b) The proof of part (b) is analogous and will be left to the reader. (ii) By the assumptions in (ii) we have, for some δ > 0,
∗ ∗ f(2k+1)(ξ)δ2k+1
f(x +δ)=x +δ+ 2k+1)! (A.1.5)
A.2 Local Stability of Oscillatory Nonhyperbolic Maps 479
for some ξ ∈ (x∗, x∗ + δ). Furthermore,
∗ ∗ f(2k+1)(ξ)δ2k+1
f(x −δ)=x −δ+ 2k+1)! . (A.1.6)
(a) If f(2k+1)(x∗) < 0, then, from (A.1.5), f(x∗+δ) < x∗+δ and, from (A.1.6), f(x∗ − δ) > x∗ − δ. Hence x∗ is asymptotically stable.
(b) The proof of part (b) is analagous and will be left to the reader.
P
Consider the map f(x) = x+(x−1)4, where x∗ = 1 is a fixed point of f with f′(x∗) = 1, f′′(x∗) = f′′′(x∗) = 0, and f(4)(x∗) = 24 > 0. Then by Theorem A.3, x∗ is semi-asymptotically stable from the left.
A.2 Local Stability of Oscillatory Nonhyperbolic Maps
We now consider the case when f′(x∗) = −1. A nice trick here is to look at the map g(x) = f(f(x)) = f2(x).
A.2.1 Results with g(x)
Sincex∗ isafixedpointoff,itmustbeafixedpointofgandg′(x∗)=1. Moreover, g′′(x∗) = 0 and g′′′(x∗) = 2Sf(x∗). Notice that x∗ is asymptot- ically stable {unstable} under g if, and only if, it is asymptotically stable {unstable} under f. This is due to the fact that |fn(x∗)| < 1 if and only if |gn(x∗)| < 1.
We can then apply the second half of Theorem A.3 to get the following result.
Theorem A.4. Suppose that f ∈ C(2k+1) and x∗ is a fixed point of f such that f′(x∗) = −1. If g′′(x∗) = ... = g(2k)(x∗) = 0 and g(2k+1)(x∗) ̸= 0, then:
(1) x∗ is asymptotically stable if g(2k+1)(x∗) < 0, and (2) x∗ is unstable if g(2k+1)(x∗) > 0.
Observe that this strategy does not use the other part of Theorem A.3– where x∗ is semi-asymptotically stable under g. That is, the case where f′(x∗) = −1, g′′(x∗) = … = g(2k−1)(x∗) = 0, and g(2k)(x∗) ̸= 0.
We now argue that this situation will never occur for analytic f. Theorem A.5. Let f be analytic with f′(x∗) = −1. Then, for some k > 1,
(1) If g′′(x∗) = . . . = g(2k−1)(x∗) = 0, then g(2k)(x∗) = 0.
480 A. Stability of Nonhyperbolic Fixed Points of Maps on the Real Line
(2) x∗ cannot be semi-asymptotically stable under g. Proof. By Taylor’s Theorem, we have some small δ with
f(x∗ +δ)=f(x∗)+δf′(x∗)+ δ2f′′(x∗) +… 2!
= x∗ − δ + 0(δ2).
Hence,forx0 =x∗+δ>x∗,wehavef(x0)
Now,iff2k(x0)→x∗ ask→∞,thenf2k+1(x0)→x∗ ask→∞.Hence either x∗ is asymptotically stable or x∗ is unstable and, more importantly, it cannot be semi-asymptotically stable. P
These results using g(x) are conclusive but not entirely satisfactory. For example, we return to f(x) = −x + 2×2 − 4×3. To determine the stability of f(x) at 0, we need to find derivatives of g(x) = −x+4×2 −8×3 + 64×5 − 192×6 + 384×7 − 384×8 + 256×9. It turns out that g5(0) = 7680; hence, by Theorem A.4, 0 is an unstable fixed point. However, this was computationally difficult, and we would like an analogue of Theorem A.4 using only the derivatives of f(x).
Remark: If f′(x∗) = 1, f(k)(x∗) = 0 for all k > 1, and f is analytic, then f(x) = x. Consequently, every point in the vicinity of x∗ is a fixed point and x∗ is thus stable but not asymptotically stable. If f′(x∗) = −1, g(k)(x∗) = 0 for all k > 1, and if f is analytic, then g(x) = x. Hence every point in the vicinity of x∗ is periodic of period 2, and x∗ is again stable but not asymptotically stable.
ExampleA.6. Considerthemapsf1(x)=x+e−x−2,f2(x)=x+xe−x−2,
f3(x) = x−xe−x−2 , with fi(0) = 0. Each of these maps has fi′(0) = −1, and
f(k)(0) = 0 for all k > 1. However, the fixed point 0 is semi-asymptotically i
stable from the left, unstable, and asymptotically stable, respectively. Example A.7. Contemplate May’s genotype selection model
x(n)eα(1−2x(n))
x(n+1)= 1−x(n)+x(n)eα(1−2x(n)), α>0, x∈(0,1). (A.2.1)
At the fixed point x∗ = 1, f′(x∗) = 1−α. The fixed point is thus asymptoti- 22
cally stable for 1 < α < 4 by Theorem 1.13. At α = 4, we have f′(x∗) = −1,
g′′(x∗) = g′′′(∗) = 0, but g′′′(x∗) = −32 < 0. Hence by Theorem A.4, the
fixed point x∗ = 1 is asymptotically stable. 2
Appendix B
The Vandermonde Matrix
The generalized Vandermonde matrix is given by
⎛1 0 ...1 0 ...⎞
⎜ λ1 1 ... λr 1 ...⎟
V =⎜ λ21 2λ1 ... λ2r 2λr ...⎟ (B.1)
⎜. . . . ⎟ ⎝....⎠
λk−1 (k−1)λk−2 ... λk−1 (k−1)λk−2 11rr
and consists of k × mi submatrices corresponding to the eigenvalues λi,
1 ≤ i ≤ r, ri=1mi = k. The first column in the k×m1 subma-
trix is c1 = (1,λ1,λ21,...,λk1)T , the second column is c2 = 1 c′1(λ1) = 1!
(0,1,2λ ,3λ2,...,kλk−1)T, ..., the sth column is c = 1 c(s−1)(λ ), 1 1 1 s (s−1)!1 1
where c(m)(λ ) is the mth derivative of column c . The extension of this 111
definition to other k × mi submatrices is done in the natural way. We are now going to prove the following result.
Lemma B.1 [76].
W(0)=detV =
1≤i
482 B. The Vandermonde Matrix
regular Vandermonde matrix (2.3.3) and thus (B.2) holds. For the example m1 = 3, m2 = 2 (λ1 = λ2 = λ3,λ4 = λ5), the sum of multiplicities which exceed 1 is 3 + 2 = 5. To illustrate the induction step, let
1 0 1 1 0
λ1
W = λ2 1
λ3 1
λ41
So that the sum of multiplicities greater than 1 is 4.
Assuming (B.2) for W yields
Note that
W(0) = 2 dt2 Wt=λ1
1 t
2λ1 t2
λ2 λ2 2 λ3
3λ2 t3
1 2
2 4 λ 32
4λ31t4 λ42
W = (t − λ1)2(λ2 − t)2(λ2 − λ1)4.
1d2
= 1(λ2 −λ1)4 d 2(t−λ1)(λ2 −t)2 −2(t−λ1)2(λ2 −t)t=λ1
1 2λ2 . 3 λ 2
2 dt
= 1(λ2 −λ1)42(λ2 −t)2 −4(t−λ1)(λ2 −t)
2 −4(t−λ1)(λ2 −t)+2(t−λ1)2
t=λ1
= (λ2 − λ1)6.
In general, W(0) is formed from W as long as there is one multiplicity mi > 1. The general case may be proved in an analogous manner. P
Appendix C
Stability of Nondifferentiable Maps
The main objective of this appendix is to prove Theorem 4.8. In fact, we will prove a more general result which appeared in Elaydi and Sacker [50]. In the sequel we will assume that the f : I → I is continuous on the closed and bounded interval I = [a, b]. Clearly if I = [a, ∞), and f is bounded and continuous, f(I) ⊂ J ⊂ I, where J is a closed and bounded interval, then f : J → J. The following lemma and its corollary are immediate sequences of the Intermediate Value Theorem.
Lemma C.1. Let J = [c, d] ⊂ [a, b] such that either: (i) f(c)>c and f(d)
(i) Assume that f(c) > c and f(d) < d. Then for the map g(x) = f(x)−x, g(c) > 0 and g(d) < 0. Hence by the Intermediate Value Theorem, there exists x∗ between c and d such that g(x∗) = 0. Hence f(x∗) = x∗, and thus x∗ is a fixed point of f.
The proof of (ii) is similar and will be left to the reader. P Corollary C.2. Suppose that J = [c,d] ⊂ I. If f(d) > d and (c,d) is fixed
point-free, then f(x) > x for all x ∈ (c,d).
We are now ready to present the main result.
483
484 C. Stability of Nondifferentiable Maps
Theorem C.3. Let f : I → I be continuous. Then the following statements are equivalent:
(i) f has no points of minimal period 2 in (a, b). (ii) For all x0 ∈ (a,b), {fn(x0)} converges in I.
Proof. (ii) ⇒ (i).
If {x ̄1,x ̄2} is a periodic orbit of period 2 in (a,b), then {fn(x ̄)} does not converge as it oscillates between x ̄1 and x ̄2.
(i) ⇒ (ii) .
Assume there exists x0 ∈ (a,b) such that {fn(x0)} does not converge. Thus x0 is not a fixed point or an eventually fixed point. Hence its orbit O(x0) = {x0,x(1),x(2),…} can be partitioned into two sequences A = {x(k)|f(x(k)) > x(k)} and B = {x(k)|f(x(k)) < x(k)}. Then A ̸= ∅ and B ̸= ∅. We claim that A is strictly monotonically increasing, i.e., i < j implies x(i) < x(j). Assume the contrary, that there exists x(i),x(j) ∈ A such that i < j but x(i) > x(j). This means that fi(x0) > fj(x0). Let j = i + r. Then fr(x(i)) < x(i). Since x(i) is not a fixed point of f, there exists a small δ > 0 such that the interval (x(i) − δ, x(i)) is free of fixed points. Thus we may conclude that there exists a largest fixed point z of f in [a,x(i)] (z may equal a). Hence the interval (z,x(i)) is fixed point-free. And since f(x(i)) > x(i), it follows by Corollary C.2 that f(x) > x for all x ∈ (z, x(i)).
Let zn be a sequence in (z,x(i)) that converges to z. Then f(zn) > zn and limn→∞ f(zn) = f(z) = z. There exists N1 ∈ Z+ such that n > N1, f(zn) ∈ (z,x(i)). For n > N1, f2(zn) > f(zn) > z and limn→∞ f2(zn) = f2(z) = z. There exists N2 ∈ Z+ such that for n > N2, f2(zn) ∈ (z,x(i)). Repeating this process, there are N3, N4, . . . , Nr such that for n > Nt, 1≤t≤r,ft(zn)∈(z,x(i)).ForN =max{Nt|1≤t≤r},ft(zn)∈ (z,x(i)), 1 ≤ t ≤ r. We conclude that there exists y ∈ (z,x(i)) such that y,f(y),…,fr(y) ∈ (z,x(i)). Hence fr(y) > fr−1(y) > ··· > f(y) > y. But fr(x(i)) < x(i) implies, by Lemma C.1, the existence of a fixed point in (y,x(i)), a contradiction which establishes our claim that A is strictly monotonically increasing.
Similarly, we may show that B is strictly monotonically decreasing. De- fine x ̄1 = supA, x ̄2 = infB. Then x ̄1 ≤ x ̄2 and hence neither is an end point. Since A ∪ B = O(x0), it follows that {x ̄1, x ̄2} = Ω(x0), the set of all limit points of O(x0). Since Ω(x0) is invariant, either:
(a) f(x ̄1) = x ̄2, f(x ̄2) = x ̄1, or (b) f(x ̄1) = x ̄1, f(x ̄2) = x ̄2, or
( c ) x ̄ 1 = x ̄ 2 ,
(d) f(x ̄1) = x ̄1 and f(x ̄2) = x ̄1 or f(x ̄1) = x ̄2 and f(x ̄2) = x ̄2.
C. Stability of Nondifferentiable Maps 485
Case (a) is excluded since there are no 2-cycles; cases (b) and (d) are also excluded since neither A nor B is invariant. Hence the only case left is case (c) which confirms the convergence of the sequence {fn(x0)}. P
As an immediate consequence of the preceding theorem, we have the following important result on global asymptotic stability.
Corollary C.4. Let x∗ be a fixed point of a continuous map on the closed and bounded interval I = [a,b]. Then x∗ is globally asymptotically stable relativeto(a,b)ifandonlyiff2(x)>xforx
Proof. The necessity is clear. To prove the sufficiency, notice that the given assumptions imply that there are no periodic points of minimal period 2. Hence by Theorem C.3, {fn(x0} converges for every x0 ∈ I. Now if x0 ∈ (a,x∗), f(x0) > x0. For, otherwise, we would have f(x0) < x0 < f2(x0), which implies, by the Intermediate Value Theorem, the presence of a fixed point of the map f in the interval (a,x∗), a contradiction. Similarly, one may show that for all x0 ∈ (x∗,b), f(x0) < x0. Thus limn→∞ fn(x0) = c, where c is an interior point in the interval (a,b). Furthermore, since c is a fixed point of the map f, it follows that c = x∗. Hence x∗ is globally attracting.
It remains to show that x∗ is stable. We have two cases to consider.
Case (i): The map f is monotonically increasing in a small neighborhood (x∗ − δ,x∗). Since f(x) > x for all x ∈ (x∗ − δ,x∗), it follows that for x0 ∈(x∗−δ,x∗),wehavex0
Appendix D
Stable Manifold and the Hartman–Grobman–Cushing Theorems
D.1 The Stable Manifold Theorem
Consider the nonlinear difference system
x(n + 1) = f (x(n))
(D.1.1)
such that f has a fixed point x∗ ∈ Rk and f ∈ C2 in an open neighborhood of x∗. Let A = Df(x∗) be the Jacobian of f at x∗. Then (D.1.1) may be written in the form
x(n + 1) = Ax(n) + g(x(n). (D.1.2) The associated linear system is given by
z(n + 1) = Az(n). (D.1.3)
The fixed point x∗ is assumed to be hyperbolic, where none of the eigen- values of A lie on the unit circle. Arrange the eigenvalues of A into two sets: S = {λ1,λ2,…,λr}, G = {λr+1,…,λk} with |λi| < 1 for λi ∈ S and |λj| > 1 for λj ∈ G. Let Es be the eigenspace spanned (gener- ated) by the generalized eigenvectors corresponding to S and let Eu be the eigenspace spanned by the generalized eigenvectors corresponding to G. Then Rk = Es Eu. The sets Es and Eu are called the stable and unstable subspaces of x∗, respectively.
The local stable manifold of x∗ in an open neighborhood G defined as
Ws(x∗,G) ≡ Ws(x∗) = {x0 ∈ G | O(x0) ⊂ G and lim fn(x0) = x∗}. n→∞
487
488
D. Stable Manifold and the Hartman–Grobman–Cushing Theorems
W u (x*)
G
Ws (x*)
x*
FIGURE D.1. Es is tangent to Ws(x∗) and Eu is tangent to Wu(x∗). W u (x*)
G
Ws (x*)
x*
FIGURE D.2. Stable and unstable manifolds Ws(x∗) and Wu(x∗) in a neighbor- hood G of x∗.
To define the unstable manifold, we need to look at negative orbits. Since f is not assumed to be invertible, we have to define a principal negative orbit O−(x0) = {x(−n)} of a point x0 as follows. We let x(0) = x0, and f(x(−n − 1)) = x(−n), n ∈ Z+. The local unstable manifold for x∗ in G is defined to be the set
W u(x∗, G) ≡ W u(x∗) = {x0 ∈ G | there exists a negative orbit, O−(x0) ⊂ G and lim x(−n) = x∗}.
n→∞
The following theorem states that Es is tangent to Ws(x∗) and Eu is
tangent to Wu(x∗) at the fixed point x∗ (see Figures D.1 and D.2).
Theorem D.1 (The Stable Manifold Theorem). Let x∗ be a hyper- bolic fixed point of a C2-map f : Rk → Rk. Then in an open neighborhood G of x∗ there exist two manifolds Ws(x∗) of dimension Es and Wu(x∗) of dimension Eu such that:
(i) Es is tangent to Ws(x∗) at x∗ and for any solution x(n) of (D.1.1) with x(0) ∈ Ws, lim x(n) = x∗.
n→∞
(ii) Eu is tangent to Wu(x∗) at x∗ and if x(0) ∈ Wu(x∗), then there exists
a principal negative solution x(−n) with lim x(−n) = x∗. n→∞
D.2 The Hartman–Grobman–Cushing Theorem 489
Proof. See Cushing [24] and Robinson [128]. P
D.2 The Hartman–Grobman–Cushing Theorem
The Stable Manifold Theorem tells us what happens to solutions that lie on either the stable manifold W s or the unstable manifold W u in a neigh- borhood of a hyperbolic fixed point. The question that we are going to address here is: What happens to solutions whose initial points do not lie on either Ws or Wu?
The answer to this question is given by the classical Hartman–Grobman Theorem in differential equations and its analogue in difference equations. However, this theorem requires that the map is a diffeomorphism, that is differentiable and a homeomorphism. Two maps f : X → X and g : Y → Y are said to be topologically conjugate if there is a homeomorphism h : Y → X such that f(h(y)) = h(g(y)) for all y ∈ Y .
Theorem D.2 (Hartman–Grobman). Let f : Rk → Rk be a Cr-diffeomorphism with hyperbolic fixed point x∗. Then there exists neigh- borhoods V of x∗ and W of 0 and a homeomorphism h:W →V such that f(h(x)) = h(Ax), where A = Df(x∗).
In other words, f is topologically conjugate in a neighborhood of the fixed point x∗ to the linear map induced by the derivative at the fixed point (see Figure D.3).
Proof. See Robinson [128]. P As pointed out in Cushing [24], this classical theorem does not hold for
noninvertible maps, as may be seen from the following example.
Example D.3. Consider the one-dimensional difference equation x(n + 1)=x2(n).TheJacobianatthefixedpointx∗ =0isA=0.Ifhisthe conjugacy homeomorphism, then f(h(x)) = h(Ax). Then f(h(x)) = h(0) = 0. Thus [h(x)]2 = 0 and h(x) = 0 for all x ∈ R, a contradiction, since h is one to one.
Cushing [24] extended the Hartman–Grobman Theorem to noninvertible maps and the new result will henceforth be called HGC (Hartman–
WAW
hh
VfV
FIGURE D.3. f is conjugate to A = Df(x∗), f(h(x)) = h(Ax).
490 D. Stable Manifold and the Hartman–Grobman–Cushing Theorems
Grobman–Cushing). But before stating the theorem, we need to introduce a few definitions.
For a sequence x(n) ∈ Rk, let ∥x∥+ = sup |x(n)|, where |x(n)| is a norm n∈Z+
on Rk. The sets
BS+ = {x(n) | ∥x∥+ < +∞},
BS0+ = {x(n) ∈ BS+ | lim |x(n)| = 0},
n→+∞ are (Banach)1 spaces under the norm ∥ · ∥+.
Similarly, we define
and
Define
∥x∥− = sup |x(n)| n∈Z−
BS− = {x(n) | ∥x∥− < +∞},
BS0− = {x(n) ∈ BS− | lim |x(n)| = 0}.
n→−∞
±
(δ)={x(n)∈BS± |∥x∥± ≤δ},
±
(δ)={x(n)∈BS0± |∥x∥± ≤δ}.
Suppose that x∗ is a hyperbolic fixed point of a map f ∈ Cr and let A = Df(x∗) be its
Jacobian. There exists constants c and δ such that the following hold:
(a) There is a one to one bicontinuous map between a (forward) solution of (D.1.1) lying in +(δ) and a (forward) solution of its linearization (D.1.3) lying in +(cδ).
(b) There is a one to one bicontinuous map between a (forward) solution of (D.1.1) lying in +0 (δ) and a (forward) solution of (D.1.3) lying in +0 (cδ).
Similar statements hold for −(δ) and −0 (δ).
Proof. See Cushing [24]. P
1A Banach space is a complete space with a norm, where every Cauchy sequence converges in the space.
Theorem D.4 (Hartman–Grobman–Cushing).
Appendix E
The Levin–May Theorem
To prove Theorem 5.2, we need the following result from Linear Algebra [68]:
P: “The k zeros of a polynomial of degree k ≥ 1 with complex coefficients depend continuously upon the coefficients.”
To make this more precise, let x ∈ Ck, and f(x) = (f1(x),f2(x),..., fk(x))T inwhichfi :Ck →C,1≤i≤k.Thefunctionf iscontinuousatx if each fi is continuous at x, i.e., for each ε > 0 there exists δ > 0 such that if ||y−x|| < δ, then |fi(x)−fi(y)| < ε where ||·|| is a vector norm on Ck. Now P may be stated intuitively by saying that the function f : Ck → Ck which takes the k coefficients (all but the leading one) of a monic polynomial of degree k to the k zeros of the polynomial, is continuous. Precisely, we have the following result.
Lemma E.1 [68]. Let k ≥ 1 and let
p(x)=xk +a1xk−1 +···+ak−1x+ak
be a polynomial with complex coefficients. Then for every ε > 0, there is a δ > 0 such that, for any polynomial,
satisfying
q(x)=xk +b1xk−1 +···+bk−1x+bk
max |ai − bi| < δ 1≤i≤k
491
492 E. The Levin–May Theorem
we have
min max |λj − μτ(i)| < ε, τ 1≤j≤k
where λ1,λ2,...,λk are the zeros of p(x) and μ1,μ2,...,μk are the zeros of q(x) in some order, counting multiplicities, and the minimum is taken over all permutations τ of 1,2,...,k.
The characteristic equation associated with equation (5.1.18) is given by λk+1 − λk + q = 0.
Since the characteristic roots may, in general, be complex, we may put λ = reiθ. This yields the equation
rk+1eiθ(k+1) − rkeiθk + q = 0. (E.1)
The general stability may be mapped out as a function of q and k as follows: if, for the dominant eigenvalues, θ = 0 and r < 1 (λ is real and |λ| < 1), then there is monotonic damping; if, for the dominant eigenvalues, θ ̸= 0, r < 1 (λ is complex, |λ| < 1), then there is oscillatory damping; and if r > 1 (|λ| > 1) for any eigenvalue, the zero solution is unstable.
The next lemma shows that for
kk
q1 = (k + 1)k+1 (E.2)
there is monotonic damping if 0 < q ≤ q1.
Lemma E.2. All solutions of (5.1.20) converge monotonically to the zero
solution if
0 < q ≤ q1.
Proof. To find the region of values of q where solutions of (5.1.20) converge monotonically to the zero solution we let θ = 0 in (E.1). This yields
rk+1−rk+q=0 or
q = rk − rk+1. (E.3)
Consider the function q = h(r) = rk − rk+1. Clearly, h(0) = h(1) = 0 and h(r) > 0 if and only if 0 < r < 1. Moreover, if q = 0, then r = 0 is of multiplicity k. Since h′(r) = rk−1(k − (k + 1)r), we conclude that:
(i) If0
q=h(r)
q1
q
E. The Levin–May Theorem 493
(iii) If r >
r-axis at r = 1.
r r1 k r2 1
k+1
FIGURE E.1. For q < q1, there are two positive solutions r1, r2 of (E.3).
(ii) If r =
k , then h′(r) = 0 and q attains its maximal value k+1
k k k k k+1 q1 =h(r)=h k+1 = k+1 − k+1
kk k = k+1 1−k+1
kk
= (k+1)k+1.
k , then h′(r) < 0 and thus q is decreasing and intersects the k+1
Hence, for every q > 0, there are two positive real solutions r1 and r2 of
(E.3)ifandonlyifq
496 E. The Levin–May Theorem
from below. Consequently, the complex pair of dominant eigenvalues can- not reenter the stable region once it leaves it, and so the zero solution of (5.1.18) is unstable for all q > q2.
Equating the real part with the real part and likewise with the imaginary part in (E.1) yields
r= sinkθ sin(k + 1)θ
and
q = rk cos kθ − rk+1 cos(k + 1)θ
(E.12)
= (sin kθ)k (sin kθ)k
cos kθ [sin(k + 1)θ]k
cos(k + 1)θ sin kθ [sin(k + 1)θ]k+1
−
= [sin(k + 1)θ]k+1 [sin(k + 1)θ cos kθ − cos(k + 1)θ sin kθ]
= (sinkθ)k sinθ . [sin(k + 1)θ]k+1
If r = 1, we obtain, for (E.12),
sinkθ = sin(k + 1)θ and coskθ = −cos(k + 1)θ
and hence
sin kθ = sin kθ cos k + cos kθ sin k, cos kθ = −coskθ cos θ + sin kθ sin θ.
Multiplying the first equation by cos kθ and the second by sin kθ and then adding yields
sin θ = sin 2kθ.
Multiplying the first equation by sin kθ and the second by cos kθ and then
subtracting yields
cos θ = −cos2kθ. From (E.12) we have, for r = 1,
dr =kcot(kθ)−(k+1)cot[(k+1)θ] dθ
(E.13)
= (2k + 1) cot kθ
and from (E.13) we have, for r = 1,
dq = [sin(k + 1)θ]k+1[k2(sin kθ)k−1 cos kθ sin θ + (sin kθ)k cos θ] dθ [sin(k + 1)θ]2k+2
− (sin kθ)k sin θ · (k + 1)2[sin(k + 1)θ]k cos(k + 1)θ [sin(k + 1)θ]2k+2
= k2 cos kθ sin θ + cos θ − (k + 1)2 sin θ cos(k + 1)θ
E. The Levin–May Theorem 497
[sin(k + 1)θ]2 sin(k + 1)θ =(2k2 +2k+1)cotkθ−cot2kθ
[sin(k + 1)θ]2 = (2k2 + 2k + 1) cot kθ + 1 (tan kθ − cot kθ)
2 1121
=2 k+2 cotkθ+2tankθ.
Clearly, dq and dr both have the same sign as cot kθ, and hence dr = dr / dq
is positive. This completes the proof. P
dθ dθ dq dθdθ
Appendix F
Classical Orthogonal Polynomials
This is a list of some classical orthogonal polynomials Qn(x), their defi- nitions, the corresponding intervals of orthogonality (a,b), and difference equations Qn+1(x) − (Anx + Bn)Qn(x) + CnQn−1(x) = 0.
Name
1. Jacobi:
2. Gegenbauer:
(ultraspherical)
3. Legendre:
4. Chebyshev: (First kind)
5. Chebyshev:
(Second kind)
6. Hermite:
7. Laguerre:
8. Charlier:
Definition P α,β (x)
(a, b) (−1, 1)
(−1, 1) (−1, 1)
(−1, 1) (−1, 1)
(−∞, ∞) (0, ∞)
(0, ∞)
Difference
Equation
see (9.5.12), (9.5.15), (9.5.16)
An=2ν+n,Bn=0 n+1
Cn = 2ν+n−1 n+1
2n+1 An=n+1,Bn=0
Cn=n n+1
An =2,Bn =0 Cn=1
An =2,Bn =0 Cn=1
An =2,Bn =0
Cn=2n
An = 2n+α+1−x
n
P nν ( x )
(see (9.4.10))
(0,0) Pn(x) = Pn
(see (9.4.9)) Tn(x) = cos nθ,
θ = cos−1(x) Un(x) = sin(n+1)θ ,
sin θ θ = cos−1(x)
Hn (x)
(see (9.4.15)) L αn ( x )
(see (9.4.13))
(α)
Cn (x)
(see (9.5.17))
n+1 Cn = n+α
Bn=0 n+1
An=1
Bn =−n−α Cn = an
499
Appendix G
Identities and Formulas
n+1nn
r =r+r−1, n
n+α n+β 2n+α+β kn−k=n.
k=0
Leibniz’s Formula
k dxn−k dxk k=0
n n
dxn
dxn (1 − x)n+α(1 + x)n+β =
dn
k Dn−k(1 − x)n+αDk(1 + x)n+β = (−1)n(1 − x)α(1 + x)βn!
n n n − k k d(uv)= nd dv,
k=0
n
n+α n+β (x−1)k(x+1)n−k. k=0 n−k k
501
Answers and Hints to Selected Problems
Exercises 1.1 and 1.2
(d) c n
3. (a) n!(2n +c−1)
(b) c + en − 1 e−1
9. 38 payments + final payment $52.29 11. (a) A(n+1) = (1+r)A(n)+T
(b) $25,000 [(1.008)n − 1]
13. $136,283.50
1 1
15. (a)r=1− (b) 2,933 years
Exercises 1.3
3. (a) α−1 β
(b) 2x(n) 1+x(n)
5. (b) μ = 3.3
1. (a) cn! n(n−1)
(b) c3
(c) cen(n−1)
2
5700
2
503
504 Answers and Hints to Selected Problems
7. (i) D(n) = −p(n) + 15 S(n + 1) = 2p(n) + 3
(iii) p∗ = 4, unstable
11. (a) p(n+1)=−1p2(n)+1 √2
(b)p∗=−1+ 3
(c) asymptotically stable
Exercises 1.4
1. 3.
5. 7. 9.
(a) y(n+1)=y(n)−ky2(n), y(0)=1
(a) y(n+1)=y(n)+0.25(1−y(n)), y(0)=2
(a) y(n+1)=y(n)+1y2(n)+1
y(n + 1) = 5y(n) 4−y(n)
Nonstandard: y(n + 1) = 5y(n)+n 5−y(n)
Euler: y(n + 1) = y(n) + hy2(n) + hn
42
Exercises 1.5
0 : asymptotically stable
1.
±1 : unstable
3. 0: asymptotically stable 5. 0: unstable
7. 0: unstable
9. Hint: Use L’Hoˆpital’s rule
11. Hint: Consider monotonic and nonmonotonic functions 16. (a) from the left
(b) from the right
Exercises 1.6
{0, 1}: asymptotically stable |b2 −3ab|<1
1 2
5. 7.
9.
11. 13.
17. 19.
Exercises 1.7
3,3
(b) unstable
f(x) = −x c=−7
4
x ̄1 = μ+1−
((μ+1)(μ−3), x ̄2 = μ+1+ ((μ+1)(μ−3), 2μ 2μ
2. Hint: Let x(n) = sin2 θ(n)
5. Hint: Show that 0 < f′(x) < 1 for x∗ < x < 1
13. c=−3 ∗4
2
Answers and Hints to Selected Problems 505
7. Hint: Show that Sfμ2(x(0)) < 0
9. Hint: Use a calculator or a computer 11. Hint:Lety=−μx+1μ
2
17. x1 is unstable
x∗2 is asymptotically stable x∗3 is asymptotically stable
19. x∗ = 0 is unstable Exercises 1.8
1. Therearefixedpointsx∗1 =0,x∗2 =1,x∗3 =9 Ws(x∗1) = B(x∗1) = (−1,1)
x∗2 is unstable, −1 is eventually fixed
Ws(x∗3) = [−3,−1)∪(1,9], B(x∗3) = (1,9]
9. Hint: Consider three cases:
(a) 0
(a) F(n+2)=F(n+1)+2F(n)
(b) 3, 5, 11
(ii) Hint:Let$10equal1unit,n=5,N=10 13.9298
9.66 × 10235
Y(n+3)−(a1 +1)Y(n+2)−(a2 −a1)Y(n+1)+a2Y(n)=h
√n √n
(b) Y(n)=c +c 1+ 5 +c 1− 5 +α +βn
124341 M (n) = M (n0 )2n−n0
3n−2n ⎞ ⎠
15. (a)
(b) c = 1
Exercises 3.1
⎛2n+1−3n 1. ⎝
2n+1 − 2(3n)
⎛ 11⎞
2(3n) − 2n 2n+1−3n −2+2n+1
− 3n
2 2 ⎟
1 1 ⎟ −2+23n⎟
⎜
⎜
3.⎜ (−2)n+3n 2−2n
⎝⎠
−2n+2 + 4(3n) 4 − 2n+2 −1 + 2(3n)
508
Answers and Hints to Selected Problems
⎛1 ⎞ ⎜⎝3(2n+1 + (−1)n)⎟⎠
2n+1 ⎛ 3−2n+1 ⎞
⎜ ⎟ ⎜⎝ 2(1−2n) ⎟⎠
2(−1 + 2n)
(a) Hint: Use (3.1.18)
Hint: If λ1 = λ2 = λ and λn = a0 +a1λ+a2λ2 +···+ak−1rk−1, differentiate to get another equation nλn−1 = a1 + 2a2 λ + · · · + (k −
5.
7.
10. 12.
13. (i)
⎝
1)ak−1λk−2 ⎛2n+1−3n
3n−2n ⎞ ⎠
2n+1 − 2(3n)
(ii) Same as Problem 3
⎛0 1 0⎞ ⎜⎝0 0 1⎟⎠ 110
15. (a)
(b) (2/5, 1/5)
2(3n) − 2n
22
Exercises 3.2
⎛11 + 3n− 115n⎞ 9.⎜16 4 16 ⎟
⎝−5 1 11 ⎠ 16 −4n−165n
15. a1(−2)n + a2(−6)n 17. a1 +a24n + 1n4n
3
√n √n 19.a 1− 5 +a 1+ 5
1222 Exercises 3.3
n+1 n 2−4
1.
⎛3 n+1 n⎞
2(4n )
⎜7[(−1) +6 ]⎟
3. ⎜⎝ 3(−1)n + 46n ⎟⎠ 77
0
Answers and Hints to Selected Problems 509
⎛⎞⎛⎞⎛⎞
011 5. c12n ⎜⎝1⎟⎠ + c2 ⎜⎝−1⎟⎠ + c33n ⎜⎝7⎟⎠
⎛
⎜ ⎜
9. ⎜ ⎜⎝
11. (a)
002
nπ nπ⎞ 2n/2 −c2sin 4 +c3cos 4 ⎟
n/2 nπ nπ⎟
−c2cos 4 −c3sin 4 ⎟ n/2 nπ nπ⎟⎠
2
c1+2 c2cos 4 +c3sin 4
n n−1 3 n3
0 3n ⎛2n n2n−1
n(n − 1)2n−3⎞ n−1 ⎟
⎜ n
(b)⎜⎝0 2 n2 ⎟⎠
0 0 2n ⎛ ⎞⎛
n(n−1) ⎞ 3n−2
2 1 −2 3n ⎜−1 0 1⎟⎜
n3n−1 3n
2 ⎟ n3n−1 ⎠
(c) ⎝ 2 ⎠ ⎝ 0 00300 3n
⎛ 3⎞ ⎜0 −1 2⎟
⎜⎝1 2 0⎟⎠ 003
⎛n2⎞ 200 0
⎜
⎜ 0 2n n2n
⎟ n(n − 1)2n−2⎟
n2n ⎟⎠
⎛c1(2n − n2n−1) + c2n2n + c3(3n22n−1 + 3n(n − 1)2n−3)⎞
(d)⎜⎝0 0 2n 000 2n
⎜ n−2 n n−4 ⎟ 13. ⎜⎝ −c1n2 +c22 (1−n)−3c3n(n−1)2 ⎟⎠
c3 2n
19. Hint: Use the similarity matrix P = diag(1, α, α2, . . . , αk−1)
Exercises 3.4
5. Hint: First change the equation to a system and then show that the monodromy matrix is equal to Φ(N)
510 Answers and Hints to Selected Problems
Exercises 3.5
3. Hint:ConsiderATξ=ξwithξ=(1,1,…,1)T 5. (i) Hint: Consider (I − A)x = 0
(ii) Hint:Use(I−A)(I+A+A2+···+An−1)=I−An ⎛⎞
5/9 7. ⎜⎝2/9⎟⎠
2/9 9. 0.25
11. 177.78; 272.22 13. 0
Exercises 4.1
1. (a)3,3,3 (b) 6, 4, 33
4
(c) 6, 7, 5.21
3. Hint: Use D = diag(1, ε, ε2)
Exercises 4.3
1.
3.
5.
(a) unstable
(b) asymptotically stable
(c) asymptotically stable (d) stable
⎛5 0 1⎞ ⎜12 2⎟
⎜−1 −1 5⎟ ⎜⎝ 4 ⎟⎠
100 3
(a) uniformly stable (b) no conclusion
(c) asymptotically stable (d) no conclusion
Exercises 4.4
1. (a) asymptotically stable (b) unstable
(c) unstable
(d) asymptotically stable
3. unstable
5. stable, but not asymptotically stable
1. 3.
exponentially stable
12 12
Hint:V =a2 x1−√ +b2 x2−√ 22
Answers and Hints to Selected Problems 511
Exercises 4.5
1. Hint:LetV(x)=x21+x2
The fixed point (0, 0) is globally asymptotically stable.
13. Hint: Let V = xy and then use Problem 11 Exercises 4.6
The equilibrium point
1 1 √ , √
22
is unstable. The equilibrium point
11
−√ ,−√ is unstable.
4. (a)
6. 8.
01 ⎜⎝0⎟⎠, ⎜⎝1⎟⎠ 00
22
⎛⎞⎛⎞
(b) undetermined, unstable
unstable
if |a| < 1 and |b| < 1, then the zero solution is asymptotically stable
N(γ + β) α − (γ + β) (N,0), α ,βN α(γ+β)
10. (a)
(b) The first point is asymptotically stable if α < 1 and unstable
γ+β
if α > 1. The second point is asymptotically stable.
Exercises 5.1 and 5.2
3. 1<α<2.62
11. Hint: Let g(z) = p1zk−1 +p2zk−2 +···+pk, and f(z) = zk on the unit
disk
12. Hint: Let f(z) = p1zk−1, g(z) = zk −p2zk−2 +···+pk, on the circle
of radius 1 + ε, for some appropriate ε > 0
Exercises 5.3
1. −1 < b < 0.78 2
6. Hint: Make the change of variable N(n) = N∗ex(n) k
12.
γ+β
Hint: Use the variation of constant formula (3.2.12) and then use Theorem 8.12
ai −b 1 k 2 k 7.x∗= i=0 + b− ai +4a bi
k k i=0 i=0 2bi 2bi
i=0 i=0
512 Answers and Hints to Selected Problems
Exercises 6.1
1. (a)
(b) (c)
z(z−cosω) , |z|>1 z2 −2zcosω+1
z(z2−1)sin2 , |z|>1 (z2 −2zcos2+1)2
z , |z|>1 (z − 1)2
3.−z+a2+a, |z|>|a| z(z − a)
5. (z+1)2zn−3 zn − 1
7. Hint: Use mathematical induction on k 9. 1
(z − a)3
15. Hint: y(n) − y(n − 1) = nx(n)
17. (a) z2sinω (z−a)(z2 −2zcosω+1)
(b) z2(z − cos ω) (z−1)(z2 −2zcosω+1)
Exercises 6.2
1. (a) 2/3[2−n − 1]
(b) −1/7(−2)n + 1/7n(−2)n + 6/7
3. (a) (−2)n−3(3n2 − n) (n−1)
(b) 2−n+1 +2sin 2 π
√ n √ n
11+51−5 5.√5 2 − 2
7. 1(n+1) 2
11. Hint: Replace n by n + 1 1−e1
2−e + 2−e (e−1)n Exercises 6.3
1. x(n) = 1x(0)[1 + 2(4n)] 3
unstable
3. Problem 1. unstable Problem 2. uniformly stable
5. unstable
Exercises 6.4
1. asymptotically stable
3. not asymptotically stable
4. Hint:∞ nan = a n=0 (1−a)2
fora<1,∞ n=0
n2an = a2+a (1−a)3
4. Hint: See Theorem 4.9
5. (a) x(n) = −1(−3)n + 1(4n)
Answers and Hints to Selected Problems 513
Exercises 6.5
3. asymptotically stable 5. uniformly stable
Exercises 6.6
⎛√77√ ⎞
⎜(1+ 2)2n−1+(1− 2)(−1)n 7.(a)⎜ 2
0 ⎟ √ √ ⎟
3− 6 3n+ 2+ 6 (−2)n⎠ 55
⎝ 0
⎛1+√2 1−√2 ⎞
(b) ⎜⎝ 2 (2n −n−1)+ ⎛1n3n⎞
−1+ 1(2n)+1(3n) 12 2
Exercises 7.1
5. Hint: Use Theorem 7.3
7. Hint: Consider the function f(λ) = λk+1 − λk + p and show that it
attains its minimum when λ = (k − 1)/k Exercises 7.2
8. Hint: Use Problem 7
9. Hint: Use Problem 7 12. Hint: Use Theorem 7.16 14. Hint: Use Problem 13
⎜−2+12(2 )+2(3 )⎟ 11. ⎜⎝ ⎟⎠
8 [(−1)n +2n−1]⎟⎠
0
514 Answers and Hints to Selected Problems
Exercises 7.3
3. Hint: Let x(n) = αey(n) and then use Theorem 7.18
4. Hint: Let x(n) = x∗ey(n) and then apply Theorem 7.19
6. Hint: Let z(n) = x(n)/x(n + 1) and then mimic the proof of Theorem
7.18
7. Hint: Follow the hint in Problem 6
Exercises 8.1
12. Hint: Use f(t) = (1 + o(1))g(t)
14. (c) Hint: Show that ∞ e−xt dt ≤ e−x
1 tn−1 n−2 15. Hint: Use integration by parts
16. Hint:Writen kk =nn[1+(n−1)n−1 +···+ 1 ] k=1 nn nn
Exercises 8.2
14. Hint: Notice first that log n−1 (1 + u(i)) = n−1 log(1 + u(i)) i=n0 i=n0
15. (c) Hint: Use the mean value theorem
(e) Hint: Substitute (8.2.19) into (8.2.17)
(f) Hint: Solve (8.2.28) and then use Problem 14
17. x1(n)∼n2,x2(n)∼ 1 ,n→∞ (n+2)!
20. Hint: Let y(n) = x(n + 1)/x(n)
Exercises 8.4
1n n−1
12. Hint: Let x(n) = −2 Exercises 8.5
p1(j) z(n)
j =n0
7. Hint: Reverse the order of summation on the left-hand side as in Figure 8.2
10. Hint: Use Problem 8 and then let A(n) = x2(n)∆y(n) − ∆x2(n)y(n) and B(n) = ∆x1(n)y(n) − x1(n)∆y(n), then mimic the proof of Problem 8
Exercises 9.3
9. (x,y)=(37,47)+m(48,61)
10. Hint: Consider the continued fraction representation of Then √l = A(m−1)+A(m)ξm+1 ,
B(m−1)+B(m)ξm+1
ξm+1=2b0+ 1 1 ...=√l+b0;
√
l
b1+ b2+
A(m)(A(m) − B(m)b0) − B(m)(B(m)l − A(m)b0) = (−1)m−1
show that
or
Answers and Hints to Selected Problems 515
A2(m) − lB2(m) = (−1)m−1
Conclude that x = A(m), y = B(m) is a solution of Pell’s equation if
m is odd, and if m is even, x = A(2m+1), y = B(2m+) is a solution 11. x=8,y=3
Exercises 9.4 and 9.5
5. (n+1)Pnν+1(x)=2(ν+n)xPnν(x)−(2ν+n−1)Pnν−1(x)
6. (n + 1)Lαn+1(x) = (2n + α + 1 − x)Lαn(x) − (n + α)Lαn−1(x)
7. Hn+1(x) = 2xHn(x) − 2nHn−1(x)
9. Jn+1(z)=(2n)Jn(z)−Jn−1(z) z
12. Hint: Use (9.5.18) and let u → x Exercises 9.6
7. Hint: Use the Cauchy integral formula: dn (1−x2)n = n! (1−t2)n dt dxn 2πi c (t−x)n+1
Exercises 10.1 and 10.2
1. W = 1 −2 , |W | = 1 ̸= 0, the system is completely controllable 01
3. Since A is diagonal and B has a row of zeros then, by inspection, the system is not completely controllable
5. rank(W ) = 4 < 5, the system is not completely controllable
11.W= 1 a11+a12 ,|W|=a21+a22−a11−a12̸=0thusa22+a21̸= 1 a21+a22
a11 + a12 Exercises 10.3
1.
3.
5.
(a) V = 0 2 ,|V|=2,x0 =V−1 y(0) = 1/4(a+b) 4 −2 y(1) 1/2b
21
21
1 a+b
W = 1 c+d ,|W| = −(a+b)+c+d ̸= 0. Thus, for a system
to be completely controllable a + b ̸= c + d, and for a system to be
10 completelyobservable,V = a b ,|V|=b̸=0
rank(V ) = 4, the system is not completely observable
(b) V =
,rank(V)=1<2,thesystemisnotobservable
516 Answers and Hints to Selected Problems
Exercises 10.4
1. K = −0.1166 −0.6982
3. K = −1.8599 0.5293 2.8599
5. Hint: Put the equation into a system form
Exercises 10.5
0.875
1. E = −1.125 3. unsolvable
Maple Programs
(I) Solution of scalar difference equations and systems of difference equations using rsolve
> rsolve({x(n+1)-x(n)/(n+1)=1/(n+1)!,x(0)=1},x);
n+1 Γ(n+1)
> rsolve({x(n+1)=2*x(n)+y(n),y(n+1)=2*y(n),x(0)=a,y(0)=b},
{x,y});
{y(n)=b2n,x(n)= 1b2n +2na} 2
> rsolve(x(n+2)-5*x(n+1)+6*x(n)=0,x);
−(2x(0) − x(1))3n − (−3x(0) + x(1))2n
517
518 Maple Programs
Cobweb Program
>#Cobweb Program
>#Call as: cobweb(F, n, s, l, u)
>#Where: F: The one parameter function
># n: The number of iterations to be performed
># s: The initial value of x with which to start
># l: The lower bound value for x and y
># u: The upper bound value for x and y
>cobweb:=proc(function, iterations, initial, lowerbound,
upperbound)
>local F, n, s, u, i, y, G, l;
>F:=function;
>n:=iterations;
>s:=initial;
>l:=lowerbound;
>u:=upperbound;
>with(plottools)
>y:=eval(subs(x=s,F));
>G:=[line([l,l], [u,u]), line([s,0], [s,y]),
plot(F,x=l..u,color=black)];
>for i from l to n do
> G:=[op(G), line([s,y], [y,y]))];
> s:=y;
> y:=evalf(subs(x=s,F));
> G:=[op(G), line([s,s], [s,y])];
>od
>plots[display](G,tickmarks=[0,0]);
>end
> # Example: draw the cobweb diagram of the function
> # F(x)=3.7*x*(1-x) with initial point 0.1.
>cobweb(3.7*x*(1-x),10,0.1,0.1);
Bifurcation Diagram Program
>bifur(c*x*(1-x), 3.57, 4, 100, 50, .01, .5, 0, 1);
furcation Diagram Program
> # Bifurcation Diagram Program
> # Call as: bifur(F, l, r, N, T, d, s, b, t)
>#
> # Where:
># F: The one parameter function in terms of c and x
># l: The left bound on the graph
># r: The right bound on the graph
># N: The number of iterations to perform
># T: The number of iterations to discard
># d: The step size of the parameter (c)
># s: The value of x
># b: The bottom bound on the graph
># t: The top bound on the graph
>#
> bifur:=proc(function, left, right, iterations, discard, step, start, bottom, top)
> local F, l, r, N, T, d, s, t, i, p, b, j, k, G;
> F:=function;
> l:=left;
> r:=right;
> N:=iterations;
> T:=discard;
> d:=step;
> s:=start;
> t:=top;
> b:=bottom; G:=[];
> with(plottools):
> for i from l by d*(r-l) to r do
> p:=s;
> for >
> od;
> for ktoNdo
j toTdo
p:=evalf(subs(x=p, c=i, F));
>
>
> od;
> od;
> plots[display](G, axes=boxed, symbol=POINT,
view=[l..r, b..t]);
> end:
p:=evalf(subs(x=p, c=i, F));
G:=[op(G), point([i,p])];
Maple Programs 519
520 Maple Programs
> # Example: Draw the graph of F(x)=c*x*(1-x), where c is
> # between 3.5 and 4 and initial point is 0.5.
> bifur(c*x*(1-x),3.5,4,200,50,.01,.5,0,1);
μ
Phase Space Diagram with Four Initial Points
> # Phase Space Diagram Program (with 4 initial points) > # Call as: phase4(A, x, y, z, v, n)
>#
> # Where:
># A: The matrix entries where f(x)=Ax
># v: The initial point (v1,v2)
># x: The initial point (x1,x2)
># y: The initial point (y1,y2)
># z: The initial point (z1,z2)
># n: The number of iterations to perform
>#
> phase4:=proc(matrix11, matrix12, matrix21, matrix22, initial1, initial2, initial3, initial4, initial5, initial6, initial7, initial8, iterations)
> local A, x, n, G, F, H, J, x1, x2, i, x3, x4, w1, w2, y3, y4, z1, z2, z3, z4, y, z, v1, v2, v3, v4, v, K;
> A:=array(1..2,1..2,[[matrix11,matrix12], [matrix21,matrix22]]);
> x:=array(1..2,1..1,[[initial1],[initial2]]);
y:=array(1..2, 1..1, [[initial3],[initial4]]);
z:=array(1..2, 1..1, [[initial5],[initial6]]);
v:=array(1..2, 1..1, [[initial7],[initial8]]);
> n:=iterations;
> x1:=x[1,1]; x2:=x[2,1]; w1:=y[1,1]; w2:=y[2,1]; z1:=z[1,1]; z2:=z[2,1]; v1:=v[1,1]; v2:=v[2,1];
> G:=[]; H:=[]; J:=[]; K:=[];
> with(plottools):
> for i from 1 to n do
> F:=array(1..2, 1..1, [[(A[1,1]*x1)+(A[1,2]*x2)],
[(A[2,1]*x1)+(A[2,2]*x2)]]);
> x3:=F[1,1]; x4:=F[2,1];
> G:=[op(G), line([x1,x2],[x3,x4])];
> x1:=x3; x2:=x4;
> F:=array(1..2, 1..1, [[(A[1,1]*w1)+(A[1,2]*w2)],
[(A[2,1]*w1)+(A[2,2]*w2)]]);
> y3:=F[1,1]; y4:=F[2,1];
> H:=[op(H), line([w1,w2],[y3,y4])];
> w1:=y3; w2:=y4;
> F:=array(1..2, 1..1, [[(A[1,1]*z1)+(A[1,2]*z2)],
[(A[2,1]*z1)+(A[2,2]*z2)]]);
> z3:=F[1,1]; z4:=F[2,1];
> J:=[op(J), line([z1,z2],[z3,z4])];
> z1:=z3; z2:=z4;
Maple Programs 521
522 Maple Programs
> F:=array(1..2, 1..1, [[(A[1,1]*v1)+(A[1,2]*v2)],
[(A[2,1]*v1)+(A[2,2]*v2)]]);
> v3:=F[1,1]; v4:=F[2,1];
> K:=[op(K), line([v1,v2],[v3,v4])];
> v1:=v3; v2:=v4;
> od;
> plots[display](G,H,J,K,tickmarks=[0,0],color=black);
> end:
> # Example: Draw the phase space diagram of the system
> # x(n+1)=A*x(n) where A=array(2,0,0,0.5).
> phase4(2,0,0,.5,1,3,-1,3,-1,-3,1,-3,6);
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References 529
Index
∼, 335
Abel’s formula, 128
Abel’s lemma, 68
Abel’s summation formula, 63 absorbing Markov chains, 163 Ackermann’s formula, 466
actual saddle, 201
actual unstable focus, 203 Adams, 377
adult breeding, 258
age structured population model,
259
algebraic multiplicity, 143
amortization, 6
analytic, 292, 479 annihilator, 85
annual mortality, 259 annual plant model, 113 antidifference operator, 61 Ap ́ery sequence, 379 Appendix A, 34
Appendix B, 77 Appendix C, 182 Appendix D, 190 Appendix E, 248 Appendix F, 425 Appendix G, 414, 415
applications, 229
asymptotic, 337
asymptotically constant system,
360
asymptotically diagonal systems,
351
asymptotically stable, 95, 177, 185 asymptotically stable attracting, 28 asymptotically stable attractor
periodic orbit, 48 asymptotically stable equilibrium
point, 12
asymptotically stable equilibrium
price, 16
asymptotically stable fixed point,
43, 46
asymptotically stable focus, 199 asymptotically stable node, 195,
197, 198 attracting, 11, 176, 180
attracting but not stable equilibrium, 181
attracting unstable fixed point, 182 autonomous, 2, 117, 118, 135, 176 autonomous linear systems, 186 autonomous systems, 131
531
532 Index
Backward difference, 281 Baker’s function, 40 Banach space, 490
basic theory, 125
basin of attraction, 50, 231 Benzaid–Lutz theorem, 358 Bessel function, 420 Beverton–Holt model, 263 bifurcation, 43, 240 bifurcation diagram, 47, 49 bifurcation diagram of Fμ., 47 bilinear transformation, 309 binomial sums, 381
biological species, 14 Birkhoff’s theorem, 377 Birkhoff–Adams, 377
block diagram, 431 block-diagonal matrices, 148 blowfly, 224
bottom-feeding fish, 262 bounded, 177
business cycle model, 233
Cannibalism, 238, 239 canonical saddle, 201 canonical unstable focus, 202 capacity of the channel, 112 carbon dating, 9
Carvalho’s lemma, 41, 42 Casoratian, 67, 69, 70, 133 Casoratian W(n), 74
Cauchy integral formula, 287 Cauchy product, 417 Cayley–Hamilton theorem, 119,
120, 124 center (stable), 200
channel capacity, 115
chaos, 243
chaotic, 24
characteristic equation, 75, 119 characteristic roots, 75 Charlier, 499
Charlier polynomials, 420, 426 Chebyshev (First kind) , 499 Chebyshev (Second kind) , 499 Chebyshev polynomials, 81, 413 Chebyshev polynomials of the first
kind, 427 Christoffel–Darboux identity, 420
Clark, 251
classical orthogonal polynomials,
413
closed-loop system, 457
closure, 208
Cobweb phenomenon, 14
Cobweb theorem of economics, 17 Coffman, 363
Coffman’s theorem, 364 companion matrix, 133 competitive species, 117 complementary solution, 84 complete controllability, 436, 452 complete observability, 452 completely controllable, 432, 433,
435, 442, 451, 461, 462 completely observable, 446, 448,
451, 468
completely observable and
controllable system, 450 complex characteristic roots, 78 complex eigenvalues, 140 complex poles, 286
confluent form, 421
conjugacy homeomorphism, 489 consistent, 192
constant solution, 9 constructibility, 452 constructible, 452
continued fractions, 397, 421 continued fractions and infinite
series, 408 control, 83, 429
controllability, 432, 433 controllability canonical forms, 439 controllability matrix, 433, 434 controllability of the origin, 436 controllability to the origin, 452 controllable canonical form, 442 controllable to the origin, 436 controlled system, 429, 430 converge, 398
convergence of continued fractions,
400 converges, 345
converges conditionally, 345 convolution, 278
convolution∗ of two sequences, 278 Costantino, 238
criterion for asymptotic stability, 182
criterion for the asymptotic stability, 27
current-controlled DC, 432 Cushing, 238, 489
cycles, 35
Dannan, 245, 250
De Moivre, 273
De Moivre’s theorem, 78 deadbeat response, 465
definite sum, 5
degenerate node, 197
Dennis, 238
density-dependent reproductive
rate, 42 Desharnais, 238
design via pole placement, 457 development of the algorithm for
An, 119 diagonalizable, 135
diagonalizable matrices, 135, 136 dichotomy, 352
difference calculus, 57
difference equation, 1
difference operator, 57 differential equation, 4 dimension, 73 Diophantine equation, 412 discrete analogue of the
fundamental theorem of
calculus, 58
discrete dynamical system, 1 discrete equivalents for continuous
systems, 431
discrete Gronwall inequality, 220 discrete Putzer algorithm, 123 discrete Taylor formula, 63 diverge, 398
diverges, 345
dominant, 161, 370, 425 dominant characteristic root, 91 dominant solution, 91, 425 drunkard’s walk, 163, 164 duality principle, 451
dyadic rational, 20
Economics application, 14
eigenvalue, 118, 119
Eigenvalue Separation Theorem,
468, 469 eigenvector, 136, 143
Elaydi, 245
Elaydi and Harris, 120
Elaydi and Jang, 226
Elaydi and Sacker, 483
electric circuits, 288 electromagnet, 470
epidemic model, 302
equation of motion, 437 equation with delay, 207 equilibrium point, 9, 11, 43, 176 equilibrium price, 15
Erbe and Zhang, 313, 327 Ergodic Poincar ́e type (EP), 390 Euclidean algorithm, 410 Euclidean norm l2, 174
Euler identity, 276
Euler’s algorithm, 21
Euler’s method, 20
eventually k periodic, 35 eventually equilibrium (fixed)
point, 9
eventually eventually eventually eventually eventually eventually
equilibrium points, 10 fixed point, 10 negative, 314 negative solution, 315 positive, 313
positive solution, 315,
329 Evgrafov, 363
explicit criteria for stability of Volterra equations, 295
exponential integral En(x), 339 exponentially stable, 177 Extension of Perron’s Second
Theorem, 387 external force, 83
Eynden, 125
Factorial polynomials, 60 Favard’s theorem, 418 feedback control, 457 feedback controller, 473 Feigenbaum, 46, 47 Feigenbaum number, 46 Feigenbaum table, 47
Index 533
534 Index
Fibonacci, 79
Fibonacci sequence, 79
final value theorem, 278 first iterate, 1
fixed point, 9
fixed points of T2, 39
fixed points of T3, 40 Floquet exponents, 156, 190 Floquet multiplier, 156–158 Floquet multipliers, 190 flour beetle, 238, 268 forcing term, 83
fully synchronous chaotic
attractors, 243 fundamental, 76
fundamental matrix, 126, 306 fundamental matrix of system, 145 fundamental recurrence formula,
397, 398, 417 fundamental set, 67, 70, 71 Fundamental Theorem, 72
Gain state matrix, 457
gambler’s ruin, 107
gamma function, 7
Gauss–Siedel iterative method, 193 Gegenbauer (or ultraspherical)
polynomials, 416 Gegenbauer (ultraspherical), 499 Gegenbauer polynomials, 416, 424,
427
general Riccati type, 100
general solution, 73, 76, 137 generalization of Poincar ́e–Perron
theorem, 372
generalized eigenvectors, 144, 149,
188
generalized Gronwall’s inequality,
375
generalized Vandermonde
determinant, 77 generalized Vandermonde matrix,
481 generalized zero, 321
generating function, 273, 427 generations, 1
genetic inheritance, 161 geometric multiplicity, 142 Gerschgorin disks, 253
Gerschgorin’s theorem, 252 global stability, 50, 261
globally asymptotically stable, 12 globally asymptotically stable
equilibrium, 210 globally attracting, 11, 182
golden mean, 80
Gronwall inequality, 220 Grove, 265
Gyori and Ladas, 313, 318
Haddock, 262
Hartman, 321
Hartman’s definition, 321 Hartman–Grobman, 489 Hartman–Grobman theorem, 489 Hartman–Grobman–Cushing, 490 Hautus and Bolis, 261
heat equation, 167, 169
heat transfer, 167
Henrici, 423
Henson, 238
hereditary, 291
Hermite, 499
Hermite polynomials, 416, 426, 427 high-order difference equations, 360 homogeneous, 2
homogeneous linear difference
system, 125 Hooker and Patula, 313
host–parasitoid systems, 232 hybrid, 161
hyperbolic, 28 hypergeometric function, 424
Ideal sampler, 431
identity operator, 58
immediate basin of attraction, 50 index for maximum property
(IMP), 391 infectives, 226
infinite products, 344
information theory, 112
inherent new reproductive number,
239
initial and final value, 277
initial value problem, 66, 130 initial value theorem, 277 inners of a matrix, 246
input, 83, 446
input–output system, 83, 446, 449 integral representation, 423 Intermediate Value Theorem, 477 invariant, 208
inventory analysis, 114
inverse Z-transform, 282
inversion integral method, 282, 287 invertible, 481
iterative methods, 192
Jacobi, 499
Jacobi iterative method, 192, 193 Jacobi polynomials, 414
Jacobian matrix, 220
Jordan block, 142–144
Jordan canonical form, 142 Jordan chain, 144
Jordan form, 135, 142, 143, 187
k-cycle, 35
k-dimensional observer, 467 k-dimensional system, 132 k-periodic, 35
k × k controllability matrix, 440 Kalman, 432
Kocic and Ladas, 261, 303 Kreuser, 372
Kronecker delta function, 413 Kronecker delta sequence, 276 kth-order linear homogeneous
difference, 66 Kuang and Cushing, 259
Kulenovic and Ladas, 261 Kuruklis, 245, 248
l∞ norm, 292
l1 norm, 174, 292
l2 or Euclidean norm, 292 ladder network, 288, 289 Lagrange identity, 193 Laguerre, 499
Laguerre polynomials, 380, 416,
426, 427 Laplace transform, 432
larvae, 238
LaSalle’s invariance principle, 207,
209
Laurent series, 287
Laurent’s theorem, 282
leading principal minors, 214 left-shifting, 277
Legendre, 499
Legendre function, 424 Legendre polynomials, 415, 426 Leibniz’s formula, 501 Leonardo di Pisa, 79
level curves, 206
Levin and May, 245, 248 Levinson’s theorem, 355
Li and Yorke, 37
Liapunov, 173, 204, 219 Liapunov equation, 215 Liapunov function, 204 Liapunov functional, 297, 301 Liapunov functions for linear
autonomous systems, 214 Liapunov stability theorem, 205 Liapunov theory, 173 Liapunov’s direct method, 204 Liapunov’s direct, or second,
method, 204 limit inferior, 314
limit point, 207
limit set Ω(x0), 208
limit superior, 314
limiting behavior, 91
limiting behavior of solutions, 91 limiting equation, 329
linear combination, 66
linear difference equations, 57 linear differential equations, 118 linear first-order difference
equations, 2
linear homogeneous equations
constant coefficients, 75 linear independence, 66
linear independent, 126
linear periodic system, 153
linear scalar equations, 246 linearity, 58, 277
linearity principal, 128 linearization, 219
linearized equation, 331
linearized stability result, 258 linearly dependent, 66
linearly independent solutions, 128
Index 535
536 Index
local stability of oscillatory nonhyperbolic maps, 479
logistic equation, 13, 43 LPA model, 243
Lucas numbers, 82 Lucas numbers L, 82 Lucilia cuprina, 224
M ̈obius transformation, 400, 406, 410
Maple, 17
marginal propensities, 166 marginal propensity to consume,
109 Markov, 159, 160
Markov chains, 159
Markov matrices, 160
matrix difference equation, 126 matrix equation, 306
matrix norms, 175
maximal invariant subset, 209 May’s genotype selection model,
480 Meschkowski, 344
metallic sphere, 471 method of successive
approximation, 353 method of undetermined coefficients, 83, 85
midpoint method, 116
minimal, 425
minimal polynomial, 445
minimal solution, 421, 425
minimal subdominant recessive, 370 minors, 214
moments, 413
monic, 413
monodromy matrix, 156
mosquito model, 270
mosquito population, 266
μ∞, 46
multiple poles, 287
multiplication by an property, 279 multiplication by nk, 279
nth iterate, 1
national income, 108, 165 Neumann’s expansion, 167 Nevai class, 424
Newton’s method of computing the square root of a positive number, 18
Newton’s theorem, 63 Newton–Puiseux diagram, 372, 373 Newton–Raphson method, 29 Nicholson–Bailey model, 235 nilpotent matrix, 145
nodes, 21
non-self-adjoint, 322 nonautonomous, 2, 118 nonautonomous linear systems, 184 nonhomogeneous, 2 nonhomogeneous differential
equation, 4 nonhomogeneous linear difference
equation, 64 nonhomogeneous system, 129 nonlinear difference equations, 327,
382
nonlinear equations transformable
to linear equations, 98 nonnegative, 160
nonobservable system, 447 nonoscillatory, 313
nonoscillatory nonhyperbolic maps,
477
nonstandard scheme, 24
norm, 174
norm of a matrix, 174
normal, 142
norms of vectors and matrices, 174 North Atlantic plaice, 262
nth approximant, 398
null sequence, 346
numerical solutions of differential
equations, 20
O, 335
o, 335
observability, 446
observability canonical forms, 453 observability matrix, 448, 453 observer, 467
one species with two age classes,
229
open-loop system, 457
(open-loop) time-invariant control system, 457
operator norm, 174, 175 ordinary dichotomy, 352, 382 orthogonal polynomials, 421 oscillate, 93, 94, 313 oscillating, 91
oscillation theory, 313 oscillatory, 313, 323 oscillatory solution, 322 output, 446
Parameters, 89
parasitoids, 235
partial denominator, 398
partial fractions method, 282, 283 partial numerator, 398
particular solution, 84, 130
Pell’s equation, 413
period three implies chaos, 37, 49 period-doubling bifurcation, 243 periodic, 13, 35, 176
periodic attractor, 49
periodic orbit, 35
periodic points, 35
periodic solution, 157
periodic system, 153, 190 permutations, 492
Perron, 160, 173, 219, 340, 372 Perron’s approach, 219
Perron’s example, 344
Perron’s First Theorem, 344 Perron’s Second Theorem, 344 Perron’s theorem, 160 perturbation, 219
perturbations of Chebyshev
polynomials, 424
perturbed diagonal system, 351 phase space, 178
phase space analysis, 194
Pielou logistic delay, 331
Pielou logistic delay equation, 224,
331
Pielou Logistic Equation, 18 Pielou logistic equation, 99 Pincherle, 402
Pituk, 388
Pochhammer symbol, 424 Poincar ́e–Perron, 425 Poincar ́e, 340
Poincar ́e type, 343
Poincar ́e type (P), 390 Poincar ́e’s theorem, 340, 343 Poincar ́e–Perron, 344 Poincar ́e–Perron theorem, 348 Poisson probability distribution,
235
polynomial operator, 85
population, 13, 42
population dynamics, 57
positive definite, 204, 214
positive definite symmetric matrix,
215
positive innerwise, 247
positive limit set, 208 positive orbit, 1
positive semidefinite, 216 positively invariant, 51 power series method, 282 power shift, 59
prime number theorem, 338 probability, 159
probability vector, 160 product, 61
product rule, 61
pro jection matrix, 359, 382 propagation of annual plants, 104,
105
properties of the Z-transform, 277 pupal, 238
Putzer algorithm, 118, 120, 131 Puu, 233
Quadratic Liapunov function, 205
Rabbit problem, 79
radius of convergence, 274, 277 rank, 433
recessive, 161
recruitment, 258
reduced-order observers, 474 regions of convergence, 274, 275 regions of divergence, 275 regular continued fractions, 409 regular Markov chains, 160, 161 relation ∼, 337
repeated poles, 284
repelling point, 28
residue theorem, 287
Riccati equation, 98
Index 537
538 Index
Riccati transformations, 322 Riccati type, 98, 99
Ricker’s equation, 43 Ricker’s map, 54, 243 Riemann zeta function, 409 right-shifting, 277
Rodrigues’ formula, 426 Rouch ́e’s theorem, 256, 295 Routh stability criterion, 310
Saddle (unstable), 196 Samuelson–Hicks model, 233 Scha ̈fli’s integral, 426 Schur–Cohn criterion, 246, 247 Schwarzian derivative, 31, 49 second-order difference equations,
369
second-order linear autonomous
(time-invariant) systems,
194 Sedaghat, 181, 261
self-adjoint, 320
self-adjoint second-order equations,
320
semi-asymptotically stable, 35, 44,
480 semisimple, 143, 187
semisimple eigenvalue, 143 semistability, 34
semistable from the left, 30 semistable from the right, 30 shift operator, 57
shifting, 277
similar, 135
similarity transformation, 440 simple, 143
simple eigenvalue, 160
simple poles, 284
skew symmetric matrices, 142 Smith, 261
solution, 3, 65
spectral radius, 175
stability, 11
stability by linear approximation,
219
stability of a k periodic point, 39 stability of linear systems, 184 stability of nondifferentiable maps,
483
stability of periodic points, 39 stability of the 2-cycle, 45
stability theory, 173
stability via linearization, 256 stabilizability, 462
stabilizable, 462, 463
stabilization by state feedback, 457 stabilization of nonlinear systems
by feedback, 463 stable, 11, 176, 184
stable and unstable manifolds, 488 stable economy, 166
stable equilibrium price, 16
Stable Manifold Theorem, 487–489 stable matrix, 191
stable subspace (manifold), 188 Stable Subspace (Manifold)
Theorem, 189 stable subspaces, 487
Stair Step (Cobweb) diagrams, 13 state feedback, 457
state feedback gain matrix, 459 state transition matrix, 127
step size, 21
Stirling numbers, 63, 64 Stirling’s formula, 338
Strong Poincar ́e type (SP), 390 Sturm separation theorem, 321 sufficient conditions for
stability, 251 sum of residues, 293
summation, 5
summation by parts formula, 62 superposition principle, 72
survival coefficient, 258
susceptible individuals, 302 susceptibles, 226
Sylvester’s criterion, 214
symbol O, 335, 337
symbol o, 337
symmetric matrices, 142 symmetric matrix, 216 synchronous, 241
synchronous 3-cycle, 243
system of first-order equations, 132 systems, 117
T2, 37 T3, 37
Taylor’s theorem, 477
tent function, 36
tent map, 10
3-cycle, 37, 48
three-term difference equations, 313 time domain analysis, 273 time-invariant, 2, 117, 118, 135 time-variant, 2, 118
Toeplitz, 168
Toeplitz matrix, 168
trade model, 165
transform method, 273 transient, 163
transition matrix, 160 transmission of information, 110 tridiagonal determinant, 82 tridiagonal Toeplitz matrix, 168 Trjitzinsky, 377
2-cycle, 39, 45
22-cycle, 46
23-cycle, 46
Ultraspherical polynomials, 416 uncontrolled system, 429, 430 uniform attractivity, 176, 185 uniformly asymptotically stable,
177, 185, 186, 294, 300 uniformly attracting, 176
uniformly stable, 176, 184, 186, 294 unique solution, 2, 66, 126 uniqueness, 66
uniqueness of solutions, 125
unit impulse sequence, 276 unitary matrices, 142 unstable, 11, 176
unstable fixed point, 43 unstable focus, 199 unstable limit cycle, 211 unstable node, 196 unstable subspaces, 487
Vandermonde determinant, 75, 82 variation of constants formula, 130,
166, 168, 305, 353, 382 variation of constants parameters,
89
variation of V , 204
vector space, 73
Volterra difference equation, 294
Volterra difference equations of convolution type, 291
Volterra integrodifferential equation, 291
Volterra system of convolution type, 299
Volterra systems, 299
Weak Poincar ́e type (WP), 390 weakly monotonic, 263 Weierstrass M -test, 354
whale populations, 258
Wimp, 346, 423 Wong and Li, 377 Wronskian, 67
Z-transform, 273, 274, 432 Z-transform pairs, 311 Z-transform of the periodic
sequence, 280 Z-transform versus the Laplace
transform, 308 zero solution, 179, 187
zero-order hold, 431
zeros of the polynomial, 491
Index 539