Based on notes by A. Annibale, R. Kuhn and H.C ... - kcl.ac.uk

Lecture notes for Introduction to Dynamical Systems: CM131A

Based on notes by A. Annibale, R. Kuhn and H.C. Rae

2018-2019

Lecturer: G. Watts

1

Contents

1 Overview of the course 4

1.1 Revision Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

I Differential Equations 10

2 First order Differential Equations 11

2.1 Basic Ideas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.2 First order differential equations . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.3 General solution of specific equations . . . . . . . . . . . . . . . . . . . . . . . 13

2.3.1 First order, explicit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.3.2 First order, variables separable . . . . . . . . . . . . . . . . . . . . . . 14

2.3.3 First order, linear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.3.4 First order, homogeneous . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.4 Initial value problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.5 Existence and Uniqueness of Solutions — Picard’s theorem . . . . . . . . . . 23

2.5.1 Picard iterates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3 Second order Differential Equations 31

3.1 Second order differential equations . . . . . . . . . . . . . . . . . . . . . . . . 31

3.1.1 Second order linear, with constant coefficients . . . . . . . . . . . . . . 32

3.2 Existence and Uniqueness of Solutions — Picard’s theorem . . . . . . . . . . 38

3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

II Dynamical Systems 44

4 Introduction to Dynamical Systems 45

Version of Mar 14, 2019

2

5 First Order Autonomous Systems 50

5.1 Trajectories, orbits and phase portraits . . . . . . . . . . . . . . . . . . . . . . 50

5.2 Termination of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

5.3 Estimating times of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

5.4 Stability — A More General Discussion . . . . . . . . . . . . . . . . . . . . . 66

5.4.1 Stability of Fixed Points . . . . . . . . . . . . . . . . . . . . . . . . . . 66

5.4.2 Structural Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

5.4.3 Stability of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

5.5 Asymptotic Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

5.5.1 Asymptotic Analysis and Dynamical Systems . . . . . . . . . . . . . . 74

5.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

6 Second Order Autonomous Systems 85

6.1 Phase Space and Phase Portraits . . . . . . . . . . . . . . . . . . . . . . . . . 85

6.2 Separable Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

6.3 The structure of orbits and Phase Space . . . . . . . . . . . . . . . . . . . . . 98

6.4 Limit cycles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

6.5 Fixed points of second-order autonomous systems . . . . . . . . . . . . . . . . 101

6.6 Linear Stability Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

6.6.1 Step 1 — Taylor Expansion of Velocity Functions . . . . . . . . . . . . 102

6.6.2 Step 2 — Finding the Jordan canonical form of the Jacobian . . . . . 104

6.6.3 Step 4 — Exploring the Consequences for Dynamics . . . . . . . . . . 105

6.7 Beyond linear stability analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 119

6.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

III Application to Classical Mechanics 132

7 Elements of Newtonian Mechanics 133

7.1 Motion of a particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

7.2 Newton’s Laws of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

7.2.1 Newton’s First Law (N1) . . . . . . . . . . . . . . . . . . . . . . . . . 138

7.2.2 Newton’s Second Law (N2) . . . . . . . . . . . . . . . . . . . . . . . . 138

7.2.3 Newton’s Third Law (N3) . . . . . . . . . . . . . . . . . . . . . . . . . 138

7.3 Newton’s Law of Gravitation . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

7.4 Motion in a Straight Line; the Energy Equation . . . . . . . . . . . . . . . . . 143


3

7.5 Equilibrium and Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

7.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

8 Hamiltonian Systems 165

8.1 Hamilton’s equations for motion in a potential . . . . . . . . . . . . . . . . . 166

8.2 Stability problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173

8.3 Summary: how to analyse motion in a potential . . . . . . . . . . . . . . . . . 176

8.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181

A Functions of two variables 187

A.1 The partial derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

A.2 Continuity of a function of two variables . . . . . . . . . . . . . . . . . . . . . 189

B Taylor’s Theorem 190

B.1 Taylor Expansion for Functions of One Variable . . . . . . . . . . . . . . . . . 190

B.2 Taylor Expansion for Functions of two variables . . . . . . . . . . . . . . . . . 191

B.3 Vector functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192

C Basic Linear Algebra 195

C.1 Fundamental ideas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195

C.2 Invariance of Eigenvalues Under Similarity Transformations . . . . . . . . . . 197

C.3 Jordan Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197

C.4 Basis Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199

C.5 Rotations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200

C.6 Area Preserving Transformations . . . . . . . . . . . . . . . . . . . . . . . . . 200

C.7 Examples and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

D Jacobians and Change of variables 205

D.1 Change of variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205


Chapter 1: Overview of the course 4

Chapter 1

Overview of the course

This course is about the study of quantities which vary in time. The quantities are meant torepresent some “system” and the variation in time is called the “dynamics”, hence the name.

These ideas can be very general - the quantities could take real values, such as the coordinatesof bodies in a classical mechanics problems, they could take integer values such as the numberof individuals in a population - and the variation in time can be continuous (as the positionof a body is defined for all values of t) or discrete (maybe we measure the population onlyonce each day). The study of variations of systems with time obviously has gone on for along time, but the development of the branch of mathematics known as “Dynamical Systems”really started in the 1890s with Poincare, inspired by the problems of the motion of planets.It is partly pure mathematics and partly applied and that is reflected in the way this courseis organised: there are some theorems, some study of properties of differential equations andsome applications.

To give a flavour of what’s to come we’ll look at three models of population size and onephysical system which are all typical examples.

The simplest model of a population is that of Thomas Malthus who proposed (in 1798) thatif there are no constraints on the resources available then the rate of growth of a populationwill be proportional to its size - the number of births in any time period will be proportionalto the number of people. If the size of the population is x and the birth rate is r per unittime then this can be modelled by the differential equations

dx

dt= r x . (1.1)

The solutions of this equation arex(t) = x0 er t . (1.2)

This exponential growth of course leads to problems, as Malthus knew, and so is not a goodmodel for long term population sizes. A better model is the logistic equation (devised byPierre-Francois Verhulst in 1844) in which limits of resources mean there is a maximum sizec for a stable population:

dx

dt= f(x) , f(x) = r x

(1− x

c

), (1.3)



for which we can also find the general solution as the function x(t)

x =c

1−Ae−r t, where A = 1− c

x0(1.4)

The exponential solution (1.2) is one we are familiar with, the solutions (1.4) may be harderto visualise. One of the aims of the course is to see how to find out properties of the dynamicsof a model – exact or approximate – using graphical methods. The simplest method is if onecan find the exact solutions and then plot these, so for the exponential growth model we canplot x vs. t or t vs. x, as in figure 1.1

-2 2 4 t

-1.0

-0.5

0.5

1.0

x

-1.0 -0.5 0.5 1.0x

-2

2

4

t

Figure 1.1: Sketches of the solutions (1.2) for r = 0.1 together with thephase portrait showing the fixed point at x = 0.

Also included is a useful guide to the way the system behaves, known as the “phase portrait”which shows whether x increases or decreases with t and also that x = 0 is a “fixed point”.

For the more complicated Verhulst model, the corresponding plots are in figure 1.2

-4 -2 2 4 t

-1.0

-0.5

0.5

1.0

1.5

2.0

x

-1.0 -0.5 0.5 1.0 1.5 2.0 x

-4

-2

2

4

t

Figure 1.2: Sketches of the solutions (1.4) for r = 0.1 and c = 1 togetherwith the phase portrait showing the fixed points at x = 0 and x = 1.

In both these cases we can solve the equations exactly but the behaviour of the solutions iscaptured in the phase portrait showing the way x changes with t and the fixed points.



These models can be generalised to a system with two species, a predator and a prey, wheresimple models for their interactions lead to two coupled equations

dx

dt= x(a− by) ,

dy

dt= y(cx− d) . (1.5)

This is the Lotka-Volterra model, also called the hare-lynx or rabbit-fox model for two situ-ations it describes. This is a case where we cannot find an exact solution in general; We canonly find exact solutions if x = 0 (no predators), if y = 0 (no prey) or if the populations takethe constant values x = d/c, y = a/b. However, we can study other aspects of the system -we can sketch the solutions (technically, sketch the phase curves) and see that there are goingto be solutions that stay close to the fixed point at (d/c, a/b), that these will have periodicoscillations and we can find the approximate period of these oscillations. There is anotherfixed point at (0, 0) which is of a different nature; it is not stable, solutions close to this pointdo not have to stay close and we can again analyse motion near here.

0 1 2 3 4 5 6

0.0

0.5

1.0

1.5

2.0

Prey

Predator

Figure 1.3: Sketch of the time-evolution of prey (x-axis) and predator (y-axis) populations in the case a = 1, b = 2, c = 1, d = 2. The fixed point at(2, 1/2) is the isolated dot, and typical time evolution is a periodic orbitaround the fixed point.

The first two examples are completely solved using the theory of differential equations; thisthird example takes us into the study of dynamical systems proper where we pay attentionto structural properties of the solutions without necessarily solving them.

Finally, as an example of a discrete dynamical system, another very well-known model is the“logistic map” which states how a variable xn changes as n increases in integer steps:

xn+1 = rxn(1− xn) . (1.6)

This is most famous because it exhibits chaotic behaviour when r > 3.56995... Not only is



there no exact solution, but the tiniest change in the starting value of x1 leads to unpredictablechanges in the time evolution.

While this discrete chaotic system is interesting, it is also requires very sophisticated mathe-matical analysis and in this course we will stick to the simpler case of time being a continuousparameter and the variables in the system having real values, so that the systems are de-scribed by one or more functions of t and the dependence on t is defined through differentialequations.

We can now describe the structure of the course: the course is split into three parts: differentialequations, dynamical systems and classical mechanics.

The first part is on differential equations and their solutions. We first consider first orderdifferential equations, give methods to solve some standard classes and state a theorem guar-anteeing the existence and uniqueness of solutions - it is easy to write down problems whichhave no solution or which have an infinite number of solutions and it is good to have a test todecide what the answer is. We then turn to second order differential equations, state a similartheorem on existence and uniqueness of solutions and consider the case of linear equationswith constant coefficients in some detail.

The second part is on dynamical systems. We introduce the ideas of the order of a dynamicalsystem, fixed points, stability, phase portraits, and then study these in turn for first andsecond order dynamical systems and how to sketch and analyse phase portraits such as figure1.3, how to classify fixed points and discuss motion in their neighbourhoods.

The third and final part is on applications of this theory in classical mechanics. We introduceNewton’s laws of motion as second order differential equations for the position of massivebodies and give his law of gravity. (1.7). We use this as an example of the idea of a potentialand analyse motion in a potential using the methods of dynamical systems.

A typical example of a classical mechanical system is that of a weight suspended from aspring. If a weight with mass m is held up by such a spring then the height z of the weightcould satisfy a differential equation such as

md2z

dt2= −kz − lz3 −mg . (1.7)

When l = 0 this is an ideal spring satisfying Hooke’s law and we will see how to find thegeneral solution

z = z0 +A sin(ωt) +B cos(ωt) , where z0 = −mg/k and ω2 =k

m. (1.8)

When l 6= 0 as in a more realistic model of a spring, we cannot solve the differential equationin terms of elementary functions and cannot find z as an explicit function of t, but thesystem shares many of the same properties as the ideal spring equation (1.7), for example thefrequency of small oscillations is approximately the same as that when l = 0.

In the case of the weight on the spring, the potential is

V (z) = mgz +k

2z2 +

l

4z4 with the equation of motion mz = −dV

dz. (1.9)

The final topic in this section is on Hamilton’s reformulation of classical mechanical systemswhich is both very closely related to the ideas of dynamical systems but also central to modern



algebraic approaches and leads on to the more advanced modules “Classical Mechanics” inyear 2 and “Quantum Mechanics” in year 3.

The equivalent reformulation by Hamilton defines a function H(p, q) where q = z and p = mzand equations of motion given by

H(p, q) =p2

2m+ V (q) ,

dq

dt=∂H

∂p,

dp

dt= −∂H

∂q, (1.10)

which has the advantage of showing immediately that the function H(p, q) is constant in time.

This formulation also makes it very easy to find the equilibrium points of the mechanicalsystem and to the sketch the phase portrait of the system. The phase portrait showing howthe values of q and p vary with time for an ideal spring and a non-ideal spring are shown infigure 1.4. They look very much the same - the differences are that the position of the fixedpoint has moved and that the orbits are circles for the ideal spring and squashed circles forthe non-ideal spring.

-3 -2 -1 0 1

-2

-1

0

1

2

-3 -2 -1 0 1

-2

-1

0

1

2

Figure 1.4: Sketches of the phase portrait for the system solutions (1.10)for m=g=k=1 with l = 0 (ideal spring) with fixed point at (−1, 0) on theleft and l = 16/27 (non-ideal spring) with fixed point at (−3/4, 0) on theright.



1.1 Revision Exercises

This problem-set is about prerequisites, based on material covered in Calculus 1 and LinearMethods/Linear Algebra and Geometry I.

Exercise 1.1 Evaluate the following integrals:

1.

∫ 1

0dx

x2

(1 + x6)2.

∫ ∞0

dxx3

(1 + x8)

3.

∫ 1

0dx

x2

(1− x2)1/24.

∫dx

1√x(1− x)

5.

∫dx x2 exp(2x) 6.

∫dx xn lnx , n a positive integer.

Exercise 1.2 Give the Taylor expansion up to third order in x about x = 0 for the followingfunctions

1. f(x) =1

1− x2. f(x) =

√1 + x 3. f(x) = ln(1 + x) 4. f(x) = sinh(x)

Exercise 1.3 Discuss the salient properties (find zeros, poles, extrema, inflection points,asymptotes ...) and produce a qualitatively correct graph of the following functions

1. f(x) = x(x− 1)(x− 2) 2. f(x) = x2(x− 1)(x− 2)

3. f(x) =x2

1− x24. f(x) =

x2

1 + x2

5. f(x) = x exp(−x2) 6. f(x) =x

tan(x)

7. f(x) =x

cosh(x)

Exercise 1.4 This exercise recalls properties of differential operators and linear maps.

Consider L = ddx defined as a map acting on differentiable functions of x,

L : f 7→ L(f) =df

dx, L(f)(x) =

df

dx(x) .

Show that L is a linear map, i.e. show that for any real constants α and β and any differentiablefunctions f(x) and g(x),

L(αf + βg) = αL(f) + βL(g) ,

Which of the following is a linear map on differentiable functions of x?

1. M1 : f 7→ f +df

dx, 2. M2 : f 7→ d2f

dx2− x2f , 3. M3 : f 7→ x+

df

dx


Part I

Differential Equations

10

Chapter 2: Differential Equations 11

Chapter 2

First order Differential Equations

This chapter starts with some elementary ideas about differential equations. We then considerfirst order differential equations, discuss different classes and give methods to solve some ofthese. We then discuss initial value problems, show that there are initial value problemswhich can have no solutions or contrastingly an infinite number of solutions and finish witha theorem which gives conditions for the existence and uniqueness of a solution to an initialvalue problem.

2.1 Basic Ideas

A differential equation, or more correctly an ordinary differential equation (ODE), is a relationbetween a function of one variable and one or more of its derivatives (evaluated at the samepoint). We will often think of a function x(t) where t we can think of as time and x as aposition, a population, a share price... We will also often use y(x). In general, we can usea dash to denote differentiation, so that y′(x) = dy/dx, y′′(x) = d2y/dx2; if we use t as thevariable then we can denote derivatives using dots, x = dx/dt, x = d2x/dt2.

Here are some examples:

dx

dt= 3x2 sin(x+ t) (2.1)

x = cos(2t) (2.2)

d2y

dx2=

(dy

dx

)3

− y (2.3)

One of the most important properties of an ODE is its order

Definition 2.1. The order of an ODE is the order of the highest derivative appearing in theODE

In the three examples above, the orders are 1, 2 and 2 respectively. Even though (dy/dx)3

appears in (2.3), the highest derivative is the second derivative d2y/dx2 and so the order ofthe ODE is two.

We shall start in this chapter by considering first order differential equations and then turnto second order differential equations in chapter 3.



2.2 First order differential equations

We start by considering first order (ordinary) differential equations of the form

d

dtx = f(t, x) , (2.4)

in which f is a suitably defined function of two real variables.

Definition 2.2. Solution of an ODE

We say that a function g is a solution of the differential equation (2.4) if equation (2.4) issatisfied identically when we substitute x(t) = g(t), i.e.

d

dtg(t) ≡ f(t, g(t)) (2.5)

This allows a lot of freedom in the choice of the solution, for example for the extremelyelementary ODE with f(t, x) = 0,

d

dtf = 0 , (2.6)

then f(x) = 0 and f(x) = 1 are both solutions and indeed f(x) = c is a solution for any valueof c.

This leads to the idea of a general solution

Definition 2.3. General Solution

We say that the family of functions h(·, C) is a general solution of equation (2.4), if

d

dth(t, C) ≡ f(t, h(t, C)) , (2.7)

whatever C. [Rem: The notation is to be understood in the sense that x = h(·, C) is afunction with parameter C, i.e. x(t) = h(t, C).]

E.g.: the ODE ddtx = ax has general solution x(t) = Ceat, with C = x(0).

One way to pick out a particular solution is to choose an initial value for x(t), that is wechoose an initial time t0 and require that x(t0) = x0. This is then called an initial valueproblem

Definition 2.4. An initial value problem for a first order ODE

We say the problem of finding a function x(t) such that

d

dtx = f(t, x) (2.8)

andx(t0) = x0 (2.9)

is an initial value problem.



Definition 2.5. A solution of an initial value problem

We say that a function g is a solution of the initial value problem (2.8,2.9) if equation (2.8)is satisfied identically when we substitute x(t) = g(t), i.e.

d

dtg(t) ≡ f(t, g(t)) (2.10)

and the initial condition (2.9)g(t0) = x0 . (2.11)

is also satisfied.

Thus the function g must satisfy the differential equation and the initial condition.

If we plot the points (t, g(t)); t ∈ IR in the Cartesian plane we generate a trajectory orsolution curve of (2.4) which passes through the point (t0, x0) determined by the initialcondition x(t0) = x0. As we change the initial condition (i.e. the value of x0, the prescribedvalue of x at t = t0, and possibly t0) we generate a family of solution curves of (2.4) alsocalled the flow corresponding to (2.4).

We will now review methods for finding the general solution of a few simple types of ODE (onlyfor a few types do general solutions in terms of elementary functions exist!), and then turnto the question of initial value problems and briefly enquire into the question of uniquenessof solutions.

2.3 General solution of specific equations

We now summarise some well known methods for finding the general solutions of certain typesof first and second order differential equations. Recall that the general solution of a first orderdifferential equation involves one parameter.

We proceed from simple to more complicated situations, trying to reduce the more complicatedsituations to the simpler ones dealt with before.

2.3.1 First order, explicit

Explicit equations are those which can be solved by direct explicit integration; they are ofthe form

dx

dt= f(t) (2.12)

so that the r.h.s. does not depend on the unknown function x; thus explicit first orderequations are not at all interesting ODEs.

Integrating with respect to time t one obtains

x =

∫dt f(t) = F (t) + C , (2.13)



with F (t) + C denoting the indefinite integral of the function f . Thus one has

x = x(t) = F (t) + C (2.14)

as the general solution of (2.13).

2.3.2 First order, variables separable

Equations of the typedx

dt= f(x) g(t) (2.15)

may be integrated formally by separation of variables:

dx

f(x)= g(t) dt ⇒

∫dx

f(x)=

∫g(t) dt , (2.16)

Denoting the indefinite integrals involved by∫

dxf(x) = F (x) + C, and

∫dt g(t) = G(t) + C

and combining the two integration constants into one, one obtains

F (x) = G(t) + C , (2.17)

which — on (formally) solving for x — gives

x = x(t) = F−1(G(t) + C) (2.18)

with arbitrary C as the general solution of (2.15). It is not always possible to find an explicitexpression for F−1 and one has sometimes just to be happy with the relation (2.17)

Example 2.1. Find the general solution of the differential equation

dx

dt= αx , (2.19)

where α is a constant. This differential equation will arise frequently in our later work. We have

dx

x= α dt , ⇒

∫dx

x=

∫α dt ⇒ ln |x| = αt+ C (2.20)

where C is a constant, so that|x| = eαt eC , (2.21)

This is an expression for |x(t)|, but choosing the free parameter differently we can find x(t) explicitlyas,

x(t) = Aeαt , (2.22)

where A = ±eC is now the free parameter.


dx

dt=x

t, (2.23)



This is clearly separable, and so

dx

x=

dt

t, ⇒

∫dx

x=

∫dt

t⇒ ln |x| = ln |t|+ C (2.24)

where C is a constant, so that changing constants,

x = At , (2.25)

for some A.


dx

dt=

2t+ 1

3x2 + exp(x), (2.26)

This is clearly separable, and so

(3x2+exp(x))dx = (2t+1)dt , ⇒∫

(3x2+exp(x))dx =

∫(2t+1)dt ⇒ x3+exp(x) = t2+t+C ,

(2.27)

where C is a constant. In this case there is no way to solve equation (2.27) to find x as a function of

t but it is still possible to plot solutions by solving for x(t) numerically as in figure 2.1

-3 -2 -1 1 2 3

-1.0

-0.5

0.5

1.0

1.5

2.0

Figure 2.1: Sketch of three solutions to equation (2.27) with C = −1 (solidline), C = 0 (dotted line) and C = 1 (dashed line).



2.3.3 First order, linear

First order, linear ordinary differential equations are of the form

dx

dt= f(t)x(t) + g(t) , (2.28)

with f and g given functions of t. They are called “linear” because, when written as

dx

dt− f(t)x(t)− g(t) = 0 , (2.29)

the left-hand side is a linear function of x and its derivative. It also means that the differentialequation can be written in terms of a differential operator L which is a linear map when actingon the vector space of differentiable functions:

L(x) = g , where L =

(d

dt− f(t)

), L(x) =

(dx

dt− f(t)x(t)

). (2.30)

We can check that L is a linear map by considering two functions x1 and x2 and arbitraryconstants α, β and seeing that

L(αx1 + βx2) = αL(x1) + βL(x2) . (2.31)

This automatically has implications for the space of solutions which we will look at in moredetail later in this section. For the moment we will just look at how to solve these equations.

Equations of the form (2.28) can be solved using the method of integrating factors. This is amethod that reduces the problem to that of solving two simpler equations, one of which canbe solved by separation of variables, the other by direct integration. It works as follows.

Multiply through by µ = µ(t) where µ(t) is a function of t which we choose later. We have

µdx

dt− µ f x = µg , (2.32)

which can be transformed into

d

dt(µx) +

(− µf − dµ

dt

)x = µg . (2.33)

By choosing µ such thatdµ

dt= −µf , (2.34)

a choice that requires solving a separable first order ODE to actually determine µ, one findsthat the transformed ODE simplifies to

d

dt(µx) = µg . (2.35)

This is a first order explicit equation for the product µx, once µ is known from the solutionof the separable ODE defining it. A solution of the ODE defining µ is

µ(t) = exp−∫f(t) dt . (2.36)



Note that the indefinite integral in the exponent is only defined up to an arbitrary additiveconstant. As we need to know µ only up to an arbitrary multiplicative constant, we can setany additive constant in the exponential to zero (also we have implicitly used this freedom toassume that µ is positive !).

With this choice of µ the explicit equation for the product µx is integrated to give

µ(t)x(t) =

∫µ(t)g(t) dt , x(t) =

1

µ

∫µ(t)g(t) dt . (2.37)

Note again that the indefinite integral here comes with an arbitrary integration constant Cas a free parameter of the general solution. This can not be ignored but is used to satisfyinitial conditions as usual. The function µ is referred to as an integrating factor for equation(2.28).

To return now to the fact that the equation can be written as

L(x) = g . (2.38)

We know from the theory of linear algebra that the solution to a linear equation such as (2.38)is only defined up to an element of the kernel of L, that is a function xcf (t) which satisfies

L(xcf ) = 0 . (2.39)

The kernel of a linear operator is a vector space so the solutions to (2.39) form a vector space- for a first order ODE this will be a one-dimensional vector space. An equation such as (2.39)is known as a homogeneous linear differential equation, a solution xcf (t) of this equation iscommonly known as a “complementary function”. The original equation (2.38) is known asan inhomogeneous linear differential equation, the term g on the right-hand-side is an calledthe inhomogeneous term and any solution xpi(t) of (2.38) is called a particular integral. Hencethe result is that the general solution of (2.38) is

x(t) = xpi(t) + xcf (t) , where L(xpi) = g and L(xcf ) = 0 . (2.40)

This also means that the difference of any two solutions of the inhomogeneous equation (2.38)will be a solution of the homogeneous equation (2.39). We will return to these ideas in chapter3 when we treat second order linear differential equations in more detail.

Now for some examples of the solution of linear differential equations.

Example 2.4. Find the general solution of the ODE

t(t+ 1)dx

dt− (t+ 2)x = t3(2t− 3) ⇔ dx

dt− (t+ 2)

t(t+ 1)x = t2

(2t− 3)

(t+ 1)(2.41)

We can see that this a first order linear ODE. The functions f(t) and g(t) can be read off from theequation as

f = − t+ 2

t(t+ 1), g = t2

(2t− 3)

t+ 1. (2.42)

The integrating factor is therefore

µ(t) = exp

(−∫

(t+ 2)

t(t+ 1)dt

)(2.43)



Using(t+ 2)

t(t+ 1)=

2(t+ 1)− tt(t+ 1)

=2

t− 1

t+ 1(2.44)

one obtains

µ(t) = exp (−2 ln t+ ln(t+ 1)) =t+ 1

t2, (2.45)

so

µx =

∫µ(t) t2

(2t− 3)

(t+ 1)dt =

∫(2t− 3) dt = t2 − 3t+ C (2.46)

Hence

x(t) =t2

t+ 1(t2 − 3t+ C) (2.47)


dx

dt=x

t, (2.48)

This is clearly linear, and if we write it as

dx

dt− x

t= 0 , (2.49)

then we read off that f = 1/t and g = 0 and so the integrating factor is

µ = exp

(−∫

dt

t

)= exp(− ln(t)) =

1

t(2.50)

and so the differential equation becomes

d(µx)

dt= 0 ⇒ µx = C , so that the general solution is x(t) =

C

µ= C t . (2.51)

This is the same differential equation as in example 2.2, the solution method is different but thesolution is the same.

2.3.4 First order, homogeneous

The ODE is called homogeneous if it is of the form

dx

dt= F (x/t) (2.52)

Homogeneous is being used in a different way to the previous section; that is just how it is.This equation can be solved by the substitution x(t) = tv(t), or v = x/t. We then have

dx

dt=

d

dt(tv) = t

d

dtv + v = F (v) ,

dv

dt=

1

t(F (v)− v) , (2.53)

which is separable with the solution

dv

F (v)− v=

dt

t⇒

∫dv

F (v)− v= ln |t|+ C . (2.54)



After carrying out the v integration we then have a relation between x(t) and t which withluck we can simplify to find x(t) - but this is often not possible.

The most common ODEs of this form we will encounter are when R(x/t) is a ratio of poly-nomials of the same degree, that is

F (x/t) =P (t, x)

Q(t, x), (2.55)

where P,Q are homogeneous polynomials of (the same) degree m, i.e. P and Q are sums ofterms of the form cxptm−p of the same total degree m.

Example 2.6. Consider the differential equation

dx

dt=

(x2 − t2)

2xt. (2.56)

The function on the right hand side is the ratio of homogeneous quadratic polynomials so we candivide top and bottom by t2 to get instead

dx

dt=

((x/t)2 − 1)

2(x/t)=

(v2 − 1)

2v. (2.57)

where x = tv(t). We then obtain

tdv

dt+ v =

(v2 − 1)

2v, (2.58)

and so

tdv

dt=

(v2 − 1)

2v− v = −1 + v2

2v. (2.59)

Separation of variables now gives∫2v

1 + v2dv = −

∫dt

t=⇒ ln(1 + v2) = − ln |t|+ C , (2.60)

where C is a parameter. Exponentiation now yields

|t|(1 + v2) = eC ⇔ t(1 + v2) = A , (2.61)

where A = ±eC is a free parameter. It follows that

t(1 + x2/t2) = A ⇔ x2 + t2 = At . (2.62)

This equation may be re-written as

(t−A/2)2 + x2 = (A/2)2 , (2.63)

from which we see that the solution curves are a family of circles in the x − t plane with centres at

(0, A/2) and radii A/2. Notice that each of the circles passes through (0, 0), whatever the value of A,

so the solution is not unique in the vicinity of (0, 0).


dx

dt=x

t, (2.64)



This is clearly homogeneous. With the substitution xv t we get

tdv

dt+ v = v , (2.65)

or simplydv

dt= 0 , (2.66)

with the general solutionv = C ⇒ x = C t . (2.67)

This is again the same differential equation as in example 2.2, the solution method is again different

but the solution is the same. As in the previous example, all the solutions go through (0, 0).

2.4 Initial value problems

So far we have been concerned with finding the general solution x(t) = h(t, C) to an ODE.Finding a solution to an initial value problem usually just amounts to find the value(s) of Cfor which the general solution satisfies the condition, that is finding C such that

x(t0) = h(t0, C) = x0 . (2.68)

For many of the examples we have looked at so far this is straightforward.

Example 2.8. Solve the initial value problem

dx

dt= αx , x(t0) = x0 . (2.69)

This is the ODE in example 2.1 with general solution x = A exp(α t). Substituting the initial conditionswe find the equation

A eαt0 = x0 ⇒ A = x0 e−αt0 , (2.70)

so that the solution to the problem is

x(t) = x0 eα(t−t0) . (2.71)

Example 2.9. Find the solution to the initial value problem

t(t+ 1)dx

dt− (t+ 2)x = t3(2t− 3) , x(t0) = x0 . (2.72)

This is the differential equation in example 2.4 which has the general solution

x(t) =t2

t+ 1(t2 − 3t+ C) (2.73)

Substituting in the initial conditions we find

C =x0(t0 + 1)

t20− t20 + 3t0 . (2.74)

However, we have already seen differential equations for which it is easy to state initial valueproblems which have either no solutions or an infinite number of solutions. One of the simplestis the ivp



Example 2.10. Consider the initial value problem (2.23),

dx

dt=x

t, x(t0) = x0 . (2.75)

This differential equation was considered many times already and the general solution is(2.51),

x(t) = C t . (2.76)

If we choose an initial time t0 6= 0 then the constant C is easily seen to be C = x0/t0 and theunique solution satisfying

x(t) = x0

(t

t0

). (2.77)

If by contrast we take t0 = 0 then x(0) = 0 for all the solutions (2.76) and so

Proposition 2.1. The initial value problem

dx

dt=x

t, x(0) = x0 , (2.78)

has no solutions if x0 6= 0 and has an infinite number of solutions if x0 = 0.

A second well-known case where an initial value problem has more than one solution is

Example 2.11. Consider the initial value problem

dx

dt= 3x2/3 , x(0) = 0 . (2.79)

This is a separable ODE and following the method in the notes gives

1

3x−2/3dx = dt ⇒

∫1

3x−2/3dx =

∫dt ⇒ x1/3 = t+ C . (2.80)

With the initial condition that x(0) = 0 we C = 0 and the solution

x = t3 . (2.81)

There is, however, another different and obvious solution to the problem (2.79), which is

x(t) = 0 . (2.82)

This means that we have found two solutions to the i.v.p.,

1. x(t) = t3 (2.83)

2. x(t) = 0 (2.84)

The situation is worse, however, since we can also consider the combined solutions

3. x(t) =

0 t ≤ 0

t3 t ≥ 0(2.85)



4. x(t) =

t3 t ≤ 0

0 t ≥ 0(2.86)

and in fact we can find an infinite set of solutions. For each positive number a then this is asolution

5. x(t) =

0 t ≤ a(t− a)3 t ≥ a

(2.87)

and so is this

6. x(t) =

(t+ a)3 t ≤ −a0 t ≥ −a

(2.88)

These functions are all continuous and their derivatives exist and are also continuous. Theycertainly satisfy the differential equation and the initial conditions.

Initial value problems for the ODE in example 2.6 can also have no solutions or an infinitenumber of solutions since, as in the case of proposition 2.1, all the solutions satisfy x(0) = 0.

So we now have found several initial value problems which either have no solution or whichhave an infinite number of solutions. It is very helpful to have a test which enables us todetermine if there is a solution and if it is unique. This is answered by Picard’s theoremwhich gives the conditions under which a solutions exists and is unique and is covered in thenext section.



2.5 Existence and Uniqueness of Solutions — Picard’s theorem

There are many theorems on the existence of solutions to an initial value problem and onwhether it is unique or not. Some of these require a lot of Analysis to be able to state, so wewill here look at one which is fairly easy to state, even if it is not the strongest:

Theorem 2.1 (Picard)

Let A denote the set(t, x) : |t− t0| ≤ δ, |x− x0| ≤ δ (2.89)

(a square centre the initial point (t0, x0) and side of length 2δ). Suppose that f : A → IR iscontinuous1 and that the partial derivative2 fx = ∂f

∂x is also continuous on A. Then the initialvalue problem

dx

dt= f(t, x) , x(t0) = x0 , (2.90)

has a unique solution which is defined on the interval [t0 − δ1, t0 + δ1], for some δ1 ≤ δ. Seefigure 2.2.

A

t0-δ1 t0+δ1t0-δ t0 t0+δ

x0-δ

x0

x0+δ

Figure 2.2: Sketch of the region A surrounding the point (t0, x0) on whichf and ∂f/∂x are continuous and the smaller interval |t− t0| ≤ δ1 on whichthe solution exists and is unique. The ivp example illustrated is example2.12 with t0 = 2.5, x0 = 2, δ = 1.5, δ1 = 0.3. The solution diverges at t = 2which is outside the dotted interval.

1The definition of continuity of a function of two variables is in appendix A2The definition of the partial derivative is also in appendix A



This isn’t the “best” version of Picard’s Theorem, there will be initial value problems whichdo not satisfy the conditions of this theorem and still have unique solutions, but this versionhas the advantage that it can be stated in a form which is likely to make reasonable sense tofirst year students.

We will not give a full proof, but indicate its heuristics (“how the theorem works”) based onPicard iterates, a practical procedure for constructing the solution.

Note that we cannot assume that the solution exists for all t. A very simple example is this:

Example 2.12. Consider the initial value problem

dx

dt= −x2 , x(t0) = x0 . (2.91)

Here the function f and its derivative are

f(t, x) = −x2 ∂f

∂x(t, x) = −2x . (2.92)

These are perfectly well behaved functions for all values of t and x and so we can take theconstant δ to be as large as we like. The solution to this problem is easy to find as the ODEis separable and it is

x(t) =x0

1 + x0(t− t0). (2.93)

This has a singularity at

t− t0 = − 1

x0, (2.94)

and so the maximum value of δ1 is |1/x0|; the ivp satisfies the conditions of Picard’s theoremfor any value of δ and so there is a unique solution around the point (t0, x0), but the maximuminterval (of the sort in the theorem) for which it is defined must satisfy

|t− t0| ≤ δ1 <1

|x0|. (2.95)

An illustration of how this works is shown in figure 2.2. Choosing the initial values to bet0 = 2.5, x0 = 2, we can easily take δ = 1.5 as f and fx are continuous for all x, t. Thesolution diverges for t = 2 and so the theorem just says there is some value of δ1 for whichthe function is defined on the interval [2− δ1, 2 + δ1]; from the explicit solution we know thatδ1 < 0.5, the figure shows the interval for δ1 = 0.3 on which the solution does exist and isunique.

2.5.1 Picard iterates

Assuming that eqs.(2.4), (2.9) have a solution we may integrate both sides with respect to tto obtain

x(t) = x0 +

∫ t

t0

dudx

du= x0 +

∫ t

t0

du f(u, x(u)) (2.96)

Eq. (2.96) is fully equivalent to the pair (2.4), (2.9). Note that we cannot really evaluate ther.h.s. as we do not know x(u) for u 6= t0; in other words Eq. (2.96) is an integral equation.

A constructive way of solving such an equation is by iteration



• Start with the best guess at our disposal:

x(u) = x0(u) = x0 , t0 ≤ u ≤ t . (2.97)

• Then define a sequence of functions xn(t) by

xn+1(t) = x0 +

∫ t

t0

du f(u, xn(u)) , n = 0, 1, 2, . . . . (2.98)

To actually prove Picard’s theorem we would have to show that, under the stated conditions,the sequence of functions xn(t) converges (uniformly) to a limit function x(t), and thenprove that x(t) is the unique solution of the differential equation.

Intuitively, convergence of the iteration rests on the observation that, for |t− t0| small,the contribution of the integral to the r.h.s. of (2.96) is small under the stated condition,because f(u, x(u)) is bounded for u ∈ [t0, t]. Because of this, a small (uniform) error

δn = maxs∈[t0,t]

|xn(s)− x(s)| , (2.99)

will be diminished under iteration. For, we have

δn+1 = maxs∈[t0,t]

∣∣∣∣∫ s

t0

du(f(u, xn(u))− f(u, x(u))

)∣∣∣∣≤ max

s∈[t0,t]

∫ s

t0

du∣∣∣(f(u, xn(u))− f(u, x(u))

)∣∣∣= max

s∈[t0,t]

∫ s

t0

du∣∣∣(fx(u, x(u)) (xn(u))− x(u)) + o(xn(u))− x(u))

)∣∣∣≤ C|t− t0| δn ,

where C is finite owing to the continuity of fx(u, x). (Inequalities above are due to

triangle-inequalities for absolute values of sums; we have exploited existence and con-

tinuity of fx(u, x)). In conclusion we have: δn+1 < δn for sufficiently small |t − t0|.By the same reasoning one would show that there cannot be two different solutions,

say x and y of the ODE with identical initial conditions x(t0) = y(t0) = x0; if they

were different, so ∆ = maxs∈[t0,t] |y(s) − x(s)| > 0, and both solutions of (2.98), then

by going through the same reasoning as for the error δn we would have to conclude

∆ < ∆, a contradiction — which finally proves the point.

Let’s see how the Picard method works in a specific case.

Example 2.13. Consider the differential equation

dx

dt= f(t, x) = x , x(0) = 1 . (2.100)

Integrating both sides of the equation with respect to t we obtain

x(t) = 1 +

∫ t

0du f(u, x(u)) = 1 +

∫ t

0du x(u) . (2.101)



Following the method described above we now define the Picard iterates by

xn+1(t) = 1 +

∫ t

0du xn(u) , x0(t) = 1 . (2.102)

Setting n = 0 we obtain

x1(t) = 1 +

∫ t

0du x0(u) = 1 + t . (2.103)

Substituting n = 1 we obtain

x2(t) = 1 +

∫ t

0du x1(u) = 1 +

∫ t

0du (1 + u) = 1 + t+

t2

2!. (2.104)

Continuing in this way, with n = 2, 3, · · · we find that

xn(t) = 1 + t+t2

2!+t3

3!+ · · ·+ tn

n!. (2.105)

We see that as n→∞ xn(t)→∑∞

r=0tr

r! = et, which is indeed the unique solution of the givenequation.

Note 1: If the conditions of Picard’s theorem are not satisfied by a particular differentialequation, the equation may not have a solution or, if it does, the solution may not be unique;in some cases there may even be an infinite number of solutions, all satisfying the same initialcondition, as we have seen. Let us look again at two of these examples, 2.10 (where thesolution does not exist) and 2.79 (where the solution is not unique).

Example 2.14. Consider the initial value problem (2.23),

dx

dt=x

t, x(0) = 1 . (2.106)

Here f(t, x) = x/t. If we try to construct the Picard iterates, we start with x0(t) = 1 thentry to define

x1(t) = x0 +

∫ t

0f(u, x0) du = 1 +

∫ t

0

1

udu , (2.107)

and the integral diverges and x1(t) does not exist. The method of constructing the Picarditerates fails. We see that f(t, x) is not defined for t = 0 and there is no region A around theinitial values (t0, x0) = (0, 1) in which f and fx are defined and continuous. The ivp does notsatisfy the conditions of Picard’s theorem, there is no guarantee that a solution exists andindeed there is no solution.

Example 2.15. Consider the initial value problem in example 2.11

dx

dt= 3x2/3 , x(0) = 0 . (2.108)

In this case we have

f(t, x) = 3x2/3 ,∂f

∂x=

2

x1/3. (2.109)

f is a well behaved function for all x and t but fx is not defined for x = 0, and in particularis not defined in any region Z surrounding the initial values (t0, x0) = (0, 0). If we try toconstruct the Picard iterates, we will actually find the solution x(t) = 0 but the proof thatthis is unique fails.



Note 2: Picard’s theorem is a local existence theorem; it says that Eqs. (2.4), (2.96) have aunique solution sufficiently near t0. It might be thought that if the function f which occurs inEq. (2.4) is well behaved and defined over the whole of IR2 that we could extend our solutionarbitrarily far from t0; for example, we might begin by computing x(t0 + δ1) and use thisvalue of x as an initial value at t = t0 + δ1; application of Picard’s theorem then allows us toextend our solution as far as t0 + δ1 + δ2, for some δ2 > 0 and by repeating the process wecould extend our solution to t = t0 + δ1 + δ2 + δ3 + · · ·+ δn, for some δi > 0 (i = 1, 2, 3, · · ·n).However, this does not prove that the solution could be extended arbitrarily far from t0, asone might find that

tn = t0 +

n∑k=1

δk −→ t∞ <∞ , as n→∞ , (2.110)

exactly as happens in example 2.12.



2.6 Exercises

Shorter Exercises

Exercise 2.1

Classify the following ODEs in terms of the classification given in class [(1) first order, explicit;(2) first order, separable (3) first order, linear; (4) first order, homogeneous; (5) none of these].An equation may belong to more than one class. You are not asked to solve these equations;occasionally, you will need some reshuffling of terms to ‘see’ which class is involved

(a)dx

dt= −10x+ 4 (b)

dy

dx= sin(x)

(c)dx

dt= 2− x2

t2(d) t

dx

dt− 3x = 2

(e)dN

dt+N = Ntet+2 (f)

dx

dt=x3 + t3

2xt2

(g)dy

dx= x5y − sinx (h) tx

dx

dt+ 3x− 4 = 0

(i) y′′ − 4y′ + 24y = 0 (j)dy

dx+ y lnx = e−x lnx

Exercise 2.2

Find the general solutions to the following differential equations:

(a) y′ = x2y2 (b)dx

dt+ x ln t = e−t ln t

(c) xdy

dx= 3 (d) y′ + xy = x

(e) cos(x) y′ − sin(x) y = 4 (f)dy

dx=x

y+y

x

Exercise 2.3

Find the solutions to the following initial value problems

(a)dx

dt=x+ 1

t+ 1with x(0) = 1 (b) x

dy

dx+ y = exp(x) with y(1) = 0

(c)dy

dx= 3x+ 5 with y(0) = 2 (d) x = t2 with x(0) = 1

(e) y′ + y = exp(x) with y(1) = 0 (f)dy

dx− y = y2 with y(0) = 1



Longer Exercises

Exercise 2.4

Find solutions of the initial value problem

dx

dt= x1/3 sin(2t), x(0) = 0

other than the trivial solution x(t) = 0 for all t.Comment on your results in the light of Picard’s theorem.

Exercise 2.5 (taken from the 2008 exam)

Suppose that a population doubles its original size in 100 years, and triples it in 200 years.Show that this population cannot satisfy the Malthusian law of population growth which is

dp(t)

dt= a p(t) with a = constant

Exercise 2.6

Find the general solution to:dx

dt=

x

2t+

t

2x

Show that this defines a family of 2-sided hyperbolae in the (t, x) plane, symmetric about thepoint (C, 0), where C is a parameter, one side always going through the point (0, 0).

Discuss the solutions in the vicinity of the point (0, 0).



Supplementary Exercises

Supplementary Exercise 2.1

We have looked at differential equations where y′(x) is independent of x,

y′(x) = f(x) ,

or a linear function of xy′(x) = f(x) + g(x)y(x) .

The next step would be quadratic functions of y,

y′(x) = f(x) + g(x)y(x) + h(x)y(x)2 .

This called the Riccati equation and cannot be solved in general. There are certain circum-stances in which it can, though, and one is when the “constant term” f(x) is zero; this iscalled the Bernoulli equation:

y′(x) = g(x)y(x) + h(x)y(x)2 .

Make the substitution u(x) = 1/y(x) and show that the resulting equation for u′(x) is linear.

Use this method to solve the initial value problem

y′(x) = y(x) + y2(x) , y(0) = 1 .

Note: this equation can also be solved directly as the equation is separable [problem 2.3(f)]

Supplementary Exercise 2.2

Suppose that we know that there is a relation between y(x) and x, which we can write as

M(x, y(x)) = 0 . (2.111)

We can differentiate this equation with respect to x using the chain rule for partial differen-tiation (see appendix A) to get

∂M

∂x+ y′(x)

∂M

∂y= 0 .

This can also be written as∂M

∂xdx+

∂M

∂ydy = 0 ,

in which case it is known as an “exact” differential equation.

Show that the differential equation

f(x, y) + g(x, y) y′(x) = 0

comes from a relation of the form (2.111) if

∂f

∂y=∂g

∂x.

The function M can then be found by the fundamental theorem of calculus applied to pathintegrals in two dimensions (see chapter 6 in the lecture notes for Calculus II), or by a suitableguess for M . Use this method to solve the initial value problem

(y + 2x) + (2y + x)y′ = 0 , y(0) = 1 .


Chapter 3: Second order differential Equations 31

Chapter 3

Second order Differential Equations

3.1 Second order differential equations

Second order ordinary differential equations have the form

d2x

dt2= g

(t, x,

dx

dt

). (3.1)

The general solution of a second order differential equation is of the form x(t) = h(t, C1, C2),involving two parameters C1, C2. These can be fixed in innumerable ways, for example forsome problems it would be natural to ask that x takes fixed values at two different times,

x(t0) = x0 and x(t1) = x1 , (3.2)

but such conditions are very hard to analyse. Much simpler, and also very relevant in manyproblems, are initial conditions, asking that x and x take particular values at some initialtime t0,

x(t0) = x0 and x(t0) = v0 . (3.3)

We will call the equation (3.1) and the conditions (3.3) an initial value problem for a secondorder differential equation. The initial conditions define a point (t0, x0, v0) ∈ IR3; it is assumedthat g is defined in some neighbourhood of this point, for example in the cube

A = (t, x, v) : |t− t0| ≤ δ, |x− x0| ≤ δ, |v − v0| ≤ δ , for some δ > 0 (3.4)

centred at (t0, x0, v0) and with side-length 2δ.

A second order ODE can always be rewritten in terms of a system of two first order ODEs.This is achieved by setting x1(t) = x(t), x2(t) = x(t) and equation (3.1) then becomes thetwo coupled equations

dx1

dt=

dx

dt= x2,

dx2

dt=

d2x

d t2= g

(t, x,

dx

dt

)= g(t, x1, x2) .

In terms of x1, x2 the initial conditions become x1(t0) = x0, x2(t0) = v0. We may thereforereplace the second order equation (3.1) and their initial conditions by the two coupled first



order differential equations and initial conditionsdx1

dt= x2 ,

dx2

dt= g(t, x1, x2)

,

x1(t0) = x0 ,

x2(t0) = v0 .(3.5)

Equations (3.5) together define an initival value problem for the two coupled first orderdifferential equations. The use of the word coupled reflects the fact that the equations for x1

and x2 do not just depends on x1 and x2 separately.

Note: The above procedure to rewrite a second order ODE in terms of two ODEs of firstorder can be generalised to n-th order ODEs of the form

dnx

dtn= g

(t, x, x(1), x(2) . . . , x(n−1)

), with x(k) ≡ dkx

dtk,

and with initial conditions of the form

x(t0) = a1 , x(1)(t0) = a2 , x

(2)(t0) = a3 , . . . , x(n−1)(t0) = an .

Introducing x1 = x, x2 = x(1), . . . , xn = x(n−1), one obtains the equivalent system of firstorder equations

d

dtx1 = x2

d

dtx2 = x3

......

d

dtxn = g(t, x1, x2, . . . , xn)

with initial conditions xk(t0) = ak, for k = 1, . . . , n.

Suppose that by some means we have found a solution of equations (3.5), x1(t) = φ(t), x2(t) =ψ(t), satisfying the initial conditions, so that φ(t0) = x0, ψ(t0) = v0. If we plot the points(φ(t), ψ(t)) in IR2 we obtain a solution curve which passes through the point (x0, v0) definedby the initial conditions; as we vary (x0, v0) we generate a family of solution curves, eachlabelled by the two parameters x0 and v0.

Second order differential equations (3.1) and systems of two coupled first order differentialequations (3.5) are typically much less likely to be solvable than first order differential equa-tions. We have seen that any linear first order ODE can be solved [just by using integration]by the method of integrating factors: this is not true for second order linear differential equa-tions. The study of such equations could be a module in its own right, and there are bookswritten on special classes of equations. We will just look at one special class which can besolved: second order linear ODEs with constant coefficients

3.1.1 Second order linear, with constant coefficients

These are ODEs of the form

ad2x

dt2+ b

dx

dt+ cx = φ(t) , (3.6)



where a, b, c are constants and φ is a given function of time. Just as with first order linearequations, these can be written in terms of a linear differential operator L:

L(x) = φ , L =

(a

d2

dt2+ b

d

dt+ c

). (3.7)

Homogeneous equations:

We will first consider the special case of the homogeneous equation,

ad2x

dt2+ b

dx

dt+ cx = 0 L(x) = 0 . (3.8)

Linear and homogeneous ODEs have the important properties that (i) a solution x(t) can bemultiplied with an arbitrary constant, and remains a solution; (ii) for any pair x1 and x2 ofindependent solutions (which are not proportional to each other), any linear combination ofthe solutions x1 and x2 with arbitrary constant coefficients is also a solution of 3.8):

L(x1) = 0 and L(x2) = 0 ⇒ L(αx1 + βx2) = 0 for any constants α, β . (3.9)

This means that the solutions of the homogeneous equation form a vector space, the kernelof the operator L. For a second order linear ODE this is a two-dimensional vector space.

Inhomogeneous equations:

The general case isL(x) = φ . (3.10)

If we have two solutions x1 and x2 of the inhomogeneous equation then their difference satisfiesthe homogeneous equation:

L(x1) = φ and L(x2) = φ ⇒ L(x1 − x2) = L(x1)− L(x2) = φ− φ = 0 . (3.11)

Conversely, if we know one solution xpi to the inhomogeneous equation then the combina-tion xpi + xcf is also a solution to the inhomogeneous equation for any solution xcf to thehomogeneous equation:

L(xpi) = φ and L(xcf ) = 0 ⇒ L(xpi + xcf ) = L(xpi) + L(xcf ) = φ+ 0 = φ . (3.12)

Solving a second order linear ODE with constant coefficients:

One can always solve the ODEs (3.6) and (3.8) by simple integration. This is definitely notthe case of linear second order ODEs where the coefficients are not constant, it is a veryspecial feature. There are several methods one can use to find the solutions: they all boildown to the same in the end, they just organise the steps differently.

One way to explain why the fact that the coefficients are constants means the equation canbe solved is that it lets one factorise the operator L into the product of two first order lineardifferential operators. It is simpler if we choose to normalise (3.6) by dividing by a so thatwe are considering the operator

L =d2

dt2+ α

d

dt+ β . (3.13)



We can then write this as a product

L =

(d

dt− λ1

)(d

dt− λ2

)=

d2

dt2− (λ1 + λ2)

d

dt+ λ1λ2 , (3.14)

where λ1 and λ2 satisfyλ1 + λ2 = −α , λ1λ2 = β , (3.15)

which are equivalent to the equation

λ2 + αλ+ β = 0 . (3.16)

This is call the auxiliary or characteristic equation. This helps us because we have now turnedour second order equation into two first order linear equations which we can solve by usingthe method of integrating factors (twice).

As the first step, we write L2(x) = u and so the original equation becomes

L(x) = L1(L2(x)) = L1(u) =

(d

dt− λ1

)u = φ , (3.17)

for which we know the solution:

u = eλ1t[∫ t

e−λ1uφ(u)du+ C1

]. (3.18)

Now, using the expression for u we then have

u = L2(x) =

(d

dt− λ2

)x = eλ1t

[∫ t

e−λ1uφ(u)du+ C1

]. (3.19)

This is again a first order linear ODE which we again solve by the method of integratingfactors to find

x = eλ2t[∫ t

e−λ2ueλ1u[∫ u

e−λ1vφ(v)dv + C1

]du+ C2

]. (3.20)

As expected, this solution depends on two arbitrary constants. Pulling out these two arbitraryconstants, we get the general solution as

x = eλ2t[∫ t

e−λ2ueλ1u[∫ u

e−λ1vφ(v)dv

]du

]+

1

λ1 − λ2C1eλ1t + C2eλ2t , (3.21)

provided λ1 6= λ2. Note that we have not said which of the two solutions of the characteristicequation is which: it will not affect the space of solutions to the problem.

If λ1 = λ2 then instead we get

x = eλt[∫ t [∫ u

e−λvφ(v)dv

]du

]+ C1 t eλt + C2eλt . (3.22)

The functions appearing with the arbitrary constants are the solutions of the homogeneousequation.

There is a simper way to find these as follows:



Solving the homogeneous equation: simpler method

The homogeneous equation is solved with the help of solutions of the so-called auxiliarypolynomial, or auxiliary equation. This method is based on the observation that Eq. (3.8) issolved by functions of the form

x(t) = eλt (3.23)

provided λ is properly chosen. The condition on λ is obtained by inserting the exponentialansatz into (3.8). Using

dx

dt= λ eλt and

d2x

dt2= λ2 eλt (3.24)

one gets (aλ2 + bλ+ c

)eλt = 0 (3.25)

Since eλt 6= 0, we see that eλt is a solution of Eq. (3.8) provided

P (λ) = aλ2 + bλ+ c = 0. (3.26)

Eq. (3.26) is referred to as the auxiliary equation. It is a quadratic equation in λ and as suchthere are three possibilities:

(i) Equ. (3.26) has two distinct real solutions λ1, λ2. It follows that x1 = eλ1t and x2 = eλ2t

are independent solutions of the given differential equation.

The general solution of Eq. (3.8) is now obtained by forming an arbitrary linear combinationof these two solutions: it is

x(t) = αeλ1t + βeλ2t, (3.27)

where α, β are parameters. One can also show that these combinations exhaust the set ofpossibilities (i.e. there are no solutions of a form other than (3.27)).

(ii) The auxiliary equation has two complex conjugate roots λ± = µ ± iν. This means thatx±(t) = e(µ±iν)t are solutions of Eq. (3.8), as are all linear combinations thereof. Now

x±(t) = e(µ±iν)t = eµt(cos νt± i sin νt).

Using special linear combinations of these, we conclude that

x1(t) =x+(t) + x−(t)

2= eµt cos νt and x2(t) =

x+(t)− x−(t)

2i= eµt sin νt (3.28)

are two (independent) solutions of Eq. (3.8). Forming linear combinations of these, we seethat the general solution of Eq. (3.8) is given by

x(t) = eµt(α cos νt+ β sin νt) (3.29)

in this case; here α, β are parameters.

(iii) The third possibility is that Eq. (3.26) has two coincident real roots, λ1,2 = λ, say. Inthis case we get initially only a single solution

x1(t) = α eλt (3.30)



where α is an arbitrary constant. An independent solution is in this case obtained by themethod of varying constants. That is one tries to find an independent solution of the formx2(t) = F (t)eλt, in which the constant α above is replaced by an as yet unknown function F .The function F is determined by inserting this ansatz into (3.8), which gives[

aF + (2aλ+ b)F + (aλ2 + bλ+ c)F]eλt = 0 (3.31)

As the coefficient of F in this equation is P (λ) and thus vanishes, and the coefficient of Fvanishes because λ = −b/2a was assumed to be the unique solution of P (λ) = 0, we are leftwith the condition

F = 0 ⇒ F (t) = α+ β t . (3.32)

The general solution of F = 0 was obtained by two explicit integrations involving two inte-gration constants α and β. Summarising: the general solution of (3.8) in this case is again anarbitrary linear combination of the two independent solutions just found, and can be writtenas

x(t) = (α+ βt) eλt (3.33)

with arbitrary parameters α and β.

Solving the Inhomogeneous equations: method of varying constants

We now consider the inhomogeneous Eq. (3.6) with φ(t) 6= 0.

The first thing to note is the following: Suppose x0(t) is any particular solution of the inho-mogeneous equation. Then a solution of the form

x(t) = x0(t) + xh(t) , (3.34)

with xh(t) a general solution of the corresponding homogeneous equation is also a solutionof the inhomogeneous equation, and this exhausts the possibilities, i.e. there is no solution ofthe inhomogeneous equation which is not of this form. We already know how to find xh(t).So it is sufficient to find a special solution of the inhomogeneous equation.

A particular solution of an inhomogeneous equation can be found by the method of varyingconstants from a solution x1(t) = α eλt of the homogeneous equation. Thus we attempt asolution x0(t) of the inhomogeneous equation of the form x0(t) = F (t)eλt with an unknownfunction F to be determined by inserting the ansatz into (3.6). Following the reasoning incase (iii) above, this gives[

aF + (2aλ+ b)F + (aλ2 + bλ+ c)F]eλt = φ(t) (3.35)

Once more the coefficient of F in this equation , being P (λ), vanishes. So we get

aF + (2aλ+ b)F = φ(t) e−λt . (3.36)

This ODE for F does only involve first and second order derivatives of F and not the functionF itself. This can be used to reduce the order of the ODE. Setting ψ(t) = F (t), one obtains

aψ + (2aλ+ b)ψ = φ(t) e−λt . (3.37)

This is a first order linear equation and we know how to solve it using integrating factors (seeSec 2.4.3 above). Once ψ(t) is obtained along those lines, F (t) follows by a further integrationw.r.t. time t, thus finally allowing to write down x0(t) = F (t)eλt.



Example 3.1. Find a particular solution of the differential equation

d2x

dt2− 9x = t, (3.38)

The characteristic equation is λ2−9 = 0 so that λ = ±3 and the solution to the homogeneous equationis

x = ae3t + be−3t . (3.39)

We can take as a starting point for the method of varying constants the function

x = F (t)e3t , (3.40)

so thatx = (F + 3F )e3t , x = (F + 6F + 9F )e3t , (3.41)

and the equations for F and for ψ = F become

x− 9x = (F + 6F )e3t = t , F + 6F = te−3t , ψ + 6ψ = te−3t . (3.42)

Using the integrating factor exp(6t), this becomes

d

dt(e6tψ) = te3t , (3.43)

which can be integrated directly to give

e6tψ =

∫te3tdt = (

t

3− 1

9)e3t + C , ψ = (

t

3− 1

9)e−3t + Ce−6t , (3.44)

which can again be integrated to give

F =

∫ψdt = − t

9e−3t + C ′e−6t +D , x = F e3t = − t

9+ C ′e−3t +De3t . (3.45)

This method will always work, but quite often one can be lucky enough to find a specialsolution faster by using inspired guesswork.

Solving the Inhomogeneous equations: guessing

The method of guessing is perfectly valid in mathematics: if we try to guess a solution andget it right, then we have solved the problem. A guess is often called a “trial function”or an “ansatz” (plural “ansatze”) to make it sound better This is often the fastest way.If the inhomogeneous term involves trigonometric functions, we can try a combination oftrigonometric functions; if it is a polynomial, we can try a polynomial.


d2x

dt2− 9x = t, (3.46)

Since the inhomogeneous term is a polynomial, we could guess

x = a+ bt+ ct2 , (3.47)

which givesx− 9x = (−9c)t2 + (−9b)t+ (−9a+ 2c) . (3.48)

The solution is clearly a = 0, b = −1/9, c = 0 or

x(t) = −1

9t . (3.49)




d2x

dt2+ x = sin(t), (3.50)

Since the inhomogeneous term is a trigonometric function, we could guess

x = a sin(t) + b cos(t) , (3.51)

but this just givesx+ x = 0 . (3.52)

We can then try a polynomial times a trigonometric function,

x = (at+ bt2) sin(t) + (ct+ dt2) cos(t) , (3.53)

which givesx = (c+ (a+ 2d)t+ bt2) cos(t) + (a+ (2b− c)t− dt2) sin(t) , (3.54)

x = (2b− 2c+ (−a− 4d)t− bt2) sin(t) + (2a+ 2d+ (4b− c)t− dt2) cos(t) , (3.55)

so thatx+ x = (2b− 2c− 4dt) sin t+ (2a+ 2d+ 4bt) cos t , (3.56)

so that the solution is a = b = d = 0, c = −1/2,

x = −1

2t cos(t) . (3.57)

3.2 Existence and Uniqueness of Solutions — Picard’s theorem

Picard’s theorem for this system of equations now takes the form:

Picard’s theorem Let A denote the cube

(t, x1, x2) : |t− t0| ≤ δ, |x1 − a1| ≤ δ, |x2 − a2| ≤ δ, for some δ > 0

and suppose that f1, f2 : A → IR are continuous. Suppose also that the partial derivativesw.r.t. the second and third arguments, ∂fi

∂x1and ∂fi

∂x2for i = 1, 2 are continuous on A. Then

the system of differential equations

dx1

dt= f1(t, x1, x2) ,

dx2

dt= f2(t, x1, x2) , x1(t0) = a1 , x2(t0) = a2

has a unique solution x1 = φ(t), x2 = ψ(t), t ∈ [t0 − δ1, t0 + δ1], for some δ1 ≤ δ.

No full proof of this result will be given here; the underlying ideas are however, the same asin the case of first order equations, relying on convergence of an iterative solution of a systemof integral equations which together are equivalent to the system of ODEs of first order withand their initial conditions. They are obtained by integration of Eqs (3.5) with respect to t,which gives

xi(t) = ai +

∫ t

t0

du fi(u, x1(u), x2(u)) , (i = 1, 2) (3.58)



These equations are solved by means of Picard iteration in full analogy to the first order case.

Define xi,n(t), for i = 1, 2 and n = 0, 1, 2, 3, . . . by

xi,0(t) = ai, (3.59)

xi,n+1(t) = ai +

∫ t

t0

du fi(u, x1,n(t), x2,n(t)) , (n = 0, 1, 2, 3, . . . ) (3.60)

Subject to the stated conditions one can then prove that the sequences of functions x1,n(t),x2,n(t) converge to the unique solutions x1 = φ(t), x2 = ψ(t) as n→∞, provided that δ issufficiently small.

Let’s see by an example how this works out in practice.

Example 3.4. Simple Harmonic Motion: The equation describing simple harmonic motion given by

d2x

dt2+ x = 0

with initial conditions

x(0) = 0 ,dx

dt(0) = 1 .

As above put x1(t) = x(t), x2(t) = dxdt so that

dx1dt

= x2 ,dx2dt

= −x1 ,

with x1(0) = 0 and x2(0) = 1. The corresponding integral equations for x1 and x2 are

x1(t) =

∫ t

0

du x2(u) , x2(t) = 1−∫ t

0

du x1(u) .

The Picard iterates are defined by

x1,n+1(t) =

∫ t

0

du x2,n(u) , x1,0(t) = 0 ,

x2,n+1(t) = 1−∫ t

0

du x1,n(u) , x2,0(t) = 1 .

Setting n = 0, 1, 2, 3, . . . in turn we find that

x1,1(t) =

∫ t

0

du x2(0, u) =

∫ t

0

du = t ,

x2,1(t) = 1−∫ t

0

du x1(0, u) = 1−∫ t

0

du 0 = 1 ,

x1,2(t) =

∫ t

0

du x2,1(u) =

∫ t

0

du 1 = t ,

x2,2(t) = 1−∫ t

0

du x1,1(u) = 1−∫ t

0

du u = 1− t2

2!,

x1,3(t) =

∫ t

0

du x2,2(u) =

∫ t

0

du(

1− u2

2!

)= t− t3

3!,

x2,3(t) = 1−∫ t

0

du x1,2(u) = 1−∫ t

0

du u = 1− t2

2!,



x1,4(t) =

∫ t

0

du x2,3(u) =

∫ t

0

du(

1− u2

2!

)= t− t3

3!,

x2,4(t) = 1−∫ t

0

du x1,3(u) = 1−∫ t

0

du(u− u3

3!

)= 1− t2

2!+t4

4!,

x1,5(t) =

∫ t

0

du x2,4(u) =

∫ t

0

du(

1− u2

2!+u4

4!

)= t− t3

3!+t5

5!,

and so on. The pattern is clear, and we see the Maclaurin series for sin(t) appearing for x1,n(t) asn→∞, so

x1,n(t)→ x(t) = t− t3

3!+t5

5!− t7

7!+t9

9!− · · · = sin(t) ,

which is the solution of the differential equation satisfying the given initial conditions.



3.3 Exercises

Shorter Exercises

Exercise 3.1Find the general solutions to the following differential equations

(a) y′′ + 3y′ + 2y = 0

(b) y′′ + y′ + y = 0

(c) y′′ + y = exp(x)

(d) y′′ − y = exp(−x)

Exercise 3.2Find the solutions to the following initial value problems

(a) y′′ + y′ + y = 0 with y(0) = 0, y′(0) = 1

(b) y′′ − 3y′ + 2y = x with y(0) = 0, y′(0) = 0

(c) y′′ − 2y′ + y = x with y(0) = 0, y′(0) = 0

(d) y′′ + 4y = sin(x) with y(0) = 0, y′(0) = 0

Longer Exercises

Exercise 3.3 (taken from the 2008 exam)Three solutions of a certain 2-nd order inhomogeneous linear differential equation, with con-stant coefficients are

ψ1(t) = t ψ2(t) = t+ et ψ3(t) = 1 + t+ et.

Find the general solution of the inhomogeneous differential equation.

Exercise 3.4The motion of an idealised pendulum is described by

y + 2y = 0 ,

in which y denotes the angle of deviation from the vertical orientation. Solve this differentialequation, with initial conditions given by

y(0) =π

12, y(0) = 0 .

Suppose there is now friction which dampens the motion, so that it is described by

y + 2y = −2y .

Solve this differential equation with the same initial conditions.



Exercise 3.5The distance d(t) that a parachutist falls satisfies the equation

d = g − αd ,

where g > 0 is the acceleration due to gravity and α > 0 represents the effects of air resistancedue to the parachute.

Solve this equation subject to the initial conditions

d(0) = 0 , d(0) = 0 .

You may find it easier to first solve for the speed, v(t) = d(t) which satisfies

v = g − αv .

What is the limiting speed (terminal velocity) of the parachutist?

How could you predict this without solving the equation?

Exercise 3.6Consider the initial value problem

x+ 3x+ 2x = 1 + t , x(0) = 0 , x(0) = 0 .

Solve this problem using the method of varying constants with trial functions

(i)x(t) = F (t)e−t , (ii)x(t) = F (t)e−2t .

Check that both methods give the same solution.




Supplementary Exercise 3.1If y1 and y2 are functions, then we can define a combination W (x) called the Wronskian asthe determinant

W = det

(y1 y2

y′1 y′2

)= y1y

′2 − y2y

′1 (3.61)

Suppose that y1 and y2 both satisfy the differential equation

y′′(x) + a(x)y′(x) + b(x)y(x) = 0 , (3.62)

(a) Show thatdW (x)

dx+ a(x)W (x) = 0 . (3.63)

This means that W (x) can be found very easily by the method of integrating factors as

W (x) = exp(−∫a(x)dx) . (3.64)

Suppose that Y (x) is known to be a solution of (3.62)(b) show that any other solution satisfies

y′ − Y ′

Yy =

W

Y. (3.65)

Consider the ODE

y′′ +

(−1− 1

x

)y′ +

1

xy = 0 (3.66)

(c) Show that exp(x) is a solution of equation (3.66).

(d) Use the Wronskian method to find a second solution of (3.66).

Supplementary Exercise 3.2The Riccati equation from problem 2.1 cannot always be solved, but it can always be turnedinto a second order linear differential equation which may make its analysis easier.Recall that the general Riccati equation is

y′(x) = f(x) + g(x)y(x) + h(x)y(x)2 .

(a) Show that the substitution

y = − u′

hu,

turns the non-linear first order equation for y(x) into a second order linear differential equa-tion for u(x).Any solution u(x) of this second order linear differential equation will give a solution y(x) ofthe non-linear Riccati equation.(b) Apply this method to find the second order linear differential equation for u correspond-ing to the Riccati equation

y′ = y + y2 ,

in problem 2.1 and show that it leads to the same solution.(c) How can it be that the two parameter families of solutions to the second order equationfor u leads to a one parameter family of solutions for y?


Part II

Dynamical Systems

44

Chapter 4: Introduction to Dynamical Systems 45

Chapter 4

Introduction to Dynamical Systems

First we give a definition of a dynamical system as we will consider it in this course.

Definition 4.1. Dynamical System (DS)

A DS is any system described by a set of variables x1, . . . , xn which depend on time t andsatisfy a set of ordinary differential equations (ODEs) of the form

d

dtxi = fi(t, x1, . . . , xn) , i = 1, . . . , n . (4.1)

The functions fi are called velocity functions. If we write the variables as a vector, thenits velocity f is given by the velocity functions,

x =

x1

x2...xn

,d

dtx = f(t,x) =

x1

x2...xn

=

f1

f2...fn

. (4.2)

If the variables x take a value x(t1) = a at time t1 then we say that the system is in the statea. The possible allowed values for the variables x is called the phase space

Definition 4.2. Phase Space ΓThe phase space Γ is defined as the set of admissible values for x. E.g., if x = (x) describessome population density, or the concentration of a chemical substance, then it cannot benegative and so Γ = IR+ = x ∈ IR|x ≥ 0. If x = (x, y) describe the percentages of apopulation of people who have either never caught a disease or who have recovered from it,then each of these must be between 0 and 100 and together cannot add up to more than 100 sothat the phase space is the triangular set Γ = (x, y)|0 ≤ 0 ≤ 100, 0 ≤ y ≤ 100, x+ y ≤ 100.We could even say that x is an angle on a circle so that the phase space is Γ = x|x ≡ x+2πwhere the angle x is identified with the angle x+ 2π.



We next define the order of a DS and whether it is autonomous or not autonomous:

Definition 4.3. Order

If the description involves n variables, the dynamical system is said to be of order n.

Definition 4.4. Autonomous and Non-Autonomous Systems

If the time t does not explicitly appear in the velocity functions fi, the system is said to beautonomous, otherwise it is non-autonomous. Thus for autonomous systems, we have

d

dtx = f(x) (4.3)

instead of (4.2). (Physically, an autonomous system is a system in which the laws governingit are time-independent, or one where there are no time-dependent ‘external’ perturbationsof the dynamics). .

An n-th order non-autonomous system is equivalent to an (n+1)-the order autonomous systemif we just introduce an extra variable x0 which is equal to t,

x0 = t ,d

dtx0 = 1 , (4.4)

so that (4.1) becomes

d

dtxi = fi(x0, x1, . . . , xn) , i = 0, . . . , n . (4.5)

with f0(x0, x1, . . . , xn) = 1.

Since we can always turn a non-autonomous system into an autonomous system (of higherorder) we will almost always deal simply with autonomous systems in this course. We treatfirst order autonomous DS in chapter 5 and second order autonomous DS in chapter 6. Firstorder systems are very simple, second order systems are more complicated but their behaviourcan still be characterised fairly easily; Third order systems can be much more complicated.

There are two useful ways to think about the way a system evolves in time. For example, if wewant to think how someone moves around in two-dimensions (e.g. walking around London),we can plot a graph of their position against time. Alternatively, we can just trace on amap where they went and in which order. The first option keeps all the information there is,including the speed they are going at; second option is easier to draw (it is a line on a map)and sometimes that is all we want to know about. This leads to the following two sets ofdefinitions:



Definition 4.5. Trajectories, Flow

The set of points (t,x(t)); t ∈ IR which solve the ODEs (4.2), and for which x(t0) = x0

is called the trajectory or solution curve of the DS passing through x0. The set of alltrajectories obtained by varying t0 and x0 through all physically allowed values is called theflow of the DS.

Definition 4.6. Orbits, Phase-flow, Phase portrait

The set of points x(t); t ∈ IR which solve the ODEs (4.2), and for which x(t0) = x0 is calledthe orbit of the DS passing through x0. We think of the DS moving along an orbit. The setof all orbits obtained by varying t0 and x0 through all physically allowed values is called thephase-flow of the DS. A picture which gives an illustrative selection of the orbits and thedirection of motion is called a phase portrait

Note the difference between phase-flow and flow; the flow contains more dynamical informa-tion than the phase-flow, but the phase-flow is easier to visualise as it is one dimension smallerand we will often use phase portraits to understand the behaviour of dynamical systems. Togive an example, we can think about the DS with the velocity functions

x = − y

x2 + y2, y =

x

x2 + y2, (4.6)

which describe circular motion where the speed depends on the distance from the origin: thefurther out, the slower the motion. The flow is the collection of trajectories (t,x(t)) whichare in IR3; the phase-flow is the collection of orbits which are in IR2; these are shown in figure4.1.

-1.0 -0.5 0.0 0.5 1.0

-1.0

-0.5

0.0

0.5

1.0

Phase-Flow

Figure 4.1: Sketches of the flow (in IR3) and a phase portrait illustratingthe phase-flow (in IR2) for the simple 2nd order autonomous system (4.6).



The phase portrait of the phase-flow shows directly that the orbits are circles, the flow showsthat the motion nearer the centre is faster but makes it harder to see that the orbits are reallycircles.

This example also shows why orbits get their name: we think of planets moving along orbits,which are curves in space.

It is obvious that if the system starts at some point on a circle then it will stay in the samecircle for all time. This is an example of an invariant set.

Definition 4.7. Invariant SetAn invariant set S is a set such that if x(t0) ∈ S then x(t) ∈ S for all t

Particularly important examples of invariant sets are when the set is just a single point, inwhich case it is called a fixed point or an equilibrium point

Definition 4.8. Fixed Point (FP) or Equilibrium point

A point a = (a1, . . . an)t is called a fixed point of the (autonomous) DS if

fi(a) = 0 ⇔ d

dtxi = 0 at x = a , i = 1, . . . , n .

A system which is at a FP will stay there forever, unless perturbed.

It is also very important to know whether a fixed point is stable or not: will motion thatstarts near a fixed point stay near the fixed point for all time or can it move away?

Definition 4.9. Stability of a FP

• A FP a of a DS is called strongly stable, if all trajectories starting (sufficiently) closeto a will approach a under the dynamics.

• A FP a of a DS is called unstable, if there are trajectories starting close to a whichevolve away from a under the dynamics. (Note that in n-th order DS, mixed situationsexist, i.e. some subset of the trajectories starting sufficiently close to a will approacha, whereas there exist others which will evolve away from a, no matter how close to a

they start.

• A FP a of a DS is called (marginally) stable, if all trajectories starting (sufficiently)close to a will neither approach a under the dynamics, nor will they evolve away fromit, but rather ‘keep circling around’.

Fig. 4.2 illustrates the definitions for 2nd order autonomous systems:



Figure 4.2: Illustration of the possible flows in the vicinity of a FP. From left to right: stronglystable, unstable, unstable (mixed), marginally stable.

We can give a formal definition of a stable fixed point and a strongly stable fixed point interms of ε and δ but will rarely if ever use this:

Definition 4.10. Stability of a FP′

A FP a of a DS is called stable, if for any ε > 0 we can find a δ > 0 such that ||x(0)−a|| < δimplies that ||x(t)− a|| < ε for all time. i.e. the motion never moves away from the FP.

A FP a of a DS is called strongly stable, if for any δ > ε > 0 we can find a time T suchthat ||x(0) − a|| < δ implies that ||x(t) − a|| < ε for all times t > T , i.e. the motion mustmove in towards the FP

These two ideas are captured in figure 4.3

Figure 4.3: Illustration of the formal definitions of stability and strong stability.On the left, stability means that whatever larger circle we define, there is a smaller circle suchthat every flow starting inside the smaller circle stay insides the large circle for all time.On the right, strong stability means that whatever smaller and larger circles we define, thereis a time after which every motion starting inside the larger circle remains inside the smallercircle.

We now turn to 1st order autonomous DS and consider how to characterise their dynamics,find the fixed points and determine their stability and decide if a flow carries on for all timeor if it terminates after a finite time.


Chapter 5: First Order Autonomous Systems 50

Chapter 5

First Order Autonomous Systems

5.1 Trajectories, orbits and phase portraits

In this chapter we look in detail at first order autonomous dynamical systems. The state attime t of a first order autonomous dynamical system is described by one variable x(t) (whichwe will assume is in IR unless we say otherwise) and which satisfies a differential equation ofthe form

dx

dt= f(x). (5.1)

The velocity function f(x) does not depend on t because the system is autonomous. Thismeans that the differential equation (5.1) is separable and so we can immediately write downthe solution as an integral:

F (x) =

∫ x

x0

dx

f(x)=

∫ t

t0

dt = t− t0 . (5.2)

The trajectories of the dynamical system are the solutions x(t) which pass through the points(t0, x0).

The fixed points of a first order autonomous system (5.1) are values of x for which f(x) = 0.

The phase portrait is a sketch of the phase space of the DS (which will be the real line orsome subset) together with arrows indicating the direction of the motion along the orbit. Thedirection is to the right if x is increasing, i.e. if f(x) > 0 and to the left if x is decreasing,i.e. f(x) < 0. It is usual to combine a plot of f(x) with the phase portrait. It is thenstraightforward to read off from the phase portrait whether any given fixed point is stable orunstable.

Sometimes we perform the integration and solve to find x(t) explicitly, sometimes we can onlyfind an implicit definition of x(t) and sometimes we cannot even perform the integral exactly– but often we do not need to perform the integral exactly to be able to work our properties ofthe dynamical system. Most of the rest of this chapter is on how to deduce properties of thedynamical system and its trajectories without even necessarily being able to do the integral.

Let us first consider a few simple examples to illustrate the concepts of trajectories, orbitsand phase portraits.



Example 5.1. Sketch the trajectories and phase portrait for the DS with phase space the realline and velocity function

x = f(x) = 1 . (5.3)

This differential equation has the general solution

x = a+ t , (5.4)

and so the trajectories are straight lines. There are no fixed points. The orbits are the valuesof x which a trajectory passes through, and these are the whole real line. To complete thephase portrait we indicate the direction of flow along the orbit by sketching arrows to theright. It is also often convenient to plot the function f(x) along with the phase portrait.

-2 -1 1 2t

-1

1

2

3

x(t)

x

1

f(x)

Figure 5.1: Sketch of the trajectories (5.4) for the DS (5.3), x = 1, togetherwith a sketch of the velocity function f(x) = 1 and the phase portrait.

Example 5.2. Sketch the trajectories and the phase-portrait for the DS with phase space thereal line and velocity function

x = f(x) = r (x− 1) . (5.5)

This has the general solutionx = 1 + x0 exp(r t) . (5.6)

There is now one fixed point, x = 1 at which f(1) = 0. The orbits are now the sets x|x < 1,x = 1 and x|x > 1 each of which is an invariant set. The nature of the trajectories andof the fixed point depends on the sign of r. If r > 0 then the trajectories move away from thepoint x = 1 and fixed point is unstable; if r < 0 then the trajectories move in towards fromthe point x = 1 and fixed point is stable. We show the flow (comprised of the trajectories)and the phase portraits (i.e. the phase-flow comprised of the orbits together with a plot ofthe velocity function) in figures 5.2 (for r > 0) and 5.3 (for r < 0)



-2 -1 1 2x

-1

1

2

3

t

0

x

f(x)

Figure 5.2: Sketch of the trajectories (5.6) for the DS (5.5), x = r(x− 1),for r > 0 together with a sketch of the velocity function f(x) = r (x − 1)and the phase portrait.

-2 -1 1 2x

-1

1

2

3

t

0

x

f(x)

Figure 5.3: Sketch of the trajectories (5.6) for the DS (5.5), x = r(x− 1),for r < 0 together with a sketch of the velocity function f(x) = r (x − 1)and the phase portrait.

Finally, a slightly more complicated example with several fixed points for which we can findthe general solution but for which the trajectories are not easy to find given the exact solution.

Example 5.3. Sketch the trajectories and the phase-portrait for the DS with phase space thereal line:

x = f(x) = (x− 1)(x− 2)(x− 3) . (5.7)

In this case the integral (5.2) can be found exactly but is not very informative: explicitly:

F (x) =

∫ x

x0

dx

(x− 1)(x− 2)(x− 3)

=

∫ x

x0

(1/2

x− 1− 1

x− 2+

1/2

x− 3

)dx

=

[1

2log |x− 1| − log |x− 2|+ 1

2log |x− 3|

]xx0



=1

2log

∣∣∣∣ (x− 1)

(x0 − 1)

(x0 − 2)2

(x− 2)2

(x− 3)

(x0 − 3)

∣∣∣∣= t− t0 (5.8)

In this case to sketch the trajectories it is much easier first to sketch the velocity functionand the phase portrait:

We can easily plot the velocity function:

1 2 3 4

-1.0

-0.5

0.5

1.0

Figure 5.4: Sketch of f(x) = (x− 1)(x− 2)(x− 3).

The velocity of the system when it is at x is given by f(x). This function is negative for x < 1and 2 < x < 3, zero at x = 1, 2, 3 and positive for 1 < x < 2 and x > 3. This means that thesystem is moving to the left if x < 1 or 2 < x < 3, it is stationary if x = 1, 2, 3 and is movingto the right if 1 < x < 2 or x > 3. If we sketch the real line and include arrows showing whichway the system is moving and show where the fixed point are, we get the phase portrait ofthe system:

1 2 3

Figure 5.5: Phase portrait of the dynamical system (5.7).

It is actually more common and also more useful to combine the graph of f(x) and the phaseportrait, so from now on we will always do this when we sketch the phase portrait. Here aretwo ways we can do this, putting the arrows on the x–axis as in figure 5.5 or displacing themabove and below to make it easier to see what is going on:



1 2 3

f(x)

1 2 3

f(x)

Figure 5.6: Phase portraits of the dynamical system (5.7).

Looking at the phase portrait in figure 5.6, we can see that the fixed points are at the zeroesof the velocity function f(x), and we can also see that x = 1 and x = 3 are unstable (theflows nearby move away) and that the fixed point at x = 2 is stable (the flows nearby movetowards it). We can also see that if the system starts anywhere inside the interval [1, 2] thenit cannot escape it.

Now that we have the phase portrait, it is easy to sketch the trajectories, if we want to:the trajectories through the fixed points are straight lines; the other trajectories move in thedirection shown by the arrows on the phase portrait. The only question is: do the trajectoriesthat head towards the fixed points or ±∞ actually reach these points in a finite time? Willthe flow starting at x = 3/2 reach the point x = 2? Will the flow starting at x = 7/2 reachx =∞? These are very important questions and the subject of the next section.

-1 0 1 2 3 4

1.0

1.5

2.0

2.5

3.0

Figure 5.7: A sketch of the flow, or the set of trajectories of the dynamicalsystem (5.7).



5.2 Termination of Motion

If a system has a stable fixed point, and initial conditions are such that the fixed point is beingapproached, one may ask whether the system will reach the fixed point in finite time. Thesame may be asked about ‘escaping’ to ±∞. These considerations lead us to the followingdefinition.

Definition 5.1. Terminating MotionThe motion of a dynamical system is said to be terminating, if it reaches a fixed point or goesto ±∞ in finite time.

Here are two elementary examples

Example 5.4. Consider the DS with phase space x ≥ 0 given by

dx

dt= f(x) , f(x) = −xα , α > 0, . (5.9)

Sketch the phase portrait and find the values of α for which the motion starting at t = 0 atx0 > 0 terminates at x = 0.

The velocity function f(x) is always negative and f(0) = 0 if α > 0, leading to the simplephase portrait in figure 5.8 The differential equation is separable and so we can immediatelyintegrate it to find

t =

∫ t

0dt =

∫ x(t)

x0

dx

f(x)= −

[1

1− αx1−α

]x(t)

x0

=1

1− α((x0)1−α − x(t)1−α) , (5.10)

or solving for x(t),

x(t) =[−(1− α)t+ (x0)1−α]1/(1−α)

. (5.11)

To see if the motion terminates, we could look at the solution (5.11) but it is more direct tosee for which values of α the integral (5.10) is finite as x(t)→ 0. The answer is that the timeto reach x = 0 is finite if 1−α > 0 i.e. α < 1. (The case α=1 is that of exponential approachwhich we already know does not terminate.)We illustrate this in figure 5.8 with plots of f(x) and x(t) in the cases α = 1/2, 1 and 2.

0.5 1.0 1.5 2.0 x

-3.0

-2.5

-2.0

-1.5

-1.0

-0.5

f(x)

(a) The velocity functions andphase portrait

0.0 0.5 1.0 1.5 2.0 x

1

2

3

4

t

(b) The trajectories

Figure 5.8: Plots of the velocity function and solutions of the DS (5.4) for α = 1/2 (solidline), α = 1 (dashed) and α = 2 (dotted) in the case x0 = 2



Example 5.5. Consider the DS with phase space x ≥ 0 given by

dx

dt= f(x) , f(x) = xα , . (5.12)

Find the values of α for which the motion starting at t = 0 at x0 > 0 terminates at x =∞.

The velocity function f(x) is now always positive. The differential equation is again separableand so we can immediately integrate it to find

t =

∫ t

0dt =

∫ x(t)

x0

dx

f(x)=

[1

1− αx1−α

]x(t)

x0

=1

1− α(x(t)1−α − (x0)1−α) . (5.13)

To see if the motion terminates, we see for which values of α the integral (5.13) is finite asx(t)→∞. The answer is that the time to reach x =∞ is finite if 1− α < 0 i.e. α > 1. (Thecase α=1 is that of exponential approach which we already know does not terminate.) Notethat this is exactly the opposite condition to the previous example.

Whether a motion is terminating or not can depend on the starting point of the motion. Asimple example showing this is the logistic equation (1.3) with the solutions (1.4).

Example 5.6. For which values of x0 does the motion given by the logistic equation and itsgeneral solution terminate?

dx

dt= r x

(1− x

c

), x =

c

1−Ae−r t, A = 1− c

x0(5.14)

If the motion starts at either of the two fixed points x = 0 or x = c it stays there for all time;by definition these motions terminate. If the motion starts at a point 0 < x0 < c or x0 > cthen it will not terminate: as t increases then the motion will tend towards x = c but willnot reach it in finite time; if however the motion starts at x0 < 0 then it will diverge to −∞at the finite time

t =1

rlog

(1− c

x0

)(5.15)

In the example of the logistic equation we can find the solution exactly, but we can also usuallyanswer the question without needing to solve the dynamical system. Given a dynamicalsystem described by (5.1),

dx

dt= f(x).

and initial condition x(t0) = x0, the time τ01 needed for the system to reach x1 can in principlebe found from (5.2):

τ01 = t1 − t0 =

∫ t1

t0

dt =

∫ x1

x0

dx

f(x). (5.16)

We have to be slightly careful about signs since the flow can be moving to the right (f(x) > 0,x1 > x0) or moving to the left (f(x) < 0, x1 < x0) but in both cases the integral will be apositive number.



To find out if a motion starting at x0 terminates at x1 in finite time all we have to do isfind out if the integral τ01 is finite (the motion terminates) or is infinite (the motion does notterminate). Sometimes we can calculate the integral exactly, sometimes we cannot and needto find other methods

To answer the questions posed earlier: will the flow for the system in example 5.3 starting atx = 3/2 reach x = 2? Will the flow starting at x = 7/2 reach x = ∞? — we can calculatethese integral exactly.

Example 5.7. Find the time it takes to go from x = 3/2 to a position ξ < 2:

τ01 =

∫ ξ

3/2

dx

(x− 1)(x− 2)(x− 3)

=

∫ ξ

3/2

(1/2

x− 1− 1

x− 2+

1/2

x− 3

)dx

=

[1

2log |x− 1| − log |x− 2|+ 1

2log |x− 3|

]ξ3/2

=1

2log |ξ − 1| − log |2− ξ|+ 1

2log |3− ξ| − 1

2log 3 . (5.17)

As ξ → 2, the term (− log |2− ξ|)→∞; the integral diverges as ξ → 2; the time it takes forthe motion to reach x = 2 is infinite; in other words, the flow does not reach x = 2 in finitetime, the flow does not terminate.

Let’s now find the time it takes to go from x = 7/2 to a position ξ > 7/2:

τ01 =

∫ ξ

7/2

dx

(x− 1)(x− 2)(x− 3)

=

∫ ξ

7/2

(1/2

x− 1− 1

x− 2+

1/2

x− 3

)dx

=

[1

2log |x− 1| − log |x− 2|+ 1

2log |x− 3|

]ξ7/2

=1

2log

∣∣∣∣(ξ − 1)(ξ − 3)

(ξ − 2)2

∣∣∣∣+1

2log(9/5) . (5.18)

As ξ → ∞, the term (ξ − 1)(ξ − 3)/(ξ − 2)2 → 1; the integral is finite as ξ → ∞; the timeit takes for the motion to reach x = ∞ is finite; in other words, the flow reaches x = ∞ infinite time 1/2 log(9/5) = 0.293893.., the flow terminates.

It is important to realise that it is only the motion very close to the fixed point that is relevant:the flow will always reach any point near the fixed point; the question is: will it be able toget that final extra part of the way from close to the fixed point right to the fixed point.



This is not always the case: one can think up velocity functions with the same phase diagramas figure 5.6 but for which the motion starting at x = 3/2 does terminate and the motionstarting at x = 7/2 does not, for example

Example 5.8. Consider the DS

f(x) = [(x− 1)(x− 2)(x− 3)]1/3 , τ01 =

∫ 2

3/2

dx

f(x)= 0.966897... (5.19)

1 2 3

Figure 5.9: Phase portrait for the DS (5.19).

The difference between the two DS, (5.7) and (5.19) is how the velocity function approacheszero near x = 2 and how it approaches ∞ as x → ∞. We shall some time working out howbest to decide if a motion terminates or not.

There is one criterion we already have which enables us to prove that under certain circum-stances a motion cannot reach a fixed point in a finite time and that is Picard’s theorem.

Corollary to Picard’s Theorem: I

Suppose that x = a is a fixed point of the DS

x = f(x) (5.20)

and that f(x) and f ′(x) are both continuous at x = a, then the only trajectory that terminatesat x = a in finite time is the constant trajectory x(t) = a.

Proof: suppose there was a second trajectory ξ(t) which satisfied xi(t0) = a. Then Picard’stheorem says the solution to the differential equation (5.20) satisfying x(t0) = a is unique.But x(t) = a is such a solution, hence ξ(t) = x(t) = a. Hence no trajectories that approachx = a can ever reach x = a in finite time apart from the constant trajectory.



Corollary to Picard’s Theorem: II

Consider the DS defined on the real line x = f(x) and suppose that f(x) and f ′(x) arecontinuous on the whole real line. Then the orbits of the system consist of fixed points andopen intervals. The invariant sets are unions of fixed points and open intervals; equivalently,the phase spaces is the union of fixed points and open invariant sets, that is open intervalswhich are invariant sets that contain no fixed points.

Proof: consider a point x = a. Consider the trajectory with x(t0) = a. By Picard’s theoremthis is unique. The orbit of this trajectory is either the point a (a is a fixed point) or an openinterval (since the motion cannot terminate). Hence all orbits are either fixed points or openintervals, which are also called open invariant sets.

We can check the examples we have looked at so far.

Example 5.3. In this case

f(x) = (x− 1)(x− 2)(x− 3) , f ′(x) = 3x2 − 12x+ 11 , (5.21)

are both continuous at x = 2 and so any motion that approaches x = 2 cannot reach x = 2in finite time, as we showed in example 5.7.

In this case the phase space (the real line) is the union of the fixed points 1, 2, 3 andthe open invariant sets (−∞, 1), (1, 2), (2, 3), (3,∞).

Example 5.8. In this case f(x) is continuous for all x but

f ′(x) =1

3

((x− 2)(x− 3)

(x− 1)2

)1/3

+1

3

((x− 1)(x− 2)

(x− 2)2

)1/3

+1

3

((x− 1)(x− 3)

(x− 3)2

)1/3

, (5.22)

diverges at x = 2. This means that it may be possible for a flow to terminate at x = 2: it isnot ruled out by Picard’s theorem.

In this case the open intervals (1, 2) and (2, 3) are not invariant sets since a trajectory startingin (1, 2) will reach the point x = 2 in finite time [hence leaving the open interval] and likewisea trajectory starting in (2, 3) will also reach the point x = 2 in finite time [again leaving theopen interval].



5.3 Estimating times of Motion

Estimates for τ01 can be obtained without giving the full solution of the ODE describing thedynamical system in question, i.e., without actually doing the x-integral in (5.16) exactly,using the following observation: If φ and ψ are real valued functions defined on an interval[a, b] such that

φ(t) ≥ ψ(t) ∀ t ∈ [a, b],

then ∫ b

aφ(t) dt ≥

∫ b

aψ(t) dt .

In terms of a dynamical system, this says that if one dynamical system moves from a to b ata faster speed then a second, it will take less time; if it moves slower then it will take moretime.

To be concrete, let us consider two velocity functions f(x) and g(x) defined on an interval[a, b] such that f(x) ≥ g(x) > 0 on (a, b). The the time taken for the two different systems tomove from a to b are

τf =

∫ b

a

dx

f(x), τg =

∫ b

a

dx

g(x). (5.23)

Since f(x) ≥ g(x) then 1/f(x) < 1/g(x) and so

τf ≤ τg . (5.24)

This is illustrated in figure 5.10

f(x)

g(x)

a bx

(a) Two velocity functionsf(x) > g(x)

x= f(x)

x= g(x)

a bt

τf

τg

x

(b) The trajectories for the twovelocity functions; the fastersystem reaches b first.

1

f (x)

1

g (x)

a bx

(c) The times taken to movefrom a to b are the areas underthe graphs of 1/f and 1/g.

Figure 5.10: Sketches illustrating the time of motion for different velocity functions.

Using these elementary facts to estimate times of motion and determine if a motion terminatesor not is a bit of an art. Given a velocity function f(x), there is an infinite choice of functionsh(x) a g(x) such that h(x) > f(x) > g(x); some will be helpful, some not. We will look atthis in the next few examples.



Example 5.9. Find a lower bound on the time taken for the DS 5.3 to move from a point3 < x0 to a point x1. Use this to find lower estimates of the time T to move from x0 = 4 tox1 =∞.

For x0 > 3, we have f(x0) > 0 and so we simply have

τ01 =

∫ x1

x0

dx

f(x)=

∫ x1

x0

dx

(x− 1)(x− 2)(x− 3)(5.25)

To find a lower bound on the time taken, we want to find a speed h(x) which is faster thanf(x) so that h(x) > f(x), 1/h(x) < 1/f(x) and we find bounds

τh =

∫ x1

x0

dx

h(x)<

∫ x1

x0

dx

f(x)= τ01 (5.26)

The idea is to find a function h(x) which gives a simpler integral. Since x > 3, we can seelook at each of the factors in f(x) and find simple inequalities

(a) 2 < (x− 1) (5.27)

(b) 1 < (x− 2) (5.28)

(c) 0 < (x− 3) (5.29)

(d) (x− 3) < (x− 2) < (x− 1) (5.30)

(i) The first three are not very helpful, but we can use (d) to obtain

(x− 1)(x− 2)(x− 3) < (x− 1)3 , (5.31)

and so

τ01 >

∫ x1

x0

dx

(x− 1)3=

1

2

(1

(x0 − 1)2− 1

(x1 − 1)2

). (5.32)

This is true and helpful. As x→∞ this is finite,

τ01 >1

2(x0 − 1)2. (5.33)

and so with x0 = 4 we get a lower estimate

T >1

18= 0.0555... . (5.34)

(ii) As another choice, we could instead have used (x− 1) < x to get

(x− 1)(x− 2)(x− 3) < x3 , (5.35)

and so

τ01 >

∫ x1

x0

dx

x3=

1

2

(1

(x0)2− 1

(x1)2

). (5.36)

This is again true and again helpful but the bound is not as good. In the limit x1 →∞we find

τ01 >1

2(x0)2, (5.37)

leading to

T >1

32= 0.03125 , (5.38)

which is smaller than the previous estimate.



Example 5.10. Find an upper bound on the time taken for the DS 5.3 to move from a point3 < x0 to a point x1. Use this to show that the motion reaches x = ∞ in finite time. Findan upper estimate for the time to reach x =∞ from x0 = 4,

For x0 > 3, we have f(x0) > 0 and so we simply have

τ01 =

∫ x1

x0

dx

f(x)=

∫ x1

x0

dx

(x− 1)(x− 2)(x− 3)(5.39)

To find an upper bound on the time taken, we want to find a speed g(x) which is slower thanf(x) so that f(x) > g(x), 1/f(x) < 1/g(x) and we find the bound

τ01 =

∫ x1

x0

dx

f(x)<

∫ x1

x0

dx

g(x)= τg (5.40)

The idea is to find a function g(x) which gives a simpler integral. Since x > 3, we can seelook at each of the factors in f(x) and find simple inequalities

(a) 2 < (x− 1) (5.41)

(b) 1 < (x− 2) (5.42)

(c) 0 < (x− 3) (5.43)

(d) (x− 3) < (x− 2) < (x− 1) (5.44)

(i) We can use (a) and (b) to obtain

(x− 1)(x− 2)(x− 3) > 2 · 1 · (x− 3) ≡ g1(x) , (5.45)

and so

τ01 <

∫ x1

x0

dx

2(x− 3)=

1

2lnx0 − 3

x1 − 3. (5.46)

As x1 →∞, this lower bound diverges, so while this inequality is true it is not helpful.

(ii) We could instead have used (d) twice to find

(x− 1)(x− 2)(x− 3) > (x− 3)3 ≡ g2(x) , (5.47)

and so

τ01 <

∫ x1

x0

dx

(x− 3)3=

1

2

(1

(x0 − 3)2− 1

(x1 − 3)2

). (5.48)

This is also true but now is helpful. As x→∞ this is finite, so we have found

τ01 <1

2(x0 − 3)2, (5.49)

and with x0 = 4 we get the upper estimate

T <1

2= 0.5 . (5.50)



(iii) Finally, as another choice, we could use (b) and (d) [once] to get

(x− 1)(x− 2)(x− 3) < (x− 3) · 1 · (x− 3) = (x− 3)2 ≡ g3(x) , (5.51)

and so

τ01 <

∫ x1

x0

dx

(x− 3)2=

(1

(x0 − 3)− 1

(x1 − 3)

). (5.52)

This is again true and again helpful. As x→∞ this is also finite,

τ01 <1

(x0 − 3), (5.53)

and so we have found another upper bound for the case x0 = 4

T < 1 . (5.54)

Again this is worse than the previous upper bound

These three choices show that there is more than one way to find an upper bound, but thatone has to use some care to preserve the important properties of the original velocity function.

We illustrate these choices in figure 5.11 for the choice x0 = 4.

f(x)

g1(x)

g2(x)

g3(x)

0 2 4 6 8 10x

20

40

60

80

100

120

(a) The velocity functions

0 5 10 15 20 25 30t

τf

τg2

τg3

x

(b) The trajectories for the ve-locity functions; three trajec-tories reach x = ∞ in finitetime, one does not.

0 2 4 6 8 10x

0.2

0.4

0.6

0.8

1.0

(c) The times taken to movefrom 4 to ∞ are the areas un-der the graphs.

Figure 5.11: Sketches illustrating the time of motion for different velocity functions: f(x)(solid), g1(x) (Dashed), g2(x) (dotted) and g3(x) (dot-dashed)

It is instructive to compare our estimates with the exact result which we foundbefore. The exact answer is

T =1

2ln

(4

3

)' 0.143841.. , (5.55)

compared with the estimates 0.0555 < T < 0.50 which we obtained above



Example 5.11. Find a lower bound on the time taken for the DS 5.3 to move from a point2 < x0 < 3 to a point x1. Use this to show that the motion cannot terminate

For 2 < x0 < 3, we have f(x0) < 0; the system will approach the stable fixed point at x = 2,so x(t) < x0 for t > t0. In this case write

τ01 =

∫ x1

x0

dx

f(x)=

∫ x0

x1

dx

|f(x)|=

∫ x0

x1

dx

(x− 1)(x− 2)(3− x)(5.56)

We want to find a speed which is faster than f(x) so that if the faster moving system does notreach x = 2 then the slower moving system cannot. This means we want to find a functionh(x) such that h(x) > |f(x)| and for which the time

τh =

∫ x0

x1

dx

h(x)(5.57)

is easy to calculate.

Since 2 < x < 3, we can see look at each of the factors in |f(x)| and find simple inequalities

(a) 1 < (x− 1) < 2 (5.58)

(b) 0 < (x− 2) < 1 (5.59)

(c) 1 > (3− x) > 0 (5.60)

(d) (x− 2) < (x− 1) (5.61)

(i) We can use (a) and (c) to obtain

(x− 1)(x− 2)(3− x) < 2 · (x− 2) · 1 , (5.62)

and so

τ01 =

∫ x1

x0

dx

(x− 1)(x− 2)(3− x)>

∫ x1

x0

dx

2(x− 2)=

1

2lnx0 − 2

x1 − 2. (5.63)

As x1 → 2, this lower bound diverges, so the system approaches 2, but will not reach itin finite time.

(ii) We could also have used (a), (b) and (d) to find

(x− 1)(x− 2)(3− x) < 2 · 1 · 1 , (5.64)

and so

τ01 =

∫ x1

x0

dx

(x− 1)(x− 2)(3− x)>

∫ x1

x0

dx

2=

1

2(x1 − x0) . (5.65)

This is true but not helpful. The time τ01 diverges because of the singularity in 1/|f(x)|at x = 2, if we remove this singularity in our estimate we will end up with a finite timeand this will not help prove that the motion does not terminate.



(iii) Finally, as another choice, we could use (c) and (d) to get

(x− 1)(x− 2)(3− x) < (x− 2) · (x− 2) · 1 , (5.66)

and so

τ01 =

∫ x1

x0

dx

(x− 1)(x− 2)(3− x)>

∫ x1

x0

dx

(x− 2)2=

1

2

[1

x0 − 2− 1

x1 − 2

]. (5.67)

This is also true and helpful. As x1 → 2 the term 1/(x1 − 2) diverges and so we canalso use this estimate to show that the motion does not terminate.

These three choices show that there is more than one way to prove a motion terminates, butthat one has to use some care to preserve the important properties of the original velocityfunction.

Example 5.12. As a final example, find upper and lower bounds on the time T for the DSin example 5.8 to move from x = 3/2 to x = 2

This is the DSf(x) = [(x− 1)(x− 2)(x− 3)]1/3 (5.68)

for which the exact result is T = 0.966897...

Since 1 < x < 2 we have the simple inequalities on the non-negative factors (x− 1), (2− x)and (3− x):

(a) 0 < (x− 1) < 1 (5.69)

(b) 0 < (2− x) < 1 (5.70)

(c) 1 < (3− x) < 2 (5.71)

(d) (2− x) < (3− x) (5.72)

We can use (a) and (c) to find

(x− 1)(2− x)(3− x) < 2 , (5.73)

and so find the simple lower bound on T

T =

∫ 2

3/2

dx

((x− 1)(2− x)(3− x))1/3>

∫ 2

3/2

dx

21/3=

1

24/3= 0.39685.. . (5.74)

To find an upper bound, (a) and (b) are not useful, but we can instead (with a bit morethought) come up with

x0 = 3/2⇒ x ≥ 3/2⇒ (x− 1) > 1/2 , (5.75)

and so combining this with (c) we get

(x− 1)(2− x)(3− x) > 1/2 · (2− x) · 1 =1

2(2− x) , (5.76)



or combining it with (d) we get

(x− 1)(2− x)(3− x) > 1/2 · (2− x) · (2− x) =3

2(2− x)2 . (5.77)

Together these give the two upper bounds

T <

∫ 2

3/2

dx

(1/2(2− x))1/3=

[3

221/3(2− x)2/3

]2

3/2

= 1.19055... ,

T <

∫ 2

3/2

dx

(1/2(2− x)2)1/3=

[3

(2− x)1/3

(1/2)1/3

]2

3/2

= 3 . (5.78)

Hence we have found

0.39685.. < T < 1.19055.. , while the exact result is T = 0.966807... (5.79)

5.4 Stability — A More General Discussion

In chapter 4 we introduced the notion of fixed points and investigated their stability graph-ically by looking at the phase portrait and investigating qualitative properties of velocityfunctions in the vicinity of fixed points. Below we look at stability of first order autonomousdynamical systems in a more systematic and quantitative way, discussing (i) linear (and non-linear) stability analysis of fixed points, (ii) the concept of structural stability of a dynamicalsystem, and (iii) the notion of stability of motion in turn.

5.4.1 Stability of Fixed Points

We consider the system described by the equation

dx

dt= f(x)

and suppose that x = a is a fixed point. Assuming that f is a suitably differentiable functionin a neighbourhood of x = a we may approximate f by its Taylor expansion about x = a,

f(x) = f(a) + f ′(a)(x− a) +f ′′(a)

2!(x− a)2 + · · · = f ′(a)(x− a) +

f ′′(a)

2!(x− a)2 + . . . (5.80)

Near x = a, i.e. for |x− a| 1, the dominant contribution to the r.h.s of (5.80) comes fromthe first non-vanishing derivative of f at a.

Case (i) f ′(a) 6= 0.In this case f(x) ' f ′(a)(x− a) for |x− a| 1, and the ODE is approximated by

dx

dt' f ′(a)(x− a) ⇔ d(x− a)

dt' f ′(a)(x− a)

The solution isx− a ' (x0 − a) exp[f ′(a)t] ;



the distance x − a thus (i) increases exponentially in time, if f ′(a) > 0, whereas (ii) itdecreases exponentially in time, if f ′(a) < 0. In the first case the fixed point is called(linearly) unstable, in the second, it is called (linearly) stable. This result is in accordwith the graphical stability analysis based on a phase portrait as in the figure shownbelow.

Case (ii) f ′(a) = 0, but f ′′(a) 6= 0.In this case f has a double zero at x = a, and

f(x) ' 1

2f ′′(a)(x− a)2

near x = a. The ODE is approximated by

dx

dt' 1

2f ′′(a)(x− a)2 ⇔ d(x− a)

dt' 1

2f ′′(a)(x− a)2

The solution (found by separation of variables), reads

x− a ' x0 − a1− t

2f′′(a)(x0 − a)

Supposing that f ′′(a) > 0 so that f has a minimum at x = a, we find

(a) If x0 > a, then x− a increases as a function of time t.

(b) If x0 < a, then |x− a| decreases as a function of time t.

Conversely, supposing that f ′′(a) < 0 so that f has a maximum at x = a, we find

(a) If x0 > a, then x− a decreases as a function of time t.

(b) If x0 < a, then |x− a| increases as a function of time t.

These results are once more in accordance with the graphical stability analysis basedon a phase portrait as in figure 5.12.

a b cx

f(x)

Figure 5.12: Phase portrait of a system with velocity function that has fixed pints at a withf ′(a) > 0, at b with f ′(b) < 0 and at c, with f ′(c) = 0, but f ′′(c) < 0.



Definition 5.2. Linear Stability AnalysisThe stability analysis based on (and requiring only) Taylor series expansions to 1st order isreferred to as linear stability analysis of a fixed point.

Note The approximate solution of the ODEs describing the dynamical system near fixedpoints based on Taylor series expansions is valid only in the vicinity of fixed points. Thequality of the approximate description will therefore improve, if a fixed point is approachedunder the dynamics, whereas the approximate description deteriorates, if the system movesaway from a fixed point.

5.4.2 Structural Stability

Definition 5.3. Structural StabilityA system described by the differential equation

dx

dt= f(x)

is called structurally unstable, if the number of fixed points is changed by some arbitrarily small(continuous) perturbation f(x)→ f(x) + εg(x), (i.e., for some continuous g and ε arbitrarilysmall), otherwise it is structurally stable (i.e., for all continuous g there is a sufficiently smallε, such that the number of fixed points remains unchanged).

The simplest systems that is structurally unstable is when f has a double zero at x = a.Close to x = a the function is approximately a parabola,

f(x) ' c(x− a)2 + . . . . (5.81)

If we add a small constant shift to f then the graph will either intersect the axis in twopoints (so the number of fixed points has gone up) or it will not intersect the axis at all (thenumber of fixed points has gone down). Since the number of fixed points changes, the systemis structurally unstable. See figure 5.13.

As another example, the function f = x3 has a single fixed point at x = 0 which is a triplezero of f . If we add a constant shift to f then the function still has a single fixed point, butif we add a small linear function then it either has one or three fixed points: the number offixed points changes so it is also structurally unstable.

More generally, if f has a zero of any multiplicity higher than 1 at x = a, the system isstructurally unstable, but we might have to add a more complicated function than just aconstant shift. (The proof is left as an exercise to the reader.)

In structurally unstable systems the collection of invariant open sets (and thus qualitativeproperties of the dynamics) can change drastically in response to small perturbations. Thismay e.g. of economic relevance as the following example demonstrates.

Example 5.13. Bistable Ecosystems

Consider a population dynamics of the form

dx

dt= rx(1− x/c)− ax2

b2 + x2,



f(x)-ϵ f(x) f(x)+ ϵ

Figure 5.13: Velocity function of a structurally unstable system with a fixed point which isa double zero [centre figure]. Vertical displacement by ±ε increases [left figure] or decreases[right figure] the number of fixed points by one as shown.

f(x)- ϵ(x-a) f(x) f(x)+ϵ(x-a)

Figure 5.14: Velocity function of a structurally unstable system with a fixed point which is atriple zero [centre figure]. Changing f(x) by ±ε(x− a) leaves the number of fixed points thesame [left figure] or increases it by two [right figure].

in which all parameters (r, c, a, b) are taken to be positive. The first contribution describes logisticgrowth, and taken by itself has x = c as a stable fixed point. The second contribution on the r.h.s.exhibits ‘threshold behaviour’. It produces a significant contribution only for x > xc ' b. This systemcould describe a population of larvae, preyed upon by a natural enemy (woodpeckers) if present insufficient quantity. For sufficiently small r the system has a stable small-x fixed point an unstable fixedpoint and a stable fixed point at large x. The stable fixed point at low x and the unstable fixed pointcoalesce and disappear as r is increased beyond a certain value rc and the system would approach astable high-x solution. Economically, this could spell disaster if the larvae destroy wood that is meantto be sold!. Only by decreasing r to a very small value r∗ rc can the situation be reverted to astable small x situation; (see the diagram).

Example 5.14. (King’s College, Summer 1994)

A first order dynamical system is described by the differential equation

x = v(x) = x(x− 1)2(x− 2)3(x− 3)5 .



x

f(x)

x

f(x)

Figure 5.15: Phase portrait of the bi-stable ecosystem for r < rc (left) and r > rc (right).

Explain briefly why the initial condition x(0) = 2 ⇒ x(t) = 2 ∀t ≥ 2. Sketch the phase diagram andwrite down the invariant open sets. Is the system structurally stable? Give your reasons. Given thatx(0) = 5/2 prove that the system will not reach x = 2 in a finite time. Suppose instead that x(0) = 4.Prove that x(t)→∞ as t→ τ, where τ < 1/10.

We note that v, v′ are both continuous in any neighbourhood of the point (t = 0, x = 2); Picard’stheorem therefore applies to the given equation. Clearly x(t) = 2 satisfies the differential equationand the initial condition x(0) = 2. It follows from Picard’s theorem that x(t) = 2, t ≥ 0.

Referring to the phase diagram (see below) we see that the invariant open sets are

(−∞, 0), (0, 1), (1, 2), (2, 3), (3,∞) .

The system is not structurally stable since the transformation v(x) → v(x) − ε(ε > 0) increases thenumber of fixed points from four to five (v(x) has a double zero at x = 1).

It is clear from the phase diagram that if x(0) = 5/2 the system moves towards x = 2 and the time τ1taken to reach x = x1 < 5/2 is

τ1 =

∫ x1

5/2

dx

x(x− 1)2(x− 2)3(x− 3)5=

∫ 5/2

x1

dx

x(x− 1)2(x− 2)3(3− x)5.

Now x(x− 1)2(3− x)5 < (5/2)(5/2− 1)2(3− 2)5 = 45/8 so

τ1 >8

45

∫ 5/2

x1

dx

(x− 2)3=

8

45

(−2 +

1

2

1

(x1 − 2)2

).

We conclude that τ1 →∞ as x1 → 2+ and that the system does not reach x = 2 in a finite time.

The time τ taken by the system to go from x(0) = 4 to infinity is given by

τ =

∫ ∞4

dx

x(x− 1)2(x− 2)3(x− 3)5.

Since x > x− 3, x− 1 > x− 3, x− 2 > x− 3, we obtain

τ <

∫ ∞4

dx

(x− 3)11=

1

10.

This motion is therefore terminating.



x

f(x)

Figure 5.16: Phase portrait of the system with velocity function v(x) = x(x−1)2(x−2)3(x−3)5.

5.4.3 Stability of Motion

Beyond looking at stability of fixed points, one may also look at stability of motion itself inthe following sense. Let

dx

dt= f(x) (5.82)

describe a first order autonomous dynamical system, and let x(t) be a solution for some giveninitial condition. Suppose now, that the solution is slightly disturbed

x(t)→ y(t) = x(t) + ε(t) (5.83)

at, say, t = t1, with |ε(t)| 1. Enquiring into stability of motion means asking whether theperturbation will increase or decrease under the dynamics. We study this issue only at linearorder.

Definition 5.4. Stability of motion

The motion described by (5.82) is stable, if a small perturbation ε(t) is diminished under thedynamics; it is called unstable, if a small perturbation is amplified.

Proposition 5.1. The motion of an autonomous first order dynamical system described bya velocity function f is linearly stable at x, if f ′(x) < 0, whereas it is linearly unstable at x,if f ′(x) > 0.

Proof Consider the ODE

dy

dt= f(y) ⇔ dx

dt+

dε

dt= f(x+ ε)

Expanding the r.h.s. to first order in ε gives

dx

dt+

dε

dt' f(x) + f ′(x)ε .

As x is a solution of (5.82), we get an ODE for ε,

dε

dt' f ′(x)ε ,



in which f ′(x) = f ′(x(t)) is a known function of time. The equation is therefore separableand solved by

ε(t) ' ε(t1) exp[ ∫ t

t1

dt′f ′(x(t′))]

(5.84)

Note that for small t− t1 the exponential may be approximated by exp[f ′(x(t1))(t− t1)

]in

the above expression.

Thus the perturbation increases at an exponential rate, and the motion is unstable wheneverf ′(x) > 0, whereas the perturbation decreases at an exponential rate, and the motion is stablewhenever f ′(x) < 0, as claimed. #

Note 1: The notion of dynamic stability is extremely important if one considers solvingan ODE numerically on a computer, as it tells whether rounding errors will amplify whenpropagating a solution or whether they will diminish.

Note 2: A first order dynamical system which is linearly unstable when integrated forwardin time, becomes linearly stable when integrated backwards (and vice versa).

5.5 Asymptotic Analysis

Asymptotic analysis is concerned with approximate analytic investigations of dynamical sys-tems in situations where either full solutions are not readily available, or where approximatesolutions are sufficient to give a reasonably precise first level of analysis of a given system.

The aim is to give a solid basis to the idea of two functions “behaving the same” or “havingthe same leading behaviour”.

We start with a key definition.

Definition 5.5. Asymptotic Equivalence

A function f is said to be asymptotically equivalent to another function g as x → x∗ — insymbols f(x) ∼ g(x), as x→ x∗ — if they become of the same order of magnitude as x→ x∗,i.e.

f(x) = g(x)(1 + ε(x)) , with ε(x)→ 0 , as x→ x∗ .

In this definition the case x∗ = ±∞ is allowed.

There are several things to note:

(i) Given a function f(x) and a point x∗, infinitely many functions will be asymptoticallyequivalent to f(x) as x→ x∗. This is guaranteed by the huge freedom in the choice ofthe function ε(x).

(ii) The asymptotic behaviour of a function can be very different at different points, so iff(x) ∼ g(x) as x→ a, this does not imply that f(x) ∼ g(x) for any other point.

(iii) If you are given a function f(x) and asked to find a simpler function which is asymp-totically equivalent to f(x) as x→ x∗, it is often usual to choose a function of the formc(x− x∗)α [if this is indeed correct], and this is often called the leading asymptotic be-haviour, but this is not a unique choice and other choices may be better for the problemin hand.



Here are some examples showing that the asymptotic behaviour of a simple function can bedifferent in different limits

Example 5.15. Consider the function f(x) = −x+x2. Find the leading asymptotic behaviourof f(x) in the limits x→ 0, x→ 1, x→∞ and x→ 2.

(a) f(x) ∼ −x as x→ 0.

Proof: f(x) = −x(1− x) and so f(x) = −x(1 + ε(x)) with ε(x) = x and the conditionsare met.

(b) f(x) ∼ (x− 1) as x→ 1.

Proof: f(x) = (x− 1)(1 + ε(x)) where ε(x) = x− 1 and so the conditions are met.

(c) f(x) ∼ x2 as x→∞.

Proof: f(x) = x2(1− 1x) and so again the conditions are met.

(d) f(x) ∼ 2 as x→ 2.

Proof: f(x) = 2(1 + ε(x)) with ε(x) = (x− 2)(x+ 1)/2 and so again the conditions aremet.

Here are some examples showing that there is no unique choice of asymptotic equivalence

Example 5.16. Givenf(x) = x− x3 .

We then have

(a) f(x) ∼ x as x→ 0.

(b) f(x) ∼ x+ x2 as x→ 0.

(c) f(x) ∼ −x3 as x→∞

(d) f(x) ∼ −x3 + 5 as x→∞

We can check the first two of these:

(a) With f(x) = x − x3, g(x) = x then (f(x)/g(x)) − 1 = 1 − x2 = 1 + ε(x) so that indeedε(x) = −x2 → 0 as x→ 0.

(b) With f(x) = x−x3, g(x) = x+x2 then (f(x)/g(x))−1 = (1−x2)/(1+x) = 1−x = 1+ε(x)so that now ε(x) = −x and again ε(x)→ 0 as x→ 0.

The proof of the other two statements according to the definitions is left as an exercise.

Asymptotic analysis is useful because if two velocity functions are similar then their trajec-tories will be similar. If two velocity functions f(x) and g(x) have the same fixed point x = aand

f(x) ∼ g(x) as x→ a , (5.85)



then if the flow terminates, the times to reach x = a from a point b near a are also asymp-totically equivalent as b→ a. This means that one can investigate whether a flow terminatesas it approaches a fixed point (or ±∞) or not by looking at a simpler velocity function. Thisis often simpler than trying to find an upper or lower bound.

5.5.1 Asymptotic Analysis and Dynamical Systems

Suppose we have a DSdx

dt= f(x) , (5.86)

which has a fixed point at x = a. We want to understand what happens as x → a and toknow whether the time to reach x = a from x0 is finite or not. We know this is given by anintegral, so we investigate

τ exact01 =

∫ x1

x0

dx

f(x). (5.87)

This can be hard, time-consuming or impossible to do exactly and again it can be hard ortime-consuming to find upper and lower bounds, so we can instead consider an alternativeintegral:

τasymp01 =

∫ x1

x0

dx

g(x), (5.88)

where f(x) ∼ g(x) and f(x) = g(x)(1 + ε(x)) and ε(x) → 0 as x → 0. This means we canwrite this asymptotic time as

τasymp01 =

∫ x1

x0

(1 + ε(x))

f(x)dx =

∫ x1

x0

1

f(x)dx︸︷︷︸

τexact01

+

∫ x1

x0

ε(x)

f(x)dx︸︷︷︸

correction

. (5.89)

We can then show that

(a) τ exact01 is finite as x1 → a if and only if τasymp01 is finite as x1 → a.The reason is that the only important part of the integral is the region close to x = aand in that part, ε(x) is small.This means we can test for termination of motion using simpler asymptotically equiva-lent functions.

(b) If the time is finite and we put x1 = a then τ exact01 ∼ τasymp01 as x0 → a.The reason is that now the integration range is only over a small region between x0 anda where ε(x) is small. This means we can get asymptotically good estimates of the timefor a motion to terminate using asymptotically equivalent velocity functions withouthaving to search for upper and lower bounds.

We can also go further. If we want a solution to the initial value problem

dx

dt= f(x) , x0 = a+ ε , (5.90)

where ε is small, then we can get a good approximation near x = a by instead solving theinitial value problem

dxasymp

dt= g(xasymp) , x0 = a+ ε , (5.91)



the reason being that g(x) and f(x) are very close if x is near a, and the solutions x(t) andxasymp(t) will also be close so long as x is close to a.

Example 5.17. We will look at this more in some examples based on the Dynamical systemwith phase space the positive real axis x ≥ 0,

dx

dt= f(x) , f(x) =

√x (x− 1)(x− 2)(x− 3) . (5.92)

If we plot f(x) we easily see that there are four fixed points 0, 1, 2 and 3, that 0and 2 are stable fixed points and 1 and 3 are unstable fixed points.

1 2 3

x

f HxL

Figure 5.17: Sketch of f(x) and the phase space for the system (5.92).

Since f and f ′ are both continuous for all x > 0, corollary II to Picard’s theorem implies thethe open intervals (1, 2) and (2, 3) are open invariant sets, but since f ′(0) diverges we cannotsay whether (0, 1) is an invariant set or not. Here are several questions we can analyse usingasymptotic analysis:

(i) Does the motion tending towards x = 0 terminate?

(ii) Does the motion tending towards x = 2 terminate? [we know the answer is “no” but wecan still check it]

(ii) Does the motion towards x =∞ terminate?

We will use asymptotic methods to answer these questions. We can compare them with theexact answer because we can find the time of motion exactly using the substitution x = u2

and the method of partial fraction to get∫dx

f(x)=

1

2log

∣∣∣∣1−√x1 +√x

∣∣∣∣+1

2√

3log

∣∣∣∣∣1−√x/3

1 +√x/3

∣∣∣∣∣+1√2

log

∣∣∣∣∣1 +√x/2

1−√x/2

∣∣∣∣∣ . (5.93)

Finding the exact answer is quite time-consuming and so the asymptotic methods are a lotfaster, which is why the method is particularly useful in this case. Note that although we can



find the time of motion explicitly, we cannot invert this to find x(t) explicitly, but we canfind the asymptotic form of the motion near a fixed point and so another questions we cananswer is

(iv) What is the asymptotic solution near the fixed points x = 0, x = 2 and x =∞?x

Example 5.17(i)Near x = 0 the factors (x − 1), (x − 2) and (x − 3) are approximately constant and so theleading behaviour of f(x) is

f(x) ∼ −6√x , (5.94)

which we can check by expanding f(x) out to see

f(x) =√x(x3 − 9x2 + 11x− 6) = −6

√x(1 +O(x)) . (5.95)

This means we can calculate the asymptotic time of motion for x near 0,

τasymp01 =

∫ x1

x0

dx

−6√x

=1

3

√x0 −

1

3

√x1 . (5.96)

This is finite as x1 → 0 and so the motion terminates. We thus also get an asymptoticestimate of the time of motion from x0 to the fixed point 0,

τasymp =1

3

√x0 . (5.97)

We can compare this with the exact answer (5.93), which we can expand for small x as

τ exact =1

3

√x

(1 +

11

18x+

17

36x2 + . . .

). (5.98)

We see, τasymp ∼ τ exact as x→ 0.

Example 5.17(ii)Near x = 2, the factors

√x, (x− 1) and (x− 3) are approximately constant, so that

f(x) ∼ −√

2(x− 2) . (5.99)

This means we can calculate the asymptotic time of motion for x near 2,

τasymp01 =

∫ x1

x0

dx

−√

2(x− 2)=

[− 1√

2log |x− 2|

]x1x0

. (5.100)

This diverges as x1 → 2 and so the asymptotic motion does not terminate in finite time andso the true motion does not terminate. We can see that the exact answer (5.93) also divergesas x→ 2 because of the term log |1−

√x/2|.

Example 5.17(iii)As x → ∞, the constant terms in the factors (x − 1), (x − 2) and (x − 3) become irrelevantand so

f(x) ∼ x7/2 . (5.101)



Again we can check explicitly since

f(x) =√x(x3 − 9x2 + 11x− 6) = x7/2

(1 +O

(1

x

)). (5.102)

This means we can calculate the asymptotic time of motion when x is large,

τasymp01 =

∫ x1

x0

dx

x7/2=

2

5(x−5/20 − x−5/2

1 ) . (5.103)

This is finite as x1 →∞ and so the motion terminates. We also have an asymptotic estimateof the time of motion from x0 to ∞,

τasymp =2

5x−5/20 . (5.104)

We can compare this with the exact answer (5.93), which we can expand for larges x as

τ exactx =2

5x−5/20

(1 +

30

7

1

x+

125

9

1

x2+ . . .

). (5.105)

We see,τasymp ∼ τ exact as x→∞ . (5.106)

Example 5.17(iv)From the previous calculations we find the following asymptotic solutions:

For x0 near x = 0, we can use (5.96) to get

xasymp(t) = (3t−√x0)2 . (5.107)

For x0 near x = 2, we can use (5.100) to get

xasymp(t) = 2 + (x0 − 2) exp(−√

2t) . (5.108)

For x0 large, we can use (5.103) to get

xasymp(t) = (5t/2− x0−5/2)−2/5 . (5.109)

It is hard to check these exactly since we cannot easily solve for the exact solution x(t) butthey are numerically good. This is left as an exercise for the reader.

Example 5.18. To illustrate further the use of asymptotic analysis in finding asymptotic solution,we now consider a case where we are able to obtain the full solution in addition to the approximateexpressions, and thereby fully assess the quality of the approximate solutions.

Consider a dynamical system of the form

dx

dt= f(x) = −x+ x2 .

It has fixed points at x = 0 (stable) and x = 1 (unstable). The solution of the ODE is obtained byseparation of variables, using partial fractions to do the x-integral:

t =

∫ x

x0

dx′

x′(x′ − 1)= ln

x− 1

x− ln

x0 − 1

x0



with x(0) = x0. Solving for x gives

x(t) =x0

x0 − (x0 − 1)et

If x0 < 1, the system will approach the fixed point at 0, if x0 > 1, the system will escape to +∞, andthe motion will terminate at time t = ln(x0/(x0 − 1)).

Let us compare these exact results with those of the corresponding asymptotic analyses.

(a) Considering the behaviour at small x, we have f(x) ' −x, as x → 0. Solving the correspondingapproximate ODE

dx

dt' −x ,

assuming x0 1, gives x(t) ' x0e−t. This is indeed just the result one would get to lowestorder in x0 in an expansion of the exact solution.

(b) Considering the behaviour at large x, we have f(x) ∼ x2, as x → ∞. Solving the correspondingapproximate ODE

dx

dt∼ x2 ,

assuming x0 1, gives x(t) ' x0

1−x0t. Asymptotic equality would be guaranteed as long as

x − x0 remains finite. In this case, as in the exact solution, we observe that x(t) diverges atfinite t. I.e., the motion terminates and the time to escape to +∞ is t = 1/x0. This is indeedjust the result one would get to lowest order in 1/x0 in an expansion of the the time at whichthe motion terminates in the exact solution.

The following figure compares exact and approximate solution in the regions of small and large x. Thequality of the approximations is remarkable even in the region of the divergence, although exact andapproximate solutions terminate at slightly different times (so that there the error is strictly speakinginfinitely large and the approximation bad). Note that the large x solution can even be continued withreasonable precision across the divergence!



0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

0.11

0 2 4 6 8 10

x(t

)

t

10

100

1000

10000

100000

1e+06

0 0.002 0.004 0.006 0.008 0.01

x(t

)

t

-20000

-15000

-10000

-5000

0

5000

10000

0 0.005 0.01 0.015 0.02

x(t

)

t

Figure 5.18: (a) Full solution (full curve) and approximate solution (dashed curve) in the region ofsmall x, with x0 = 0.1. (b) Full and approximate solution in the region of large x, with x0 = 100,approaching the divergence. (c) Full and approximate solution in the region of large x crossing thedivergence.



5.6 Exercises

Shorter exercises

Exercise 5.1Consider the following velocity functions:

(i) v(x) = x3 (ii) v(x) = x−x2 (iii) v(x) = x−x4/3 (iv) v(x) = x1/3−x4/3

In each case answer the following questions:

(a) Sketch the graph of v(x), find the fixed points of the dynamical system and state whetherthey are stable or unstable.

(b) For each fixed point x = a,

(i) Does v(x) have a Taylor expansion around x = a to all orders in (x− a)?

(ii) Can you write down a linear approximation to v(x) around x = a?

(iii) Can you decide if the fixed point is stable or unstable based on the linear analysis?

Exercise 5.2Prove the following statements on asymptotic behaviour:

(i) f(x) = x2 + x3 , g(x) = x3 , f(x) ∼ g(x) , x→∞

(ii) f(x) = x2 + x3 , g(x) = x2 , f(x) ∼ g(x) , x→ 0

(iii) f(x) = cosh(x) , g(x) =ex

2, f(x) ∼ g(x) , x→∞

(iv) f(x) = cos(x)− 1 , g(x) = −x2

2, f(x) ∼ g(x) , x→ 0

(v) f(x) = tan(x) , g(x) =1

π2 − x

, f(x) ∼ g(x) , x→ π

2

Longer exercises

Exercise 5.3Consider the first order autonomous dynamical system

dx

dt= f(x) = x2(x− 1)(x− 2)2

(a) Identify the fixed points, and the invariant open sets.

(b) Sketch the phase portrait and characterize the stability of the fixed points using yoursketch.

(c) Draw qualitative solution curves, one for each invariant open set. State briefly the mainfeatures of these curves, referring to properties of the phase portrait.



Exercise 5.4Investigate the first order autonomous dynamical system described by

dx

dt= v(x) = (x− 1)(x− 2) .

Sketch the phase portrait including the velocity function and the phase flow. Write down thefixed points of the motion and the invariant open sets. Give a sketch of qualitative solutioncurves, one for each invariant open set. Describe them briefly, referring to properties of thephase portrait. Show – without using the exact solution of the differential equation – that, ifx(0) = 3/2, the system will not reach the nearest stable fixed point in finite time. Show alsothat the exact solution of this equation, with initial condition x(0) = 3/2, is

x(t) =2 + exp(t)

1 + exp(t).

Exercise 5.5Discuss the system described by

dx

dt= v(x) = (x− 1)(x− 2)(x− 3)(x− 4) .

Sketch the phase portrait. Write down the fixed points of the motion and discuss theirstability. Suppose that x(0) = 5. Prove that the motion is terminating. Give upper andlower limits for the time to reach infinity. Show that, if x(0) = 7/2, the system will not reachx = 3 in a finite time.

Exercise 5.6Consider the dynamical system of problem 5.5

dx

dt= v(x) = (x− 1)(x− 2)(x− 3)(x− 4) .

For each fixed point find the corresponding linearised system and state the stability of eachfixed point based on the linearised system. Check your results by comparing them with thephase portrait.



Exercise 5.7 (part of 2008 exam)Consider the dynamical system described by

dx

dt= f(x) = (x+ 1)(2− x)(x− a).

where a is a parameter.

(a) State for which values of a the system is structurally stable and for which values of a thefixed point at x = −1 is stable. Give reasons for your answer.

(b) In what follows, assume a = 1.

(i) Identify the fixed points, characterize their stability, and give the invariant opensets of the dynamics.

(ii) Sketch the phase portrait.

(iii) Draw qualitative solution curves for the initial conditions (i) x(0) = −3/2, (ii)x(0) = 0, and (iii) x(0) = 3/2. Justify your results.

(iv) Write the linearised equation about the fixed points x = 1 and x = 2. Solve theseequations and comment on how your results are related to the stability of the twofixed points respectively.

Exercise 5.8Consider the first order dynamical system described by

dx

dt= v(x) = −

√x+ x , x ≥ 0

(a) Sketch the phase portrait, identify the fixed point(s) and discuss their stability.

(b) Write down a simpler asymptotic equation of motion in the vicinity of x = 0, and solveit with initial condition x(0) = a with 0 < a 1. Show that the asymptotic systemwill reach the nearest stable fixed point in finite time and calculate this time. Does thisshow that the full system will also reach the fixed point in finite time?

(c) Solve the differential equation exactly, and compute the time needed to reach the nearestfixed point from x(0) = a with 0 < a 1 using the exact solution. Verify that to lowestorder in a the approximate and the exact results agree.

(d) Show that if 0 < x < 1 then |v(x)| <√x. Show that this gives a lower bound on the

time for the system to reach x = 0 starting from x(0) = a with 0 < a < 1. Comparethis with the exact time required to reach x = 0 and the asymptotic approximation.




dx

dt= f(x) =

x5

1 + x2

(a) Sketch the phase portrait, identify the fixed point(s) and discuss their stability.

(b) Write down a simpler asymptotic equation of motion as x→∞ and solve it with initialcondition x(0) = a with a 0. Show that the asymptotic system will reach x =∞ infinite time and calculate this time. Does this show that the full system will also reachx =∞ in finite time?

(c) Solve the differential equation exactly, and compute the time needed to reach x = ∞from x(0) = a with a 0 using the exact solution. Verify that to lowest order in 1/athe approximate and the exact results agree.

(d) Show that if x > 1 then |f(x)| > x3/2. Show that this gives an upper bound on the timefor the system to reach x = ∞ starting from x(0) = a with a > 1. Compare this withthe exact time required to reach x = 0 and with the asymptotic approximation.

Exercise 5.10Consider a first order dynamical system evolving in the phase space Γ = x ≥ 0 accordingto the equation

dx

dt= xγ(x− 2), γ ≥ 0

(a) Write the asymptotic equation of motion valid for x 1 and solve it for the initialcondition x(0) = x0, with x0 1. Use this solution to show that the motion isterminating for γ > 0.

(b) Write the asymptotic equation of motion valid in the vicinity of x = 0 and solve it forthe initial condition x(0) = x0, 0 < x0 1, assuming γ 6= 1. Use this result to showthat the motion is terminating for γ < 1 whereas it is not terminating for γ > 1.

Supplementary exercises

Supplementary Exercise 5.1Sketch the phase portraits for the three systems defined on the phase space −1 ≤ x ≤ 1:

(i) x =√

1− x2

(ii) x = 1− x2

(iii) x = (1− x2)3/2

For which of these is the solution with x(0) = 0 terminating? Can you relate this to theconditions for Picard’s theorem to hold? Can you find the exact solution for x(t) in eachcase?



Supplementary Exercise 5.2Consider the dynamical system defined for x ∈ R

x = exp(−1/x2) .

Sketch the phase portrait for this system. Does it have a fixed point? Is the flow startingfrom x(0) = −1 terminating?

Hint: Consider a change of variables from x to u = −1/x and use the fact that exp(a) > an/n!for all a > 0.

Supplementary Exercise 5.3The time evolution of the magnetization m of a ferromagnetic material in the vicinity of thecritical temperature Tc is well described by the first order autonomous dynamical system

dm

dt= v(m) = am−m3 ,

where a = (Tc − T )/Tc and T is the temperature.(i) Sketch the phase portrait including the velocity function and the phase flow using thearrow representation of the flow, for the cases T > Tc and T < Tc. Write down the fixedpoints of the motion and the invariant open sets of the dynamics for both T > Tc and T < Tc;discuss the stability of the fixed points in both cases.(ii) Show, for T < Tc – without using the exact solution of the differential equation – that, ifm(0) =

√a/2, the system will not reach m =

√a in finite time.

(iii) Show also that for T < Tc the exact solution of this equation, with initial conditionm(0) =

√a/2, is

m(t) =

√a

1 + 3 exp(−2at)

(iv) For T > Tc and for T = Tc give solutions of the asymptotic equation of motion in thevicinity of m = 0, assuming m(0) = m0 > 0, but small.


Chapter 6: Second Order Autonomous Systems 85

Chapter 6

Second Order Autonomous Systems

A second order autonomous system describes the time evolution of the state of a system insome two-dimensional phase-space Γ through two differential equations. We have alreadyseen one of these in the introductory chapter, the Lotka-Volterra (or fox-rabbit or lynx-hare)system (1.5) where the phase space is the quadrant x > 0, y > 0 of IR2. This system alreadyexhibits many of the features of second order autonomous systems that are not possible forfirst order systems, for example it has periodic orbits and two new kinds of fixed points. Wehave also already seen its phase portrait in figure 1.3. It turns out that the possible behavioursof second order systems are much richer than that of first order systems.

In this chapter we look at qualitative methods and at approximate analytic methods to studysecond order systems, in particular we will look at methods to help sketch phase portraitsand to analyse the nature of fixed points.

In chapter 4 we said that a second order DS Will describe the time evolution of a systemdescribed by two variables (x1, x2) through differential equations of the form

dxidt

= fi(x1, x2) , i = 1, 2 , (6.1)

where the functions f1 and f2 are referred to as velocity functions. These functions do notdepend on t as that is the definition of autonomous

In this chapter we often find it convenient to use conventional x, y notation instead of x1, x2

and sometimes to write f1, f2 as F,G, so the we could write these equations asdx

dt= f1(x, y)

dy

dt= f2(x, y)

or

dx

dt= F (x, y)

dy

dt= G(x, y)

, (6.2)

or in vector notation

dr

dt≡ d

dt

(xy

)= f(x, y) =

(f1(x, y)f2(x, y)

)or

(xy

)=

(F (x, y)G(x, y)

). (6.3)

6.1 Phase Space and Phase Portraits

The terminology introduced in chapter 4 applies in an obvious way to second order au-tonomous systems.



Definition 6.1. Phase Space ΓThe phase space Γ is defined as the set of admissible values for the coordinates x and y. Thepair (x, y) denotes a point in IR2, so generally Γ ⊆ IR2. For some systems, Γ may be a propersubset of IR2: e.g., if x and y denote concentrations of two substances in a binary chemicalreaction, then x ≥ 0 and y ≥ 0.

Definition 6.2. Trajectory/Solution Curve (through x0)The set of points (t, x(t), y(t)); t ∈ IR with (x, y) a solution of (6.3) (and x(t0) = x0 andy(t0) = y0 for some t0) is called a trajectory or solution curve (through (x0, y0)).

Definition 6.3. Fixed PointsA point (a, b) ∈ IR2 is a fixed point of the dynamical system (6.3) if f f1(a, b) = 0 andf2(a, b) = 0. This means that the constant trajectory x(t) = a, y(t) = b is a solution of thedynamical system (6.2).

Definition 6.4. Orbit or Phase Curve (through (x0, y0))The set of points (x(t), y(t)); t ∈ IR with (x, y) a solution of (6.3) (and x(t0) = x0 andy(t0) = y0 for some t0) is called orbit or phase curve (through (x0, y0)).

If we parametrise the phase curve as y(x) then we have

dy

dx=y

x=G(x, y)

F (x, y). (6.4)

It is sometimes possible to solve this differential equation to find the phase curves even ifthe t dependence of x(t) and y(t) cannot be found. It is also sometimes useful to solve theequations for the phase curve first and then substitute y = y(x) in the equation for x to solvefor x(t) explicitly.

Note: Provided the functions F and G satisfy the conditions of Picard’s theorem in a regionaround a point then there is precisely one trajectory that starts at that point and hence onephase curve through that point. This also means that a trajectory starting at a fixed pointremains there for all time; if there are several distinct phase curves passing through the samepoint, or if a trajectory can reach a fixed point in finite time then the Picard conditions mustfail at that point.

Definition 6.5. Phase FlowThe collection of all possible phase curves is called the phase flow of the dynamical system(6.3). Velocity information, i.e., f(x, y) = f1(x, y)e1 + f2(x, y)e2 drawn at each (x, y) ∈ Γ isoften included in representations of the phase flow.

The vector f(x, y) is called the velocity of the flow (at (x, y)) The set of vectors f(x, y); (x, y) ∈Γ is called the velocity field.

Definition 6.6. Phase PortraitA graphical representation of the the phase space Γ, including a graph of the phase flowand/or the velocity f(x, y) evaluated at representative points. There are several ways topresent the information in a phase portrait which we show in the examples below.

There is one new concept which is very helpful especially in drawing phase portraits, and thatis



Definition 6.7. Null-ClinesThe set of points (x, y) ∈ IR2 for which f1(x, y) = 0 is called null-cline of f1.

The equation of the null-cline, f1(x, y) = 0, is a relation that must hold between x and y forpoints (x, y) to be on the null-cline. This generally defines a curve in IR2. For instance iff(x, y) = x2 + y2 − r2, then the null-cline of f is a circle of radius r centred at the origin ofIR2.

Note: On a null-cline of f1, the x-component of the velocity vanishes so that the flow isvertical; similarly, on a null-cline of f2, it is the y-component of the velocity which is zero sothat the flow is horizontal. Fixed points are intersections of the null-clines of f1 and f2: Itis common to include the null-clines of the velocity functions in the phase portrait since thenull clines divide the phase space into regions on which the velocity field points in differentgeneral directions which we can call by the points of the compass: NW, NE, SW, SE:

NE If f1 > 0, f2 > 0 then the flow is in the general direction or in the direction NE.

NW If f1 < 0, f2 > 0 then the flow is in the general direction or in the direction NW.

SW If f1 < 0, f2 < 0 then the flow is in the general direction or in the direction SW.

SE If f1 > 0, f2 < 0 then the flow is in the general direction or in the direction SE.

The velocity field points N, S (vertical) or E, W (horizontal) on the null-clines themselves.

We now go through several examples in increasing complexity before turning to some of theideas they highlight.

Example 6.1. Consider the second order autonomous systems with phase space the planedefined by

(a)

x = 2

y = −1(b)

x = 2

y = x(c)

x = x

y = y(d)

x = y

y = −x(e)

x = 2xy

y = y2 − x2

The general procedure we shall use is the following:

(i) We first plot the null clines. These are the sets on which f1(x, y) and f2(x, y) vanish. Thevelocity field is vertical where f1 = 0 and horizontal where f2 = 0.

(ii) The null-clines divide the phase space into 8 regions on which the velocity field is inone of the directions NW, NE, SE, SW. The velocity field points N, S, E or W on thenull-clines themselves. We can label each of the 8 regions by the sign of the componentsfi and in this way come to an understanding of the direction of the velocity field.

(iii) Looking at magnitude of f , we can see where the velocity is larger or smaller; i.w. wherethe motion is faster or slower.



(iv) We can sketch arrows at representative points to show the direction, or the directionand magnitude, of the velocity field1

(v) We try to sketch the orbits by joining up the arrows. We can check this by solving forthe equations of the phase curves exactly, if that is possible

Example 6.1(a)This is the system

x = F (x, y) = 2 , y = G(x, y) = −1 ,

(xy

)= f =

(2−1

). (6.5)

(i), (ii) The velocity functions F (x, y) and G(x, y) are never zero, so we could say that thenull-clines are the empty sets.(iii) The velocity function is a constant so it has the same magnitude everywhere.(iv) We can sketch the velocity field by plotting the vector f at representative points on atypical part of the phase plane (see figure 6.1).(v) If we join up the arrows we get straight lines. We can check by solving the equationsexactly,

x = 2⇒ x = 2t+ a , y = −1⇒ y = −t+ b , so y = −1

2x+ c , (6.6)

or solving the equation for the phase curves,

dy

dx=G(x, y)

F (x, y)= −1

2, y = −x/2 + y0 . (6.7)

-2 -1 0 1 2

-2

-1

0

1

2

(a) A set of arrows represent-ing the velocity field.

-2 -1 0 1 2

-2

-1

0

1

2

(b) A set of orbits joining upthe arrows of the velocity field.

Figure 6.1: Two examples of a phase portrait for the system (6.5).

1Note: If the magnitudes of the velocities vary in a large range, one may choose to plot the direction fieldn(x, y) = f (x, y)/|f (x, y)| (instead of the velocity field) in a phase diagram; The direction field is a field ofunit vectors pointing in the direction of the velocity. As with the flow, visual impression of the actual directionfield can be more faithful, if the arrows representing the direction field n(x, y) are not drawn such that theyare rooted in (x, y) (and as they strictly should) but rather with their midpoint at (x, y). This is the typicalresult of the Mathematica command VectorPlot.



Example 6.1(b)This is the system

x = F (x, y) = 2 , y = G(x, y) = x ,

(xy

)= f =

(2x

). (6.8)

(i) The velocity function F (x, y) is never zero but G(x, y) = 0 on the y–axis, x = 0.(ii) F (x, y) > 0 always but G(x, y) < 0 for x < 0 and G(x, y) > 0 for x > 0, hence the flowpoints SE for x < 0 and NE for x > 0.(iii) The velocity function has |f | =

√2 + x2 and so |f | is bigger the further it is from the

y–axis.(iv) We can sketch the velocity field by plotting the vector f at representative points on atypical part of the phase plane (see figure 6.2).(v) If we join up the arrows we get curves. We can check by solving the equation for thephase curves,

dy

dx=G(x, y)

F (x, y)=x

2, y =

1

4x2 + y0 , (6.9)

and we see that these are actually parabolae, which we can also sketch (see figure 6.2).

-2 -1 0 1 2

-2

-1

0

1

2

(a) The null-clines showingwhere the velocity field is hor-izontal and vertical.

f1>0, f2<0: SE f1>0, f2>0: NE

-2 -1 0 1 2

-2

-1

0

1

2

(b) The regions in which thevelocity field points SE andNE.

-2 -1 0 1 2

-2

-1

0

1

2

(c) A set of arrows represent-ing the velocity field.

-2 -1 0 1 2

-2

-1

0

1

2

(d) A set of orbits joining upthe arrows of the velocity field.

Figure 6.2: Constricting the phase portrait for the system (6.8).



Example 6.1(c)This is the system

x = F (x, y) = x , y = G(x, y) = y ,

(xy

)= f =

(xy

). (6.10)

(i) The velocity function F (x, y) = 0 on the y–axis and G(x, y) = 0 on the x–axis. Theseintersect at the point (0, 0) and so this is a fixed point.(ii) The velocity field points NE in the first quadrant, NW in the second, SW in the thirdand SE in the fourth.(iii) The velocity function has |f | =

√x2 + y2 and so the field is bigger and the flow faster

the further from the origin.(iv) The velocity fields points radially away from the origin (see figure 6.3).(v) If we join up the arrows it looks like we get radial straight lines. We can check by solvingthe equations for the flows,

x = x⇒ x = aet , y = y ⇒ y = bet , so y = (b/a)x , (6.11)

or solving directly for the phase curves, as the equation is separable:

dy

dx=G(x, y)

F (x, y)=y

x, log y = log x+ c , y = Ax , (6.12)

and we see that these are indeed straight lines.



-2 -1 0 1 2

-2

-1

0

1

2


f1<0, f2>0: NW f1>0, f2>0: NE

f1<0, f2<0: SW f1>0, f2<0: SE

-2 -1 0 1 2

-2

-1

0

1

2


-2 -1 0 1 2

-2

-1

0

1

2


-2 -1 0 1 2

-2

-1

0

1

2


Figure 6.3: Constructing the phase portrait for the system (6.10).

Example 6.1(d)This is the system

x = F (x, y) = −y , y = G(x, y) = x ,

(xy

)= f =

(−yx

). (6.13)

(i) The velocity function F (x, y) = 0 on the x–axis and G(x, y) = 0 on the y–axis. Theseintersect at the point (0, 0) and so this is a fixed point.(ii) The velocity field points NW in the first quadrant, SW in the second, SE in the third andNE in the fourth.(iii) The velocity function has |f | =

√x2 + y2 and so the field is bigger and the flow faster

the further from the origin.(iv) The velocity fields points radially away from the origin (see figure 6.4).(v) If we join up the arrows it looks like we get radial straight lines. We can check by solvingthe equations for the flows,

x = −y , y = x , ⇒ x = −x , ⇒ x = A cos t+B sin t , y = A sin t−B cos t ,

so x2 + y2 = A2 +Ba , (6.14)



or solving directly for the phase curves, as the equation is separable

dy

dx=G(x, y)

F (x, y)= −x

y,

1

2y2 = −1

2x2 + c , x2 + y2 = A , (6.15)

and we see that these are indeed circles.

-2 -1 0 1 2

-2

-1

0

1

2


f1<0, f2<0: SW f1<0, f2>0: NW

f1>0, f2<0: SE f1>0, f2>0: NE

-2 -1 0 1 2

-2

-1

0

1

2


-2 -1 0 1 2

-2

-1

0

1

2


-2 -1 0 1 2

-2

-1

0

1

2


Figure 6.4: Constructing the phase portrait for the system (6.13).

Example 6.1(e)Describe the phase portrait associated with the velocity field given by

f(x, y) = 2xye1 + (y2 − x2)e2 , f(x, y) =

(f1(x, y)f2(x, y)

), f1 = 2xy , f2 = y2 − x2 . (6.16)

(i) f1(x, y) = 2xy vanishes on the lines x = 0 and y = 0 so that the velocity field is verticalon these lines; f1(x, y) = y2 − x2 vanishes on the lines x = y and y = −x so that the velocityfield is horizontal on these lines. The null-clines intersect at one point (0, 0) which is thereforea fixed point of the DS. (ii) The null-clines divide the phase space into 8 regions on which thevelocity field is in one of the directions NW, NE, SE, SW as in figure 6.5. (iii) Looking at thefi we see that the velocity field has larger magnitude the further it is from the origin. If we



want to be more accurate, the velocity field at (x, y) makes an angle θ with the x-axis where

tan θ =f2

f1=y2 − x2

2xy

and has magnitude proportional to

|f(x, y)| = ((4x2y2 + (y2 − x2)2)1/2 = (x2 + y2) .

(iv) The velocity field appears to be making circles symmetrically either side of the y–axis.(v) We cannot easily solve the equations for the trajectories, (6.16), but we have alreadysolved the equation of the phase curves

dy

dx=y2 − x2

2xy, (6.17)

in example 2.6 with the general solution

(x−A/2)2 + y2 = (A/2)2 . (6.18)

This describes a family of circles, with centres at (A/2, 0) and radii A/2. Note that they allpass through (0, 0), but since this is a fixed point the orbits are actually the circles with theorigin omitted.



-2 -1 0 1 2

-2

-1

0

1

2


f1>0, f2>0: NE

f1>0, f2>0: NE

f1>0, f2<0: SE

f1>0, f2<0: SE

f1<0, f2>0: NW

f1>0, f2>0: NW

f1<0, f2<0: SW

f1<0, f2<0: SW

-2 -1 0 1 2

-2

-1

0

1

2

(b) The regions in which thevelocity field points NE, NW,SE, SW.

-2 -1 0 1 2

-2

-1

0

1

2


-2 -1 0 1 2

-2

-1

0

1

2


Figure 6.5: Steps in a construction of a sketch of the phase-portrait of velocity field (6.16).



6.2 Separable Systems

In second order dynamical systems the dynamical evolutions of the variables x and y are quitegenerally coupled: both velocity functions depend on x and y.

If we have the special situation

x = F (x) , y = G(y) , (6.19)

then we say that the system is separable, the differential equations for x(t) and y(t) are alsoseparable and can be solved easily to find

t− t0 =

∫ x(t)

x0

dx

F (x)=

∫ y(t)

y0

dy

G(y). (6.20)

If a system is not separable in the original coordinates (x, y) it may be possible to find newcoordinates (u, v) in which it is2. We have already met a system which is separable in newcoordinates, which is example 6.1(d). The orbits are circles and so it could be helpful toconsider polar coordinates (r, θ) defined by

x = r cos θ

y = r sin θ,

r =

√x2 + y2

θ = arctan( yx

) . (6.21)

To find r and θ it can be simplest to differentiate the equations

r2 = x2 + y2 , tan θ =y

x, (6.22)

leading to

2rr = 2xx+ 2yy ⇒ r =xx+ yy

r=xF (x, y) + yG(x, y)

r, (6.23)

sec2 θ θ = (1 + tan2 θ)θ =xy − yxx2

⇒ θ =xy − yxx2 + y2

=xG(x, y)− yF (x, y)

x2 + y2. (6.24)

We now turn to example 6.1(d)

Example 6.2. Consider the following DS in polar coordinates

x = F (x, y) = −y , y = G(x, y) = x . (6.25)

We find

r =xF (x, y) + y G(x, y)

r= 0 , θ =

xG(x, y)− y F (x, y)

x2 + y2= 1 . (6.26)

These equations are both separable and very simple with the solutionsr = r0

θ = t+ θ0

,

x = r0 cos(t+ θ0)

y = r0 sin(t+ θ0). (6.27)

The phase portrait is in figure 6.4.

2It is always possible to find coordinates which cover a small region of phase space around any point whichis not a fixed point in which the equations are separable, see supplementary exercise 6.3, but these do not haveto extend to the whole of phase space.



Example 6.3. Another system separable in polar coordinates

Consider the system described by the equations

x = αx− ωy, y = ωx+ αy, (6.28)

where α, ω are constants.

Changing to polar coordinates (r, θ) we find

r =xF (x, y) + y G(x, y)

r= α

x2 + y2

r= αr , θ =

xG(x, y)− y F (x, y)

x2 + y2= ω . (6.29)

These are again very simple and can be solved to find the trajectories

r = r0eαt , θ = ωt+ θ0 . (6.30)

The orbits are spirals. If α > 0 these are spiralling away form the origin and the origin is an unstablefixed point; if α < 0 they are spiralling in towards the origin which is a stable fixed point.

-2 -1 0 1 2

-2

-1

0

1

2

(a) The case α = 1, ω = 4: anunstable fixed point.

-2 -1 0 1 2

-2

-1

0

1

2

(b) The case α = −1, ω = 4: astable fixed point.

Figure 6.6: Phase portraits of two examples of the DS (6.28).

A more complicated example separable in polar coordinates.

Example 6.4. Show that the following system is separable in polar coordinates and sketchthe phase portrait

x = F = x− x3 − xy2 − y , y = G = y − y3 − yx2 + x . (6.31)

Following the previous procedure we find

r =xF + y G

r= r − r3 , θ =

xG− xFr2

= 1 . (6.32)

The solution for θ = θ0 + t; the equation for r can be analysed as first order autonomoussystem (with phase space r ≥ 0) and we see that it has two fixed points, an unstable fixedpoint at r = 0 and a stable fixed point at r = 1 (corresponding to a periodic trajectory ofthe full system with r = 1 and a closed orbit, the unit circle). The remaining orbits of the



full system either spiral out from the fixed point at (0, 0) and tend towards the unit circle orspiral in from infinity again tending towards the unit circle. The unit circle is an example ofa limit cycle. This is illustrated in figure 6.7.

1r

f HrL

(a) The phase portrait of thefirst order DS r = r(1− r2).

-2 -1 0 1 2

-2

-1

0

1

2

(b) The phase portrait of thefull system.

Figure 6.7: Phase portraits of the DS (6.31).

A final example separable in different coordinates

Example 6.5. Show that the following DS is separable in coordinates (u, v) where x =u cosh v, y = u sinh v:

x = y , y = x . (6.33)

We can find u and v from differentiating u2 = x2 − y2 and tanh v = y/x, or instead by directsubstitution

x = u cosh v + uv sinh v =x

uu+ yv = y , y = u sinh v + uv cosh v =

y

uu+ xv = x ,

u = 0 , v = 1 , (6.34)

which is separable with trajectories

u = u0 , v = t+ v0 . (6.35)

The orbits are hyperbolae, as in figure 6.8 since

x2 − y2 = u2(cosh2 v − sinh2 v) = u20 . (6.36)

Note that the coordinates (u, v) only cover the region |x| ≥ |y|. Can you suggest a change ofcoordinates to cover the other region?



-2 -1 0 1 2

-2

-1

0

1

2

Figure 6.8: Phase portraits of the system (6.33). The coordinates (u, v) cover the shadedregion only.

6.3 The structure of orbits and Phase Space

We have seen that for a well-behaved velocity function (obey the requirements of Picard’stheorem) the orbits of a first order dynamical system are either fixed points or open intervals.Each trajectory in a first order system either stays at a fixed point or approaches a fixedpoint (or ±∞). There are only three kinds of fixed point, a stable fixed point and two kindsof unstable fixed point as in figure 5.12.

The situation is much more complicated for second order autonomous systems. In the fewexamples so far we have already met several different types of stable and unstable fixed point,and a new kind of limiting behaviour, the limit cycle.

It is in fact possible to show that the limiting behaviour of any trajectory is either (a) afixed point (b) a limit cycle or (c) a union of fixed points and open intervals interpolatingthem but this is too complicated to even state carefully here, quite apart from contemplateproving. The example (6.137) in supplementary exercise 6.2 with α > 0 exhibits just such acomplicated behaviour; the phase portrait is in figure 6.8. There are fixed points at (0, 0),(−1,−1) and (1,−1) and orbits between these last two; every orbit that starts close to (0, 0)ends up approaching the other two fixed points and the orbits between them ever more closely.This might sound complicated, but it is still simple compared to the behaviour for third orhigher order systems where far more random and chaotic behaviour is possible. The reasonis it is still relatively simple for second order systems is that the phase space is in the planeand an orbit is a line on the plane which divides the space up into the part on one side of theline and the part on the other. A line in three dimensions does not split three-dimensionalspace up at all.



Α=

1

4

-2 -1 0 1 2

-2

-1

0

1

2

Figure 6.9: Phase portraits of the system (6.137) with α = 1/4..

In this course we will look more deeply into two aspects of the structure of second ordersystems, namely the stability of limit cycles and the stability of fixed points. The first topicis relatively straightforward as it is essentially a one-dimensional problem and we summarisethe results in section 6.4; the second is more complicated. We shall look in detail into fixedpoints in 6.5 and undertake a linear stability analysis and classify the possible fixed pointsthat can arise in this way in section 6.6.

6.4 Limit cycles

A limit cycle of a second order dynamical system is a periodic trajectory corresponding to aclosed orbit that does not pass through any fixed points. The orbits that pass close to thelimit cycle can either all stay close to, or be attracted in towards, the limit cycle, in whichcase it is a stable limit cycle, they can all be repelled (it is an unstable limit cycle) or some canbe attracted and some repelled; formally this is also unstable but can be called “half-stable”.

We have already such a system in example 6.4 for which the unit circle is a stable limit cycle- all nearby trajectories spiral ever closer to the unit circle. We shall give some examples ofthe other types of behaviour for systems separable in polar coordinates.

Example 6.6. Consider the systems in polar coordinates with

r = f(r) , θ = 1 . (6.37)

The condition for there to be a limit cycle is that the constant trajectory r = a is a solution,which requires f(a) = 0. From our analysis in section 5.4.1 on the stability of fixed points,if f ′(a) > 0 then this is an unstable fixed point of the first order system for r and hence anunstable limit cycle; the first order system and the limit cycle is “half-stable”. Three simplemodels of this are

(a) f(r) = r − r3: the fixed point at r = 1 has f(1) = 0 and f ′(1) = −2, hence it is stable.



This is precisely example 6.4 and the phase portrait of the first order system for r andthe full second order system are in figure 6.7

(b) f(r) = −r + r3: the fixed point at r = 1 has f(1) = 0 and f ′(1) = +2, hence it isunstable.

(c) f(r) = r(1− r2)2: now f(1) = f ′(1) = 0 but f ′′(1) = 8, so the cycle is “half-stable”.

We illustrate cases (b) and (c) in figures 6.10 and 6.11 with plots of f(r), the phase portraitof the first order system corresponding to r and the phase portrait of the full second ordersystem.

1r

f HrL

(a) The phase portrait of thefirst order DS r = −r(1− r2).

-2 -1 0 1 2

-2

-1

0

1

2


Figure 6.10: Phase portraits of the DS (6.6)(b).

1r

f HrL

(a) The phase portrait of thefirst order DS r = r(1− r2)2.

-2 -1 0 1 2

-2

-1

0

1

2


Figure 6.11: Phase portraits of the DS (6.6)(c).



6.5 Fixed points of second-order autonomous systems

Let’s consider a second order autonomous system described by

dx

dt= F (x, y) ,

dy

dt= G(x, y) , (6.38)

or in vector notationd

dt

(xy

)= f(x, y) =

(F (x, y)G(x, y)

). (6.39)

Let’s assume that the point (a, b) is a fixed point, i.e. F (a, b) = G(a, b) = 0 and (a, b) is anintersection point of the null-clines of F and G. The constant trajectory x = a, y = b is asolution of the dynamical system.

This means that we can perform a Taylor expansion of the functions F and G around thefixed point and the constant terms will be zero. The Taylor expansion in two dimensions is

f(x, y) = f(a+ x− a, b+ y − b)) = f(a, b) +∂f

∂x(x− a) +

∂f

∂y(y − b)

+1

2!

[∂2f

∂x2(x− a)2 + 2

∂2f

∂x∂y(x− a)(y − b) +

∂2f

∂y2(y − b)2

]+ . . . , (6.40)

where it is understood that all partial derivatives are evaluated at (x, y) = (a, b). Only termsup to second order in the small quantities x− a and y − b are explicitly shown.

Applying this expansion idea to both velocity functions F and G, and assuming that (x, y) =(a, b) is a fixed point of the system, so that the zeroth order terms vanish, F (a, b) = G(a, b) =0, we get

F (x, y) = (x− a)∂F

∂x+ (y − b)∂F

∂y+ . . .

G(x, y) = (x− a)∂G

∂x+ (y − b)∂G

∂y+ . . . (6.41)

We saw in section 5.4.1 that if f ′(a) 6= 0 for a first order system at a fixed point f(a) = 0then the nature of the fixed point is completely determined by the sign of f .

We can try the same analysis here for a second order system: if we neglect the second orderand higher terms in 6.41 and approximate the full system by a linear system of equationswe can ask: under what conditions is the nature of the fixed point determined by the linearsystem, and what quantities are necessary to determine the nature?

This is the subject of the next section.

6.6 Linear Stability Analysis

The idea is to approximate a second order DS near a fixed point by a linear DS. Sometimesthis is all that is needed to determine its stability - sometimes not.

We will investigate the stability of the fixed point by performing a linear stability analysis,based on a two-variable Taylor expansion (to first order) of the velocity functions F and Gin the vicinity of the fixed point. This procedure involves three steps.



(1) Taylor expansion of the two velocity functions in the vicinity of the fixed point (a, b) inquestion. At first order, the terms in this expansion can be arranged into a matrix, theso-called Jacobian J of the system.

(2) Computing the associated Jordan canonical forms C, corresponding to J ; the Jordancanonical form is determined by the eigenvalues. We will see how to (almost) determinedthe form from analysing simple quantities calculated from the matrix J without needingto find the eigenvalues themselves.

(3) Exploring the consequences for dynamics: The linearised equations of motion formulatedin a coordinate system corresponding to the Jordan forms can all be solved. Thesolutions allow a complete classification of the fixed points (and the nature of the phaseflow in their vicinity).

6.6.1 Step 1 — Taylor Expansion of Velocity Functions

We start from the differential equations (6.38) and substitute F (x, y) and G(x, y) by theirfirst order approximations from (6.41) and obtain the linearised system

x = (x− a)∂F

∂x+ (y − b)∂F

∂y, y = (x− a)

∂G

∂x+ (y − b)∂G

∂y. (6.42)

It is no much more useful to write this in vector notation as(x

y

)=

((x− a)∂F∂x + (y − b)∂F∂y(x− a)∂G∂x + (y − b)∂G∂y

)=

(∂F∂x

∂F∂y

∂G∂x

∂G∂y

)︸︷︷︸

The Jacobian J

(x− ay − b

), (6.43)

where J denotes the Jacobian matrix given by

J =

(∂F∂x

∂F∂y

∂G∂x

∂G∂y

), (6.44)

evaluated at the fixed point (a, b). Writing x = x− a, y = y − b Eq (6.43) becomes

d

dt

(xy

)= J

(xy

). (6.45)

The change of variable (x, y) → (x, y) represents a change of origin; the origin of the x-ycoordinates is in fact the fixed point (x = a, y = b).

Linear systems of the sort (6.45) are straight forward to solve; we have already solved severalin examples 6.1(a), (b), (c) and (d). Let us consider another useful example:

Example 6.7. Find the solutions of the following linear dynamical system

(xy

)=

(2x− yy

)=

(2 −10 1

)(xy

). (6.46)



If we define X = x− y and Y = y then we find

X = x− y = 2(x− y) = 2X , Y = y = y = Y , (6.47)

or (X

Y

)=

(2 00 1

)(XY

). (6.48)

The equations have decoupled and we find the solutions

X = a e2t , Y = b et . (6.49)

We see that (0, 0) is a strongly unstable fixed point of the type known as an “unstablenode” (will give all the various types names later on). The orbits in the new coordinates areparabolae,

X =a

b2Y 2 . (6.50)

We can plot the orbits in both the new coordinates (X,Y ) and the old coordinates (x, y) infigure 6.12

-2 -1 0 1 2

-2

-1

0

1

2

-2 -1 0 1 2

-2

-1

0

1

2

Figure 6.12: Sketch of the trajectories (5.4) for the DS (6.46) in original coordinates(x, y) (on the left) and redefined (X,Y ) coordinates (on the right).

If we think about what we have done, we took a system of the form(xy

)= M

(xy

), (6.51)

where M is a constant matrix and found new coordinates (X,Y ), defined in terms of a newconstant matrix P by (

xy

)= P

(XY

),

(XY

)= P−1

(xy

). (6.52)

so that the system (X

Y

)= P−1M P︸︷︷︸

C

(XY

), (6.53)



is simpler. It turns out that the analysis works well if we take the matrix C to the Jordancanonical form of J — this is why we have called it C, for canonical. It is time to revise somelinear algebra and see how to find the canonical form C of a matrix J .

6.6.2 Step 2 — Finding the Jordan canonical form of the Jacobian

The Jordan canonical form of a matrix M is determined by the eigenvalues of the matrix M .If they are both real and distinct, as in example 6.7, then the canonical form is diagonal. Oneway to find the eigenvalues of a 2×2 matrix M we can assume the existence of an eigenvectorand solve the resulting equations. If

M =

(a bc d

), (6.54)

and we assume that the vector v =

(1α

)is an eigenvector with eigenvalue λ then from the

equationM v = λ v , (6.55)

we can eliminate α and obtain the quadratic equation for λ

p(λ) = λ2 − tλ+ ∆ = 0 , (6.56)

where t is the trace of the matrix M and ∆ is the determinant of M :

t = (a+ d) = Tr(M) , ∆ = ad− bc = Det(M) . (6.57)

The equation (6.56) is called the characteristic equation for λ.

It is a Theorem, the Cayley-Hamilton theorem, that every matrix M obeys its own charac-teristic equation, i.e.

p(M) = M2 − tM + ∆ I = 0 , (6.58)

where I is the identity matrix.

Given the characteristic equation (6.56), we can find the eigenvalues as

λpm =1

2t± 1

2

√D , D = t2 − 4∆ . (6.59)

D is called the discriminant of the matrix M . We now write the characteristic equation as

p(λ) = (λ− λ+)(λ− λ−) = λ2 − (λ+ + λ−)λ+ (λ+λ) = 0 , (6.60)

and comparing this with (6.56) we see that

t = λ+ + λ− , ∆ = λ+λ− . (6.61)

Looking at the form of the eigenvalues (6.59), we see that it is the sign of the discriminant Dwhich determines if they are real or complex. We distinguish three cases:



(1) D > 0, λ± are both real and the Jordan canonical form C is diagonal. It is possible tofind a real matrix P such that the canonical form C is diagonal,

C = P−1MP =

(λ+ 00 λ−

). (6.62)

(2) D < 0, λ± are complex conjugate complex numbers, λ± = α ± iβ. It is possible to finda real matrix P such that the canonical form is

C = P−1MP =

(α β−β α

). (6.63)

and the Jordan canonical form is not diagonal (It has a complex diagonal canonicalform but this is the real canonical form).

(3) D = 0, there is only one solution for λ but two choices for the Jordan canonical form C.

(3i) If M is diagonal, then it is already in canonical form,

C = M =

(λ 00 λ

). (6.64)

(3i) If M is not diagonal, then it is possible to find a real matrix P such that

C = P−1MP =

(λ 10 λ

). (6.65)

It is sometimes necessary to find a suitable matrix P – this is covered in section C.3 inappendix C. We note that P is not unique: there is an infinite choice of suitable matricesP , there is a procedure in appendix C which finds one of which works.

6.6.3 Step 4 — Exploring the Consequences for Dynamics

It turns out that the the resulting differential equations, when formulated in the X,Y coor-dinate system can be readily solved, and the stability of the fixed point determined for eachof the three different variants of Jordan canonical form.

Note that the origin of the new coordinates (X,Y ) is in fact the fixed point (x = a, y = b). Weassume that the stability of the fixed point (a, b) of the system described by equation 6.3 maybe determined by examining the behaviour of the linearised system 6.53 near X = 0, Y = 0;this is valid provided the matrix J is non-singular. If one of the eigenvalues of J were zero,and hence J singular, one would in fact have to consider higher orders in the Taylor expansionof the velocity functions to finally decide the stability of a fixed point.

We are looking atd

dt

(XY

)= C

(XY

). (6.66)

Referring to the three different types of Jordan canonical form introduced above we canrewrite this in terms of two coupled equations for the components X and Y . We now gothrough the various cases systematically.



D > 0 there are two real and distinct eigenvalues. λ1 6= λ2 ∈ IR we get the two equations

d

dtX = λ1X ,

d

dtY = λ2Y

which are readily solved to give

X(t) = X0eλ1t , Y (t) = Y0eλ2t

All that is left is to work out the signs of the eigenvalues which is fixed by t and ∆

D > 0,∆ > 0 This means λ1λ2 > 0 and so both eigenvalues are of the same sign. Thereare two choices:

D > 0,∆ > 0, t > 0 This means both eigenval-ues are positive, λ1, λ2 > 0.The motion is away fromthe fixed point: we get a so-called unstable node.

D > 0,∆ > 0, t < 0 This means both eigenval-ues are negative, λ1, λ2 < 0.The motion is towards thefixed point: we get a so-called stable node.

D > 0,∆ < 0 This means λ1λ2 < 0 and so the eigen-values are of opposite same sign. Thismeans the motion is out along one axisand in along the other. This fixed pointis unstable but it is given a special name,it is a hyperbolic fixed point.

D < 0 This means there are a pair of complex conjugate eigenvalues: λ1,2 = µ ± iν withµ ∈ IR and ν ∈ IR. In this case, the equations of motion for the X and Y coordinateare

d

dtX = µX + νY ,

d

dtY = −νX + µY

This system of equations is separable in polar coordinates X = R cos θ, Y = R sin θ.Following the line of reasoning in Section 6.2, we find a solution of the form

R(t) = R0 eµt , θ(t) = θ0 − νt



which translates back into the X,Y coordinate system as

X(t) = R0 eµt cos(θ0 − νt) , Y (t) = R0 eµt sin(θ0 − νt)

The nature of the fixed point now depends on whether the motion is spiralling awayfrom the fixed point (µ > 0) or spiralling in towards the fixed point (µ < 0). Sincet = 2µ, we can decide this on the sign of t.

D < 0, t > 0 This means µ > 0 and so the motionis spiralling away from the fixed point.This is known as an unstable focus.

D < 0, t < 0 This means µ < 0 and so the motion isspiralling in to towards from the fixedpoint. This is known as a (strongly)stable focus. The word strongly isused to differentiate it from the ellipticalfixed point below.

D < 0, t = 0 This means µ = 0 and so the motionis circular in the coordinates (X,Y ).When transformed back into the coor-dinates (x, y) it would be an ellipse andso this fixed point is known as an ellip-tical fixed point. It is also mid-waybetween the stable and unstable foci - asmall change in the system would turnit into one or the other and so it is alsoknown as a marginally stable focus

Finally we come to the case

D = 0 This means there is only one solution to the characteristic equation p(λ) = 0. Ctakes one of two forms: (

λ 00 λ

),

(λ 10 λ

)(6.67)

In the first case C = λI and so M = λI as well. This means that if M 6= λI thenC must be of the second form. If the matrix C = λI then the flow is diagonalisablethen it is very similar to the case of two distinct real eigenvalues except that theorbits are straight lines. It only depends on whether λ > 0 or λ < 0 to decide ifthe fixed point is unstable or stable.

If, however, C is not diagonalisable and instead takes the form

C =

(λ 10 λ

)



then we need to solve the equations

d

dtX = λX + Y ,

d

dtY = λY

Note that the Y dynamics is independent of X; the solution is Y (t) = Y0eλt.Inserting this into the ODE for X, one obtains a first order linear equation whichis readily solved (using integrating factors). Overall, we get

X(t) = (X0 + Y0 t)eλt , Y (t) = Y0eλt

Note that the ratio of X and Y as a function of t is

Y (t)

X(t)=

Y0

X0 + Y0 t,

and it approaches 0 for t → ±∞. This type of fixed point is called an impropernode and again it only depends on the sign of λ to determine if it is stable orunstable.

Since λ = t/2, we need only look at the sign of t to decide which the case is.

This gives us four cases

D = 0, t > 0, M , diagonal This is an unstable node.

D = 0, t > 0, M not diagonal This is an unstable improper node

D = 0, t < 0, M , diagonal This is a stable node.

D = 0, t < 0, M not diagonal This is a stable improper node]



The properties of the phase flow in the vicinity of the fixed point are finally obtained fromthe properties of the phase flow in the X,Y coordinate system by transforming back into thex, y system via (

xy

)= P

(XY

); (6.68)

see Eq. (6.52). This transformation is linear (and non-singular). It may involve rotations,reflections, and scaling of axes.

We summarise the eight types of flow pattern and the various fixed points in figure 6.13

Δ<0, D>0

Hyperbolic fixed point

Δ>0, D>0

Unstable node

Δ>0, D>0

Stable node

D<0, t>0

Strongly unstable focus

D<0, t<0

Strongly stable focus

D=0, t>0

Unstable nodeor unstable improper node

D=0, t<0Stable nodeor stable improper node

D<0, t=0

Marginally stable focus

t2 =4Δ, D=0

or

or

Δ

t

Figure 6.13: Summary of the eight types of fixed point in the ∆–t plane

We have given a way to find the nature of the fixed point based on the values of t, ∆ andD = t2 − 4∆ but of course this is just a way to summarise the various possibilities for theeigenvalues. We could just have easily given the method:



(1) Find the eigenvalues of M

(2) Knowing the eigenvalues of M , find the Jordan normal form C

(3) Solve the equations X = CX to find the flow pattern near the fixed point

The ultimate results will be the same.



We illustrate the various fixed points in a series of examples.

Example 6.8. Find the fixed points of the second order autonomous system described by the differ-ential equations

x = y − x2 + 2 , y = 2(x2 − y2) , (6.69)

and investigate their stability.

The fixed points are given by solutions of

y − x2 + 2 = 0, x2 − y2 = 0 . (6.70)

The second of these equations gives y = ±x and we soon find that there are four fixed points, (-1,-1),(2,2), (1,-1), (-2,2). We have, in the above notation,

f1(x, y) = y − x2 + 2 , f2(x, y) = 2(x2 − y2) ,

and the Jacobian matrix J , evaluated at arbitrary (x, y) is given by

J =

(∂f1∂x

∂f1∂y

∂f2∂x

∂f2∂y ,

)=

(−2x 14x −4y

), (6.71)

We look at these fixed points in turn.

Case 1 The fixed point (−1,−1) has

J(−1,−1) =

(2 1−4 4

). (6.72)

The matrix J(−1,−1) has trace t = 6, determinant ∆ = 12 and discriminant D = −12. From theclassification we find it is a focus since D < 0 and unstable since t > 0, so it is an unstablefocus.

Let us now find the eigenvalues exactly. J(−1,−1) has eigenvalues λ given by

(λ− 2)(λ− 4) + 4 = 0 ⇒ λ = 3± i√

3. (6.73)

It follows that there is a matrix P such that

P−1J(−1,−1)P =

(3

√3

−√

3 3

). (6.74)

Near (−1,−1) the linearised equations thus take the form

d

dt

(XY

)=

(3

√3

−√

3 3

)(XY

), (6.75)

where (XY

)= P−1

(x+ 1y + 1

). (6.76)

This system of equations is separable in polar coordinates. Thus we change the variables from(X,Y ) to (r, θ) where X = r cos θ, Y = r sin θ. We know from the arguments following Eq (6.28)that our equations take the form

r = 3r, θ = −√

3 .

These equations have integrals

r(t) = r0e3t , θ(t) = θ0 −

√3t ,



where r0 = r(0), θ0 = θ(0). The phase curves have the form of ‘outgoing spirals’ indicated inthe diagram and we deduce that (x = −1, y = −1) is an unstable focus.

Note In discussing the stability we didn’t need to know the form of P , merely that it existed.However, let’s calculate P explicitly so as to exhibit the change of variable we’ve used. Recallthe procedure for finding Jordan forms: We consider the equation(

2 1−4 4

)(uv

)= (3 + i

√3)

(uv

). (6.77)

These equations aren’t independent and we need only look at v = (1 + i√

3)u. Taking u = 1 wesee that an eigenvector is (

1

1 + i√

3

)=

(11

)+ i

(0√3

). (6.78)

With respect to the basis

f1 =

(11

), f2 =

(0√3

), (6.79)

we haveJ(−1,−1)(f1 + if2) = (3 + i

√3)(f1 + if2), (6.80)

so thatJ(−1,−1)f1 = 3f1 −

√3f2 , J(−1,−1)f2 =

√3f1 + 3f2 . (6.81)

It follows that

P−1J(−1,−1)P =

(3

√3

−√

3 3

), (6.82)

where

P =

(1 0

1√

3

). (6.83)

We see that

detP =√

3, P−1 =1√3

(√3 0−1 1

). (6.84)

The change of variables (x, y) 7→ (X,Y ) which we used in the above argument is therefore givenby (

XY

)= P−1

(x+ 1y + 1

), (6.85)

from which we derive

X = x+ 1, Y =y − x√

3. (6.86)

Case 2 The fixed point (2, 2): A similar calculation shows the Jacobian is

J(2,2) =

(−4 18 −8

), (6.87)

which has trace t = −12, determinant ∆ = 24 and discriminant D = 48. Hence the eigenvaluesare real (D > 0), of the same sign (∆ > 0) and negative (t < 0) and the fixed point is a strongstable node.

We now repeat the analysis in detail: The linearised equations take the form

d

dt

(XY

)= P−1

(−4 18 −8

)P

(XY

), X = x− 2, Y = y − 2 (6.88)



where we choose P so as to bring the matrix(−4 18 −8

)(6.89)

into Jordan form. The matrix in question has eigenvalues λ given by

(λ+ 4)(λ+ 8)− 8 = 0 ⇒ λ = −6± 2√

3 , (6.90)

so thatλ = λ1 = −6 + 2

√3 < 0 , λ = λ2 = −6− 2

√3 < 0 . (6.91)

It follows that there exists a P such that

P−1(−4 18 −8

)P =

(λ1 00 λ2

). (6.92)

With this choice of P our equations become

X = λ1X, Y = λ2Y , (6.93)

from which we deriveX(t) = X0e

λ1t, Y (t) = Y0eλ2t , (6.94)

where X0 = X(0), Y0 = Y (0). We see that (x = 2, y = 2) is a strongly stable fixed point (astrongly stable node).

We can, of course, eliminate t to obtain the linearised form of the equations of the phase curves.We find that

Y = Y0

(X

X0

)λ2/λ1

, λ2/λ1 = 2 +√

3 . (6.95)

Case 3 The fixed point at (1,−1): Here we find that

J(1,−1) =

(−2 14 4

), t = 2,∆ = −12, D = 52 . (6.96)

Since ∆ < 0 we know immediately that the eigenvalues are real and of opposite sign and thefixed point is hyperbolic.

To go through this in detail again: the system is equivalent to

d

dt

(XY

)= P−1

(−2 14 4

)P

(XY

), X = x− 1, Y = y + 1 (6.97)

where we choose P so as to bring the matrix(−2 14 4

)(6.98)

into Jordan form. This matrix has eigenvalues λ given by

(λ+ 2)(λ− 4)− 4 = 0 ⇒

λ = λ1 = 1 +√

13 > 0 , λ = λ2 = 1−√

13 < 0.

The argument then proceeds as in Case 2 and we obtain the equations

X = λ1X, Y = λ2Y , (6.99)

which givesX(t) = X0e

λ1t , Y (t) = Y0eλ2t , (6.100)

in a now familiar notation. Since λ1 > 0, eλ1t → ∞ as t → ∞. We see that (x = 1, y = −1) isan unstable node, more specifically a hyperbolic fixed point.



Case 4 The fixed point at (−2, 2). We see that

J(−2,2) =

(4 1−8 −8

), t = −4,∆ = −24, D = 64 (6.101)

and so since ∆ < 0 the fixed point is hyperbolic. Explicitly, the eigenvalues are

λ = λ1 = −2 + 2√

7 > 0, λ = λ2 = −2− 2√

7 < 0 . (6.102)

The argument used in Case 3 leads to the conclusion that (x = −2, y = 2) is also an unstable(hyperbolic) fixed point.

Here we give a numerical plot of the phase portrait and we see that indeed we have found thecorrect nature for each fixed point.

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

Figure 6.14: Phase-portrait of DS (6.69) showing that the fixed points are hyperbolic at(1,−1) and (−1, 1), a stable node at (2, 2) and an unstable focus at (−2,−2).

Example 6.9. Find the fixed point of the system of equations

x = f1(x, y) = 3x+ y + 1 , y = f2(x, y) = −x+ y + 6 , (6.103)

and investigate its stability.

The fixed points are given as solutions of

3x+ y + 1 = 0 , −x+ y − 6 = 0 , (6.104)

so that (−7/4, 17/4) is the only fixed point of the system. The Jacobian matrix J , evaluated at (x, y),is given by

J =

(∂f1∂x

∂f1∂y

∂f2∂x

∂f2∂y

)=

(3 1−1 1

), (6.105)

which has t = 4, ∆ = 4, D = 0. Since D = 0 this is either a node or an improper node, but since J isnot diagonal it must be an improper node and since t > 0 this is an unstable improper node.



Our equations may be written

d

dt

(xy

)=

(3 1−1 1

)(x+ 7/4y − 17/4

). (6.106)

Note that this equation is exact, because the original equations are linear ; we have not made anyapproximations, since the second and higher order derivatives of f1, f2 are all zero. Proceeding in theusual way we obtain

d

dt

(XY

)= P−1

(3 1−1 1

)P

(XY

), (6.107)

where (XY

)= P−1

(x+ 7/4y − 17/4

), (6.108)

and P is chosen so that the matrix (3 1−1 1

), (6.109)

is reduced to Jordan form.

We find that the eigenvalues of this matrix are given by

(λ− 1)(λ− 3) + 1 = 0 ,⇒ (λ− 2)2 = 0 , (6.110)

so that λ = 2 (twice). We can therefore choose P so that (see Eq (C.3))

C = P−1(

3 1−1 1

)P =

(2 10 2

). (6.111)

In the coordinate system (X,Y ) determined by P the differential equations become

X = 2X + Y , Y = 2Y . (6.112)

The second of these equations has general solution

Y (t) = Y0e2t , (6.113)

and the equation for X becomesX − 2X = Y0e

2t . (6.114)

This equation is of linear type with integrating factor µ(t) = exp∫−2 dt = e−2t. We therefore obtain

d

dt

(e−2tX

)= Y0e

2te−2t = Y0 (6.115)

from which we easily deriveX(t) = te2tY0 + e2tX0 . (6.116)

Since e2t → ∞ as t → ∞ we conclude that (x = −7/4, y = 17/4) is an unstable fixed point, morespecifically an unstable improper node. This is confirmed by eliminating t: we find that the phasecurves are given by

X =1

2Y ln

(Y

Y0

)+

(X0

Y0

)Y . (6.117)

For each choice of the initial values (X0, Y0) we obtain a phase curve. A plot of the phase portrait isin figure 6.15



-10 -5 0 5 10

-10

-5

0

5

10

Figure 6.15: Phase-portrait of DS (6.103) showing that the fixed point at (−7/4, 17/4) is anunstable improper node.

Example 6.10. Consider the second order system described by the differential equations

x = v1(x, y) = ax− bxy + ε sin(x− d/c) ,y = v2(x, y) = cxy − dy + ε sin(y − a/b) (6.118)

where ε is constant and a, b, c, d are positive constants.

We note that v1(d/c, a/b) = 0, v2(d/c, a/b) = 0 so that (d/c, a/b) is a fixed point.

The Jacobian matrix J , evaluated at (x, y) is given by

J =

(∂v1∂x

∂v1∂y

∂v2∂x

∂v2∂y ,

)=

(a− by + ε cos(x− d/c) −bx

cy cx− d+ ε cos(y − a/b)

), (6.119)

so that

J(d/c,a/b) =

(ε − bdcacb ε

). (6.120)

Near (x = d/c, y = a/b) the linearised equations become

d

dt

(XY

)= P−1J(d/c,a/b)P

(XY

), (6.121)

where (XY

)= P−1

(x− d/cy − a/b

), (6.122)

and the matrix P is chosen so as to bring J(d/c,a/b) into Jordan form. A standard calculation showsthat the eigenvalues of J(d/c,a/b) are given by

λ = ε± i(ad)1/2 , (6.123)



and our equations become

d

dt

(XY

)=

(ε (ad)1/2

−(ad)1/2 ε

)(XY

). (6.124)

This system is separable in polar coordinates and we write

X = r cos θ, Y = r sin θ , (6.125)

to obtainr = εr, θ = −(ad)1/2 . (6.126)

If ε = 0 we get r(t) = r0 which describes a family of circles with centre (X = 0, Y = 0), the fixedpoint. We see that if ε = 0 the fixed point is stable.

Note When ε = 0 the equations reduce to the Lotka-Volterra (“hare-lynx”) equations of chapter 1.The Lotka-Volterra equations are not structurally stable in the sense that if ε 6= 0 we can move froma situation of stable equilibrium to one of unstable equilibrium, as we shall now see. If ε < 0 we have

r(t) = r0eεt → 0 as t→∞,

and (x = d/c, y = a/b) is strongly stable. However, if ε > 0 we see that

r(t) = r0eεt →∞ as t→∞,

so that (x = d/c, y = a/b) is a position of unstable equilibrium.

Example 6.11. Harmonic motion. Consider the differential equation of simple harmonic motion

x+ x = 0 . (6.127)

Setting y = x we obtaind

dt

(xy

)=

(v1(x, y)v2(x, y)

)=

(y−x

)= J

(xy

), (6.128)

where

J =

(0 1−1 0

). (6.129)

Notice that Eq (6.128) is exact — no approximations have been made. Moreover, the matrix J isalready in the Jordan form appropriate to real 2× 2 matrices with eigenvalues ±i.We note that Eq (6.128) has one fixed point, namely (x = 0, y = 0). If we interpret the equationx + x = 0 in terms of the motion of a Newtonian particle (see following chapter), the fixed pointcorresponds to the particle at rest at x = 0. We now demonstrate that the fixed point is stable asfollows. Changing to polar coordinates (r, θ) where x = r cos θ, y = r sin θ we find in the usual waythat

r = 0 , θ = −1.

which have the integralsr(t) = r0 , θ(t) = θ0 − t,

in the usual notation. These are a family of circles centred on the origin (x = 0, y = 0) with radiusgiven by the parameter r0. The fixed point is therefore stable. We can also show this directly formEq (6.128) by noting that

x = y , y = −x.Eliminating t we obtain

dy

dx= −x

y⇒ x dx+ y dy = 0.

It follows thatx2 + y2 = r0

2,

where r0 is a parameter. In other words, the phase curves are indeed a family of circles, with centresat (0, 0); the point (0, 0) is therefore a stable fixed point.



Example 6.12. The damped harmonic oscillator

In physical situations there are often forces such as friction which lead the oscillations todecay. Adding such a term to equation (6.127) gives the differential equation

x+ λx+ x = 0 . (6.130)

With y = x this is the second order system(xy

)=

(0 1−1 −λ

) (xy

), (6.131)

a linear system and hence equivalent to one of the linearised systems studied in section 6.6with ∆ = 1, t = −λ and D = λ2 − 4. Hence we see that for 0 < λ < 2 this is a stablefocus; for λ = 2 it is a stable improper node; and for λ > 2 it is stable node. We can sketchthe graph of x(t) for typical trajectories by looking at the phase portrait and following thefeatures along an orbit.

Firstly, for λ = 1/4, this is a stable focus. Starting at the point A on the phase portrait withx = 1, x = 0, the motion is for x to oscillate but with ever decreasing amplitude, as in figure6.16. This is called under damping. The exact solution is

x = e−t/8 cos(ωt) , ω = 3√

7/8 (6.132)

ω is sometimes called the damped angular frequency or pseudofrequency of the motion.

For λ = 2, this is a stable improper node. The motion will tend to zero after at most oneoscillation as in figure 6.17 This is called critical damping.

For λ = 4, the motion is very similar to the critically damped solution but tends to zerorather slower in the final stages as in figure 6.18; this is called over damping. One can showthat the motion tends to zero faster for critical damping – see exercise 6.20.

A

B

C

D

E

F

-2 -1 0 1 2

-2

-1

0

1

2

(a) The phase portrait.

A

B

C

D

E

F

5 10 15 t

-1.0

-0.5

0.5

1.0

xHtL

(b) A typical solution.

Figure 6.16: The under-damped harmonic oscillator with λ = 1/4.



A

B

C D

-2 -1 0 1 2

-2

-1

0

1

2


A

B

C

D

1 2 3 4 5 6 t

-1.0

-0.5

0.5

1.0

xHtL

(b) Two typical solutions.

Figure 6.17: The critically damped harmonic oscillator with λ = 2.

A

B

-2 -1 0 1 2

-2

-1

0

1

2


A

B

1 2 3 4 5 6 t

-0.5

0.5

1.0

xHtL

(b) A typical solution.

Figure 6.18: The over-damped harmonic oscillator with λ = 4.

6.7 Beyond linear stability analysis

The methods of section 6.6 do not always work, exactly as the linear stability analysis of afirst order system dies not work in f ′(a) = 0. Looking at figure 6.13, we see that the secondorder analysis fails exactly when ∆ = detJ = 0. In this case one or both of the eigenvaluesof J is zero and we need to go to higher orders in the Taylor expansion to understand thenature of the fixed point.

We have already met a case where the fixed point is not one of the eight types in figure 6.13,and that is example 6.1(e). The fixed point at (0, 0) in figure 6.5 is not one of the eight types,and the reason is that the Jacobian vanishes at the fixed point:

F = 2xy , G = y2 − x2 , J(x, y) =

(∂Fdx

∂F∂y

∂Gdx

∂G∂y

)=

(2y 2x−2x 2y

), J(0, 0) =

(0 00 0

).

(6.133)There are more examples in supplementary exercise 6.4.



6.8 Exercises

Short Exercises

Exercise 6.1 Sketch the phase diagram associated with the uniform velocity field given by

f(x, y) = −2e1 + e2 ,

Find the equation of the phase curves.

Do the same for the velocity field given by

f(x, y) = −2e1 + xe2 ,

Exercise 6.2For each of the following second order dynamical systems of the form

dx

dt= v1(x, y) ,

dy

dt= v2(x, y)

you should

(i) Find the null-clines of v1 and v2 and hence find the fixed points.

(ii) Give a qualitative sketch of the phase portrait, including the null-clines and using theusual arrow representation of the phase flow.

(iii) Find an equation for dy/dx and solve it to find the phase curves explicitly

(a) v1 = y , v2 = 1

(b) v1 = x , v2 = −y

(c) v1 = −y , v2 = 2x

(d) v1 = y , v2 = x+ x2

(e) v1 = x , v2 = y (1− x)

Exercise 6.3For each of the following second order dynamical systems of the form

dx

dt= v1(x, y) ,

dy

dt= v2(x, y)

you should

(i) Find the null-clines of v1 and v2 and hence find the fixed points.

(ii) Give a qualitative sketch of the phase portrait, including the null-clines and using theusual arrow representation of the phase flow.

(a) v1 = 2xy , v2 = x2 + y2

(b) v1 = 2xy , v2 = 1− x2 − y2



Exercise 6.4 Find the fixed points of the system described by

x = −3y + xy − 4, y = y2 − x2

Exercise 6.5 Find the fixed point of the system of differential equations

x = exp (x+ y)− y, y = xy − x


Exercise 6.6 Find the fixed point of the system of differential equations

x = x+ 3y, y = −6x+ 5y


Longer Exercises

Exercise 6.7Consider second order linear dynamical systems of the form(

xy

)= J

(xy

),

where J is one of the following constant matrices. In each case you should, in turn,

(a) Find the trace, t, the determinant, ∆, of J .

(b) Find the discriminant, D = t2 − 4∆, of the characteristic equation for J .

(c) State the nature of the fixed point (0, 0).

(d) Find the eigenvalue(s) of J

(e) Give the Jordan normal form, C, of J .

(f) Find a matrix P such that P−1JP is in Jordan normal form. Is it the same as the matrixyou stated in (e)? If not, why not?

(You do not need to do part (d) to do part (c), etc.)

(i)

(1 31 −1

)(ii)

(1 −30 −2

)(iii)

(2 2−1 0

)(iv)

(1 −11 3

)(v)

(−3 −1−1 −3

)(vi)

(−5 10−10 11

)(vii)

(−8 13−5 8

)(viii)

(11 −518 −8

)



Exercise 6.8Consider the following second order dynamical systems. In each case, linearise the systemaround the stated fixed point(s), find the trace and determinant of the corresponding constantmatrix and state the nature of the fixed point of the linear system. In which cases does thisdetermine the nature of the fixed point of the full system? Compare with the phase portraitsgiven in the solution sets.

(i) Exercise 6.2(b): the fixed point at (0, 0) of the system

x = x , y = −y

(ii) Exercise 6.2(c): the fixed point at (0, 0) of the system

x = −y , y = 2x

(iii) Exercise 6.2(d): the fixed points at (0, 0) and (−1, 0) of the system

x = y , y = x+ x2

(iv) Exercise 6.2(e): the fixed point at (0, 0) of the system

x = x , y = y(1− x)

(v) Exercisr 6.3(b): the fixed points at (1, 0) and (0,−1) of the system

x = 2xy , y = 1− x2 − y2

(vi) Supplementary exercise 6.2: the fixed point at (0, 0) in the three cases α = −1/4,α = 1/4 and α = 0 of the system

x = y + x2 + αx(1− y + 2x2) , y = −2x(1 + y)

Exercise 6.9Write the equation

x+ x = f(x)

in standard form by putting y = x. Assuming that f is suitably differentiable and thatf(0) = 0 discuss the stability of the fixed point (x = 0, y = 0), distinguishing the casesf ′(0) > 1, f ′(0) < 1.

Note We could have guessed the answer by the following argument. Near x = 0f(x) ' f(0) + xf ′(0) + · · · so that the differential equation is approximated by

x+ (1− f ′(0))x ' 0.

If f ′(0) < 1 the motion is essentially Simple Harmonic and therefore the fixed point shouldbe stable.



Exercise 6.10Write the differential equation

x+ 2λx+ ω2x = 0

(the equation of damped harmonic motion —see Exercise 5.13) in standard form by puttingy = x. Find the fixed point and discuss its stability.

Exercise 6.11A second order dynamical system is described by the differential equations

x = −y + x sinπr, y = x+ y sinπr,

where r = (x2 + y2)1/2. Prove that the system is separable in polar coordinates and show inparticular that

r = r sinπr.

Sketch the phase diagram associated with this equation write down the fixed points, andindicate which are stable and which are unstable.

Describe qualitatively, without computation, how the representative point (x, y) behaves as afunction of time t given the initial conditions r(0) = 3/2, θ(0) = 0. What is your conclusionif r(0) = 5/2, θ(0) = 0.?

Exercise 6.12 Consider the system of differential equations

x = (ay − b)x+ f(x), y = (c− dx)y + g(y),

where a, b, c, d are positive constants, and f, g are differentiable functions such thatf(c/d) = 0, g(b/a) = 0. Notice that the “fox-rabbit” equations are a particular case of these.

Show that (c/d, b/a) is a fixed point of this system. Show that linearisation of these equationswith respect to this fixed point gives

d

dt

(XY

)=

(α ac/d

−bd/a β

)(XY

)≡ J

(XY

)where α = f ′(c/d), β = g′(b/a), X = x− c/d, Y = y − b/a.

Show that the matrix J has eigenvalues λ given by

λ = µ±√ν2 − bc , µ =

α+ β

2, ν =

α− β2

.

Assuming that det J 6= 0, discuss the stability of the fixed point (c/d, b/a) in the followingcases:

(i) µ > 0, ν2 ≥ bc, (ii) µ > 0, ν2 < bc,

(iii) µ < 0, ν2 > bc, αβ > −bc, (iv) µ < 0, ν2 ≤ bc,

(v) µ = 0, ν2 > bc, (vi) µ = 0, ν2 < bc.

Why has the case µ = 0, ν2 = bc been omitted?



Exercise 6.13 A second order dynamical system is described by the equations

d

dt

(xy

)= J

(xy

), J =

(5 1−1 1

).

Solve these equations, subject to the initial conditions x(0) = 0, y(0) = 1.

Can you think of another way of doing this without finding the Jordan form of the matrix J?

Hint Try substituting x(t) = aeλt, y(t) = beλt and determine the constraints on a, b, λ. Youshould find that

(λI − J)

(ab

)= 0 ,

where I is the 2× 2 unit matrix—this should ring a bell! Consider taking appropriate linearcombinations of solutions. In this context note that if the column vectors X1, X2 are solutionsof

LX = JX, L =d

dt,

then (αX1 + βX2) is also a solution since

L(αX1 + βX2) = αLX1 + βLX2 = αJX1 + βJX2 = J(αX1 + βX2).

Exercise 6.14 A second order dynamical system is described by the system of differentialequations

x = y + xf(R), y = x+ yf(R), R = (x2 − y2)1/2.

Show that the system is separable in the coordinates (R,φ) where

x = R coshφ, y = R sinhφ.

Suppose now thatf(R) = (R− 1)(R− 2)(R− 3).

Show that the rectangular hyperbolae Hn, n = 1, 2, 3, where

Hn = (x, y) : x2 − y2 = n2, n = 1, 2, 3,

are invariant sets for the system.

Exercise 6.15A second order dynamical system with phase space (x, y) ∈ R2 = Γ is described by thedifferential equations

x = y + xf(x2 − y2), y = x+ yf(x2 − y2) .

(i) Show that the system is separable in hyperbolic coordinates (r, θ), where

x = r cosh θ, y = r sinh θ,

and in particular that r = rf(r2) and θ = 1. Which part of the phase space does thiscover?[hint, use cosh2 θ − sinh2 θ = 1]



(ii) Show that the system is also separable in different hyperbolic coordinates (ρ, φ), where

x = ρ sinhφ, y = ρ coshφ,

and in particular that ρ = ρf(−ρ2) and φ = 1. Which part of the phase space does thiscover?

Exercise 6.16Consider the first order dynamical system

dx

dt= −x− y , dy

dt= −y + x , (6.134)

wherex = r cos θ , y = r sin θ .

(i) Show that the system is separable in the coordinates (r, θ).

(ii) Sketch the phase portrait for the 1st order dynamical system satisfied by r(t), find thefixed points and discuss their nature.

(iii) Solve the 1st order dynamical systems for r and for θ exactly. Does the motion termi-nate?

(iv) Find the exact solution of the system (6.134) with initial conditions x(0) = 1, y(0) = 0.Sketch the phase portrait for the dynamical system (6.134) including a sketch of theexact solution.


dx

dt=x

rf(r)− 2y ,

dy

dt=y

rf(r) +

x

2, (6.135)

wherex = 2r cos θ , y = r sin θ , f(r) = r(r − 1)2(r − 2)(r − 3)

(i) Show that the system is separable in the coordinates (r, θ).

(ii) Sketch the phase portrait for the 1st order dynamical system satisfied by r(t), find thefixed points and discuss their nature.

(iii) Find the limit cycles for the dynamical system (6.135). Which are stable, which areunstable and which are half-stable?

(iv) Sketch the phase portrait for the dynamical system (6.135).



Exercise 6.18 [taken from 2008 exam]Suppose that a second order autonomous dynamical system has a fixed point at (x, y) = (0, 0),and that the linearised dynamics in the vicinity of this fixed point is given by

d

dt

(xy

)= J

(xy

)where

J =

(1 1c 1

)for some real constant c.

(a) Determine the eigenvalues of J , and give the Jordan canonical form C corresponding toJ for (i) c < 0, (ii) c = 0, (iii) c > 0.

(b) The Jordan canonical form C is related to J via a similarity transformation C = P−1JP ,with a suitable non-singular matrix P . You are not asked to find P .

Now we pass to the transformed variables

(XY

)= P−1

(xy

)and consider the equations

of motiond

dt

(XY

)= C

(XY

).

For the regimes (i) c < 0, (ii) c = 0 and (iii) c > 0, solve these equations and therebyinvestigate the nature of the fixed point.

Exercise 6.19 [Taken from 2009 exam (the exam was slightly easier than this version)]A particle of mass m is moving on a straight line under the influence of a force F (x), but alsoexperiences friction, so that the equation of motion is

md2x

dt2+ 2γ

dx

dt= F (x)

with γ > 0.

(a) Rewrite the equation of motion as a system of two first order equations by setting x = y.

(b) State the conditions necessary to have a fixed point of the dynamics. Let x? be thecoordinate of such a point. Calculate the Jacobian J of the system dynamics in x?, andshow that the eigenvalues are given by

λ1,2 = − γm±√γ2

m2+F ′(x?)

m

(c) Assume F ′(x?) > 0. Give the Jordan canonical form C of J .

(d) The Jordan canonical form C is related to J via a similarity transformation C = P−1JP ,with a suitable non-singular matrix P . You do not need to find P .



Now we pass to the transformed variables

(XY

)= P−1

(x− x?y

)and consider the

equations of motiond

dt

(XY

)= C

(XY

).

Solve these equations and thereby investigate the nature of the fixed point.

(e) Find P .

(f) Discuss briefly the nature of the fixed point when F ′(x?) < 0 for the relevant regimes ofγ.

Exercise 6.20 Consider the equation of damped harmonic motion,

x+ λx+ ω2x = 0 , (6.136)

where λ and ω are real positive constants.

Find the general solutions for the case of under-damping (0 < λ < 2ω), critical damping(λ = 2ω) and over-damping (λ > ω)

Consider the exponentially decaying part of the solution and show that the decay is fastestfor critical damping.




Supplementary Exercise 6.1Consider the second order dynamical system

x = (1− r2)x− (1− x/r)y

y = (1− r2)y − (1− x/r)x

where r =√x2 + y2.

(a) Find the equations for r and θ in polar coordinates, x = r cos θ, y = r sin θ.

(b) Show that the system has two fixed points.

(c) Sketch the phase portrait of the system.

Note - all points in a small disk around the point (1, 0) eventually flow to (1, 0), but someof them go all the way round the unit circle to get there. The point (1, 0) is clearly anunstable fixed point (compare it with problem 4s.1) but all flows in the neighbourhood of(1, 0) eventually end up there, so this is again an interesting example when coming up withdefinitions of stability.

Supplementary Exercise 6.2Consider the dynamical system

x = y + x2 , y = −2x(1 + y)

(a) Find the fixed points of this system

(b) Show that

d

dt(1 + y) = −2x(1 + y) ,

d

dt(1− y − 2x2) = 2x(1− y − 2x2)

(c) Show that the lines y = −1 and y = 1− 2x2 are preserved by the dynamical system.

(d) Consider

H =1

2y2 + x2(1 + y) .

Show thatH = 0 .

This mean that lines of H =constant are also phase curves, or unions of phase curves.

(e) Sketch the phase portrait of the dynamical system

(f) Now consider the new system (where |α| < 1/2)

x = y + x2 + αx(1− y − 2x2) , y = −2x(1 + y) (6.137)

Show thatH = 2αx2(1 + y)(1− y − 2x2) .



This last equation can be used to show that the lines y = −1 and y = 1− 2x2 remain unionsof phase curves for the new system but that all orbits inside the region bounded by thesecurves all spiral out from the origin if α > 0, and all spiral in if α < 0.

Supplementary Exercise 6.3A problem on separability and construction on separable variables.

The idea: Give new coordinates to each point (x, y) by (i) finding the trajectories through(x, y) (ii) Consider the trajectory that passes through (0, v) at time t = 0 and through (x, y)at time t = u Let the new coordinates on phase space be (u, v). The equations satisfied by(u, v) will be

v = 0 , u = 1 .

Try this for the dynamical system

x = 1 + x , y =1

x− 1.

(a) Sketch the phase portrait of the system.

(b) Solve the equation for x(t) as a function of t such that x(0) = 0. [ans: x(u) = eu − 1 ]

(c) Solve the equation for the phase curves through (0, v)

dy

dx=

1

x2 − 1,

[ans: y = v + (1/2) log |(1− x)/(1 + x)|]

(d) Choose as new variables (v, u). State the relation between (x, y) and (u, v)[ans: u = log(1 + x), v = y − (1/2) log |(1− x)/(1 + x)|]

(e) Show that in these variables the dynamical system becomes

v = 0 , u = 1 .

Supplementary Exercise 6.4Problems when linearisation is not possible

Consider the dynamical systems

(i) x = x2 − y2 , y = 2xy

(ii) x = x3 − 3xy2 , y = 3x2y − y3

Sketch the phase portraits of these systems.

Show that these are examples of complex dynamical systems, i.e

d

dt(x+ iy) = f(x+ iy) ,

for some function f .

Solve these two systems exactly under the conditions x(0) = 1, y(0) = 1.



Supplementary Exercise 6.5A problem on structural stability of linearisations

Consider the linear dynamical system(xy

)= M

(xy

).

Consider changing the matrix M by a small amount εN .

Show (by explicit computation) that

Tr(M + εN) = Tr(M) + εTr(N) ,

det(M + εN) = det(M) + ε det(M)Tr(M−1N) + . . . .

Consider D(M) = [Tr(M)]2 − 4 det(M)

Show that

D(M + εN) = D(M) + ε(2Tr(M)Tr(N)− 4 det(M) Tr(M−1N)) + . . .

Check these statements for the particular cases

(a) A centre or marginally stable focus: M =

(0 1−1 0

), N =

(1 00 1

).

What is the nature of the system for ε > 0 and ε < 0?

(b) An improper node: M =

(1 10 1

), N =

(0 01 0

)What is the nature of the system for ε > 0 and ε < 0?

This shows that by adding a suitable small matrix εN , if D(M) = 0 then one can chooseD(M + εN) > 0 or D(M + εN) < 0, i.e. an improper node can be charged into a proper nodeor a focus.



Supplementary Exercise 6.6Principal directions for focii are harder to define than for nodes, but can be found as follows.

We start with a marginally stable focus in normal form

r =

(xy

)= C

(xy

)= C r , where r =

(xy

), C =

(0 −ββ 0

).

This is a rotationally symmetric flow with the orbits being circular. To make the orbitsellipses, we can stretch the flows in the x direction so that the orbits are ellipses with (majorand minor) axes parallel to the x– and y– axes.

(i) Show that changing coordinates from (x, y) to (X, y) with X = αx leads to the system(Xy

)= C ′

(Xy

), C ′ =

(0 −αββ/α 0

).

(ii) show that a rotation through an angle θ from (X, y) to (u, v) with(Xy

)= R

(uv

), R =

(cos(θ) sin(θ)− sin(θ) cos(θ)

),

leads to the system (uv

)= C ′′

(uv

), C ′′ = R−1C ′R ,

and calculate C ′′ explicitly.

(iii) Suppose that the fixed point at the origin of the dynamical system(uv

)=

(a bd −a

) (uv

),

is a marginally stable focus. By comparing with your results in part (ii), show that the orbitsare ellipses with axes at an angle θ to the (x, y) axes where

tan(2θ) = − 2a

b+ d.

(iv) Show that this agrees with the definition that the principal axes are in the direction rsuch that r · v(r) = 0.


Part III

Application to Classical Mechanics

132

Chapter 7: Elements of Newtonian Mechanics 133

Chapter 7

Elements of Newtonian Mechanics

In this present and the following chapter we are going to apply the general methods introducedto study dynamical systems to problems of classical mechanics. We begin by introducingelementary facts of Newtonian Mechanics, starting with kinematics — i.e., the description ofparticle motion — thereafter continuing with proper dynamics, i.e. the investigation of thedynamical laws governing particle motion in mechanics. In particular, we look at Newton’slaw of gravitation, and we study motion on a straight line as a special realization of secondorder dynamical systems. For autonomous systems in which forces depend only on position,energy is conserved, and this aspect can be used to help analysing properties of particlemotion in one dimension. Fixed points and their stability are discussed in terms of potentialenergy. It turns out that mechanical systems in one dimension which have a conserved energycan only have two types of fixed point — unstable hyperbolic ones and (marginally) stableelliptic ones.

7.1 Motion of a particle

What is a particle? We regard a particle as a physical object, possessing a mass, whosedimensions are negligible in relation to other lengths arising in the description of its motion.For example, we may treat the Earth as a particle for the purposes of obtaining a goodapproximation to its path round the sun. Someone may object: “This isn’t very satisfactory;an undefined term, mass, has crept in!” It must be admitted at this stage that mass is indeedan intuitive term, but we shall see how the concept may be made precise once we have statedNewton’s Laws of motion.

We assume that space is Euclidean and that it can be described in terms of a system ofthree dimensional Euclidean position vectors measured relative to some conveniently chosenorigin. In describing the motion of a particle the term ‘position’ has no absolute significance;it is only meaningful to talk about the particle’s position relative to an observer. In order todescribe a particle’s motion an observer O may choose a reference frame consisting of a righthanded system of Cartesian axes Oxyz, usually fixed relative to the observer; the positionof a particle P at a particular instant is then made precise by assigning its position vectorr = OP.

It is assumed that the observer O possesses a clock with which to measure time t. Generallyspeaking, the particle’s position will vary with time and we may indicate this by writingr = r(t). In classical Newtonian theory it is assumed that time is absolute in the sense thattwo different observers, whatever their state of relative motion, can synchronise their clocks



in such a way that they both assign the same value of time t to a particular event. (Thisassumption was shown to be untenable by Einstein in 1905 when he enunciated the SpecialTheory of Relativity.)

We continue in the above notation. O denotes an observer, Oxyz an associated referenceframe, and r(t) the position vector of a particle P at time t. We give the following definitions:

Definition 7.1. The velocity v(t) of the particle (as measured by O) at time t is defined by

v(t) =dr

dt= lim

δt→0

r(t+ δt)− r(t)

δt. (7.1)

The velocity is the rate of change of the position vector.

Definition 7.2. The speed of the particle (as measured by O) at time t is the magnitude ofits velocity i.e. |v| = |r|.

We note that whereas velocity is a vector, speed is a scalar.

As the particle moves it traces out a curve Γ, say. Let Pt denote the particle’s position attime t, and Pt+δt its position at time t+ δt, so that r(t), r(t+ δt) are the position vectors ofPt, Pt+δt respectively. We note that

r(t+ δt)− r(t)

δt=

PtPt+δt

δt.

Proceeding to the limit as δt → 0 we see that the velocity vector v(t) (assumed non-zero)points along the direction of the forward tangent to the curve Γ at Pt.

Definition 7.3. The acceleration a(t) (as measured by O) of the particle at time t is definedby

a(t) =dv

dt=d2r

dt2. (7.2)

The acceleration is the rate of change of the velocity vector with respect to time.

Notation It is traditional in mechanics to employ dot notation to denote differentiation of ascalar or vector function with respect to time. We write

f ≡ df

dt, r ≡ dr

dt, r ≡ d2r

dt2,

and so on.

If the Cartesian coordinates of the particle P at time t are x(t), y(t), z(t) we may write

r = xe1 + ye2 + ze3,



where e1, e2, e3 denote the unit vectors parallel to the axes Ox, Oy, Oz respectively. Theparticle’s velocity is then

v = xe1 + ye2 + ze3.

(It is assumed that e1, e2, e3 are fixed relative to the observer O—so that their time derivativesare all zero.) Similarly the acceleration of the particle at time t is

a = xe1 + ye2 + ze3.

We may calculate the distance travelled by the particle between times t = t0 and t = t1 asfollows. Using s to denote arc length measured from a fixed point on Γ we have(

ds

dt

)2

=

(dx

dt

)2

+

(dy

dt

)2

+

(dz

dt

)2

.

Taking square roots we see that the required distance is given by the integral∫ t1

t0

|r| dt =

∫ t1

t0

|v| dt.

In other words, the distance travelled by the particle between times t0 and t1 is obtained byintegrating the speed between these limits. Note that the distance travelled is (generally) notequal to |r(t1)− r(t0)|.

Note We emphasise that the concepts of velocity, speed and acceleration have been definedwith respect to an observer O; different observers will assign different values to these functions,depending on their state of relative motion.

The following example shows that it is possible for a particle to move with constant speed(so that the magnitude of its velocity is constant) yet have non-zero acceleration.

Example 7.1. Circular Motion

A particle P moves in the plane xOy in such a way that its position vector with respect to O is

r(t) = b cosωte1 + b sinωte2,

where b, ω are positive constants. Find the velocity, speed and acceleration of the particle at time t.

Clearly |r(t)| = b for all t; it follows that the particle moves on a circle, centre O and radius b. Attime t the position vector OP of the particle makes an angle θ = ωt with the x–axis, and since θ = ωwe see that the particle moves round the circle in an anti–clockwise sense at a constant angular speedθ = ω.

We note that the particle’s velocity is given by

v(t) = r = −bω sinωte1 + bω cosωte2.

It is clear that v.r = 0 so that the vector v is indeed tangential to the circle. The speed of the particleis equal to |v| = (b2ω2cos2ωt+ b2ω2sin2ωt)1/2 = bω. Thus the speed is constant, and equal to bω.



The particle’s acceleration is given by

a = v = −bω2 cosωte1 − bω2 sinωte2 = −ω2r.

We conclude that the acceleration vector is always directed from P towards the centre O of the circle,and that it has magnitude bω2 6= 0.

We should not be too surprised by this result since, although the magnitude of the velocityvector is constant, its direction is continuously changing as the particle moves round the circle.This is why the acceleration (rate of change of the velocity vector) is non-zero, despite thefact that the speed (magnitude of the velocity vector) is constant.

Example 7.2. Uniform Motion

A particle moves with constant velocity V relative to an observer O. Find its path, given that at t = 0its position vector with respect to O is r0.

We haver = V ,

and integration givesr(t) = tV + C,

where C is a constant vector. Since r(0) = r0 we see that C = r0 and therefore

r(t) = tV + r0.

It is clear that the path of the particle, as viewed by O is a straight line which passes through thepoint with position vector r0, and whose direction is given by the vector V .

Example 7.3. Motion with constant acceleration

A particle moves so that its coordinates (x, y, z) with respect to Cartesian axes Oxyz are given by thedifferential equations

x = 0, y = 0, z = −g,

where g is a positive constant. The initial conditions are such that

x(0) = U, y(0) = 0, z(0) = V, x(0) = 0, y(0) = 0, z(0) = 0,

where U, V, are positive constants.

Find the coordinates of the particle at time t.

An observer O′ is coincident with O at time t = 0, and moves with constant velocity Ue1 relative toO. O′ describes the motion of the particle in terms of coordinates x′, y′, z′ with respect to Cartesianaxes O′x′y′z′, where O′x′ is parallel to Ox, O′y′ is parallel to Oy, and O′z′ is parallel to Oz.

Write down a set of equations relating x, y, z and x′, y′, z′ at time t. Describe geometrically the motionof the particle as viewed by O and O′.

Integrating the given equations, and using the stated initial conditions to evaluate the constants ofintegration gives:

x(t) = A = x(0) = U, x(t) = Ut+A′ = Ut (x(0) = 0),



y(t) = B = y(0) = 0, y(t) = B′ = 0 (y(0) = 0),

z(t) = −gt+ C = −gt+ V, (z(0) = V ),

z(t) = V t− 1

2gt2 + C ′ = V t− 1

2gt2 (z(0) = 0).

Summarising we have

x(t) = Ut, y(t) = 0, z(t) = V t− 1

2gt2.

Eliminating t now gives

z =V x

U− gx2

2U2= − g

2U2(x− UV/g)2 +

V 2

2g.

Writing Z = z−V 2/(2g), X = x−UV/g, we see that the path is given by Z = − g2U2X

2. We concludethat the path of the particle as viewed by observer O is a parabola. (See the diagram)

zZ

X

xO O O’

z z’

Px

x’

U tx’

Figure 7.1: Relativity of description of motion

The observer O′ see things rather differently. Since

dOO′

dt= Ue1,

and O,O′ coincide at time t = 0, it follows that

OO′(t) = tUe1.

Referring to the diagram we see that the equations relating x, y, z and x′, y′, z′ are

x′ = x− Ut, y′ = y, z′ = z,

so that

x′(t) = 0, y′(t) = 0, z′(t) = V t− 1

2gt2.

The path of the particle as viewed by O′ is therefore a straight line.



7.2 Newton’s Laws of motion

We now discuss the laws which govern the motion of a Newtonian particle. Newton assumedthe existence of a preferred class of observers and associated reference frames, called inertialframes of reference, with respect to which Newton’s laws of motion were supposed to hold.

First, every particle P possesses an inertial mass m > 0 which is normally assumed to beconstant, whatever the state of motion of the particle. If r(t) is the position vector of Prelative to an observer O its momentum p is defined by the equation

p = mr = mv,

where v is the particle’s velocity.

7.2.1 Newton’s First Law (N1)

Newton’s First Law states that with respect to an inertial frame of reference a particle moveswith constant velocity in a straight line unless constrained to depart from this state by a forceacting on it. The nature of the term force is made precise through the Second and ThirdLaws.

7.2.2 Newton’s Second Law (N2)

Newton’s Second Law states that with respect to an inertial frame of reference the rateof change of momentum of a particle is equal to the total force acting on the particle. IfF 1,F 2, . . . ,F n are the forces acting on a particle P, whose position vector is r(t) with respectto the origin O of an inertial frame of reference, we have

d

dt(mr) = F ,

where the total force F is given by

F =n∑i=1

F i.

Assuming that m is constant we may write

mr = F .

In words, the mass times the acceleration of the particle is equal to the total force acting onthe particle.

7.2.3 Newton’s Third Law (N3)

Newton’s Third Law is often stated in the form “action and reaction are equal (in magnitude)but opposite (in direction)”. To see more clearly the implication of this law suppose that the



force exerted on a particle i by a particle j is F ij ; correspondingly the force exerted by particlei on particle j is F ji and

F ij + F ji = 0,

orF ij = −F ji.

The forces in question could be gravitational or, in the case of two particles connected by ataut string, the tension force provided by the string.

We now show, very briefly, that Newton’s laws determine a scale of mass, once an arbitraryunit of mass has been chosen. Consider two particles 1, 2 with inertial masses m1,m2 respec-tively. Suppose that particle 2 exerts a force F 12 on particle 1 and that particle 1 exerts aforce F 21 on particle 2. If a1,a2 denote the accelerations of the two particles with respect tosome inertial frame of reference we have, using N2,

m1a1 = F 12, m2a2 = F 21.

Adding these equations, and using Newton’s Third Law, we obtain

m1a1 +m2a2 = F 12 + F 21 = 0. (7.3)

It follows thatm1

m2=|a2||a1|

.

By measuring the accelerations a1,a2 we can determine the ratio m1/m2, and having chosenparticle 2 (say) to have unit (inertial) mass, the mass m1 of particle 1 is then determined.Newton’s laws therefore determine, in principle, a mass scale.

Returning to equation 7.3 we note that if r1, r2 denote the position vectors of the two particleswith respect to the origin O of an inertial frame of reference then

m1r1 +m2r2 = 0,

and therefored2

dt2(m1r1 +m2r2) = 0. (7.4)

The centre of mass of the two particle system has position vector R defined by

R =m1r1 +m2r2

m1 +m2.

It now follows from equation 7.4 that

d2R

dt2= 0.

We may interpret this result by saying that the centre of mass of the two particle systemis unaccelerated with respect to any inertial frame of reference, and therefore moves withconstant velocity in a straight line with respect to such a frame. All this supposes, of course,that F 12,F 21 are the only forces acting.



We note that similar arguments can be applied to show that if two particles with inertialmasses m1,m2 coalesce to form a third particle with inertial mass m3, then m3 = m1 +m2,so that its inertial mass is the sum of the inertial masses of its component particles.

Throughout this discussion we have used the term “inertial frame of reference” and one mayreasonably ask “Are there any such frames?” Einstein and Infeld in their book “The Evolutionof Physics” show that this question poses serious difficulties and it is not possible to give anexample of a frame of reference which is strictly inertial! Well, one may ask: “Would areference frame with origin on the Earth’s surface, and fixed relative to the Earth, constitutean inertial frame of reference?” Because of the Earth’s rotation this is not strictly the case;nevertheless it is a reasonable approximation—in the sense that an application of Newton’slaws in such a frame leads to fairly satisfactory predictions. By making a better choice ofreference frame it is possible to take account of effects which arise from the Earth’s rotation.

Nevertheless, this is not an entirely satisfactory state of affairs; we have a set of laws but weare unable to exhibit a reference frame in which they are strictly valid! However, if one suchinertial frame F1 exists, then all frames moving with constant velocity relative to F1 are alsoinertial. (This point is considered in more detail in an exercise at the end of this chapter.)

In order to use Newton’s laws of motion to predict the motion of a particle, given a knowledgeof certain initial conditions (its initial position and velocity), we require explicit mathematicalformulae for the force(s) acting on the particle; of particular importance are gravitationalforces.

7.3 Newton’s Law of Gravitation

Consider two particles of masses m1,m2 respectively and suppose that the position vector ofm2 with respect to m1 is r. The the gravitational force F exerted on m2 by m1 is given by

F = −Gm1m2

r3r, r = |r|, (7.5)

where G is a universal constant known as Newton’s constant of gravitation.

The force F has magnitude |F | = Gm1m2/r2 so that the magnitude of the gravitational force

varies inversely as the square of the distance between the particles; the - sign indicates thatm2 is attracted towards m1.

If we now work out the acceleration of m2 due to the gravitational force we get

m2 r2 = F = −Gm1m2

r3r , r2 = −Gm1

r3r . (7.6)

This acceleration is independent of the mass m2 and is called the “gravitational field” due tom1.



Strictly speaking, the masses which occur in equation 7.5 are gravitational masses whereasthe mass appearing in N2 is the inertial mass. The fact that the acceleration (7.6) is thesame for all particles m2 is an experimental fact, first demonstrated in a famous experimentconducted from the leaning tower of Pisa by Galileo and subsequently tested to great accuracyby Eotvos. As a consequence, we can deduce that the ratio of inertial mass to gravitationalmass is the same for all particles, and by an appropriate choice of units we can assume thatthe inertial and gravitational masses of a particle are equal. In what follows, therefore, weshall not make any distinction between inertial and gravitational mass.

In some of the problems which we study the motion is due to the gravitational force arisingfrom the whole Earth. If we assume that the Earth is a uniform sphere of radius a, and totalmass M, then it can be shown that the gravitational force experienced by a particle at a pointabove the Earth’s surface is the same as the gravitational force which would be exerted onthe particle by a particle situated at the centre of the Earth, with a mass equal to the totalmass of the Earth. This means that the gravitational force F exerted on a particle of massm, whose position vector is r relative to O, the centre of the Earth, is given by

F = −GmMr3

r. (7.7)

and the gravitational field of the Earth is

g = −GMr3

r. (7.8)

If the particle remains near the Earth’s surface throughout the motion, so that r/a is alwaysclose to 1, we may replace equations (7.7) and (7.8) by

F = −GmMa2

er = −mger , g = −ger (7.9)

where er is the unit vector parallel to r and g = GM/a2. In the approximation which we havedescribed the gravitational force on the mass m has magnitude mg and is directed along thedownward vertical towards the centre of the Earth. We shall employ this notation in whatfollows without further explanation.

Example 7.4. Motion near the Earth’s Surface

A particle P of mass m is projected vertically upwards from a point A on the Earth’s surface withspeed V. Discuss the subsequent motion, assuming that the particle remains near the Earth’s surfacethroughout the motion.

We assume that we can neglect air resistance and that the only force acting on the particle is in factgravitational.

Let O denote the Earth’s centre. We choose axes Oxyz which are fixed relative to the Earth and whichare such that the z-axis points vertically upwards, through the point of projection; as noted previouslysuch a reference frame is approximately inertial. Suppose that the particle’s coordinates at time t are(0, 0, z(t)); the initial conditions then demand that z(0) = a, z(0) = V.

Application of Newton’s Second Law (N2) gives

mze3 = −mge3, z = −g.



Integrating with respect to time t we find that

z = −gt+ C,

where C is a constant. Since z(0) = V , it follows that z = −gt+V. A further integration with respectto t immediately gives

z(t) = a+ V t− 1

2gt2.

The maximum height of the particle above the Earth’s surface is attained when z = 0 i.e. whent = V/g. The corresponding value of z is then given by

z = a+V 2

g− 1

2gV 2

g2= a+

V 2

2g.

We conclude that the particle rises to a height V 2/(2g) above the Earth’s surface in a time V/g beforefalling back towards the point of projection.

We have assumed throughout this calculation that the gravitational force exerted on m bythe Earth is constant; if this is to be a reasonable assumption, we would expect that themaximum height which we have just calculated should be a good deal less than a, the Earth’sradius i.e.

V 2

2g a, V 2 2ag.

We shall see below that V = (2ag)1/2 does indeed have a special significance.

Example 7.5. Motion Leaving the Vicinity of Earth’s Surface

Suppose that our assumption is not tenable and that the speed of projection V is such that V 2 iscomparable with 2ag. In this case we have to use the exact expression for the gravitational force,without any approximations. The equation of motion is

mze3 = −GmMz2

e3, z = −GMz2

.

Since z = −g when z = a it follows that GM = ga2 and we may write

z = −ga2

z2.

Multiplying this equation by z and noting that

d

dt(1

2z2) = zz

we obtaind

dt(1

2z2) = −ga

2

z2dz

dt.

Integration with respect to t yields

1

2z2 = −

∫ga2

z2dz + C =

ga2

z+ C,

where C is constant. Using the initial conditions z(0) = a, z(0) = V we obtain

C = 12V

2 − ag and the equation for z becomes

z2 =2ga2

z+ V 2 − 2ag. (7.10)



The maximum value of z, z = zmax, say, is obtained by setting z = 0 and is given by

zmax =2ga2

2ag − V 2.

We see that zmax →∞ as V → (2ag)1/2 from below; this value of V is referred to as the escape speedof the particle. Substituting the relevant values of a, g we find that the escape speed is approximately11.2km/sec.

For values of V which are less than (2ag)1/2 the particle rises to z = zmax before falling back to thepoint of projection. Re-writing equation 7.10 in terms of zmax we find that

z2 =2ga2(zmax − z)

zzmax.

If we require to compute the time τ for the particle to reach zmax we extract the square root andseparate the variables to obtain

τ =

∫ τ

0

dt =

(zmax2ga2

)1/2 ∫ zmax

a

(z

zmax − z

)1/2

dz.

This integral can be evaluated by means of the substitution z = zmaxsin2θ.

7.4 Motion in a Straight Line; the Energy Equation

Let x(t) denote the coordinate at time t of a particle of mass m moving in a straight linealong the x-axis under the influence of a force Fe1. The equation of motion is

mxe1 = Fe1, mx = F.

We now make the special assumption that F depends only on x, the position coordinate ofthe particle so that F = F (x) and the equation of motion becomes

mx = F (x).

Multiplying by x, and recalling that

d

dt

(1

2x2

)= xx , (7.11)

we obtain1

2mx2 =

∫F (x)

dx

dtdt+ C =

∫F (x) dx+ C , (7.12)

where C is a constant of integration. We write

T =1

2mx2, V (x) = −

∫F (x) dx (7.13)

and refer to T as the kinetic energy of the particle and to V as the potential energy functionof the particle. In terms of these definitions the last equation may be written

T + V = C = constant. (7.14)



T + V, the sum of the kinetic and potential energies of the particle, is called the total energyof the particle and equation 7.14 then states that the total energy is constant throughoutthe motion—it is a constant of the motion. We emphasise that this result depends on theassumption that F is a function of x only.

Note that V is defined in terms of an indefinite integral of F (x); it follows that the potentialenergy V is only unique up to a constant. We note that equation 7.13 implies

F (x) = −dV

dx. (7.15)

We may say that the force acting on the particle is minus the gradient of the potential energyfunction—even though this is slightly sloppy, since the actual force is not F (x) but F (x)e1.

Example 7.6. Potential of Constant Gravitational Force

Suppose that the axis Ox points along the upward vertical and that the particle is moving along Oxunder constant gravity so that the force acting is −mge1.

The potential energy function V is given by

−dVdx

= −mg, V (x) = mgx+ C.

If we arbitrarily require that V (0) = 0 we may set C = 0 and write V (x) = mgx. In this case theenergy equation 7.14 reads

1

2mx2 +mgx = constant.

Given a set of initial conditions i.e. the values of x and x at t = 0 we could evaluate the constantexplicitly.

Example 7.7. Simple Harmonic Motion

Suppose that the particle is moving along the x-axis under the action of a force −mω2xe1, where ω isconstant.

First note that, whatever the sign of x, the force is always directed towards the origin O. The potentialenergy V is given by

−dVdx

= −mω2x, V (x) =1

2mω2x2.

(We have arbitrarily required that V (0) = 0 in order to set the constant of integration equal to zero.)

The energy equation reads1

2mx2 +

1

2mω2x2 = constant,

which we can re-write in the formx2 + ω2x2 = a2,

where a is a constant. We could integrate this equation to find x as a function of t but it is perhapseasier to go back to Newton’s Second Law:

mx = −mω2x, x+ ω2x = 0.



We know from our work on differential equations that this equation has general solution

x(t) = α cosωt+ β sinωt,

where α, β are constants. It is well known that the right hand side of this equation can be expressedin the form A sin(ωt+ ε), where a, ε are constants and we may therefore write the general solution as

x(t) = A sin(ωt+ ε). (7.16)

We may assume, without loss of generality that A is positive.

Bearing in mind the properties of the sine function it is clear that the particle executes an oscillatorymotion between the points x = ±A. A is called the amplitude of the motion. The centre of the motionis the point x = 0.

The time τ taken by the particle to complete one full oscillation is given by ωτ = 2π so that

τ =2π

ω.

τ is called the period of the motion. The number of complete oscillations per unit time is called thefrequency of the oscillation, and is usually denoted by ν. We see that

ν =1

τ=

ω

2π.

The constant ε which appears in equation 7.16 is sometimes referred to as the phase angle. The motionwhich we have just described is said to be Simple Harmonic and the equation x + ω2x = 0 is calledthe differential equation of Simple Harmonic motion.

Example 7.8. Simple Harmonic Motion II

Suppose instead that the force acting is −mω2(x− a)e1, where a is constant.

In this case the equation of motion becomes

mx = −mω2(x− a), x+ ω2(x− a) = 0.

We write z = x−a and obtain the equation z+ω2z = 0 (The differential equation of Simple Harmonicmotion). This equation has general solution

z(t) = A sin(ωt+ ε),

where A > 0, ε are constants. Expressing this result in terms of x we obtain

x(t) = a+A sin(ωt+ ε).

We see that this motion is also simple harmonic with amplitude A, and period τ = 2π/ω. The oscillationtakes place between the points x = a±A and the centre of the motion is x = a.

How does Simple Harmonic motion arise in practice?

Example 7.9. Elastic Strings and Springs (Hooke’s Law)

One end of a light elastic string, of unstretched length b, is secured to a fixed point O, whilst a particleof mass m is attached to the other end, and the system hangs in equilibrium under (constant) gravity.Subsequently the particle is set in motion by a blow which imparts to the particle an initial speed



u in the sense of the downward vertical. Assuming that the string remains stretched, determine thesubsequent motion, neglecting air resistance; you may assume Hooke’s Law.

Choose the x-axis so that Ox points along the downward vertical and let x(t) denote the x-coordinateof the mass m at time t. The forces acting on the particle are

(i) The force of gravity, mge1,

(ii) The force −Te1 due to the tension in the string— which acts on the particle in the sense of theupward vertical. It is necessary to postulate this force - it is the force which balances gravity whenthe particle is hanging in equilibrium, and it continues to act throughout the subsequent motion.

Application of N2 givesmxe1 = mge1 − Te1.

In order to make progress we have to make some assumption as to how T depends on x. If we assumeHooke’s law we may write

T = k(x− b),

where k is a positive constant which depends on the nature of the elastic material. Hooke’s law assumesthat the tension is proportional to the extension of the string beyond its natural length. The equationof motion now becomes

mxe1 = mge1 − k(x− b)e1.

Writing z = x− b our equation becomes

z + ω2z = g, ω2 = k/m. (7.17)

The homogeneous equation z + ω2z = 0 has general solution of the form z = A sin(ωt + ε). We notethat z = g/ω2 is a particular solution of equation 7.17 and we conclude that its general solution is

z(t) = A sin(ωt+ ε) + g/ω2.

Initially the system is in equilibrium so that z = 0; equation 7.17 shows that z(0) = g/ω2. The systemis set in motion with initial speed u in the sense of the downward vertical so z(0) = u. Imposing theseinitial conditions on the general solution we deduce that

A sin ε = 0, Aω cos ε = u.

These equations are satisfied by ε = 0, A = u/ω. Hence

z(t) =u

ωsinωt+

g

ω2,

or, written in terms of x

x(t) = b+u

ωsinωt+

g

ω2.

We conclude that the particle executes Simple Harmonic motion about x = b + gω2 as centre (this is

the equilibrium value of x, before the blow is struck); the period is

τ = 2π/ω = 2π√

mk and the amplitude of the oscillation is u

ω .

Note that the potential energy function V is given by

−dVdx

= mg − k(x− b), V (x) = −mgx+k

2(x− b)2 + C,



where C is a constant.

Note that the least value of x(t) throughout the motion is equal to b− uω + g

ω2 and that for the stringto be in a state of tension throughout the entire motion this must be greater than b. This implies thatu < g/ω—a constraint on the magnitude of the initial speed of the particle.

7.5 Equilibrium and Stability

Consider a particle of mass m, moving along the x-axis under a potential V (x), so that theforce F acting on m is given by

F = −dVdx

e1 = −V ′(x)e1 .

By N2 the equation of motion is

mxe1 = −V ′(x)e1, x = −V′(x)

m(7.18)

Putting x1 = x, x2 = x we find that

x1 = x2, x2 = −V′(x1)

m. (7.19)

Let x = a be a stationary point of the potential function V, so that V ′(a) = 0. Suppose alsothat the initial conditions are x(0) = a, x(0) = 0; this means that at t = 0 the particle isinstantaneously at rest at x = a. In terms of x1, x2 the initial conditions become

x1(0) = a, x2(0) = 0. (7.20)

Assuming that V has continuous second order derivatives we deduce from Picard’s Theoremthat the unique solution of equations 7.19 subject to the initial conditions expressed byequations 7.20 is

x1(t) = a, x2(t) = 0, ∀t ≥ 0.

Reverting to x, x this means that the particle remains at rest at x = a for all t ≥ 0. We saythat x = a is an equilibrium position for the particle; the possible equilibrium positions arethe stationary points of V.

Suppose now that the particle, at rest at the equilibrium point x = a, is set in motion withvelocity ue1. In the subsequent motion energy is conserved and we have

1

2mu2 + V (a) =

1

2mx2 + V (x),

so that1

2mx2 =

1

2mu2 + V (a)− V (x). (7.21)

First suppose that x = a is a minimum of V . Since 12mx

2 ≥ 0 we deduce from equation 7.21that

V (x) ≤ 1

2mu2 + V (a). (7.22)



a a

V(a)

1 2a

x

V(x)

Figure 7.2: Stable periodic motion in the vicinity of a minimum of the potential energy.

Referring to the diagram we see that, if u is sufficiently small, the particle’s motion will beconfined to the interval [a1, a2], where a1, a2 are solutions of the equation

V (x) =1

2mu2 + V (a).

Perhaps it will be helpful to describe the motion in a little more detail. To be definite, supposethat u > 0 so that the particle moves towards a2. As it does, V (x) increases, and the particleloses kinetic energy, coming to instantaneous rest at x = a2. It cannot move to the right ofa2, otherwise 1

2mx2 would become negative! Having reached x = a2 the particle starts back

towards x = a, passing through x = a with speed u, and comes to a state of instantaneousrest once more at the point x = a1, before moving back to x = a2. We see that the particleoscillates between x = a1 and x = a2.

In conclusion, we see that if |u| is small enough, the particle will remain “near” x = a, in thesense described, and we are justified in saying that x = a is a position of stable equilibrium.

Suppose, on the other hand, that x = a is a maximum of the potential function V and thatthe particle, initially at rest at x = a is set in motion with velocity ue1.As the particle movesaway from x = a the potential V (x) decreases and therefore 1

2mx2 increases by virtue of

equation 7.21. The particle does not remain “near” x = a and we are justified in saying thatx = a is a position of unstable equilibrium.

The above arguments are based on the energy equation and graphical considerations. Alter-natively, we may proceed as follows. Since x = a is an equilibrium point,

V ′(a) = 0. The equation of motion is (see equation 7.18)

x = −V′(x)

m

Near x = a Taylor’s formula gives

V (x) = V (a) +V ′(a)(x− a)

1!+V ′′(a)(x− a)2

2!+ · · ·



Assuming that x is “near” a, and that V ′′(a) 6= 0, we have

V (x) ' V (a) +1

2V ′′(a)(x− a)2

and the equation of motion approximates to

x = − 1

m

d

dx

(V (a) +

1

2V ′′(a)(x− a)2

)= −V

′′(a)

m(x− a).

Writing z = x− a we obtain

z +V ′′(a)

mz = 0.

We have to deal with two cases, depending on the sign of V ′′(a).

Case 1 Suppose that V ′′(a) > 0, so that x = a is a minimum of V. Writing

ω2 =V ′′(a)

m> 0

we obtainz + ω2z = 0.

This equation has general solution

z = A sin(ωt+ ε),

where A, ε are constants. Expressing z in terms of x we now have

x(t) = a+A sin(ωt+ ε).

The initial conditions x(0) = a, x(0) = u require that A sin ε = 0, Aω cos ε = u from whichit follows that ε = 0, A = u/ω. Therefore

x(t) = a+u

ωsinωt.

In the approximation considered, the particle executes Simple Harmonic motion about x = awith amplitude u/ω and period 2π/ω. This result agrees with our earlier conclusions and wesee that a minimum stationary point of V is a position of stable equilibrium.

Case 2 Now suppose that V ′′(a) < 0, so that x = a is a maximum of V. Writing

Ω2 = −V′′(a)

m> 0

we obtainz − Ω2z = 0.


z(t) = αeΩt + βe−Ωt,



x

V(x)

21

Figure 7.3: The potential that allows stable periodic motion in the vicinity of x = 2; theminimum energy needed to escape to −∞ is shown as a dashed line.

x(t) = a+ αeΩt + βe−Ωt,

where α, β are constants. Imposing the initial conditions x(0) = a, x(0) = u we find that

x(t) = a+u

Ωsinh Ωt.

Since sinh Ωt→∞ as t→∞ we see that our assumption that x remains “near” a is invalid.This conclusion is in agreement with our graphical arguments using the energy equation: amaximum of V is a position of unstable equilibrium.

Example 7.10. Motion in a Cubic Potential A particle of unit mass moves along the axis Ox undera potential

V (x) = 2x3 − 9x2 + 12x.

Find the values of x for which the particle can remain in equilibrium. Show that if (x−2)2 is regardedas negligible in comparison with (x − 2) then the period τ of small oscillations about x = 2 is givenby τ = 2π/

√6.

Suppose that x(0) = 2, x(0) = −u, where u >√

2. Describe (qualitatively) the subsequent motion.

Hint: Sketch the graph of V (x) and appeal to the principle of energy conservation.

We haveV ′(x) = 6x2 − 18x+ 12 = 6(x2 − 3x+ 2) = 6(x− 1)(x− 2).

The possible equilibrium positions are given by V ′(x) = 0 i.e. by x = 1, x = 2. The equation ofmotion is

x = −V ′(x), x+ 6(x− 1)(x− 2) = 0.

Writing z = x− 2 we derivez + 6z(z + 1) = 0.

Assuming that terms in z2 are negligible in comparison with the linear terms in z we obtain as anapproximation the differential equation

z + 6z = 0.

This describes Simple Harmonic motion about the point z = 0 i.e. about x = 2 and the period τ ofthe oscillations is given by τ = 2π/

√6.



Referring to the graph of V we see that if the particle is projected from x = 2 with velocity −ue1 itwill move to the left of x = 2. As it does so energy is conserved:

1

2x2 + V (x) =

1

2u2 + V (2).

If the particle still has positive kinetic energy when it reaches x = 1 (a maximum of V ) then it willmove off to −∞, since V decreases (rapidly) for x < 1. The condition for this to happen is

1

2u2 + V (2)− V (1) > 0.

Evaluating V (1), V (2) we find that the condition for the particle to move off to −∞ is u >√

2.

Example 7.11. Motion in a Non-Linear Potential

A particle of mass m moves along the axis Ox under the potential V given by

V (x) =cx

(a2 + x2),

where a, c are positive constants. Sketch the graph of V ; show that V has a minimum at x = −a anda maximum at x = a.

Given that x(0) = −a, x(0) = u, where u is suitably small, discuss the subsequent motion. Byreference to your graph describe what will happen to the particle if, instead, the initial conditions arex(0) = a, x(0) = u, where u can be either positive or negative.

x

V(x)

−a a

Figure 7.4: Potential that allows stable periodic motion in the vicinity of x = −a; dependingon initial conditions, escape to ±∞ is possible.

We find that

V ′(x) =c(a2 − x2)

(a2 + x2)2, V ′′(x) =

2cx(x2 − 3a2)

(a2 + x2)3.

Now V ′(x) = 0 when x = ±a. It is clear from consideration of the graph of V that x = −a is aminimum of V (and therefore a position of stable equilibrium), and that x = a is a maximum of V,and therefore a position of unstable equilibrium.

Near x = −a the equation of motion is

mx = −dVdx

= − d

dx

(V (−a) +

V ′(−a)(x+ a)

1!+V ′′(−a)(x+ a)2

2!+ · · ·

)= −V ′′(−a)(x+ a),



on the assumption that the higher powers of (x+ a) are negligible. Writing z = x+ a we obtain

z +V ′′(−a)

mz = 0, z +

c

2ma3z = 0.


z = A sin(ωt+ ε), x(t) = −a+A sin(ωt+ ε),

whereω2 =

c

2ma3.

In the approximation which we have described the motion is Simple Harmonic about x = −a as centre,and the period is

τ = 2π

√2ma3

c.

Imposing the initial conditions x(0) = −a, x(0) = u, we find that

x(t) = −a+u

ωsinωt.

Reference to the graph of V suggests that our approximation is likely to be reasonable if the amplitudeu/ω a i.e. if u ωa.

Suppose now that the initial conditions are x(0) = a, x(0) = u. The energy equation yields

1

2mx2 + V (x) =

1

2mu2 + V (a).

We see that in the subsequent motion 12mx

2 > 12mu

2 since V (x) < V (a). We deduce that if u > 0 theparticle will move off to +∞ and that if u < 0 the particle will move off to −∞.



7.6 Exercises

Some physical data

Newton’s constant GN 6.67× 10−11 m3 kg−1s−2

Mass of the Sun Msun 1.99× 1030 kg

Mass of the Earth Mearth 5.97× 1024 kg

Mass of the Moon Mmoon 7.35× 1022 kg

Radius of the Sun rsun 6.96× 108 m

Radius of the Earth rearth 6.37× 106 m

Radius of the Moon rmoon 1.74× 106 m

Distance of Earth from Sun dearth−sun 1.50× 1011 m

Distance of Moon from Earth dearth−moon 3.84× 108 m

Short Exercises

Exercise 7.1A particle moves along a path

r(t) =

cos(f(t))sin(f(t))

0

.

Calculatev = r , a = v .

Show (by explicit calculation) that

r · r = 1 , v · r = 0 , r · a ≤ 0 .

Find condition(s) on f(t) for r · a = 0 to be true. What does this imply for the particle’smotion?

Find condition(s) on f(t) for v · a = 0 to be true.

Exercise 7.2For each of the potentials V (x) below, find the maxima, minima and points of inflection [i.e.all points where V ′(x) = 0] and sketch the graph of V (x).

(a) V (x) = (x2 − 1)2 (b) V (x) = −sech2(x) (c) V (x) =x

x2 + 1

(d) V (x) = x4 − 4

3x3 (e) V (x) = x2e−x



Longer Exercises

Exercise 7.3A particle moves so that its position vector with respect to the origin O of a reference frameOxyz is

r(t) = b cosωt e1 + b sinωt e2 + V te3,

where e1, e2, e3 are unit vectors parallel to the axes Ox,Oy,Oz; b, ω, V are positive constants.Find(i) The velocity and speed of the particle,(ii) Prove that the particle moves on a circular cylinder, whose axis is Oz.(iii) Find the particle’s acceleration, and indicate its direction in a diagram in the case b =ω = V = 1.

Exercise 7.4A particle has position vector r(t) = e1 + te2 + t2e3 relative to the origin O of a frame Oxyz.Find the velocity, the speed, and the acceleration of the particle.Find the distance travelled by the particle between t = 0 and t = 1.What is the instantaneous direction of the particle’s motion at t = 1? (Calculate the relevantunit vector.)

Exercise 7.5Two particles 1, 2 move in such a way that their position vectors r1, r2, with respect to theorigin O of an inertial frame of reference, are given by

r1(t) = e1 + te2 + t2e3,

r2(t) = te1 + t2e2 + t3e3,

respectively. Write down expressions for the velocity, speed and acceleration of the particles attime t and show that, at the instant when they collide, the angle α between their instantaneousdirections of motion is given by

cosα =8√70.

Exercise 7.6A particle moves so that its position vector r with respect to the origin O of an inertial frameof reference is

r = b cosωte1 + b sinωte2 + V t2e3,

where b, ω, V are positive constants.Verify that the acceleration vector is orthogonal to r when t = bω/(

√2V ). Prove also that

the acceleration vector makes a constant angle β with the direction e3, where

cosβ =2V

(b2ω4 + 4V 2)1/2.

Exercise 7.7O,O′ are the origins of two reference frames F, F ′. At time t = 0

OO′ = a. Suppose that O′ moves with a constant velocity V relative to O. At time t a particleP has position vector r(t) relative to O and r′(t) relative to O′. Use a suitable vector diagramto write down an equation relating r(t) and r′(t). Deduce that if Newton’s second Law (N2)applies in the frame F, it also applies with respect to the frame F ′.



Exercise 7.8Suppose that two particles 1, 2 of masses m1,m2 act on each other in such a way that theforce on particle 1 due to particle 2 is F 12 and that on particle 2 due to particle 1 is F 21;then

F 12 + F 21 = 0,

by Newton’s Third Law (N3). Write down N2 for each particle in some inertial frame ofreference and deduce that the centre of mass of the 2 particle system moves with constantvelocity with respect to the chosen frame.(Recall that the position vector r of the centre of mass is given by

r =m1r1 +m2r2

(m1 +m2),

where r1, r2 are the position vectors of m1,m2 with respect to the chosen origin.)

Prove that the above formula for the centre of mass defines the same point in space, irrespec-tive of the choice of origin.

Hint Choose origins O,O′ and suppose that the two particles have position vectors ri, ri′

(i = 1, 2) with respect to O,O′ respectively. Then

ri = OO′ + ri′, (i = 1, 2).

The centre of mass has position vector R,R′ with respect to the origins O,O′ where

R =m1r1 +m2r2

(m1 +m2),

R′ =m1r1

′ +m2r2′

(m1 +m2).

Now verify thatR = OO′ + R′.

Exercise 7.9A particle of mass m moves along the x-axis under a constant force Ce1. Let x(t) denote itscoordinate at time t. Given the initial conditions

x(0) = a, x(0) = u,

prove that

x(t) = u+ ft, x(t) = a+ ut+1

2ft2, x2 = u2 + 2f(x− a),

where f = C/m.

Note These formulae are only applicable if the force acting on the particle is constant; this isseldom the case, yet many students treat these formulae as though they had universal validityand could be used to find the answer to any problem in mechanics! Remember, solving aproblem in Newtonian mechanics involves setting up the equation of motion N2 with therelevant set of initial conditions; if you proceed in this way you can’t go wrong!



Exercise 7.10A particle is projected vertically upwards from the surface of the Earth with speed V =(ag)1/2, where a is the Earth’s radius. Recall that the force of gravity, when the particle isat (0, 0, z) is F = −(mga2/z2)e3, m being the particle’s mass.

(We’re following the notation used earlier in this chapter, choosing the origin O at the Earth’scentre, and the axis Oz vertically upwards through the point of projection (0, 0, a).)Show that the particle rises a height a above the Earth’s surface before it starts to fall,and prove that the time to fall from this position of maximum height, back to the point ofprojection, is τ where

τ = (ag)−1/2

∫ 2a

az1/2 dz/(2a− z)1/2,

and evaluate this integral by the substitution z = 2asin2θ.

(The answer is (π/2 + 1)√

ag .)

Exercise 7.11A particle is projected vertically upwards from the Earth’s surface with speed (3ag/2)1/2,where a denotes the Earth’s radius, regarded as a uniform sphere. Use the energy equationto show that the particle rises to a height 3a above the Earth’s surface before returning tothe point of projection.

Exercise 7.12A particle of unit mass moves along the axis Ox of an inertial frame of reference under theaction of a force F = (k/x3)e1, where k is a positive constant. Given that the particle startsfrom rest at x = a show that its speed at the point x = 2a is (3k/4a2)1/2. Compute the timetaken by the particle to move from x = a to x = 2a.

Exercise 7.13A particle of unit mass moves along the x−axis under the action of a force derived from apotential V (x) = − cosx. The particle, initially in equilibrium at x = 0, is set in motion withspeed u in the sense of the positive x−axis. Write down the equation of motion of the particleand, assuming that |x(t)| remains small ( π/2) in the subsequent motion, find x(t) in termsof u.(You may assume that the replacement sinx ≈ x is justified.)What is the period τ of the oscillation?Sketch V (x) and show, by considering the energy equation, that if u exceeds a certain criticalvalue u0 (which should be found), the particle will move off to +∞.

Exercise 7.14A particle of mass m moves along the x-axis under the potential

V (x) =cx

(a2 + x2)2,

where a, c are positive constants. Find the period of small oscillations about the position ofstable equilibrium.



Exercise 7.15Referring to Example 7.9 suppose that we attempt to model the effect of air resistance bysupposing that the air exerts a force −2mλxe1 on the particle, where λ is a positive constant.The modified equation of motion is then

mxe1 = mge1 − k(x− b)e1 − 2mλxe1,

or, writing z = x− b,z + 2λz + ω2z = g, ω2 = k/m.

Find the general solution of this equation on the assumption that λ2 − ω2 < 0. (This meansthat damping due to air resistance is small).Suppose that at time t = 0 the particle is hanging in equilibrium under gravity when it isstruck a blow which causes it to move with initial velocity ue1 in the sense of the downwardvertical. Find the displacement of the particle at time t, assuming that the string remainstaut throughout the motion.We say that the particle is executing Damped Harmonic Motion; the equation

z + 2λz + ω2z = 0, λ > 0

is sometimes referred to as the equation of Damped Harmonic motion.

Exercise 7.16 (King’s College, Summer 1995 exam)Consider the motion of a particle of mass m when it moves along a straight line under theaction of forces Fe1 derivable from potentials V (x).(a) Use Newton’s equation of motion to prove that the total energy of the particle is a constantof the motion.(b) When F = m(11x−2 − 36x−3) compute V (x). If the particle is initially at x = 1 and hasinitial velocity 4e1 find the maximum value of x which it can attain.(c) If x = a is an equilibrium point of a motion show that, when |x − a| is small, Newton’sequation of motion is approximated by

md2z

dt2+ V ′′(a)z = 0,

where z = x− a and V ′′ = d2V/dx2.Let V (x) = x3 + βx2 − γx, where β and γ are positive real numbers. Identify all the stableand unstable equilibrium points and describe the motion near any stable equilibrium points.

Exercise 7.17 [Taken from 2004 exam]A particle is moving in an inertial frame of reference in such a way that its position vectorr(t) with respect to the origin O is given by

r(t) = cos(ωt) e1 + sin(ωt) e2 +(h− at2

2

)e3 ,

where the eα, α = 1, . . . , 3, are orthogonal unit basis vectors of a Cartesian coordinate system.

(a) Find the velocity, speed and acceleration of the particle, and compute the cosine of theangle between its velocity and acceleration as functions of time.



(b) Assume h > 0 and a > 0. Determine, at which time t the particle will hit the e1-e2

plane. Assume that, when hitting that plane, the particle is reflected by instantaneouslychanging the sign of the e3-component of the velocity. Determine the cosine of the anglebetween the directions of motion immediately before and after the reflection.

Exercise 7.18Consider a system of 2 particles each of which exerts a force on the other. Show that thecentre of mass moves uniformly with respect to an inertial frame. Show further that

µr = F12

where F12 is the force on particle 1 due to particle 2, r = r1 − r2 and the reduced mass µ isgiven by

µ =m1m2

m1 +m2

Consider a system of 3 particles each of which exerts a force on the other two. Denote theforce on particle i due to particle j by Fij . Use Newton’s laws to show that the centre ofmass of the system moves uniformly.

Exercise 7.19Consider a particle moving in R3 with position

r =

R cos(ωt)R sin(ωt)

0

,

where R and ω are positive constants.

Calculate the velocity v and acceleration a and their magnitudes v and a.

Show that

r = R , v = Rω , a = −ω2r , a = ω2R , a =v2

r.

These formulae show that constant velocity does not mean constant acceleration and are alsovery useful when dealing with motion in a circle at constant speed.

Exercise 7.20Suppose that the Earth is at the origin of an inertial coordinate system and the moon as atposition r.

Where is the centre of mass of the Earth-moon system?

Is it inside or outside the Earth?

Exercise 7.21A particle of mass m is propelled from the surface of the earth, with an initial velocityv0 = (a, b, c), with c > 0. Choose the origin O of the coordinate system Oxyz to be on thesurface of the earth with z axis pointing vertically upward, and x and y axes parallel to theearth’s surface. Let O be the point from which the particle is propelled.

(a) State Newton’s equation of motion for this system, assuming a constant gravitationalforce in the negative z direction, F = (0, 0,−mg).



(b) Solve these equations for the given initial conditions and find the position, velocity, speedand acceleration of the particle as functions of time.

(c) Find the time at which the particle reaches maximum height, and give the position,velocity, speed and acceleration of the particle at that time.

(d) Find the time at which the particle returns to the earth’s surface, and give the position,velocity, speed and acceleration of the particle at that time.

(e) State briefly which aspects of your solution you find remarkable, and why.

Exercise 7.22A particle moves under the influence of a constant gravitational field (in the vicinity of earth’ssurface) and of the air resistance (which is velocity-dependent). Newton’s equation can thenbe written as

mx = −m g − 2mλ x ,

in which x denotes the height above ground, and λ is a positive constant.

(a) Use the equation of motion to demonstrate that energy E = (mx2/2 + mgx) is notconserved in this system. More specifically, show that energy is a non-increasing functionof time (i.e. E ≤ 0)

(b) Solve the equation of motion. Give the full solution for the case that at t = 0 the particleis propelled upwards with initial speed u > 0.

Exercise 7.23A particle of unit mass moves along the x-axis under the action of a force derived from apotential V (x) = 2x exp(−x2/2).

(a) Sketch V (x), write down the equation of motion, find the fixed points of the motion andclassify them as either elliptic or hyperbolic.

(b) The particle, initially in equilibrium at the stable equilibrium point, is set in motion withspeed v(0) = u > 0 in the positive x-direction. Assuming that deviations from stableequilibrium remain small in the subsequent motion, find x(t) in terms of u. What isthe period τ of the oscillation?

(c) Show that there is critical value uo such that the particle will move off to x = +∞ ifu > u0 but will not it u < u0.

(d) If u > u0, show that its speed v(t) tends to a limit, v(t)→ v∞ as t→∞.

Show that the limiting speed has a lower bound, u∞ such that v∞ > u∞ > 0 for allinitial speeds v(0) > u0. Find the best (largest) lower bound on the limiting speed.

Exercise 7.24A particle of unit mass moves along the x-axis under the action of a force derived from apotential V (x) = x4

4 −x2

2 .

(a) Write down the equation of motion of the particle and find the equilibrium points.



(b) Sketch V . Assume that the particle is released from a position x0 with speed x0 suchthat its total energy is E = −ε2, with 0 < ε < 1/2. Find the turning points of thedynamics.

(c) For the initial condition given in (b), find the ε-dependent coordinate x?(ε) at which thesystem reverses its motion for the first time assuming that x0 > 0 and x0 > 0. Answerthe same question for the case x0 < 0 and x0 > 0.

Exercise 7.25A particle of unit mass moves along the x-axis under the action of a force derived from apotential V (x) = − cosx. Write down the equation of motion of the particle and find theequilibrium points. The particle, initially in equilibrium at x = 0, is set in motion with speedu > 0 in the positive x-direction. Assuming that |x(t)| remains small ( π

2 ) in the subsequentmotion, find x(t) in terms of u.

What is the period τ of the oscillation?. Sketch V and show that if u exceeds a certain criticalvalue uo (which should be found), the particle will move off to x = +∞.

Exercise 7.26Consider the motion of a particle of mass m when it moves along a straight line under theaction of a force derivable from a potential.

(a) If the potential V (x) = x3 + βx2 − γx where β and γ are positive real numbers, identifyall the equilibrium points and describe the motion near the stable ones.

(b) In the case that F = m(11x−2 − 36x−3), compute and sketch V (x). If the particle isinitially at x = 1 and has initial speed u = 4 find the maximum value of x it can attain.




Supplementary Exercise 7.1Consider a particle of mass 1 moving along the real axis in the potential

V (x) = x sin(x) .

Sketch the phase portrait.

Repeat this for the potentialV (x) = x cos(x) .

Supplementary Exercise 7.2(a)Consider a particle moving along a path (x(t), y(t)) in the (x, y) plane.

Let its coordinates in polar coordinates be (r(t), θ(t)) where

x = r cos(θ) , y = r sin(θ) .

Show that its speed is given by v where

v2 = r2 + r2θ2 .

(b)Consider a particle moving along a path (x(t), y(t), z(t)) in R3.

Let its coordinates in spherical polar coordinates be (r(t), θ(t), φ(t)) where

x = r cos(φ) sin(θ) , y = r sin(φ) sin(θ) , z = r cos(θ) .

Show that its speed is given by v where

v2 = r2 + r2θ2 + r2 sin2(θ)φ2 .

Supplementary Exercise 7.3Consider a particle moving in the (x, z) plane with the motion satisfying(

xz

)=

(0a

),

(x(0)z(0)

)=

(00

),

(x(0)z(0)

)=

(v0

),

where a ≥ 0 and v ≥ 0.

Show that the distance s travelled by the particle at time t satisfies

vt+1

2at2 ≥ s ≥ 1

2at2 .

Supplementary Exercise 7.4Calculate the gravitational field at the surface of the moon due to the moon.

Calculate the gravitational forces at the centre of the Earth due to the moon and due to theSun. Which is stronger?



Calculate the difference between the gravitational forces due to the moon on the side of theEarth nearest the moon compared to the force at the centre of the earth (see figure)

rearth

dearth−moon

dearth−moon − rearth

MoonEarth

Repeat this calculation for difference between the field at the surface of the Earth and at thecentre of the the Earth due to the Sun.

Which difference is stronger?

It is the difference between the forces at the surface of the Earth (where the oceans are)and on the centre of the Earth (determining the planet’s motion) which is responsible for thetides, so although the gravitational pull of the Sun on the Earth is far stronger than that ofthe moon, it is the Moon which is principally responsible for the tides.

Supplementary Exercise 7.5The inwards acceleration in a circular path of radius r at constant speed v is v2/r (problem8.1)

Suppose that a planet in orbit around the sun follows a circular path of radius r with the sunat the centre, show that its speed satisfies

v2 =GN Msun

r.

Show that the period of the orbit satisfies

T 2 =4π2

GN Msunr3 .

This is an example of Kepler’s third law (T 2 ∝ r3) that you will study more generally inIntermediate Dynamics.

Supplementary Exercise 7.6(a)Consider a particles of moving along the x-axis in the gravitational fields of a body of massM1 at x = −1 and another of mass M2 at x = 1.

Write down the gravitational potential energy of the particle of mass m due to each of thetwo bodies separately and the total gravitational potential energy of the particle of mass m.

Hence, write down the equation of motion of the particle.

Sketch the two separate contributions to the gravitational potential energy and the totalgravitational potential energy of the particle.

Sketch the phase portrait of the associated dynamical system.



Are there any equilibrium points? Are they elliptic (stable) or hyperbolic (unstable)? Ifstable, find the period of small oscillatory motion around the stable point?

(b)A particle of charge q and mass m = 1 moves in the region −1 < x < 1 in the electric field oftwo static charges, one of charge Q1 at position x = −1 and the other of charge Q2 at x = 1.Its motion is determined by Coulomb’s law for the total potential energy

V =qQ1

4πε0

1

|x+ 1|+qQ2

4πε0

1

|x− 1|.

Here ε0 is a constant.

Note that the potential energy depends on the relative signs of the electric charges as forcecan be either attractive or repulsive (gravitational masses are always positive and the forceis always attractive).

Consider the case where |Q1| = |Q2| = 4πε0/q.

How many different cases do you have to consider?

For each of the separate cases in turn:

Sketch the phase portrait of the associated dynamical system.

Are there any equilibrium points?

Are they elliptic (stable) or hyperbolic (unstable) or can you not decide?

What is the period of the motion around any stable point of equilibrium.

Supplementary Exercise 7.7Consider a massive particle attached to a pivot by a string of length l such that the stringmakes an angle θ with the downward vertical.

Find the initial speed v0 at which the particle has to be moving when it is vertically beneaththe pivot for the string to go slack when the string makes an angle 2π/3 with the downwardvertical.

Find the position of the particle when the string goes taut again. (see diagram below)

Θ

v0

?

2 Π

3

String goes slack at

Θ =2 Π

3



Supplementary Exercise 7.8Photons moving in the gravitational field of a black hole can be described by a function r(θ)where r is a radial coordinate and θ is an angular coordinate.

With u = 1/r, the function u(θ) satisfies

d2u

dθ2= −u+

3

2rSu

2 ,

where rS is the Schwarzschild radius, the location of the event horizon of the black hole. [seeequation (293) of the lecture notes for module 6ccm334a]

Find the radius at which photons can move in a periodic orbit around the black hole. Is thisa stable or unstable orbit?

Find a potential function V (u) for which the u(θ) evolves as though moving in the potentialV (u).

Sketch the phase portrait and describe the possible paths a photon can take.


Chapter 8: Hamiltonian Systems 165

Chapter 8

Hamiltonian Systems

Hamilton’s formulation of mechanics is a way to deal with mechanics more abstractly. It wasformulated by Sir William Rowan Hamilton (1805-1865) (who also invented quaternions). Itlets you treat more complicated systems easily, greatly simplifies the consideration of sym-metries and is of key importance in the development of quantum mechanics.

It is not the first reformulation that is still in use – that was by Joseph Louis Lagrange (1788).This is based on the Lagrangian L, a function of the system’s coordinates and velocities. Fora system with one degree of freedom x, L is a function of x and x with the equations ofmotion being given by

d

dt

(∂L

∂x

)=∂L

∂x, (8.1)

which come from a “least action principle”. The standard way to construct L is as

L = T − V , (8.2)

where T is the kinetic energy and V is the potential energy. For simple harmonic motion,

mx = −mω2x , V =1

2mω2x2 , (8.3)

this gives

L =1

2mx2︸︷︷︸T

− 1

2mω2x2︸︷︷︸V

,∂L

∂x= −mω2x ,

∂L

∂x= mx , (8.4)

and the equations of motion are correctly given through (8.1)

d

dt(mx) = −mω2x . (8.5)

There is a lot more on this in the module Intermediate Dynamics/Classical Dynamics.

Hamilton reformulated this by replacing the velocity x by a new variable, the momentum p,and the Lagrangian L by a new function H(x, p), the Hamiltonian, defined by

p =∂L

∂x, H = p x− L . (8.6)

The equations of motion are then given by Hamilton’s equations

x =∂H

∂p, p = −∂H

∂q. (8.7)



Example 8.1. The Hamiltonian for the simple harmonic oscillator

The Lagrangian is

L =1

2mx2 − 1

2mω2x2 , (8.8)

and so the momentum p and Hamiltonian are given by

p =∂L

∂x= mx , H = px− L =

p2

m− 1

2

p2

m+

1

2mω2x2 =

p2

2m︸︷︷︸T

+1

2mω2x2︸︷︷︸V

. (8.9)

In this case H is the total energy and this is in fact its general interpretation. We can checkthat Hamilton’s equations give back the correct dynamics for x:

x =∂H

∂p=

p

m

p = −∂H∂x

= −mω2x⇒ x =

p

m= −ω2x . (8.10)

More generally a Hamiltonian system of one degree of freedom may be described in terms oftwo functions q, p which evolve in time according to Hamilton’s Eqs (8.7) for some HamiltonianH(q, p, t). In this wider sense the system need not be mechanical, and even in a mechanicalsituation q need not necessarily be a linear displacement—for example, it could be the anglewhich a pendulum makes with the vertical. The variable q is referred to as a generalisedcoordinate and p is the momentum conjugate to q. q, p are said to be conjugate variables. Ofcourse, in the wider sense which we have described p, the momentum conjugate to q, neednot be a physical momentum.

In what follows we shall usually be dealing with an autonomous Hamiltonian system; in thiscase H = H(q, p) and H doesn’t depend on time t explicitly.

The Lagrangian and Hamiltonian formulations of mechanics allow one to find and studythe equations of motions of complicated systems such as spinning bodies or interconnectedparticles easily, but in such cases the definition of the momentum conjugate to a generalisedcoordinate can be complicated and one usually first sets up the Lagrangian formalism tofind the momentum and then constructing the Hamiltonian. There is one simple situation inwhich we can directly write down the Hamiltonian and that is for motion in a potential aswe studied in the previous chapter.

8.1 Hamilton’s equations for motion in a potential

Consider a particle of mass m moving in a straight line along the q-axis (it is traditional inHamiltonian theory to use q rather than x) in a potential V (q). If we conjecture that theHamiltonian is given by H = T + V and that p = mx, we are led to

H(q, p) =p2

2m+ V (q) . (8.11)



Hamilton’s equations (8.7) are thenx =

∂H

∂p=

p

m

p = −∂H∂x

= −dV

dx

⇒ mx = p = −V ′(x) . (8.12)

This is the simplest possible application of Hamilton’s formulation. The only essential ingre-dients are the coordinate q, the conjugate momentum p, and the Hamiltonian H(q, p, t). Themechanical system which we are describing is said to have one degree of freedom (representedby the coordinate q) but in terms of our earlier definitions is a second order dynamical systemwhich is autonomous provided V doesn’t depend on t explicitly. We note in passing thatHamilton’s equations may be expressed in the form

d

dt

(qp

)= σ(∇H),

where ∇ is the vector differential operator given by

∇ =∂

∂qe1 +

∂

∂pe2

and σ is the 2× 2 anti-symmetric matrix given by

σ =

(0 1−1 0

).

Suppose now that q(t), p(t) satisfy Hamilton’s equations with H = H(q, p). Then H(q, p) isconstant throughout the motion—we say that H is a constant of the motion. To prove thisresult we note that

H =∂H

∂qq +

∂H

∂pp,

by the chain rule. Using Hamilton’s equations we obtain

H =∂H

∂q

∂H

∂p+∂H

∂p

(−∂H∂q

)= 0.

We conclude that H is indeed a constant of the motion. Such Hamiltonian systems are said tobe conservative; the Hamiltonian is constant throughout the motion. The phase curve whichpasses through the point (q0, p0) of phase space is given by

H(q, p) = H(q0, p0).

In this case we have no complicated differential equations to solve!

Returning to case of motion in a potential, suppose that V = V (q) so that

H = H(q, p) =p2

2m+ V (q).



Note that p2

2m = 12mq

2, so the first term in the expression for H is the kinetic energy; V isthe potential energy and H is the total energy of the system in this case.

We note that the velocity v(q, p) of the flow associated with a Hamiltonian H(q, p) (seeChapter 8, Section 1.1) is given by

v(q, p) =

(∂H∂p

−∂H∂q

).

A flow which is governed by Hamilton’s equations is referred to as a Hamiltonian flow.

Example 8.2. A Non-Standard Hamiltonian System

Prove that the system of one degree of freedom whose motion is governed by the second order differ-ential equation

q +G(q)q2 − F (q) = 0

is Hamiltonian.

We note that the given equation implies that

d

dq

(1

2q2)

+G(q)q2 − F (q) = 0,

so thatd

dq

(q2)

+ 2G(q)q2 − 2F (q) = 0.

Using the integrating factor

µ(q) = exp

∫2G(q) dq

we obtaind

dq

(µ(q)q2

)− 2µ(q)F (q) = 0.

Writing

µ(q)F (q) = −dVdq

we see thatd

dq

(µq2 + 2V (q)

)= 0,

and therefore1

2µq2 + V (q)

is constant throughout the motion. If we write q = p/µ, so that

p2

2µ+ V (q)

is constant, it is reasonable to guess that the given differential equation is equivalent to Hamilton’sequations with Hamiltonian H(q, p) given by

H(q, p) =p2

2µ+ V (q).



First note that

q =p

µ(q)=

∂

∂p

(p2

2µ(q)

)=∂H

∂p,

so that the first of Hamilton’s equations is satisfied.

Moreover, since p = µq

p = µq +dµ

dqq2 = µ(−Gq2 + F (q)) +

dµ

dqq2

= −dVdq− µGq2 +

dµ

dqq2.

Now dµdq = 2Gµ. It follows that

p = −dVdq

+1

2

dµ

dqq2 = −dV

dq+

1

2

dµ

dq

p2

µ2

= −dVdq− ∂

∂q

(p2

2µ

)= −∂H

∂q.

Summarising, the given equation is equivalent to the Hamiltonian equations

q =∂H

∂p, p = −∂H

∂q,

where

H(q, p) =p2

2µ+ V (q)

and

µ(q) = exp

∫2G(q) dq, V (q) = −

∫µ(q)F (q) dq + C.

small

Example 8.3. Divergence-Free Velocity Fields and Hamiltonian Systems

Consider the equationd

dt

(xy

)= v(x, y) =

(v1(x, y)v2(x, y)

). (8.13)

In order for this system to be Hamiltonian with generalised coordinate x and conjugatemomentum y there must exist H(x, y) such that

x =∂H

∂y, y = −∂H

∂x. (8.14)

This requires that

v1(x, y) =∂H

∂y, v2(x, y) = −∂H

∂x. (8.15)

In order for these equations to be consistent v1, v2 must satisfy

∂v1

∂x=

∂2H

∂x∂y= −∂v2

∂y, (8.16)



∂v1

∂x+∂v2

∂y= 0, i.e. ∇ · v = 0; (8.17)

in other words, the vector v has to be solenoidal1. One can prove that the converse is alsotrue i.e. if v is solenoidal, then the system is Hamiltonian and one can find the HamiltonianH.

If we know that a system is Hamiltonian then we can write

∇H =

(∂H∂x∂H∂y

)=

(−v2(x, y)v1(x, y)

)= W , (8.18)

and we can use the fundamental theorem of calculus for line integrals to write

H(x, y) =

∫ (x,y)

(x0,y0)W · dx , (8.19)

that is we can find the Hamiltonian by integration of the vector field W along a path fromsome fixed point (x0, y0). This path can be chosen freely – the result of the integration(8.19) is independent of the path. The choice of the starting point (x0, y0) is also irrelevantas changing the starting point only adds a constant to H which does not affect Hamilton’sequations.

As an example: Consider the system(qp

)= v =

(p− q2

2qp

). (8.20)

We have

∇ · v =∂

∂q

(p− q2

)+

∂

∂p(2qp) = −2q + 2q = 0 , (8.21)

and hence the system is Hamiltonian. We can reconstruct H as the line integral

H(Q,P ) =

∫ (Q,P )

(q0,p0)(p− q2)dp− 2qpdq . (8.22)

One way is to choose (q0, p0) = (0, 0) and the path to consist of the straight line

r(t) =

(QλPλ

), q(λ) = Qλ , p(λ) = Pλ , 0 ≤ λ ≤ 1 . (8.23)

The integral then becomes

H(Q,P ) =

∫ 1

λ=0(p(λ)− q(λ)2)

dp

dλdλ − 2q(λ)p(λ)

dp

dλdλ

=

∫ 1

0

((Pλ−Q2λ2)P − 2QPλ2

)dλ

=

∫ 1

0(P 2λ− 3Q2Pλ2)dλ

1The term solenoidal comes from the magnetic field B of a solenoid. This field satisfies ∇ · B = 0, andhence any field satisfying the same equation is called solenoidal



=1

2P 2 −Q2P . (8.24)

We can check this does indeed give back the correct equations of motion:

H(q, p) =1

2p2 − q2p⇒ q =

∂H

∂p= p− q2 , p = −∂H

∂q= 2qp . (8.25)

Example 8.4. Pendulum

A simple pendulum consists of an inelastic string of length a with a particle P of mass m attachedto one end; the other end is attached to a fixed point O and the system is free to move in a verticalplane under gravity. Discuss the motion and show in particular that the system is Hamiltonian.

mg

T

y

x

ψ

P (x,y)

Figure 8.1: Coordinate system of the pendulum discussed in the present example.

Referring to the diagram suppose that the y-axis points in the direction of the downward verticalthrough O. Let T denote the tension in the string, and suppose that the string makes an angle ψ(t)with the downward vertical.

The particle P has coordinates (x, y) given by

x = a sinψ, y = a cosψ.

The equation of motion of the particle is given by N2:

m(xe1 + ye2) = (−T sinψe1 + (mg − T cosψ)e2),

mx = −T sinψ, my = mg − T cosψ.

It follows thatmx cosψ −my sinψ = −mg sinψ, (8.26)

T = mg cosψ −mx sinψ −my cosψ. (8.27)

Nowx = aψ cosψ, x = aψ cosψ − aψ2 sinψ,

y = −aψ sinψ, y = −aψ sinψ − aψ2 cosψ.

We deduce from Eqs (8.26) and (8.27) that

ψ +g

asinψ = 0, T = mg cosψ +maψ2.



The first of these equations is a differential equation for ψ and the second gives the tension T in termsof ψ. We note that when ψ is small (i.e. near 0), so that the string is always close to the downwardvertical, the equation for ψ approximates to

ψ +g

aψ = 0.

In this case the motion is approximately Simple Harmonic with period τ given by

τ = 2π

√a

g.

Returning to the exact equation for ψ we note that

ma2ψ = −mga sinψ.

We write p = ma2ψ so that

ψ =p

ma2=

∂

∂p

(p2

2ma2

).

Moreover,

p = ma2ψ = −mga sinψ = − ∂

∂ψ

(−mga cosψ

).

We conclude that

ψ =∂H

∂p, p = −∂H

∂ψ,

where

H(ψ, p) =p2

2ma2−mga cosψ.

The system is therefore Hamiltonian in the generalised coordinate ψ, p being the momentum conjugateto ψ; H(ψ, p) is of course the Hamiltonian. It is interesting to note that the kinetic energy T of theparticle is given by

T =1

2m(x2 + y2) =

1

2m(cos2ψ + sin2ψ)ψ2 =

1

2m

p2

m2a4=

p2

2ma2

so that the first term in the expression for H is just the kinetic energy of the particle. The secondterm is in fact the potential energy V due to gravity since

mge2 = −dVdy

e2,

and thereforeV = −mgy = −mga cosψ.

It follows that the Hamiltonian is the total energy of the system; since H doesn’t depend on t explicitlyit is a constant of the motion and

H(ψ, p) =p2

2ma2−mga cosψ = C,

where C is constant; the value of C depends on the initial conditions.

We note that ψ can range over the interval [0, 2π], the ends 0, 2π being identified, so that the domainof ψ is essentially the unit circle S1. On the other hand, p can range over the whole of R. The phasespace of this system is the Cartesian product R× S1 which we can picture as a circular cylinder.



8.2 Stability problems

Suppose that a particle of mass m moves along the q-axis under a potential V (q). As notedabove, the Hamiltonian for the motion is

H(q, p) =p2

2m+ V (q).

The Hamilton equations are (qp

)=

(p/m

−V ′(q)

).

We see that the fixed points are those points (q, p) for which p = 0 and V ′(q) = 0. Thisagrees with our earlier conclusion that the equilibrium points of such a system correspond tothe stationary points of the potential energy function V. Suppose that V ′(q0) = 0. Then nearq0 we can approximate V by

V (q) = V (q0) + (q − q0)V ′(q0) +1

2(q − q0)2V ′′(q0) + · · ·

so that the linearised Hamiltonian equations become(qp

)=

(0 1/m

−V ′′(q0) 0

)(q − q0

p

).

The matrix (0 1/m

−V ′′(q0) 0

).

has eigenvalues λ given by λ2 = − 1mV

′′(q0). If V ′′(q0) > 0, so that q0 is a minimum of V , wehave

λ = ±iβ, β =√V ′′(q0)/m.

We deduce that there is a matrix P such that(X

Y

)=

(0 β−β 0

)(XY

),

where (XY

)= P−1

(q − q0

p

).

The fixed point (q = q0, p = 0) is the origin of the new (X,Y ) coordinates. Following standardprocedures we write

X = r cos θ, Y = r sin θ

and obtainr = 0, θ = −β, r(t) = r0, θ(t) = θ0 − βt.

These equations describe a family of circles and the fixed point is therefore stable; the term‘elliptic’ is sometimes used to describe such fixed points. The time τ taken by the systemto move round one of these circles is precisely the time taken by the system to execute onecomplete Simple Harmonic oscillation about q = q0 and is given by

τ =2π

β= 2π

√m

V ′′(q0),



as found previously by a different method.

In a similar manner we find that if V ′′(q0) < 0, so that q0 is a maximum of the potentialfunction V , that the linearised Hamiltonian equations may be written as

X = λ1X, Y = −λ1Y, λ1 =

√−V

′′(0)

m> 0,

with integralsX(t) = X0e

λ1t, Y (t) = Y0e−λ1t.

The fixed point is clearly unstable, the phase curves being a family of rectangular hyperbolasgiven by XY = X0Y0; such unstable fixed points are sometimes said to be ‘hyperbolic’ fixedpoints.

Example 8.5. Hamiltonian with Quartic Potential

The Hamiltonian H of a particle of unit mass moving in a straight line along the q-axis is given by

H(q, p) =1

2p2 +

1

4q4 − 1

2q2.

Discuss the motion.

q

q

Separatrix

Figure 8.2: Quartic potential described in the text and construction of the phase curves atdifferent energies.



This is the Hamiltonian for a particle moving under the potential V (q) given by

V (q) =1

4q4 − 1

2q2.

We note that for small q2 the force acting on the particle is approximately equal to

− d

dq(−1

2q2)e1 = qe1;

it follows that the potential generates a short range repulsion. On the other hand, for large values ofq2 the term 1

4q4 is dominant and this generates a force equal to

− d

dq(1

4q4)e1 = −q3e1.

This means that the potential V produces a long range attraction which tends to pull the particle backtowards the origin. The over-all effect is seen from the graph of V.

Hamilton’s equations give (qp

)=

(p

q − q3).

The fixed points are given by p = 0, q−q3 = 0 i.e. by p = 0, q = 0, q = ±1. Linearising the equationsin the usual way we have to consider the Jacobian J(q, p) at the fixed points. We have

J(q, p) =

(0 1

1− 3q2 0

).

We see that

J |0,0 =

(0 11 0

).

This matrix has eigenvalues λ = ±1 and the now familiar arguments lead us to conclude that (0, 0) isan unstable fixed point. Similarly we find that

J |(±1,0) =

(0 1−2 0

).

This matrix has eigenvalues λ = ±i√

2, so that we are dealing with a stable (elliptic type) fixed point.The Jordan form is given by the matrix (

0√

2

−√

2 0

).

A standard calculation leads to the conclusion that the system makes a complete circuit (in phasespace) of the fixed point point (q = ±1, 0) in time τ given by τ = 2π/

√2, precisely the time taken by

the particle to execute one complete Simple Harmonic oscillation about q = ±1.

The phase curves are given by

H(q, p) =1

2p2 +

1

4q4 − 1

2q2 = E,

where E is a parameter.

A phase curve which passes through a hyperbolic (unstable) fixed point is known as a separatrix; itsimportance lies in the fact that it marks the boundary between motions having different properties.



In this example the phase curve which passes through the hyperbolic point (q = 0, p = 0) is given byE = 0 so that the separatrix has equation given by

p = ± q√2

(2− q2)1/2.

The accompanying diagram showing the phase curves and the potential function V exhibits the fol-lowing symmetries:

V (−q) = V (q), H(q, p) = H(q,−p), H(−q, p) = H(q, p).

As an example, suppose that the particle starts from P1 with coordinates (q = q1, p = 0) (so that itstarts from rest at q = q1). It moves round the phase curve in the sense indicated, passing through P2

with coordinates (q = −q1, p = 0) before returning to P1. The time τ taken for one complete oscillationis just four times the time taken to go from q = 0 to q = q1. The relevant phase curve has equationgiven by

1

2p2 +

1

4q4 − 1

2q2 =

1

4q1

4 − 1

2q1

2

and Hamilton’s equations give

q =∂H

∂p= p

so that

τ = 4

∫ q1

0

dq

p= 2√

2

∫ q1

0

dq

( 14 (q14 − q4)− 1

2 (q12 − q2))1/2

= 4√

2

∫ q1

0

dq

(q12 − q2)1/2(q12 + q2 − 2)1/2.

8.3 Summary: how to analyse motion in a potential

We can now summarise the analysis of motion in a potential using all the tools we havedeveloped in the course so far.

A particle of mass m = moving in a potential V (x) corresponds to the Hamiltonian

H =p2

2m+ V (x) (8.28)

with Hamilton’s equations being

x =∂H

∂p=

p

m, p = −∂H

∂q= −V ′(x) , (8.29)

⇒ mx = −V ′(x) . (8.30)

1. As a Hamiltonian system with H independent of t, the phase curves are H(x, p) =constant.

2. Treating (8.29) as a second order autonomous dynamical system, we can sketch the phaseportrait using the standard methods: the null clines are p = 0 on which x = 0 and theflow is vertical; x such that V ′(x) = 0 on which p = 0 and the flow is horizontal. Theselines divide the plane into regions on which we can deduce the general direction of theflow.



3. We can find the fixed points and classify their nature. A fixed point of the system (8.29) is(x = x∗, p = 0) where V ′(x∗) = 0. This will be either elliptic or hyperbolic, dependingon the sign of V ′′(x∗).

If V ′′(x∗) > 0 then this is a minimum of V (x), a point of stable equilibrium of (8.30)and an elliptic (marginally stable) fixed point of (8.29);

If V ′′(x∗) < 0 then this is a maximum of V (x), a point of unstable equilibrium of(8.30) and a hyperbolic (unstable) fixed point of (8.29);

4. If there is a hyperbolic fixed point (x∗, 0) then the separatrix will go through this fixedpoint and its equation with be H(x, p) = H(x∗, 0).

5. Knowing H(x, p) = p2/(2m)+V (x), we can sketch V (x) and then choosing suitable valuesof the total energy E which satisfies H(x, p) = E, we can deduce the ranges of x towhich the motion is satisfied

6. We can find the exact period of periodic orbits using the reduction of the second ordersystem (8.29) to a first order system for x by use of the equation of conservation ofenergy:

x = ±√

2/m(E − V (x)) ,

∫dx√

2/m(E − V (x))=

∫dt . (8.31)

We can find the approximate period of small periodic orbits around a fixed point bylinear stability analysis resulting in

T ∼ 2π√V ′′(x∗)/m

. (8.32)

7. We can sketch typical trajectories by following the orbits on the phase portrait.

We will apply this now to two examples.

Example 8.6. Analyse the motion in the potential of a particle of mass m = 1

V (x) =1

4x4 − 1

3x3 − x2 . (8.33)

and sketch the graph of x vs t for a particle starting at x = 0 with initial momentum p = −1.

First we plot V (x). For large x, V (x) ∼ x4/4. The equilibrium points are solutions ofV ′(x) = 0. We have

V ′(x) = x3 − x2 − 2x = x(x+ 1)(x− 2) , (8.34)

so the points of equilibrium are x = 0, x = −1 and x = 2. To find their nature we calculateV ′′(x∗). We have

V ′′(x) = 3x2 − 2x− 2 , (8.35)

so V ′′(0) = −2, an unstable point of equilibrium/a hyperbolic fixed point of the Hamiltoniansystem; V ′′(−1) = 3 so this is a stable point of equilibrium/an elliptic fixed point of theHamiltonian system; V ′′(2) = 6 so this is also a stable point of equilibrium/an elliptic fixedpoint of the Hamiltonian system. At the fixed points, V (0) = 0, V (−1) = −5/12 andV (2) = −8/3. Since p = −V ′(x), the flow is horizontal for these values of x.



We can now plot V (x) and choose four representative values for the total energy, which arelabelled A, B, C, and D. The phase curves are H(x, p) = E; If the total energy is E thensince E = p2/2 + V (x), the motion of the particle is confined to regions in which V (x) ≤ E.

Given our choices, a particle with energy A will move in an orbit around the fixed point atx = 2; a particle with energy B will orbit around either the fixed point at x = −1 or thefixed point at x = 2 but since V (0) > B it cannot pass through the point x = 0; a particlewith energy C=0 moves along the separatrix which is given by p2/2 + V (x) = 0. It will tendasymptotically to the fixed point at x = 0 but will not reach it in finite time.

A particle with energy D will move in a periodic orbit enclosing both fixed points.

This is summarised in figure 8.3. Note that the flow is vertical on the null-cline p = 0 andhorizontal on the null-clines x = −1, x = 2 and x = 2.

x

VHxL

A

BCD

x

p

X

Figure 8.3: Analysis of the potential (8.33).

A particle starting at x = 0 with negative momentum (such as the starting point X) willinitially move off to the left then stop, return to some (larger) positive value of x and carryon in a periodic motion, something like the plot in figure 8.4. Note that the speed slows



Figure 8.4: Sketch of the motion of a particle starting at point X.

slightly as the particle passes through x = 0 as this is a local maximum of the potentialenergy and hence a local minimum of the kinetic energy.

Example 8.7. Analyse the motion of a particle of mass m = 1 moving in the potential

V (x) =x3 − 5x

x2 + 3. (8.36)

and describe the shape of the phase curves for large negative x. Can a particle escape tox = −∞ in finite time?

First we plot V (x). For large x, V (x) ∼ x, for small x, V (x) ∼ −5x/3. The equilibriumpoints are solutions of V ′(x) = 0. We have

V ′(x) =(x2 − 1)(x2 + 15)

(x2 + 3)2, (8.37)

so the points of equilibrium are x = −1 and x = 1. To find their nature we calculate V ′′(x∗).We have

V ′′(−1) = −2 , V ′′(1) = 2 , (8.38)

so x = −1 is a local maximum of the potential/an unstable point of equilibrium/a hyperbolicfixed point of the Hamiltonian system and x = 1 is a local minimum of the potential/a stablepoint of equilibrium/an elliptic fixed point of the Hamiltonian system. At the fixed points,V (−1) = 1 and V (1) = −1. Since p = −V ′(x), the flow is horizontal for these values of x.

We can now plot V (x) and choose four representative values for the total energy, which arelabelled A, B, C, and D.

Given our choices, a particle with energy A will move in from x = −∞ but will not haveenough energy to pass over the local maximum at x = −1 and will return to x = −∞; Aparticle with energy B will either perform a similar motion or will move in a periodic orbitaround the fixed point at x = 1, depending on its initial position and momentum; a particlewith energy C=1 moves along the separatrix. It will either tend asymptotically to the fixedpoint at x = −1 but not reach it in finite time or it will escape to x = −∞.

A particle with energy D will move in from x = −∞, pass through x = −1 and x = 0, reacha maximum value of x and then change direction and escape to x = −∞. Its speed will be alocal minimum at x = −1 and a local maximum at x = 0 before reaching zero at which pointit will change direction.



This is summarised in figure 8.5. Note that the flow is vertical on the null-cline p = 0 andhorizontal on the null-clines x = −1 and x = 1.

For large negative x, V (x) ∼ x and so H(x, p) = E = p2/2 + V (x) ∼ p2/2 + x. The curvesx = E − p2/2 are parabolae.

We find p = x = −√

2√E − x ∼ −

√2√|x| for large negative x and so the time to reach a

point x1 is asymptotically

τ =

∫ x1 dx

x∼∫ x1 dx

−√

2|x|=√

2|x1|

This diverges as x1 → −∞ and so the particle cannot escape to infinity in finite time.

x

VHxL

A

B

C

D

x

p

Figure 8.5: Analysis of the potential (8.36).



8.4 Exercises

Shorter exercises

Exercise 8.1For each of these Hamiltonian systems: Write down and solve Hamilton’s equations, writedown the equations of the phase curves and sketch the phase curves (i.e. orbits in the phasespace), find the equations of any separatrices, find all fixed points and describe their nature.

(i) H =p2

2+q2

2(ii) H =

p2

2− q2

2(iii) H = pq .

Show that the last two systems can be related by q = (p+ q)/√

2 and p = (p− q)/√

2.

Exercise 8.2Consider the following equations of motion. In each case decide whether they can be seen asHamilton’s equations for some Hamiltonian H(q, p). If they can, find the Hamiltonian andcheck that this leads to the correct equations of motion; if they cannot, explain why

(a) q = 1 , p = 1

(b) q = 1 , p = −1

(c) q = q , p = −p

(d) q = q , p = p

(e) q = 2p+ 2q , p = −2p− 1

(f) q = ep , p = −peq

(g) q = 1 + qep , p = −ep

Recall that a necessary condition for the equations of motion q = v(q, p) , p = w(q, p) bederived from a Hamiltonian is that ∂v/∂q + ∂w/∂p = 0, in which case H(q, p) = H(q0, p0) +∫ (q,p)

(q0,pq)wdp− vdq .



Longer exercises

Exercise 8.3Consider the system described by the Hamiltonian

H =1

2p2 − 1

4q4

(a) Write down Hamilton’s equations and find the fixed points of the dynamics. Can youclassify them as either elliptic or hyperbolic?

(b) Derive the equations for the the phase curves, in particular that of the separatrix. Solvethe equation of motion along the separatrix.

(c) Sketch the phase portrait.

Exercise 8.4The motion of a particle of unit mass moving along the x-axis is described by the Hamiltonian

H(x, p) =1

2p2 +

x

x2 + 1,

where p is the momentum conjugate to x.

Write down Hamilton’s equations, find the fixed points of the system, and classify them aseither hyperbolic or elliptic fixed points.

Find the equation of the separatrix and sketch the phase diagram. The particle is releasedfrom rest at x = −3. Explain briefly, in the context of your diagram, why the subsequentmotion is oscillatory. Find the maximum speed of the particle and obtain an expression forthe period τ of one complete oscillation.

Exercise 8.5Consider the system described by the Hamiltonian

H =1

2p2 − q2

(1 + q2)2

.

(a) Write down Hamilton’s equations, find the fixed points of the dynamics and classifythem as either elliptic or hyperbolic.

(b) Derive the equations for the the phase curves, in particular that of the separatrix. Solvethe equation of motion along the separatrix.




Exercise 8.6The motion of a particle of unit mass along the x axis is governed by the Hamiltonian

H(x, p) =1

2p2 + V (x)

with

V (x) =

sinx , |x| ≤ 3π

2

−sign(x) , |x| > 3π2

(a) Write down Hamilton’s equations of motion for this system. Find the fixed points for|x| < 3π

2 and classify them as either elliptic or hyperbolic.

(b) Derive the equation for the phase curves of the system, in particular that of the separatrix.


(d) The particle is released from rest at x = −1. Describe the subsequent motion for thiscase. Give the coordinate(s) at which the maximum speed is attained and find the valueof the maximum speed.

(e) Suppose now that the particle is released from x = 0 with initial velocity u > 0 inthe positive x-direction. Give the minimum value of u for which the motion will beunbounded. Give the limiting velocity that the particle will have as x → ∞, if themotion is unbounded.

(f) Find the fixed points for |x| ≥ 3π/2. Can you classify them into elliptic or hyperbolicby considering the linearisation of Hamilton’s equations? Are the fixed points stableor unstable - recall our definition of stable was that changing their initial position andvelocity by a small amount would still leave their future motion close to the fixed point.

[Note sign(x) is the sign function, which evaluates to 1 when the argument of the functionis positive, i.e. for x > 0 and evaluates to −1 when x < 0. It is discontinuous at x = 0.]

Exercise 8.7 Throughout the following question you may assume that two positive electriccharges, of magnitude c1, c2 respectively, distance z apart, repel each other with a forcec1c2/z

2, the direction of the force being along the straight line join of the charges.

Fixed positive electric charges, each of magnitude e1 are placed on the x-axis at x = 0, x =2a (a > 0). A third positive electric charge of magnitude e2 and mass m is constrained tomove on the x-axis between the first two charges so that its coordinate is x(t) at time t. Writedown N2 for the moving charge e2, and show that it can be cast in Hamiltonian form with

H(x, p) =p2

2m+µ

x+

µ

2a− x,

where µ = e1e2.

Find the fixed points of the system and sketch the phase diagram for 0 < x < 2a. The chargee2 starts from rest at x = a(1 +k), 0 < k < 1. Find the equation of the phase curve on which



it moves in the form f(x, p) = constant (*) and show by consideration of the phase diagramthat it will subsequently oscillate with period τ given by

τ = 4

∫ a(1+k)

a

mdx

p(x),

where p(x) is given by (*). Verify that the substitution x = a+ak sinφ leads to the expression

τ = 2

(ma3(1− k2)

µ

)1/2 ∫ π/2

0(1− k2sin2φ)1/2 dφ.



Supplementary exercises

Supplementary Exercise 8.1This problem is about a particle (of mass m) moving smoothly (with no friction) on a wirein the x–z plane described by a function z = f(x) in a constant vertical gravitational field ofstrength g.

There are several ways to tackle the problem. At the most basic level, we can work out theforces on the particle and set up Newton’s equations that way.

We could also work out the kinetic and potential energy and set up a Lagrangian for thesystem, which is well outside this course, and then from that obtain a Hamiltonian anddescribe the system in terms of the Hamiltonian and Hamilton’s equations.

The test will be whether we get the same equations from the two methods.

Having set the system up, we can then solve the equations - which from the Hamiltonianpoint of view is a simple matter of saying H = constant.

(A)

To begin with: obtaining the equations of motion from Newton’s equations. If the particlehas speed v, then its acceleration is v and Newton’s equations say that mv is given by thecomponent of the gravitational force parallel to the motion of the particle, denoted F‖. Thecomponent perpendicular to the motion, F⊥ keeps the particle on the wire. If the inclinationof the wire to the horizontal is an angle θ as in the figure below, show that

(a) |F‖| = mg sin θ ,

(b) v =x

cos θ,

(c) v = −g sin θ ,

(d) tan θ = f ′(x) ,

(e) x = −x2 f ′f ′′

1 + (f ′)2− g f ′

1 + (f ′)2.

θ

mg

F‖

F⊥

v

x

z

z = f(x)



(B)

We can then consider the construction of the Hamiltonian. Just to show how it goes, thekinetic energy T , the gravitational potential energy V and Lagrangian of the particle are

T =1

2mv2 =

1

2mx2(1 + (f ′)2) , V = mgf(x) , L = T − V =

1

2mx2(1 + (f ′)2)−mgf .

This means the generalised momentum p conjugate to x and the Hamiltonian H(p, x) are

p =∂L

∂x= mx(1 + (f ′)2) , H(p, x) = px− L =

1

2m

p2

1 + (f ′)2+mgf .

(a) Using this expression for the Hamiltonian, show that Hamilton’s equations are

x =p

m(1 + (f ′)2) , p =

p2

m

f ′f ′′

(1 + (f ′)2)2−mgf ′ .

(b) Hence show that

x+ x2 f ′f ′′

1 + (f ′)2= −g f ′

1 + (f ′)2.

This is exactly the sort of system considered in lectures and in notes and this is the sameequation as found in part (A) from Newton’s laws.

(C)

(a) Show that if f(x) = h is constant, this reduces to motion in the absence of a gravitationalfield.

(b) Show that if f(x) = αx corresponding to a slope of constant gradient then the particle’sacceleration is constant.

(c) Consider the motion on a circular wire of radius 1, ie f(x) = 1−√

1− x2. Let the angleφ be defined by x = sin(φ) as in the figure below.

φ

Show that this reduces to the standard equation for a pendulum,

φ = −g sinφ .

[Hint: substitute x = sin(φ) into the equation for x. You will need to work out that x =cos(φ)φ, etc. ]


Appendix A: Functions of two variables 187

Appendix A

Functions of two variables

A.1 The partial derivative

The partial derivatives of a function of two variables f(t, x) are defined as

∂f

∂tf(t, x) = lim

h→0

f(t+ h, x)− f(t, x)

h, (A.1)

∂f

∂xf(t, x) = lim

h→0

f(t, x+ h)− f(t, x)

h. (A.2)

They are also often written using subscripts, so that

ft =∂f

∂t, fx =

∂f

∂x. (A.3)

This definition means that you can calculate a partial derivative in exactly the same way asa normal derivative just by assuming that any other variables are just constants, with all theusual rules such as the product and chain rules.

For example,

∂

∂t(exp(xt2)) = 2tx exp(xt2) ,

∂

∂x(exp(xt2)) = t2 exp(xt2) . (A.4)

One can also take the derivative of a function more than once, and so there are in principlefour second partial derivatives:

∂2f

∂x2=

∂

∂x

(∂f

∂x

),

∂2f

∂x∂t=

∂

∂x

(∂f

∂t

),

∂2f

∂t∂x=

∂

∂t

(∂f

∂x

),∂2f

∂t2=

∂

∂t

(∂f

∂t

), (A.5)

For well-behaved functions, it actually does not matter in which order you take the two mixedderivatives: you get the same result,

∂

∂x

(∂f

∂t

)=

∂

∂t

(∂f

∂x

). (A.6)

It is possible to come up with functions for which this is not true, but they have to be carefullycontrived and will not make an appearance in this course.



The reason for a different notation is that sometimes one wants to think of some variables dependingon others. So, for example, if x is a function of t, then f(t, x(t)) depends on t both through the explicitt dependence in the first argument but also on the t dependence of x in the second argument. Thismeans one can defined the “total derivative” of f(t, x(t)) with respect to t and gets the result

d

dt

(f(t, x(t)

)= lim

h→0

f(t+ h, x(t+ h))− f(t, x(t))

h

=∂f

∂t(t, x(t)) +

dx

dt(t)

∂f

∂x(t, x(t)) . (A.7)

This will not play any role in this course, but it goes some way to explaining why we need two different

notations d/dt and ∂/∂t for the two different concepts.



A.2 Continuity of a function of two variables

There are many equivalent definitions of the continuity of a function of one variable f(x) atx = a, for example:

f(x) is continuous at x = a if the two directional limits exist and are equal to f(a):

limx→a−

f(x) = limx→a+

f(x) = f(a) . (A.8)

f(x) is continuous at x = a if, for any ε > 0, one can find a δ > 0 such that

|x− a| < δ ⇒ |f(x)− f(a)| < ε . (A.9)

These definitions are necessary and useful but it is also important to understand that thefundamental idea is that one the function f is continuous if one can draw the graph of f(x)as a line. If there is a jump at x = a or if the function is not defined there then it is notcontinuous at x = a.

For a function of two variables f(x, y), the line is replaced by a surface. The graph of afunction of two variables is the surface with height z = f(x, y). The function f is continuousat (a, b) if one can approach (a, b) from any direction, in any manner, and the limit alongthat path is the same as the value of the function.

This can again be expressed in many ways, of which just two are

f(x, y) is continuous at (a, b) if the limit exists and is equal to f(a):

lim(x,y)→(a,b)

f(x, y) = f(a, b) . (A.10)

f(x, y) is continuous at (a, b) if, for any ε > 0, one can find a δ > 0 such that

|(x− a)2 + (y − b)2| < δ ⇒ |f(x)− f(a)| < ε . (A.11)

A simple heuristic test is that any function for which the graph is a smooth surface will becontinuous. This will apply to almost all functions of two variables we consider.

The simplest way for a function of two variables to not be continuous at a point is for it notto be defined there (i.e. it is “infinite there”). There are other ways which will be highlightedif they occur.

For example:

— The function f(x, y) = 1/(x2 + y2) is continuous everywhere except at the point (0, 0)where it is not defined.

— The function arctan(y/x) is continuous everywhere except along the line x = 0.


Appendix B: Taylor’s Theorem 190

Appendix B

Taylor’s Theorem

B.1 Taylor Expansion for Functions of One Variable

We recall Taylor’s Theorem1 for functions of one real variable: Suppose that φ is a suitablydifferentiable real valued function on some interval [x− δ, x+ δ]. Then

φ(x+ h) = φ(x) +h

(1!)φ(1)(x) +

h2

(2!)φ(2)(x) + · · ·

+hn−1

(n− 1)!φ(n−1)(x) +Rn, h ∈ [−δ, δ]

where Rn, the remainder after n terms, has the form

Rn =hn

(n!)φ(n)(ξ),

and ξ is some point between x and x+h. In applications we are often interested in situationswhere h is close to 0 and it may be justifiable to neglect the term Rn, for some suitable valueof n. This amounts to saying that we can approximate φ(x + h) by a polynomial in h oforder (n− 1). Notice that in the computation of φ(x+ h) we require to know the value of φand its derivatives at the point x. The particular case where x = 0 goes under the name ofMacLaurin’s2 Theorem; for reasons which will shortly be apparent, we use t rather than h todenote the variable:

φ(t) = φ(0) +t

(1!)φ(1)(0) +

t2

(2!)φ(2)(0) + · · ·+ tn−1

(n− 1)!φ(n−1)(0) +Rn, t ∈ [−δ, δ]

where Rn, the remainder after n terms, has the form

Rn =tn

(n!)φ(n)(ξ),

and ξ is some point between 0 and t.

1Brook Taylor (1685-1731). In 1715 he published his most important work which contained a proof of thetheorem which now bears his name.

2Colin MacLaurin (1698-1746) entered Glasgow University at the age of 11; professor of mathematics atAberdeen at the age of 19!



B.2 Taylor Expansion for Functions of two variables

Suppose now that f is a real valued function defined in some neighbourhood of the point (x, y)(for example in a square of side 2δ, centre (x, y), whose sides are parallel to the axes) andthat f has partial derivatives to any required order. Motivated by our knowledge of Taylor’sTheorem for functions of one variable we aim to obtain an expression for f(x + h, y + k) interms of powers of h, k and the partial derivatives of f evaluated at the point (x, y). Let Adenote the point (x, y) and B the point (x+ h, y + k). Any point P on the straight line ABhas coordinates (x + ht, y + kt), 0 ≤ t ≤ 1; A corresponds to t = 0 and B corresponds tot = 1. Let φ(t) = f(x + ht, y + kt), where x, y, h, k are regarded as fixed and t is variable,with 0 ≤ t ≤ 1. We note that φ(0) = f(x, y) and that φ(1) = f(x+ h, y + k). An applicationof MacLaurin’s formula gives

φ(t) = φ(0) +t

(1!)φ(1)(0) +

t2

(2!)φ(2)(0) + · · ·+ tn−1

(n− 1)!φ(n−1)(0) +Rn.

Let’s write φ(t) = f(u, v), u = x+ht, v = y+kt. By the Chain Rule of partial differentiation

φ(1)(t) =∂f

∂u

∂u

∂t+∂f

∂v

∂v

∂t= h

∂f

∂u+ k

∂f

∂v.

Thus

φ(1)(0) = h∂f

∂u+ k

∂f

∂v,

where the partial derivatives are evaluated at t = 0. A little thought shows that

φ(1)(0) = h∂f(x, y)

∂x+ k

∂f(x, y)

∂y,

where the derivatives of f are to be evaluated at (x, y). We see that symbolically

d

dt= h

∂

∂x+ k

∂

∂y,

where the t derivative is to be computed at t = 0 and the x, y partial derivatives at (x, y). Itis clear that

φ(n)(0) =

(h∂

∂x+ k

∂

∂y

)nf(x, y),

where (h∂

∂x+ k

∂

∂y

)nf(x, y)

is to be interpreted as(h∂

∂x+ k

∂

∂y

)(h∂

∂x+ k

∂

∂y

)· · ·(h∂

∂x+ k

∂

∂y

)f(x, y),

there being n factors in all. Setting t = 1 in the MacLaurin expansion of φ we obtain

f(x+ h, y + k) = f(x, y) +1

1!

(h∂

∂x+ k

∂

∂y

)f(x, y) +

1

2!

(h∂

∂x+ k

∂

∂y

)2

f(x, y) + · · ·



· · ·+ 1

(n− 1)!

(h∂

∂x+ k

∂

∂y

)n−1

f(x, y) +Rn.

It frequently happens in applications that (x+ h, y+ k) is near to (x, y) so that (h, k) is nearto (0, 0) and that it is justifiable to neglect R3. In this approximation we are retaining onlythe linear and quadratic terms in h, k in our expansion of f(x + h, y + k) about the point(x, y).

Note: The calculation of (h∂

∂x+ k

∂

∂y

)nf(x, y)

is straightforward. For example, with n = 2 we have(h∂

∂x+ k

∂

∂y

)2

f(x, y) =

(h∂

∂x+ k

∂

∂y

)(h∂

∂x+ k

∂

∂y

)f(x, y)

= h2∂2f

∂x2+ hk

∂2f

∂x∂y+ kh

∂2f

∂y∂x+ k2∂

2f

∂y2

= h2∂2f

∂x2+ 2hk

∂2f

∂x∂y+ k2∂

2f

∂y2,

using the commutative property of partial differentiation. It’s therefore clear that we canformally square out (

h∂

∂x+ k

∂

∂y

)2

provided we make the following identifications:(∂

∂x

)(∂

∂x

)=

∂2

∂x2,

(∂

∂y

)(∂

∂y

)=

∂2

∂y2,

(∂

∂x

)(∂

∂y

)=

∂2

∂x∂y,

and so on. Using this principle we see, for example, that(h∂

∂x+ k

∂

∂y

)3

f = h3∂3f

∂x3+ 3h2k

∂3f

∂x2∂y+ 3hk2 ∂3f

∂x∂y2+ k3∂

3f

∂y3.

Nevertheless, calculations of this sort require care; the formal expansion procedure works herebecause h, k are constants as far as the x, y differentiations are concerned.

B.3 Vector functions

In applications to dynamical systems we are often concerned with vector valued functions oftwo real variables (x, y). With each point (x, y) we associate a 2–vector

f(x, y) =

(f1(x, y)f2(x, y)

),



where f1, f2 are scalar (i.e. real valued functions) of the real variables (x, y). In applicationswe frequently require to expand f(x+h, y+k) about the point (x, y) retaining only the linearterms in h, k. We have

fj(x+ h, y + k) = fj(x, y) + h∂fj∂x

(x, y) + k∂fj∂y

(x, y) + · · · (j = 1, 2).

The dots indicate that we have neglected higher order terms in h, k. We conclude that

f(x+ h, y + k) = f(x, y) +

(h∂f1∂x + k ∂f1∂yh∂f2∂x + k ∂f2∂y

)+ · · · = f(x, y) + J

(hk

)+ · · ·

where J denotes the Jacobian matrix defined by f1, f2, evaluated at (x, y), i.e.

J =∂(f1, f2)

∂(x, y)=

(∂f1∂x

∂f1∂y

∂f2∂x

∂f2∂y

).

Notation Given a function f of two real variables (x, y) we frequently use subscript notationto denote its partial derivatives:

fx ≡ ∂f

∂x, fy ≡

∂f

∂y,

fxx ≡ ∂2f

∂x2, fxy ≡

∂2f

∂x∂y, fyy ≡

∂2f

∂y2

Example B.1. Expand f(x, y) = sin(πxy) about the point (1, 1) neglecting third and higher orderterms in (x− 1), (y − 1).

In this case h = x− 1, k = y − 1 and Taylor’s formula gives

sin(πxy) = f(1, 1) + (x− 1)fx(1, 1) + (y − 1)fy(1, 1)

+1

2!

[(x− 1)2fxx(1, 1)

+2(x− 1)(y − 1)fxy(1, 1) + (y − 1)2fyy(1, 1)]

+ · · ·

Now,fx = πy cos(πxy) , fy = πx cos(πxy),

fxx = −(πy)2 sin(πxy) , fyy = −(πx)2 sin(πxy),

fxy =∂

∂yfx = −(πy)(πx) sin(πxy) + π cos(πxy) =

∂

∂xfy = fyx

Sincef(1, 1) = 0, fx(1, 1) = −π, fy(1, 1) = −π,fxx(1, 1) = 0, fyy(1, 1) = 0, fxy(1, 1) = −π

it follows thatsin(πxy) ' −π(x− 1)− π(y − 1)− π(x− 1)(y − 1) + · · ·

Exercise B.1 Find the expansion of

f(x, y) =

(arctan(xy)x+ sin(xy)

)about the point (0, 0), as far as the linear terms in x, y.



Exercise B.2 Find the (two variable) Taylor expansion of ln(1 + x2 + y2) about the point(1, 1), up to and including the quadratic terms.

Let

f(x, y) =

(x/(x2 + y2)

ln(1 + x2 + y2)

)Expand f about the point (1, 1), neglecting quadratic and higher order terms.

Exercise B.3 Suppose that f is a real valued function of the real variables (x, y) with con-tinuous partial derivatives to all orders. Suppose that fx(a, b) = 0, fy(a, b) = 0, so that (a, b)is a stationary point of f. Use Taylor’s Theorem to show that near (a, b) f is given by

f(x, y)− f(a, b) =1

2XTHX + · · · , X =

(x− ay − b

),

T denotes transpose, and H is the Hessian matrix given by(fxx fxyfxy fyy

),

evaluated at (a, b).

We may argue that near (a, b) the sign of f(x, y) − f(a, b) is determined by the sign of theleading term in the Taylor expansion i.e. by XTHX, if this term is not identically zero. A2 × 2 real symmetric matrix H is said to be positive definite, negative definite dependingon whether XTHX > 0 (X 6= 0), XTHX < 0 (X 6= 0), respectively. We conclude thatif H is positive definite then f(x, y) − f(a, b) > 0 for all (x, y) inside some circle centre(a, b), (x, y) 6= (a, b), and therefore f has a (local) minimum at (a, b). On the other hand,if H is negative definite then f(x, y) − f(a, b) < 0 for all (x, y) inside some circle centre(a, b), (x, y) 6= (a, b), and in this case f has a (local) maximum at (a, b). We learn from bookson algebra that H is positive definite provided its determinant is positive and the two entrieson the leading diagonal are positive; likewise H is negative definite provided its determinantis positive and the two entries on the leading diagonal are negative. We therefore concludethat

Case 1 f has a local minimum at the stationary point (a, b), if

fxx(a, b)fyy(a, b)− (fxy(a, b))2 > 0 , fxx(a, b) > 0 , and

fyy(a, b) > 0 .

Case 2 f has a local maximum at the stationary point (a, b), if

fxx(a, b)fyy(a, b)− (fxy(a, b))2 > 0 , fxx(a, b) < 0 , and

fyy(a, b) < 0 .


Appendix C: Basic Linear Algebra 195

Appendix C

Basic Linear Algebra

C.1 Fundamental ideas

We consider the space IR2 of column vectors

X =

(x1

x2

), x1, x2 ∈ IR.

With the usual rule of addition of vectors and multiplication of vectors by a real scalar IR2

becomes a linear vector space. It is assumed that the geometrical interpretation of suchvectors is well known. Any two linearly independent vectors constitute a basis and any vectorcan be written as a linear combination of basis vectors. Geometrically, any two (non–zero)vectors, which are neither parallel nor anti–parallel, are linearly independent, and thereforeconstitute a basis. We use e1, e2 to denote the standard basis so that

e1 =

(10

), e2 =

(01

),

A linear transformation a of the vector space IR2 is a map

a : IR2 → IR2

such thata(X + Y ) = aX + aY , a(λX) = λaX,

for all vectors X, Y ∈ IR2 and for every scalar λ ∈ IR. The action of any linear transformationa of IR2 is determined by its action on any set of basis vectors. We note that ae1, ae2 may bewritten as a linear combination of e1, e2 (because e1, e2 constitute a basis). We may write

aei =2∑

k=1

akiek, i = 1, 2 (C.1)

for some unique coefficients aik. These coefficients determine the matrix A = (aik) of thetransformation a with respect to the standard basis given by e1, e2. We note that

Ae1 =

(a11 a12

a21 a22

) (10

)=

(a11

a21

)= a11e1 + a21e2,



Ae2 =

(a11 a12

a21 a22

) (01

)=

(a12

a22

)= a12e1 + a22e2 .

Reference to equation C.1 shows that Aei = aei, (i = 1, 2) so that we may calculate the actionof the transformation a on any vector X simply by calculating AX, where A is the matrixof the transformation with respect to the standard basis as defined by equations C.1. Weconsider below the matrix of the linear transformation a with respect to some non-standardbasis, and show how it relates to the matrix of a with respect to the standard basis. Supposethat f1, f2 is another basis. We can therefore express f1, f2 as linear combinations of e1, e2 :

f i =2∑

k=1

pkiek, i = 1, 2 (C.2)

The coefficients pki define a matrix P = (pki).

Because f1, f2 is a basis we can express e1, e2 as linear combinations of f1, f2. Let’s write

en =2∑

m=1

qmnfm, n = 1, 2, (C.3)

for some unique coefficients qmn which define a matrix Q = (qmn).

We note that

en =2∑

m=1

qmnfm =∑

qmn∑

prmer =∑

(PQ)rner.

(All the sums run from 1 to 2)

Equating coefficients we see that

(PQ)rn = δrn, δrn = 1 (r = n), δrn = 0 (r 6= n).

Hence PQ = I, where I denotes the 2× 2 unit matrix. It follows that Q = P−1, as expected.

Just as we’ve considered the matrix A of the linear transformation a with respect to the basise1, e2 (see equation C.1) we may consider its matrix A′ = (a′ki) (say) with respect to the basisf1, f2. Referring to equation C.1 this is defined by

af i = Af i =

2∑k=1

a′kifk i = 1, 2 (C.4)

Let’s calculate the matrix A′.

We have

Af i = A

2∑k=1

pkiek =∑

pkiAek =∑

pki∑

anken =

2∑k,n=1

ankpki

2∑m=1

qmnfm



=

2∑k,n,m=1

qmnankpkifm =

2∑m=1

(QAP )mifm

We therefore see that the matrix A′ of the linear transformation a with respect to the basisf1, f2 is given by

A′ = QAP = P−1AP.

This is a simple, but fundamental result. We note in passing that the arguments are identicalin IRn.

Given a matrix A one aims to choose a basis f1, f2 which makes A′ = P−1AP as simple aspossible. Before obtaining the Jordan canonical forms for 2×2 real matrices A we prove thefollowing simple result.

C.2 Invariance of Eigenvalues Under Similarity Transformations

Let A be any n× n matrix, and let P be any non-singular n× n matrix. Then the matricesA and P−1AP have the same eigenvalues.

Proof

We note that

det(λI − P−1AP ) = det(P−1(λI −A)P ) = detP−1 det(λI −A) detP

= det(P−1P ) det(λI −A) = det I det(λI −A) = det(λI −A).

It follows thatdet(λI − P−1AP ) = 0 ⇐⇒ det(λI −A) = 0

and therefore the matrices P−1AP and A have the same eigenvalues.

C.3 Jordan Forms

Now let A be any real 2 × 2 matrix, and regard A as the matrix of a linear transformationa of IR2. Then A is similar to one of only three fundamentally different forms of matrix— the so-called Jordan normal form corresponding to A. Which of the three Jordan formscorresponds to a given matrix A is decided by the eigenvalues of A.

Thus to obtain the Jordan forms we start by calculating the eigenvalues of A. Three differentcases may occur.

(a) The eigenvalues λ1, λ2 of A are real and distinct, λ1 > λ2.

Let f1, f2 be the eigenvectors corresponding to λ1, λ2 respectively, so that

Af1 = λ1f1, Af2 = λ2f2.



Using the same notation as employed above we see that the matrix A′ of the transforma-tion a with respect to the basis f1, f2 (it is a basis because eigenvectors correspondingto distinct real eigenvalues are linearly independent – proved in CM113A) is given by

A′ =

(λ1 00 λ2

),

and therefore

P−1AP =

(λ1 00 λ2

), where P = (f1 : f2) . (C.5)

(P is expressed in partitioned form)

(b) The eigenvalues λ1, λ2 are real and coincident: λ1 = λ2 = λ, say.

There are two possibilities. (i) First let’s suppose that A has two linearly independenteigenvectors f1, f2 so that

Af1 = λf1, Af2 = λf2,

and f1, f2 form a basis.

With respect to the basis f1, f2 a has matrix

A′ =

(λ 00 λ

)= λI, P−1AP = λI, P = (f1 : f2) .

In this case A = λI, so the matrix was of Jordan canonical form (diagonal) to beginwith.

(ii) The second possibility is that A has only one independent eigenvector, f1 (say).We extend this to a basis of IR2 by picking another vector f2 (f2 isn’t an eigenvectorof A.) We can write Af2 = µf1 + νf2, for some real µ, ν.

We haveAf1 = λf1, Af2 = µf1 + νf2

so, with respect to the basis f1, f2, a has matrix

A′ = P−1AP =

(λ µ0 ν

).

Imposing the condition that A and P−1AP have the same eigenvalues (see Section C.2above) we immediately obtain ν = λ.

Writing f1 = 1µF 1 we see that

M−1AM =

(λ 10 λ

), M = (F 1 : f2) . (C.6)



(c) The remaining possibility is that A has complex eigenvalues λ = α+ iβ, α− iβ.

Suppose that the eigenvalue equation has been solved in the form

A(f1 + if2) = (α+ iβ)(f1 + if2).

Equating real and imaginary parts we then have

Af1 = αf1 − βf2, Af2 = βf1 + αf2.

One can prove without too much difficulty that f1, f2 are linearly independent andtherefore constitute a basis of IR2.

With respect to this basis a has matrix A′ where

A′ = P−1AP =

(α β−β α

), (C.7)

and P = (f1 : f2) .

We conclude by showing how the components of a vector with respect to the bases (f1, f2)and (e1, e2) are related.

C.4 Basis Transformation

Suppose that (f1, f2) and (e1, e2) are bases related by equations C.2 and let y1, y2, x1, x2,denote the components of some vector with respect to the bases (f1, f2), (e1, e2), respec-tively.

Writing

X =

(x1

x2

)Y =

(y1

y2

),

we have2∑i=1

xiei =

2∑i=1

yif i =

2∑i,j=1

yipjiej

and, equating coefficients, we obtain∑2

i=1 pkiyi = xk; or, in column vector notation,

X = PY , Y = P−1X .

We frequently use such a linear change of variable in our discussions of the stability of fixedpoints.



C.5 Rotations

Consider a set of Cartesian axes xOy, with origin O; the vectors e1, e1 may be regarded asunit vectors parallel to the x and y axes respectively. We aim to calculate the matrix A(θ)(with respect to the standard basis (e1, e2)) of the transformation which rotates a vectorthrough an angle θ about the origin O. Let r denote the position vector of the point P withcoordinates (x1, x2) so that r = OP = x1e1 + x2e2. Suppose that under the rotation r = OPmaps to r′ = OP ′, where P ′ has coordinates (x′1, x

′2). Regarding the Cartesian plane as a

copy of the Argand diagram we may write

x′1 + ix′2 = eiθ(x1 + ix2) = (cos θ + i sin θ)(x1 + ix2).

Equating real and imaginary parts we obtain

x′1 = x1 cos θ − x2 sin θ, x′2 = x1 sin θ + x2 cos θ,

or, in matrix notation, r′ = A(θ)r, where

A(θ) =

(cos θ − sin θsin θ cos θ

).

Note: Reference to equation C.7 shows that A(θ) has complex eigenvalues cos θ ± i sin θ, afact which may be confirmed by direct calculation.

C.6 Area Preserving Transformations

In this section we determine the class of linear transformations of the Cartesian plane whichpreserve areas. We employ the notation of section C.5. First we note that the origin O is leftinvariant by any linear transformation of the Cartesian plane. We also note that any non-singular linear transformation of the Cartesian plane maps parallel lines into parallel lines.To see this consider the straight line which passes through the point with position vector r0

and whose direction is specified by the vector n; this line has equation

r = r0 + λn, (C.8)

where λ is a real parameter. Let a be a non-singular linear transformation of the Cartesianplane whose matrix with respect to the standard basis is A. The image of the straight linedescribed by equation C.8 is given by

Ar = Ar0 + λ(An), which has the form

r′ = r0′ + λn′. (C.9)

We conclude that all straight lines parallel to the vector n map to straight lines which areparallel to n′ = An;An is not the zero vector because A is non-singular. It is clear thereforethat non-singular transformations of the Cartesian plane map parallelograms to parallelo-grams. Let P,R be points with coordinates (x1, y1), (x2, y2) respectively. The positionvectors OP, OR of P,R are given by

OP = x1e1 + y1e2, OR = x2e1 + y2e2.



Consider the parallelogram OPQR. Under the action of a linear transformation a of theCartesian plane this parallelogram maps to an image parallelogram OP ′Q′R′, where P ′, R′

have coordinates (x1′, y1

′), (x2′, y2

′) respectively, say. If A is the matrix of the linear trans-formation a, with respect to the standard basis, where

A =

(α βγ δ

), (C.10)

then we have the relations(x1′

y1′

)=

(α βγ δ

) (x1

y1

),

(x2′

y2′

)=

(α βγ δ

) (x2

y2

). (C.11)

An elementary calculation shows that the area of parallelogram OPQR is

|OP×OR| = |x1y2 − x2y1|; similarly the area of the image parallelogram O′P ′Q′R′ is

|x1′y2′ − x2

′y1′|.

It now follows from equation C.11 that

x1′y2′ − x2

′y1′ = v(αx1 + βy1)(γx2 + δy2)− (γx1 + δy1)(αx2 + βy2)

and after a little simplification we find that

x1′y2′ − x2

′y1′ = (αδ − βγ)(x1y2 − x2y1) = detA(x1y2 − x2y1).

We conclude that a parallelogram and its image under the linear transformation a have thesame area provided | detA| = 1. Of particular interest are those transformations for whichdetA = 1; of course, det I = 1, and the transformations with detA = 1 are those which can bereached continuously from the identity. We shall see later that area preserving transformationsare significant in Hamiltonian mechanics.

Note: Under the change of variable (x1, y1) 7→ (x1′, y1

′) described by equation C.11 areaelements transform according to the rule

dx1′ dy1

′ = |J |dx1 dy1, where J =∂(x1

′, y1′)

∂(x1, y1).

Referring to equation C.11 and carrying out the differentiations we soon find that J = detA;as found above, the condition for areas to be preserved is therefore |detA| = 1.

C.7 Examples and Exercises

Example C.1. For each of the matrices Ai (1 = 1, 2, 3) given by

A1 =

(1 21 1

), A2 =

(2 1−2 4

), A3 =

(3 −11 1

)find the appropriate Jordan form Ji (i = 1, 2, 3) and a matrix Pi (i = 1, 2, 3) such that Pi

−1AiPi =Ji, (i = 1, 2, 3).



Consider the third of these matrices, A3. A vector has components x1, x2 with respect to the standardbasis e1, e2 and components y1, y2 with respect to the basis f1,f2 in which the linear transformationcorresponding to A3 has the Jordan matrix J3. Write down the linear equations which connect x1, x2and y1, y2.

The eigenvalues of A1 are given by

det(λI −A1) = 0, so that λ = 1±√

2.

Solving the eigenvalue equation (1 21 1

)(xy

)= λ

(xy

),

we find that x = (λ− 1)y and we may take as eigenvectors(2√2

), λ = 1 +

√2 ,

(2

−√

2

), λ = 1−

√2 .

It follows that

P1−1A1P1 =

(1 +√

2 0

0 1−√

2

), where P1 =

(2 2√2 −

√2

).

Similarly, the eigenvalues of A2 are given by

det(λI −A2) = 0, λ = 3± i.

The complex eigenvector of A2 corresponding to the eigenvalue 3 + i is given by(2 1−2 4

) (xy

)= (3 + i)

(xy

).

We find that2x+ y = (3 + i)x, y = (1 + i)x.

Taking x = 1 we obtain y = 1 + i and the corresponding complex eigenvector is

f1 + if2 =

(1

1 + i

)so that

f1 =

(11

), f2 =

(01

).

Clearly, f1,f2 are linearly independent, and since

A2f1 = 3f1 − f2 , A2f2 = f1 + 3f2

we conclude that

P2−1A2P2 =

(3 1−1 3

), P2 = (f1 : f2) =

(1 01 1

).

Finally, A3 has eigenvalues given by

det(λI −A3) = 0, λ = 2 (twice) .



To obtain an eigenvector consider the equation(3 −11 1

) (xy

)= 2

(xy

).

It follows that3x− y = 2x, x = y,

so that we may take as an eigenvector

f1 =

(11

).

Extend this to a basis of IR2 by the vector

f2 =

(10

).

Now

Af2 =

(31

)= αf1 + βf2 ,

from which we easily find that α = 1, β = 2.

Hence, A3f1 = 2f1, A3f2 = f1 + 2f2 and choosing f1,f2 as a basis

P3−1A3P3 =

(2 10 2

), P3 = (f1 : f2) =

(1 11 0

).

The connection between the components x1, x2 of a vector with respect to the standard basis and itscomponents y1, y2 with respect to the basis f1,f2 is given by(

x1x2

)=

(1 11 0

)(y1y2

).

Exercise C.1 Find the image of the rectangle OBCD with (Cartesian) coordinates

(0, 0), (1, 0), (1, 2), (0, 2) under the linear transformation of the Cartesian plane whose matrixwith respect to the standard basis is (

1 11 2

).

Sketch your result; it should look like a parallelogram!

Exercise C.2 Describe the transformation of the Cartesian plane whose matrix with respectto the standard basis is

1

2

(√3 −1

1√

3

).

Write down the matrix of the linear transformation which describes a rotation about Othrough an angle of π/4.

Exercise C.3 Show that the rotation matrix

A(θ) =

(cos θ − sin θsin θ cos θ

),



is orthogonal i.e. that A(θ)TA(θ) = I.

Show, by direct calculation, that A(θ1)A(θ2) = A(θ1+θ2).What is the geometrical significanceof this result? Write down, without calculation the matrix A(θ)n, where n is a positive integer.

Verify that A(θ) has eigenvalues cos θ ± i sin θ.

Exercise C.4 Find the Jordan forms appropriate to the matrices(1 11 2

),

(2 1−4 4

),

(3 1−1 1

).

Exercise C.5 Show that the set G given by

G =

A =

(a bc d

), a, b, c, d ∈ IR, detA 6= 0

forms a group under matrix multiplication. Prove also that the set H ⊆ G which consists ofthose matrices which have detA = 1 is a normal sub-group of G.

Exercise C.6 Recall that a 2× 2 real matrix is orthogonal if ATA = I. Show that any suchmatrix represents an area preserving transformation. As noted above, the rotation matricesA(θ) are orthogonal.

Exercise C.7 Suppose that a linear transformation a of the Cartesian plane has matrix Awith respect to the standard basis e1, e2. Suppose that under the transformation a OP 7→OP′, where P has coordinates (x1, x2), P ′ has coordinates (x1

′, x2′), and that a leaves lengths

invariant i.e.x1

2 + x22 = x1

′2 + x2′2,

for every choice of (x1, x2). Prove that A is orthogonal i.e. that ATA = I. Deduce thatdetA = ±1.


Appendix D: Jacobians 205

Appendix D

Jacobians and Change of variables

D.1 Change of variables

Suppose we have a second order autonomous dynamical system for two variables x(t) andy(t) which has a fixed point at (x∗, y∗). If the DS is

dx

dt= f(x, y) ,

dy

dt= g(x, y) , (D.1)

then we havef(x∗, y∗) = (x∗, y∗) = 0 . (D.2)

We know that the nature of the fixed point is determined by the normal form of the Jacobianmatrix J evaluated at the fixed point,

J =

(∂f∂x

∗ ∂f∂y

∗

∂g∂x

∗ ∂g∂y

∗

), (D.3)

where ∂f∂x

∗is the partial derivative evaluated at the fixed point.

If we change from the variables (x, y) to new variables (u, v) then the nature of the fixed pointshouldn’t change. If the DS in the new variables is

du

dt= F (u, v) ,

dv

dt= G(u, v) , (D.4)

then we know that the nature of the fixed point is determined by the Jacobian at the fixedpoint in these coordinates,

J ′ =

(∂F∂u

∗ ∂F∂v

∗

∂G∂u

∗ ∂G∂v

∗

). (D.5)

We now see how to relate J ′ to J .

The first step is to find F and G. We can find these from the chain rule for partial derivatives

du

dt=

∂u

∂x

dx

dt+∂u

∂y

dy

dt=∂u

∂xf +

∂u

∂yg ≡ F (u, v) ,

dv

dt=

∂v

∂x

dx

dt+∂v

∂y

dy

dt=∂v

∂xf +

∂v

∂yg ≡ G(u, v) . (D.6)

We now need to calculate the matrix elements of J ′, for example

∂F

∂u=

∂

∂u

(∂u

∂xf +

∂u

∂yg

)Version of Mar 14, 2019

Appendix D: Jacobians 206

=∂

∂u

(∂u

∂x

)f +

∂u

∂x

∂f

∂u+

∂

∂u

(∂u

∂y

)g +

∂u

∂y

∂g

∂u

=∂

∂u

(∂u

∂x

)f +

∂u

∂x

(∂f

∂x

∂x

∂u+∂f

∂y

∂y

∂u

)+

∂

∂u

(∂u

∂y

)g +

∂u

∂y

(∂g

∂x

∂x

∂u+∂g

∂y

∂y

∂u

)(D.7)

At the fixed point, f(x∗, y∗) = g(x∗, y∗) = 0 and so

∂F

∂u

∗=

∂u

∂x

∂f

∂x

∗∂x

∂u+∂u

∂x

∂f

∂y

∗ ∂y

∂u+∂u

∂y

∂g

∂x

∗∂x

∂u+∂u

∂y

∂g

∂y

∗ ∂y

∂u(D.8)

This is the top left entry of the matrix(∂F∂u

∗ ·· ·

)=

(∂u∂x

∂u∂y

· ·

)(∂f∂x

∗ ∂f∂y

∗

∂g∂x

∗ ∂g∂y

∗

)(∂x∂u ·∂y∂u ·

)(D.9)

Calculating the top right entry, we get

∂F

∂v=

∂

∂v

(∂u

∂xf +

∂u

∂yg

)=

∂

∂v

(∂u

∂x

)f +

∂u

∂x

∂f

∂v+

∂

∂v

(∂u

∂y

)g +

∂u

∂y

∂g

∂v

=∂

∂v

(∂u

∂x

)f +

∂u

∂x

(∂f

∂x

∂x

∂v+∂f

∂y

∂y

∂v

)+

∂

∂v

(∂u

∂y

)g +

∂u

∂y

(∂g

∂x

∂x

∂v+∂g

∂y

∂y

∂v

)⇒ ∂F

∂v

∗=

∂u

∂x

∂f

∂x

∗∂x

∂v+∂u

∂x

∂f

∂y

∗∂y

∂v+∂u

∂y

∂g

∂x

∗∂x

∂v+∂u

∂y

∂g

∂y

∗∂y

∂v, (D.10)

and so on, so that all together we get

J ′ =

(∂F∂u

∗ ∂F∂v

∗

∂G∂u

∗ ∂G∂v

∗

)=

(∂u∂x

∂u∂y

∂v∂x

∂v∂y

)︸︷︷︸

M

(∂f∂x

∗ ∂f∂y

∗

∂g∂x

∗ ∂g∂y

∗

)︸︷︷︸

J

(∂x∂u

∂x∂v

∂y∂u

∂y∂v

)︸︷︷︸

N

(D.11)

However, the matrices M and N are inverses, MN = 1, as we can check

MN =

(∂u∂x

∂u∂y

∂v∂x

∂v∂y

)(∂x∂u

∂x∂v

∂y∂u

∂y∂v

)

=

(∂u∂x

∂x∂u + ∂u

∂y∂y∂u

∂u∂x

∂x∂v + ∂u

∂y∂y∂v

∂v∂x

∂x∂u + ∂v

∂y∂y∂u

∂v∂x

∂x∂v + ∂v

∂y∂y∂v

)

=

(∂u∂u

∂u∂v

∂v∂u

∂v∂v

)=

(1 00 1

)(D.12)

This means that J ′ and J are similar,

J ′ = M J M−1 , (D.13)

and hence have the same normal form. This means that the nature of the fixed point isunchanged when we change coordinates, as of course should be the case.


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