402 Notes Bernard

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Dynamical Systems and ChaosAn Introduction

Bernard DeconinckDepartment of Applied Mathematics

University of WashingtonCampus Box 352420

Seattle, WA, 98195, USA

April 10, 2009


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i

Prolegomenon

These are the lecture notes for Amath 402: Dynamical Systems. This is the rst year these

notes are typed up, thus it is guaranteed that these notes are full of mistakes of all kinds,both innocent and unforgivable. Please point out these mistakes to me so they may be

corrected for the benet of your successors. If you think that a different phrasing of something would result in better understanding, please let me know.

These lecture notes are not meant to supplant the textbook used with this course. Themain textbook is Steven Strogatz Nonlinear Dynamics Systems and Chaos (Perseus

Book Group, 2001).

These notes are not copywrited by the author and any distribution of them is highlyencouraged, especially without express written consent of the author.


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Contents

1 Resources 11.1 Internet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Software . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Books . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2 Introduction 32.1 Continuous time t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Discrete time n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.3 So, whats the deal with dynamical systems? . . . . . . . . . . . . . . . . . . 4

3 Flows on the line 73.1 General considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.2 The phase portrait . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83.3 Stability of equilibrium solutions . . . . . . . . . . . . . . . . . . . . . . . . 113.4 The impossibility of oscillatory solutions . . . . . . . . . . . . . . . . . . . . 123.5 Another example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.6 Example: population growth . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.7 Linear stability analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.8 Solving differential equations using a computer . . . . . . . . . . . . . . . . . 20

3.8.1 The Euler method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.8.2 The improved Euler method . . . . . . . . . . . . . . . . . . . . . . . 223.8.3 Runge-Kutta methods . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4 Bifurcations in one-dimensional systems 274.1 The saddle-node bifurcation . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.1.1 A more complicated example . . . . . . . . . . . . . . . . . . . . . . 294.1.2 The normal form of a saddle-node bifurcation . . . . . . . . . . . . . 314.2 The transcritical bifurcation . . . . . . . . . . . . . . . . . . . . . . . . . . . 324.3 The pitchfork bifurcation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354.5 Example: insect outbreak . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

4.5.1 Hysteresis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414.5.2 Varying k . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

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iv CONTENTS

5 Flows on the circle 455.1 The uniform oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 465.2 The nonuniform oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

6 Two-dimensional linear systems 516.1 Set-up . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 516.2 A rst example... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 526.3 Some denitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 576.4 Classication of linear systems . . . . . . . . . . . . . . . . . . . . . . . . . . 586.5 The eigenvalue plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 656.6 Application: Love affairs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

6.6.1 A responsive and a ckle lover . . . . . . . . . . . . . . . . . . . . . . 676.6.2 Two cautious lovers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

7 Nonlinear systems in the phase plane 737.1 Generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 737.2 Fixed points and linearization . . . . . . . . . . . . . . . . . . . . . . . . . . 737.3 Application: competing species . . . . . . . . . . . . . . . . . . . . . . . . . 787.4 Conservative systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 817.5 Application: the pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . 847.6 Application: the damped pendulum . . . . . . . . . . . . . . . . . . . . . . . 87

8 Limit cycles 918.1 The van der Pol oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . 928.2 Bendixsons theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 948.3 Weakly nonlinear oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . 968.4 Regular perturbation theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 978.5 Two-timing: the method of multiple scales . . . . . . . . . . . . . . . . . . . 100

8.5.1 The linear example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1018.5.2 The nonlinear example: back to the van der Pol oscillator . . . . . . . 103

8.6 A digression: Fourier series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1068.7 The method of averaging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1088.8 The Poincare-Bendixson theorem . . . . . . . . . . . . . . . . . . . . . . . . 110

9 Bifurcations in two-dimensional systems 113

9.1 Saddle-node bifurcations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1139.2 Pitchfork bifurcations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1159.3 Hopf bifurcations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

10 Difference equations 12710.1 Linear constant-coefficient difference equations . . . . . . . . . . . . . . . . . 12810.2 Nonlinear difference equations and linearization . . . . . . . . . . . . . . . . 131

10.2.1 Fixed points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131


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CONTENTS v

10.2.2 Stability of xed points . . . . . . . . . . . . . . . . . . . . . . . . . . 13210.3 Poincare sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

11 The logistic map 13911.1 Fixed points and their stability . . . . . . . . . . . . . . . . . . . . . . . . . 13911.2 Visualizing the logistic map . . . . . . . . . . . . . . . . . . . . . . . . . . . 14111.3 Periodic solutions of difference equations . . . . . . . . . . . . . . . . . . . . 14311.4 Self-similarity of the bifurcation diagram . . . . . . . . . . . . . . . . . . . . 14611.5 The existence of a period 3 windows . . . . . . . . . . . . . . . . . . . . . . 14711.6 Lyapunov exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15211.7 Universality in unimodal maps . . . . . . . . . . . . . . . . . . . . . . . . . . 154

12 Fractals 15712.1 Countable vs. uncountable sets . . . . . . . . . . . . . . . . . . . . . . . . . 15712.2 Cantor sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15912.3 Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

12.3.1 The Similarity dimension . . . . . . . . . . . . . . . . . . . . . . . . . 16112.3.2 The Box-counting dimension . . . . . . . . . . . . . . . . . . . . . . . 163

13 The Lorenz equations 16513.1 Volume contraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16613.2 The Lorenz map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16813.3 The box-counting dimension of the Lorenz attractor . . . . . . . . . . . . . . 171

14 Strange attractors 173

14.1 Stretching and folding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17314.2 Strange attractors: denition . . . . . . . . . . . . . . . . . . . . . . . . . . 17314.3 The Bakers map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17414.4 The Henon map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177


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Chapter 1

Useful resource materials

1.1 Internet

pplane : A useful java tool for drawing phase portraits of two-dimensional systems.See http://math.rice.edu/~dfield/dfpp.html . Caltechs Chaos course : an entire on-line course with many cool applets anddemonstrations. The link here is to a box-counting dimension applet. See

http://www.cmp.caltech.edu/~mcc/Chaos_Course/Lesson5/Demos.html .

Strogatz video : this collection of videos has the author of our text-book illustrating different aspects and applications of dynamical systemsand chaos. All are instructive, and most are very amusing. Seehttp://dspace.library.cornell.edu/handle/1813/97 .

1.2 Software

Dynamics Solver. A software package for the simulation of dynamical sys-tems. Tons of features. Not always intuitive, but pretty easy to start with. Seehttp://tp.lc.ehu.es/jma/ds/ds.html . Freeware.

Phaser. A software package for the simulation of dynamical systems. Looks great,but non-intuitive steep learning curve. See http://www.phaser.com/ . A single licenseis $60.

1.3 Books

Dynamics, the Geometry of Behavior by Ralph H. Abraham and ChristopherD. Shaw. See http://www.aerialpress.com/aerial/DYN/ . Theres some historyhere: this book is commonly referred to as the picture book of dynamical systems.

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2 CHAPTER 1. RESOURCES

Ralph Abraham is one of the masters of the subject. His exposition together with thedrawings of Chris Shaw makes a lot of sometimes difficult concepts very approachableand intuitive. A true classic. Unfortunately, the printed edition is unavailable. Anebook edition is available from http://www.aerialpress.com/aerial/DYN/ . I havethe ebook if anyone wants to take a look at it.

Differential Equations, Dynamical Systems & An Introduction to Chaos byMaurice W. Hirsch, Stephen Smale, and Robert L. Devaney (Elsevier Academic Press,2004). A classic introductory text, also covering many standard aspects of differentialequations. The authors are true pioneers of our subject matter.


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Chapter 2

Introduction

This course presents an introduction to dynamical systems . So, what are dynamical

systems? A dynamical system is a recipe that tells us how to evolve certain variables forwardin time t. Well encounter two distinct ways of doing this.

2.1 Continuous time tIf we are interested in systems where the variables depend on a time variable t

R , dynamicalsystems are systems of differential equations

y = f (y, t ),

where

y = y(t).

In other words, we want to determine the t-dependence of the vector function y =(y1, y2, . . . , yN )T . Here N is referred to as the dimension of the system. Note that theright-hand side of the dynamical system is specied by the vector function f (y, t ) =(f 1(y1, . . . , yN , t ), . . . , f N (y1, . . . , yN , t ))T . Thus, in essence, we are looking at solving a (vec-tor) differential equation. So, why not call this course Differential Equations, pt. 2?Because well adopt a different point of view, thats why!

2.2 Discrete time nIf instead we are interested in systems where the variables depend on a discrete time variablen

Z , dynamical systems take the form

yn +1 = F (yn , n).

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4 CHAPTER 2. INTRODUCTION

It should be noted that the index n here is supposed to be thought of as a (discrete)independent variable, like t was for continuous-time systems. It is merely a matter of con-vention to write the dependence on discrete variables like n as an index, instead of usingparentheses. As before, yn may be a vector function, yn = ( y1,n , . . . , yN,n ). In that case theright-hand side also becomes a vector function, of course. The types of equations we en-counter are known as difference equations , and you are probably less familiar with them thanyou are with differential equations. Difference equations occur often when we are samplingcontinuous-time processes at discrete time intervals (for instance, once per week), or whenwe investigate processes that are inherently discrete, such as the standings of sports leaguesafter every day of game play. So, we could just study difference equations, right? Not right:again well have a different point of view.

2.3 So, whats the deal with dynamical systems?For the sake of argument, lets assume were looking at continuous-time systems. As a matterof fact, well spend more time with this case than with the discrete-time case anyways. Whatis the main difference between the theory of differential equations and the theory of dynamicalsystems?

Are you sitting down? Heres some bad news:

Fact 1. Most differential equations cannot be solved. Lets be more precise: mostdifferential equations cannot be solves in terms of functions that we know. In fact, almostno differential equation can be solved in such terms. For instance, the simple rst-orderequation

y = y2 + t2

for a function y = y(t) cannot be solved. Bummer!Having taken a course on differential equations you now feel cheated. Hey, dont shoot

the messenger! If we had told you this when you took your differential equations course,you wouldnt have been very motivated, right 1? Why then, do we put you through an entirecourse on differential equations, presenting lots of solution techniques for the few equationswe can solve? Well, we have to start somewhere: solving those few equations provides lotsof intuition on which we can build.

How do we deal with this observation? Theres two ways:

We could solve differential equations numerically, using a computer, or Instead of trying to nd a formula for the solution, we could restrict ourselves to justtrying to nd important characteristics of the solutions.

1 Im assuming you were a bundle of energy, constantly paying attention and providing feedback in yourdifferential equations course. . .


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2.3. SO, WHATS THE DEAL WITH DYNAMICAL SYSTEMS? 5

Both approaches are viable, and they actually complement each other: the rst approachproduces numbers, which are required for many applications. The second approach providesinsight, and it provides some checks for the answers produced by the rst approach. So thisis the good news2!

Fact 2. We dont care about all the details of the solutions (usually). Whatwe do care about are properties of the solutions: periodicity, asymptotic values and limits,etc . Returning to the above example, we can easily see that all solutions approach innityas t , since y 0 and y = 0 only for y = 0 and t = 0. Thus solutions increase withoutbound faster and faster.

In the theory of dynamical systems, we are interested in the qualitativebehavior of solutions, not necessarily their quantitative properties.

2 You can stand up now.


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6 CHAPTER 2. INTRODUCTION


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Chapter 3

Flows on the line

3.1 General considerationsIn this chapter we study dynamical systems consisting of one differential equation with acontinuous time variable that are autonomous:

dydt

= y = f (y),

where y is a function of the independent variable t. The equation is called autonomousbecause the right-hand side does not depend on t explicitly: the only t dependence is throughy = y(t). This dynamical system is called one dimensional because for any value of t,y

R 1 = R , or, in other words, y is a scalar.

This differential equation is separable, so we can actually solve it:

dydt

= f (y)

1

f (y)dy = dt

1f (y) dy = dt + c

1

f (y)dy = t + c.

Depending on f (y), we may or may not be able to do the integral on the left. In many caseswe cant. Even if we can do it, we have obtained only an implicit solution. In order to obtainan explicit solution, we have to solve the remaining equation for y, which again may or maynot be possible. This is not the approach we will follow.

Heres the specic question we want to answer: for a given initial condition, deter-

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8 CHAPTER 3. FLOWS ON THE LINE

mine the behavior of the solution as t .Lets look at an example.

Example. Consider the equation

y = sin y,

with initial condition

y(0) = y0.

Using separation of variables, as above, and using lots of calculus, we nd (check this!)

t = ln |cosecy + cot y|+ c.At t = 0 we want y = y0, thus

0 = ln |cosecy0 + cot y0|+ c c = ln |cosecy0 + cot y0|,

so that

t = lncosecy0 + cot y0cosecy + cot y

.

This equation can actually be solved for y, but it is not easy (try it!). Using the above, we

return to our question: as t , what can we say about y, for different y0? You noticethis is a hard question to answer. Is there another way of getting at this information in amore convenient way? Well, you know I wouldnt be asking if the answer was No!

3.2 The phase portraitThe phase portrait of a dynamical system contains a lot of information about the solutions of the system, without all the detail one nds in a solution formula. Usually the phase portraitcan be obtained much more easily, and it often contains lots of the interesting informationwe want. Since phase portraits are new to us now, lets take it step by step. Well use our

rst example above to illustrate the steps as we encounter them.1. Draw the phase space: the phase space is the collection of all possible y values. For

our example, since we have only one differential equation, this is R . Thus, the phasespace looks like

y


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3.2. THE PHASE PORTRAIT 9

For higher-dimensional systems, we have to draw as many coordinate axes as thereare components of y. Draw is in quotes, because once we get beyond 2 or 3 compo-nents, were doomed by our measly human capabilities of not being able to see morethan three dimensions 1.

2. Find the equilibrium points: Next, we nd the equilibrium points or xed points .These are the values of y for which nothing happens: if we are at one of these values,nothing changes. In other words, for the equilibrium points y = 0. This implies that

f (y) = 0 ,

which is an algebraic equation for y. Granted, this may be a difficult algebraic problemto solve for y, depending on how complicated the function f (y) is, but its better thanhaving to solve a differential equation. Having found all solutions of this algebraic

problem, weve determined all the constant solutions of the differential equation. Theycorrespond to points in the phase space. So, having found these points, we draw themin the phase space.

Going back to our example, we have to solve

sin y = 0,

which results in

y = 0,

,

2 , . . . .

Drawing these points in the phase space gives the picture below.

y0 22At this point, let us remark why we restricted ourselves to autonomous equations.Suppose that our equation was not autonomous:

y = f (y, t ).

Looking for equilibrium solutions, as before, we need y = 0, which implies we have tosolve f (y, t ) = 0, for constant values of y. But this is not possible: solving f (y, t ) =0 results in values of y that will depend on t, thus y (t) = 0 and we cannot ndany equilibrium solutions. In general, non-autonomous systems are signicantly morecomplicated that autonomous ones. Well do very little with non-autonomous systemsin this course.

1 Four, if youre moving, but thats a different course.


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3. Determine the motion in between any two equilibrium points: At this point,all that is left to do is determine the motion around the xed points. For one-dimensional systems, theres not many possibilities. This question becomes more in-teresting once we consider higher-dimensional systems. But, hold your horses, werenot there yet.Since we have determined all the xed points, i.e., all values for which f (y) = 0, thefunction f (y) is either strictly positive or negative in between any two xed points.Wherever this function is positive, we have y > 0, meaning y is increasing there. Weindicate this by a right arrow between the two xed points. Elsewhere the functionf (y) is negative, implying y < 0 and y is decreasing. Well put a left arrow. Thisresults in the picture below, which is the phase portrait for the system were currentlystudying.

y0 22+ ++

What do we do with the phase portrait? Heres a start: plunk down an initial conditionanywhere you want. Where will it go? How will it evolve? Theres three possibilities.

If we manage, somehow, to select one of the xed points as an initial condition, nothingchanges. The time evolution does not lead to different values. If we choose an initial condition in a region with an arrow to the right (increasing yvalues), the solution starting from the chosen initial value will evolve to higher values.

The solution cannot go past any equilibrium points: at these points there is no change.As the equilibrium points are approached, the value of y , which is also f (y), decreases,meaning that the increase in y is less, and thus the equilibrium point is only approachedas t . Note that the approach to the equilibrium point is monotone, as the arrowalways points to the right.

If we choose an initial condition in a region with an arrow to the left (decreasing yvalues), the solution starting from the chosen initial value will evolve to lower values.As before, the solution cannot go past any equilibrium points. As the equilibriumpoints are approached, the value of y , which is also the value of f (y), approacheszero, meaning that the decrease in y is slower, and thus the equilibrium point is only

approached as t . Note that the approach to the equilibrium point is monotone,as the arrow always points to the left.We deduce that the knowledge of the phase portrait allows us to infer the graph of

the solutions as functions of t, for different initial conditions. This graph is indicated inFigure 3.1. Although we can always infer this graph from the phase portrait, we will hardlyever bother. The information we want is in the phase portrait and we will focus on thatinstead.


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3.3. STABILITY OF EQUILIBRIUM SOLUTIONS 11

Figure 3.1: The solutions of the system y = sin y, for varying initial conditions. Here thehorizontal axis represents t, while the vertical axis represents y. In essence, the vertical axisis the phase portrait turned on its side. The gure was produced using dfield .

3.3 Stability of equilibrium solutions

From the phase portrait we easily infer the behavior of the dynamical system near the xedpoints. For instance, for our example, we immediately see that values near y = 0 end upfarther away from it, as time goes on. We call such xed points unstable . Similarly, we seethat values that start near y = evolve to values even closer to y = . such a xed point iscalled stable . Formally, we have the following denitions.

Denition. A xed point is called stable if solutions that start near it stay near it.Denition. A xed point is called asymptotically stable if solutions that start near it,

approach the xed point as t .Denition. A xed point is called unstable if solutions that start near it, end up

wandering away from it.

Thus, in our example, the xed point y = 0 is unstable, whereas the xed points at


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y = are asymptotically stable.

3.4 The impossibility of oscillatory solutionsThe equation were working with is

y = f (y).

It follows from the discussion in the previous section that the phase space is separated indifferent regions, with the equilibrium points as their boundaries. In these regions either

f (y) > 0,or

f (y) < 0,leading to

y > 0or

y < 0,

respectively. This implies that solutions are monotonely increasing or decreasing, until theyreach one of the stable equilibrium points, if there are any. In particular, it is not possible

for solutions of equations of the formy = f (y),

to oscillate.

3.5 Another exampleBefore we move on to other things, lets do another example to illustrate all the conceptsweve introduced, to ensure it all makes sense. Consider the one-dimensional system

y = 4 y2,

with y R . Lets draw the phase portrait and determine which, if any, xed points arestable, unstable, etc .1. The phase space: This is the real line, as drawn below.

y


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3.6. EXAMPLE: POPULATION GROWTH 13

2. Fixed points or equilibrium solutions: to nd these we solve

f (y) = 0 4y2 = 0 y2 = 4 y = 2.These are indicated in the phase space below.

y22

3. The phase portrait: lastly, we determine whether f (y) = 4y2 is positive or negativein between any xed points. This gives the phase portrait below.

y22

We see that the xed point y = 2 is unstable, whereas the xed point at y = 2 isasymptotically stable. If we wanted, we could sketch the actual solutions as a function of t.This is done in Figure 3.2.

3.6 Example: population growthAs a more extended example, we consider a simple model for the evolution of a population.For argument sake, lets assume were talking about a population N (t) of rabbits in a large,but nite meadow. The meadow is assumed very large, but its niteness implies there is alimit to the resources available to the rabbit colony. For starters, we may assume that thechange in the population will depend on the size of the population, but it will not dependon when that population is obtained. The consequence of this simple assumption is that weconsider an autonomous model for the population dynamics:

dN dt

= f (N ),

for some function f (N ), to be determined. What kind of function should f (N ) be? Wedont have much information. Lets expand f (N ) in its power series. We get the dynamicalsystem

dN dt

= a + bN + cN 2 + dN 3 + . . . .

Here a, b, c, d, . . . are the expansion coefficients of the power series. Lets nd out what wecan say about them.

1. The term a. For very small populations our model implies that the most importantcontribution comes from the rst term in the power series, thus we consider


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Figure 3.2: The solutions of the system y = 4 y2, for varying initial conditions. Here thehorizontal axis represents t, while the vertical axis represents y. As before, the vertical axisis the phase portrait turned on its side. The gure was produced using dfield .

N = a N = N 0 + at.what can we say about this? The above equation says, among other things, that if westart with no initial population ( N 0 = 0), then the population grows linearly in time.This is surely peculiar. To avoid this unrealistic phenomenon, we are forced to equatea to zero:

a 0,reducing our model to

dN dt

= bN + cN 2 + . . . .


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3.6. EXAMPLE: POPULATION GROWTH 15

2. The term bN . We proceed as before. For small N , the dominant contribution is dueto the rst term bN , since the term with a is gone. Thus we study

N = bN.The solution of this simple equation is

N = N 0ebt .

We nd that the population changes exponentially as long as N is small (meaning,small enough, so that the next term ( cN 2) is not important yet). Now, do we expectthe population to increase or decrease? We are dealing with a small population withan abundance of resources, so we expect it to increase. We conclude that

b > 0,

establishing exponential population growth for initially small populations. The con-stant b is referred to as the growth rate.

3. The term cN 2. Next, we incorporate the quadratic term. Ignoring the cubic andhigher-order terms, we have

N = bN + cN 2,

which well rewrite as

N = b 1 + cN

bN.

This allows us to reinterpret b(1+ cN/b ) as an instantaneous growth rate. The quotesare there to reect that its not really a growth rate as its not constant, due to theexplicit dependence on N . Thus this growth rate changes in time, because N does.

For small N our growth rate equals b, as expected. However, as N gets bigger, thenite amount of resources should have an impact on the dynamics, and we see that itdoes: the dependence of the growth rate on N becomes more and more important.We can also argue at this point that c should be negative,

c < 0,

since otherwise the growth rate would increase with N , resulting in ever faster ratesof growth, despite the nite amount of resources. With negative c however, we see thatthe growth rate will decrease, and it can even be zero or negative for sufficiently large


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N . This is to be expected: if we start with a rabbit population that is too large tobe sustained by the resources of the meadow, the population should decrease. Forconvenience we rewrite

c = bK

,

where K is a positive constant. Our equation becomes

N = b 1 N K

N.

4. Higher-order terms. We could continue along these lines, including more and moreterms. However, at this stage we have already incorporated the two effects we wishedto model: the growth of small populations and the negative effect of the nite amountof resources. So, well stick with the quadratic model, which is known as the logisticequation .

Next, we draw the phase portrait for the logistic model of the population dynamics.First, the phase space is given by the positive or zero real numbers. We can exclude thenegative real axis, since were looking at population numbers, which are inherently positive.The xed points are found by solving

b 1 N K

N = 0,

which gives

N = 0 or N = K.

The rst xed point is expected: nothing happens if we start without a population. Thesecond xed point is more interesting: it is the value of the population that balances growthand resources. We expect it to be stable, as the arrows of the phase portrait in Figure 3.3below conrm. The value K is referred to as the carrying capacity of the population. Thesame gure also shows the behavior of different solutions as a function of time t.

3.7 Linear stability analysisIt is possible to determine the stability of a xed point from the phase portrait as wevedone before. However, theres an even easier way of doing this. Better still, this approachgeneralizes easily to higher-dimensional systems as well see.

Consider the system

y = f (y).


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3.7. LINEAR STABILITY ANALYSIS 17

N0

Figure 3.3: The phase portrait (top) for the logistic equation, with b = 1 and K = 2. Below,several solutions as functions of t are displayed, using dfield .

Lets examine a part of the phase space near a xed point y0. At the xed point we havef (y0) = 0. Thus there are two possibilities, shown in Figure 3.4. Note that we are assumingthat y0 is a simple root of the equation f (y0) = 0. Roots of higher multiplicity result inmore complicated scenarios that well comment on briey in what follows. That case alsoplays an important role in the study of bifurcations, which well get to in the next chapter.

We see from the gure that the xed point is unstable if f (y) is increasing at the xedpoint, whereas it is stable if f (y) is decreasing there. We can characterize this as follows:

f (y 0 ) > 0 implies y 0 is unstable , and

f (y 0 ) < 0 implies y 0 is stable .

Another way of coming to this same conclusion, and relying less on pictures, is thefollowing. Lets linearize our equation near y = y0. Equate


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y0 y0

f(y)f(y)

Figure 3.4: The two possible phase portraits near a xed point that is a simple root of f (y) = 0. On the left we have an unstable xed point, on the right a stable one.

y = y0 + x,

where we think of x as a small quantity, compared to y0. Then

y = f (y)

y0 + x = f (y0 + x)

x = f (y0) + xf (y0) + x2

2 f (y0) + . . . ,

where we have used the Taylor expansion of f (y) near y0, and the fact that y0 = 0. Further,since f (y0) = 0, we have

x = f (y0)x + x2

2 f (0) + . . . .

If the rst term is not zero, it dominates the other terms on the right-hand side, and weget (approximately)

x = f (y0)x,

which gives

x = x0ef (y0 )t .

Here x0 is the initial deviation from y0. To investigate stability, we need to see what thedynamics of this deviation is as t . We see that


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3.7. LINEAR STABILITY ANALYSIS 19

limt

x = 0, if f (y0) < 0 (stability) ,

and

limt

x = , if f (y0) > 0 (instability) ,where the is determined by x0. In any case, we see that we nd the same conclusion thisway as using the graphical method. Note that these calculations show that

= 1

|f (y0)|is a characteristic time for the behavior near the xed point: it is the time it takes thedynamics to change by a factor of e = 2.718 . . ..

A few remarks are in order. The linearization method requires that f (y) can be expandedin a Taylor series near y = y0. Specically, if f (y) has a vertical tangent at the xed point,the linearization method cannot be used, since there is no Taylor series. Also, if f (y) has ahorizontal tangent at y = y0, the method cannot be used. In that case y0 is a higher-orderroot of the equation and the linear term vanishes since f (y0) = 0. Thus we have to takethe quadratic and higher-order terms into account. In a case like that, it is possible that thegraphical method still gives the right result, even if the linearization method is inconclusive.As an example, you should examine the system

y = y3 = f (y).

Youll nd easily that y = 0 is a xed point, but with f (y) = 3 y2 is zero there. Thus thelinear stability analysis is inconclusive. The graphical method shows that the xed point isunstable, since y3 > 0 for y > 0 and y3 < 0 for y < 0. Thus y = 0 is an unstable xed pointof the system y = y3.

Example. Consider the logistic equation. We have

N = bN 1 N K

= f (N ).

As weve seen, there are two xed points:

N 1 = 0, N 2 = K.

We calculate the derivative of f (N ):

f (N ) = bN 1 N K

= bN bK

N 2 f (N ) = b 2bK

N.

evaluating this at the xed points gives

f (N 1) = f (0) = b > 0 N 1 = 0 is unstable ,


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and

f (N 2) = f (K ) = b < 0 N 1 = K is stable,in accordance with our earlier results. 3.8 Solving differential equations using a computerWe wont say much about using numerical solutions in this course, as we are more interestedin the qualitative behavior of solutions. Nevertheless, using numerical methods is unavoid-able and very helpful in real-life applications. Further, this section will give us a rst glimpseof dynamical systems with a discrete time variable.

The ideas well outline below also work for non-autonomous equations, but to ease thepresentation we will restrict ourselves to autonomous systems. Thus we study an equationof the form

y = f (y).

Observe that for every pair ( t, y) this allows us to draw an arrow in the ( t, y)-pair indicatingthe slope to the solution of the differential equation at that point. The collection of thesearrows is known as the direction eld. One is drawn in Figure 3.5.

So, how do we numerically construct the solution to the equation passing through a givenpoint (the initial condition)? Well, one idea would be to just follow the direction of the arrowfor a bit. This is exactly what the Euler method does.

3.8.1 The Euler methodUsing the idea of following the arrows, Eulers scheme simply says that the derivative of thesolution y(t0) is given by

y (t0) = f (y(t0)) y(t0 + t) y(t0)

t ,

where on the right-hand side weve used the differential quotient of y(t0) and the neighboringpoint y(t0 + t). We get

y(t0 + t) y(t0) + f (y(t0)) t.This is known as the explicit Euler method with time step t. If we let t1 = t0 + t, theny1 = y0 + f (y0) t,

where y0 = y(t0), y1 = y(t1), etc . In general,

yn +1 = yn + f (yn ) t,


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3.8. SOLVING DIFFERENTIAL EQUATIONS USING A COMPUTER 21

Figure 3.5: A direction eld for an ordinary differential equation. The plot was producedusing dfield .

which is a rst-order (since theres only one shift appearing) difference equation . Howevernaive, Eulers method is not bad. It is clear that to obtain a good approximation, one needsto choose a sufficiently small t.

One measure of the accuracy of a numerical scheme is its local truncation error :

E n +1 = |yn +1 y(tn + t)| .This is the error made by moving from the point we obtained for the nth position to the(n + 1)th point where the Euler method takes us. We can say more by using a Taylorexpansion:

E n +1 = |yn +1 y(tn + t)|= yn + f (yn ) t y(tn ) + ty (tn ) +

( t)2

2 y (tn ) + . . .


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= yn + f (yn ) t yn + tf (y(tn )) + ( t)2

2 y (tn ) + . . .

=

( t)2

2 y (tn ) + . . .

= O( t)2,where O(( t)2) indicates that the local truncation error is of second order in t. Thus, atevery time step were making an error which is quadratic in the small time-step t.

3.8.2 The improved Euler methodThe ideas that went in to the Euler method are pretty rudimentary. Can we do somethingfancier, something that might result in an approximation scheme that has a better localtruncation error, for instance? It turns out there are many ways of doing this. The simplestway to improve Eulers method is to use the improved Euler method 2!

Heres how it works. For Eulers method we use derivative information from the currentpoint to get to the next point. But the derivative changes all along the solution curve thatwere trying to approximate. So, perhaps, we could do better by incorporating derivativeinformation from the next point. Why dont we do this? Thats an easy one to answer: wedont do this, because we dont know the next point on the solution curve. On the otherhand, we do know a decent approximation to this next point, namely what we get fromEulers method. Lets use that. Heres what well do.

1. We use Eulers method to get to a new point yn +1 :

yn +1 = yn + f (yn ) t.

This is illustrated by the black arrow in Figure 3.6.

2. The derivative information from this predictor point is f (yn +1 ), indicated by the redarrow in Figure 3.6.

3. Using this new direction, we calculate a new, updated guess for our new point, byaveraging the two directions:

yn +1 = yn + t

2 (f (yn ) + f (yn +1 )).

This gives a new scheme to numerically solve differential equations. Lets examine thelocal truncation error to conrm that this scheme is actually better than Eulers method.We have

2 No, seriously!


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tn tn+1 t

yn

yn+1

y

t

^n+1y

Figure 3.6: An illustration using too many colors of the improved Euler method. The blackarrow shows the prediction from Eulers method. The red arrow shows the derivative at

the prediction point. The blue arrow gives the average direction of the black and the redarrow, resulting in a nal point for the improved Euler method, which might be a betterapproximation.

E n +1 = |yn +1 y(tn + t)|= yn +

t2

(f (yn ) + f (yn +1 )) y(tn ) + ty (tn ) + ( t)2

2 y (tn ) +

( t)3

6 y (tn ) + . . .

= t2

f (yn ) + t2 f (yn + tf (yn ) ty (tn ) ( t)

2

2 y (tn ) ( t)

3

6 y (tn ) + . . .)

= t2

f (yn ) + t

2f (yn ) + tf (yn )f (yn ) +

( t)2f (yn )2

2 f (yn ) + . . .

ty (tn ) ( t)2

2 y (tn )

( t)3

6 y (tn ) + . . .

= t + ( t)2 + ( t)3 + . . . ,


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where

= 12

f (yn ) + 12

f (yn ) y (tn ) = 0 ,and we have used that

y (tn ) = f (y(tn )) = f (yn ).

Next, we have

= 12

f (yn )f (yn ) 12

y (tn ) = 0 ,

where we have used that

y (tn ) = f (y(tn )) = f (yn )

y (tn ) = f (y(tn ))y (tn ) = f (yn )f (yn ),

by the chain rule. Similarly, you can show that generically (meaning, almost always, unlessf (y) is chosen in a very special way) = 0. This implies that the improved Euler method doesindeed appear to be a more accurate method than Eulers method, as its local truncationerror is given by

E n +1 = O(( t)3),which is one order of magnitude better than Eulers method.

3.8.3 Runge-Kutta methodsMethods like the above can be more rened. At some point, we have to pay a price inefficiency (read: speed of the computation) compared to accuracy of the method. It isgenerally agreed that the 4th order Runge-Kutta method is one of the best methods when itcomes to balancing these two effects. The method is given by

yn +1 = yn + t

6 (k1 + 2 k2 + 2 k3 + k4) ,

where

k1 = f (yn ),k2 = f (yn + k1/ 2),k3 = f (yn + k2/ 2),k4 = f (yn + k3).


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One can show (do not necessarily try this at home!) that the local truncation methodfor this method is

E n +1 =

O(( t)5).


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Chapter 4

Bifurcations in one-dimensionalsystems

Up to this point, we have considered systems of the form

x = f (x),

where x is one-dimensional. We have seen that the behavior of solutions of the these equa-tions is rather restricted: solutions change in a monotone way. Either they approach a xedpoint, or else they head off to one of .

So, what can be interesting about systems like this? The answer is their dependenceon parameters .

In this chapter, we will study bifurcations. Bifurcation theory is the study of how solu-tions of dynamical systems change as a parameter is changed. As an example, consider thegrowth of mosquito populations. As we all know, this heavily depend not only on the size of the population and the available resources 1, but also on the temperature. If the temperatureT is so that T < T c (a certain threshold) then the mosquitos are immobilized. On the otherhand, if T > T c, they become very active. Thus, if we expect to have a good model for themosquito evolution, the temperature should show up as a parameter. Further, for the specialvalue T = T c, we expect something special to happen in the system. If the phase portraitsfor T < T c and T > T c are qualitatively different, we say that a bifurcation occursat T = T c.

4.1 The saddle-node bifurcationConsider the system

x = + x2,1 That would be us.. .

27


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28 CHAPTER 4. BIFURCATIONS IN ONE-DIMENSIONAL SYSTEMS

()1/2

()1/2

x x0

x

(a) (b) (c)

Figure 4.1: The three different phase portraits for the system x = + x2, for different valuesof . On the left (Figure (a)), < 0. In the middle (Figure (b)), = 0, while > 0 on the

right in Figure (c).

where is a real-valued parameter. The xed points of this system are given by

+ x2 = 0 x2 = .We see immediately that there are three different possibilities:

1. < 0: In this case is positive, meaning we can take a square root. Thus there aretwo distinct xed points (which we indicate with a super-index ):x = .

As shown in Figure 4.1a, the phase portraits do not change qualitatively as changesin this range. The xed point at is always stable, and the xed point at is unstable.

2. = 0 : This is easy. Now theres only one xed point,

x = 0 ,

which is semi-stable.

3. > 0: In this case there are no xed points, and the value of x increases forever. Thisscenario is illustrated in Figure 4.1c.

Thus, in the bifurcation of changing from a negative value to a positive value, two xedpoints disappear and instead of all solutions with initial condition less than (whichceases to be dened for > 0) converging to a xed point, they end up drifting off to innity.

In a saddle-node bifurcation, two xed points are destroyed or appear.

We wont explain where the name of this bifurcation comes from, until we start talkingabout bifurcations in higher-dimensional systems. Somehow well survive until then. The


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4.1. THE SADDLE-NODE BIFURCATION 29

x

x

x>0

=0


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x*

Figure 4.3: The bifurcation diagram for x = + x2, displaying a saddle-node bifurcation.

Lets see what these two equations give for our current example. The rst conditionweve already written down above. The second one gives

f (x) = 0 1 + e x = 0 .

In general, the two equations

f (x) = 0 , f (x) = 0 ,

give two conditions to be satised at the bifurcation point ( c, xc ). Thus, even if we cannotsolve these equations with pen and paper, we still have two equations for two unknowns,which we could solve using a computer. Sweet!

So, what do we get here? The second equation gives

e x = 1 x = 0.Using this in the second equation gives

= 1.

Thus, a saddle-node bifurcation takes place at ( c, xc ) = (1 , 0).


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4.1. THE SADDLE-NODE BIFURCATION 31

x

Figure 4.4: A schematic bifurcation diagram for x = + x2, displaying a saddle-nodebifurcation, using a dashed line for a curve of unstable xed points and a solid line for acurve of stable xed points.

4.1.2 The normal form of a saddle-node bifurcationHaving determined the bifurcation point, we can now put our system in a familiar form, atleast in a neighborhood of the bifurcation point. Again, consider

x = x e x = F (, x ).

We have made the dependence of the right-hand side on the parameter explicit. Near anypoint ( 0, x0) in (, x ) plane, we may expand the function F (, x ) using a two-dimensionalTaylor series as

F (, x ) = F (0, x0) + ( 0) F ( 0 ,x 0 )

+ ( x x0) F x ( 0 ,x 0 )

+

12

( 0)2 2F 2 ( 0 ,x 0 )

+ ( 0)(x x0) 2F x ( 0 ,x 0 )

+

12

(x x0)2 2F x 2 ( 0 ,x 0 )

+ . . . ,


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where the . . . terms denote terms that are at least of cubic order. Lets work out what thisexpansion gives for our F (, x ), near ( c, xc ) = (1 , 0). We have

F ( c, xc ) = 0 ,F ( c ,x c )

= 1 ,

F x ( c ,x c )

= 0 ,

2F 2 ( c ,x c )

= 0 ,

2F x

( c ,x

c )

= 0 ,

2F x 2 ( c ,x c )

= e x

(( c ,x c )) = 1.

Here the rst and third result are by construction. Indeed, these are the conditions weimposed to locate the bifurcation point. Putting all this together, we get

x = 0 + ( c) 1 + ( x x

c ) 0 + 12

( c)2 0 + ( c) 0 + 12

(x x

c )2 (1) + O(3)= (

1)

1

2x2 +

O(3),

where O(3) denotes terms of third order. This description allows us to draw a bifurcationdiagram easily, valid near the bifurcation point ( c, xc ) = (1 , 0). This is shown in Figure 4.5.One should remember that this description, or any description of a bifurcation, is only

valid close to the bifurcation point. This example illustrates that the original system

x = + x2

may be considered as the standard system with a saddle-node bifurcation. We (or atleast, I) refer to it as the normal form for the saddle-node bifurcation.

4.2 The transcritical bifurcationAnother type of bifurcation exists for the following prototypical system:

x = x x2,where

R . The xed points are given by


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4.3. THE PITCHFORK BIFURCATION 33

x

1

Figure 4.5: A schematic bifurcation diagram for x = x e x , near the bifurcation point( c, xc ) = (1 , 0).

x x2 = 0 x = 0 , x = .

Thus, for every value of , there are two xed points. Well, not exactly: if = 0 these twoxed points collide. Thus we have a good guess that something happens at ( c, xc ) = (0 , 0).The bifurcation diagram is displayed below, in Figure 4.6. A schematic (no arrows) is shownin Figure 4.7.

It follows from these gures that for < 0 x = 0 is a stable xed point, whereas x = is unstable. For > 0, the roles are reversed. Now x = is stable, and x = 0 is unstable.

In a transcritical bifurcation, two families of xed points collide and exchangetheir stability properties. The family that was stable before the bifurcation isunstable after it. The other xed point goes from being unstable to being stable .

4.3 The pitchfork bifurcationAs a set-up example, lets start by considering the example of a solid elastic beam (thinksteel), which is supporting a weight. This situation is illustrated in Figure 4.8, for twodifferent weights. If the weight W does not exceed a critical weight W c, the beam supportsthe weight. Thats nice. If the weight exceeds a critical weight W c, then it will buckle, eitherto the left or to the right (because we have a two-dimensional beam; three-dimensional


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x

0

Figure 4.6: The bifurcation diagram for x = x x2, showing the transcritical bifurcation.

beams have more options). The smallest perturbation (for instance in the composition of the weight) may determine which way the beam will buckle. The possibility of the beamremaining upright is still there, but this possibility is unstable. This set-up, which involvesthree xed points in the latter scenario, cannot be modeled by a saddle-node or transcriticalbifurcation. Thats where the pitchfork bifurcation comes in.

The normal form of a pitchfork bifurcation is given by the system

x = x x3.The xed points are given by

x = 0 , x = ,where the latter two are only dened for > 0. Case-by-case, we get

1. < 0: x = 0 is the only xed point. It is stable, as an elementary linear stabilityanalysis (for instance) shows.

2. = 0: x = 0 is the only xed point, but it has triple multiplicity. It remains stable,as a graphical analysis shows. Note that the linear stability analysis is inconclusivehere.

3. > 0: x = 0 and x = are all xed points. The xed point x = 0 has lost itsstability (sob), but the other two xed points are stable.


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4.4. EXERCISES 35

x

0

Figure 4.7: A schematic bifurcation diagram for x = x x2.

The bifurcation diagram is given in Figure 4.9. We see that in a pitchfork bifurcationone family of xed points transfers its stability properties to two families after/before thebifurcation. Note that this is very similar to the example of the steel beam we discussedabove. The bifurcation of this example is called supercritical , because the nontrivial ( i.e. ,nonzero) families of xed points occur for values of the parameter larger than the bifurcationvalue.

Similarly, a pitchfork bifurcation is called subcritical if the nontrivial xed points occur forvalues of the parameter lower than the bifurcation value. Such a case is shown in Figure 4.10.It can be obtained from the system

x = x + x3.

Now the xed points are given by

x = 0 , x = ,where the latter two are only dened for values of < 0, i.e. , before the bifurcation whichtakes place at = 0. In that case it appears that the stability of a xed point (at zero, inthis case) is destroyed through its collision with two families of unstable xed points.

4.4 ExercisesTo practice the concepts weve learned so far, you should try to classify the bifurcations inthe two following problems.

Consider the problem


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WW c

Figure 4.8: A beam. Actually, two beams. With weights on top. So there.

x = ( R x2)(R 2 + x2).

Find all the xed points. How many bifurcations are there, and of what type? Drawthe bifurcation plane. Answer: there are two transcritical bifurcations, and two saddle-node bifurcations .

Consider the problem

x = Rx x2

1 + x2.

How many bifurcations are there, and of what type? Draw the bifurcation plane.Answer: there is one transcritical bifurcations, and two saddle-node bifurcations. There

is also a singular ( i.e., at x

= ) bifurcation at R = 0.

4.5 Example: insect outbreakWell examine a model for the dynamics of the spruce budworm, which is a real pest inEastern Canada, and other parts of Canada. When an outbreak occurs, the budworm candefoliate and kill the r trees in an entire forest in about four years! Similar pests have


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4.5. EXAMPLE: INSECT OUTBREAK 37

x

Figure 4.9: The bifurcation diagram for a supercritical pitchfork bifurcation

had a major impact on some of the national forests of Washington, especially in the Entiat,Chiwawa and Wenatchee areas.

Well examine a model where we assume that the forest variables are constant, especiallywhen compared to the fast changing spruce worm population variable. At the end of ouranalysis we will take the potential change of the forest parameters into account, which leadsto bifurcations.

Lets look at which model we will use. Well start with a logistic model (which takes into

account exponential growth for small populations, and nite resources), augmented with adeath rate term, due to the presence of predators, mainly birds. This leads to a model of the form

N = RN 1 N K p(N ),

where p(N ) is the announced death rate term. What would be a good model for the deathrate? Taking into account that for very small populations N , the death rate should be small,as predators have a hard time locating prey, and that any predator can only eat so muchprey, we expect the term p(N ) to behave as in Figure 4.11.

It does not really matter what the functional form of p(N ) is, as long as it has the abovefeatures: monotone, quadratically small for small N and saturating as N . We can use

p(N ) = BN 2

A2 + B 2,

where A and B are constants. Thus we have

dN dt

= RN 1 N K

BN 2

A2 + N 2.


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x

Figure 4.10: The bifurcation diagram for a subcritical pitchfork bifurcation.

There are 4 parameters in this problem: A, B, R and K . Lets see if we can rescale our variables to reduce this number. Let

N = N 0x, t = t0.

Expressing the equation using the new variables and x introduces two new constants in theequation, t0 and N 0. We get to choose these at will, to simplify our life as much as possible.We get

N 0t0

dxdt

= RN 0x 1 N 0x

K BN 20 x2

A2 + N 20 x2.

We can rewrite this as

dxd

= Rt 0x 1 N 0x

K Bt 0x2

N 0(A2/N 20 + x2).

There are different choices we can make to simplify our life. If we wish to make the last termparameter free, we can pick

N 0 = A, Bt0 = N 0, N 0 = A, t 0 = A/B.We can now rewrite our equation as

dxd

= x 1 xk

x2

1 + x2,

where we have dened

= RA

B , k =

K A

,


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p(N)

N

Figure 4.11: A likely prole for the death rate p(N ).

two parameters dened in terms of the old ones. The consequence of this rescaling businessis that we have reduced the number of parameters in our system from 4 to 2! Thats a gooddays work! We see that represents a rescaled growth rate, whereas k represents a rescaledcarrying capacity.

The concept of rescaling variables to reduce the number of parameters in an equation isan important one, which well use many more times in this course. Especially for equationswith units, this allows us rewrite everything in terms of dimensionless variables, which isimportant as it allows us to make sense out of statements about large and small, etc . Wellcome back to this when we need it.

We now analyze this system. We start by looking for the xed points. They satisfy

x 1 x

k=

x2

1 + x2,

which implies there is one xed point x = 0, which is hardly exciting. The rest is moreenticing:

1 x

k=

x

1 + x2.

This (after getting rid of denominators) is a cubic equation for x. We could use Cardanos

formulae for the solutions of the cubic, but does anyone really want to do that? Didnt thinkso. Lets do a graphical analysis instead.

1. For very small values of , (i.e. , < 1, see below) we see (Figure 4.12, upper left)that there is one xed point, x = a, which is a relatively small value. Since the curveis below the straight line it follows that the xed point is stable, thus the value x = ais stable. It is referred to as the refuge level. A budworm population of this size in aforest does not pose a threat.


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b=c

a=b

k x k x


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1 2

x

x

a

b

c

Figure 4.13: The bifurcation diagram for the budworm problem, for a xed value of k.

4. For = 2, the xed points x = a and x = b collide, and they annihilate in asaddle-node bifurcation. At this value of , x = a = b is semi-stable, whereas theoutbreak level x = c persists and is stable.

5. Finally, for > 2, the outbreak level is the only xed point remaining. The point isstable and all initial conditions lead to outbreak. Sad but true 2.

Putting all of this together, results in the bifurcation diagram shown in Figure 4.13. Asdiscussed there are two saddle-node bifurcations.

4.5.1 Hysteresis

Question: what can happen if the forest parameters changed for some externalreason? Lets examine this. Suppose we start in the regime where < 1, but close to it.Since x = a (refuge) is the only stable point, thats where well end up. Now suppose wereto increase beyond 1. The budworm population will remain on the x = a (refuge) branchof xed points, since it is stable. Now, when reaches the point where 2, somethinghappens, since beyond there, the refuge level does not exist anymore. The only stable xed

2 (c) Metallica, 1991.


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point is given by the outbreak level, thus the population will see a sharp rise, from x = ato x = b. For larger values of , the budworm population remains at outbreak levels, sincethat is the only stable scenario. This is all bad news.

Suppose we, somehow, someway, somewhere, were to reduce the forest parameters. Wouldwe be able to undue these effects? Suppose we start from outbreak, but decreases to levelsbelow 2. Since the outbreak branch is stable, this is where the population level remains.Doh3! Lets see what happens as we increase more. As decreases below 1, we nallyget the sharp decrease in the budworm population we were hoping for. This drop happensbecause the outbreak level ceases to exist, and the refuge level is the only level that is stable.

We see that the system has executed the yellow loop of Figure 4.13, known as a hysteresisloop. Such phenomena are known from electromagnetism, but they occur more widely in alarge variety of dynamical systems. We see that the presence of two saddle-node bifurcations(one subcritical, one supercritical) leads to hysteresis.

4.5.2 Varying k

What happens if we also consider variations in k? It is clear that the values of 1 and 2 willvary as k is changed, but it also appears that lots will remain as discussed above. In moregenerality, lets determine what behavior is possible in our two-dimensional parameter space.To this end, lets determine at what values of the parameters ( k, ) bifurcations occur.

We have already seen that the only bifurcations in this problem are of saddle-node type.And we know how to determine saddle-node bifurcations: we need a xed point, and avertical tangent at the xed point. Recall that our system looks like

x = x 1 xk x2

1 + x2 = F (k,,x ).

At a bifurcation point, we need

F (k,,x ) = 0 , F (k,,x )

x = 0 .

Writing out these equations, we get

x 1 xk x2

1 + x2 = F (k,,x ) = 0 ,

x

x 1 xk

x2

1 + x2 = F (k,,x ) = 0 .

The rest is tedious algebra:

3 (c) H. Simpson, American philosopher.


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k

refuge

outbreak

refuge+outbreak

Figure 4.14: The bifurcation curves for the budworm problem in the parameter ( k, )-plane.


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Chapter 5

Flows on the circle

Instead of considering the dynamical system

x = f (x)

with xR , we could consider it on the circle. By this we mean that we let x[0, L), for somepositive L. Further, whenever x crosses one of the boundaries of its interval of denition, we

assume it comes back in on the other side. Mathematically, this implies were considering xmod L. This amounts to considering x like an angle on the circle. To emphasize this more,we write

= f ()

instead, with

S L , the circle with circumference L. Thus denotes the rate of change of the angle , or the angular velocity. The main difference with ows on the line is that nowthe ow can return to where it was, by going around the circle. Thus, periodic solutions are possible !

Example. In this rst example we show that periodic solutions do not always occur forsystems dened on the circle. Consider

= sin , S 2

First we look for xed points:

sin = 0

= N ,

for any integer N . Note that only N = 0 and N = 1 give rise to distinct points on S 2 .Thus, we have two xed points:

1 = 0,

2 = .

Using , we determine the phase portrait of the system, which is displayed in Figure 5.1.We see that the xed point 2 = is (asymptotically) stable, whereas

1 = 0 is unstable.

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46 CHAPTER 5. FLOWS ON THE CIRCLE

1 =0

*

2 =

*

Figure 5.1: The phase portrait for = sin on S 2 .

We have encounted this same example before, but we treated it as a system on the line.It is far more natural to treat it as a system on the circle! Whether that is the correct thingto do or not has to be determined by the specic application that leads one to the system.

One should also realize that not every system can be considered as a system on the circle:for instance the system

= 2

cannot be regarded as a system on S 2 . Indeed, the ow gives conicting information at = 0 and at = 2, which should correspond to the same point. The system says thatthe rate of change at that point is either = 0, or else = 4 2, which are very different.Such inconsistencies imply that this system cannot be dened on the circle. Somehow wellsurvive.

5.1 The uniform oscillatorThe simplest ow on the circle is given by the system

= ,

where is a real nonzero constant and S 2 . This system is easily solved:

= t + 0,


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5.2. THE NONUNIFORM OSCILLATOR 47

where 0 represents the initial angle. We see that the entire circle is rotating with angularvelocity . After

T = 2 ,

(the period) all points are back where they started.

5.2 The nonuniform oscillatorNext we consider the more complicated system

= a sin ,with S 2 . For the sake of argument we consider both and a to be positive. Lookingfor the xed points, we nd

= a sin .There are three different cases.

1. If a < then the xed point equation has no solutions (see Figure 5.2, left). The wholecircle rotates towards increasing angles. The velocity is higher for angles in the thirdand fourth quadrant, as is clear from Figure 5.2. In this case, the motion is periodic:after some amount of time, the circle will be back to its initial position. The phaseportrait is shown on the right of Figure 5.2. It is actually pretty easy to gure out theperiod T of the motion. We have

= a sin

T

0dt =

2

0

d a sin

T = 2

0

d a sin

.

The bounds of integration have been picked as follows: we can choose any initialcondition we want: the whole circle is rotating, so all solutions will be periodic. Now,after one period = 2, namely the circle has rotated once. In this case we can work outthe integral. Dandy! The important part is to realize that even if this is not possible,we have obtained an expression for the period, in terms of an integral which could beeasily evaluated numerically. Here we get, using the substitution = 2 arctan u (checkthis!)

T = 2

2 a2.


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a

0

0

fast

slow

Figure 5.2: The right-hand side of = a sin , for a < , on the left. The phase portraitfor this case is shown on the right.

0

0

/2

Figure 5.3: The right-hand side of = a sin , for a = , on the left. The phase portraitfor this case is shown on the right.

2. When a = , there is exactly one xed point at / 2, but it is a semi-stable xed point,as observed from Figure 5.3. Note that all initial conditions eventually lead to thexed point at / 2. We say that the xed point is attracting. This does not imply thatthe xed point is stable though.

3. For a > , there are two distinct xed points, one stable and one unstable. These areborn out of a saddle-node bifurcation at the critical value a = .


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5.2. THE NONUNIFORM OSCILLATOR 49

0

0

Figure 5.4: The right-hand side of = a sin , for a > , on the left. The phase portraitfor this case is shown on the right.


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Chapter 6

Two-dimensional linear systems

We have seen that the behavior of linear systems is very constrained. Things got a bit more

interesting when we discussed bifurcations. This kind of lured us to a second dimension,but not really. Lets go there now. Well start by studying linear systems, to build someintuition before we go to nonlinear systems.

6.1 Set-upA two-dimensional linear system is a system of the form

x1 = ax1 + bx2x2 = cx1 + dx2.

Here a, b, c, and d are constants. We conveniently rewrite this using vector-matrix notationas

x = Ax,

with

x = x1x2,

and

A = a bc d .

One immediate consequence of the system being linear is that if x(1) and x(2) are solutions,then so is

x = c1x(1) + c2x(2) ,

where c1 and c2 are arbitrary constants.

51


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52 CHAPTER 6. TWO-DIMENSIONAL LINEAR SYSTEMS

A nice consequence of considering two-dimensional systems is that we can start workingwith mechanical systems, since Newtons law is a differential equation of second order. Forinstance, consider the case of a particle suspended from an elastic spring, governed by Hookeslaw:

my = ky,where m denotes mass, k is Hookes spring constant and y is the displacement away fromequilibrium. Our standard approach is to rewrite this as a rst-order system. Since theequation is of second order, we introduce two variables:

x1 = y, x2 = y .

What are the differential equations for these quantities? We start with x1. Taking a deriva-tive, we get

x1 = y = x2.

That was easy! Now for x2:

x2 = y = km

y = km

x1.

Thus

x1 = x2x2 =

k

mx1

.

Rewriting this in matrix form, we obtain

ddt

x1x2

= 0 1

k/m 0 x1x2

.

6.2 A rst example...Well consider the system

ddt x1x2 = a 00 1 x

1x2 ,

where a is a real parameter. Note that this system isnt really worthy of the name system,as its really two decoupled scalar equations:

x1 = ax1, x2 = x2.

These equations are easily solved:


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6.2. A FIRST EXAMPLE... 53

x1

x10

x20

x2

t

Figure 6.1: A plot of x1(t) and x2(t) as functions of t, for specic initial conditions x10 andx20 .

x1 = x10eat , x2 = x20et ,

where x10 and x20 are initial conditions for x1 and x2 respectively.How do we graphically represent these solutions? We could plot x1(t) and x2(t) as

functions of t for different initial conditions. Such a plot is shown in Figure 6.1. Thedisadvantage of this approach is that wed need a different plot for different initial conditions.More importantly, in order to use this approach, we need to be able to solve the differentialequations explicitly. More often than not, we wont be able to do this, so wed be stuck.Further, often we dont care much for the detail of these plots. As before we are moreinterested in qualitative features.

Instead, we can opt for the same approach we used for one-dimensional systems, wherewe drew the phase portrait. The phase space (the collection of permissible pairs ( x1, x2))will be the plane R R , or a part thereof. The equations

x1 = x10eat

x2 = x20et

dene a parametric plot in the phase plane: for a given initial condition, every value of tdenes a point ( x1, x2) in the phase plane which moves around as t is varied. The phaseportrait is the collection of the parametric plots so obtained. You may object: it appears westill require an explicit form of the solutions to be able to draw the phase portrait. Correct.But well see later how we can draw a relatively correct phase portrait without having theactual solution in hand. This is what were going for.


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For the example below, well distinguish different cases: a > 1, a = 1, 0 < a < 1, a = 0,and a < 0. Note that for all cases the point (0 , 0) is a xed point: (x1, x2) is identically zerothere, thus this point does not move.

1. Case 1: a > 1. Lets start by seeing what happens on the axes: since the equationsare decoupled, we see that points on the axes stay on them. Their dynamics movesthem along the axes, except for the origin. Further, since

x1 = ax1,

we see that the motion on the x1 axis is away from the origin. Similarly, for the x2axis, we have

x2 = x2,also leading to motion away from the origin. In general, for points that may be off theaxes, we have the following two limit results:

limt

x1(t) = , limt x2(t) = ,and

limt

x1(t) = 0 , limt

x2(t) = 0 .

In other words, both x1 and x2 tend to as time goes on, and they both tend to zero,as we go back in time. In order to draw an accurate phase portrait, we need to ndout which of x1 or x2 dominates as t . First, lets examine t . Since a > 1,x1 grows faster than x2, and it follows that the curve through points far away from theorigin will go predominantly in the direction of x1. Second, as t , since a > 1,it follows that x1 decays faster than x2, implying that x2 will dominate. Thus, closeto the origin, the solutions will follow the x2 direction. All the information we havegathered results in the phase portrait shown in Figure 6.2. As stated, the origin is axed point. It is clearly unstable. It is referred to as an unstable node .

2. Case 2: a = 1 . Now we have

x1 = x10et , x2 = x20et .

Thus x1 and x2 grow (or decay) at exactly the same rate as t (as t ). Youmay guess that this will result in straight-line motion in the phase plane, and this iscorrect. We can easily show this: using the above two equations, we have


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6.2. A FIRST EXAMPLE... 55

Figure 6.2: The phase portrait for the system x1 = ax 1, x2 = x2, with a > 1. Here x1 is onthe horizontal axis, x2 on the vertical axis. The plot was produced using pplane .

x2x1

= x20x10

x2 = x20x10

x1.

In the phase plane, this is the equation of a straight line through the origin, with slopex20/x 10 determined by the initial conditions. As before, all motion is away from theorigin. The phase portrait is shown in Figure 6.3. The origin is still an unstablenode . Sometimes it is also called an unstable star .

3. Case 3: 0 < a < 1. Since a is still positive, but smaller than 1, we can repeat theanalysis from the rst case, but with the roles of x1 and x2 switched: now x2 dominatesas t, whereas x1 is dominant close to the origin ( t ). Again, the origin isan unstable node . The phase portrait is shown in Figure 6.4.

4. Case 4: a = 0 . This case is special. The xed point at the origin is not unique

anymore: the entire x1 axis consists of xed points. On this axis, our system is givenby

x1 = 0, x2 = x2.

Thus, if we start with initial conditions on the x1 axis: (x10 , x20) = ( x10 , 0), then thereis no dynamics. Thus the entire x1 axis consists of xed points. Thus, with a = 0, thexed points are not isolated! Further, the system is easily solved:


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Figure 6.3: The phase portrait for the system x1 = ax 1, x2 = x2, with a = 1. Here x1 is onthe horizontal axis, x2 on the vertical axis. The plot was produced using pplane .

x1 = x10 , x2 = x20et

.

These are equations for vertical lines, as shown in Figure 6.5. You can regard thisphase portrait as a limit case of that in Figure 6.4, as a 0.

5. Case 5: a < 0. This case is different from all the others. Now points on the x1axis approach the origin. Other points are drawn closer to the x2 axis, but ultimately,

when they get close to it, they drift off towards innity. The phase portrait is shownin Figure 6.6. The xed point at the origin is called a saddle point . This name isused because the phase portrait obtained is similar to the top view of the trajectoriesa ball would follow when put on a saddle-surface in three-dimensional space. A saddlepoint is unstable, because the probability of ending up at the xed point is zero: inan experimental setting you would never be able to pick initial conditions that get tothe origin: the slightest error would put you ever so slightly off the x1 axis, resultingin eventual drift away from the xed point.


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6.3. SOME DEFINITIONS 57

Figure 6.4: The phase portrait for the system x1 = ax1, x2 = x2, with 0 < a < 1. Here x1 ison the horizontal axis, x2 on the vertical axis. The plot was produced using pplane .

6.3 Some denitions

In this section we present some denitions that we have already been using somewhat loosely.

Denition. (Lyapunov stability) A xed point x is said to be Lyapunov stable if alltrajectories that start close to it remain close to it for all time t > 0. Otherwise the xedpoint is said to be Lyapunov unstable .

Usually well drop the Lyapunov bit and just say stable or unstable. Note that inall cases of our example in the previous section, the xed point at the origin was Lyapunovunstable.

Denition. (Attractive xed point) A xed point is called attractive if all trajectoriesthat start near it approach it as t .

It may be tempting to assume that attractive xed points are stable. This is not true,and we have already seen an example. Here it is again.

Example. Consider the dynamical system on the circle given by

= (1 sin ).with phase portrait given in Figure 5.3. In this case the xed point at = / 2 is attractive,


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Figure 6.5: The phase portrait for the system x1 = ax 1, x2 = x2, with a = 0. Here x1 is onthe horizontal axis, x2 on the vertical axis. The plot was produced using pplane .

but not stable. Denition. (Asymptotic stability) A xed point is said to be asymptotically stable

if it is both attractive and stable.

6.4 Classication of linear systemsLets return to the general linear system

x = Ax,

where A is an N N matrix and x is an N -dimensional vector. Since this equation hasconstant coefficients, we guess a solution of the form

x = et

v,where is a constant, and v is a constant vector. We hope to be able to determine bothof these in order to obtain solutions of this form for the differential equation. To do so, wesubstitute this ansatz in the equation. We need

x = et v.

Substitution gives


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6.4. CLASSIFICATION OF LINEAR SYSTEMS 59

Figure 6.6: The phase portrait for the system x1 = ax 1, x2 = x2, with a < 0. Here x1 is onthe horizontal axis, x2 on the vertical axis. The plot was produced using pplane .

et v = Aet v.

Since et is scalar, we can move it in front of the matrix A, and cancel it from the equation.

Thus we have

Av = v.

This equation tells us that the s we seek are the eigenvalues of the matrix A, whereas thevs are their corresponding eigenvectors. To nd , we write down the eigenvalue equation:

det( A I ) = 0 ,where det denotes the determinant of the matrix, and I is the N N identity matrix. Thisequation is called the characteristic equation of the system. It is a polynomial equationin of degree N . Finding its solutions is the hard part of this entire solution procedure.But, in the end, weve reduced the problem of nding solutions of these constant-coefficientlinear dynamical systems to nding roots of a polynomial. This is progress!

Thus, with every eigenvalue k and its corresponding eigenvector v(k ) we get a solution

x(k ) = ekt v(k ) .

Since the equation is linear, we can get the general solution by taking linear combinationsof all of these solutions:


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x = x(1) + . . . + x(N ) = e 1 t v(1) + . . . e N t v(N ) .

Honestly, this is not the full story: sometimes we do not get enough solutions this way,

because some of the eigenvalues have higher multiplicity and not enough eigenvectors arefound. This will not be a big concern for us in this course, but you should be aware it maybe an issue. Next, well examine some of the different possibilities.

Two real eigenvalues: nodes. Lets assume that both 1 and 2 are real, and 1 >2 > 0. Then

x = c1e 1 t v(1) + c2e 2 t v(2) .

First, lets assume we start with an initial condition so that c2 = 0. Then

x = c1e 1 t v(1) ,

which corresponds to a straight line in the phase plane. Indeed, this formula implies that x isalways a multiple of the eigenvector v(1) . As t , we approach as the exponential factorbecomes bigger and bigger. The value of c1 (determined by the initial condition), determineswhether we end up in the direction of the eigenvector ( c1 positive), or the opposite direction(c1 negative). More explicitly, we may obtain the equation for the straight line:

x2x1

= v(1)2v(1)1

x2 = v(1)2v(1)1

x1,

which is the equation for a straight line through the origin with slope v(1)2 /v(1)1 . A similar

analysis holds for x(2)

: it represents a straight-line solution, in the direction of the secondeigenvector. The eigenvalue is positive again, thus on this line we move away from the origin.Other solutions are linear combinations of these fundamental ones. To draw the phase

portrait, we need to follow the arrows. The hard part is to gure out which arrows to follow.

As t the solution e 1 t v(1) dominates since 1 > 2. Thus for large time solutionswill predominantly follow the direction of the rst eigenvector. As t the solution e 2 t v(2) dominates since 1 > 2. Thus for large negative timethe rst solution damps faster than the second one. Thus close to the origin (which

is where we are as t ) will predominantly follow the direction of the secondeigenvector.In this case the origin is called an unstable node. A representative phase portrait is

shown in Figure 6.7. A stable node is similar, but with the directions of all arrows reversed.This occurs when both eigenvalues are stricly negative.

Two real eigenvalues: saddle points. If we have two real eigenvalues, but oneis positive and the other is negative, we end up with a saddle point. Lets assume that1 > 0 > 2. Then we have


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6.4. CLASSIFICATION OF LINEAR SYSTEMS 61

Figure 6.7: The phase portrait a stable node. Here x1 is on the horizontal axis, x2 on thevertical axis. The plot was produced using pplane .

x = c1e 1 t v(1) + c2e 2 t v(2) .

As before, the two fundamental solutions give rise to straight-line solutions in the directionsof the eigenvectors. In the direction of v(1) , the motion is away from the origin, since 1 > 0.On the other hand, in the direction of v(2) , the motion is towards the origin, since 2 < 0.For the other solutions, we just follow the arrows. The phase portrait for a saddle point isgiven in Figure 6.8. Note that, using Lyapunovs denition, a saddle point is unstable.

Denition. (Stable manifold) The stable manifold of a xed point is the collectionof all trajectories that approach the xed point as t .

As examples, for the saddle point, the stable manifold consists of the straight line in thedirection of v(2) . For the unstable node, the stable manifold consists of only the xed point.On the other hand, for a stable node, the stable manifold is the entire phase plane.

Denition. (Unstable manifold) The unstable manifold of a xed point is the collec-tion of all trajectories that approach the xed point as t .

Thus, the unstable manifold is the stable manifold as we go back in time. As examples,for the saddle point, the unstable manifold consists of the straight line in the direction of v(1) . For the unstable node, the unstable manifold consists of the entire phase plane.

Two complex eigenvalues: spirals and centers. If the eigenvalues are complex, we


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