Introduction to Dynamical Systems
CM131A
R. Kühn
Based on Lecture Notes by H. C. Rae
Department of Mathematics King’s College London
January 2005
2
Contents
I
Introduction and Overview
5
General Dynamics
9
1 Introduction
11
2 Differential Equations
2.1 Basic Ideas – First Order Equations . . . . . . . . .
2.2 Picard’s theorem . . . . . . . . . . . . . . . . . . . .
2.2.1 Picard iterates . . . . . . . . . . . . . . . . .
2.3 Second order differential equations . . . . . . . . . .
2.4 General solution of specific equations . . . . . . . . .
2.4.1 First order, explicit . . . . . . . . . . . . . . .
2.4.2 First order, variables separable . . . . . . . .
2.4.3 First order, linear . . . . . . . . . . . . . . .
2.4.4 First order, homogeneous . . . . . . . . . . .
2.4.5 Second order linear, with constant coefficients
3 First Order Autonomous Systems
3.1 Phase Space and Phase Portraits . . . . . . .
3.2 Estimating Times of Motion . . . . . . . . . .
3.3 Stability — A More General Discussion . . .
3.3.1 Stability of Fixed Points . . . . . . . .
3.3.2 Structural Stability . . . . . . . . . . .
3.3.3 Stability of Motion . . . . . . . . . . .
3.4 Asymptotic Analysis . . . . . . . . . . . . . .
3.4.1 Elements of Asymptotic Analysis . . .
3.4.2 Asymptotic Analysis and Approximate
4 Second Order Autonomous Systems
4.1 Phase Space and Phase Portraits . .
4.1.1 Phase Curves . . . . . . . . .
4.2 Separable systems . . . . . . . . . .
4.3 Stability of fixed points . . . . . . .
4.3.1 Linear Stability Analysis . . .
.
.
.
.
.
3
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
15
15
16
16
19
22
22
22
23
24
26
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
Solutions of ODEs
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
29
29
30
33
33
35
38
39
39
40
.
.
.
.
.
45
45
48
49
50
50
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
4
CONTENTS
4.4
II
4.3.2 Transformation to Jordan Canonical Forms . . . . . . . . . . . . . . . . .
Limit cycles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Application to Classical Mechanics
51
58
63
5 Elements of Newtonian Mechanics
5.1 Motion of a particle . . . . . . . . . . . . . . .
5.2 Newton’s Laws of motion . . . . . . . . . . . .
5.2.1 Newton’s First Law (N1) . . . . . . . .
5.2.2 Newton’s Second Law (N2) . . . . . . .
5.2.3 Newton’s Third Law (N3) . . . . . . . .
5.3 Newton’s Law of Gravitation . . . . . . . . . .
5.4 Motion in a Straight Line; the Energy Equation
5.5 Equilibrium and Stability . . . . . . . . . . . .
.
.
.
.
.
.
.
.
65
65
69
70
70
70
72
75
78
6 Hamiltonian Systems
6.1 Hamilton’s equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Stability problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
89
89
95
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
7 Lagrangian Theory
7.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2 Lagrange’s equation . . . . . . . . . . . . . . . . . . . . . .
7.2.1 Equivalence with Newton’s Equations of Motion . .
7.2.2 Invariance under Coordinate Transformations . . . .
7.2.3 Exploiting the Invariance of Lagrange’s Equations .
7.3 Further Developments of the Theory . . . . . . . . . . . . .
7.4 Equivalence of Lagrangian and Hamiltonian Mechanics . . .
7.4.1 Lagrange’s Equations Imply Hamilton’s Equations of
7.4.2 Hamilton’s Equations Imply Lagrange’s Equations of
7.5 Worked examples . . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
Motion
Motion
. . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
101
101
102
103
104
105
106
107
108
109
111
A Taylor’s Theorem
123
A.1 Taylor Expansion for Functions of One Variable . . . . . . . . . . . . . . . . . . . 123
A.2 Taylor Expansion for Functions of two variables . . . . . . . . . . . . . . . . . . . 124
A.3 Vector functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
B Basic Linear Algebra
B.1 Fundamental ideas . . . . . . . . . . . . .
B.2 Invariance of Eigenvalues Under Similarity
B.3 Jordan Forms . . . . . . . . . . . . . . .
B.4 Basis Transformation . . . . . . . . . . . .
B.5 Rotations . . . . . . . . . . . . . . . . . .
B.6 Area Preserving Transformations . . . . .
B.7 Examples and Exercises . . . . . . . . . .
. . . . . . . . . .
Transformations
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
129
129
131
131
133
134
134
135
Introduction and Overview
The present course is about dynamical systems i.e. systems described by a set of n variables
{x1 , . . . , xn }, each depending on time t, governed by dynamical laws given in terms of ODEs of
the form
d
xi = fi (t, x1 , . . . , xn )
i = 1, . . . , n .
(1)
dt
We shall learn about methods to analyse such systems. Available methods can generally be
classified as (i) qualitative, (ii) analytical, and (iii) numerical. We shall mainly treat the
qualitative and analytical methods. The outline of the course is as follows.
Part I: General Dynamics
• Qualitative methods: flow diagrams, direction fields, analysis of fixed points and their
stability
• Methods for solving (some standard) ODEs
• Approximate ODEs and their solutions (fixed points, asymptotic analysis)
• Methods for systems of order 1 (1-D autonomous)
• Methods for systems of order 2 (1-D non-autonomous, 2-D autonomous)
• Systems with and without conservation laws.
• Asymptotic stability
Part II: Applications to Physical Systems; Classical Mechanics
• Newton’s laws
• Gravitation
• 1-D mechanical systems as 2nd order dynamical systems
• Conservative Systems, Hamiltonian Systems
• Elements of Lagrangian Mechanics
We illustrate some of the methods we shall be learning about, and their potential use, with an
example.
5
6
CONTENTS
Example 0.1 Population Dynamics and Harvesting Strategies
Population of a fish-species without fishing described by (logistic) growth equation
d
x = f0 (x) = r0 x(1 − x/c) .
dt
(2)
Modelling aspects: r0 is the difference between death- and birth-rates. The (1 − x/c) contribution takes
into account that food, space and other resources are not in unlimited supply, so unlimited exponential
growth at rate r0 would be unsustainable in the long run.
Qualitative analysis: Inspect the velocity function f0 (x). Its sign will determine whether ẋ < 0 (for
x > c) or ẋ > 0 (for x < c). Can be done graphically by drawing f0 (x) and an arrow representation
indicating the ’direction of the flow’.
f (x)
0
0
c
x
Figure 1: Velocity function f0 (x) and arrow representation of the flow.
The parameter c is the capacity of the eco-system; it is a stable fixed point of dynamics: x∗ = c. All
solutions are attracted by it (from above if x(t0 ) > c, from below if x(t0 ) < c
With fishing at rate r the dynamics is modified to
d
x = fr (x) = r0 x(1 − x/c) − rx .
dt
(3)
This system is analysed in the same way as the original one without fishing. One has x∗ (r) = c(1 − r/r0 )
as new stable FP.
Assume that revenues from fishing created at rate Ṙ(r, t) = a r x(t) − b, in which the first term represents
the income generated by selling fish, while the second term represents the running costs of the fishing
business. At stationarity, i.e. at the FP one has
Ṙ(r) = a r x∗ − b .
This has a unique maximum — representing the optimal sustainable fishing strategy — at ropt = r0 /2,
with Ṙ(ropt ) = ar0 c/4 − b. Whether sustainable fishing will be profitable, Ṙ(ropt ) > 0, then depends on
the parameters a, r0 , c, and b of the system.
Further details about the system may be learnt of course by actually solving the ODE. One finds that
the solution of (3)is of the form
x(t) = x0
c (1 − r/r0 ) e(r0 −r)t
.
c (1 − r/r0 ) − x0 + x0 e(r0 −r)t
(4)
With the full solution in hand, one can ask more refined questions, such as comparing optimal sustainable
fishing strategies with strategies that create revenues in the short run, but are not sustainable. To
7
CONTENTS
investigate these, one has to look at the total revenues created up to some finite time t, i.e. at R(a, b, t) =
Rt
dt Ṙ(a, b, t). The integral can be done, and the result is displayed in the following figure.
0
R(a,b,t)
4
3
2
1
0
-1
-2
20
15
0
10
0.5
1
r/r0
5
1.5
r0 t
2 0
Figure 2: Total revenues created up to time t as a function of r/r0 , for a system with c = 1,
a = 1 and b = 0.1. The figure clearly shows that in the short run, fishing rates larger than r0
are more profitable than the optimal sustainable rate, but they soon generate losses (negative
total revenues)
8
CONTENTS
Part I
General Dynamics
9
Chapter 1
Introduction
Definition 1.1 Dynamical System (DS)
A DS is any system described by a set of variables {x1 , . . . , xn } — n taken to be finite throughout
the course —- each depending on time t, in such a way that the laws governing the change are
given in terms of ordinary differential equations (ODEs) of the form
d
xi = fi (t, x1 , . . . , xn )
, i = 1, . . . , n .
(1.1)
dt
We shall refer to the functions fi as velocity functions, as they describe the rate of change of
the xi
Note: No variables other than the xi (and possibly t) appear on the r.h.s of (1.1). In this
sense the description of a dynamical system in terms of the xi , and ODEs (or dynamical laws)
involving only these, is complete. The dynamical laws are formulated locally, which is the level
where we may have insight into the behaviour of a system and thus be able to model it. That
is, we are able to formulate the laws governing infinitesimal changes of the system (i.e. the
ODEs) but we are not necessarily able to see immediately what such laws would imply about
the system behaviour over extended time spans. Describing methods available to elucidate the
long-term behaviour of DSs is the main aim of this course
The description of the dynamics of natural or artificial systems in terms of DS à la (1.1) covers
a very wide spectrum of (interesting) possibilities, as we shall see in what follows. Not covered
are descriptions of the dynamics of continua (fluids, fields, . . . ).
Notation We shall use vector notation x = (x1 , . . . , xn )t , and f = (f1 , . . . , fn )t , to rewrite (1.1)
as
d
x = f (t, x)
(1.2)
dt
Definition 1.2 Order
If the description involves n variables, the dyn. system is said to be of order n.
Definition 1.3 Autonomous and Non-Autonomous Systems
If the time t does not explicitly appear in the velocity functions fi , the system is said to be
autonomous, otherwise it is non-autonomous. Thus for autonomous systems, we have
d
x = f (x)
dt
11
(1.3)
12
CHAPTER 1. INTRODUCTION
instead of (1.2). (Physically, an autonomous system is a system in which the laws governing it
are time-independent, or one where there are no time-dependent ‘external’ perturbations of the
dynamics). .
We shall in what follows mainly deal with autonomous DS. This is not a severe restriction as
a non-autonomous system of order n can be transformed into an autonomous system of order
n + 1 by introducing an additional variable x0 satisfying
d
x0 = 1 ,
dt
(1.4)
so that (1.1) becomes
d
xi = fi (x0 , x1 , . . . , xn )
dt
, i = 0, . . . , n .
(1.5)
with f0 (x0 , x1 , . . . , xn ) = 1.
Definition 1.4 Orbits, Phase-flow
The set of points {x(t); t ∈ IR} which solve the ODEs (1.2), and for which x(t0 ) = x0 is called
the orbit of the DS passing through x0 . The set of all orbits obtained by varying t0 and x0
through all physically allowed values is called the phase-flow of the DS.
Definition 1.5 Trajectories, Flow
The set of points {(t, x(t)); t ∈ IR} which solve the ODEs (1.2), and for which x(t0 ) = x0 is called
the trajectory or solution curve of the DS passing through x0 . The set of all trajectories
obtained by varying t0 and x0 through all physically allowed values is called the flow of the DS.
Note the difference between phase-flow and flow; the latter contains more dynamical information
than the former.
Definition 1.6 Fixed Point (FP)
A point a = (a1 , . . . an )t is called a fixed point of the (autonomous) DS if
fi (a) = 0
⇔
d
xi = 0
dt
at x = a
, i = 1, . . . , n .
A system which is at a FP will stay there forever, unless perturbed.
Definition 1.7 Stability of a FP
• A FP a of a DS is called strongly stable, if all trajectories starting (sufficiently) close to
a will approach a under the dynamics.
• A FP a of a DS is called unstable, if trajectories starting close to a exist which will evolve
away from a under the dynamics. (Note that in n-th order DS, mixed situations exist,
i.e. some subset of the trajectories starting sufficiently close to a will approach a, 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 from it, but
rather ‘keep circling around’.
13
Figure 1.1: Illustration of the possible flows in the vicinity of a FP: strongly stable, unstable,
unstable (mixed), marginally stable (from left to right).
We shall make these notions more precise later on. Fig. 1.1 illustrates the definitions.
Example 1.1 Population Dynamics
Population dynamics with unlimited resources (Malthus, 1798)
d
x = f0 (x) = Bx − Dx = r0 x .
dt
r0 is the difference between birth- and death-rates B and D. Solution is of the form
x(t) = x0 exp(r0 t) .
(i) exponential (unlimited) growth for r0 > 0 (B > D)
⇒ x = 0 is unstable fixed point.
(ii) exponential decay for r0 < 0 (B < D)
⇒ x = 0 is stable fixed point.
(iii) stationary situation for r0 = 0
⇒ x0 (arbitrary) is marginally stable
Example 1.2 Population Dynamics
Dynamics of interacting populations (predator-prey system) (Lotka-Volterra, 1925/26)
d
xi = fi (x) = (Bi − Di )xi
dt
, i = 1, 2 .
with (x2 population of predators preying on prey population x1 ), thus
D1
=
d1 x2
B1 = b1
B2
=
b2 x1
D2 = d2
This gives
d
x1
dt
d
x2
dt
= f1 (x) = (b1 − d1 x2 )x1
= f2 (x) = (b2 x1 − d2 )x2
Fixed points are a1 = (0, 0)t and a2 = (d2 /b2 , b1 /d1 )t . a1 is unstable, a2 is marginally stable. Leads
to oscillatory behaviour in predator and prey population, which could be exploited economically (e.g.
hare–lynx ⇒ price of lynx-fur high when number of lynx is low!).
14
CHAPTER 1. INTRODUCTION
In order to study DS, we obviously need to know about ODEs.
Your previous exposure to ODEs may suggest that they are usually solvable in terms of elementary function, analytically or by inspired guesswork. This is not the case. E.g. for the
Lotka-Volterra system, no closed form expression for the solution is known. But: one can still
learn a lot about such systems using qualitative methods (drawing diagrams of the velocity
functions, of their null-clines, of the the phase flow, searching for FPs, i.e., intersections of all
null-clines), or by analytic methods that fall short of finding exact solutions (investigating
stability properties of fixed points, solving approximate equations that hold close to fixed points
or for very small or very large xi etc.) We begin, however, with a reminder of elementary theory
for ODEs.
Chapter 2
Differential Equations
This chapter reviews a few elementary facts about ODEs. We (i) define basic notions using
ODEs of first order, (ii) briefly look into the issue of existence and uniqueness of solutions, (iii)
discuss how ODEs of higher order can be reduced to systems of first order, and (iv) review
methods to find general methods for a few important types of ODEs.
2.1
Basic Ideas – First Order Equations
We start by considering first order (ordinary) differential equations of the form
d
x = f (t, x) ,
dt
(2.1)
in which f is a suitably defined function of two real variables, and an initial condition is given
as
x(t0 ) = x0 .
(2.2)
Definition 2.1 Solution
We say that x = g is a solution of the differential equation (2.1) with initial condition (2.2) if
equation (2.1) is satisfied identically when we substitute x(t) = g(t), i.e.
d
g(t) ≡ f (t, g(t))
dt
and
g(t0 ) = x0 .
Thus x = 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 or solution curve of (2.1) which passes through the point (t0 , x0 ) determined by the initial condition
x(t0 ) = x0 . As we change the initial condition (i.e. the value of x0 , the prescribed value of x
at t = t0 , and possibly t0 ) we generate a family of solution curves of (2.1) also called the flow
corresponding to (2.1).
In elementary courses on differential equations much emphasis is placed on methods for finding
the general solution of (2.1).
15
16
CHAPTER 2. DIFFERENTIAL EQUATIONS
Definition 2.2 General Solution
We say that x = h(·, C) is a general solution of Eq. (2.1), if
d
h(t, C) ≡ f (t, h(t, C))
dt
whatever C. The value of C may be chosen so as to make x = h(·, C) satisfy any reasonable
initial condition x(t0 ) = x0 i.e. h(t0 , C) = x0 .
E.g.: the ODE
d
dt x
= ax has general solution x(t) = Ceat , with C = x(0).
Before reviewing methods for finding the general solution of a few simple types of ODE (only
for few types do general solutions in terms of elementary functions exist!), we briefly enquire
into the question of uniqueness of solutions.
Indeed, even if we have to recognise that usually we cannot find a general solution in terms of
the elementary functions of analysis, it is important to know whether a given equation possesses
a solution and, if so, whether the solution is unique. In the next section we state without proof
Picard’s theorem.
2.2
Picard’s theorem
Picard’s theorem states the condition under which the solution of (2.1), (2.2)
d
x = f (t, x)
dt
,
x(t0 ) = x0
is unique. In colloquial terms the condition is: the derivative f (t, x) must be sufficiently smooth
in both variables t and x in the neighbourhood of (t0 , x0 ). More precisely we have the
Theorem 2.1 (Picard)
Let A denote the set
{(t, x) : |t − t0 | ≤ δ, |x − x0 | ≤ δ}
(a square centre the initial point (t0 , x0 ) and side of length 2δ). Suppose that f : A → IR is
continuous and that fx = ∂f
∂x is also continuous on A. Then the differential equation (2.1) has
a unique solution (satisfying the initial condition x(t0 ) = x0 ) which is defined on the interval
[t0 − δ1 , t0 + δ1 ], for some δ1 ≤ δ.
This isn’t the best version of Picard’s Theorem, but it has been stated in a form which is likely
to make reasonable sense to first year students.
We will not give a full proof, but indicate its heuristics based on Picard iterates, a practical
procedure for constructing the solution.
2.2.1
Picard iterates
Assuming that eqs.(2.1), (2.2) have a solution we may integrate both sides with respect to t to
obtain
Z t
Z t
dx
x(t) = x0 +
du
= x0 +
du f (u, x(u))
(2.3)
du
t0
t0
17
2.2. PICARD’S THEOREM
Eq. (2.3) is fully equivalent to the pair (2.1), (2.2). Note that we cannot really evaluate the
r.h.s. as we do not know x(u) for u 6= t0 ; in other words Eq. (2.3) 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 .
• Then define a sequence of functions {xn (t)} by
xn+1 (t) = x0 +
Z
t
t0
du f (u, xn (u)) ,
n = 0, 1, 2, . . . .
(2.4)
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 then prove
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 (2.3) 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 = max |xn (s) − x(s)|
s∈[t0 ,t]
will be diminished under iteration. For, we have
Z s
du f (u, xn (u)) − f (u, x(u)) δn+1 = max s∈[t0 ,t]
t0
Z s
du f (u, xn (u)) − f (u, x(u)) ≤ max
s∈[t0 ,t] t0
Z s
= max
du fx (u, x(u)) (xn (u)) − x(u)) + o(xn (u)) − x(u)) s∈[t0 ,t]
≤
t0
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 continuity 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 (2.4), 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 2.1 Consider the differential equation
dx
=x,
dt
x(0) = 1.
Integrating both sides of the equation with respect to t we obtain
Z t
x(t) = 1 +
du x(u) .
0
18
CHAPTER 2. DIFFERENTIAL EQUATIONS
Following the method described above we now define the Picard iterates by
Z t
du xn (u) , x0 (t) = 1 .
xn+1 (t) = 1 +
0
Setting n = 0 we obtain
x1 (t) = 1 +
Substituting n = 1 we obtain
x2 (t) = 1 +
Z
Z
t
du x0 (u) = 1 + t .
0
t
du x1 (u) = 1 +
0
Z
t
du (1 + u) = 1 + t +
0
t2
.
2!
Continuing in this way, with n = 2, 3, · · · we find that
xn (t) = 1 + t +
We see that as n → ∞ xn (t) →
P∞
tr
r=0 r!
t3
tn
t2
+ + ··· + .
2! 3!
n!
= et , which is indeed the unique solution of the given equation.
Note 1: If the conditions of Picard’s theorem are not satisfied by a particular differential
equation, 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 initial
condition.
Example 2.2 Consider the differential equation
dx
= x1/2 ,
x(0) = 0 .
dt
It is clear that x(t) = 0 is a solution, since it satisfies the equation and the initial condition. A second
solution which satisfies the equation and the initial condition is x = t2 /4. Why is this? In the above
notation f (t, x) = x1/2 , fx = 21 x−1/2 , so that fx → ∞ as x → 0+; the second condition of Picard’s
theorem is not satisfied. In this example the uniqueness aspect of the theorem has been lost.
Note 2: Picard’s theorem is a local existence theorem; it says that Eqs. (2.1), (2.3) have a
unique solution sufficiently near t0 . It might be thought that if the function f which occurs in
Eq. (2.1) is well behaved and defined over the whole of IR2 that we could extend our solution
arbitrarily far from t0 ; for example, we might begin by computing x(t0 + δ1 ) and use this value
of x as an initial value at t = t0 + δ1 ; application of Picard’s theorem then allows us to extend
our solution as far as t0 + δ1 + δ2 , for some δ2 > 0 and by repeating the process we could 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 , as one might find that
tn = t0 +
n
X
k=1
δk −→ t∞ < ∞ ,
as n → ∞ .
For example, the equation
dx
= f (t, x) = x2 ,
dt
has the solution
x(t) =
x(t0 ) = x0 ,
1
(t0 +
1
x0 )
−t
,
and it is clear that x(t) → ∞ as t → t0 + 1/x0 ; the procedure which we have just described
would not allow us to extend the solution as far as t = t0 + 1/x0 .
19
2.3. SECOND ORDER DIFFERENTIAL EQUATIONS
2.3
Second order differential equations
Second order ordinary differential equations have the form
d2 x
dx
,
= g t, x,
dt2
dt
with initial conditions of the form
(2.5)
dx
(t0 ) = v0 .
dt
The initial conditions define a point (t0 , x0 , v0 ) ∈ IR3 ; it is assumed that g is defined in some
neighbourhood of this point, for example in the cube
x(t0 ) = x0 ,
A = {(t, x, v) : |t − t0 | ≤ δ, |x − x0 1| ≤ δ, |v − v0 | ≤ δ, for some δ > 0
(2.6)
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
d
x = x2 (t) in Eq (2.5). Then
is achieved by writing x(t) = x1 (t), dt
d
d
x1 =
x = x2 ,
dt
dt
d
d2
d
x2 =
x = g t, x, x = g(t, x1 , x2 ) .
dt
dt2
dt
In terms of x1 , x2 the initial conditions become x1 (t0 ) = x0 , x2 (t0 ) = v0 . We may therefore
replace the second order equation (2.5) by two coupled first order differential equations
dx1
= x2 , ,
dt
dx2
= g(t, x1 , x2 ),
dt
(2.7)
with initial conditions
x1 (t0 ) = x0 , x2 (t0 ) = v0 .
(2.8)
The use of the word coupled signifies that the equation for x1 depends on x2 and the equation
for x2 depends on x1 , as well as on x2 .
Note: This procedure can be generalized to n-th order ODEs of the form
dn x
= g t, x, x(1) , x(2) . . . , x(n−1)
dtn
and with initial conditions of the form
,
with x(k) ≡
dk x
,
dtk
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 first order
equations
d
x1 = x2
dt
d
x2 = x3
dt
..
..
.
.
d
xn = g(t, x1 , x2 , . . . , xn )
dt
20
CHAPTER 2. DIFFERENTIAL EQUATIONS
with initial coditions xk (t0 ) = ak , for k = 1, . . . , n.
Suppose that by some means we have found a solution of Eqs (2.7),
x1 (t) = φ(t), x2 (t) = ψ(t), satisfying the initial conditions, so that φ(t0 ) = a1 ,
ψ(t0 ) = a2 . If we plot the points (φ(t), ψ(t)) in IR2 we obtain a solution curve which passes
through the point (a1 , a2 ) defined by the initial conditions; as we vary (a1 , a2 ) we generate a
family of solution curves, each labelled by two parameters ( corresponding to (a1 , a2 )).
Eqs (2.7), (2.8) have the general form
dx1
dt
dx2
dt
= f1 (t, x1 , x2 ) ,
x1 (t0 ) = a1 ,
= f2 (t, x1 , x2 ) ,
x2 (t0 ) = a2 .
(2.9)
Picard’s theorem for this system of equations now takes the form:
Picard’s theorem Let A denote the cube
{(t, x, v) : |u − t0 | ≤ δ, |x − a1 | ≤ δ, |v − a2 | ≤ δ, for some δ > 0}
and suppose that f1 , f2 : A → IR are continuous. Suppose also that the partial derivatives w.r.t.
the second and third arguments, (fj )v , (fj )w , (j = 1, 2) are continuous on A. Then the system
of differential equations
dx2
= f2 (t, x1 , x2 ) ,
dt
dx1
= f1 (t, x1 , x2 ) ,
dt
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 as in
the case of first order equations, relying on convergence of an iterative solution of a system of
integral equations which together are equivalent to the system of ODEs of first order with and
their initial conditions. They are obtained by integration of Eqs (2.9) with respect to t, which
gives
Z
t
xi (t) = ai +
t0
du fi (u, x1 (u), x2 (u)) ,
(i = 1, 2)
(2.10)
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 ,
xi,n+1 (t) = ai +
(2.11)
Z
t
t0
du fi (u, x1,n (t), x2,n (t)) ,
(n = 0, 1, 2, 3, . . .)
(2.12)
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 is
sufficiently small.
Let’s see by an example how this works out in practice.
21
2.3. SECOND ORDER DIFFERENTIAL EQUATIONS
Example 2.3 Simple Harmonic Motion: The equation describing simple harmonic motion given by
d2 x
+x=0
dt2
with initial conditions
x(0) = 0 ,
As above put x1 (t) = x(t), x2 (t) =
dx
dt
dx
(0) = 1 .
dt
so that
dx1
= x2 ,
dt
dx2
= −x1 ,
dt
with x1 (0) = 0 and x2 (0) = 1. The corresponding integral equations for x1 and x2 are
Z t
Z t
x1 (t) =
du x2 (u) ,
x2 (t) = 1 −
du x1 (u) .
0
0
The Picard iterates are defined by
x1,n+1 (t)
=
Z
t
du x2,n (u) ,
x1,0 (t) = 0 ,
0
x2,n+1 (t)
= 1−
Z
t
du x1,n (u) ,
x2,0 (t) = 1 .
0
Setting n = 0, 1, 2, 3, . . . in turn we find that
Z t
Z t
du = t ,
du x2 (0, u) =
x1,1 (t) =
0
0
Z t
Z t
x2,1 (t) = 1 −
du x1 (0, u) = 1 −
du 0 = 1 ,
0
0
Z t
Z t
x1,2 (t) =
du x2,1 (u) =
du 1 = t ,
0
0
Z t
Z t
t2
x2,2 (t) = 1 −
du x1,1 (u) = 1 −
du u = 1 −
,
2!
0
0
Z t
Z t
t3
u2 =t−
x1,3 (t) =
,
du x2,2 (u) =
du 1 −
2!
3!
0
0
Z t
Z t
t2
,
x2,3 (t) = 1 −
du x1,2 (u) = 1 −
du u = 1 −
2!
0
0
Z t
Z t
u2 t3
x1,4 (t) =
du x2,3 (u) =
du 1 −
=t−
,
2!
3!
0
0
Z t
Z t
t2
t4
u3 =1− +
x2,4 (t) = 1 −
,
du x1,3 (u) = 1 −
du u −
3!
2! 4!
0
0
Z t
Z t
u4 t5
t3
u2
+
,
=t− +
x1,5 (t) =
du x2,4 (u) =
du 1 −
2!
4!
3! 5!
0
0
and so on. The pattern is clear, and we see the Maclaurin series for sin(t) appearing for x1,n (t) as n → ∞,
so
t3
t5
t7
t9
x1,n (t) → x(t) = t − + − + − · · · = sin(t) ,
3! 5! 7! 9!
which is the solution of the differential equation satisfying the given initial conditions.
22
CHAPTER 2. DIFFERENTIAL EQUATIONS
The so-called general solution of a second order differential equation of the form
d2 x
dx
= g t, x,
2
dt
dt
is a solution of the form x(t) = h(t, C1 , C2 ), involving two parameters C1 , C2 which can be
chosen so as to satisfy the two initial conditions x(t0 ) = x0 , dx
dt (t0 ) = v0 .
2.4
General solution of specific equations
We now summarise some well known methods for finding the general solutions of certain types
of first and second order differential equations. Recall that the general solution of a first order differential equation involves one parameter, whilst the general solution of a second order
differential equation will involve two parameters.
We proceed from simple to more complicated situations, trying to reduce the more complicated
situations to the simpler ones dealt with before.
2.4.1
First order, explicit
Explicit equations are those which can be solved by direct explicit integration; they are of the
form
dx
= f (t)
(2.13)
dt
so that the r.h.s. does not depend on the unknown function x; thus explicit first order equations
are not strictly ODEs.
Integrating w.r.t. time t one obtains
x=
Z
dt f (t) = F (t) + C ,
(2.14)
with F (t) + C denoting the indefinite integral of the function f . Thus one has
x = x(t) = F (t) + C
(2.15)
as the general solution of (2.14). The constant C, may be fixed by inserting an initial condition
into (2.15), giving x(t0 ) = x0 = F (t0 ) + C, hence C = x0 − F (t0 ).
2.4.2
First order, variables separable
Equations of the type
dx
= f (x) g(t)
dt
may be integrated formally by separation of variables:
dx
= dt g(t)
f (x)
⇒
Z
dx
=
f (x)
(2.16)
Z
dt g(t)
where C is a parameter. Denoting the indefinite integrals involved by fdx
(x) = F (x) + C, and
R
dt g(t) = G(t) + C and combining the two integration constants into one, one obtains
R
F (x) = G(t) + C ,
(2.17)
2.4. GENERAL SOLUTION OF SPECIFIC EQUATIONS
23
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 (2.16). The constant C, may be fixed by inserting
an initial condition into (2.17), giving C = F (x0 ) − G(t0 ).
Example 2.4 Find the general solution of the differential equation
dx
= αx ,
dt
where α is a constant. This differential equation will arise frequently in our later work. We have
dx
= dt α ,
x
⇒
ln |x| = αt + C
by separating variables and integrating, so that
|x| = eαt eC = Beαt ,
so
x(t) = Aeαt ,
where A = ±B = ±eC is an arbitrary parameter. It can be fixed by initial conditions as usual.
2.4.3
First order, linear
First order, linear equations are of the form
dx
+ f (t)x = g(t) ,
dt
(2.19)
with f and g given functions of t. Equations of the form (2.19) are solved using the method of
integrating factors. This is a method that reduces the problem to that of solving two simpler
equations, one of which can be 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
µ
dx
+ µ f x = µg ,
dt
which can be transformed into
d
dµ
(µx) + µf −
dt
dt
By choosing µ such that
x = µg .
dµ
=0;
dt
a choice that requires solving a separable first order ODE to actually determine µ, one finds that
the transformed ODE simplifies to
d
(µx) = µg .
dt
µf −
24
CHAPTER 2. DIFFERENTIAL EQUATIONS
This is a first order explicit equation for the product µx, once µ is known from the solution of
the separable ODE defining it. A solution of the ODE defining µ is
µ(t) = exp
Z
dt f (t)
.
(2.20)
Note that the indefinite integral in the exponent is only defined up to an arbitrary additive
constant. As we need to know µ only up to an arbitrary multiplicative constant, we can set
any additive constant in the exponential to zero (also we have implicitly used this freedom to
assume that µ is positive !).
With this choice of µ the explicit equation for the product µx is integrated to give
µ(t)x(t) =
Z
dt µ(t)g(t) .
Note again that the indefinite integral here comes with an arbitrary integration constant C as
a free parameter of the general solution. This can not be ignored but is used to satisfy initial
conditions as usual. The function µ is referred to as an integrating factor for Eq. (2.19).
Example 2.5 Consider the ODE
t(t + 1)
dx
− (t + 2)x = t3 (2t − 3)
dt
⇔
dx
(t + 2)
(2t − 3)
−
x = t2
dt
t(t + 1)
(t + 1)
It is a first order linear ODE. The functions f (t) and g(t) can be read of from the equation. The
integrating factor is therefore
Z
(t + 2)
µ(t) = exp − dt
t(t + 1)
Using
2(t + 1) − t
2
1
(t + 2)
=
= −
t(t + 1)
t(t + 1)
t
t+1
one obtains
µ(t) = exp (−2 ln t + ln(t + 1)) =
so
µx =
Hence
Z
dt µ(t) t2
(2t − 3)
=
(t + 1)
x(t) =
2.4.4
Z
t+1
,
t2
dt (2t − 3) = t2 − 3t + C
t2
(t2 − 3t + C)
t+1
First order, homogeneous
Consider the equation
P (t, x)
dx
=
,
dt
Q(t, x)
where P, Q are homogeneous functions of (the same) degree m in the sense that
P (t, λt) = tm P (1, λ),
Q(t, λt) = tm Q(1, λ),
(2.21)
2.4. GENERAL SOLUTION OF SPECIFIC EQUATIONS
25
for some m. The general solution of Eq. (2.21) may be found by making the substitution
x = tv(t) where v has to be determined. We have
t
dv
P (t, tv)
tm P (1, v)
P (1, v)
+v =
= m
=
.
dt
Q(t, tv
t Q(1, v)
Q(1, v)
Hence
t
P (1, v)
dv
=
− v ≡ R(v) .
dt
Q(1, v)
Separation of variables now gives
dv
= ln |t| + C,
R(v)
Z
where C is a parameter. After carrying out the v integration we can express v in terms of x, t
using v = x/t; hopefully we can solve the resulting equation for x in terms of t and the parameter
C, but often this is not the case.
Example 2.6 Consider the differential equation
dx
(x2 − t2 )
=
.
dt
2xt
Writing x = tv(t) we obtain
t
t2 (v 2 − 1)
(v 2 − 1)
dv
+v =
=
2
dt
2t v
2v
and so
t
dv
1 + v2
=−
.
dt
2v
Separation of variables now gives
Z
Z
2v
dt
dv = −
1 + v2
t
=⇒
ln(1 + v 2 ) = − ln |t| + C,
where C is a parameter. Exponentiation now yields
|t|(1 + v 2 ) = eC
⇔
t(1 + v 2 ) = A ,
where A = ±eC is a parameter. It follows that
t(1 + x2 /t2 ) = A
⇔
x2 + t2 = At .
This equation may be re-written
(t − A/2)2 + x2 = (A/2)2 ,
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). This does not contradict Picard’s theorem because the
function f given by
v 2 − u2
f (u, v) =
2uv
doesn’t satisfy the conditions of Picard’s theorem in any neighbourhood of (0, 0).
26
2.4.5
CHAPTER 2. DIFFERENTIAL EQUATIONS
Second order linear, with constant coefficients
These are ODEs of the form
d2 x
dx
+b
+ cx = φ(t) ,
2
dt
dt
where a, b, c are constants and φ is a given function of time.
a
Homogeneous equations:
mogeneous,
(2.22)
We will first consider the special case where the equation is hoa
d2 x
dx
+b
+ cx = 0 ,
2
dt
dt
(2.23)
i.e. where φ(t) ≡ 0.
Linear and homogeneous ODEs have the important properties that (i) a solution x(t) can be
multiplied with an arbitrary constant, and remains a solution; (ii) for any pair x1 and x2 of
independent solutions (which are not proportional to each other), any linear combination of the
solutions x1 and x2 with arbitrary constant coefficients is also a solution of (2.23). This fact will
be exploited below.
The homogeneous equation is solved with the help of solutions of the so-called auxiliary polynomial, or auxiliary equation. This method is based on the observation that Eq. (2.23) is solved
by functions of the form
x(t) = eλt
provided λ is properly chosen. The condition on λ is obtained by inserting the exponential
ansatz into (2.23). Using
dx
= λ eλt
dt
and
d2 x
= λ2 eλt
dt2
one gets
(aλ2 + bλ + c)eλt = 0
Since eλt 6= 0, we see that eλt is a solution of Eq. (2.23) provided
L(λ) = aλ2 + bλ + c = 0.
(2.24)
Eq. (2.24) is referred to as the auxiliary equation. It is a quadratic equation in λ and as such
there are three possibilities:
(i) Equ. (2.24) has two distinct real solutions λ1 , λ2 . It follows that x1 = eλ1 t and x2 = eλ2 t are
independent solutions of the given differential equation.
The general solution of Eq. (2.23) is now obtained by forming an arbitrary linear combination
of these two solutions: it is
x(t) = αeλ1 t + βeλ2 t ,
(2.25)
where α, β are parameters. One can also show that these combinations exhaust the set of
possibilities (i.e. there are no solutions of a form other than (2.25)).
(ii) The auxiliary equation has two complex conjugate roots λ± = µ ± iν. This means that
x± (t) = e(µ±iν)t are solutions of Eq. (2.23), as are all linear combinations thereof. Now
x± (t) = e(µ±iν)t = eµt (cos νt ± i sin νt).
27
2.4. GENERAL SOLUTION OF SPECIFIC EQUATIONS
Using special linear combinations of these, we conclude that
x1 (t) =
x+ (t) + x− (t)
= eµt cos νt
2
and
x+ (t) − x− (t)
= eµt sin νt
2i
x2 (t) =
are two (independent) solutions of Eq. (2.23). Forming linear combinations of these, we see that
the general solution of Eq. (2.23) is given by
x(t) = eµt (α cos νt + β sin νt)
(2.26)
in this case; here α, β are parameters.
(iii) The third possibility is that Eq. (2.24) has two coincident real roots, λ1,2 = λ, say. In this
case we get initially only a single solution
x1 (t) = α eλt
where α is an arbitrary constant. An independent solution is in this case obtained by the
method of varying constants. That is one tries to find an independent solution of the form
x2 (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 (2.23), which gives
h
i
aF̈ + (2aλ + b)Ḟ + (aλ2 + bλ + c)F eλt = 0
As the coefficient of F in this equation is L(λ) and thus vanishes, and the coefficient of Ḟ
vanishes because λ = −b/2a was assumed to be the unique solution of L(λ) = 0, we are left
with the condition
F̈ = 0
⇒
F (t) = α + β t .
The general solution of F̈ = 0 was obtained by two explicit integrations involving two integration
constants α and β. Summarising: the general solution of (2.23) in this case is again an arbitrary
linear combination of the two independent solutions just found, and can be written as
x(t) = (α + βt) eλt
(2.27)
with arbitrary parameters α and β.
Inhomogeneous equations: We now consider the inhomogeneous Eq. (2.22) with φ(t) 6= 0.
The first thing to note is the following: Suppose x0 (t) is any particular solution of the inhomogeneous equation. Then a solution of the form
x(t) = x0 (t) + xh (t) ,
with xh (t) a general solution of the corresponding homogeneous equation is also a solution of
the inhomogeneous equation, and this exhausts the possibilities, i.e. there is no solution of the
inhomogeneous equation which is not of this form. We already know how to find the 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 varying
constants from a solution x1 (t) = α eλt of the homogeneous equation. Thus we attempt a
28
CHAPTER 2. DIFFERENTIAL EQUATIONS
solution x0 (t) of the inhomogeneous equation of the form x0 (t) = F (t)eλt with an unknown
function F to be determined by inserting the ansatz into (2.22). Following the reasoning in case
(iii) above, this gives
h
i
aF̈ + (2aλ + b)Ḟ + (aλ2 + bλ + c)F eλt = φ(t)
Once more the coefficient of F in this equation , being L(λ), vanishes. So we get
aF̈ + (2aλ + b)Ḟ = φ(t) e−λt .
This ODE for F does only involve first and second order derivatives of F and not the function
F itself. This can be used to reduce the order of the ODE. Setting ψ(t) = Ḟ (t), one obtains
aψ̇ + (2aλ + b)ψ = φ(t) e−λt .
This is a first order linear equation and we know how to solve it using integrating factors (see
Sec 2.4.3 above). Once ψ(t) is obtained along those lines, F (t) is follows by a further integration
w.r.t. time t, thus finally allowing to write down x0 (t) = F (t)eλt .
Occasionally one can be lucky to find a special solution faster by using inspired guesswork as in
the following
Example 2.7 Find the solution of the differential equation
d2 x
+ 9x = t,
dt2
subject to the initial conditions x(0) = 0,
dx
dt (0)
(2.28)
= 0.
The homogeneous equation corresponding to Eq (2.28) is
d2 x
+ 9x = 0 .
dt2
(2.29)
The corresponding auxiliary equation (obtained by substituting the trial solution eλt ) is
λ2 + 9 = 0
with solution λ = ±3i. It follows that the general solution of equation 2.29 is
x(t) = α cos 3t + β sin 3t,
where α, β are parameters. A particular solution of equation 2.28 is clearly x0 (t) = t/9, as follows by
inspection. We conclude that the general solution of Eq (2.28) is
t
x(t) = α cos 3t + β sin 3t + ,
9
(2.30)
where α, β are parameters.
We now pick the solution which satisfies the initial conditions. Imposing the conditions x(0) = 0,
dx
dt (0) = 0 then gives α = 0 and β = −1/27. The required solution is therefore
x(t) = −
t
1
sin 3t + .
27
9
Chapter 3
First Order Autonomous Systems
In the present chapter we specialise to first order autonomous dynamical systems. We begin by
preparing terminology in terms of which such systems are usefully characterised. We look at
qualitative methods and at approximate analytic methods to study them. The concept of fixed
points and their stability will be analysed in detail. A brief introduction to asymptotic analysis
is also given.
A first order autonomous dynamical system is described by a differential equation of the
form
dx
= f (x).
(3.1)
dt
The function f in this equation is referred to as the velocity function. Unless stated otherwise,
it is assumed that x ∈ IR and f : IR → IR.
Note that the independent variable t does not appear in the velocity function f . This is the
defining property of an autonomous system. The future evolution of the system thus depends
solely on the present value of x. If x is such that f (x) > 0, then x will increase as a function of
t; conversely if x is such that f (x) < 0, then x will decrease as a function of t.
3.1
Phase Space and Phase Portraits
Definition 3.1 Phase Space Γ
The phase space Γ is defined as the set of admissible values for x. E.g., if x denotes some
population density,
or the concentration of a chemical substance, then Γ = IR+ = {x ∈ IR; x ≥
p
0}. If f (x) = g(x) for some real function g, then Γ ⊆ {x; g(x) ≥ 0}.
Definition 3.2 Fixed Points
A point a ∈ IR is a fixed point of the dynamical system (3.1), iff f (a) = 0. If a system is at a
fixed point a — x(t) = a for some t — it will stay there forever, as dx
dt = 0 by (3.1)
Definition 3.3 Orbit/Phase Curve (through x0 )
The set of points {x(t); t ∈ IR} with x a solution of (3.1) (and x(t0 ) = x0 for some t0 ) is called
orbit or phase curve (through x0 ).
Definition 3.4 Trajectory/Solution Curve (through x0 )
The set of points {(t, x(t)); t ∈ IR} with x a solution of (3.1) (and x(t0 ) = x0 for some t0 ) is
called trajectory or solution curve (through x0 ).
29
30
CHAPTER 3. FIRST ORDER AUTONOMOUS SYSTEMS
Definition 3.5 Flow
The collection of all possible phase curves, including velocity information, i.e., with f (x) =
f (x)e1 drawn at each x ∈ Γ, is called the flow of the dynamical system (3.1). The vector f (x)
is called velocity of the flow (at x).
Definition 3.6 Phase Portrait
A graphical representation of the the phase space Γ, including a graph of the velocity function
f and a graph of the flow is called phase portrait (also referred to as phase diagram) of the
system.
Definition 3.7 Invariant Open Set
An invariant open set is a set S = {x(t); −∞ < t < +∞} ⊆ Γ. These open sets are minimal
open sets in the sense that they contain no proper subset with this property. Note that an
invariant set of points is not left in finite time, if the system starts in this set.
Example 3.1 First order dynamical system described by the velocity function
f (x) = (x − 1)(x − 2)(x − 3) .
(3.2)
The phase portrait is shown in the figure below. For this system Γ = IR; the fixed points are at x = 1,
x = 2, and x = 3. The invariant open sets of this system are the open intervals S1 = (−∞, 1), S2 = (1, 2),
S3 = (2, 3), and S3 = (3, +∞). By inspection, we identify x = 1 and x = 3 as unstable fixed points,
whereas x = 2 is a stable fixed point.
f(x)
1
2
3
x
Figure 3.1: Phase portrait discussed in the example above.
Note 1: The phase space can always be decomposed into (i) fixed points and (ii) invariant open
sets.
Note 2: The ODEs, in terms of which first order autonomous systems are described are always
of the separable type.
3.2
Estimating Times of Motion
If a system has a stable fixed point, and initial conditions are such that the fixed point is being
approached, one may ask, whether the system will reach the fixed point in finite time. The same
may be asked about ‘escaping’ to ±∞. These considerations lead us to the following definition.
31
3.2. ESTIMATING TIMES OF MOTION
Definition 3.8 Terminating Motion
The motion of a dynamical system is said to be terminating, if it reaches ±∞ (or a fixed point)
in finite time.
Given a dynamical system described by (3.1),
dx
= f (x).
dt
and initial condition x(t0 ) = x0 . The time τ01 needed for the system to reach x1 can in principle
be computed by solving the (separable) ODE,
τ01 =
t1
Z
dt =
t0
Z
x1
x0
dx
.
f (x)
(3.3)
Clearly, if f (x0 ) > 0, then x will increase as a function of time, and x1 > x0 for τ01 > 0;
conversely f (x0 ) < 0, then x will decrease as a function of time, and x1 < x0 for τ01 > 0.
Estimates for τ01 can be obtained without giving the full solution of the ODE describing the
dynamical system in question, i.e., without actually doing the x-integral in (3.3) 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
Z
b
a
φ(t) dt ≥
Z
b
ψ(t) dt .
a
We use Example 3.1 to explain, how times of motion can be estimated based on this observation.
We consider only the cases 2 < x0 < 3 and x0 > 3.
Case: 2 < x0 < 3
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 =
Z
x1
x0
dx
=
f (x)
Z
x0
x1
dx
=
|f (x)|
Z
x0
x1
dx
(x − 1)(x − 2)(3 − x)
A lower bound for τ01 is obtained by using
2<x<3
⇒ (x − 1)(3 − x) < 2 · 1 ⇒
Hence
τ01
1
>
2
Z
x0
x1
1
1
> .
(x − 1)(3 − x)
2
dx
1 x0 − 2
= ln
.
(x − 2)
2 x1 − 2
As x1 → 2, this lower bound diverges, so the system approaches 2, but will not reach it in
finite time.
Case: 3 < x0 < +∞
In this case the system will escape to +∞, and we will show that +∞ is reached in finite
time.
32
CHAPTER 3. FIRST ORDER AUTONOMOUS SYSTEMS
(a) Upper bound for τ01 : Use (x − 1)(x − 2)(x − 3) > (x − 3)3 . Hence
τ01 <
Z
x1
x0
"
#
1
1
1
1
dx
=
−
≤
(x − 3)3
2 (x0 − 3)2 (x1 − 3)2
(x0 − 3)2
(b) Lower bound for τ01 : Use (x − 1)(x − 2)(x − 3) < (x − 1)3 . Hence
τ01 >
Z
x1
x0
"
dx
1
1
1
=
−
3
2
(x − 1)
2 (x0 − 1)
(x1 − 1)2
#
Thus the time to reach x1 = +∞ satisfies
1
1
< τ01 <
,
2
(x0 − 1)
(x0 − 3)2
i.e. the approach to +∞ is terminating.
You are encouraged to use the above reasoning to obtain appropriate estimates for times of
motion for the other cases x0 < 1 and 1 < x0 < 2 as well.
It is instructive to compare our estimates with exact results. We may integrate Eq (3.2)
exactly by separation of variables. The method of partial fractions gives
1
1 1
1
1 1
=
−
+
(x − 1)(x − 2)(x − 3)
2x−1 x−2 2x−3
and integration gives
(x − 1)(x − 3)(x − 2)−2 = Ee2t ,
where E is a constant. Assuming x0 = 4 this gives E = 3/4 and therefore
(x − 1)(x − 2)−2 (x − 3) =
As x → ∞, t → τ01 , where
τ01
3 2t
e .
4
4
1
= ln
' 0.1438,
2
3
compared with the estimate 0.056 < τ01 < 0.50 which we obtained above for the same x0 .
Exercise 3.1 Consider the motion of the system described by the differential equation
dx
= (x − 1)(x − 2).
dt
Find the fixed points, and discuss their stability. Show that if x(0) = 3/2 the system will not
reach x = 1 in a finite time.
Exercise 3.2 Discuss the motion given by
dx
= (x − 1)(x − 2)(x − 3)(x − 4).
dt
Sketch the phase diagram. Write down the fixed points, and discuss their stability. What are the
invariant open sets?
Suppose that x(0) = 5; prove that the motion is terminating and find upper and lower bounds
for the time τ taken by the system to reach infinity.
Show that if, instead, x(0) = 7/2 the system will not reach x = 3 in a finite time.
3.3. STABILITY — A MORE GENERAL DISCUSSION
33
Exercise 3.3 Consider the motion described by
dx
= xα , x > 0, α > 0.
dt
For which values of α is the motion terminating?
Exercise 3.4 Write down the fixed points of the logistic equation of population growth
dx
= r0 x(1 − x/c) ,
dt
in which r0 > 0 and c > 0, and investigate their stability. What is the significance of your
conclusions for the population described by the logistic equation?
3.3
Stability — A More General Discussion
In Ch. 3.1 we introduced the notion of fixed points and investigated their stability graphically
by looking at the phase portrait and investigating qualitative properties of velocity functions
in the vicinity of fixed points. Below we look at stability of first order autonomous dynamical
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 dynamical system, and (iii)
the notion of stability of motion in turn.
3.3.1
Stability of Fixed Points
We consider the system described by the equation
dx
= f (x)
dt
and suppose that x = a is a fixed point. Assuming that f is a suitably differentiable function in
a neighbourhood of x = a we may approximate f by its Taylor expansion about x = a,
f (x) = f (a) + f 0 (a)(x − a) +
f 00 (a)
f 00 (a)
(x − a)2 + . . . = f 0 (a)(x − a) +
(x − a)2 + . . . (3.4)
2!
2!
Near x = a, i.e. for |x − a| 1, the dominant contribution to the r.h.s of (3.4) comes from the
first non-vanishing derivative of f at a.
Case (i) f 0 (a) 6= 0.
In this case f (x) ' f 0 (a)(x − a) for |x − a| 1, and the ODE is approximated by
dx
' f 0 (a)(x − a)
dt
⇔
d(x − a)
' f 0 (a)(x − a)
dt
The solution is
x − a ' (x0 − a) exp[f 0 (a)t] ;
the distance x − a thus (i) increases exponentially in time, if f 0 (a) > 0, whereas (ii) it
decreases exponentially in time, if f 0 (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 accord
with the graphical stability analysis based on a phase portrait as in the figure shown
below.
34
CHAPTER 3. FIRST ORDER AUTONOMOUS SYSTEMS
Case (ii) f 0 (a) = 0, but f 00 (a) 6= 0.
In this case f has a double zero at x = a, and
1
f (x) ' f 00 (a)(x − a)2
2
near x = a. The ODE is approximated by
1
d(x − a)
1
dx
' f 00 (a)(x − a)2
⇔
' f 00 (a)(x − a)2
dt
2
dt
2
The solution (found by separation of variables), reads
x0 − a
x−a'
1 − 2t f 00 (a)(x0 − a)
Supposing that f 00 (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 00 (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 result are once more in accordance with the graphical stability analysis based on a
phase as in the figure shown below.
f(x)
a
b
x
c
Figure 3.2: Phase portrait of a system with velocity function that has fixed pints at a with
f 0 (a) > 0, at b with f 0 (b) < 0 and at c, with f 0 (c) = 0, but f 00 (c) < 0.
Definition 3.9 Linear Stability Analysis
The stability analysis based on (and requiring only) Taylor series expansions to 1st order is
referred to as linear linear stability analysis of a fixed point.
Note The approximate solution of the ODEs describing the dynamical system near fixed points
based on Taylor series expansions is valid only in the vicinity of fixed points. The quality of
the approximate description will therefore improve, if a fixed point is approached under the
dynamics, whereas the approximate description deteriorates, if the system moves away from a
fixed point.
35
3.3. STABILITY — A MORE GENERAL DISCUSSION
3.3.2
Structural Stability
Definition 3.10 Structural Stability
A system described by the differential equation
dx
= f (x)
dt
is called structurally unstable, if the number of fixed points is changed by some arbitrarily small
(continuous) perturbation f (x) → f (x) + g(x), (i.e., for some continuous g and arbitrarily
small), otherwise it is structurally stable (i.e., for all continuous g there is a sufficiently small ,
such that the number of fixed points remains unchanged).
Note If f has a double zero at x = a the system is structurally unstable since replacing f (x) by
f (x) + increases the number of fixed points by one if < 0, and reduces the number of fixed
points by one if > 0 (see the diagram). More generally, if f has a zero of any multiplicity
higher than 1 at x = a, the system is structurally unstable. (The proof is left as an exercise to
the reader.)
f(x)
x
Figure 3.3: Phase portrait of a structurally unstable system, with a double zero which is a
minimum of the velocity function (full curve). Vertical displacement by ± (dashed curves)
decreases or increases the number of fixed points by one as shown.
In structurally unstable systems the collection of invariant open sets (and thus qualitative properties of the dynamics) can change drastically in response to small perturbations. This may e.g.
of economic relevance as the following example demonstrates.
Example 3.2 Bistable Ecosystems
Consider a population dynamics of the form
dx
ax2
= rx(1 − x/c) − 2
,
dt
b + x2
in which all parameters (r, c, a, b) are taken to be positive. The first contribution describes logistic growth,
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 system could
describe a population of larvae, preyed upon by a natural enemy (woodpeckers) if present in sufficient
36
CHAPTER 3. FIRST ORDER AUTONOMOUS SYSTEMS
quantity. For sufficiently small r the system has a stable small-x fixed point an unstable fixed point and
a stable fixed point at large x. The stable fixed point at low x and the unstable fixed point coalesce
and disappear as r is increased beyond a certain value rc and the system would approach a stable high-x
solution. Economically, this could spell disaster if the larvae destroy wood that is meant to be sold!. Only
by decreasing r to a very small value r∗ rc can the situation be reverted to a stable small x situation;
(see the diagram).
f(x)
f(x)
x
1
x
2
x
x
3
x
3
x
Figure 3.4: Phase portrait of the bi-stable ecosystem for r < rc (left) and r > rc (right).
Example 3.3 (King’s College, Summer 1994)
A first order dynamical system is described by the differential equation
ẋ = v(x) = x(x − 1)2 (x − 2)3 (x − 3)5 .
Explain briefly why the initial condition x(0) = 2 ⇒ x(t) = 2 ∀t ≥ 2. Sketch the phase diagram and
write down the invariant open sets. Is the system structurally stable? Give your reasons. Given that
x(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 0 are both continuous in any neighbourhood of the point (t = 0, x = 2); Picard’s theorem
therefore applies to the given equation. Clearly x(t) = 2 satisfies the differential equation and 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) − ( > 0) increases the
number 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 τ1
taken to reach x = x1 < 5/2 is
Z 5/2
Z x1
dx
dx
=
.
τ1 =
2 (x − 2)3 (x − 3)5
2 (x − 2)3 (3 − x)5
x(x
−
1)
x(x
−
1)
x1
5/2
Now x(x − 1)2 (3 − x)5 < (5/2)(5/2 − 1)2 (3 − 2)5 = 45/8 so
Z 5/2
8
1
dx
8
1
τ1 >
−2
+
.
=
45 x1 (x − 2)3
45
2 (x1 − 2)2
37
3.3. STABILITY — A MORE GENERAL DISCUSSION
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
Z ∞
dx
τ=
.
2 (x − 2)3 (x − 3)5
x(x
−
1)
4
Since x > x − 3, x − 1 > x − 3, x − 2 > x − 3, we obtain
Z ∞
dx
1
τ<
=
.
(x − 3)11
10
4
This motion is therefore terminating.
v(x)
1
2
3
x
Figure 3.5: Phase portrait of the system with velocity function v(x) = x(x − 1)2 (x − 2)3 (x − 3)5 .
Example 3.4 Motion Terminating at Finite x.
Suppose that v is a real valued function with continuous derivative v 0 , and that v(x) > 0, x 6= b, v(b) =
0, v 0 (b) < 0. Prove that the motion described by the differential equation
dx p
= v(x)
dt
is terminating at x = b.
First we note that v takes negative values just to the right of x = b; this follows from the fact that v 0 is
continuous and v(b) = 0, v 0 (b) < 0. The motion is therefore undefined for x > b; the phase space is the
region x ≤ b.
Suppose that x(0) = x0 < b. Since v 0 is continuous there exists a δ > 0 such that
v 0 (x) <
v 0 (b)
, x ∈ [b − δ, b + δ];
2
this follows from the definition of continuity with = −v 0 (b)/2(> 0).
The time τ1 for the system to go from x0 to b − δ is clearly finite because the integral
Z
b−δ
x0
dx
p
v(x)
38
CHAPTER 3. FIRST ORDER AUTONOMOUS SYSTEMS
is finite. Between b − δ and b we need to be more careful. For x ∈ [b − δ, b] we have, by the Mean Value
Theorem,
v(x) − v(b)
= v 0 (ξ) ,
x−b
for some ξ ∈ (b − δ, b) so that
v(x) = (b − x)(−v 0 (ξ)) > (b − x)(−v 0 (b)/2) = K(b − x) ,
where K is a positive number. The time τ2 taken by the system to go from b − δ to b is
r
Z b
Z b
dx
1
δ
dx
p
τ2 =
.
< √
=2
1/2
K
K b−δ (b − x)
v(x)
b−δ
The total time taken by the system to reach b, τ1 +τ2 is therefore finite. We see that the motion terminates
at x = b.
3.3.3
Stability of Motion
Beyond looking at stability of fixed points, one may also look at stability of motion itself in the
following sense. Let
dx
= f (x)
(3.5)
dt
describe a first order autonomous dynamical system, and let x(t) be a solution for some given
initial condition. Suppose now, that the solution is slightly disturbed
x(t) → y(t) = x(t) + ε(t)
(3.6)
at, say, t = t1 , with |ε(t)| 1. Enquiring into stability of motion means asking whether the
perturbation will increase or decrease under the dynamics. We study this issue only at linear
order.
Definition 3.11 Stability of motion
The motion described by (3.5) is stable, if a small perturbation ε(t) is diminished under the
dynamics; it is called unstable, if a small perturbation is amplified.
Proposition 3.1 The motion of an autonomous first order dynamical system described by a
velocity function f is linearly stable at x, if f 0 (x) < 0, whereas it is linearly unstable at x, if
f 0 (x) > 0.
Proof Consider the ODE
dy
= f (y)
dt
⇔
dx dε
+
= f (x + ε)
dt
dt
Expanding the r.h.s. to first order in ε gives
dx dε
+
' f (x) + f 0 (x)ε .
dt
dt
As x is a solution of (3.5), we get an ODE for ε,
dε
' f 0 (x)ε ,
dt
39
3.4. ASYMPTOTIC ANALYSIS
in which f 0 (x) = f 0 (x(t)) is a known function of time. The equation is therefore separable and
solved by
ε(t) ' ε(t1 ) exp
hZ
t
i
dt0 f 0 (x(t0 ))
t1
(3.7)
Note that for small t − t1 the exponential may be approximated by exp [f 0 (x(t1 ))(t − t1 )] in the
above expression.
Thus the perturbation increases at an exponential rate, and the motion is unstable whenever
f 0 (x) > 0, whereas the perturbation decreases at an exponential rate, and the motion is unstable
whenever f 0 (x) < 0, as claimed.
#
Note 1: The notion of dynamic stability is extremely important if one considers solving an ODE
numerically on a computer, as it tells whether rounding errors will amplify when propagating a
solution ore whether they will diminish.
Note 2: A first order dynamical system which is linearly unstable when integrated forward in
time, becomes linearly stable when integrated backwards (and vice versa).
3.4
Asymptotic Analysis
Asymptotic analysis is concerned with approximate analytic investigations of dynamical systems
in situations where either full solutions are not readily available, or where approximate solutions
are sufficient to give a reasonably precise first level of analysis of a given system.
Asymptotic analysis exploits simplifying features of a first order dynamical system described by
dx
= f (x) ,
dt
(3.8)
when (i) the velocity function diverges at a certain point, f (x) → ±∞, as x → x∗ , or (ii) the
velocity function approaches zero at a certain point, f (x) → 0, as x → x∗ . Here x∗ may be
finite, or infinite.
The case, where f (x) → 0 , as x → x∗ with x∗ finite describes a fixed point of the dynamics.
The corresponding asymptotic analysis has been described in Ch 3.3.1 and constitutes just a
special case of the considerations to be detailed now.
3.4.1
Elements of Asymptotic Analysis
We start with a couple of key definitions.
Definition 3.12 Asymptotic Equality
A function f is said to be asymptotically equal to another function g as x → x∗ — in symbols
f (x) ' g(x), as x → x∗ — if their difference vanishes as x → x∗ , i.e., iff
f (x) = g(x) + ε(x) ,
with
ε(x) → 0
and
ε(x)
→0,
g(x)
as
x → x∗ .
The second condition here is to make sure that in cases where f (hence g) become small, as
x → x∗ , the dominant contribution to f is identified
40
CHAPTER 3. FIRST ORDER AUTONOMOUS SYSTEMS
Definition 3.13 Identical Asymptotic Behaviour
A function f is said to behave asymptotically like another function g as x → x∗ — in symbols
f (x) ∼ g(x), as x → x∗ — if they become of the same order of magnitude as x → x∗ , i.e., iff
f (x) = g(x)(1 + ε(x)) ,
with ε(x) → 0 ,
as
x → x∗ .
In both definitions, the case x∗ = ±∞ is permitted.
Note Asymptotic equality f (x) ' g(x) always implies identical asymptotic behaviour f (x) ∼
g(x); the reverse is not true in general. It is true, if both functions tend to zero as x → x∗ , but
it need not be true if f diverges, because for f (x) ∼ g(x) we have |f (x) − g(x)| = |ε(x)g(x)| and
this difference may diverge if f , hence g diverge at x∗ , despite the fact that ε(x) → 0.
This is also illustrated in the following
Example 3.5 Given
f (x) = x − x3 .
We then have
(a) f (x) ∼ x as well as f (x) ' x, as x → 0.
(b) f (x) ∼ −x3 as x → ∞, but not f (x) ' −x3 .
The detailed verification of these statements according to the foregoing definitions is left as an exercise
to the reader.
3.4.2
Asymptotic Analysis and Approximate Solutions of ODEs
The virtue of asymptotic analysis lies in the fact that it can be used to obtain approximate
solutions of ODEs in situations where full solutions may not be obtainable, or obtainable only
at the cost of significantly larger effort.
To see this consider a system described by (3.8). This equation is solved by separation of
variables,
Z x
Z t
dx0
=
dt0
0)
f
(x
t0
x0
R
Assuming t0 = 0, and putting
dx
f (x)
= F (x) + C, one obtains the formal solution
F (x) − F (x0 ) = t
⇔
x = x(t) = F −1 (t + F (x0 )) .
This solution may be difficult to work out, if either the integration or the evaluation of the
inverse function F −1 involved are difficult.
Approximate solution using asymptotic equality of functions: Let us first explore possible simplifications assuming that f (x) ' g(x). In this case we have
Z
x
x0
dx0
=
f (x0 )
Z
x
x0
dx0
=
g(x0 ) + ε(x0 )
Z
t
dt0
t0
Assuming once more t0 = 0, and exploiting the smallness of ε(x) and ε(x)/g(x), we can expand
and obtain
Z x
Z x
dx0
dx0 0
0
0
0 2
=
1
−
ε(x
)/g(x
)
+
(ε(x
)/g(x
))
−
.
.
.
=t
(3.9)
0
0
x0 f (x )
x0 g(x )
41
3.4. ASYMPTOTIC ANALYSIS
As long as x − x0 remains finite, we see that all correction terms in this integral are negligible
compared to the zeroth-order term of the expansion, so to lowest order we get
t=
x
Z
x0
R
from which, on putting
dx
g(x)
dx0
'
f (x0 )
Z
dx0
g(x0 )
x
x0
= G(x) + C, one obtains
G(x) ' t + G(x0 )
⇔
x ' G−1 (t + G(x0 )) ,
where the last equivalence holds under fairly general conditions on G. It may turn out that the
present problem is solvable whereas the original one is not (or difficult to solve).
Approximate solution using identical asymptotic behaviour of functions: Suppose
now that we have the weaker condition of identical asymptotic behaviour, f (x) ∼ g(x). In this
case, instead of the full solution, we have
Z
x
x0
dx0
=
f (x0 )
Z
x
x0
dx0
=
g(x0 )(1 + ε(x0 ))
Z
t
dt0
t0
Assuming t0 = 0 as before, and exploiting the smallness of ε(x), we can expand and obtain
Z
x
x0
dx0
=
f (x0 )
Z
x
x0
dx0 0
0 2
1
−
ε(x
)
+
ε(x
)
−
.
.
.
=t
g(x0 )
(3.10)
As long as x − x0 remains finite, all correction terms in this integral are once more negligible
compared to the zeroth-order term of the expansion; so as before to lowest order we get
t=
Z
x
x0
from which, on putting
R
dx
g(x)
dx0
'
f (x0 )
Z
x
x0
dx0
g(x0 )
= G(x) + C, one obtains the same approximate result as above,
G(x) ' t + G(x0 )
⇔
x ' G−1 (t + G(x0 )) ,
that is, approximate equality of x(t) with the solution obtained from a velocity function which
has only the same asymptotic behaviour as the original one and is not necessarily approximately
equal.
Possible corrections to the lowest order results: We may look a bit more carefully in order to
assess the size of possible corrections to the above lowest order results. In both cases (3.9) and (3.10)
above, integrating the equation of motion using separation of variables and asymptotic analysis leads to
an expression of the form
Z x
Z x
dx0
dx0 0
1
−
ε̃(x
)
=t,
(3.11)
=
0
0
x0 f (x )
x0 g(x )
where
ε̃(x0 ) = ε(x0 )/g(x0 ) − (ε(x0 )/g(x0 ))2 + . . . =
ε(x0 )
→0,
+ ε(x0 )
g(x0 )
as x0 → x∗
in the case (3.9) of approximate asymptotic equality of the functions f and g, while
ε̃(x0 ) = ε(x0 ) − ε(x0 )2 + . . . =
ε(x0 )
→0,
1 + ε(x0 )
as x0 → x∗
42
CHAPTER 3. FIRST ORDER AUTONOMOUS SYSTEMS
in the case (3.10) of identical asymptotic behaviour of the functions f and g.
R x dx0
0
Using the fact that that the integral x0 g(x
0 ) ε(x ) can be bounded in the form
Z x
Z x
Z x
dx0
dx0
dx0
0
≤
ε(x
)
≤
ε
(x,
x
)
εl (x, x0 )
u
0
0
0
0
x0 g(x )
x0 g(x )
x0 g(x )
(in the case where g(x0 ) and ε(x0 ) are positive, one would take εl (x, x0 ) = minx0 ∈[x0 ,x] ε(x0 ) and εl (x, x0 ) =
maxx0 ∈[x0 ,x] ε(x0 ), the cases where g or ε (or both) are negative can be handled in similar ways), with
both |εl (x, x0 )| 1 and |εu (x, x0 )| 1 in the region where the asyptotic analysis applies, we know
that there exists a function ε̂(x, x0 ) — also small — such that the result of the integration (3.11) can be
written in the form
Z x
dx0 0
1
−
ε̃(x
)
=
G(x)
−
G(x
)
1
−
ε̂(x,
x
)
=t
0
0
0
x0 g(x )
Solving formally for x = x(t), and abbreviating ε̂ = ε̂(x, x0 ), gives
!
!
t
tε̂
−1
−1
x(t) = G
+ G(x0 ) = G
t + G(x0 ) +
1 − ε̂
1 − ε̂
which upon performing a Taylor expansion results in
−1
x(t) = G
t + G(x0 ) + (G−1 )0 t + G(x0 )
tε̂
+O
1 − ε̂
tε̂
1 − ε̂
2 !
We can rewrite this result using the inverse function rule (G−1 )0 (z) = 1/G0 (G−1 (z)) and the fact that
G0 (z) = 1/g(z), to get
2 !
tε̂
tε̂
−1
−1
x(t) = G
(3.12)
t + G(x0 ) + g G (t + G(x0 ))
+O
1 − ε̂
1 − ε̂
The first contribution to the r.h.s of (3.12) is recognized as the lowest order result obtained earlier.
Corrections to this lowest order result will be small as long as
tε̂
tε̂
= g(x̃(t))
g G−1 (t + G(x0 ))
1 − ε̂
1 − ε̂
in which x̃(t) = G−1 (t + G(x0 )) is the approximate solution to lowest orderm, is small.
Example 3.6 To illustrate the use of asymptotic analysis, we consider a case where we are able obtain
the full solution in addition to the approximate expressions, and thereby fully assess the quality of the
approximate solutions.
Consider a dynamical system of the form
dx
= f (x) = −x + x2 .
dt
It has fixed points at x = 0 (stable) and x = 1 (unstable). The solution of the ODE is obtained by
separation of variables, using partial fractions to do the x-integral:
Z x
dx0
x−1
x0 − 1
t=
= ln
− ln
0
0 − 1)
x
(x
x
x0
x0
with x(0) = x0 . Solving for x gives
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 +∞, and
the motion will terminate at time t = ln(x0 /(x0 − 1)).
Let us compare these exact results with those of the corresponding asymptotic analyses.
x(t) =
43
3.4. ASYMPTOTIC ANALYSIS
(a) Considering the behaviour at small x, we have f (x) ' −x, as x → 0. Solving the corresponding
approximate ODE
dx
' −x ,
dt
assuming x0 1, gives x(t) ' x0 e−t . This is indeed just the result one would get to lowest order
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 corresponding
approximate ODE
dx
∼ x2 ,
dt
x0
. Asymptotic equality would be guaranteed as long as x − x0
assuming x0 1, gives x(t) ' 1−x
0t
remains finite. In this case, as in the exact solution, we observe that x(t) diverges at finite t. I.e.,
the motion terminates and the time to escape to +∞ is t = 1/x0 . This is indeed just the result one
would get to lowest order in 1/x0 in an expansion of the the time at which the motion terminates
in the exact solution.
The following figure compares exact and approximate solution in the regions of small and large x. The
quality of the approximations is remarkable even in the region of the divergence, although exact and
approximate solutions terminate at slightly different times (so that there the error is strictly speaking
infinitely large and the approximation bad). Note that the large x solution can even be continued with
reasonable precision across the divergence!
0.11
1e+06
0.1
0.09
100000
0.08
0.07
10000
x(t)
x(t)
0.06
0.05
1000
0.04
0.03
100
0.02
0.01
0
10
0
2
4
6
8
10
0
0.002
0.004
t
0.006
0.008
0.01
t
10000
5000
x(t)
0
-5000
-10000
-15000
-20000
0
0.005
0.01
0.015
0.02
t
Figure 3.6: (a) Full solution (full curve) and approximate solution (dashed curve) in the region of small
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 the divergence.
44
CHAPTER 3. FIRST ORDER AUTONOMOUS SYSTEMS
Chapter 4
Second Order Autonomous Systems
In this chapter we look at second order autonomous dynamical systems. As before we begin by
preparing terminology in terms of which such systems are usefully characterised. Much of this
terminology in fact carries over from the case of first order systems in fairly obvious ways.
We look at qualitative methods and at approximate analytic methods to study second order
systems. It turns out that the spectrum of possible behaviours of second order systems is much
richer than that of first order systems. The concept of fixed points and their stability will be
analysed in detail.
A second order autonomous dynamical system is described by differential equations of the
form
dxi
= fi (x1 , x2 ) ,
i = 1, 2.
(4.1)
dt
The functions f1 and f2 are referred to as velocity functions.
In this chapter we often find it convenient to use conventional x, y notation instead of x1 , x2 .
We therefore write these equations as
dy
= f2 (x, y) ,
dt
dx
= f1 (x, y) ,
dt
or in vector notation
dr
d
≡
dt
dt
4.1
x
y
!
= f (x, y) =
f1 (x, y)
f2 (x, y)
!
(4.2)
Phase Space and Phase Portraits
The terminology introduced in Chapter 3 extends in an obvious way to second order autonomous
systems.
Definition 4.1 Phase Space Γ
The phase space Γ is defined as the set of admissible values for the coordinates x and y. The pair
(x, y) denotes a point in IR2 , so generally Γ ⊆ IR2 . For some systems, Γ may be a proper subset
of IR2 : e.g., if x and y denote concentrations of two substances in a binary chemical reaction,
then x ≥ 0 and y ≥ 0.
45
46
CHAPTER 4. SECOND ORDER AUTONOMOUS SYSTEMS
Definition 4.2 Fixed Points
A point (a, b) ∈ IR2 is a fixed point of the dynamical system (4.2), iff f1 (a, b) = 0 and f2 (a, b) = 0.
If a system is at a fixed point, i.e., (x(t), y(t)) = (a, b) for some t, it will stay there forever, as
dy
dx
dt = 0 and by dt = 0 (4.2)
Definition 4.3 Null-Clines
The set of points (x, y) ∈ IR2 for which f1 (x, y) = 0 is called null-cline of f1 .
The defining equation of the null-cline, f1 (x, y) = 0, establishes a relation that must hold between
x and y for points (x, y) to be on the null-cline. This relation generally defines a curve in IR2 .
For instance if f (x, y) = x2 + y 2 − r 2 , then the null-cline of f is a circle of radius r centred at
the origin of IR2 .
Note: On a null-cline of f1 , the x-component of the velocity vanishes; similarly, on a null-cline
of f2 , it is the y-component of the velocity which is zero. Fixed points are intersections of the
null-clines of f1 and f2 :
Definition 4.4 Orbit or Phase Curve (through (x0 , y0 ))
The set of points {(x(t), y(t)); t ∈ IR} with (x, y) a solution of (4.2) (and x(t0 ) = x0 and
y(t0 ) = y0 for some t0 ) is called orbit or phase curve (through (x0 , y0 )).
Note: Provided the functions f1 and f1 satisfy the conditions of Picard’s theorem there is
precisely one phase curve passing through each point of phase space; if there are several distinct
phase curves passing through the same point then the Picard conditions fail at that point.
Definition 4.5 Trajectory/Solution Curve (through x0 )
The set of points {(t, x(t), y(t)); t ∈ IR} with (x, y) a solution of (4.2) (and x(t0 ) = x0 and
y(t0 ) = y0 for some t0 ) is called trajectory or solution curve (through (x0 , y0 )).
Definition 4.6 Flow
The collection of all possible phase curves, including velocity information, i.e., with f (x, y) =
f1 (x, y)e1 + f2 (x, y)e2 drawn at each (x, y) ∈ Γ, is called the flow of the dynamical system (4.2).
The vector f (x, y) is called velocity of the flow (at (x, y)).
Definition 4.7 Phase Portrait
A graphical representation of the the phase space Γ, including a graph of the flow (the velocity
field f (x, y) evaluated at representative points, e.g. on a equidistant grid of points) is called
phase portrait (also referred to as phase diagram) of the system.
Example 4.1 Example of a Phase Diagram Describe the phase diagram associated with the velocity
field given by
f (x, y) = 2xye1 + (y 2 − x2 )e2 .
We choose a suitable grid, as described above. The portion of the phase diagram pertaining to the
half-space x > 0 is shown below. The arrow at (x, y) makes an angle θ with the x-axis where
tan θ = (y 2 − x2 )/(2xy)
and has magnitude proportional to
|f (x, y)| = ((4x2 y 2 + (y 2 − x2 )2 )1/2 = (x2 + y 2 ) .
47
2
2
1.5
1.5
1
1
0.5
0.5
0
0
y
y
4.1. PHASE SPACE AND PHASE PORTRAITS
-0.5
-0.5
-1
-1
-1.5
-1.5
-2
-2
-1.5
-1
-0.5
0
x
0.5
1
1.5
2
2
2
1.5
1.5
1
1
0.5
0.5
0
0
y
y
-2
-0.5
-0.5
-1
-1
-1.5
-1.5
-2
-2
-1.5
-1
-0.5
0
x
0.5
1
1.5
2
-2
-1.5
-1
-0.5
0
x
0.5
1
1.5
2
-2
-2
-1.5
-1
-0.5
0
x
0.5
1
1.5
2
Figure 4.1: Phase Portrait of the system discussed in Example 4.1, including null-clines of f1 and
f2 and an arrow representation of the flow (upper left panel). The upper right panel contains
the same data, but arrows representing the velocity field f (x, y) are not drawn such that they
are rooted in (x, y) as in the left panel (and as they strictly should); rather we chose the midpoint
of the arrows to be at (x, y) which gives a better visual impression of the actual flow. The lower
two panels show the corresponding versions of the direction fields n(x, y).
48
CHAPTER 4. SECOND ORDER AUTONOMOUS SYSTEMS
Note: If the magnitudes of the velocities vary in a large range, one may choose to plot the
direction field n(x, y) = f (x, y)/|f (x, y)| (instead of the velocity field) in a phase diagram; The
direction field is a field of unit vectors pointing in the direction of the velocity. As with the flow,
visual impression of the actual direction field can be more faithful, if the arrows representing
the direction field n(x, y) are not drawn such that they are rooted in (x, y) (and as they strictly
should) but rather with their midpoint at (x, y)
4.1.1
Phase Curves
In the example discussed above, we obtain a clear impression from the phase diagram that the
phase curves are a family of circles and we can confirm this by integrating the equations
dx
= 2xy ,
dt
exactly. Eliminating t we see that
dy
= 2xy = y 2 − x2
dt
dy
y 2 − x2
=
.
dx
2xy
This equation is of the first order homogeneous type, and we can integrate it by means of the
substitution y = xv(x). We find that
2vdv
dx
=−
v2 + 1
x
⇒
ln[x(1 + v 2 )] = B ,
where B is a parameter and, after exponentiation,
y2
x 1+ 2
x
=A,
A = eB .
We may re-write this equation as
(x − A/2)2 + y 2 = (A/2)2 .
This equation describes a family of circles, with centres at (A/2, 0) and radii A/2. Note that
they all pass through (0, 0). (What do you conclude from this?)
The method just explained is generally applicable to obtain phase curves of second order autonomous system such as described by Eq. (4.2). Eliminating t from that equation one obtains
dy
f2 (x, y)
=
.
dx
f1 (x, y)
(4.3)
This is an ODE describing a 1st order system, with x as the independent variable, though it is in
general not an autonomous system; all methods for 1st order systems described in the previous
chapters can be applied to obtain qualitative, approximate analytical, or exact information about
phase curves y = y(x). [Note: it may sometimes be easier to investigate phase curves taking y
f1 (x,y)
as the independent variable, i.e. finding functions x = x(y) as solutions of dx
dy = f2 (x,y) .]
Exercise 4.1 Sketch the phase diagram associated with the uniform velocity field given by
f (x, y) = k1 e1 + k2 e2 ,
49
4.2. SEPARABLE SYSTEMS
where k1 , k2 are constants. Find the equation of the phase curves.
Do the same for the velocity field given by
f (x, y) = k3 e1 + xe2 ,
where k3 is constant.
4.2
Separable systems
In second order dynamical systems the dynamical evolutions of the variables x and y are quite
generally coupled: both velocity functions depend on x and y.
Sometimes it is possible to change variables from (x, y) to new variables (x1 , y1 ) in such a way
that the differential equations describing the system take the form
x˙1 = v1 (x1 ) ,
x˙2 = v2 (x2 ).
We say that the system is separable in the variables (x1 , x2 ). In essence the second order system
has been de-coupled into two first order systems described by the first order differential equations
x˙1 = v1 (x1 ) and x˙2 = v2 (x2 ) respectively. We note that if x1 = a is a fixed point of the first
system (so that v1 (a) = 0) and if x2 = b is a fixed point of the second system (so that v2 (b) = 0)
then (a, b) is a fixed point of the second order system; for v (a, b) = v1 (a)e1 + v2 (b)e2 = 0. An
important example of a separable second order autonomous dynamical system is the following.
Example 4.2 A System Separable in Polar Coordinates
Consider the system described by the equations
ẋ = αx − ωy, ẏ = ωx + αy,
where α, ω are constants.
Changing to polar coordinates (r, θ) we write
x = r cos θ, y = r sin θ.
We have
ẋ = −rθ̇ sin θ + ṙ cos θ,
and
ẏ = rθ̇ cos θ + ṙ sin θ.
Substituting in the given equations we obtain
(ṙ − αr) cos θ − r sin θ(θ̇ − ω) = 0,
(ṙ − αr) sin θ + r cos θ(θ̇ − ω) = 0.
Bearing in mind the identity cos2 θ + sin2 θ = 1 we find that
ṙ = αr, θ̇ = ω.
We conclude that the system is separable in the polar coordinates (r, θ).
(4.4)
50
4.3
CHAPTER 4. SECOND ORDER AUTONOMOUS SYSTEMS
Stability of fixed points
The stability problem for second order autonomous systems is likely to be more complicated
than for first order systems; the extra dimension allows more possibilities.
Suppose that (a, b) is a fixed point of the system described by equations 4.2. We make the
following definitions:
Definition 4.8 Attractor
We say that the fixed point (a, b) is an attractor for a particular motion (x(t), y(t)) satisfying
Eqs (4.2) if (x(t), y(t)) → (a, b) as t → ∞.
Definition 4.9 Strongly Stable Fixed Point
We say that a fixed point (a, b) is strongly stable if it is an attractor for all phase curves which
enter some neighbourhood of (a, b).
This conveys the notion that all phase curves which pass sufficiently close to (a, b) are “sucked
in” to (a, b) as t → ∞. The following definition provides a weaker concept of stability.
Definition 4.10 Stable Fixed Point
We say that a fixed point (a, b) is stable if for every neighbourhood N1 of (a, b) there exists a
neighbourhood N2 of (a, b), N2 ⊆ N1 , such that (x(0), y(0)) ∈ N2 ⇒ (x(t), y(t)) ∈ N1 for all
t ≥ 0.
Roughly speaking this definition says that if (a, b) is a stable fixed point then any motion of
the system which starts close enough to (a, b) remains close to (a, b). Notice that this definition
does not require that the system tends to (a, b) as t → ∞. If (a, b) is strongly stable then it is
stable, but the converse is not true.
4.3.1
Linear Stability Analysis
Expanding the functions f1 , f2 about (a, b) by Taylor’s formula Eqs (4.2) become, in the linear
approximation,
!
!
d
x
x−a
=J
+ ···,
(4.5)
y−b
dt y
where J denotes the Jacobian matrix given by
J=
∂f1
∂x
∂f2
∂x
∂f1
∂y
∂f2
∂y ,
!
(4.6)
evaluated at (a, b). Writing x̃ = x − a, ỹ = y − b Eq (4.5) becomes
d
dt
x̃
ỹ
!
=J
x̃
ỹ
!
.
(4.7)
The change of variable (x, y) → (x̃, ỹ) represents a change of origin; the origin of the x̃-ỹ
coordinates is in fact the fixed point (x = a, y = b).
51
4.3. STABILITY OF FIXED POINTS
4.3.2
Transformation to Jordan Canonical Forms
The linearity of Eq. (4.7) and the constancy of J at a fixed point under considerations allows to
perform a further linear transformation on (4.7) in such a way that the transformed linear equations of motion are described by one of the three possible Jordan canonical forms corresponding
to J (See Appendix B).
Suppose that P is any 2 × 2 non-singular constant matrix. Multiplying Eq (4.7) by P −1 we
obtain
!
!
d −1 x̃
x̃
−1
−1
,
(4.8)
= P JP P
P
ỹ
ỹ
dt
which we re-write as
d
dt
X
Y
!
X
Y
!
where
=P
−1
=P
X
Y
JP
−1
x̃
ỹ
!
!
,
.
(4.9)
(4.10)
We note that the origin of the new coordinates (X, Y ) is in fact the fixed point (x = a, y = b).
We assume that the stability of the fixed point (a, b) of the system described by equation 4.2
may be determined by examining the behaviour of the linearised system 4.9 near X = 0, Y = 0;
this is valid provided the matrix J is non-singular.
We now choose the matrix P so that the matrix J is brought into Jordan form (see Appendix B).
It turns out that the the resulting differential equations can be readily solved and the stability
of the fixed point determined for each of the three different variants of Jordan canonical form.
We illustrate the above reasoning by a series of examples.
Example 4.3 Find the fixed points of the second order autonomous system described by the differential
equations
ẋ = y − x2 + 2 ,
ẏ = 2(x2 − y 2 )
and investigate their stability.
The fixed points are given by solutions of
y − x2 + 2 = 0, x2 − y 2 = 0.
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 − y 2 ) ,
and the Jacobian matrix J, evaluated at arbitrary (x, y) is given by
! ∂f1
∂f1
−2x
1
∂x
∂y
=
,
J=
∂f2
∂f2
4x −4y
∂x
∂y ,
We look at these fixed points in turn.
Case 1 The fixed point (−1, −1) has
J(−1,−1) =
2 1
−4 4
.
(4.11)
52
CHAPTER 4. SECOND ORDER AUTONOMOUS SYSTEMS
Y
r
θ
X
Figure 4.2: Representative phase-curve in the vicinity of the fixed point, shown in the X, Y
coordinate system.
The matrix J(−1,−1) has eigenvalues λ given by
(λ − 2)(λ − 4) + 4 = 0
√
λ = 3 ± i 3.
⇒
It follows that there is a matrix P such that
P
−1
J(−1,−1) P =
√3
− 3
√ 3
..
3
Near (−1, −1) the linearised equations thus take the form
√ d
X
X
3
√3
,
=
Y
Y
− 3 3
dt
where
X
Y
=P
−1
x+1
y+1
(4.12)
.
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 (4.4)
that our equations take the form
√
ṙ = 3r, θ̇ = − 3 .
These equations have integrals
r(t) = r0 e3t ,
θ(t) = θ0 −
√
3t ,
where r0 = r(0), θ0 = θ(0). The phase curves have the form of ‘outgoing spirals’ indicated in the
diagram and we deduce that (x = −1, y = −1) is an unstable fixed point.
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. Recall the
procedure for finding Jordan forms: We consider the equation
√
2 1
u
u
.
= (3 + i 3)
−4 4
v
v
53
4.3. STABILITY OF FIXED POINTS
√
These equations aren’t independent and we need only look at v = (1 + i 3)u. Taking u = 1 we see
that an eigenvector is
0
1√
1
√
+i
=
.
1
1+i 3
3
With respect to the basis
f1 =
we have
1
1
, f2 =
√0
3
√
J(−1,−1) (f 1 + if 2 ) = (3 + i 3)(f 1 + if 2 ),
so that
J(−1,−1) f 1 = 3f 1 −
√
3f 2 ,
It follows that
P −1 J(−1,−1) P =
where
P =
We see that
√
3f 1 + 3f 2 .
J(−1,−1) f 2 =
1
1
√
3
3
√3
− 3
√0
3
√
1
det P = 3, P −1 = √
3
,
.
√
3 0
.
−1 1
The change of variables (x, y) 7→ (X, Y ) which we used in the above argument is therefore given by
X
x+1
−1
=P
Y
y+1
from which we derive
y−x
X = x + 1, Y = √ .
3
Case 2 The fixed point (2, 2): A similar calculation shows that the linearised equations take the form
d
X
−4
1
X
−1
=P
P
,
X = x − 2, Y = y − 2
Y
8 −8
Y
dt
where we choose P so as to bring the matrix
−4
1
8 −8
into Jordan form. The matrix in question has eigenvalues λ given by
(λ + 4)(λ + 8) − 8 = 0
so that
√
λ = λ1 = −6 + 2 3 < 0 ,
⇒
√
λ = −6 ± 2 3,
√
λ = λ2 = −6 − 2 3 < 0.
It follows that there exists a P such that
λ1
−4
1
−1
P
P =
0
8 −8
With this choice of P our equations become
Ẋ = λ1 X, Ẏ = λ2 Y,
0
λ2
.
54
CHAPTER 4. SECOND ORDER AUTONOMOUS SYSTEMS
Y
X
Figure 4.3: Representative phase-curves in the vicinity of a stable node, shown in the X, Y
coordinate system.
from which we derive
X(t) = X0 eλ1 t , Y (t) = Y0 eλ2 t ,
where X0 = X(0), Y0 = Y (0). The phase curves have the form shown in the diagram and we see
that (x = 2, y = 2) is a strongly stable fixed point (a stable node).
We can, of course, eliminate t to obtain the linearised form of the equations of the phase curves.
We find that
λ2 /λ1
√
X
Y = Y0
, λ2 /λ1 = 2 + 3.
X0
Case 3 The fixed point at (1, −1): Here we find that
d
X
X
−2 1
,
= P −1
P
Y
Y
4 4
dt
where we choose P so as to bring the matrix
−2 1
4 4
X = x − 1, Y = y + 1
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
Ẋ = λ1 X, Ẏ = λ2 Y,
which gives
X(t) = X0 eλ1 t ,
Y (t) = Y0 eλ2 t ,
in a now familiar notation. Since λ1 > 0,eλ1 t → ∞ as t → ∞. We conclude that (x = 1, y = −1)
is an unstable fixed point, and reference to the phase diagram confirms this; the phase curves have
the form shown. In the present case. there is one stable direction, and one unstable direction; a
fixed point of this type is called hyperbolic.
55
4.3. STABILITY OF FIXED POINTS
Y
X
Figure 4.4: Representative phase-curves in the vicinity of a hyperbolic fixed point, shown in the
X, Y coordinate system.
Case 4 The fixed point at (−2, 2). We see that
J(−2,2) =
4
1
−8 −8
and this matrix has eigenvalues
√
√
λ = λ1 = −2 + 2 7 > 0, λ = λ2 = −2 − 2 7 < 0.
The argument used in Case 3 leads to the conclusion that (x = −2, y = 2) is an also an unstable
(hyperbolic) fixed point.
Example 4.4 Find the fixed point of the system of equations
ẋ = f1 (x, y) = 3x + y + 1 ,
ẏ = f2 (x, y) = −x + y + 6
and investigate its stability.
The fixed points are given as solutions of
3x + y + 1 = 0 ,
−x + y − 6 = 0,
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
! ∂f1
∂f1
3 1
∂x
∂y
J=
=
,
(4.13)
∂f2
∂f2
−1 1
∂x
∂y
so that our equations may be written
d
x
3 1
x + 7/4
=
.
y
−1 1
y − 17/4
dt
Note that this equation is exact, because the original equations are linear ; we have not made any approximations, since the second and higher order derivatives of f1 , f2 are all zero. Proceeding in the usual
way we obtain
d
X
3 1
X
= P −1
P
,
Y
−1 1
Y
dt
56
CHAPTER 4. SECOND ORDER AUTONOMOUS SYSTEMS
where
and P is chosen so that the matrix
is reduced to Jordan form.
X
Y
= P −1
3
−1
x + 7/4
y − 17/4
1
1
,
We find that the eigenvalues of this matrix are given by
(λ − 1)(λ − 3) + 1 = 0 , ⇒ (λ − 2)2 = 0,
so that λ = 2 (twice). We can therefore choose P so that (see Eq (B.3))
2 1
3 1
−1
.
P =
P
0 2
−1 1
In the coordinate system (X, Y ) determined by P the differential equations become
Ẋ = 2X + Y ,
Ẏ = 2Y.
The second of these equations has general solution
Y (t) = Y0 e2t
and the equation for X becomes
Ẋ − 2X = Y0 e2t .
R
This equation is of linear type with integrating factor µ(t) = exp −2 dt = e−2t . We therefore obtain
d
e−2t X = Y0 e2t e−2t = Y0
dt
from which we easily derive
X(t) = te2t Y0 + e2t X0 .
Since e2t → ∞ as t → ∞ we conclude that (x = −7/4, y = 17/4) is an unstable fixed point. Eliminating
t we find that the phase curves are given by
1
Y
X0
X = Y ln
+
Y.
2
Y0
Y0
For each choice of the initial values (X0 , Y0 ) we obtain a phase curve.
Example 4.5 Consider the second order system described by the differential equations
ẋ = v1 (x, y) = ax − bxy + sin(x − d/c),
ẏ = v2 (x, y) = cxy − dy + sin(y − a/b),
where 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
! ∂v1
∂v1
a − by + cos(x − d/c)
−bx
∂x
∂y
J=
=
,
∂v2
∂v2
cy
cx − d + cos(y − a/b)
∂x
∂y ,
(4.14)
57
4.3. STABILITY OF FIXED POINTS
so that
J(d/c,a/b) =
− bd
c
ac
b
.
Near (x = d/c, y = a/b) the linearised equations become
d
X
X
−1
,
= P J(d/c,a/b) P
Y
Y
dt
where
X
Y
= P −1
x − d/c
y − a/b
and the matrix P is chosen so as to bring J(d/c,a/b) into Jordan form. A standard calculation shows that
the eigenvalues of J(d/c,a/b) are given by
λ = ± i(ad)1/2
and our equations become
d
dt
X
Y
=
−(ad)1/2
(ad)1/2
X
Y
.
This system is separable in polar coordinates and we write
X = r cos θ, Y = r sin θ
to obtain
ṙ = r, θ̇ = −(ad)1/2 .
If = 0 we get r(t) = r0 which describes a family of circles with centre (X = 0, Y = 0), the fixed point.
We see that if = 0 the fixed point is stable.
Note When = 0 the equations reduce to the Lotka-Volterra (“hare-lynx”) equations of Example 1.2
of Chapter 1. The Lotka-Volterra equations are not structurally stable in the sense that if 6= 0 we can
move from a situation of stable equilibrium to one of unstable equilibrium, as we shall now see. If < 0
we have
r(t) = r0 et → 0 as t → ∞,
and (x = d/c, y = a/b) is strongly stable. However, if > 0 we see that
r(t) = r0 et → ∞ as t → ∞,
so that (x = d/c, y = a/b) is a position of unstable equilibrium.
Example 4.6 Harmonic motion. Consider the differential equation of simple harmonic motion
ẍ + x = 0.
Setting y = ẋ we obtain
d
dt
x
y
=
v1 (x, y)
v2 (x, y)
where
J=
=
0 1
−1 0
y
−x
=J
x
y
,
(4.15)
.
Notice that Eq (4.15) is exact — no approximations have been made. Moreover, the matrix J is already
in the Jordan form appropriate to real 2 × 2 matrices with eigenvalues ±i.
58
CHAPTER 4. SECOND ORDER AUTONOMOUS SYSTEMS
We note that Eq (4.15) has one fixed point, namely (x = 0, y = 0). If we interpret the equation ẍ + x = 0
in terms of the motion of a Newtonian particle (see following chapter), the fixed point corresponds to
the particle at rest at x = 0. We now demonstrate that the fixed point is stable as follows. Changing to
polar coordinates (r, θ) where x = r cos θ, y = r sin θ we find in the usual way that
ṙ = 0 ,
θ̇ = −1.
which have the integrals
r(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 radius given
by the parameter r0 . The fixed point is therefore stable. We can also show this directly form Eq (4.15)
by noting that
ẋ = y ,
ẏ = −x.
Eliminating t we obtain
dy
x
=−
dx
y
⇒
x dx + y dy = 0.
It follows that
x2 + y 2 = r0 2 ,
where r0 is a parameter. In other words, the phase curves are indeed a family of circles, with centres at
(0, 0); the point (0, 0) is therefore a stable fixed point.
4.4
Limit cycles
Consider the following example.
Example 4.7 Show that the second order system described by the differential equations
ẋ = y + xf (r), ẏ = −x + yf (r),
(4.16)
where r = (x2 + y 2 )1/2 , is separable in polar coordinates (r, θ).
Now suppose that f (r) = (1 − r 2 ). Sketch the phase diagram associated with equation 4.16 and
explain why the circle S1 with equation x2 + y 2 = 1 is an invariant set.
Given that r(0) = 4, θ(0) = 0, show that in the subsequent motion the system approaches the
configuration S1 , but that it will not attain S1 in a finite time. Given the same initial conditions,
would these conclusions hold if f (r) = (1 − r 2 )2 ?
Writing
x = r cos θ, y = r sin θ
we obtain
−r θ̇ sin θ + ṙ cos θ = r sin θ + r cos θf (r),
r θ̇ cos θ + ṙ sin θ = −r cos θ + r sin θf (r).
Eliminating θ using the identity cos2 θ + sin2 θ = 1 we find that
ṙ = rf (r)
59
4.4. LIMIT CYCLES
and similarly
θ̇ = −1.
If f (r) = 1 − r 2 we have
ṙ = v(r) = r(1 − r 2 ).
The phase diagram for this differential equation is as shown. We observe that r = 1 is a stable
fixed point; if r(0) = 1, then r(t) = 1 ∀t ≥ 0. Interpreted 2-dimensionally this means that if the
system starts on the circle S1 it remains on this circle for all time, the point (x, y) representing
the system moving round S1 in a clockwise sense given by θ̇ = −1. The circle S1 is an invariant
cycle of the system.
v(r)
r
1
Figure 4.5: Phase-portrait of the radial motion described by the velocity function v(r) = rf (r) =
r(1 − r 2 ), exhibiting a stable fixed point at r = 1, which corresponds to a stable limit cycle of
the 2D motion.
We see from the phase diagram associated with the differential equation for r that if r(0) = 4
the system moves towards r = 1, but it will not actually reach r = 1 in a finite time, as we show
below. The point (x, y) representing our system winds round the origin at a constant rate given
by θ̇ = −1 and as it does so gets closer and closer to the configuration S1 .
That the time to reach the circle S1 is infinite may be seen as follows. The time τ (r1 ) for the
system to go from r = r0 > 1 to r = r1 is given by
τ (r1 ) =
so that
τ (r1 ) ≥
Z
r0
r1
Z
r1
r0
dr
=
r(1 − r)(1 + r)
Z
r0
r1
dr
r(r − 1)(r + 1)
dr
1
=
(− ln(r1 − 1) + ln(r0 − 1)).
r0 (r0 + 1)(r − 1)
r0 (r0 + 1)
We see that τ (r1 ) → ∞ as r1 → 1; the system does not reach the configuration S1 in a finite
time.
60
CHAPTER 4. SECOND ORDER AUTONOMOUS SYSTEMS
If f (r) = (1 − r 2 )2 the phase diagram has the form shown. If r(0) = 4 it is clear that, although
the circle S1 is still an invariant cycle, the system moves off to r = ∞ very rapidly. Viewed 2dimensionally the representative point (x, y) spirals round the circle S1 but moves out to infinity
very quickly.
v(r)
r
1
Figure 4.6: Phase-portrait of the radial motion described by the velocity function v(r) = rf (r) =
r(1 − r 2 )1 , exhibiting an unstable fixed point at r = 1, which corresponds to an unstable limit
cycle of the 2D motion.
Exercise 4.2 Find the fixed points of the system described by
ẋ = −3y + xy − 4, ẏ = y 2 − x2
Exercise 4.3 Find the fixed point of the system of differential equations
ẋ = exp (x + y) − y, ẏ = xy − x
and investigate its stability.
Exercise 4.4 Find the fixed point of the system of differential equations
ẋ = x + 3y, ẏ = −6x + 5y
and investigate its stability.
Exercise 4.5 Write the equation
ẍ + x = f (x)
in standard form by putting y = ẋ. Assuming that f is suitably differentiable and that f (0) = 0
discuss the stability of the fixed point (x = 0, y = 0), distinguishing the cases f 0 (0) > 1, f 0 (0) < 1.
61
4.4. LIMIT CYCLES
Note We could have guessed the answer by the following argument. Near x = 0
f (x) ' f (0) + xf 0 (0) + · · · so that the differential equation is approximated by
ẍ + (1 − f 0 (0))x ' 0.
If f 0 (0) < 1 the motion is essentially Simple Harmonic and therefore the fixed point should be
stable.
Exercise 4.6 Write the differential equation
ẍ + 2λẋ + ω 2 x = 0
(the equation of damped harmonic motion —see Exercise 5.13) in standard form by putting
y = ẋ. Find the fixed point and discuss its stability.
Exercise 4.7 A second order dynamical system is described by the differential equations
ẋ = −y + x sin πr, ẏ = x + y sin πr,
where r = (x2 + y 2 )1/2 . Prove that the system is separable in polar coordinates and show in
particular that
ṙ = r sin πr.
Sketch the phase diagram associated with this equation write down the fixed points, and indicate
which are stable and which are unstable.
Describe qualitatively, without computation, how the representative point (x, y) behaves as a
function of time t given the initial conditions r(0) = 3/2, θ(0) = 0. What is your conclusion if
r(0) = 5/2, θ(0) = 0.?
Exercise 4.8 Consider the system of differential equations
ẋ = (ay − b)x + f (x), ẏ = (c − dx)y + g(y),
where a, b, c, d are positive constants, and f, g are differentiable functions such that
f (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 equations
with respect to this fixed point gives
d
dt
X
Y
=
α
−bd/a
ac/d
β
X
Y
≡J
X
Y
where α = f 0 (c/d), β = g0 (b/a), X = x − c/d, Y = y − b/a.
Show that the matrix J has eigenvalues λ given by
λ=µ±
p
ν 2 − bc, µ =
α+β
α−β
, ν=
.
2
2
62
CHAPTER 4. SECOND ORDER AUTONOMOUS SYSTEMS
Assuming that det J 6= 0, discuss the stability of the fixed point (c/d, b/a) in the following cases:
(i) µ > 0, ν 2 ≥ bc, (ii) µ > 0, ν 2 < bc,
(iii) µ < 0, ν 2 > bc, αβ > −bc, (iv) µ < 0, ν 2 ≤ bc,
(v) µ = 0, ν 2 > bc, (vi) µ = 0, ν 2 < bc.
Why has the case µ = 0, ν 2 = bc been omitted?
Exercise 4.9 A second order dynamical system is described by the equations
d
dt
x
y
=J
x
y
, 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. You
should find that
a
= 0,
(λI − J)
b
where I is the 2 × 2 unit matrix—this should ring a bell! Consider taking appropriate linear
combinations of solutions. In this context note that if the column vectors X1 , X2 are solutions
of
d
LX = JX, L = ,
dt
then (αX1 + βX2 ) is also a solution since
L(αX1 + βX2 ) = αLX1 + βLX2 = αJX1 + βJX2 = J(αX1 + βX2 ).
Exercise 4.10 A second order dynamical system is described by the system of differential equations
ẋ = y + xf (R), ẏ = x + yf (R), R = (x2 − y 2 )1/2 .
Show that the system is separable in the coordinates (R, φ) where
x = R cosh φ, y = R sinh φ.
Suppose now that
f (R) = (R − 1)(R − 2)(R − 3).
Show that the rectangular hyperbolae Hn , n = 1, 2, 3, where
Hn = {(x, y) : x2 − y 2 = n2 , n = 1, 2, 3},
are invariant sets for the system.
Part II
Application to Classical Mechanics
63
Chapter 5
Elements of Newtonian Mechanics
In the present and the following chapters, we are going to apply the general methods introduced to study dynamical systems to problems of classical mechanics. We begin by introducing
elementary facts of Newtonian Mechanics, starting with kinematics — i.e., the description of
particle motion — thereafter continuing with dynamics proper, i.e. the investigation of the
dynamical laws governing particle motion in mechanics. In particular, we look at Newtons law
of gravitation, and we study motion on a straight line as a special realisation of second order
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 particle motion in one
dimension. Fixed points and their stability are discussed in terms of potential energy. It turns
out, that autonomous mechanical systems in one dimension can only have two types of fixed
point — unstable hyperbolic ones, and (marginally) stable elliptic ones.
5.1
Motion of a particle
What is a particle? We regard a particle as a physical object, possessing a mass, whose dimensions 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 good approximation 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 indeed an intuitive
term, but we shall see how the concept may be made precise once we have stated Newton’s Laws
of motion.
We assume that space is Euclidean and that it can be described in terms of a system of three
dimensional Euclidean position vectors measured relative to some conveniently chosen origin.
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 to describe
a particle’s motion an observer O may choose a reference frame consisting of a right handed
system of Cartesian axes Oxyz, usually fixed relative to the observer; the position of a particle
P at a particular instant is then made precise by assigning its position vector r = OP.
It is assumed that the observer O possesses a clock with which to measure time t. Generally
speaking, the particle’s position will vary with time and we may indicate this by writing r = r (t).
In classical Newtonian theory it is assumed that time is it absolute in the sense that two different
observers, whatever their state of relative motion, can synchronise their clocks in such a way that
65
66
CHAPTER 5. ELEMENTS OF NEWTONIAN MECHANICS
they both assign the same value of time t to a particular event. (This assumption was shown to
be untenable by Einstein in 1905 when he enunciated the Special Theory of Relativity.)
We continue in the above notation. O denotes an observer, Oxyz an associated reference frame,
and r(t) the position vector of a particle P at time t. We make the following definitions:
Definition 5.1 The velocity v (t) of the particle (as measured by O) at time t is defined by
v (t) =
r(t + δt) − r(t)
dr
= lim
.
δt→0
dt
δt
(5.1)
The velocity is the rate of change of the position vector.
Definition 5.2 The speed of the particle (as measured by O) at time t is the magnitude of its
velocity i.e. |v | = |ṙ |.
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 at time
t, and Pt+δt its position at time t + δt, so that r(t), r (t + δt) are the position vectors of Pt , Pt+δt
respectively. We note that
r(t + δt) − r(t)
Pt Pt+δt
=
.
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 5.3 The acceleration a(t) (as measured by O) of the particle at time t is defined by
a(t) =
d2 r
dv
= 2.
dt
dt
(5.2)
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 a
scalar or vector function with respect to time. We write
df
dr
d2 r
f˙ ≡ , ṙ ≡
, r̈ ≡ 2 ,
dt
dt
dt
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 + y e2 + z e3 ,
where e1 , e2 , e3 denote the unit vectors parallel to the axes Ox, Oy, Oz respectively. The
particle’s velocity is then
v = ẋe1 + ẏ e2 + ż e3 .
67
5.1. MOTION OF A PARTICLE
(It is assumed that e1 , e2 , e3 are fixed relative to the observer O—so that their time derivatives
are all zero.) Similarly the acceleration of the particle at time t is
a = ẍe1 + ÿ e2 + z̈ e3 .
We may calculate the distance travelled by the particle between times t = t0 and t = t1 as
follows. 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
Z
t1
t0
|ṙ | dt =
Z
t1
t0
|v | dt.
In other words, the distance travelled by the particle between times t0 and t1 is obtained by
integrating the speed between these limits. Note that the distance travelled is (generally) not
equal to |r (t1 ) − r(t0 )|.
Note We emphasise that the concepts of velocity, speed and acceleration have been defined
with 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.
small
Example 5.1 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 cos ωte1 + b sin ωte2 ,
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. At
time t the position vector OP of the particle makes an angle θ = ωt with the x–axis, and since
θ̇ = ω we see that the particle moves round the circle in an anti–clockwise sense at a constant
angular speed θ̇ = ω.
We note that the particle’s velocity is given by
v (t) = ṙ = −bω sin ωte1 + bω cos ωte2 .
It is clear that v .r = 0 so that the vector v is indeed tangential to the circle. The speed of the
particle is equal to |v | = (b2 ω 2 cos2 ωt + b2 ω 2 sin2 ωt)1/2 = bω. Thus the speed is constant, and
equal to bω.
68
CHAPTER 5. ELEMENTS OF NEWTONIAN MECHANICS
The particle’s acceleration is given by
a = v̇ = −bω 2 cos ωte1 − bω 2 sin ωte2 = −ω 2 r .
We conclude that the acceleration vector is always directed from P towards the centre O of the
circle, and that it has magnitude bω 2 6= 0.
We should not be too surprised by this result since, although the magnitude of the velocity
vector 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 the fact
that the speed (magnitude of the velocity vector) is constant.
Example 5.2 Uniform Motion
A particle moves with constant velocity V relative to an observer O. Find its path, given that at t = 0
its position vector with respect to O is r 0 .
We have
ṙ = V ,
and integration gives
r(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 the point
with position vector r0 , and whose direction is given by the vector V .
Example 5.3 Motion with constant acceleration
A particle moves so that its coordinates (x, y, z) with respect to Cartesian axes Oxyz are given by the
differential equations
ẍ = 0, ÿ = 0, z̈ = −g,
where g is a positive constant. The initial conditions are such that
ẋ(0) = U, ẏ(0) = 0, ż(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 O0 is coincident with O at time t = 0, and moves with constant velocity U e1 relative to O.
O0 describes the motion of the particle in terms of coordinates x0 , y 0 , z 0 with respect to Cartesian axes
O0 x0 y 0 z 0 , where O0 x0 is parallel to Ox, O0 y 0 is parallel to Oy, and O0 z 0 is parallel to Oz.
Write down a set of equations relating x, y, z and x0 , y 0 , z 0 at time t. Describe geometrically the motion
of the particle as viewed by O and O0 .
Integrating the given equations, and using the stated initial conditions to evaluate the constants of
integration gives:
ẋ(t) = A = ẋ(0) = U, x(t) = U t + A0 = U t (x(0) = 0),
ẏ(t) = B = ẏ(0) = 0, y(t) = B 0 = 0 (y(0) = 0),
69
5.2. NEWTON’S LAWS OF MOTION
1
1
ż(t) = −gt + C = −gt + V, (ż(0) = V ), z(t) = V t − gt2 + C 0 = V t − gt2 (z(0) = 0).
2
2
Summarising we have
1
x(t) = U t, y(t) = 0, z(t) = V t − gt2 .
2
Eliminating t now gives
Vx
gx2
g
V2
2
z=
−
=
−
(x
−
U
V
/g)
.
+
U
2U 2
2U 2
2g
Writing Z = z − V 2 /(2g), X = x − U V /g, we see that the path is given by Z = − 2Ug 2 X 2 . We conclude
that the path of the particle as viewed by observer O is a parabola. (See the diagram)
z
Z
z
z’
X
x
P
x’
Ut
x
O
O
O’
x’
Figure 5.1: Relativity of description of motion
The observer O0 see things rather differently. Since
dOO0
= U e1 ,
dt
and O, O0 coincide at time t = 0, it follows that
OO0 (t) = tU e1 .
Referring to the diagram we see that the equations relating x, y, z and x0 , y 0 , z 0 are
x0 = x − U t, y 0 = y, z 0 = z,
so that
1
x0 (t) = 0, y 0 (t) = 0, z 0 (t) = V t − gt2 .
2
The path of the particle as viewed by O0 is therefore a straight line.
5.2
Newton’s Laws of motion
We now discuss the laws which govern the motion of a Newtonian particle. Newton assumed the
existence of a preferred class of observers and associated reference frames, called inertial frames
of reference, with respect to which Newton’s laws of motion were supposed to hold.
70
CHAPTER 5. ELEMENTS OF NEWTONIAN MECHANICS
First, every particle P possesses an inertial mass m > 0 which is normally assumed to be
constant, whatever the state of motion of the particle. If r(t) is the position vector of P relative
to an observer O its momentum p is defined by the equation
p = mṙ = mv,
where v is the particle’s velocity.
5.2.1
Newton’s First Law (N1)
Newton’s First Law states that with respect to an inertial frame of reference a particle moves
with constant velocity in a straight line unless constrained to depart from this state by a force
acting on it. The nature of the term force is made precise through the Second and Third Laws.
5.2.2
Newton’s Second Law (N2)
Newton’s Second Law states that with respect to an inertial frame of reference the rate of change
of momentum of a particle is equal to the total force acting on the particle. If F 1 , F 2 , . . . , F n
are the forces acting on a particle P, whose position vector is r(t) with respect to the origin O
of an inertial frame of reference, we have
d
(mṙ ) = F ,
dt
where the total force F is given by
F =
n
X
F i.
i=1
Assuming that m is constant we may write
mr̈ = F .
In words, the mass times the acceleration of the particle is equal to the total force acting on the
particle.
5.2.3
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 particle
i on particle j is F ji and
F ij + F ji = 0,
or
F ij = −F ji .
The forces in question could be gravitational or, in the case of two particles connected by a taut
string, the tension force provided by the string.
We now show, very briefly, that Newton’s laws determine a scale of mass, once an arbitrary unit
of mass has been chosen. Consider two particles 1, 2 with inertial masses m1 , m2 respectively.
71
5.2. NEWTON’S LAWS OF MOTION
Suppose that particle 2 exerts a force F 12 on particle 1 and that particle 1 exerts a force F 21
on particle 2. If a1 , a2 denote the accelerations of the two particles with respect to some inertial
frame of reference we have, using N2,
m1 a1 = F 12 , m2 a2 = F 21 .
Adding these equations, and using Newton’s Third Law, we obtain
m1 a1 + m2 a2 = F 12 + F 21 = 0.
It follows that
(5.3)
m1
|a2 |
=
.
m2
|a1 |
By measuring the accelerations a1 , a2 we can determine the ratio m1 /m2 , and having chosen
particle 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 5.3 we note that if r1 , r 2 denote the position vectors of the two particles
with respect to the origin O of an inertial frame of reference then
m1 r¨1 + m2 r¨2 = 0,
and therefore
d2
(m1 r1 + m2 r 2 ) = 0.
dt2
The centre of mass of the two particle system has position vector R defined by
r=
(5.4)
m1 r 1 + m2 r 2
.
m1 + m2
It now follows from equation 5.4 that
d2 r
= 0.
dt2
We may interpret this result by saying that the centre of mass of the two particle system is
unaccelerated with respect to any inertial frame of reference, and therefore moves with constant
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 inertial masses
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 may
reasonably ask “Are there any such frames?” Einstein and Infeld in their book “The Evolution
of Physics” show that this question poses serious difficulties and it is not possible to give an
example of a frame of reference which is strictly inertial! Well, one may ask: “Would a reference
frame with origin on the Earth’s surface, and fixed relative to the Earth, constitute an inertial
frame of reference?” Because of the Earth’s rotation this is not strictly the case; nevertheless
72
CHAPTER 5. ELEMENTS OF NEWTONIAN MECHANICS
it is a reasonable approximation—in the sense that an application of Newton’s laws in such a
frame leads to fairly satisfactory predictions. By making a better choice of reference 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 we are
unable to exhibit a reference frame in which they are strictly valid! However, if one such inertial
frame F1 exists, then all frames moving with constant velocity relative to F1 are also inertial.
(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 knowledge
of certain initial conditions (its initial position and velocity), we require explicit mathematical
formulae for the force(s) acting on the particle; of particular importance are gravitational forces.
5.3
Newton’s Law of Gravitation
Consider two particles of masses m1 , m2 respectively and suppose that the position vector of m2
with respect to m1 is r. The the gravitational force F exerted on m2 by m1 is given by
F =−
Gm1 m2
r, r = |r |,
r3
(5.5)
where G is a universal constant known as Newton’s constant of gravitation.
The force F has magnitude |F | = Gm1 m2 /r 2 so that the magnitude of the gravitational force
varies inversely as the square of the distance between the particles; the - sign indicates that m2
is attracted towards m1 .
Strictly speaking, the masses which occur in equation 5.5 are gravitational masses; they are a
measure of the “strength” of the gravitational field which the particle generates. However, in a
famous experiment conducted from the leaning tower of Pisa, Galileo showed that all particles
experience the same acceleration in a gravitational field. As a consequence, we can deduce that
the ratio of inertial mass to gravitational mass is the same for all particles, and by an appropriate
choice of units we can assume that the inertial and gravitational masses of a particle are equal.
In what follows, therefore, we shall 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 arising
from the whole Earth. If we assume that the Earth is a uniform sphere of radius a, and total
mass M, then it can be shown that the gravitational force experienced by a particle at a point
above the Earth’s surface is the same as the gravitational force which would be exerted on the
particle by a particle situated at the centre of the Earth, with a mass equal to the total mass
of the Earth. This means that the gravitational force F exerted on a particle of mass m, whose
position vector is r relative to O, the centre of the Earth, is given by
F =−
GmM
r.
r3
(5.6)
5.3. NEWTON’S LAW OF GRAVITATION
73
If the particle remains near the Earth’s surface throughout the motion, so that r/a is always
close to 1, we may replace equation 5.6 by
F =−
GmM
er = −mg er ,
a2
(5.7)
where er is the unit vector parallel to r and g = GM/a2 . In the approximation which we have
described the gravitational force on the mass m has magnitude mg and is directed along the
downward vertical towards the centre of the Earth. We shall employ this notation in what
follows without further explanation.
Example 5.4 Motion near Earth’s Surface
A particle P of mass m is projected vertically upwards from a point A on the Earth’s surface with speed
V. Discuss the subsequent motion, assuming that the particle remains near the Earth’s surface throughout
the motion.
We assume that we can neglect air resistance and that the only force acting on the particle is in fact
gravitational.
Let O denote the Earth’s centre. We choose axes Oxyz which are fixed relative to the Earth and which
are such that the z-axis points vertically upwards, through the point of projection; as noted previously
such 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, ż(0) = V.
Application of Newton’s Second Law (N2) gives
mz̈e3 = −mge3 , z̈ = −g.
Integrating with respect to time t we find that
ż = −gt + C,
where C is a constant. Since ż(0) = V , it follows that z = −gt + V. A further integration with respect to
t immediately gives
1
z(t) = a + V t − gt2 .
2
The maximum height of the particle above the Earth’s surface is attained when ż = 0 i.e. when t = V /g.
The corresponding value of z is then given by
z =a+
V2
1 V2
V2
− g 2 =a+
.
g
2 g
2g
We conclude that the particle rises to a height V 2 /(2g) above the Earth’s surface in a time V /g before
falling back towards the point of projection.
We have assumed throughout this calculation that the gravitational force exerted on m by the
Earth is constant; if this is to be a reasonable assumption, we would expect that the maximum
height which we have just calculated should be a good deal less than a, the Earth’s radius i.e.
V2
a, V 2 2ag.
2g
We shall see below that V = (2ag)1/2 does indeed have a special significance.
74
CHAPTER 5. ELEMENTS OF NEWTONIAN MECHANICS
Example 5.5 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 is
comparable with 2ag. In this case we have to use the exact expression for the gravitational force, without
any approximations. The equation of motion is
mz̈e3 = −
GmM
GM
e3 , z̈ = − 2 .
2
z
z
Since z̈ = −g when z = a it follows that GM = ga2 and we may write
ga2
.
z2
z̈ = −
Multiplying this equation by ż and noting that
d 1 2
( ż ) = ż z̈
dt 2
we obtain
ga2 dz
d 1 2
( ż ) = − 2
.
dt 2
z dt
Integration with respect to t yields
1 2
ż = −
2
ga2
ga2
dz
+
C
=
+ C,
z2
z
Z
where C is constant. Using the initial conditions z(0) = a, ż(0) = V we obtain
C = 12 V 2 − ag and the equation for z becomes
ż 2 =
2ga2
+ V 2 − 2ag.
z
(5.8)
The maximum value of z, z = zmax , say, is obtained by setting ż = 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 speed
of the particle. Substituting the relevant values of a, g we find that the escape speed is approximately
11.2km/sec.
For values of V which are less than (2ag)1/2 the particle rises to z = zmax before falling back to the point
of projection. Re-writing equation 5.8 in terms of zmax we find that
ż 2 =
2ga2 (zmax − z)
.
zzmax
If we require to compute the time τ for the particle to reach zmax we extract the square root and separate
the variables to obtain
τ=
Z
0
τ
dt =
zmax
2ga2
1/2 Z
a
zmax
z
zmax − z
1/2
dz.
This integral can be evaluated by means of the substitution z = zmax sin2 θ.
5.4. MOTION IN A STRAIGHT LINE; THE ENERGY EQUATION
5.4
75
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 line along
the x-axis under the influence of a force F e1 . The equation of motion is
mẍe1 = F e1 , mẍ = F.
We now make the special assumption that F depends only on x, the position coordinate of the
particle so that F = F (x) and the equation of motion becomes
mẍ = F (x).
Multiplying by ẋ, and recalling that
d 1 2
( ẋ ) = ẋẍ,
dt 2
we obtain
dx
1
mẋ2 = F (x) dt + C =
2
dt
where C is a constant of integration. We write
Z
F (x) dx + C,
1
T = mẋ2 , V (x) = −
2
Z
F (x) dx
Z
(5.9)
and refer to T as the kinetic energy of the particle and to V as the potential energy function of
the particle. In terms of these definitions the last equation may be written
T + V = C = constant.
(5.10)
T + V, the sum of the kinetic and potential energies of the particle, is called the total energy
of the particle and equation 5.10 then states that the total energy is constant throughout the
motion—it is a constant of the motion. We emphasise that this result depends on the assumption
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 potential
energy V is only unique up to a constant. We note that equation 5.9 implies
F (x) = −
dV
.
dx
(5.11)
We may say that the force acting on the particle is minus the gradient of the potential energy
function—even though this is slightly sloppy, since the actual force is not F (x) but F (x)e1 .
Example 5.6 Potential of Constant Gravitational Force
Suppose that the axis Ox points along the upward vertical and that the particle is moving along Ox
under constant gravity so that the force acting is −mge1 .
The potential energy function V is given by
−
dV
= −mg, V (x) = mgx + C.
dx
76
CHAPTER 5. ELEMENTS OF NEWTONIAN MECHANICS
If we arbitrarily require that V (0) = 0 we may set C = 0 and write V (x) = mgx. In this case the energy
equation 5.10 reads
1
mẋ2 + mgx = constant.
2
Given a set of initial conditions i.e. the values of x and ẋ at t = 0 we could evaluate the constant
explicitly.
Example 5.7 Simple Harmonic Motion
Suppose that the particle is moving along the x-axis under the action of a force −mω 2 xe1 , where ω is
constant.
First note that, whatever the sign of x, the force is always directed towards the origin O. The potential
energy V is given by
1
dV
= −mω 2 x, V (x) = mω 2 x2 .
−
dx
2
(We have arbitrarily required that V (0) = 0 in order to set the constant of integration equal to zero.)
The energy equation reads
1
1
mẋ2 + mω 2 x2 = constant,
2
2
which we can re-write in the form
ẋ2 + ω 2 x2 = a2 ,
where a is a constant. We could integrate this equation to find x as a function of t but it is perhaps
easier to go back to Newton’s Second Law:
mẍ = −mω 2 x, ẍ + ω 2 x = 0.
We know from our work on differential equations that this equation has general solution
x(t) = α cos ωt + β sin ωt,
where α, β are constants. It is well known that the right hand side of this equation can be expressed in
the form A sin(ωt + ), where a, are constants and we may therefore write the general solution as
x(t) = A sin(ωt + ).
(5.12)
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 oscillatory
motion between the points x = ±A. A is called the amplitude of the motion. The centre of the motion is
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 the
frequency of the oscillation, and is usually denoted by ν. We see that
ν=
ω
1
=
.
τ
2π
The constant which appears in equation 5.12 is sometimes referred to as the phase angle. The motion
which we have just described is said to be Simple Harmonic and the equation ẍ + ω 2 x = 0 is called the
differential equation of Simple Harmonic motion.
5.4. MOTION IN A STRAIGHT LINE; THE ENERGY EQUATION
77
Example 5.8 Simple Harmonic Motion II
Suppose instead that the force acting is −mω 2 (x − a)e1 , where a is constant.
In this case the equation of motion becomes
mẍ = −mω 2 (x − a), ẍ + ω 2 (x − a) = 0.
We write z = x − a and obtain the equation z̈ + ω 2 z = 0 (The differential equation of Simple Harmonic
motion). This equation has general solution
z(t) = A sin(ωt + ),
where A > 0, are constants. Expressing this result in terms of x we obtain
x(t) = a + A sin(ωt + ).
We see that this motion is also simple harmonic with amplitude A, and period τ = 2π/ω. The oscillation
takes 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 5.9 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 particle
of 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 the subsequent
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-coordinate of
the mass m at time t. The forces acting on the particle are
(i) The force of gravity, mge1 ,
(ii) The force −T e1 due to the tension in the string— which acts on the particle in the sense of the
upward vertical. It is necessary to postulate this force - it is the force which balances gravity when the
particle is hanging in equilibrium, and it continues to act throughout the subsequent motion.
Application of N2 gives
mẍe1 = mge1 − T e1 .
In order to make progress we have to make some assumption as to how T depends on x. If we assume
Hooke’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 assumes
that the tension is proportional to the extension of the string beyond its natural length. The equation of
motion now becomes
mẍe1 = mge1 − k(x − b)e1 .
Writing z = x − b our equation becomes
z̈ + ω 2 z = g, ω 2 = k/m.
(5.13)
The homogeneous equation z̈ + ω 2 z = 0 has general solution of the form z = A sin(ωt + ). We note that
z = g/ω 2 is a particular solution of equation 5.13 and we conclude that its general solution is
z(t) = A sin(ωt + ) + g/ω 2 .
78
CHAPTER 5. ELEMENTS OF NEWTONIAN MECHANICS
Initially the system is in equilibrium so that z̈ = 0; equation 5.13 shows that z(0) = g/ω 2 . The system
is set in motion with initial speed u in the sense of the downward vertical so ż(0) = u. Imposing these
initial conditions on the general solution we deduce that
A sin = 0, Aω cos = u.
These equations are satisfied by = 0, A = u/ω. Hence
z(t) =
u
g
sin ωt + 2 ,
ω
ω
or, written in terms of x
u
g
sin ωt + 2 .
ω
ω
We conclude that the particle executes Simple Harmonic motion about x = b +
equilibrium value
p of x, before the blow is struck); the periodu is
τ = 2π/ω = 2π m
k and the amplitude of the oscillation is ω .
x(t) = b +
g
ω2
as centre (this is the
Note that the potential energy function V is given by
−
k
dV
= mg − k(x − b), V (x) = −mgx + (x − b)2 + C,
dx
2
where C is a constant.
Note that the least value of x(t) throughout the motion is equal to b − ωu + ωg2 and that for the string
to be in a state of tension throughout the entire motion this must be greater then b. This implies that
u < g/ω—a constraint on the magnitude of the initial speed of the particle.
5.5
Equilibrium and Stability
Consider a particle of mass m, moving along the x-axis under a potential V (x), so that the force
F acting on m is given by
dV
F =−
e1 = −V 0 (x)e1 .
dx
By N2 the equation of motion is
mẍe1 = −V 0 (x)e1 , ẍ = −
V 0 (x)
m
(5.14)
Putting x1 = x, x2 = ẋ we find that
x˙1 = x2 , x˙2 = −
V 0 (x1 )
.
m
(5.15)
Let x = a be a stationary point of the potential function V, so that V 0 (a) = 0. Suppose also
that the initial conditions are x(0) = a, ẋ(0) = 0; this means that at t = 0 the particle is
instantaneously at rest at x = a. In terms of x1 , x2 the initial conditions become
x1 (0) = a, x2 (0) = 0.
(5.16)
Assuming that V has continuous second order derivatives we deduce from Picard’s Theorem that
the unique solution of equations 5.15 subject to the initial conditions expressed by equations 5.16
is
x1 (t) = a, x2 (t) = 0, ∀t ≥ 0.
79
5.5. EQUILIBRIUM AND STABILITY
V(x)
V(a)
x
a1
a
a
2
Figure 5.2: Stable periodic motion in the vicinity of a minimum of the potential energy.
Reverting to x, ẋ this means that the particle remains at rest at x = a for all t ≥ 0. We say
that x = a is an equilibrium position for the particle; the possible equilibrium positions are the
stationary points of V.
Suppose now that the particle, at rest at the equilibrium point x = a, is set in motion with
velocity ue1 . In the subsequent motion energy is conserved and we have
1
1
mu2 + V (a) = mẋ2 + V (x),
2
2
so that
1
1
mẋ2 = mu2 + V (a) − V (x).
2
2
(5.17)
First suppose that x = a is a minimum of V . Since 12 mẋ2 ≥ 0 we deduce from equation 5.17
that
1
V (x) ≤ mu2 + V (a).
(5.18)
2
Referring to the diagram we see that, if u is sufficiently small, the particle’s motion will be
confined to the interval [a1 , a2 ], where a1 , a2 are solutions of the equation
1
V (x) = mu2 + V (a).
2
Perhaps it will be helpful to describe the motion in a little more detail. To be definite, suppose
that u > 0 so that the particle moves towards a2 . As it does, V (x) increases, and the particle
loses kinetic energy, coming to instantaneous rest at x = a2 . It cannot move to the right of a2 ,
otherwise 21 mẋ2 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 instantaneous rest once
more at the point x = a1 , before moving back to x = a2 . We see that the particle oscillates
between x = a1 and x = a2 .
In conclusion, we see that if |u| is small enough, the particle will remain “near” x = a, in the
sense described, and we are justified in saying that x = a is a position of stable equilibrium.
80
CHAPTER 5. ELEMENTS OF NEWTONIAN MECHANICS
Suppose, on the other hand, that x = a is a maximum of the potential function V and that the
particle, initially at rest at x = a is set in motion with velocity ue1 .As the particle moves away
from x = a the potential V (x) decreases and therefore 21 mẋ2 increases by virtue of equation 5.17.
The particle does not remain “near” x = a and we are justified in saying that x = a is a position
of unstable equilibrium.
The above arguments are based on the energy equation and graphical considerations. Alternatively, we may proceed as follows. Since x = a is an equilibrium point,
V 0 (a) = 0. The equation of motion is (see equation 5.14)
ẍ = −
V 0 (x)
m
Near x = a Taylor’s formula gives
V (x) = V (a) +
V 0 (a)(x − a) V 00 (a)(x − a)2
+
+ ···
1!
2!
Assuming that x is “near” a, and that V 00 (a) 6= 0, we have
1
V (x) ' V (a) + V 00 (a)(x − a)2
2
and the equation of motion approximates to
1
V 00 (a)
1 d
(x − a).
V (a) + V 00 (a)(x − a)2 = −
ẍ = −
m dx
2
m
Writing z = x − a we obtain
V 00 (a)
z = 0.
m
We have to deal with two cases, depending on the sign of V 00 (a).
z̈ +
Case 1 Suppose that V 00 (a) > 0, so that x = a is a minimum of V. Writing
ω2 =
V 00 (a)
>0
m
we obtain
z̈ + ω 2 z = 0.
This equation has general solution
z = A sin(ωt + ),
where A, are constants. Expressing z in terms of x we now have
x(t) = a + A sin(ωt + ).
The initial conditions x(0) = a, ẋ(0) = u require that A sin = 0, Aω cos = u from which it
follows that = 0, A = u/ω. Therefore
x(t) = a +
u
sin ωt.
ω
81
5.5. EQUILIBRIUM AND STABILITY
In the approximation considered, the particle executes Simple Harmonic motion about x = a
with amplitude u/ω and period 2π/ω. This result agrees with our earlier conclusions and we see
that a minimum stationary point of V is a position of stable equilibrium.
Case 2 Now suppose that V 00 (a) < 0, so that x = a is a maximum of V. Writing
Ω2 = −
V 00 (a)
>0
m
we obtain
z̈ − Ω2 z = 0.
This equation has general solution
z(t) = αeΩt + βe−Ωt ,
x(t) = a + αeΩt + βe−Ωt ,
where α, β are constants. Imposing the initial conditions x(0) = a, ẋ(0) = u we find that
x(t) = a +
u
sinh Ωt.
Ω
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: a
maximum of V is a position of unstable equilibrium.
Example 5.10 Motion in a Cubic Potential
A particle of unit mass moves along the axis Ox under a 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 regarded
as negligible
√ in comparison with (x − 2) then the period τ of small oscillations about x = 2 is given by
τ = 2π/ 6.
√
Suppose that x(0) = 2, ẋ(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 have
V 0 (x) = 6x2 − 18x + 12 = 6(x2 − 3x + 2) = 6(x − 1)(x − 2).
The possible equilibrium positions are given by V 0 (x) = 0 i.e. by x = 1, x = 2. The equation of motion
is
ẍ = −V 0 (x), ẍ + 6(x − 1)(x − 2) = 0.
Writing z = x − 2 we derive
z̈ + 6z(z + 1) = 0.
Assuming that terms in z 2 are negligible in comparison with the linear terms in z we obtain as an
approximation the differential equation
z̈ + 6z = 0.
This describes Simple Harmonic
√motion about the point z = 0 i.e. about x = 2 and the period τ of the
oscillations is given by τ = 2π/ 6.
82
CHAPTER 5. ELEMENTS OF NEWTONIAN MECHANICS
V(x)
x
1
2
Figure 5.3: The potential that allows stable periodic motion in the vicinity of x = 2; the
minimum energy needed to escape to −∞ is shown as a dashed line.
Referring to the graph of V we see that if the particle is projected from x = 2 with velocity −ue1 it will
move to the left of x = 2. As it does so energy is conserved:
1
1 2
ẋ + V (x) = u2 + V (2).
2
2
If the particle still has positive kinetic energy when it reaches x = 1 (a maximum of V ) then it will move
off to −∞, since V decreases (rapidly) for x < 1. The condition for this to happen is
1 2
u + V (2) − V (1) > 0.
2
Evaluating V (1), V (2) we find that the condition for the particle to move off to −∞ is u >
√
2.
Example 5.11 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 and a
maximum at x = a.
Given that x(0) = −a, ẋ(0) = u, where u is suitably small, discuss the subsequent motion. By reference
to your graph describe what will happen to the particle if, instead, the initial conditions are x(0) =
a, ẋ(0) = u, where u can be either positive or negative.
We find that
c(a2 − x2 )
2cx(x2 − 3a2 )
00
V 0 (x) = 2
,
V
(x)
=
.
(a + x2 )2
(a2 + x2 )3
Now V 0 (x) = 0 when x = ±a. It is clear from consideration of the graph of V that x = −a is a minimum
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
V 0 (−a)(x + a) V 00 (−a)(x + a)2
dV
d
V (−a) +
mẍ = −
=−
+
+ · · · = −V 00 (−a)(x + a),
dx
dx
1!
2!
83
5.5. EQUILIBRIUM AND STABILITY
V(x)
x
−a
a
Figure 5.4: Potential that allows stable periodic motion in the vicinity of x = −a; depending on
initial conditions, escape to ±∞ is possible.
on the assumption that the higher powers of (x + a) are negligible. Writing z = x + a we obtain
z̈ +
V 00 (−a)
c
z = 0, z̈ +
z = 0.
m
2ma3
This equation has general solution
z = A sin(ωt + ), x(t) = −a + A sin(ωt + ),
where
c
.
2ma3
In the approximation which we have described the motion is Simple Harmonic about x = −a as centre,
and the period is
r
2ma3
.
τ = 2π
c
ω2 =
Imposing the initial conditions x(0) = −a, ẋ(0) = u, we find that
x(t) = −a +
u
sin ωt.
ω
Reference to the graph of V suggests that our approximation is likely to be reasonable if the amplitude
u/ω a i.e. if u ωa.
Suppose now that the initial conditions are x(0) = a, ẋ(0) = u. The energy equation yields
1
1
mẋ2 + V (x) = mu2 + V (a).
2
2
We see that in the subsequent motion 12 mẋ2 > 21 mu2 since V (x) < V (a). We deduce that if u > 0 the
particle will move off to +∞ and that if u < 0 the particle will move off to −∞.
Exercise 5.1 A particle moves so that its position vector with respect to the origin O of a
reference frame Oxyz is
r (t) = b cos ωt e1 + b sin ωt e2 + V te3 ,
where e1 , e2 , e3 are unit vectors parallel to the axes Ox, Oy, Oz; b, ω, V are positive constants.
84
CHAPTER 5. ELEMENTS OF NEWTONIAN MECHANICS
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.
Exercise 5.2 A particle has position vector r(t) = e1 + te2 + t2 e3 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 relevant
unit vector.)
Exercise 5.3 Two particles 1, 2 move in such a way that their position vectors r1 , r2 , with
respect to the origin O of an inertial frame of reference, are given by
r 1 (t) = e1 + te2 + t2 e3 ,
r2 (t) = te1 + t2 e2 + t3 e3 ,
respectively. Write down expressions for the velocity, speed and acceleration of the particles at
time t and show that, at the instant when they collide, the angle α between their instantaneous
directions of motion is given by
8
cos α = √ .
70
Exercise 5.4 A particle moves so that its position vector r with respect to the origin O of an
inertial frame of reference is
r = b cos ωte1 + b sin ωte2 + V t2 e3 ,
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
cos β =
(b2 ω 4
2V
.
+ 4V 2 )1/2
Exercise 5.5 O, O0 are the origins of two reference frames F, F 0 . At time t = 0
OO0 = a. Suppose that O0 moves with a constant velocity V relative to O. At time t a particle P
has position vector r(t) relative to O and r 0 (t) relative to O0 . Use a suitable vector diagram to
write down an equation relating r (t) and r0 (t). Deduce that if Newton’s second Law (N2) applies
in the frame F, it also applies with respect to the frame F 0 .
85
5.5. EQUILIBRIUM AND STABILITY
Exercise 5.6 Suppose that two particles 1, 2 of masses m1 , m2 act on each other in such a way
that the force 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 of reference
and deduce that the centre of mass of the 2 particle system moves with constant velocity with
respect to the chosen frame.
(Recall that the position vector r of the centre of mass is given by
r=
m1 r 1 + m2 r 2
,
(m1 + m2 )
where r1 , r 2 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, irrespective
of the choice of origin.
Hint Choose origins O, O0 and suppose that the two particles have position vectors r i , ri 0
(i = 1, 2) with respect to O, O0 respectively. Then
r i = OO0 + ri 0 , (i = 1, 2).
The centre of mass has position vector R, R0 with respect to the origins O, O0 where
R=
m1 r 1 + m2 r 2
,
(m1 + m2 )
R0 =
m1 r 1 0 + m2 r 2 0
.
(m1 + m2 )
Now verify that
R = OO0 + R0 .
Exercise 5.7 A particle of mass m moves along the x-axis under a constant force C e1 . Let x(t)
denote its coordinate at time t. Given the initial conditions
x(0) = a, ẋ(0) = u,
prove that
1
ẋ(t) = u + f t, x(t) = a + ut + f t2 , ẋ2 = u2 + 2f (x − a),
2
where f = C/m.
Note These formulae are only applicable if the force acting on the particle is constant; this is
seldom the case, yet many students treat these formulae as though they had universal validity
and could be used to find the answer to any problem in mechanics! Remember, solving a problem
in Newtonian mechanics involves setting up the equation of motion N2 with the relevant set of
initial conditions; if you proceed in this way you can’t go wrong!
86
CHAPTER 5. ELEMENTS OF NEWTONIAN MECHANICS
Exercise 5.8 A 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 is
at (0, 0, z) is F = −(mga2 /z 2 )e3 , m being the particle’s mass.
(We’re following the notation used earlier in this chapter, choosing the origin O at the Earth’s
centre, 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 of projection, is τ
where
Z
τ = (ag)−1/2
2a
a
z 1/2 dz/(2a − z)1/2 ,
and evaluate this integral by the substitution z = 2asin2 θ.
q
(The answer is (π/2 + 1)
a
g .)
Exercise 5.9 A 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
equation to show that the particle rises to a height 3a above the Earth’s surface before returning
to the point of projection.
Exercise 5.10 A particle of unit mass moves along the axis Ox of an inertial frame of reference
under the action of a force F = (k/x3 )e1 , where k is a positive constant. Given that the particle
starts from rest at x = a show that its speed at the point x = 2a is (3k/4a2 )1/2 . Compute the
time taken by the particle to move from x = a to x = 2a.
Exercise 5.11 A particle of unit mass moves along the x−axis under the action of a force
derived from a potential V (x) = − cos x. The particle, initially in equilibrium at x = 0, is set
in motion with speed u in the sense of the positive x−axis. Write down the equation of motion
of the particle and, assuming that |x(t)| remains small ( π/2) in the subsequent motion, find
x(t) in terms of u.
[You may assume that the replacement sin x ≈ 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 critical
value u0 (which should be found), the particle will move off to +∞.
Exercise 5.12 A particle of mass m moves along the x-axis under the potential
V (x) =
(a2
cx
,
+ x2 )2
where a, c are positive constants. Find the period of small oscillations about the position of stable
equilibrium.
87
5.5. EQUILIBRIUM AND STABILITY
Exercise 5.13 Referring to Example 5.9 suppose that we attempt to model the effect of air
resistance by supposing that the air exerts a force −2mλẋe1 on the particle, where λ is a positive
constant. The modified equation of motion is then
mẍe1 = mge1 − k(x − b)e1 − 2mλẋe1 ,
or, writing z = x − b,
z̈ + 2λż + ω 2 z = g,
ω 2 = k/m.
Find the general solution of this equation on the assumption that λ2 − ω 2 < 0. (This means that
damping due to air resistance is small).
Suppose that at time t = 0 the particle is hanging in equilibrium under gravity when it is struck
a blow which causes it to move with initial velocity ue1 in the sense of the downward vertical.
Find the displacement of the particle at time t, assuming that the string remains taut throughout
the motion.
We say that the particle is executing Damped Harmonic Motion; the equation
z̈ + 2λż + ω 2 z = 0, λ > 0
is sometimes referred to as the equation of Damped Harmonic motion.
Exercise 5.14 (King’s College, Summer 1995) Consider the motion of a particle of mass m
when it moves along a straight line under the action of forces F e1 derivable from potentials
V (x).
(a) Use Newton’s equation of motion to prove that the total energy of the particle is a constant
of the motion.
(b) When F = m(11x−2 − 36x−3 ) compute V (x). If the particle is initially at x = 1 and has
initial 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’s
equation of motion is approximated by
m
d2 z
+ V 00 (a)z = 0,
dt2
where z = x − a and V 00 = d2 V /dx2 .
Let V (x) = x3 + βx2 − γx, where β and γ are positive real numbers. Identify all the stable and
unstable equilibrium points and describe the motion near any stable equilibrium points.
88
CHAPTER 5. ELEMENTS OF NEWTONIAN MECHANICS
Chapter 6
Hamiltonian Systems
6.1
Hamilton’s equations
Consider a particle of mass m moving in a straight line along the q-axis (it is traditional in
Hamiltonian theory to use q rather than x) under a force F (q, t)e1 . Writing
F (q, t) = −
∂V
∂q
and p = mq̇ for the particle’s momentum (strictly the momentum is pe1 ) we have
∂V
p
∂ p2
q̇ =
, ṗ = mq̈ = F (q, t) = −
=
m
∂p 2m
∂q
so that
q̇
ṗ
!

=
p2
2m
− ∂V
∂q
∂
∂p

.
(6.1)
The mechanical system which we are describing is said to have one degree of freedom (represented
by the coordinate q) but in terms of our earlier definitions is a second order dynamical system
which is autonomous provided V doesn’t depend on t explicitly. The form of Eqs (6.1) suggests
that we write
p2
H(q, p, t) =
+ V (q, t).
(6.2)
2m
Eqs (6.1) may be written in terms of H as
q̇ =
∂H
∂H
, ṗ = −
.
∂p
∂q
(6.3)
Eqs (6.3) are called Hamilton’s equations and H is referred to as the Hamiltonian of the
system.
We note in passing that Hamilton’s1 equations may be expressed in the form
d
dt
1
q
p
!
= σ(∇H),
Sir William Rowan Hamilton (1805-1865); also invented quaternions.
89
90
CHAPTER 6. HAMILTONIAN SYSTEMS
where ∇ is the vector differential operator given by
∇=
∂
∂
e1 +
e2
∂q
∂p
and σ is the 2 × 2 anti-symmetric matrix given by
σ=
0 1
−1 0
!
.
More generally a Hamiltonian system of one degree of freedom may be described in terms of
two functions q, p which evolve in time according to Hamilton’s Eqs (6.3) for some Hamiltonian
H(q, p, t). In this wider sense the system need not be mechanical, and even in a mechanical situation q need not necessarily be a linear displacement—for example, it could be the angle which
a pendulum makes with the vertical. The variable q is referred to as a generalised coordinate
and p is the momentum conjugate to q. q, p are said to be conjugate variables. Of course, in the
wider sense which we have described p, the momentum conjugate to q, need not be a physical
momentum.
In what follows we shall usually be dealing with an autonomous Hamiltonian system; in this
case H = H(q, p) and H doesn’t depend on time t explicitly.
Suppose now that q(t), p(t) satisfy Hamilton’s equations with H = H(q, p). Then H(q, p) is
constant throughout the motion—we say that H is a constant of the motion. To prove this
result we note that
∂H
∂H
q̇ +
ṗ,
Ḣ =
∂q
∂p
by the chain rule. Using Hamilton’s equations we obtain
∂H ∂H
∂H
∂H
Ḣ =
−
+
∂q ∂p
∂p
∂q
= 0.
We conclude that H is indeed a constant of the motion. Such Hamiltonian systems are said to
be conservative; the Hamiltonian is constant throughout the motion. The phase curve which
passes 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 the mechanical example with which we began our discussion, suppose that V =
V (q) so that
p2
+ V (q).
H = H(q, p) =
2m
2
p
Note that 2m
= 12 mq̇ 2 , so the first term in the expression for H is the kinetic energy; V is the
potential energy and H is the total energy of the system in this case.
91
6.1. HAMILTON’S EQUATIONS
We note that the velocity v (q, p) of the flow associated with a Hamiltonian H(q, p) (see Chapter
8, Section 1.1) is given by
!
∂H
∂p
− ∂H
∂q
v (q, p) =
.
A flow which is governed by Hamilton’s equations is referred to as a Hamiltonian flow.
Example 6.1 A Non-Standard Hamiltonian System
Prove that the system of one degree of freedom whose motion is governed by the second order differential
equation
q̈ + G(q)q̇ 2 − F (q) = 0
is Hamiltonian.
We note that the given equation implies that
d 1 2
q̇ + G(q)q̇ 2 − F (q) = 0,
dq 2
so that
d
q̇ 2 + 2G(q)q̇ 2 − 2F (q) = 0.
dq
Using the integrating factor
Z
µ(q) = exp
we obtain
2G(q) dq
d
µ(q)q̇ 2 − 2µ(q)F (q) = 0.
dq
Writing
µ(q)F (q) = −
we see that
dV
dq
d
µq̇ 2 + 2V (q) = 0,
dq
and therefore
1 2
µq̇ + V (q)
2
is constant throughout the motion. If we write q̇ = p/µ, so that
p2
+ V (q)
2µ
is constant, it is reasonable to guess that the given differential equation is equivalent to Hamilton’s
equations with Hamiltonian H(q, p) given by
H(q, p) =
First note that
q̇ =
p2
+ V (q).
2µ
2 p
∂
p
∂H
=
,
=
µ(q)
∂p 2µ(q)
∂p
so that the first of Hamilton’s equations is satisfied.
92
CHAPTER 6. HAMILTONIAN SYSTEMS
Moreover, since p = µq̇
ṗ = µq̈ +
dµ 2
dµ 2
q̇ = µ(−Gq̇ 2 + F (q)) +
q̇
dq
dq
=−
Now
dµ
dq
dV
dµ 2
− µGq̇ 2 +
q̇ .
dq
dq
= 2Gµ. It follows that
1 dµ 2
dV
1 dµ p2
dV
+
q̇ = −
+
dq
2 dq
dq
2 dq µ2
∂ p2
∂H
dV
−
.
=−
=−
dq
∂q 2µ
∂q
ṗ = −
Summarising, the given equation is equivalent to the Hamiltonian equations
q̇ =
∂H
∂H
, ṗ = −
,
∂p
∂q
where
H(q, p) =
and
µ(q) = exp
Z
p2
+ V (q)
2µ
2G(q) dq, V (q) = −
Z
µ(q)F (q) dq + C.
small
Example 6.2 Divergence-Free Velocity Fields and Hamiltonian Systems
Consider the equation
!
!
d
x
v1 (x, y)
= v (x, y) =
.
v2 (x, y)
dt y
In order for this system to be Hamiltonian with generalised coordinate x and conjugate momentum y there must exist H(x, y) such that
ẋ =
∂H
∂H
, ẏ = −
.
∂y
∂x
This requires that
v1 (x, y) =
∂H
∂H
, v2 (x, y) = −
.
∂y
∂x
In order for these equations to be consistent v1 , v2 must satisfy
∂v1
∂2H
∂v2
=
=−
,
∂x
∂x∂y
∂y
∂v1 ∂v2
+
= 0, i.e. ∇ · v = 0;
∂x
∂y
in other words, the vector v has to be solenoidal. One can prove that the converse is also true
i.e. if v is solenoidal, the the system is Hamiltonian.
93
6.1. HAMILTON’S EQUATIONS
x
ψ
T
P (x,y)
y
mg
Figure 6.1: Coordinate system of the pendulum discussed in the present example.
Example 6.3 Pendulum
A simple pendulum consists of an inelastic string of length a with a particle P of mass m attached to
one end; the other end is attached to a fixed point O and the system is free to move in a vertical plane
under gravity. Discuss the motion and show in particular that the system is Hamiltonian.
Referring to the diagram suppose that the y-axis points in the direction of the downward vertical through
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 sin ψ, y = a cos ψ.
The equation of motion of the particle is given by N2:
m(ẍe1 + ÿe2 ) = (−T sin ψe1 + (mg − T cos ψ)e2 ),
mẍ = −T sin ψ, mÿ = mg − T cos ψ.
It follows that
mẍ cos ψ − mÿ sin ψ = −mg sin ψ,
(6.4)
T = mg cos ψ − mẍ sin ψ − mÿ cos ψ.
(6.5)
Now
ẋ = aψ̇ cos ψ, ẍ = aψ̈ cos ψ − aψ̇ 2 sin ψ,
ẏ = −aψ̇ sin ψ, ÿ = −aψ̈ sin ψ − aψ̇ 2 cos ψ.
We deduce from Eqs (6.4) and (6.5) that
ψ̈ +
g
sin ψ = 0, T = mg cos ψ + maψ̇ 2 .
a
The first of these equations is a differential equation for ψ and the second gives the tension T in terms of
ψ. We note that when ψ is small (i.e. near 0), so that the string is always close to the downward vertical,
the equation for ψ approximates to
g
ψ̈ + ψ = 0.
a
94
CHAPTER 6. HAMILTONIAN SYSTEMS
In this case the motion is approximately Simple Harmonic with period τ given by
τ = 2π
r
a
.
g
Returning to the exact equation for ψ we note that
ma2 ψ̈ = −mga sin ψ.
We write p = ma2 ψ̇ so that
ψ̇ =
2 ∂
p
p
=
.
ma2
∂p 2ma2
Moreover,
∂
−mga cos ψ .
ṗ = ma ψ̈ = −mga sin ψ = −
∂ψ
2
We conclude that
ψ̇ =
∂H
∂H
, ṗ = −
,
∂p
∂ψ
where
H(ψ, p) =
p2
− mga cos ψ.
2ma2
The system is therefore Hamiltonian in the generalised coordinate ψ, p being the momentum conjugate
to ψ; H(ψ, p) is of course the Hamiltonian. It is interesting to note that the kinetic energy T of the
particle is given by
T =
1
1
p2
p2
1
m(ẋ2 + ẏ 2 ) = m(cos2 ψ + sin2 ψ)ψ̇ 2 = m 2 4 =
2
2
2 m a
2ma2
so that the first term in the expression for H is just the kinetic energy of the particle. The second term
is in fact the potential energy V due to gravity since
mge2 = −
dV
e2 ,
dy
and therefore
V = −mgy = −mga cos ψ.
It follows that the Hamiltonian is the total energy of the system; since H doesn’t depend on t explicitly
it is a constant of the motion and
H(ψ, p) =
p2
− mga cos ψ = C,
2ma2
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 domain of
ψ is essentially the unit circle S1 . On the other hand, p can range over the whole of R. The phase space
of this system is the Cartesian product R × S1 which we can picture as a circular cylinder.
95
6.2. STABILITY PROBLEMS
6.2
Stability problems
Suppose that a particle of mass m moves along the q-axis under a potential V (q). As noted
above, the Hamiltonian for the motion is
H(q, p) =
p2
+ V (q).
2m
The Hamilton equations are
q̇
ṗ
!
!
p/m
−V 0 (q)
=
.
We see that the fixed points are those points (q, p) for which p = 0 and V 0 (q) = 0. This
agrees with our earlier conclusion that the equilibrium points of such a system correspond to
the stationary points of the potential energy function V. Suppose that V 0 (q0 ) = 0. Then near q0
we can approximate V by
1
V (q) = V (q0 ) + (q − q0 )V 0 (q0 ) + (q − q0 )2 V 00 (q0 ) + · · ·
2
so that the linearised Hamiltonian equations become
q̇
ṗ
!
0
1/m
−V 00 (q0 )
0
=
The matrix
0
1/m
00
−V (q0 )
0
!
!
q − q0
p
!
.
.
1 00
V (q0 ). If V 00 (q0 ) > 0, so that q0 is a minimum of V , we
has eigenvalues λ given by λ2 = − m
have
q
λ = ±iβ, β = V 00 (q0 )/m.
We deduce that there is a matrix P such that
Ẋ
Ẏ
!
where
X
Y
=
!
0 β
−β 0
=P
−1
!
q − q0
p
X
Y
!
!
,
.
The fixed point (q = q0 , p = 0) is the origin of the new (X, Y ) coordinates. Following standard
procedures we write
X = r cos θ, Y = r sin θ
and obtain
ṙ = 0, θ̇ = −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 system to
96
CHAPTER 6. HAMILTONIAN SYSTEMS
move round one of these circles is precisely the time taken by the system to execute one complete
Simple Harmonic oscillation about q = q0 and is given by
2π
= 2π
τ=
β
s
m
V
00 (q
0)
,
as found previously by a different method.
In a similar manner we find that if V 00 (q0 ) < 0, so that q0 is a maximum of the potential function
V , that the linearised Hamiltonian equations may be written as
Ẋ = λ1 X, Ẏ = −λ1 Y, λ1 =
s
−
V 00 (0)
> 0,
m
with integrals
X(t) = X0 eλ1 t , Y (t) = Y0 e−λ1 t .
The fixed point is clearly unstable, the phase curves being a family of rectangular hyperbolas
given by XY = X0 Y0 ; such unstable fixed points are sometimes said to be ‘hyperbolic’ fixed
points.
Example 6.4 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 2 1 4 1 2
p + q − q .
2
4
2
Discuss the motion.
This is the Hamiltonian for a particle moving under the potential V (q) given by
V (q) =
1 4 1 2
q − q .
4
2
We note that for small q 2 the force acting on the particle is approximately equal to
−
1
d
(− q 2 )e1 = qe1 ;
dq 2
it follows that the potential generates a short range repulsion. On the other hand, for large values of q 2
the term 41 q 4 is dominant and this generates a force equal to
−
d 1 4
( q )e1 = −q 3 e1 .
dq 4
This means that the potential V produces a long range attraction which tends to pull the particle back
towards the origin. The over-all effect is seen from the graph of V.
Hamilton’s equations give
q̇
p
=
.
ṗ
q − q3
The fixed points are given by p = 0, q − q 3 = 0 i.e. by p = 0, q = 0, q = ±1. Linearising the equations
in the usual way we have to consider the Jacobian J(q, p) at the fixed points. We have
0
1
J(q, p) =
.
1 − 3q 2 0
97
6.2. STABILITY PROBLEMS
q
Separatrix
q
Figure 6.2: Quartic potential described in the text and construction of the phase curves at
different energies.
We see that
J|0,0 =
0
1
1
0
.
This matrix has eigenvalues λ = ±1 and the now familiar arguments lead us to conclude that (0, 0) is an
unstable fixed point. Similarly we find that
0 1
J|(±1,0) =
.
−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
√ 2
√0
.
− 2 0
A standard calculation leads to the conclusion that the system√makes a complete circuit (in phase space)
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 2 1 4 1 2
p + q − q = E,
2
4
2
98
CHAPTER 6. HAMILTONIAN SYSTEMS
where E is a parameter.
A phase curve which passes through a hyperbolic (unstable) fixed point is known as a separatrix; its
importance 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 by E = 0
so that the separatrix has equation given by
q
p = ± √ (2 − q 2 )1/2 .
2
The accompanying diagram showing the phase curves and the potential function V exhibits the following
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 it starts
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 oscillation is just
four times the time taken to go from q = 0 to q = q1 . The relevant phase curve has equation given by
1 2 1 4 1 2
1
1
p + q − q = q1 4 − q1 2
2
4
2
4
2
and Hamilton’s equations give
q̇ =
so that
τ =4
∂H
=p
∂p
q1
√ Z q1
dq
dq
=2 2
1
4 − q 4 ) − 1 (q 2 − q 2 ))1/2
p
(
(q
0
0
4 1
2 1
Z
q
1
√
dq
.
=4 2
2
2
1/2
(q1 − q ) (q1 2 + q 2 − 2)1/2
0
Z
Exercise 6.1 The motion of a particle of unit mass moving along the x-axis is described by the
Hamiltonian
1
x
H(x, p) = p2 + 2
,
2
x +1
where p is the momentum conjugate to x.
Write down Hamilton’s equations, find the fixed points of the system, and classify them as either
hyperbolic or elliptic fixed points.
Find the equation of the separatrix and sketch the phase diagram. The particle is released from
rest at x = −3. Explain briefly, in the context of your diagram, why the subsequent motion is
oscillatory. Find the maximum speed of the particle and obtain an expression for the period τ
of one complete oscillation.
Throughout the following question you may assume that two positive electric charges, of magnitude c1 , c2 respectively, distance z apart, repel each other with a force c1 c2 /z 2 , the direction
of the force being along the straight line join of the charges.
99
6.2. STABILITY PROBLEMS
Exercise 6.2 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
to move on the x-axis between the first two charges so that its coordinate is x(t) at time t. Write
down N2 for the moving charge e2 , and show that it can be cast in Hamiltonian form with
µ
µ
p2
+ +
,
2m x 2a − x
H(x, p) =
where µ = e1 e2 .
Find the fixed points of the system and sketch the phase diagram for 0 < x < 2a. The charge e2
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 diagram that
it will subsequently oscillate with period τ given by
τ =4
Z
a(1+k)
a
m dx
,
p(x)
where p(x) is given by (*). Verify that the substitution x = a + ak sin φ leads to the expression
τ =2
ma3 (1 − k2 )
µ
1/2 Z
0
π/2
(1 − k2 sin2 φ)1/2 dφ.
100
CHAPTER 6. HAMILTONIAN SYSTEMS
Chapter 7
Lagrangian Theory
7.1
Motivation
In our description of mechanical systems we have so far considered two different approaches
Newtonian Mechanics For conservative 1D mechanical systems, equations of motion according to Newtonian mechanics are of the form
mẍ = F (x) = −V 0 (x) ,
where F is the force field and V the potential energy from which the force can be derived.
Note that
(i) Newtons equation of motion for 1D system is a 2nd order ODE, hence generally difficult
to analyse
(ii) To specify a solution x0 and u0 = ẋ(0) must be specified as initial conditions.
(iii) The equation of motion is directly derived from physical principles
— easy to formulate in Cartesian coordinates (also for many-particle or multi-dimensional
system)
— difficult to set up in higher dimensions, when motion is constrained (eg. on surface
of sphere...)
(iv) A problem: Coordinate transformations do not leave the structure of Newtons equations invariant. E.g. under the (invertible) transformation f : X → x, x = f (X), the
equation of motion
mẍ = F (x) = −V 0 (x) ,
transforms into
m[f 0 (X)Ẍ + f 00 (X)Ẋ 2 ] = F (f (X))
an equation which has a structure different from the original one (exhibiting apparent
X dependent mass, and a velocity dependent contribution).
(v) In situations of constrained motion in higher dimensions, it is difficult to quantify the
forces associated with constraints (although it may be important to know them from
engineering points of view).
101
102
CHAPTER 7. LAGRANGIAN THEORY
Hamiltonian Mechanics For conservative 1D mechanical systems, equations of motion according to Hamiltonian mechanics are of the form
ẋ =
p
∂H
=
,
m
∂p
ṗ = −F (x) = −V 0 (x) =
with
H(p, x) =
∂H
∂x
p2
+ V (x)
2m
For 1D conservative mechanical systems, we have obtained these equations by transforming
Newton’s 2nd order ODE into a system of two 1st order ODEs.
(i) Hamilton’s equations are strictly equivalent to Newton’s, but the concept of a Hamiltonian system is not restricted to Mechanics. Equations are first order, thus easier
to analyse and visualise (qualitatively in terms of flow, phase curves, fixed points...)
(ii) To specify a solution x0 and p0 must be specified as initial conditions.
(iii) If equations are obtained from Newton’s equations in terms of physical principles, all
difficulties of setting up equations — in particular with constraints on the motion in
higher dimensions — are inherited.
(iv) There is a large class of transformations — the so-called canonical transformations —
which leave Hamilton’s equations formally invariant. However, these transformations
do not allow of interpretations as immediate and simple as coordinate transformations. Within the present course we have not dealt with canonical transformations,
however (although it should be remarked that they can be an extremely powerful
tool). They can help solving the equations of motion but not setting them up.
What is needed is a method for describing and analysing mechanical systems, which (initially)
has Cartesian components of positions and velocities as independent variables (just as Newton’s
method), but formulated in such a way that a large class of coordinate transformations leaves
the structure of the equations of motion invariant, so that these transformations may be used
to actually simplify setting up equations of motion, in particular in systems with constraints.
Lagrange’s formulation of mechanics provides just such a method.
We will not follow the historical path that led to the formulation of Lagrangian mechanics, nor
provide a full account of the underlying theory in the present course. Rather, we introduce the
Lagrangian formulation mechanics initially only as a recipe to set up equations of motion. However, we do present the key results necessary to appreciate the usefulness of this reformulation
of mechanics.
7.2
Lagrange’s equation
To introduce Lagrange’s formulation of mechanics, we start out from Newton’s equation of
motion for a conservative mechanical system
mα q̈α = Fα ({qα }) = −
∂V ({qα })
,
∂qα
α = 1, . . . , n ,
(7.1)
in which q = (qα ) is the vector of coordinates of, say, N particles moving in IR3 , so n = 3N , and
q̇ = (q̇α ) is the vector of velocities.
103
7.2. LAGRANGE’S EQUATION
For this system it is known that Energy E,
E =T +V =
X mα
α
2
q̇α2 + V ({qα })
is conserved.
According to Lagrange, Newton’s equation of motion for this system can be derived from the
so-called Lagrangian function
L = L({qα }, {q̇α }) = T − V
(7.2)
via Lagrange’s equations of motion, which take the form
d ∂L
∂L
−
=0,
α = 1, . . . , n .
(7.3)
dt ∂ q̇α ∂qα
Note: In the Lagrangian L, the coordinates qα and the velocities q̇α are considered as independent variables.
The next steps then will be
(i) to verify that Lagrange’s equations of motion (7.3) are indeed equivalent to Newton’s
equations of motion (7.1);
(ii) to show that Lagrange’s equations of motion (7.3) are structurally invariant under coordinate transformations;
(iii) to demonstrate how this can be exploited to simplify setting up equations of motion using
generalised coordinates and momenta, which are in many cases better adapted to the
problems under study than Cartesian coordinates and momenta.
7.2.1
Equivalence with Newton’s Equations of Motion
To verify that Lagrange’s equations of motion (7.3) are equivalent to Newton’s equations of
motion (7.1), note first that the Lagrangian (7.2 is given by
X mα
q̇α2 − V ({qα })
L = L({qα }, {q̇α }) = T − V =
2
α
From this Lagrangian, we get
∂L
= mα q̇α ,
∂ q̇α
Combining this with
thus
d ∂L
= mα q̈α
dt ∂ q̇α
α = 1, . . . , n .
(7.4)
∂L
∂V ({qα })
=−
,
∂qα
∂qα
we immediately conclude that Lagrange’s equations of motion (7.3) coincide with Newtons equations of motion (7.1) as claimed.
Note: The equivalence persists for non-conservative systems, in which forces derive from time
dependent potentials
∂V ({qα }, t)
Fα = −
.
(7.5)
∂qα
In that case the Lagrangian attains an explicit time dependence
X mα
L = L({qα }, {q̇α }, t) =
q̇α2 − V ({qα }, t)
(7.6)
2
α
104
7.2.2
CHAPTER 7. LAGRANGIAN THEORY
Invariance under Coordinate Transformations
We now demonstrate that Lagrange’s equation is invariant under coordinate transformations
We do this for simplicity for a system with one degree of freedom. The generalisation is left as
a straightforward exercise to the reader.
Let
L = L(q, q̇, t)
be a Lagrangian in original coordinates, and let
q↔Q,
q = f (Q) .
be a coordinate transformation Note, for this transformation to be invertible the function f
must be monotone, i.e. f 0 (Q) 6= 0.
The transformation entails
q̇ = f 0 (Q)Q̇ ,
so the Lagrangian may be rewritten as
L(q, q̇, t) = L f (Q), f 0 (Q)Q̇, t ≡ L̃(Q, Q̇, t) .
The invariance of Lagrange’s equations under coordinate transformation follows from considering
∂ L̃
∂L ∂ q̇
∂L 0
=
=
f (Q)
∂ q̇ ∂ Q̇
∂ q̇
∂ Q̇
∂ L̃
∂L ∂q
∂L ∂ q̇
∂L 0
∂L 00
=
+
=
f (Q) +
f (Q)Q̇ ,
∂Q
∂q ∂Q
∂ q̇ ∂Q
∂q
∂ q̇
and
from which
d ∂ L̃
dt ∂ Q̇
=
∂L 00
d ∂L 0
f (Q)Q̇
f (Q) +
dt ∂ q̇
∂ q̇
and thus
d ∂ L̃
dt ∂ Q̇
d ∂L
∂ L̃
=
−
∂Q
dt ∂ q̇
∂L 0
−
f (Q) .
∂q
As f 0 (Q) 6= 0, it follows that
d ∂ L̃
dt ∂ Q̇
−
∂ L̃
=0
∂Q
⇔
d ∂L
dt ∂ q̇
−
∂L
=0
∂q
(7.7)
and we conclude that Lagrange’s equation is invariant under the transformation q ↔ Q.
The reasoning for multidimensional coordinate transformations
{qα } ↔ {Qα } ,
qα = fα ({Qβ })
is entirely analogous and will not be repeated here. For this transformation to be everywhere
invertible, it is required that det({∂Qα /∂Qβ }) 6= 0, hence that the Jacobian J of the transformation J = {∂Qα /∂Qβ } be invertible. Use of the chain rule as above in combination with the
invertibility of the Jacobian leads to the desired result.
105
7.2. LAGRANGE’S EQUATION
7.2.3
Exploiting the Invariance of Lagrange’s Equations
The invariance of Lagrange’s equations of motion under arbitrary coordinate transformation
provides a very useful tool to simplify setting up equations of motion. As the Lagrangian
L = T − V is a scalar quantity, it is invariant under coordinate transformations; in fact T and V
are separately scalars and invariant under coordinate transformations. In standard mechanical
systems where T and V can be associated which kinetic and potential energy, these are separately
invariant and have the meaning of kinetic and potential energy in any coordinate system.
We shall demonstrate that this can prove to be extremely useful for setting up equations of
motion in systems with constraints.
Note: For a conventional mechanical system described in Cartesian coordinates, we know that
pα = mα q̇α =
∂L
∂ q̇α
is a Cartesian component of the momentum of a particle. The invariance of Lagrange’s equation
of motion leads us to the following
Definition 7.1 Generalised Coordinates and Conjugate Momenta
Given a Lagrangian expressed in some general coordinate system of the form
L = L({qα }, {q̇α }, t) ,
we refer to the qα as generalised coordinates and to the pα defined via
pα =
∂L
∂ q̇α
as the generalised momenta conjugate to the qα .
Using the above definition we can relate the momenta p, P conjugate to q, Q respectively for the
special case q = f (Q). We have
p=
∂L
,
∂ q̇
P =
∂ L̃
∂L ∂ q̇
=
= pf 0 (Q) .
∂ q̇ ∂ Q̇
∂ Q̇
In other words,
P = pf 0 (Q) .
Note also that the coordinate systems need not be Euclidean, orthogonal, or even stationary
(nothing in our reasoning forbids that the coordinate transformations considered above depend
on time).
We demonstrate the relative ease by which equations of motion can be set up using the Lagrangian approach on an example.
Example 7.1 Pendulum in a Gravitational Field
The example of a pendulum in a gravitational field has been considered in Example 6.3 above. Referring
to Figure 6.1, we have the particle coordinates
x = a sin ψ ,
y = a cos ψ
⇒
ẋ = aψ̇ cos ψ,
ẏ = −aψ̇ sin ψ
106
CHAPTER 7. LAGRANGIAN THEORY
The kinetic energy is
T =
m 2
m
(ẋ + ẏ 2 ) = a2 ψ̇ 2
2
2
and the potential energy
V (y) = −mg(y − a) = mga(1 − cos ψ) ,
so
m 2 2
a ψ̇ − mga(1 − cos ψ)
2
The motion is effectively one-dimensional as the string-length a is kept fixed. The only generalised
coordinate is the angle ψ. The equation of motion according to Lagrange is
d ∂L
g
∂L
=0
⇔
ma2 ψ̈ + mga sin ψ = 0
⇔
ψ̈ + sin ψ = 0
−
dt ∂ ψ̇
∂ψ
a
L = T − V = L(ψ, ψ̇) =
which is just the equation of motion obtained in 6.3
7.3
Further Developments of the Theory
In this section we collect and prove a number of further results about Lagrangian theory which
are useful in applying it to specific cases.
Proposition 7.1 Lagrangians differing by additive or multiplicative constants.
Two Lagrangians differing only by a multiplicative (L̃ = cL) or additive (L̃ = L + c) constant
give rise to identical equations of motion.
The result is trivial; no proof is given.
Proposition 7.2 Lagrangians differing by a time-derivative of a function.
If two Lagrangians differ only by a total derivative of some function f (q, t) with respect to time,
they give rise to identical Lagrangian equations of motion.
Proof: Let
L̃(q, q̇, t) = L(q, q̇, t) +
∂f
∂f
d
f (q, t) = L(q, q̇, t) +
q̇ +
≡ L + ∆L
dt
∂q
∂t
Using
∂f
∂∆L
=
∂ q̇
∂q
Hence
d ∂∆L
dt ∂ q̇
∂∆L
Moreover :
∂q
⇒
d ∂∆L ∂∆L
−
≡0
dt ∂ q̇
∂q
=
=
∂2f
q̇ +
∂q 2
∂2f
q̇ +
∂q 2
∂2f
∂t∂q
∂2f
∂t∂q
irrespective of f .
#
Proposition 7.3 Symmetries and Conservation Laws
If a generalised coordinate does not explicitly appear in the Lagrangian of the system, the
corresponding conjugate momentum is conserved, i.e. a constant of the motion.
7.4. EQUIVALENCE OF LAGRANGIAN AND HAMILTONIAN MECHANICS
107
Proof: Let L = L({qα }α6=ρ , {q̇α }, t), so that the generalised coordinate qρ does not explicitly
appear in L. Writing down the corresponding Euler-Lagrange equation one gets
d ∂L
∂L
=
=0
dt ∂ q̇ρ
∂qρ
so
pρ =
∂L
= const.
∂ q̇ρ
as claimed.
#
Note: For a conventional mechanical system, the fact that a coordinate does not explicitly
appear in the Lagrangian means that the potential energy is invariant under (or symmetric
w.r.t.) changes of this coordinate. Thus the present proposition establishes a deep relation
between symmetries and conservation laws.
We demonstrate that this profound result can be an extremely useful tool for simplifying equations of motion, using motion in a central potential as an example.
Example 7.2 Motion in a Central Potential
Consider motion of a particle in a central potential, which depends only on the distance from the origin.
Without restricting generality, the motion may be thought to be constrained to the x, y-plane. Introducing
polar coordinates
x = r cos ϕ ,
y = r sin ϕ,
one obtains
m 2
(ṙ + r2 ϕ̇2 ) − V (r)
2
for the Lagrangian. The variable ϕ does not explicitly appear in L so the corresponding conjugate
momentum (called angular momentum) is conserved
L=T −V =
pϕ =
∂L
= mr2 ϕ̇ = ` = const.
∂ ϕ̇
The equation of motion for the radial coordinate follows from
d ∂L ∂L
−
=0
⇔
r̈ − mrϕ̇2 + V 0 (r) = 0
dt ∂ ṙ
∂r
One may use the value ` of the constant angular momentum to eliminate ϕ̇ from this equation so as to
finally obtain
`2
`2
0
mr̈ =
− V 0 (r) ≡ −Veff
(r)
with
Veff (r) = V (r) +
3
mr
2mr2
Thus the problem is essentially reduced to describing one-dimensional radial motion in an effective potential that depends on the value of the angular momentum. Note: for gravitational motion V (r) =
−GM m/r the effective potential has a minimum: You are invited to discuss what this implies for planetary motion!
7.4
Equivalence of Lagrangian and Hamiltonian Mechanics
Having discussed the basic elements of Lagrangian mechanics, we now demonstrate that Lagrangian Mechanics and Hamiltonian Mechanics are in fact equivalent, in the sense that the two
versions of Mechanics can be transformed into each other, and in particular, that Lagrange’s
equations of motion imply Hamilton’s equation of motion, and vice versa.
Here we demonstrate this only for systems with one degree of freedom, but the relation is more
general.
108
CHAPTER 7. LAGRANGIAN THEORY
7.4.1
Lagrange’s Equations Imply Hamilton’s Equations of Motion
Given a Lagrangian L = L(q, p, t) and Lagrange’s equation of motion
d ∂L ∂L
−
=0
dt ∂ q̇
∂q
which may be rewritten as
d
∂L
p=
dt
∂q
with
∂L
∂ q̇
the momentum canonically conjugate to q. The Hamiltonian H(q, p, t) which has q and p (rather
than q and q̇ as independent variables is defined via a so called Legendre transformation of L
that eliminates q̇ in favour of p:
p=
H = H(q, p, t) = q̇p − L(q, q̇, t)
(7.8)
in which p = ∂L
∂ q̇ and q̇ = q̇(q, p, t) is expressed in terms of q, p and t.
Using the Hamiltonian thus defined, we now demonstrate that Lagrange’s equations of motion
imply Hamilton’s equations of motion.
(i) Consider first
∂H
∂q .
We have
∂ q̇
∂L ∂L ∂ q̇
∂L
∂H
=
p−
−
=−
,
∂q
∂q
∂q
∂ q̇ ∂q
∂q
where we exploited the definition of p =
of p once more we finally get
∂L
∂ q̇ .
Using Lagrange’s equation and the definition
ṗ = −
∂H
∂q
which is one of Hamilton’s equations.
(ii) Next look at
∂H
∂p :
∂H
∂ q̇
∂L ∂ q̇
= q̇ +
p−
∂p
∂p
∂ q̇ ∂p
which, on investing the definition of p gives
q̇ =
∂H
,
∂p
i.e. Hamilton’s other equation of motion.
#
Note1 : The condition for being able to solve p = p(q, q̇, t) defined by
p(q, q̇, t) =
∂L
∂ q̇
for q̇ so as to obtain q̇ = q̇(q, p, t) is that
∂2L
6= 0 .
∂ q̇ 2
(7.9)
Note 2: The Legendre transformation from L to H has a simple geometrical interpretation in
terms of a tangent to L as shown in Fig 7.1.
7.4. EQUIVALENCE OF LAGRANGIAN AND HAMILTONIAN MECHANICS
t,q fixed
109
t,q fixed
.
L(q,q,t)
H(q,p,t)
− H(q,p,t)
{
.
− L(q,q,t)
.
q
{
.
q
p
p
Figure 7.1: Geometric interpretation of the Legendre transforms. (i) Left panel: L → H. The
tangent at q̇ to L having slope p = ∂L
∂ q̇ , with −H given as in the figure, leads to L(q, q̇, t) =
−H(q, p, t) + pq̇ which is just the equation of the Legendre transform. (ii) Right panel: H → L.
The tangent at p to H having slope q̇ = ∂H
∂p , with −L given as in the figure, leads to H(q, p, t) =
−L(q, q̇, t) + q̇p which is just the equation of the Legendre transform.
7.4.2
Hamilton’s Equations Imply Lagrange’s Equations of Motion
The Legendre transformation (7.8), when interpreting it with q̇ defined via Hamilton’s equation,
q̇ ==
∂H
,
∂p
could also be read as
L(q, q̇, t) = pq̇ − H(q, p, t) ,
(7.10)
that is — with p = p(q, q̇, t) expressed in terms of q, q̇, and t — as a Legendre transformation
that eliminates p in favour of q̇.
Using the Lagrangian thus defined, we now demonstrate that Hamilton’s equations of motion
imply Lagrange’s equations of motion.
(i) Consider first
∂L
∂ q̇ .
We have
∂p ∂H ∂p
∂L
= p + q̇
−
=p
∂ q̇
∂ q̇
∂p ∂ q̇
where we used Hamilton’s equation q̇ = ∂H
∂p . This equation reproduces the definition of p
as momentum conjugate to q in Lagrangian mechanics.
(ii) Next look at
∂L
∂q :
∂p ∂H
∂H ∂p
∂L
= q̇
−
−
∂q
∂q
∂q
∂p ∂q
which, on once more investing Hamilton’s equation q̇ =
∂L
∂H
=−
,
∂q
∂q
∂H
∂p
first gives
110
CHAPTER 7. LAGRANGIAN THEORY
from which, using Hamilton’s equation ṗ = − ∂H
∂q and p as obtained in (i), we recover
Lagrange’s equation of motion
d ∂L ∂L
−
=0
dt ∂ q̇
∂q
as claimed.
#
Note 1: The condition for being able to solve q̇ = q̇(q, p, t) defined by
q̇(q, p, t) =
∂H
∂ ṗ
for p so as to obtain p = p(q, q̇, t) is that
∂2H
6= 0 .
∂p2
(7.11)
Note 2: The Legendre transformation from H to L has a simple geometrical interpretation in
terms of a tangent to H as shown in Fig 7.1.
Finally to round off the picture we need the following result.
Proposition 7.4 Conditions for Existence of Legendre Transforms
The condition (7.9) needed to perform the Legendre transform (7.8) is satisfied if and only if
the condition (7.11) needed to perform the Legendre transform (7.10) holds. In other words, if
a Legendre transformation is well defined, so is its inverse.
Proof: We have
∂L
=p
∂ q̇
⇒
and similarly
∂H
= q̇
∂p
These combine to
⇒
∂p
∂2L
=
2
∂ q̇
∂ q̇
∂ q̇
∂2H
=
.
2
∂p
∂p
∂2L ∂2H
=1
∂ q̇ 2 ∂p2
from which the assertion follows.
#
Note 1: For a conventional mechanical system describing motion of a particle of mass m under
the action of a potential V the Lagrangian is of the form
L = L(x, ẋ) = T − V (x) =
m 2
ẋ − V (x) .
2
The corresponding Hamiltonian, obtained via the Legendre transform
H(x, p) = ẋp − L(x, ẋ)
with p = ∂L/∂ ẋ is
H(x, p) = ẋmẋ −
m 2
p2
ẋ + V (x) = T + V (x) =
+ V (x) .
2
2m
111
7.5. WORKED EXAMPLES
Thus the Lagrangian is the difference of kinetic and potential energy, the Hamiltonian is the
sum.
Note 2: All the above considerations can be generalised to systems of more than one degree of
freedom. For a system with Lagrangian
L = L({qα }, {q̇α }, t)
the Hamiltonian as the Legendre transform w.r.t. all q̇α is
H({qα }, {pα }, t) =
with pα =
∂L
∂ q̇α .
7.5
α
q̇α pα − L({qα }, {q̇α }, t)
The inverse Legendre transform is defined in an analogous way,
L({qα }, {q̇α }, t) =
with q̇α =
X
X
α
pα q̇α − H({qα }, {pα }, t)
∂H
∂pα .
Worked examples
As the concept of a Hamiltonian system is not restricted to description of particle motion, a
Hamiltonian that is not of the standard form of a sum of kinetic energy (depending only on
momenta) and potential energy (depending only on positions) makes perfect sense.
Via a Legendre transform, Lagrangian descriptions of the same non-mechanical systems can be
derived, and the corresponding Lagrangian equations of motion are equivalent to the Hamiltonian equations of motion.
The following example provides an illustration.
Example 7.3 A Non-Conventional Hamiltonian
Consider the Hamiltonian given by
H(q, p) =
p2
+ p sin q.
2
Find the corresponding Lagrangian.
We have
∂H
= p + sin q, so that p = (q̇ − sin q)
q̇ =
∂p
and
(q̇ − sin q)2
1
− sin q(q̇ − sin q) = (q̇ − sin q)2
2
2
It is a straightforward matter to verify that both Hamilton’s equations and Lagrange’s equation leads to
the equation of motion in the form
q̈ − sin q cos q = 0.
L = pq̇ − H = q̇(q̇ − sin q) −
The basis of the following examples on Lagrangian mechanics is the assumption that the Lagrangian is given by
L = T − V,
112
CHAPTER 7. LAGRANGIAN THEORY
where T is the kinetic energy, and V is the potential energy of the system. Recall that if a
particle of mass m has position vector
r = xe1 + y e2 + z e3 ,
where e1 , e2 , e2 are unit Cartesian vectors, then
T =
m 2
(ẋ + ẏ 2 + ż 2 ).
2
For a particle moving along the x-axis under a force F (x)e1 the potential energy function V is
given by
dV
F (x) = −
.
dx
More generally, for 3-dimensional motion under a force F for which curl F = 0, the potential V
is given by
∂V
∂V
∂V
e1 −
e2 −
e3 .
F = −grad V = −
∂x
∂y
∂z
Example 7.4 Pendulum
The Hamiltonian for the motion of the bob of a simple pendulum moving in a vertical plane is given by
(see Example 6.3)
p2
H(ψ, p) =
− mga cos ψ,
2ma2
where a is the length of the pendulum, ψ is the angle which the pendulum makes with the downward
vertical, and m is the mass of the bob. (Note that the ‘angular momentum’ p was introduced in a more
or less ad hoc way in that example.) Find the corresponding Lagrangian and deduce the equation of
motion using Lagrange’s equation.
In familiar notation we have
∂H
p
ψ̇ =
=
,
∂p
ma2
L(ψ, ψ̇) = pψ̇ − H = ma2 ψ̇ 2 −
=
It follows that
m2 a4 ψ̇ 2
+ mga cos ψ
2ma2
1
ma2 ψ̇ 2 + mga cos ψ.
2
∂L
∂L
= ma2 ψ̇,
= −mga sin ψ.
∂ψ
∂ ψ̇
The Lagrange equation
∂L
d ∂L
−
=0
dt ∂ ψ̇
∂ψ
therefore becomes
ma2 ψ̈ + mga sin ψ = 0, ψ̈ +
g
sin ψ = 0,
a
as found previously.
Example 7.5 Motion on Cycloid in a Gravitational Field
A particle of mass m is constrained to move in a vertical plane under the force of gravity on the smooth
cycloid with parametric equations
z = A(1 − cos ψ), x = A(ψ + sin ψ), −π < ψ < π,
113
7.5. WORKED EXAMPLES
z
R
x
mg
Figure 7.2: Schematic view of the cycloid; the force of constraint is also indicated.
where A > 0 is constant and the z-axis is vertically upwards. Show that if s is the arc length measured
from the point ψ = 0 on the cycloid,
L(s, ṡ) =
1 2 mgs2
ṡ −
.
2
8A
Find the equation of motion, and show that this motion is oscillatory with a period which is independent
of the amplitude.
The potential energy V due to gravity is given by
−mge3 = −
dV
e3 , V = mgz = mgA(1 − cos ψ).
dz
The element of arc length ds is given by
(ds)2 = (dx)2 + (dz)2 = A2 (1 + cos ψ)2 (dψ)2 + A2 sin2 ψ(dψ)2
ψ
= 2A2 (1 + cos ψ)(dψ)2 = 4A2 cos2 (dψ)2 .
2
We see that
ds = 2A cos
ψ
ψ
dψ, s = 4A sin ,
2
2
since s(ψ = 0) = 0.
The kinetic energy T of the particle is given by
T =
1
1
m(ẋ2 + ż 2 ) = mṡ2 .
2
2
Notice that in this problem there is a force of constraint acting on the particle, namely the normal reaction
R(ψ) exerted on the particle by the smooth wire (see the diagram). For the Lagrangian
L=T −V
we obtain
L(s, ṡ) =
1 2
1
ψ
1
mgs2
mṡ − mgA(1 − cos ψ) = mṡ2 − 2mgAsin2 = mṡ2 −
.
2
2
2
2
8A
Lagrange’s equation is
∂L
d ∂L
−
=0
dt ∂ ṡ
∂s
114
CHAPTER 7. LAGRANGIAN THEORY
and carrying out the relevant differentiations we obtain
mgs
ms̈ +
= 0, s̈ + Ω2 s = 0,
4A
g
2
Ω =
.
4A
The oscillation has period τ given by
2π
= 4π
τ=
Ω
s
A
g
which is independent of the amplitude of the motion.
Note that
dz
dz/dψ
A sin ψ
2A sin ψ/2 cos ψ/2
=
=
=
= tan ψ/2.
dx
dx/dψ
A(1 + cos ψ)
2Acos2 ψ/2
It follows that the tangent at the point ψ on the cycloid actually makes an angle ψ/2 with the x-axis.
Example 7.6 Motion on a Rotating Parabolic Wire
A particle of mass m is constrained to slide under gravity on a smooth parabolic wire with equation
z = 21 a2 x2 , where the z-axis is vertically upwards. The wire is made to rotate about the z-axis with
constant angular speed Ω. Show that the Lagrangian is given by
L(R, Ṙ) =
mg 2 2
1
m(Ṙ2 (1 + a4 R2 ) + R2 Ω2 ) −
a R ,
2
2
and that the corresponding Hamiltonian is
H(R, p) =
mR2
p2
+
(ga2 − Ω2 ),
2m(1 + a4 R2 )
2
where (R, θ, z) denote the cylindrical coordinates of the particle.
z
.
Ω=θ
mg
θ
y
x
Figure 7.3: Rotating parabolic wire described in the text.
Let (x, y, z) be the Cartesian coordinates of the particle at time t. Referring to the diagram we see that
x = R cos θ, y = R sin θ
115
7.5. WORKED EXAMPLES
so that
ẋ = −Rθ̇ sin θ + Ṙ cos θ = −RΩ sin θ + Ṙ cos θ,
ẏ = Rθ̇ cos θ + Ṙ sin θ = RΩ cos θ + Ṙ sin θ,
since θ̇ = Ω.
It follows that the kinetic energy T of the particle is given by
T =
1
1
m(ẋ2 + ẏ 2 + ż 2 ) = m(Ṙ2 + R2 Ω2 + ż 2 ).
2
2
Now, when the wire has rotated through an angle θ, z = 21 a2 R2 , R playing the role of x. It follows that
ż = a2 RṘ and that
T =
1
1
m(Ṙ2 + R2 Ω2 + R2 a4 Ṙ2 ) = m(Ṙ2 (1 + R2 a4 ) + R2 Ω2 ).
2
2
The potential energy V due to gravity is given by
−mge3 = −
mg 2 2
dV
e3 , V = mgz =
a R .
dz
2
We deduce that the Lagrangian L = T − V is given by
L(R, Ṙ) =
1
1
mṘ2 (a4 R2 + 1) − mR2 (ga2 − Ω2 ).
2
2
The momentum p conjugate to R is given by
p=
∂L
∂L
≡
= mṘ(a4 R2 + 1),
∂u
∂ Ṙ
so that
Ṙ =
p
m(a4 R2
+ 1)
.
The Hamiltonian is given by
H(R, p) =
=
=
pṘ − L
p2
1
p2
1
4 2
2
2
2
−
m(a
R
+
1)
−
mR
(ga
−
Ω
)
m(a4 R2 + 1)
2
m2 (a4 R2 + 1)2
2
2
2
mR
p
+
(ga2 − Ω2 ) .
2m(1 + a4 R2 )
2
Example 7.7 Moving Pendulum
Referring to the diagram, one end O0 of a light string of length a is constrained to move horizontally so
that its displacement from a fixed point O is γ(t), where γ̈ = βg, and β is a positive constant. Attached
to the other end of the string is a particle of mass m, and the system moves in a vertical plane in such a
way that the string makes an angle ψ with the downward vertical at time t.
Use N2 to show that the equation of motion of the mass m is
g
ψ̈ + f (ψ) = 0, where f (ψ) = (β cos ψ + sin ψ).
a
Use this equation to determine the period of small oscillations about the point ψ = ψ0 = − arctan β.
Write down an expression for the Lagrangian of the system and determine the corresponding expression
for the Hamiltonian. Write down Hamilton’s equations, discuss the nature of the fixed points of the
system, and show that your conclusions agree with the results derived from Newton’s equations.
116
CHAPTER 7. LAGRANGIAN THEORY
O
γ (t)
O’
y
ψ
T
P (x,y)
x
mg
Figure 7.4: Moving Pendulum.
The mass m has coordinates (x, y) where
x = a cos ψ, y = γ(t) + a sin ψ
from which we derive
ẋ = −aψ̇ sin ψ, ẏ = γ̇ + aψ̇ cos ψ,
ẍ = −aψ̈ sin ψ − aψ̇ 2 cos ψ,
ÿ = γ̈ + aψ̈ cos ψ − aψ̇ 2 sin ψ.
Application of N2 to the motion of the mass m gives
m(ẍe1 + ÿe2 ) = mge1 + (−T sin ψe2 − T cos ψe1 ),
mẍ = mg − T cos ψ, mÿ = −T sin ψ,
where T is the tension in the string.
Eliminating T from the equations of motion we obtain
mẍ sin ψ − mÿ cos ψ = mg sin ψ.
Writing this equation in terms of ψ we soon derive
g
ψ̈ + f (ψ) = 0, f (ψ) = β cos ψ + sin ψ.
a
We observe that the mass m can exist in equilibrium at the points ψ where ψ̈ = 0 i.e. where
ψ = ψ0 , ψ = π − ψ0 , where ψ0 = − arctan β.
Let’s examine what happens near ψ0 . We have
f (ψ) = f (ψ0 ) + f 0 (ψ0 )(ψ − ψ0 ) + · · · .
The equation of motion approximates to
g
ψ̈ + f 0 (ψ0 )(ψ − ψ0 ) = 0.
a
117
7.5. WORKED EXAMPLES
A straightforward calculation shows that
f 0 (ψ0 ) = (1 + β 2 )1/2 .
We conclude that the mass m executes oscillations about the point ψ0 with period τ given by
r
τ0
a
1
τ = 2π
=
,
2
1/4
g (1 + β )
(1 + β 2 )1/4
where τ0 is the period of the oscillations when O0 is unaccelerated.
We note that the kinetic energy T of the mass m is given by
T =
1
1
m(ẋ2 + ẏ 2 ) = m(γ̇ 2 + 2aγ̇ ψ̇ cos ψ + a2 ψ̇ 2 ).
2
2
The potential energy V of the mass m is given by
−
dV
e1 = mge1 , V = −mgx = −mga cos ψ.
dx
It follows that the Lagrangian is
L=T −V =
1
m(γ̇ 2 + 2aγ̇ ψ̇ cos ψ + a2 ψ̇ 2 ) + mga cos ψ.
2
As γ̇ 2 is a prescribed function of time, it follows that an equivalent Lagrangian is
L∗ =
1
m(2aγ̇ ψ̇ cos ψ + a2 ψ̇ 2 ) + mga cos ψ;
2
this is because we can express γ̇ 2 as the derivative of another function of t (namely of
1
d
∗
2 2
L = ma ψ̇ + ma
(γ̇ sin ψ) − γ̈ sin ψ + mga cos ψ.
2
dt
R
γ̇ 2 dt). Note that
By the same reasoning, therefore, L̃ is equivalent to the Lagrangian with which we started, where
L̃ =
1
ma2 ψ̇ 2 + ma(g cos ψ − γ̈ sin ψ).
2
The corresponding Hamiltonian is
H = pψ̇ − L̃,
where
p=
∂ L̃
∂ ψ̇
= ma2 ψ̇, so that ψ̇ =
p
.
ma2
Expressing H in terms of the variables (ψ, p) and using γ̈ = βg we find that
H(ψ, p) =
p2
+ mga(β sin ψ − cos ψ).
2ma2
Hamilton’s equations now give
ψ̇ =
ṗ = −
∂H
p
=
,
∂p
ma2
∂H
= −mga(β cos ψ + sin ψ).
∂ψ
118
CHAPTER 7. LAGRANGIAN THEORY
The system has fixed points given by
p = 0, ψ = ψ0 , ψ = π − ψ0 , where ψ0 = − arctan β.
To investigate the stability of the fixed point (ψ0 , 0) we follow now familiar procedures and obtain the
linearised equations in the form
d
ψ
ψ − ψ0
=J
,
p
p
dt
where
J=
0
−mga(1 + β 2 )1/2
1/ma2
0
.
The matrix J has eigenvalues λ given by
λ = ±iωσ, ω =
1/2
g
, σ = (1 + β 2 )1/4 .
a
It follows that there is a matrix P such that
d
X
0
=
Y
ωσ
dt
−ωσ
0
X
Y
,
where
X
Y
=P
−1
ψ − ψ0
p
.
The origin of the (X, Y ) coordinates is the fixed point ψ = ψ0 , p = 0. We write X = r cos θ, Y = r sin θ
and obtain
ṙ = 0, θ̇ = ωσ,
r(t) = r0 , θ(t) = θ0 + ωσt.
The phase curves are a family of circles and we deduce that we are dealing with a stable fixed point. The
system moves round one of these phase curves in time τ given by
τ = 2π/(ωσ) =
τ0
,
(1 + β 2 )1/4
where τ0 is the time taken when the end O0 is unaccelerated. Physically, τ is the period of small oscillations
of the mass m about ψ = ψ0 .
Exercise 7.1 A particle of mass m moves under the influence of gravity along the smooth spiral
defined, through the parameter ψ, by
x = a cos ψ, y = a sin ψ, z = kψ,
where a and k are positive constants and the z-axis is vertically upwards. Assuming that the
Lagrangian L has the form L = T − V, where T is the kinetic energy and V is the potential
energy due to gravity, find the Hamiltonian and use Hamilton’s equations to show that
ψ(t) = −
gk
t2
+ tψ̇(0) + ψ(0).
(a2 + k2 ) 2
119
7.5. WORKED EXAMPLES
z
.
Ω=θ
mg
θ
y
x
Figure 7.5: Rotating straight wire described in the text.
Exercise 7.2 A bead of mass m is constrained to slide under gravity on a smooth straight wire
which passes through the origin and makes an angle π/4 with the z-axis, where the z-axis is
vertically upwards. The wire is made to rotate about the z-axis so that the configuration at
time t is as shown in the accompanying diagram; the bead P has coordinates (x, y, z) and we
may express x, y in terms of z, θ by x = z cos θ, y = z sin θ, where θ, z are as indicated in the
diagram.
Given that θ̇ = Ω = constant, express ẋ, ẏ, ż in terms of z, ż, Ω and deduce that the Lagrangian
is given by
1
L = m(z 2 Ω2 + 2ż 2 ) − mgz.
2
Hence show that the Hamiltonian H is given by
H(z, p) =
1
p2
− mz 2 Ω2 + mgz,
4m 2
where p is the momentum conjugate to z.
Write down Hamilton’s equations, and deduce that if z(0) = 0, ż(0) = 0, then
2g
σt
z = − 2 sinh2
, where σ 2 = Ω2 /2.
Ω
2
Exercise 7.3 The motion of a particle of mass m is described by the Lagrangian L where
1
1
L(R, Ṙ) = mṘ2 (a4 R2 + 1) − mR2 (ga2 − Ω2 ),
2
2
and a, Ω are positive constants. Show that the Hamiltonian H is given by
H(R, p) =
1
p2
+ mR2 (ga2 − Ω2 ),
4
2
2m(1 + a R ) 2
120
CHAPTER 7. LAGRANGIAN THEORY
where p is the momentum conjugate to R.
Given the initial conditions p = m, R = 0 show by a suitable application of a conservation law
that, in the subsequent motion, R never exceeds (ga2 − Ω2 )−1/2 , provided Ω2 < ga2 .
Exercise 7.4 (King’s College, Summer 1995) Given a Lagrangian L(q, q̇, t) show that Lagrange’s equation
d ∂L
∂L
−
=0
dt ∂ q̇
∂q
implies Hamilton’s equations
q̇ =
∂H(q, p, t)
,
∂p
ṗ = −
∂H(q, p, t)
,
∂q
where p = ∂L/∂ q̇ ; the relation between L and H should be clearly stated.
A bead of mass m is constrained to slide under gravity on a smooth wire in the shape of a
vertical circle of radius R. The wire is made to rotate about the vertical axis with constant
angular velocity Ω, so that the configuration at time t is as shown in the accompanying figure.
z
.
Ω=θ
φ
R
mg
θ
y
x
Figure 7.6: Rotating circular wire described in the text.
The bead P has coordinates (x, y, z) given by
x = R sin φ cos θ,
y = R sin φ sin θ,
z = R cos φ,
121
7.5. WORKED EXAMPLES
where θ is the angle between the x-axis and the plane of the wire and where φ is the angle between
the z-axis and the vector OP . Given that θ̇ = Ω = constant, express ẋ, ẏ, ż in terms of R, Ω
and φ̇, and deduce that the Lagrangian is given by
1 L = m R2 φ̇2 + R2 Ω2 sin2 φ − mgR cos φ.
2
Hence show that the Hamiltonian is given by
H(φ, p) =
1 p2
1
− mR2 Ω2 sin2 φ + mgR cos φ
2 mR2 2
and find the fixed points of the system.
122
CHAPTER 7. LAGRANGIAN THEORY
Appendix A
Taylor’s Theorem
A.1
Taylor Expansion for Functions of One Variable
We recall Taylor’s Theorem1 for functions of one real variable: Suppose that φ is a suitably
differentiable real valued function on some interval [x − δ, x + δ]. Then
φ(x + h) = φ(x) +
h2 (2)
hn−1 (n−1)
h (1)
φ (x) +
φ (x) + · · · +
φ
(x) + Rn , h ∈ [−δ, δ]
(1!)
(2!)
(n − 1)!
where Rn , the remainder after n terms, has the form
Rn =
hn (n)
φ (ξ),
(n!)
and ξ is some point between x and x + h. In applications we are often interested in situations
where h is close to 0 and it may be justifiable to neglect the term Rn , for some suitable value of n.
This amounts to saying that we can approximate φ(x + h) by a polynomial in h of order (n − 1).
Notice that in the computation of φ(x + h) we require to know the value of φ and its derivatives
at the point x. The particular case where x = 0 goes under the name of MacLaurin’s2 Theorem;
for reasons which will shortly be apparent, we use t rather than h to denote the variable:
φ(t) = φ(0) +
t (1)
t2 (2)
tn−1
φ (0) +
φ (0) + · · · +
φ(n−1) (0) + Rn , t ∈ [−δ, δ]
(1!)
(2!)
(n − 1)!
where Rn , the remainder after n terms, has the form
Rn =
tn (n)
φ (ξ),
(n!)
and ξ is some point between 0 and t.
1
Brook Taylor (1685-1731). In 1715 he published his most important work which contained a proof of the
theorem which now bears his name.
2
Colin MacLaurin (1698-1746) entered Glasgow University at the age of 11; professor of mathematics at
Aberdeen at the age of 19!
123
124
APPENDIX A. TAYLOR’S THEOREM
A.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) and
that f has partial derivatives to any required order. Motivated by our knowledge of Taylor’s
Theorem for functions of one variable we aim to obtain an expression for f (x + h, y + k) in
terms of powers of h, k and the partial derivatives of f evaluated at the point (x, y). Let A
denote the point (x, y) and B the point (x + h, y + k). Any point P on the straight line AB has
coordinates (x + ht, y + kt), 0 ≤ t ≤ 1; A corresponds to t = 0 and B corresponds to t = 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 application of MacLaurin’s
formula gives
φ(t) = φ(0) +
t2 (2)
tn−1
t (1)
φ (0) +
φ (0) + · · · +
φ(n−1) (0) + Rn .
(1!)
(2!)
(n − 1)!
Let’s write φ(t) = f (u, v), u = x + ht, v = y + kt. By the Chain Rule of partial differentiation
∂f ∂u ∂f ∂v
∂f
∂f
+
=h
+k .
∂u ∂t
∂v ∂t
∂u
∂v
φ(1) (t) =
Thus
∂f
∂f
+k ,
∂u
∂v
where the partial derivatives are evaluated at t = 0. A little thought shows that
φ(1) (0) = h
φ(1) (0) = h
∂f (x, y)
∂f (x, y)
+k
,
∂x
∂y
where the derivatives of f are to be evaluated at (x, y). We see that symbolically
d
∂
∂
=h
+k ,
dt
∂x
∂y
where the t derivative is to be computed at t = 0 and the x, y partial derivatives at (x, y). It is
clear that
∂
∂ n
φ(n) (0) = h
+k
f (x, y),
∂x
∂y
where
h
∂
∂
+k
∂x
∂y
n
f (x, y)
is to be interpreted as
∂
∂
h
+k
∂x
∂y
∂
∂
∂
∂
h
··· h
f (x, y),
+k
+k
∂x
∂y
∂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
∂
+k
+k
h
f (x, y) +
h
1! ∂x
∂y
2! ∂x
∂y
2
f (x, y) + · · ·
125
A.3. VECTOR FUNCTIONS
∂
1
∂
+k
··· +
h
(n − 1)! ∂x
∂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 near
to (0, 0) and that it is justifiable to neglect R3 . In this approximation we are retaining only the
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
∂
∂ n
h
+k
f (x, y)
∂x
∂y
is straightforward. For example, with n = 2 we have
h
∂
∂
+k
∂x
∂y
2
= h2
f (x, y) = h
∂
∂
+k
∂x
∂y
h
∂
∂
f (x, y)
+k
∂x
∂y
2
∂2f
∂2f
∂2f
2∂ f
+
hk
+
kh
+
k
∂x2
∂x∂y
∂y∂x
∂y 2
∂2f
∂2f
∂2f
+ 2hk
+ k2 2 ,
2
∂x
∂x∂y
∂y
using the commutative property of partial differentiation. It’s therefore clear that we can formally square out
∂ 2
∂
+k
h
∂x
∂y
provided we make the following identifications:
= h2
∂
∂x
∂
∂x
=
∂2
,
∂x2
∂
∂y
∂
∂y
=
∂2
,
∂y 2
∂
∂x
∂
∂y
=
∂2
,
∂x∂y
and so on. Using this principle we see, for example, that
∂
∂
+k
h
∂x
∂y
3
f = h3
3
3
∂3f
∂3f
2
3∂ f
2 ∂ f
+
3h
k
+
k
.
+
3hk
∂x3
∂x2 ∂y
∂x∂y 2
∂y 3
Nevertheless, calculations of this sort require care; the formal expansion procedure works here
because h, k are constants as far as the x, y differentiations are concerned.
A.3
Vector functions
In applications to dynamical systems we are often concerned with vector valued functions of two
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 applications
we frequently require to expand f (x + h, y + k) about the point (x, y) retaining only the linear
terms in h, k. We have
fj (x + h, y + k) = fj (x, y) + h
∂fj
∂fj
(x, y) + k
(x, y) + · · · (j = 1, 2).
∂x
∂y
126
APPENDIX A. TAYLOR’S THEOREM
The dots indicate that we have neglected higher order terms in h, k. We conclude that
f (x + h, y + k) = f (x, y) +
∂f1
1
h ∂f
∂x + k ∂y
∂f2
2
h ∂f
∂x + k ∂y
!
+ · · · = f (x, y) + J
h
k
!
+ ···
where J denotes the Jacobian matrix defined by f1 , f2 , evaluated at (x, y), i.e.
∂(f1 , f2 )
J=
=
∂(x, y)
∂f1
∂x
∂f2
∂x
∂f1
∂y
∂f2
∂y
!
.
Notation Given a function f of two real variables (x, y) we frequently use subscript notation
to denote its partial derivatives:
fx ≡
fxx ≡
∂f
,
∂x
∂2f
,
∂x2
fy ≡
fxy
∂f
,
∂y
∂2f
≡
,
∂x∂y
fyy ≡
∂2f
∂y 2
Example A.1 Expand f (x, y) = sin(πxy) about the point (1, 1) neglecting third and higher order terms
in (x − 1), (y − 1).
In this case h = x − 1, k = y − 1 and Taylor’s formula gives
sin(πxy)
Now,
=
f (1, 1) + (x − 1)fx (1, 1) + (y − 1)fy (1, 1)
1
+ (x − 1)2 fxx (1, 1) + 2(x − 1)(y − 1)fxy (1, 1) + (y − 1)2 fyy (1, 1) + · · ·
2!
fx = πy cos(πxy) ,
fy = πx cos(πxy),
2
fxy
fxx = −(πy) sin(πxy) ,
fyy = −(πx)2 sin(πxy),
∂
∂
fx = −(πy)(πx) sin(πxy) + π cos(πxy) =
fy = fyx
=
∂y
∂x
Since
f (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 that
sin(πxy) ' −π(x − 1) − π(y − 1) − π(x − 1)(y − 1) + · · ·
Exercise A.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 A.2 Find the (two variable) Taylor expansion of ln(1 + x2 + y 2 ) about the point (1, 1),
up to and including the quadratic terms.
Let
f (x, y) =
x/(x2 + y 2 )
ln(1 + x2 + y 2 )
!
Expand f about the point (1, 1), neglecting quadratic and higher order terms.
127
A.3. VECTOR FUNCTIONS
Exercise A.3 Suppose that f is a real valued function of the real variables (x, y) with continuous
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
1
f (x, y) − f (a, b) = X T HX + · · · ,
2
X=
x−a
y−b
!
,
T denotes transpose, and H is the Hessian matrix given by
fxx fxy
fxy 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 the
leading term in the Taylor expansion i.e. by X T HX, if this term is not identically zero. A 2 × 2
real symmetric matrix H is said to be positive definite, negative definite depending on whether
X T HX > 0 (X 6= 0), X T HX < 0 (X 6= 0), respectively. We conclude that if 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 books on algebra that H is positive definite
provided its determinant is positive and the two entries on the leading diagonal are positive;
likewise H is negative definite provided its determinant is positive and the two entries on the
leading diagonal are negative. We therefore conclude that
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 .
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 ,
128
APPENDIX A. TAYLOR’S THEOREM
Appendix B
Basic Linear Algebra
B.1
Fundamental ideas
We consider the space IR2 of column vectors
x1
x2
X=
!
, 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 such vectors
is well known. Any two linearly independent vectors constitute a basis and any vector can 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 therefore constitute a
basis. We use e1 , e2 to denote the standard basis so that
1
0
e1 =
!
, e2 =
0 1
A linear transformation a of the vector space IR2 is a map
a : IR2 → IR2
such that
a(X + Y ) = aX + aY , a(λX ) = λaX ,
2
for all vectors X , Y ∈ IR and for every scalar λ ∈ IR. The action of any linear transformation
a of IR2 is determined by its action on any set of basis vectors. We note that ae1 , ae2 may be
written as a linear combination of e1 , e2 (because e1 , e2 constitute a basis). We may write
aei =
2
X
aki ek , i = 1, 2
(B.1)
k=1
for some unique coefficients aik . These coefficients determine the matrix A = (aik ) of the transformation a with respect to the standard basis given by e1 , e2 . We note that
Ae1 =
a11 a12
a21 a22
!
1
0
!
=
129
a11
a21
!
= a11 e1 + a21 e2 ,
130
APPENDIX B. BASIC LINEAR ALGEBRA
Ae2 =
a11 a12
a21 a22
!
!
0
1
a12
a22
=
!
= a12 e1 + a22 e2 .
Reference to equation B.1 shows that Aei = aei , (i = 1, 2) so that we may calculate the action
of the transformation a on any vector X simply by calculating AX , where A is the matrix of
the transformation with respect to the standard basis as defined by equations B.1. We consider
below the matrix of the linear transformation a with respect to some non-standard basis, and
show how it relates to the matrix of a with respect to the standard basis. Suppose that f 1 , f 2
is another basis. We can therefore express f 1 , f 2 as linear combinations of e1 , e2 :
fi =
2
X
pki ek , i = 1, 2
(B.2)
k=1
The coefficients pki define a matrix P = (pki ).
Because f 1 , f 2 is a basis we can express e1 , e2 as linear combinations of f 1 , f 2 . Let’s write
en =
2
X
qmn f m , n = 1, 2,
(B.3)
m=1
for some unique coefficients qmn which define a matrix Q = (qmn ).
We note that
en =
2
X
qmn f m =
m=1
X
qmn
X
prm er =
X
(P Q)rn er .
(All the sums run from 1 to 2)
Equating coefficients we see that
(P Q)rn = δrn , δrn = 1 (r = n), δrn = 0 (r 6= n).
Hence P Q = 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 basis
e1 , e2 (see equation B.1) we may consider its matrix A0 = (a0ki ) (say) with respect to the basis
f 1 , f 2 . Referring to equation B.1 this is defined by
af i = Af i =
2
X
a0ki f k i = 1, 2
(B.4)
k=1
Let’s calculate the matrix A0 .
We have
Af i = A
2
X
k=1
pki ek =
X
pki Aek =
X
pki
X
ank en =
2
X
k,n=1
ank pki
2
X
m=1
qmn f m
B.2. INVARIANCE OF EIGENVALUES UNDER SIMILARITY TRANSFORMATIONS 131
=
2
X
k,n,m=1
qmn ank pki f m =
2
X
(QAP )mi f m
m=1
We therefore see that the matrix A0 of the linear transformation a with respect to the basis
f 1 , f 2 is given by
A0 = QAP = P −1 AP.
This is a simple, but fundamental result. We note in passing that the arguments are identical
in IRn .
Given a matrix A one aims to choose a basis f 1 , f 2 which makes A0 = P −1 AP as simple as
possible. Before obtaining the Jordan canonical forms for 2 × 2 real matrices A we prove the
following simple result.
B.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 matrices A
and P −1 AP have the same eigenvalues.
Proof
We note that
det(λI − P −1 AP ) = det(P −1 (λI − A)P ) = det P −1 det(λI − A) det P
= det(P −1 P ) det(λI − A) = det I det(λI − A) = det(λI − A).
It follows that
det(λI − P −1 AP ) = 0 ⇐⇒ det(λI − A) = 0
and therefore the matrices P −1 AP and A have the same eigenvalues.
B.3
Jordan Forms
Now let A be any real 2 × 2 matrix, and regard A as the matrix of a linear transformation a
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 forms corresponds
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 different
cases may occur.
(a) The eigenvalues λ1 , λ2 of A are real and distinct, λ1 > λ2 .
Let f 1 , f 2 be the eigenvectors corresponding to λ1 , λ2 respectively, so that
Af 1 = λ1 f 1 , Af 2 = λ2 f 2 .
132
APPENDIX B. BASIC LINEAR ALGEBRA
Using the same notation as employed above we see that the matrix A0 of the transformation
a with respect to the basis f 1 , f 2 (it is a basis because eigenvectors corresponding to
distinct real eigenvalues are linearly independent – proved in CM113A) is given by
λ1 0
0 λ2
A0 =
and therefore
P
−1
λ1 0
0 λ2
AP =
!
!
, where P = (f 1 : f 2 ).
(B.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 independent
eigenvectors f 1 , f 2 so that
Af 1 = λf 1 , Af 2 = λf 2 ,
and f 1 , f 2 form a basis.
With respect to the basis f 1 , f 2 a has matrix
0
A =
λ 0
0 λ
!
= λI, P −1 AP = λI, P = (f 1 : f 2 ).
In this case A = λI, so the matrix was of Jordan canonical form (diagonal) to begin with.
(ii) The second possibility is that A has only one independent eigenvector, f 1 (say). We
extend this to a basis of IR2 by picking another vector f 2 (f 2 isn’t an eigenvector of A.)
We can write Af 2 = µf 1 + ν f 2 , for some real µ, ν.
We have
Af 1 = λf 1 , Af 2 = µf 1 + ν f 2
so, with respect to the basis f 1 , f 2 , a has matrix
0
A =P
−1
AP =
λ µ
0 ν
!
.
Imposing the condition that A and P −1 AP have the same eigenvalues (see Section B.2
above) we immediately obtain ν = λ.
Writing f 1 = µ1 F 1 we see that
M
−1
AM =
λ 1
0 λ
!
, M = (F 1 : f 2 ).
(B.6)
133
B.4. BASIS TRANSFORMATION
(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(f 1 + if 2 ) = (α + iβ)(f 1 + if 2 ).
Equating real and imaginary parts we then have
Af 1 = αf 1 − β f 2 , Af 2 = β f 1 + αf 2 .
One can prove without too much difficulty that f 1 , f 2 are linearly independent and therefore constitute a basis of IR2 .
With respect to this basis a has matrix A0 where
0
A =P
−1
α β
−β α
AP =
!
,
(B.7)
and P = (f 1 : f 2 ).
We conclude by showing how the components of a vector with respect to the bases (f 1 , f 2 ) and
(e1 , e2 ) are related.
B.4
Basis Transformation
Suppose that (f 1 , f 2 ) and (e1 , e2 ) are bases related by equations B.2 and let y1 , y2 , x1 , x2 ,
denote the components of some vector with respect to the bases (f 1 , f 2 ), (e1 , e2 ), respectively.
Writing
x1
x2
X=
!
, Y =
y1
y2
!
we have
2
X
xi ei =
i=1
2
X
yi f i =
i=1
and, equating coefficients, we obtain
P2
i=1 pki yi
2
X
yi pji ej
i,j=1
= xk ; or, in column vector notation,
X = P Y , Y = P −1 X .
We frequently use such a linear change of variable in our discussions of the stability of fixed
points.
134
B.5
APPENDIX B. BASIC LINEAR ALGEBRA
Rotations
Consider a set of Cartesian axes xOy, with origin O; the vectors e1 , e1 may be regarded as unit
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 vector through an
angle θ about the origin O. Let r denote the position vector of the point P with coordinates
(x1 , x2 ) so that r = OP = x1 e1 + x2 e2 . Suppose that under the rotation r = OP maps to
r 0 = OP 0 , where P 0 has coordinates (x01 , x02 ). Regarding the Cartesian plane as a copy of the
Argand diagram we may write
x01 + ix02 = eiθ (x1 + ix2 ) = (cos θ + i sin θ)(x1 + ix2 ).
Equating real and imaginary parts we obtain
x01 = x1 cos θ − x2 sin θ, x02 = x1 sin θ + x2 cos θ,
or, in matrix notation, r 0 = A(θ)r , where
A(θ) =
cos θ − sin θ
sin θ
cos θ
!
.
Note: Reference to equation B.7 shows that A(θ) has complex eigenvalues cos θ ± i sin θ, a fact
which may be confirmed by direct calculation.
B.6
Area Preserving Transformations
In this section we determine the class of linear transformations of the Cartesian plane which
preserve areas. We employ the notation of section B.5. First we note that the origin O is
left invariant by any linear transformation of the Cartesian plane. We also note that any nonsingular 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 = r 0 + λn,
(B.8)
where λ is a real parameter. Let a be a non-singular linear transformation of the Cartesian plane
whose matrix with respect to the standard basis is A. The image of the straight line described
by equation B.8 is given by
Ar = Ar 0 + λ(An), which has the form
r0 = r0 0 + λn0 .
(B.9)
We conclude that all straight lines parallel to the vector n map to straight lines which are
parallel to n0 = An; An is not the zero vector because A is non-singular. It is clear therefore
that non-singular transformations of the Cartesian plane map parallelograms to parallelograms.
Let P, R be points with coordinates (x1 , y1 ), (x2 , y2 ) respectively. The position vectors OP, OR
of P, R are given by
OP = x1 e1 + y1 e2 , OR = x2 e1 + y2 e2 .
135
B.7. EXAMPLES AND EXERCISES
Consider the parallelogram OP QR. Under the action of a linear transformation a of the Cartesian plane this parallelogram maps to an image parallelogram OP 0 Q0 R0 , where P 0 , R0 have coordinates (x1 0 , y1 0 ), (x2 0 , y2 0 ) respectively, say. If A is the matrix of the linear transformation a,
with respect to the standard basis, where
α β
γ δ
A=
!
,
(B.10)
then we have the relations
x1 0
y1 0
!
=
α β
γ δ
!
x1
y1
!
,
x2 0
y2 0
!
=
α β
γ δ
!
x2
y2
!
.
(B.11)
An elementary calculation shows that the area of parallelogram OP QR is
|OP × OR| = |x1 y2 − x2 y1 |; similarly the area of the image parallelogram O0 P 0 Q0 R0 is
|x1 0 y2 0 − x2 0 y1 0 |.
It now follows from equation B.11 that
x1 0 y2 0 − x2 0 y1 0 = v(αx1 + βy1 )(γx2 + δy2 ) − (γx1 + δy1 )(αx2 + βy2 )
and after a little simplification we find that
x1 0 y2 0 − x2 0 y1 0 = (αδ − βγ)(x1 y2 − x2 y1 ) = det A(x1 y2 − x2 y1 ).
We conclude that a parallelogram and its image under the linear transformation a have the same
area provided | det A| = 1. Of particular interest are those transformations for which det A = 1;
of course, det I = 1, and the transformations with det A = 1 are those which can be reached
continuously from the identity. We shall see later that area preserving transformations are
significant in Hamiltonian mechanics.
Note: Under the change of variable (x1 , y1 ) 7→ (x1 0 , y1 0 ) described by equation B.11 area elements transform according to the rule
dx1 0 dy1 0 = |J|dx1 dy1 , where J =
∂(x1 0 , y1 0 )
.
∂(x1 , y1 )
Referring to equation B.11 and carrying out the differentiations we soon find that J = det A; as
found above, the condition for areas to be preserved is therefore | det A| = 1.
B.7
Examples and Exercises
Example B.1 For each of the matrices Ai (1 = 1, 2, 3) given by
1 2
2 1
3
A1 =
, A2 =
, A3 =
1 1
−2 4
1
−1
1
find the appropriate Jordan form Ji (i = 1, 2, 3) and a matrix Pi (i = 1, 2, 3) such that Pi −1 Ai Pi =
Ji , (i = 1, 2, 3).
136
APPENDIX B. BASIC LINEAR ALGEBRA
Consider the third of these matrices, A3 . A vector has components x1 , x2 with respect to the standard
basis e1 , e2 and components y1 , y2 with respect to the basis f 1 , f 2 in which the linear transformation
corresponding to A3 has the Jordan matrix J3 . Write down the linear equations which connect x1 , x2 and
y1 , y2 .
The eigenvalues of A1 are given by
det(λI − A1 ) = 0, so that λ = 1 ±
√
2.
Solving the eigenvalue equation
1 2
1 1
x
y
=λ
x
y
we find that x = (λ − 1)y and we may take as eigenvectors
√
√2
, λ = 1 + 2,
2
√2
− 2
, λ=1−
√
2.
It follows that
P1 −1 A1 P1 =
√
0√
1+ 2
√2 .
√2
, where P1 =
2 − 2
0
1− 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
x
x
= (3 + i)
.
−2 4
y
y
We find that
2x + y = (3 + i)x, y = (1 + i)x.
Taking x = 1 we obtain y = 1 + i and the corresponding complex eigenvector is
1
f 1 + if 2 =
1+i
so that
f1 =
1
1
, f2 =
0
1
.
Clearly, f 1 , f 2 are linearly independent, and since
A2 f 1 = 3f 1 − f 2 , A2 f 2 = f 1 + 3f 2
we conclude that
P2 −1 A2 P2 =
3 1
−1 3
, P2 = (f 1 : f 2 ) =
Finally, A3 has eigenvalues given by
det(λI − A3 ) = 0, λ = 2 (twice).
1 0
1 1
.
137
B.7. EXAMPLES AND EXERCISES
To obtain an eigenvector consider the equation
x
x
3 −1
.
=2
y
y
1
1
It follows that
3x − y = 2x, x = y,
so that we may take as an eigenvector
f1 =
1
1
.
f2 =
1
0
.
Extend this to a basis of IR2 by the vector
Now
Af 2 =
3
1
= αf 1 + βf 2 ,
from which we easily find that α = 1, β = 2.
Hence, A3 f 1 = 2f 1 , A3 f 2 = f 1 + 2f 2 and choosing f 1 , f 2 as a basis
P3 −1 A3 P3 =
2
0
1
2
, P3 = (f 1 : f 2 ) =
1 1
1 0
.
The connection between the components x1 , x2 of a vector with respect to the standard basis and its
components y1 , y2 with respect to the basis f 1 , f 2 is given by
x1
x2
=
1
1
1
0
y1
y2
.
Exercise B.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 matrix
with respect to the standard basis is
!
1 1
.
1 2
Sketch your result; it should look like a parallelogram!
Exercise B.2 Describe the transformation of the Cartesian plane whose matrix with respect to
the standard basis is
!
√
1
3 √
−1
.
1
3
2
Write down the matrix of the linear transformation which describes a rotation about O through
an angle of π/4.
138
APPENDIX B. BASIC LINEAR ALGEBRA
Exercise B.3 Show that the rotation matrix
cos θ − sin θ
sin θ
cos θ
A(θ) =
!
is orthogonal i.e. that A(θ)T A(θ) = I.
Show, by direct calculation, that A(θ1 )A(θ2 ) = A(θ1 + θ2 ). What is the geometrical significance
of 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 B.4 Find the Jordan forms appropriate to the matrices
1 1
1 2
!
,
2 1
−4 4
!
,
3 1
−1 1
!
.
Exercise B.5 Show that the set G given by
G=
(
A=
a b
c d
!
)
, a, b, c, d ∈ IR, det A 6= 0
forms a group under matrix multiplication. Prove also that the set H ⊆ G which consists of
those matrices which have det A = 1 is a normal sub-group of G.
Exercise B.6 Recall that a 2 × 2 real matrix is orthogonal if AT A = I. Show that any such
matrix represents an area preserving transformation. As noted above, the rotation matrices A(θ)
are orthogonal.
Exercise B.7 Suppose that a linear transformation a of the Cartesian plane has matrix A with
respect to the standard basis e1 , e2 . Suppose that under the transformation a OP 7→ OP0 , where
P has coordinates (x1 , x2 ), P 0 has coordinates (x1 0 , x2 0 ), and that a leaves lengths invariant i.e.
2
2
x1 2 + x2 2 = x1 0 + x2 0 ,
for every choice of (x1 , x2 ). Prove that A is orthogonal i.e. that AT A = I. Deduce that det A =
±1.