Applied Dynamical Systems
Study Guide for APM2614
by
Prof A Batubenge-Tshidibi
revised by
Mr F Nyathi
Department of Mathematical Sciences
University of Souh Africa, Pretoria.
©
2022
Open Rubric
ii
.
Contents
Preface
v
1 OVERVIEW
1
2 LINEAR AUTONOMOUS DYNAMICAL SYSTEMS
7
3 LINEAR CONTROL THEORY
53
4 AUTONOMOUS NON–LINEAR SYSTEMS
82
5 ADVANCED TOPICS
125
6 APPENDIX
.
134
iii
iv
.
CONTENTS
Preface
In writing this study guide, we leant heavily on two books:
James T. Sandefur: Discrete Dynamical Systems: theory and applications. Clarendon Press,
Oxford, 1990. First Edition.
David G. Luenberger: Introduction to Dynamic Systems: theory models and applications.
John Willey, New York, 1979. First Edition.
The following books were also useful in some places:
Dennis G. Zill and Michael R. Cullen: Differential equations. PWS Kent, Boston, 1992.
Third Edition.
K.J. Falconer: Fractal geometry: mathematical foundations and applications.
This second–year module requires a background including first–year mathematics, specifically calculus and linear algebra. Some exposure to mathematical modelling is useful but not essential.
Prof A Batubenge-Tshidibi
Revised by Mr F Nyathi
v
Chapter 1
OVERVIEW
1. Introduction
First-year calculus courses can give the impression that all problems involving differential equations are soluble in the sense that the solution can be expressed as a simple formula. In fact, this
impression is quite wrong: most differential equations do not have such solutions. The reason the
impression arises, of course, is that the exercises in a first calculus course are carefully chosen so as
to have simple solutions.) So, if faced with a problem that cannot, in the above sense, be solved,
what can we do other than give up? We could use the following:
(a) Numerical methods
There are two modules on this topic in the Applied Mathematics course. Numerical methods form
an important and useful tool, but they do have limitations : all the coefficients in an equation must
have precise numerical values. This is fine if we want a solution to a specific problem, but it is not
very helpful for obtaining a general understanding of the behaviour of the solutions. For this, we
turn to dynamical systems theory.
(b) Dynamical Systems theory
The use of dynamical systems theory avoids “not seeing the wood for the trees”. In other words, we
try and understand the overall behaviour of a system, without getting bogged down in the details.
This is perhaps best illustrated by an example:
A woman jumps out of an aeroplane with a parachute. The equation describing her motion is
dv
k
= g − v 1.9 ,
dt
m
v>0
(1)
where v is her downward speed, g is the acceleration due to gravity (approximately 9.8ms−2 ), m
is the mass of the woman and parachute, and k is a constant associated with the parachute. For
safety, the woman should hit the ground at not more than 5 ms−1 . What is the least possible value
for the ratio k/m?
1
2
CHAPTER 1. OVERVIEW
Solution 1 (which gets bogged down in detailed calculus)
At t = 0, v = 0, and suppose at t = T we have v = V . Then integrating equation (1) gives
Z v=V
v=0
Z t=T
dv
k
v
g− m
=
1.9
dt.
t=0
The integral on the right is easy (it equals T ), but the first integral is difficult. Trying all the tricks
learnt in MAT102, plus some others, bring us nowhere so we resort to the following:
Solution 2 (Dynamical systems approach)
dv
= 0) at
dt
Equation (1) has a so-called fixed point (i.e. a point where
v = v∗ ≡
mg 1/1.9
k
.
(The symbol “ ≡ ” means “defined to be”.)
For v > v∗ we have dv/dt < 0 and for v < v∗ we get dv/dt > 0. Thus for this problem, whatever
the initial value of v, as time evolves v → v∗ . This is illustrated in the following diagram:
We call v∗ a stable fixed point. Thus v∗ is a good estimate of the speed at which the woman hits
the ground. Further, since we are not given the time or height of the fall, it is in fact the only
estimate that can be made. The condition v∗ < 5ms−1 implies
mg
< 51.9 ,
k
and therefore
9.8
k
≥ 1.9 ≈ 0.46045 in S.I. units.
m
5
2. What is a dynamical system?
The essential property of a dynamical system is that it involves one or more functions of time. For
example,
x(t), (x(t), y(t)), x(t) = (x1 (t), ..., xk (t)).
The dynamical system can be discrete or continuous, which means that the time t is regarded
as discrete or continuous. A continuous dynamical system is usually described by a differential
equation or a system of differential equations. For example,
dx
= cos x,
dt
3
or, in the case of a system of differential equations,
dx1
dt
dx2
dt
= x1 − x1 x2
= x2 −
x1 x2
.
2
In a discrete dynamical system the time is regarded as a non-negative integer (t = 0, 1, 2, 3, ...), and
the evolution of the system is usually described by a difference equation or a system of difference
equations. For example, we might have
x(n + 1) = 2x(n),
or the system of difference equations
p(n + 1)
=
q(n)p(n)
q(n + 1)
=
1 − p(n).
[Note that the recommended textbooks for this module use the above notation, but other textbooks
might write the difference equations as
xn+1 = 2xn
and
pn+1
= qn pn ,
qn+1
=
1 − pn .]
In the module APM114–V, Mathematical Modelling, some of the following ideas of dynamical systems theory were introduced in the context of a single differential equation or difference equation:
(a) Autonomous systems
If the right-hand side of the differential or difference equation does not involve time explicitly, the
system is called autonomous. For example, the systems
dx
= x2 − x
dt
and
x(n + 1) =
1
(x(n)2 − x(n))
2
are autonomous, but
dx
= x2 − t
dt
and
1
(x(n)2 − n)
2
are NOT autonomous. Much of the work in dynamical systems is for autonomous systems because
then the fixed points do not depend on time.
x(n + 1) =
4
CHAPTER 1. OVERVIEW
(b) Fixed points also called singular or equilibrium points.
We call x∗ a fixed point if at x = x∗ we have
dx
=0
dt
or, in the case of a difference equation, if at x(n) = x∗ we have
x(n + 1) = x(n) = x∗ .
As seen in the parachute example in section 1, fixed points can be an extremely effective tool.
(c) Stability or instability of fixed points
A formal definition and precise methods for determining the nature of fixed points are given later.
For now, we just illustrate the ideas with examples. Note that stable points are sometimes called
attractors, or attracting points, and unstable points are called repellors, or repelling points.
Examples
1.
dx
= x2 − x
dt
We solve the equation x2∗ − x∗ = 0 to get the fixed points. This gives x∗ = 0 or x∗ = 1. If x < 0
then dx/dt > 0, and if 0 < x < 1 then dx/dt < 0. If x > 1 then dx/dt > 0. We therefore get the
following phase line for this system:
(See the module APM114–V for a discussion of phase lines.) The point x∗ = 0 is a stable fixed
point, and x∗ = 1 is an unstable fixed point.
2.
dx
= x2
dt
Solving the equation x2∗ = 0 gives x∗ = 0 as the only fixed point. We see that dx/dt > 0 if x < 0
or if x > 0, so we get the following phase line:
This system is neither stable nor unstable. The reasons for this are discussed later – we include
this example to show some of the subtleties that can arise in dynamical systems theory.
1
3. x(n + 1) = − x(n) + 6
2
5
To find the fixed point(s) we substitute x∗ for x(n) and x(n + 1) in the difference equation to get
x∗ = − 21 x∗ + 6. Solving this equation for x∗ gives x∗ = 4. Plotting x(n) against n, starting at
x(0) = 2, gives the following graph:
(x(0) = 2, x(1) = 5, x(2) = 7/2, x(3) = 17/4, x(4) = 31/8, ...) No matter what starting value is
chosen for x(0), the same pattern results. Thus the fixed point at x = 4 is stable.
3. Higher order differential and difference equations
Everything described so far has been for first-order differential and difference equations. You may
wonder whether the methods of dynamical systems theory can be applied to higher-order equations.
The answer is “YES”, because one higher-order equation is equivalent to a system of first-order
equations. The following two examples illustrate this:
1. Consider the third-order differential equation
y
d3 y
+
dt3
dy
dt
2
+ ey = 3.
2
d y
Let x1 = y, x2 = dy
dt , and x3 = dt2 . Then we get an equivalent first-order system of equations
dx3
dt
dx2
dt
dx1
dt
=
1
3 − ex1 − x22
x1
=
x3
=
x2 .
2. Consider the difference equation
y(n + 1) = y(n)y(n − 1) − y(n)2 + a.
The order of a difference equation is the difference between the largest time and the smallest time
appearing in the equation. Here, the times appearing are (n + 1), n and (n − 1), so the order is
6
CHAPTER 1. OVERVIEW
(n + 1) − (n − 1) = 2. Thus the equation will be written as a system of two first-order difference
equations. First, it is convenient to rewrite the difference equation so that the smallest time value
appearing in it is n. We do this by replacing n − 1 by n, n by n + 1 and n + 1 by n + 2. Thus the
difference equation can be written as
y(n + 2) = y(n + 1)y(n) − y(n + 1)2 + a.
Now let x1 (n) = y(n), x2 (n) = y(n + 1). Then the above difference equation is equivalent to
x1 (n + 1)
=
x2 (n)
x2 (n + 1)
=
x2 (n)x1 (n) − x2 (n)2 + a.
There is more about this matter, and some more examples, in Luenberger [1979.96-97].
4. Plan for the rest of the module
The plan for the other chapters in this module is as follows: In Chapter 2 we discuss linear autonomous dynamical systems. This subject is important in its own right, and also because it forms
the basis for the material in Chapters 3 and 4.
Chapter 3 is about linear control theory. From a mathematical point of view, in the discrete case,
the system is linear but not autonomous, the system equation being
x(n + 1) = Ax(n) + u(n)
where the vector u(n) is called the control. The idea is to use the control u(n) to steer the system
state vector x to some desired value.
Chapter 4 discusses non-linear autonomous dynamical systems by linearising about the fixed points.
Particular attention is paid to systems of two first-order equations and the construction of phase
portraits. We also investigate non-linear autonomous dynamical systems using fully non-linear
techniques. We describe the Liapunov theory of stability, and also limit cycles for continuous systems and n cycles for discrete systems.
In Chapter 5 we touch on some more advanced topics: chaos, bifurcation theory and fractals.
Throughout the module we will illustrate the theory with applications from physics, biology, economics and the social sciences.
Chapter 2
LINEAR AUTONOMOUS
DYNAMICAL SYSTEMS
Linear Autonomous Dynamical systems are described by the equations
ẋ = Ax + f or x(n + 1) = Ax(n) + f
(1)
where x and f are vectors and A is a matrix. Neither A nor f varies with time. We consider in
section 1 the simplest of these systems, namely those where only one dimension is involved.
1. One-dimensional systems
In this case there is only one equation, so that A and f are just numbers a and b. Then equations
(1) become
ẋ = ax + f and x(n + 1) = ax(n) + f.
(2)
There are standard methods for writing down a general solution of (2), but here we proceed in a
slightly different way that is more in keeping with the dynamical systems approach.
In the continuous case ẋ = ax + f , the fixed point x∗ is found by solving
f
ẋ = 0 ⇒ ax∗ + f = 0 ⇒ x∗ = − .
a
(3a)
In the discrete case x (n + 1) = ax (n) + f , we replace x (n) and x (n + 1) with x∗ yielding
x∗ = ax∗ + f ⇒ x∗ =
f
.
1−a
(3b)
Let
X(t) = x(t) − x∗ or X(n) = x(n) − x∗ .
Then the continuous case in equation (2) becomes
ẋ (t) = Ẋ (t)
=
a (X (t) + x∗ ) + f
=
aX (t) + ax∗ + f
=
aX (t) ,
7
(4)
8
CHAPTER 2. LINEAR AUTONOMOUS DYNAMICAL SYSTEMS
since x∗ = − fa .
Similarly, for the discrete case in equation (2), we have
x (n + 1) = X (n + 1) + x∗ = a (X (n) + x∗ ) + f.
Hence
X (n + 1)
= aX (n) + x∗ (a − 1) + f
= aX (n) ,
f
since x∗ = 1−a
.
Equation (2) therefore becomes
Ẋ = aX or X(n + 1) = aX(n).
(5)
We see that the substitution (4) has got rid of the constant term f . Equations (5) can be solved
immediately. For the continuous case we have
Z
Z
Ẋ (t)
dt =
a dt
X (t)
ℓnX (t) = at + k (k is a constant)
so that
X (t) = Ceat (C = ek ).
From the initial condition X (0) it follows that
C = X (0)
so that the solution of Ẋ = aX is therefore given by
X (t) = X (0) eat .
(6a)
For the discrete case in equation (5), it follows from the initial condition X (0) that
X (1) = aX (0)
X (2) = aX (1) = a (aX (0)) = a2 X (0)
X (3) = aX (2) = a a2 X (0) = a3 X (0)
..
.
clearly yielding
X (n) = an X (0) .
Using (4), (3a) and (3b), the solutions of the original equation (2) are therefore
f
f
at
x(t) = − + e
x(0) +
a
a
(6b)
9
or
(7)
x(n) =
f
f
+ an x(0) −
1−a
1−a
.
[The only problem that can arise with (7) is if a = 1 (discrete case), or if a = 0 (continuous case).
Then (2) is
ẋ = f or x(n + 1) = x(n) + f,
which have solutions
x(t) = x(0) + f t and x(n) = x(0) + f n.
(8)
In this situation there are no fixed points unless f = 0, in which event every value of x is a fixed
point. However, these cases are not of much practical importance, so we will not dwell on them
further.]
From (6a) and (6b), it is easy to see under what conditions this dynamical system has a stable fixed
point. If |a| < 1 (discrete case), or a < 0 (continuous case) then an → 0 as n → ∞ or eat → 0 as
t → ∞, and therefore from equations (6a) and (6b) we see that X → 0 so that x → x∗ and the
fixed point is stable. On the other hand, if |a| > 1 (discrete case) or a > 0 (continuous case) then
X = x − x∗ → ∞ and the fixed point is unstable.
To summarize the nature of the fixed-point x∗ in the case of one-dimensional systems, we have the
following table:
discrete
case
continuous
case
|a| < 1
|a| > 1
a<0
a>0
stable
unstable
stable
unstable
We conclude this section with some examples to illustrate the above theory.
Example 1
Consider the discrete system
x(n + 1) =
1
x(n) + 2
2
with initial condition
x(0) = 2.
Then, replacing x (n + 1) and x (n) with the fixed point x∗ , we get
x∗ =
1
x∗ + 2
2
and therefore
x∗ = 4.
10
CHAPTER 2. LINEAR AUTONOMOUS DYNAMICAL SYSTEMS
Let X(n) = x(n) − 4. (This of course also means that X (n + 1) = x (n + 1) − 4.) In terms of the
new variable X the difference equation becomes
X (n + 1) + 4 =
1
(X (n) + 4) + 2,
2
clearly yielding
1
X(n)
2
n
1
X(0).
X(n) =
2
X(n + 1) =
with solution
Transforming back to the original variable x gives
n
1
x(n) =
(x(0) − 4) + 4
2
n−1
n
1
1
(−2) + 4 = −
+ 4.
=
2
2
Since (1/2)n → 0 as n → ∞ we see that x(n) → 4 as n → ∞ and the fixed point x∗ = 4 is stable.
Plotting a graph we get the following:
Example 2
Consider the discrete system
x(n + 1) = 2x(n) + 1
with initial condition
x(0) = 3.
Then
x∗ = 2x∗ + 1
and therefore
x∗ = −1.
11
Let X(n) = x(n) + 1. In terms of the new variable X the difference equation becomes
X(n + 1) = 2X(n)
with solution
X(n) = 2n X(0).
Transforming back to the original variable x gives
x(n) = 2n (x(0) + 1) − 1 = 2n+2 − 1.
Therefore x(n) → ∞ as n → ∞ and the fixed point is not stable.
Example 3
Consider the discrete system
2
x(n + 1) = − x(n), x(0) = 3.
3
This is easy to solve. The solution is x(n) = 3(−2/3)n .
Since (−2/3)n → 0 as n → ∞ the system is stable.
Example 4
Consider the system
dx
= −2x + 1, x(0) = −1.
dt
Since the system is continuous, the fixed point is found by solving
dx
= 0 ⇒ 0 = −2x∗ + 1,
dt
which gives
x∗ =
1
.
2
Making the substitution
1
X (t) = x (t) − ,
2
the differential equation becomes
dX
= −2X,
dt
in terms of the new variable X which yields the solution
X (t) = X (0) e−2t .
Transforming back to the original variable x gives
1
1 −2t
x (t) − = x (0) −
e ,
2
2
12
CHAPTER 2. LINEAR AUTONOMOUS DYNAMICAL SYSTEMS
which, together with the initial condition x (0) = −1, leads to the solution
1
1 −2t
x (t) =
+ −1 −
e
2
2
1 3 −2t
=
− e .
2 2
Example 5
Consider the system
3
dx
= x − 3,
dt
2
The solution is
x(0) = 0.
3
x(t) = −2e 2 t + 2.
2. Two-dimensional systems
We now move on to the two-dimensional case, that is, in equation (1) x and f are two-vectors and
A is a 2 × 2 matrix. The theory is described for the continuous case (ẋ = Ax + f ), and the
corresponding equations for the discrete case are summarised in section 5. The dynamical system
ẋ = Ax + f
(9)
has a fixed point provided ẋ = 0, so the fixed point is given by
Ax + f = 0.
Therefore
x∗ = −A−1 f .
(10)
X(t) = x(t) − x∗
(11)
Ẋ = AX.
(12)
We assume that det A) ̸= 0.
Then, as in the one-dimensional case, let
so that (9) becomes
Put in words, the transformation (11) moves the fixed point from −A−1 f to the origin. In order to
understand the properties of equation (9), it is sufficient to understand the simpler equation (12).
Even so, the solution of (12) is rather more complicated than solving equation (5).
Since the solution of the one-dimensional equation (5) is X = constant × exp(constant ×t), we try
as a solution of (12)
X = Keλt .
(13)
Note: This form of solution may appear to be just a lucky guess, which to some extent is.
Substituting (13) into (12) gives
Kλeλt = AKeλt .
13
Therefore
AK = λK.
(14)
Equation (14) may be familiar as the equation defining an eigenvector K, with eigenvalue λ, of a
matrix A. Eigenvalues and eigenvectors are covered in APM113 and MAT211, but not in MAT103.
Thus students taking this module may not have come across eigenvalues and eigenvectors. The
notes that follow are aimed at students with no prior knowledge of eigenvalue theory, and where
necessary standard results will simply be stated.
Equation (14) may be rewritten:
where I is the identity matrix
1
0
(A − λI)K = 0
(15)
0
. We require that K should be non-trivial (i.e. K ̸= 0),
1
which implies that
det(A − λI) = 0.
(16)
[WHY? Because if det(A − λI) ̸= 0, then (A − λI)−1 exists, and multiplying (15) by (A − λI)−1
gives K = (A − λI)−1 .0 = 0, which is not wanted.]
Equation (16) is an equation that determines λ and can be regarded as a generalisation of the
one-dimensional case in which λ = a. Writing
a b
A=
(17)
c d
then equation (16) becomes
a−λ
c
b
d−λ
= 0.
That is
(a − λ)(d − λ) − bc = 0.
(18)
which is a quadratic equation for λ. It can be rewritten as
λ2 − λ(a + d) + (ad − bc) = 0
or
λ2 − λσ + ∆ = 0
(19)
where σ = a + d is the sum of the diagonal components of A and is called the trace of A, and
∆ = ad − bc is det(A). The solution of (19) is
p
p
σ − σ 2 − 4△
σ + σ 2 − 4△
, λ2 =
.
(20)
λ1 =
2
2
In general, the procedure for obtaining
the solution of equation (12) is as follows: First, we use (15)
Kx
to find the eigenvector K(=
):
Ky
(A − λI)K = 0
(15)
14
CHAPTER 2. LINEAR AUTONOMOUS DYNAMICAL SYSTEMS
Now, because det(A − λI) = 0, (15) leads to the following two equations for Kx , Ky that are
multiples of each other:
aKx + bKy
=
0
caKx + cbKy
=
0
[For a 2 × 2 matrix det = 0 means that the two rows of the matrix are multiples of each other.]
Thus K is defined in direction only, and we represent this by writing down the solution of (15) as
cK for some undetermined constant c. Of course there are two eigenvalues λ1 , λ2 so there are two
corresponding eigenvectors
c1 K1 , c2 K2 .
Then the general solution of (12) is
X(t) = c1 K1 eλ1 t + c2 K2 eλ2 t .
(21)
For any choice of X(0) we can find c1 and c2 such that
X(0) = c1 K1 + c2 K2 .
Further, these values of c1 , c2 are unique. This is because eigenvector theory shows that if all the
eigenvalues are distinct, then the eigenvectors form a basis of the linear space. [The special case of
(16) having equal roots is considered later.] It is convenient to give examples of the above general
procedure under three sub-headings:
(A) σ 2 − 4∆ > 0: There are two distinct real solutions for λ.
(B) σ 2 − 4∆ < 0: There are two distinct complex solutions for λ.
(C) σ 2 − 4∆ = 0: “Equal roots”.
Case (A)
Once λ1 and λ2 are known, equation (15) is solved for K1 and K2 .
Example 1
Consider the system
dx
=
dt
dy
=
dt
Let X =
x
y
x + 3y
5x + 3y.
1
, then Ẋ = AX with A =
5
3
3
(22)
. Thus equation (19) becomes
λ2 − λ(4) − 12 = 0.
Thus λ1 = +6, λ2 = −2. Let K1 and K2 be the eigenvectors corresponding to eigenvalues λ1 , λ2
respectively. Then equation (15), for K1 is
−5
3
0
K1 =
.
(23)
5 −3
0
15
Note that although (23) looks like two equations, one is a multiple of the other so there is in effect
only one equation. As remarked earlier, this always happens because λ has been chosen so that
det(A − λI) = 0, and for a 2 × 2 matrix a zero determinant means that one row is a multiple of the
other. For example let
a
K1 =
.
b
Then (23) becomes
−5a + 3b =
0
...(A)
5a − 3b =
0
...(B)
[Note that (B) = −1 × (A).] Choose for example, a = 3c1 , then b = 5c1 , where c1 is an arbitrary
constant. The solution of (23) can therefore be written:
3
K1 = c1
5
for some constant c1 . Note that the choice of a and b are not unique. We could also have chosen
a = c1 (say). Then b = 35 c1 and our vector K1 would be given by
K1 = c1
1
5
3
.
However, this is just a constant multiple of the previous solution of K1 :
1
1
5
c1 5 = c1
!
3
5
3
Similarly for K2 , equation (15) is
3
5
3
5
so that
K2 = 0
K 2 = c2
1
−1
.
Thus from (13), and using the principle of superposition for linear systems, the general solution of
(22) is
x
3
1
6t
X=
= c1
e + c2
e−2t .
(24)
y
5
−1
[Reminder: This means
Note that X → ∞ as t → ∞.
x =
3c1 e6t + c2 e−2t
y
5c1 e6t − c2 e−2t ].
=
16
CHAPTER 2. LINEAR AUTONOMOUS DYNAMICAL SYSTEMS
The above solution is the general solution because it has two arbitrary constants. This is correct
as (21) is a system of two first-order differential equations. The solution curves for various values
of c1 , c2 are shown in Figure 1. The fixed point at (0, 0) is called a saddle point.
Example 2
Consider the system
dx
=2x + 8y,
dt
dy
=4y,
dt
25
so that clearly
A=
2
0
8
4
.
Then (19) becomes
λ2 − 6λ + 8 = 0.
a
Solving this equation we get λ1 = 2, λ2 = 4. Let K1 =
be the eigenvector corresponding to
b
the eigenvalue λ1 = 2. Then, from equation (15) we need to solve
(A − 2I) K1 = 0,
17
that is
0
0
8
2
a
b
=
0
0
yielding the two equations
8b
=
0
2b
=
0.
This clearly gives b = 0, and since there is no restriction on the unknown a, we can choose a = c1 ,
where c1 is an arbitrary constant. Hence, K1 is given by
1
K1 = c1
.
0
c
Next, let K2 =
be the eigenvector corresponding to the eigenvalue λ2 = 4. Then from
d
(A − λ2 I) K1 = 0 (see equation (15)), we get
−2 8
c
0
=
0 0
d
0
that is
−2c + 8d = 0 ⇒ 2c = 8d ⇒ c = 4d.
Again we have only one equation with two unknowns. [The second equation is zero multiplied by
the first one.] Clearly we can solve one unknown in terms of the other, that is, let d = c2 , then
c = 4d = 4c2 giving the vector
4
K2 = c2
1
with c2 an arbitrary constant.
Therefore
X=
x
y
= c1
1
0
e2t + c2
4
1
e4t .
(26)
The solution curves for various values of c1 , c2 are shown in Figure 2. The fixed point at (0, 0) is
called an unstable node. Clearly from (26) it follows that X → ∞ as t → ∞.
18
CHAPTER 2. LINEAR AUTONOMOUS DYNAMICAL SYSTEMS
Figure 2: Unstable node
Example 3
Consider the system
dx
= − x − 2y + 4
dt
dy
=x − 4y + 2.
dt
27
First we need to find the location of the fixed point (x∗ , y∗ ) by solving the equations
0
=
−x∗ − 2y∗ + 4,
0
=
x∗ − 4y∗ + 2.
These are easily solved to give x∗ = 2, y∗ = 1. Let X = (x − x∗ , y − y∗ ). Then
dX
−1 −2
=
X.
1 −4
dt
Equation (19) becomes
λ2 + 5λ + 6 = 0,
with solutions λ1 = −2, λ2 = −3. Substituting into equation (15) we get
1 −2
2 −2
K1 = 0,
K2 = 0.
1
2
1 −1
19
2
1
The solutions are K1 = c1
, K2 = c2
. Then
1
1
x − x∗
2
1
X=
= c1
e−2t + c2
e−3t ,
y − y∗
1
1
or
x =2c1 e−2t + c2 e−3t + 2
y =c1 e−2t + c2 e−3t + 1.
28
The solution curves for various values of c1 , c2 are shown in Figure 3. The fixed point at (2, 1) is
called a stable node. Also note from (28) that (x, y) → (2, 1) as t → ∞.
Case (B)
In this case equation (19) has two distinct complex solutions for λ. Writing β =
the solutions are
σ
σ
λ1 = + iβ, λ2 = − iβ.
2
2
Note that λ1 and λ2 are complex conjugates of each other.
Example 1
p
4 △ −σ 2 /2,
(29)
20
CHAPTER 2. LINEAR AUTONOMOUS DYNAMICAL SYSTEMS
Consider the system
dx
=6x − y,
dt
dy
=5x + 4y.
dt
Then
30
A=
6
5
−1
4
so that equation (19) becomes
λ2 − 10λ + 29 = 0.
The roots λ1 , λ2 are
λ1 = 5 + 2i, λ2 = 5 − 2i.
Thus equation (15) for K1 becomes
(1 − 2i)
−1
K1 = 0.
5
(−1 − 2i)
(31)
(32)
[Note. As in case (A), the second row of the matrix is linearly dependent on the first – just multiply
row one by (1 + 2i) to see this.] Solving (32) gives
1
K1 = c1
.
(33)
1 − 2i
Similarly, we find the other eigenvector to be
K2 = c2
1
1 + 2i
.
(34)
Note. If the matrix A is real, the vector K1 is the complex conjugate of K2 . It is simple to prove
that this is always the case, but we will not do it here. Thus the solution of (30) can be written as
x
1
1
X=
= c1
e(5+2i)t + c2
e(5−2i)t .
(35)
y
1 − 2i
1 + 2i
However, it would be useful to be able to interpret (35) in terms of real functions only. Using
e2it = cos(2t) + i sin(2t), equation (35) can be written as
x =e5t [(c1 + c2 ) cos 2t + (c1 i − c2 i) sin 2t],
y =e5t [(1 − 2i)(c1 cos 2t + ic1 sin 2t)
+ (1 + 2i)(c2 cos 2t − ic2 sin 2t)]
=e5t [(c1 + c2 ) − 2(c1 i − c2 i)] cos 2t
+ e5t [2(c1 + c2 ) + (c1 i − c2 i)] sin 2t.
36
21
Let
C1 = c1 + c2 and C2 = c1 i − c2 i.
(37)
Then
X=
x
y
cos 2t
e5t
cos 2t + 2 sin 2t
sin 2t
+ C2
e5t .
−2 cos 2t − sin 2t
=C1
38
Since we want C1 , C2 to be real, (37) implies that c1 and c2 should be complex conjugates of each
other.
Equation (38) is then an interpretation of (35) in terms of real functions.
There is a general and straightforward procedure for deriving the general, real solution (38) from
(31) (solution for λ) and (33) (solution for the first eigenvector K1 ). Use the notation K1 to mean
the complex conjugate of K1 , so K1 = K2 , and define
1
(K1 + K1 )
2
1
B2 =Im(K1 ) = (K1 − K1 ),
2
where the constant c1 multiplying K1 is taken as 1. Then
B1 =Re(K1 ) =
39
X = [C1 (B1 cos βt − B2 sin βt) + C2 (B2 cos βt + B1 sin βt)]e(σ/2)t .
(40)
The general proof of (40) is not difficult, but it involves some slightly intricate complex number
manipulations, and is omitted.
IMPORTANT: In the above formula the eigenvector K1 must be worked out for the eigenvalue
λ1 , which has a positive imaginary part.
In Example 1, K1 is given by (33), so that from
1
1
0
=
+i
1 − 2i
1
−2
we have that
B1 = Re (K1 ) =
1
1
, B2 = Im (K1 ) =
0
−2
.
Also σ/2 = 5 and β = 2. Thus the solution is
1
0
5t
X = e C1
cos 2t −
sin 2t
1
−2
0
1
5t
+e C2
cos 2t +
sin 2t ,
−2
1
which is the same as (38).
The solution curves are shown in Figure 4. The fixed point at (0,0) is called an unstable focus.
22
CHAPTER 2. LINEAR AUTONOMOUS DYNAMICAL SYSTEMS
Example 2
Consider the system
d
dt
x
y
=
−1
− 21
2
−1
x
y
+
2
1
.
(41)
The fixed point (x∗ , y∗ ) is found by solving
−1
2
x∗
2
=
−
y∗
1
− 21 −1
so that x∗ = 2, y∗ = 0. Then the eigenvalue equation (19) is
λ2 + 2λ + 2 = 0,
which has solutions λ1 = −1 + i, λ2 = −1 − i. Thus σ/2 = −1, β = 1.Then K1 =
from solving
−i
− 21
2
−i
a
b
=
0
0
,
a
b
is found
23
that is
−ia + 2b
1
− a − ib
2
=
0 ...(1)
=
0 ...(2).
From (1), choose a = 1 (say), then b = 2i so that
K1
=
1
i
2 0
1
=
+i 1
0
2
and therefore
B1 =
1
0
and B2 =
0
1
2
.
Hence
X =
0
x − x∗
1
−t
=e
C1
cos t− 1 sin t
y − y∗
0
2
0
1
+C2
cos t +
sin t .
1
0
2
Thus
x =2 + e−t (C1 cos t + C2 sin t)
1
1
y =e−t C1 sin t + C2 cos t .
2
2
The solution curves are shown in Figure 5. The fixed point at (0, 0) is a stable focus.
42
24
CHAPTER 2. LINEAR AUTONOMOUS DYNAMICAL SYSTEMS
Example 3
Consider the system
Ẋ =
2
−1
8
−2
X.
(43)
Equation (19) is
λ2 + 4 = 0
so that
λ1 = 2i, λ2 = −2i.
The equation for K1 is
(2 − 2i)
8
−1
(−2 − 2i)
which has a solution
K1 =
2 + 2i
−1
=
so that
and
B2 =
2
−1
2
−1
2
0
B1 =
K1 = 0
.
+i
2
0
25
Since σ = 0 the solution is
X = C1
2 cos 2t − 2 sin 2t
− cos 2t
+ C2
2 cos 2t + 2 sin 2t
− sin 2t
.
(44)
The solution curves are shown in Figure 6. The fixed point (0,0) is called a centre.
Case (C)
Here we have “equal roots”, so σ 2 − 4∆ = 0. This is a special case, but it is not important for the
purposes of this module. Let’s just briefly summarise the solutions. There are really two subcases:
C − 1:
Here
Ẋ =
which, for X =
x (t)
y (t)
a
0
0
a
a
0
0
a
X
(45)
, can be written as
ẋ (t)
ẏ (t)
=
x (t)
y (t)
that is
ẋ (t) = ax (t)
and ẏ = ay (t) .
26
CHAPTER 2. LINEAR AUTONOMOUS DYNAMICAL SYSTEMS
These two equations clearly have the solutions
x (t) = c1 eat and y (t) = c2 eat
so that
X=
x
y
= c1
1
0
e
at
+ c2
0
1
eat .
(46)
C − 2:
All other cases in which σ 2 − 4∆ = 0. It turns out that there is only one eigenvector K. So one
solution is
X = c1 Keλt
(47)
but the second solution still needs to be found. It is
X = c2 (Kteλt + Peλt ).
(48)
(AK − λK)teλt + (AP − λP − K)eλt = 0
(49)
(A − λI)K = 0
(50)
(A − λI)P = K.
(51)
Substituting (48) into (12) yields
so that
and
Equation (50) is the usual eigenvector equation for K and (51) then determines P.
Example
Consider the system
Ẋ =
−18
−9
3
2
X.
(52)
The eigenvalue equation is
λ2 + 6λ + 9 = 0
or
(λ + 3)2 = 0,
so
λ = −3.
Then the equation for K is
which has a solution K =
3
1
6
2
−18
−6
K = 0,
. Then from (51) the equation for P is
6
2
−18
−6
P=
3
1
27
that is
2a − 6b = 1.
Now choose (say) b = 0, then a = 21 so that P =
X = c1
3
1
e−3t + c2
1 2
. Thus the general solution of (52) is
0
3
1
te−3t +
1 2
0
e−3t .
(53)
3. Nature of fixed points: stability
The question of stability of a fixed point is rather more complicated than for the case of one firstorder equation. There are the following possibilities determined by the parameters σ(= trace A)
and ∆(= det(A)):
1. σ < 0, 0 < ∆ < σ 2 /4. This means that both eigenvalues λ1 , λ2 are distinct, real and negative.
This type of fixed point is called a stable node and was illustrated earlier in Figure 3.
2. σ > 0, 0 < ∆ < σ 2 /4. This means that both eigenvalues λ1 , λ2 are distinct, real and positive.
This is called an unstable node and is illustrated in Figure 2.
3. ∆ < 0. This implies that the eigenvalues λ1 , λ2 are distinct, real and of opposite sign. The
fixed point is then a saddle point, and is illustrated in Figure 1.
4. ∆ > σ 2 /4, σ > 0. The eigenvalues λ1 , λ2 are complex and with positive real part. The fixed
point is an unstable focus (also called an unstable spiral -point) and is illustrated in Figure 4.
5. ∆ > σ 2 /4, σ < 0. The eigenvalues λ1 , λ2 are complex and with negative real part. The fixed
point is a stable focus (also called a stable spiral point) and is illustrated in Figure 5.
6. σ = 0, ∆ > 0. The eigenvalues λ1 , λ2 are purely imaginary. The fixed point is a centre and is
illustrated in Figure 6.
The above description of fixed points is summarised in the following diagram:
28
CHAPTER 2. LINEAR AUTONOMOUS DYNAMICAL SYSTEMS
For stability we require
∆ > 0 and
σ < 0.
4. Higher dimensional systems
Next we investigate the general case for continuous systems
ẋ = Ax + f
(54)
where the dimensionality is taken as m. As before, assuming A−1 exists,
ẋ = 0 at x∗ = −A−1 f .
(55)
X = x − x∗
(56)
Ẋ = AX.
(57)
Let
then
For simplicity we assume that the eigenvalues of A are distinct, although they may be complex:
λ1 , ..., λm . The eigenvectors Ki of A are found by solving
(A − λi I)Ki = 0 (i = 1, ..., m).
(58)
Then a general solution of (54) is
x=
m
X
(ci Ki eλi t ) − A−1 f .
(59)
i=1
For a large system, carrying out the procedure to obtain (59) from (54) is tedious if not impractical.
Is there any useful information that can be obtained from (54) without finding the complete general
solution? The stability of the solution about the fixed point x∗ is often of great importance, and is
29
determined entirely by the eigenvalues λ1 , ..., λm . More precisely, it is obvious from (59) that (54)
is stable at x = x∗ if and only if
Re(λi ) < 0 (i = 1, ..., m).
60
[Although not part of this module, we should mention that there are powerful and efficient computer methods for calculating the eigenvalues of a large matrix; it is horribly inefficient to form
det(A − λI) = 0 and then solve a polynomial equation in λ.]
Example 1
Consider the system

−4 1
Ẋ =  1 5
0 1

1
−1  X.
−3
(61)
Thus det(A − λI) is

−4 − λ
1
det 
0
1
5−λ
1

1
−1  ,
−3 − λ
which is equal to
(−4 − λ) [(5 − λ) (−3 − λ) − (−1)] − 1 [(−3 − λ) − 1]
= (−4 − λ) λ2 − 2λ − 14 − (−λ − 4)
= (−4 − λ) λ2 − 2λ − 15
=
(−4 − λ) (λ + 3) (λ − 5) .
Thus the eigenvalues are –3, –4 and 5, and the general solution is
X = c1 K1 e−3t + c2 K2 e−4t + c3 K3 e5t
(62)
where the eigenvectors K1 , K2 , K3 have not been calculated. Even so, it is clear that the solution
(62) is unstable [because of the term e5t ] and is some form of three-dimensional generalisation of a
saddle point.
Example 2
Consider the system

−1
ẋ =  1
1



1 −2
−1
−1
0  x+  3  .
0 −1
2
First, the fixed point x∗ is the solution of 0 = Ax∗ +f , and after some linear algebra we find


−1
x∗ =  2  .
1
(63)
30
CHAPTER 2. LINEAR AUTONOMOUS DYNAMICAL SYSTEMS
Then putting X = x − x∗ gives Ẋ = AX, and the eigenvalues are found from 0 = det(A − λI):


−1 − λ
1
−2

1
−1 − λ
0
det 
1
0
−1 − λ
=
(−1 − λ)[(−1 − λ)2 − 1] + 1(−(−2)(−1 − λ))
= −(λ + 1)3 + 1 + λ − 2 − 2λ
= −λ3 − 3λ2 − 4λ − 2
Thus we seek the solutions of
λ3 + 3λ2 + 4λ + 2 = 0.
One solution is λ = −1, so the above equation becomes
(λ + 1)(λ2 + 2λ + 2) = 0
which is straightforward to solve, yielding
λ1
=
−1,
λ2
=
−1 + i,
λ3
=
−1 − i.
The general solution is therefore
X = c1 K1 e−t + c2 K2 e(−1+i)t + c3 K3 e(−1−i)t .
(64)
It is clear that the solution is stable, since all eigenvalues have negative real part. The term e±it can
be expressed in terms of sin t and cos t, and therefore the solution has a component which oscillates
with frequency 1/2π. [Recall that sin ωt or cos ωt has period T = 2π/ω and frequence ω/2π.] In
some applications the frequency or frequencies may be important.
5. Discrete systems
Let’s first we describe the general situation and then investigate some aspects of 2 − D systems in
more detail.
The equation describing the system is
x(n + 1) = Ax(n) + f .
(65)
A fixed point is x∗ given by solving
x∗ = Ax∗ + f .
Therefore
x∗ = (I − A)−1 f .
(66)
[We assume that (I − A) has an inverse.] Then let X = x − x∗ so that
X(n + 1) = AX(n).
(67)
31
Of course one can write down the solution of (67) as
X(n) = An X(0).
(68)
Unfortunately (68) does not give much insight into how X(n) evolves in [discrete] time. Rather, we
use eigenvalue theory again. For simplicity, we assume that A has m distinct eigenvalues λ1 , ..., λm ,
with corresponding eigenvectors K1 , ..., Km . Then, as discussed earlier, the eigenvectors form a
basis of the linear space and for any X(0) we can write
X(0) =
m
X
ci Ki
(69)
i=1
where the constants ci are unique. Putting (69) into (68) gives
X(n) =
m
X
ci An Ki =
i=1
m
X
ci λni Ki =
i=1
m
X
λni ci Ki .
(70)
i=1
From (70) it is clear that if |λi | < 1(i = 1, ..., m) then X(n) → 0 as n → ∞ for any given initial
condition X(0). Such a system is stable. On the other hand if any |λi | > 1 then, under certain
initial conditions, X(n) → ∞ as n → ∞ and the system is unstable.
Thus the long term behaviour of a discrete dynamical system (65) is determined by the eigenvalues
of A. We illustrate this and some of the subtleties that can arise with examples.
Example 1
Consider the system
y(n + 1) = −x(n) + 4y(n)
x(n + 1) = x(n) + 2y(n),
(71)
with
x(0) = 1,
Let
y(0) = 2.
X (n) =
x (n)
y (n)
1
−1
.
Then equation (71) can be written as
X (n + 1) =
with
X (0) =
2
4
1
2
X (n)
.
We have calculated eigenvalues and eigenvectors for a 2 × 2 matrix many times in this chapter, and
we just state the result:
2
1
λ1 = 2, K1 =
, λ2 = 3, K2 =
1
1
32
CHAPTER 2. LINEAR AUTONOMOUS DYNAMICAL SYSTEMS
We must now express X(0) in terms of K1 and K2 , that is we find constants c1 and c2 such that
X(0) = c1 K1 + c2 K2 .
This gives
1
=
2c1 + c2
2
= c1 + c2 .
Solving we get c1 = −1, c2 = 3. Thus the solution to the problem is
2
1
X(n) = c1 K1 λn1 + c2 K2 λn2 = −1
2n + 3
3n .
1
1
Thus
x(n) =
−2n+1 + 3n+1
y(n) =
−2n + 3n+1 .
(72)
Example 2
Consider the system
X (n + 1) =
−0.25
0.5
0.5
1
with
X (0) =
1
2
X (n)
.
(73)
The eigenvalue equation is
λ2 − λ + 0.5 = 0
so that
λ1
=
0.5 + 0.5i,
λ2
=
0.5 − 0.5i.
K1
=
K2
=
The eigenvectors are
i
2
−i
2
,
.
Thus the general solution is
X(n) = c1 (0.5 + 0.5i)n
i
2
The constants c1 , c2 are determined from X(0) =
+ c2 (0.5 − 0.5i)n
1
2
.
−i
2
.
33
We get
=
i (c1 − c2 )
2 (c1 + c2 )
c1
=
0.5 − 0.5i
c2
=
0.5 + 0.5i.
1
2
which are solved to give
Therefore
i
(0.5 − 0.5i)(0.5 + 0.5i)n
2
−i
n
+ (0.5 + 0.5i) (0.5 − 0.5i)
.
2
X(n) =
(74)
There is one important point to make about equation (74). The second term is the complex conjugate of the first term, therefore their sum is real. This type of behaviour must always occur when
the matrix A is real, because it is obvious that if X(0) is real, then so is X(n).
One can also interpret (74) in terms of sine and cosine functions. Using the (r, θ) representation of
complex numbers z = a + ib, that is
z = reiθ
where
r
=
θ
=
p
a 2 + b2
b
tan−1 ,
a
we write
(0.5 − 0.5i)
=
(0.5 + 0.5i)
=
1
√ e−iπ/4 ,
2
1 iπ/4
√ e
.
2
Then (74) becomes
n+1
iπ/2 1
e
√
eiπ/4(n−1)
2
2
n+1
−iπ/2 1
e
+ √
eiπ/4(1−n)
2
2
n+1 iπ/4(n−1)+iπ/2
1
e
+ eiπ/4(1−n)−iπ/2
√
2eiπ/4(n−1) + 2eiπ/4(1−n)
2
n+1 1
2 cos(π/2 + (n − 1)π/4)
√
4 cos((n − 1)π/4)
2
n−1 1
− sin((n − 1)π/4)
√
,
2 cos((n − 1)π/4)
2
X(n) =
=
=
=
75
34
CHAPTER 2. LINEAR AUTONOMOUS DYNAMICAL SYSTEMS
where the second last line of (75) was derived from
eit + e−it
.
2
cos t =
Example 3
Consider the system
x (n + 1) =
2
0
1
2
x (n) +
2
−1
.
(76)
Then the fixed point is given by
x∗ =
Therefore
1
0
2
0
1
1
1
2
x∗ +
−2
−1
x∗ +
and hence
x∗ =
1
1
−2
−1
.
=0
.
Let
X(n) = x(n) − x∗ .
Then
X (n + 1) =
2
0
1
2
X (n) .
The eigenvalue equation is
(λ − 2)2 = 0.
There is therefore a repeated root at λ1 = 2. As in the continuous system case, we describe
the procedure for dealing with this situation,
but we do not prove it here. Let K1 be the single
1
eigenvector, [so that here K1 = c1
]. Then define P by
0
(A − λI)P = K1 .
[Here P = c2
0
1
(76)
.] Then the general solution is
X(n) = c1 λn1 K1 + c2 (λn1 P + nλn−1
K1 )
1
(77)
so that for this example
X(n) = c1 2n
Thus
x(n) = c1 2n
1
0
1
0
0
1
+ c2 2n
+ n2n−1
.
1
0
0
1
1
+ c2 2n
+ n2n−1
+
.
1
0
1
For discrete systems the fixed points can be classified as follows:
(78)
35
(a) λ1 , λ2 are complex, and further λ2 = λ1 . In this case, as for continuous systems, there is a
spiralling effect towards the fixed point if |λ| < 1 (stable case), and away from it if |λ| > 1
(unstable case).
(b) If λ1 , λ2 real and both are greater than 1, the behaviour is analogous to an unstable node.
(c) If λ1 , λ2 are real and 0 < λ1 , λ2 < 1, the behaviour is analogous to a stable node.
(d) If λ1 , λ2 are real and 0 < λ1 < 1 < λ2 , the behaviour is analogous to a saddle point.
(e) If one or both λ1 , λ2 are negative, there is no analogy with the continuous situation. The
solution keeps jumping from one side of the fixed point to the other (see section 1, Example
3). If |λ1 | < 1 and |λ2 | < 1 then the behaviour is stable, and if either |λ1 | > 1 or |λ2 | > 1
then it is unstable.
6. Some Applications
(i) Lanchester model of combat. [These models are named after F.W. Lanchester who investigated
mathematical models of combat situations during World War I.]
Suppose that two military forces are engaged in combat. Let x(t) and y(t) be the strengths
of the two forces in some appropriate units. For example, the units could be number of soldiers, number of aircraft, or number of ships, depending on the type of battle being fought.
Strictly speaking x(t), y(t) are integers, but it is convenient to regard them as continuous,
differentiable functions of time.
The key assumption in the model is that each force has a “hitting strength” that is proportional to its size; this “hitting strength” reduces the other force as follows:
dx
= −αy,
dt
dy
= −βx
dt
(79)
where α and β are constants that depend upon the level of technology, the level of training
of the soldiers, etc. We now investigate the consequences of the model.
Clearly, there is a fixed point at the origin. The eigenvalue equation (16) is
λ2 − αβ = 0.
which gives
p
p
λ1 = − aβ, λ2 = + aβ.
The eigenvector equations are then
√
αβ
−β
√−α
αβ
K1x
K1y
=0
(80)
36
CHAPTER 2. LINEAR AUTONOMOUS DYNAMICAL SYSTEMS
and
√
− αβ
−β
−α
√
− αβ
K2x
K2y
= 0.
These equations give
K1 = c1
√ √ √α , K2 = c2 −√α .
β
β
(81)
Since λ1 , λ2 are of opposite signs, the fixed point is a saddle point. We can therefore draw
the following phase plane diagram:
Of course there cannot be a negative number of soldiers, etc., so on physical grounds x(t) ≥ 0
and y(t) ≥ 0. Thus the only part of the phase plane diagram that is relevant is the positive
quadrant:
37
In the sketch above you will see a dividing line between those solutions which, as time increases,
lead to a total elimination of x (above the line) and those which lead to a total elimination
of y (below the line). If you start anywhere on the line, both x and y will eventually be
eliminated. Whether you start on, above or below the line clearly depends on the initial
conditions x (0), y (0), so in order to derive the relationship between x (0) and y (0), which
determines which force will be eliminated, we calculate for the solution
√ √
√ √
α
−√α
e− αβt + c2
e αβt ,
X (t) = c1 √
β
β
the values of the constants c1 and c2 in terms of the initial conditions x (0), y (0) as follows:
√
x (0)
α (c1 − c2 )
= √
,
y (0)
β (c1 + c2 )
yielding
c1
=
c2
=
√
√
β + y (0) α
√
2 αβ
√
√
y (0) α − x (0) β
√
.
2 αβ
x (0)
Since we want the relationship between x (0) and y (0) when both x and y are eliminated as
time increases, we solve
x (t)
0
lim
=
,
y (t)
0
t→∞
that is
√ √ √
√
α
−√ α
0
− αβt
αβt
√
+ c2
=
c1
lim e
lim e
.
β t→∞
β
0
t→∞
38
CHAPTER 2. LINEAR AUTONOMOUS DYNAMICAL SYSTEMS
This clearly gives
c2 = 0
yielding the desired relationship
y (0) = x (0)
p
β/α.
We can therefore clearly summarise the outcome of the battle in terms of the initial conditions
as follows:
q
β
then x is totally eliminated.
(a) If y(0) > x(0) α
q
β
(b) If y(0) < x(0) α
then y is totally eliminated.
p
(c) If the two sides are equally matched, y(0) = x(0) β/α, then the two sides eliminate
each other.
(ii) The cohort model of age distribution in a population
We investigate the growth of a population that is broken into age groups. We let x1 (n), x2 (n), ..., xm (n)
represent the number in the m age groups at the beginning of time period n. Depending on the
species, the time period could be anything from a decade (perhaps appropriate for humans)
to a day (e.g. certain insects). Next we introduce the birthrates so that
x1 (n + 1) = b1 x1 (n) + b2 x2 (n) + ... + bm xm (n).
(82)
The idea is that only the first cohort is increased due to birth, and the birthrate depends
very much on the age of the parents. The other cohorts are not affected by birth but by the
survival rate from the previous cohort:
xi (n + 1) = Si−1 xi−1 (n).
(i = 2, ..., m)
Of course 0 < Si < 1. We can then write the above equations in matrix form


b1 b2 · · · · · ·
···
· · · · · · bm
 S1 0 · · · · · ·
···
0 



.. 
 0 S2

0
.


x (n + 1) =  .
 x (n) .
.
.
.
.
.
.
 ..
.
.
0
. 


 .
..
.. 
..
..
 ..
.
.
.
. 
0
0
0
Sm−1
0
(83)
(84)
This model is often used to answer the question: What is the age distribution of the population? To give the answer, we need to know the dominant eigenvalue and corresponding
eigenvector. The term “dominant eigenvalue” means the eigenvalue of greatest absolute magnitude. It can be shown [Luenberger, 1979:170–173], under reasonable assumptions about
the birthrates bi , that the dominant eigenvalue λ1 is positive, real and simple (i.e. it is not a
39
repeated root). Then if K1 is the corresponding eigenvector, it is also real. Thus the solution
is of the form
c1 K1 λn1 +
x(n) =
m
X
ci Ki λni
i=2
"
λn1
=
c1 K1 +
m
X
ci Ki
i=2
Now because λ1 > |λi | and n → ∞, the
λi
λ1
n #
.
85
Pm
i=2 term tends to zero. Thus, as n → ∞,
x(n) → λn1 c1 K1 .
(86)
Thus for large n the population distribution among the various age cohorts is given by K1 .
Note that technically this conclusion assumes that the initial population is such that c1 ̸= 0.
However, even if c1 = 0 in the initial population, the situation is not exactly described by
equation (84), or indeed by any other equation. In practice there is a certain amount of
randomness in the evolution of populations. This randomness would ensure that c1 would
become non-zero, and so equation (86) still applies.
Example
Consider the fictional species S. Suppose that, measuring n in years, the population can be
divided into three age groups: x1 , x2 , x3 . Let the birth rates be b1 = 0.5, b2 = 5, b3 = 3, and
let the survival rates be S1 = 0.5, S2 = 23 . Then equation (84) is


0, 5 5 3
x(n + 1) =  0, 5 0 0  x(n).
0 2/3 0
(85)
Then equation (16) is
(0.5 − λ)(−λ)2 − 5(0.5 × (−λ)) + 3 × 0.5 ×
Therefore
and hence
2
= 0.
3
1
1
−λ3 + λ2 + 2 λ + 1 = 0
2
2
1
−(λ + )(λ + 1)(λ − 2) = 0.
2
(86)
The dominant eigenvalue is therefore λ1 = 2 and the corresponding eigenvector is found from


−1 21
5
3
 1
−2 0  K1 = 0.
(87)
2
0
2/3 −2
40
CHAPTER 2. LINEAR AUTONOMOUS DYNAMICAL SYSTEMS
Let


K11
K1 =  K12  .
K13
Suppose that K13 = 1, then the last row of (87)
2
K1 − 2K13 = 0
3 2
gives
K12 = 3,
so that from the middle row
1
K1 − 2K13 = 0
2 2
we have
K11 = 12.
Thus


12
K1 =  3  .
1
(88)

 
12
x1
 x2  ∝  3  .
1
x3
(89)
and for large n

(iii) Safety of buildings
Mechanical structures (bridges, buildings, etc.) are built from materials that have some
elasticity and are therefore dynamical systems. To determine their response to forces such
as wind, earthquakes, etc. it is important to calculate the natural frequencies of vibration.
Consider a four-storey building, as illustrated below, and assume that the elements of the
frame are inextensible and that the mass is concentrated in the floors. The floors can be
displaced laterally with respect to each other, but the bending elasticity of the building frame
then generates restoring forces. Assuming that all frame members have the same elasticity,
the force vector f is related to the displacement vector x by f = Ax, where A is the stiffness
matrix. By Newton’s laws the force also satisfies −f = Mẍ(t), where M is the mass matrix.
Thus
ẍ(t) = −M−1 Ax = −Bx
(90)
where

A=
M=

1 −1
0
2 −1  ,
a  −1
0 −1
2


1 0 0
m 0 1 0 .
0 0 2
91
41
[m and B = M−1 A are constants associated with the material of the building; m is mass and
a is a coefficient of elasticity].
First, equation (90) is not in the usual form because it involves derivatives of second-order
(ẍ) rather than first-order (ẋ). There are two different, but ultimately equivalent ways of
handling equation (90).
(a) We look for solutions of the form
x = Ke±iωt .
(92)
Equation (92) is inspired by the solution of ẍ + ω 2 x = 0, x = e±iωt . Substituting (92)
into (90) gives
−ω 2 Ke±iωt = −Be±iωt K.
Therefore
ω 2 K = BK.
(93)
Thus ω 2 is an eigenvalue, and K is the corresponding eigenvector, of the matrix B.
Now B = M−1 A, thus

1
a 
−1
B=
m
0
−1
2
− 12

0
−1  ,
1
42
CHAPTER 2. LINEAR AUTONOMOUS DYNAMICAL SYSTEMS
so that the eigenvalue equation is
0
=
det B − w2 I
=
a
2
m −w
a
−m
=
=
=
=
a
−m
2a
2
m −w
a
− 2m
0
a
−m
a
2
0
m −w
a
2a
a2
a
− w2
− w2
− w2 −
m
m
m
2m2
h
i
a
a a
+
−
− w2
m m m a 3
1
(1 − λ) (2 − λ) (1 − λ) −
− (1 − λ)
m
2
a 3
1
(1 − λ) λ2 − 3λ + 2 − − 1
m
2
a 3
1
2
(1 − λ) λ − 3λ −
m
2
2
where λ = m
aw .
Hence we solve
1
(1 − λ) λ − 3λ −
2
2
=0
(94)
to find the solutions
λ1
=
1,
λ2
=
(3 +
λ3
=
(3 −
√
√
7)/2,
7)/2,
so that
ω1 =
a 1/2
m
ω2 =1.68
ω3 =0.42
,
a 1/2
m
a 1/2
m
,
.
95
Thus the lowest value of ω is ω3 , therefore the lowest natural frequency (frequency
ν = ω/2π) is
a 1/2
ν = 0.067
.
(96)
m
Problems occur when the lowest frequency is small, meaning in practice when ν ≤
10 (cycles/second). Thus one requires
2
m
10
≥
= 22277.
(97)
a
0.067
43
The elastic coefficient to mass must be high. The analysis described above is very important in Engineering design. It was not applied to the design of the Tocoma Narrows
bridge in the USA, and in 1940 that bridge shook itself to pieces during a storm.
(b) An alternative way of deriving (95) is to write (90) as a system of first-order equations.
Let


 
 
 
ẋ1
y1
x1
y4
 ..   ..   ..   .. 
(98)
 .  =  . , .  =  . .
y3
x3
x3
y6
Then the system (90) is ẏ = Cy where
C=
0
−B
I
0
.
(99)
It can be shown that, provided B is invertible (which it is in this case), then the eigenvalues of C are
p
p
p
±i λ1 , ±i λ2 . ± i λ3
(100)
where λ1 , λ2 , λ3 are the eigenvalues of B. Thus the solution (92), with ω given by (95),
is also found.
NATURE OF ROOTS OF POLYNOMIALS
(∆ = a4 (r1 − r2 )2 (r1 − r3 )2 (r2 − r3 )2 )
If the coefficient of a polynomial are real numbers, and its discrimant ∆ is not zero, there are
two cases:
ˆ If ∆ > 0, the Cubic has three distinct real roots.
ˆ If ∆ < 0, the cubic has one real root and two non-roots Complex conjugate
Firstly; if r is a root of a polynomial with real coefficients, then its complex conjugate is also a
root. So the non-real roots, if any, occur as pairs of conjugate roots. As a cubic polynomial has
three roots (not necessarily distinct) at least one roots must be real according to the fundamental
theorem of algebra.
As stated above, if r1 , r2 , r3 are the three roots of the cubic ax3 + bx2 + cx + d; then then discriminant is ∆ = a4 (r1 − r2 )2 (r1 − r3 )2 (r2 − r3 )2 . If only one root, say r1 , is real then r2 and r3
are complext conjugates, which implies that r2 − r3 is a purely imaginary number, and thus that
(r2 − r3 )3 is real and negative. On the other hand, r1 − r2 and r1 r3 are complex conjugates; and
their product is real and positive. Thus the discriminant is the product of a single negative ones.
That is ∆ < 0.
44
CHAPTER 2. LINEAR AUTONOMOUS DYNAMICAL SYSTEMS
Thus, depending on the types and sign of the eigenvalues, there are a few interesting cases in the
classification of equilibrium points:
ˆ Node when all eigenvalues are real and have the same sign. The node is stable (unstable)
when the eigenvalues are negative (positive).
ˆ Saddle when all eigenvalues are real and at least one of them is positive and at least one is
negative; Saddles are always unstable.
ˆ Focus-Node when it has one real eigenvalue and a pair of complex-conjugate eigenvalues,
and all eigenvalues have real parts of the same sign; The equilibrium is stable when the sign
is negative and unstable when the sign is positive.
ˆ Saddle-Focus when it has one real eigenvalue with the sign opposite to the sign of the real
part of a pair of complex-conjugate eigenvalues. This type of equilibrium is always unstable.
EXERCISES
1. For each of the following systems find the location of the singular point, and its nature
(a)
dy
dx
= x + 2y + 1,
= 4x + 2y − 1
dt
dt
(b)
dx
dy
= 5x + y − 1,
= −2x + 3y − 2
dt
dt
dy
dx
= −2x + y + 4,
= −2x + 4y + 6
dt
dt
−1 2
(d) X(n + 1) =
X(n)
−7 4
1 2
−4
(e) X(n + 1) = 1 1 X(n) +
−1
2
2
(c)
2. Find the general solution of each of the following systems, and state the location and nature
of the singular point
dx
dy
= 3x + 3y − 4,
= −x − 3y + 5
dt
dt
dX
−2 3
(b)
=
X
−3 5
dt
(a)
dx
dy
= 6x − y − 5,
= 5x + 2y − 3
dt
dt


−1 −1
0
dX  3
− 32
3 X
(d)
=
4
dt
1
1
− 12
8
4
(c)
45
3. For each of the following systems, find the solution that satisfies the given initial condition;
state the location and nature of the singular point:
dX
−1 −2
3
3
(a)
=
X+
subject to X(0) =
3
4
3
4
dt
dX
4 −5
6
(b)
=
X subject to X(0) =
5 −4
2
dt
(c)
dx
= −x + y, with x(0) = 2
dt
dy
= x + 2y + z − 5, with y(0) = 3
dt
dz
= 3y − z − 1, with z(0) = 4
dt
4. Find the general solution of each of the following discrete dynamical systems and then find
the particular solution for the given initial condition. State the nature of the singular point.
−2 12
2
(a) X(n + 1) =
X(n), with X(0) =
5
2
2 −5
5
(b) X(n + 1) =
X(n), with X(0) =
1 −2
1
(c) x(n + 1) = − 12 x(n) + 12 y(n),
y(n + 1) = − 12 x(n) + 34 y(n) with x(0) = y(0) = 3
13
25
(d) X(n + 1) =
−9 −17
2
with X(0) =
2
X(n) +
1
0
5. Make the following simplifications in the cohort model of age distribution:
– Women have children between the ages of 13 and 38 inclusive.
– Each woman has exactly one female child.
– Each woman lives to be at least 39 years old.
Using a time interval of 13 years with these assumption and considering only those females
less than 39 years old give the following model:

 


x1 (k + 1)
0 α 1−α
x1 (k)
 x2 (k + 1)  =  1 0
0   x2 (k) 
x3 (k + 1)
0 1
0
x3 (k)
where 0 ≤ α ≤ 1
46
CHAPTER 2. LINEAR AUTONOMOUS DYNAMICAL SYSTEMS
(a) Find the eigenvalues. Find the dominant eigenvalue and the corresponding eigenvector.
(b) Determine the growth rate.
(c) How quickly does the population move to equilibrium?
[The answer depends on the value of α.]
6. Assume that the population of a country is divided into two distinct segments: rural and urban. The natural yearly growth factors, due to procreation in both segments are assumed to
be identical and equal to α [i.e, the population at year k + 1 would be α times the population
at year k]. However, the population distribution is modified by migration between the rural
and urban segments. The rate of this migration is influenced by the need for a base of rural
activity that is adequate to support the total population of the country - the optimal rural
base being a given fraction γ of the total population. The yearly level of migration from rural to urban areas is proportional to the excess of rural population over the optimal rural base.
If the rural and urban populations at year k are denoted by r(k) and u(k), respectively, then
the total population is r(k) + u(k), the optimal rural base is γ[r(k) + u(k)], and thus the
excess rural population is {r(k) − γ[r(k) + u(k)]}. A simple dynamic model of the migration
process based on the above assumptions is then
r(k + 1)
=
αr(k) − β{r(k) − γ[r(k) + u(k)]}
u(k + 1)
=
αu(k) + β{r(k) − γ[r(k) + u(k)]}.
In this model, the growth factor α is positive and usually greater than unity. The migration
factor β is positive and is assumed to be less than α. The parameter γ is the ideal fraction of
the total population that would be rural in order to support the total population. This parameter is a measure of rural productivity. Each of these parameters might normally change
with time, but they are assumed constant for purposes of this example.
The model can easily be put in the state vector form
x(k + 1) = Ax(k)
where
A=
α − β (1 − γ)
βγ
β (1 − γ)
α − βγ
and
x(k) =
Show that the condition
r(k)
u(k)
0 ≤ β ≤ min
.
α
α
,
1−γ γ
is necessary and sufficient in order to guarantee that both urban and rural populations remain
non-negative given any non-negative initial conditions.
47
7. Find the general solution of each of the following systems, and state the location and nature
of singular point.
7 −1
(a) X (n + 1) =
X (n)
5
3
6 1
1
(b) X (n + 1) =
X (n) −
−2 4
2
5 −5
(c) X (n + 1) =
X (n)
5 −3
(d) x (n + 1) = 4x(n − y (n) − 2
y (n + 1) = 9x (n) − 2y (n) − 6
(e) x (n + 1) = 3y (n) − 4
y (n + 1) = −3x (n) + 6y (n) − 4
(f) x (n + 1) = −5x (n) + 5y (n)
y (n + 1) = −5x (n) + 5y (n)
8. Find the nature of the singular point in ech of the following systems


2 −1 2
2 0  X (n)
(a) X (n + 1) =  −1
−1
0 2


3
5 1
(b) X (n + 1) =  −5 −5 4  X (n)
0
0 3


6 −4 0
1 2  X (n)
(c) X (n + 1) =  1
0
2 6


2 0
0
(d) X (n + 1) =  2 3 −1  X (n)
0 1
1
9. Find the general solution of each of the following systems, and state the location and nature
of the singular point:
dy
dz
dx
=z
= −z
=y
dt
dt
dt




−1 1
0
0
dX 
1 2
1  X+  −4 
(b)
=
dt
0 3 −1
−2


1 1 4
dX 
(c)
= 0 2 0 X
dt
1 1 1
(a)
48
CHAPTER 2. LINEAR AUTONOMOUS DYNAMICAL SYSTEMS
10. Write each of the following as a first-order system and then determine the nature of the
singular point:
(a) x (n + 2) = 3x (n + 1) − 2x (n)
(b) x (n + 2) = 14 x (n)
2
(c) ddt2x + 4 dx
dt + 3 = 0
2
(d) ddt2x − 5 dx
dt + 6 = 0
SOLUTIONS
1. (a) The singular point is 32 ; − 54 .
1 2
, σ = 3, ∆ = 6.
Let A =
4 2
Since σ = 3 > 0; ∆ = −6, the fixed point is a saddle point.
1 12
, 17 .
(b) The singular point is solved to be 17
5 1
Let A =
, σ = 8, ∆ = 17.
-2 3
The fixed point is an unstable focus or unstable spiral point.
(c) The singular point is solved to be
-2 1
σ=2
Let A =
-2 4
∆ = −6.
− 13 ; − 22 .
The fixed point is a saddle point
(d) The singular
pointis solved to be (0; 0).
-1 2
σ = 3; ∆ = −10.
Let A =
-7 4
The fixed point is an unstable point from (also called an unstable spiral point).
(e) The singular
point is solved to be (4; 2).
1 2
σ = 32 ; ∆ = − 12
Let A =
1
1
.
2
2
The fixed point is a saddle point.
49
1 11 3
3
σ = 0; ∆ = −6.
. Let A =
2. (a) The singular point is solved to be − ,
-1 -3
2 6
The fixed point is a saddle point.
Thus the general solution is a given by
√
√
1 √3 e(− 6)t + c2
√3 e( 6)t + −112 .
X(t) = c1
−3 − 6
−3 + 6
6
(b) The singular point is solved to be (0, 0).
The fixed point is called a saddle point since, ∆ = −1, σ = 3.
General solution:
"
#
"
#
√
√
√
√
7+ 13
3+ 13
3− 13
− 7+6 13
6
χ(t) = c1
e 2 + c2
e 2 .
1
1
(c) The singular point
8; ∆ = 17.
13
17
−
7
. The fixed point is called unstable focus since σ =
17
General solution:
" − 12 #
" -4 #
0
1
(d) X(t) = c1
13 2
− 17
− 5 − 5i
e(4+i)t + c2
e(4−i)t +
7
− 17
1
− 52 + 2i
1
X(t) = c1
e−t + c2
5
− 65
" −4 #
e
− 21 t
+ c3
1
3
e− 2 t
−2
1
The singular point is called Stable node.
-1
2
-9
-1
t
2t
3. (a) X(t) = −32
e − 10
e +
X(t) = −32
et
1
-3
6
1
The singular point is called stable node.
(b) X(t) =
1 + 11i
3
4
3i
5 − 5
1
4
3i
5 − 5
1 − 11i
e−3it +
3
1
e3it
The singular point is called saddle point.
(c) X(t) =
33
20
"
1
3
4
3
1
#
e
"
3t
− 41
-1
0
1
#
e
"
−t
+ 35
The singular point is called a saddle point.
1
3
− 13
1
#
"
−2t
e
+
1
1
2
#
50
CHAPTER 2. LINEAR AUTONOMOUS DYNAMICAL SYSTEMS
4 (a) Cancelled
(b) X(n) =
1 3i
+
2
2
2−i
1
!n
!
-i
1 3i
+
−
2
2
!
2+i
1
!n
!
i
The singular point is called unstable.
1 n
1
+2
(c) X(n) = 2
1
2
2
1
!
1
−
4
!n
The singular point is called unstable.
"
(d) X(n) = 3
− 35
1
#
"
n
−2
− 45
− 91
0
!n
!
−2
+n
Hence the singular point is called unstable.
5. (a) The dominant eigenvalue = λ1 = 1.
(b) The growth rate = 1.
(c) The population will move to equilibrium provided that
0<α<1
or
4
≤α<1
3
or
0<α<
4
.
3
6. (a) The dominant eigenvalue = λ1 = 1.
(b) The growth rate = 1.
(c) The population will move to equilibrium provided that
0<α<1
or
4
≤α<1
3
or
0<α<
4
.
3
− 53
1
!
!n−1 #
−2
+
2
−1
!
51
7. (a) The singular point = (0, 0).
#
"
"
n
2
i
−
X(n) = c1 5 5
5−i
+ C2
1
2
i
5 + 5
#
1
5+i
The singular point is called unstable.
(b) The singular point =
"
X(n) = c1
1 12
,
.
17 17
#
#
"
"
n
n
− 12 − 5i
5−i
+ C2
5+i
+ C2
1
− 12 + 2i
1
1
17
12
17
#
The singular point is called an unstable spiral.
(c) The singular point = (0, 0)
#
"
"
n
4
− 3i
5
5
1 − 3i
+ C2
X(n) = c1
1
4
3i
5 + 5
#
n
1 + 3i
1
The singular point is called unstable.
(d) The singular point cannot be obtained since the two lines are parallel.
"
X(n) = c1
− 31
1
# 1
"
n
+ C2
−1
2
!
!n
1
+n
1
2
1
!n−1 #
!
n
The singular point = (2, 2).
(e) The singular point = (2, 2).
√
√
√
√
n
n 3+ 57
−9+ 57
9+ 57
3− 57
2
6
2
6
2
X(n) = c1
+ c2
+
2
1
2
1
2
The singular point is called unstable.
(f) The singular point = (0, 0).
X(n) = c1
1
1
" n
n−1 #
n
1
0
0
+ c2
0
0
+n
1
1
5
52
CHAPTER 2. LINEAR AUTONOMOUS DYNAMICAL SYSTEMS
8. λ1 = 2, λ2 = 2 − i, λ3 = 2 + i,
(a) The singular points are unstable. The singular points are (0, 0, 0).
(b) λ1 = 3, λ2 = −1 − 3i, λ3 = −1 + 3i
The singular points are (0, 0, 0).
(c) λ1 = 1, λ2 = −6, λ3 = 6
The singular points are (x = −2z, y = 5z
2 , z = z, z ∈ R
(d) λ1 = λ2 = λ3 = 2.
The singular points are (0, 0, 0).


 



0
1
−i
i
9. (a) X(t) = c1  −i  e−it + c2  i  e−it + c3  0   0 
0
0
1
1
The singular point is called a saddle point.
 1 

 1 


1
−1
3
 e3t + c2  0  e−t + c3  − 1  e−2t +  1 
(b) X(t) = c1 
3
1
1
1
1
The singular point is called a saddle point.
3
4
3


 5 




2
−2
0
2
(c) X(t) = c1  0  e3t + c2  − 32  e2t + c3  0  e−t +  0 
1
1
0
1
The singular point is called a saddle point.

Chapter 3
LINEAR CONTROL THEORY
1. Introduction
It was not possible to change the state of the dynamical systems we studied in Chapter 2 by means
of an “outside” control – the changes in state were entirely determined by the equations describing
the dynamical system and the initial conditions. In many practical situations we do have some
control over the system (e.g, turning a tap on or off to control the level of water in a tank, or
turning a heater on or off to control the temperature in a room). In this chapter we consider some
aspects of linear dynamical systems that are subject to linear controls.
Mathematically such systems are described by the equation
ẋ(t) = Ax(t) + Bu(t)
for continuous systems, and
x(k + 1) = Ax(k) + Bu(k)
for discrete system. If there are n state variables and m control variables then, in both cases, x is
an n × 1 column vector, u is an m × 1 column vector, A is an n × n matrix with constant entries
and B is an n × m matrix with constant entries. We will call the vector x the state vector of the
system, or simply the state of the system, and the vector u the control vector, or sometimes the
input vector to the system. In many cases some kind of measurement is performed on the system
and the results are given in the form of an output vector y. We assume that y depends linearly
on the state vector x and input vector u. The equation for y for continuous systems will then be
y(t) = Cx(t) + Du(t)
where y is an p × 1 column vector, C is an p × n matrix with constant entries and D is an p × m
matrix with constant entries.
For discrete systems the output equation is
y(k) = Cx(k) + Du(k).
Our aim in this chapter is to study two important properties of the above control systems:
53
54
CHAPTER 3. LINEAR CONTROL THEORY
(i) Controllability
Here we want to know whether we can always choose an input u to change the state vector x from
any initial value to any final value. Roughly how well we can control the system. This is dealt with
in sections 3 and 4.
(ii) Observability
If we know the output vector y and the input u, can we determine the state vector x?. In other
words, can we use the output and input to observe the state of the system? Sections 5 and 6 are
concerned with this question.
Lastly, in section 7 we investigate what happens when we connect the output of the system in some
way to the input. This is called feedback.
First, however, we will give some examples of linear control systems and show how certain diagrams
can represent such systems. This visual aid will make it easier to understand and discuss the control
system concepts described above.
2. Dynamic diagrams
These diagrams are built up from five elementary components shown in Figure 1:
Figure 1
The summer (a) adds whatever comes into it, instantaneously producing the sum. Any number of
lines can come into a summer and one line comes out. The transmission (b) multiplies what comes
in by the number written in the box. Diagram (c) is a splitter – it connects an incoming line with
any number of outgoing lines, each with the value of the incoming line. The delay (d) is the basic
dynamic component of discrete-time systems. If x(k) comes enters a delay diagram, then the value
of x one time unit earlier goes out - that is, x(k − 1) goes out. So if x(k + 1) goes in, x(k) goes
out, and if x(k − 1) goes in x(k − 2) goes out, and so on. The integrator (e) is the basic dynamic
component of continuous-time systems. If ẋ enters the integrator then x goes out. The way in
which these diagrams are put together to form linear control systems is described by the following
examples.
Example 1 (Pure integrator)
The equation for this simple system is
ẋ(t) = u(t).
55
The output x(t) is the integral of the input function u(t). The diagram is therefore as follows:
Figure 2
Example 2
Consider the control system
ẋ1
ẋ2
=
y (t)
3
1
0
2
x1
x2
+
1
0
u(t)
= x2 (t) .
Then ẋ1 = 3x1 + u. In that part of the diagram corresponding to this equation we need an
integrator for ẋ1 , a transmission box for 3x1 and a summer for the “ + ”. The equation tells us
these components are connected as follows:
Figure 3
Similarly the diagram corresponding to the equation
ẋ2 = x1 + 2x2
is as follows:
Figure 4
56
CHAPTER 3. LINEAR CONTROL THEORY
If we put these two diagrams together, we note that y = x2 gives the following diagram for the
whole system:
Figure 5
Example 3
The diagram for the one-dimensional discrete system
x(k + 1) = ax(k) + bu(k) as follows:
Figure 6
Example 4 (A very simple model for driving a car)
Suppose a car is driven along a straight road, its distance from an initial point O being s(t) at time
t. Assume the car is controlled by the accelerator, which provides a force of u1 (t) per unit mass,
and the brake, which produces a retarding force of u2 (t) per unit mass. We are interested in the
car’s position given by
x1 (t) = s(t)
and speed
x2 (t) = ṡ(t).
Then
ẋ1 = x2
and by Newton’s 2nd law
ẋ2 = u1 − u2 .
In matrix notation
ẋ1
ẋ2
=
0
0
1
0
x1
x2
The dynamic diagram for this system is as follows:
+
0
1
0
−1
u1
u2
.
57
Figure 7
Example 5 (Stick balancing)
Balancing a stick on one’s hand can be formulated as a control problem as follows:
Suppose the stick has length L and the mass M of the stick is concentrated at the top. Let the
variables u, θ, and x be as shown in the figure below:
Figure 8
It follows from Newton’s laws that the system is governed by the equation
ü(t) cos θ(t) + Lθ̈(t) = g sin θ(t)
where g is the gravitational constant.
We also have
x(t) = u(t) + L sin θ(t).
If we assume that the stick is nearly at rest close to the vertical position, then we can put θ̇2 = 0,
cos θ = 1 and sin θ = θ, and the above equations can be written as
ẍ(t) =
g
g[x(t) − u(t)].
L
Put
v(t) = ẋ(t)
58
CHAPTER 3. LINEAR CONTROL THEORY
then the matrix equation for the system is
#
#
"
"
0 1
0
ẋ (t)
x
(t)
g
g u(t).
=
+
0
−
v̇ (t)
v (t)
L
L
The dynamic diagram for the system is as follows:
Figure 9
Example 6 (Pendulum system)
Consider the pendulum system shown in the figure below:
Figure 10
We take the angle θ to be the output of the system. The equation for the system is
mL2 θ̈ = −mgL sin θ
or
θ̈ +
g
sin θ = 0.
L
Define
x1
= θ
x2
= θ̇.
Then we get
ẋ1
= x2
ẋ2
= −
g
sin x1 .
L
59
The output y is the angle θ so
y = x1 .
If the angle θ is small, we can put sin x1 = x1 . The matrix equations for the system are then
#
"
0 1
ẋ1
x1
g
=
0
−
ẋ2
x2
L x1
1 0
y =
.
x2
The dynamic diagram is as follows:
Figure 11
The system has no input.
Example 7
Consider a system consisting of an interconnected pair of water tanks, as shown in Figure 12:
Figure 12
A1 and A2 represent the cross-sectional areas of the two tanks. The single pump motor pumps
water at a controlled rate of u litres/sec into each tank. The outflow is proportional to the depth
of water in each tank (constant of proportionality µ). The flow in the interconnecting pipe is
proportional to the difference of depths in the tanks (constant of proportionality λ).
Let V1 , V2 denote the volume of water in each tank. Then
dV1
V2
V1
= u − y1 + λ
−
,
dt
A2
A1
dV2
V1
V2
= u − y2 + λ
−
,
dt
A1
A2
V1
A1
V2
y2 = µ .
A2
y1 = µ
60
CHAPTER 3. LINEAR CONTROL THEORY
Thus
dV1
dt
dV2
dt
V1
V2
+λ
+ u,
A1
A2
V1
V2
= λ
− (λ + µ)
+ u.
A1
A2
= −(λ + µ)
The final combined outflow is
y=µ
V1
V2
+
A1
A2
.
The system equations in matrix form are therefore

V̇1
V̇2
=




y
=
−
µ
A1

λ
A2 
1
 V1
+
u

V
1

2
λ
λ+µ
−
A1
A2
µ
V1
.
V2
A2
λ+µ
A1
We will return to some of these examples later.
3. Controllability for discrete-time systems
We discussed the idea of controllability very briefly in the introduction. In the section we give a precise definition of discrete systems and show how to determine whether such a system is controllable.
Definition
The n-dimensional system
x(k + 1) = Ax(k) + Bu(k)
is said to be completely controllable if for x(0) = 0 and any given vector w there exists a positive
integer N and a sequence of inputs u(0), u(1), u(2), ..., u(N − 1) such that this input sequence,
applied to the system, yields x(N ) = w.
Thus if the system is completely controllable we can “steer” the zero vector to any other vector
we choose in a finite number of steps. The choice of the zero vector as our initial vector in this
definition is only for convenience. We will show later that it follows that any vector can be steered
to any other in a finite number of steps – in fact in n or fewer steps.
Here are some examples to illustrate this definition:
Example 8
Consider the system with the following dynamic diagram:
61
Figure 13
The equations for this system are
x1 (k + 1)
=
ax1 (k)
x2 (k + 1)
=
x1 (k) + bx2 (k) + u(k).
It is clear from both the diagram and equations that the control u cannot have any effect on the
state variable x1 . Hence this system is obviously not completely controllable.
Example 9
Consider the one-dimensional discrete system with equation
x(k + 1) = x(k) + u(k).
To steer x(0) = 0 to any given number b in one step take N = 1 and choose u(0) = b. Then
x(1) = x(0) + u(0) = b.
Hence this system is by definition completely controllable.
Of course it is seldom as easy as it was in the above two examples to decide whether a system
is completely controllable. However, in the next theorem we derive a fairly simple condition on
the n × n matrix A and the n × m matrix B that is equivalent to complete controllability. You
might need to revise the concept of the rank of a matrix, which is discussed in the appendix, before
working through this theorem. We also need the following lemma, Lemma 1. We omit the proof –
if you are interested, the proof on be found in Luenberger [1979.28].
Lemma 1
Suppose A is an n × n matrix and B is an n × m matrix. Then, for any integer N ≥ n ≥ 1, the
rank of the matrix
[B, AB, ..., AN −1 B]
is equal to the rank of the matrix
[B, AB, ..., An−1 B].
The square bracket [...] notation used here is explained in the appendix.
Theorem 1
62
CHAPTER 3. LINEAR CONTROL THEORY
The discrete-time system
x(k + 1) = Ax(k) + Bu(k)
is completely controllable if and only if the n × m controllability matrix
M =[B, AB, A2 B, A3 B, ..., An−1 B]
has rank n.
[As usual, A is an n × n matrix and B is an n × m matrix.]
Proof
Suppose a sequence of inputs u(0), u(1), u(2), ..., u(N − 1) is applied to the system
x(k + 1) = Ax(k) + Bu(k)
with x(0) = 0. Then
x(1)
= Ax(0) + Bu(0) = Bu(0),
x(2)
= Ax(1) + Bu(1) = ABu(0) + Bu(1),
x(3)
= Ax(2) + Bu(2) = A2 Bu(0) + ABu(1) + Bu(2),
and so on, finally giving
x(N ) = AN −1 Bu(0) + AN −2 Bu(1) + ... + Bu(N − 1).
The expression on the right is simply a linear combination of the columns of the matrix
h
i
K = B, AB, A2 B, ..., AN −1 B .
Thus all points of the n−dimensional state space can be reached if and only if K has rank n. By
Lemma 1
rank K = rank M,
for N ≤ n, and hence the theorem is proved.
Remark
It follows from the above proof that N ≤ n. Thus we can reach any vector from the zero vector in
n or fewer steps if the system is completely controllable. In fact we can transfer the state between
two arbitrary vectors within n steps. To see this, suppose x(0) and x(n) are specified arbitrarily.
With zero input the system would move to An x(0) at time n. Then the desired input sequence is
the one that would transfer the state from the zero vector to x(n) − An x(0) at time n.
The following examples show how to apply this important theorem.
Example 10
Consider the control system in Example 8. We have
a 0
A=
1 b
63
and
B=
0
1
.
Since n = 2, we have that the controllability matrix M is given by M = [B, AB], which in this
case is
0 0
M=
.
1 b
The columns of M are linearly dependent and therefore
rank M = 1 < 2.
Hence, as we concluded earlier, this system is not completely controllable.
Example 11
Suppose we modify the system in Example 10 by shifting the input to the first stage rather than
the second. Then we get the following dynamic diagram:
Figure 14
The equations for this modified system are
x1 (k + 1)
=
ax1 (k) + u(k)
x2 (k + 1)
=
x1 (k) + bx2 (k).
Then
A=
and
B=
a
1
0
b
1
0
.
The controllability matrix M is therefore
M = [B, AB] =
1
0
a
1
.
Therefore rank M = 2, hence the system is completely controllable.
64
CHAPTER 3. LINEAR CONTROL THEORY
Example 12
Consider the simple one-dimensional system in Example 9. Then A = 1 and B = 1.
Hence
M = [B] = [1] .
Clearly rank M = 1, therefore the system is completely controllable.
Example 13
We investigate the discrete system with scalar control u and equation
x(k + 1) = Ax(k) + Bu(k)
(1)
where
−2
2
A =
1 −1
1
B =
0
and n = 2. Then
M = [B, AB] =
,
−2
1
1
0
.
Clearly rank M = 2 = n, so the system is completely controllable. Let’s find a control sequence
that transfers the zero vector to the vector
1
.
2
We know we can do this in at most two steps. Let u(0), u(1) be a control sequence such that
0
x(0) =
0
and
x(2) =
1
2
.
Then from equation (1) it follows that
x(1)
= Ax (0) + Bu(0)
x (2)
= Ax (1) + Bu (1)
= A [Ax (0) + Bu (0)] + Bu (1)
= A2 x (0) + ABu (0) + Bu (1)
= ABu (0) + Bu (1)
since x (0) = 0.
65
This gives
1
2
=
−2
1
u(0) +
1
0
u(1).
Therefore
u(1) − 2u(0)
=
1
u(0)
=
2.
Solving these equations gives u(0) = 2; u(1) = 5.
Is it possible to do the transfer in one step? We would require from
x (1) = Ax (0) + Bu (0) = Bu (0)
[since x (0) = 0] that
1
2
=
1
0
u(0).
This is obviously not possible, therefore the transfer cannot be accomplished in one step.
4. Controllability of continuous–time systems
The definition of complete controllability for these systems is analogous to our definition for discrete
systems:
Definition
The system
ẋ(t) = Ax(t) + Bu(t)
is said to be completely controllable if for x(0) = 0 and any given state x1 there exists a finite
number t1 and a piecewise continuous input u(t), 0 ≤ t ≤ t1 , such that x(t1 ) = x1 .
Piecewise continuous means that there are at most a finite number of points of discontinuity of u
in the interval [0, t1 ].)
As with the discrete case there is a simple test for complete controllability. The test is exactly the
same as for the discrete case but the proof is different. Some of the mathematical results used will
most likely be new to you, so you should study sections 3–5 of the appendix before working through
the next theorem, whose proof uses the following lemma:
Lemma 2
Suppose A is a constant n × n matrix, B is a constant n × m matrix and
rank B, ABA2 B, . . . , An−1 B = n.
Then for any t1 > 0 the matrix
Z t1
K=
0
T
e−At BBT e−A t dt
66
CHAPTER 3. LINEAR CONTROL THEORY
has an inverse K−1 .
[Remember BT denotes the transpose of B.]
We omit the proof – to give it would be too much of a digression.
Theorem 2
The n dimensional continuous-time system
ẋ(t) = Ax(t) + Bu(t)
is completely controllable if and only if the controllability matrix
M = [B, AB, A2 B, ..., An−1 B]
has rank n.
Proof
We first show that if rank M < n, then the system is not completely controllable. From the results
given in paragraph 5 in the appendix, we get for any t1 and any input function u(t), 0 ≤ t ≤ t1 ,
Z t1
x(t1 ) =
eA(t1 −t) Bu(t)dt
(1)
0
A(t1 −t)
Then, using the definition of e
,
Z t1 A2
2
x(t1 ) =
(t1 − t) ... Bu(t)dt
I + A(t1 − t) +
2!
0
or
Z t1
x(t1 )
=
B
Z t1
(t1 − t)u(t)dt
u(t)dt + AB
0
0
+A2 B
Z t1
0
2
(t1 − t)
u (t) dt + ... .
2!
When evaluated, the integrals in the above expression are simply constant m-dimensional column
vectors. Therefore the above expression shows that x(t1 ) is a linear combination of the columns of
B, AB A2 B, ... . By Lemma 1, if rank M < n, then even the infinite set of vectors formed from
the columns of B, AB, A2 B, ... does not contain n linearly independent vectors. Therefore there
is a vector x1 ∈ Rn that cannot be expressed as a linear combination of any of these vectors, and
so cannot be reached by any choice of control.
Next we show that if rank M = n then the system is completely controllable.
We do this by explicitly constructing a control u that transfers the zero vector to a given vector x1 .
By Lemma 2 for any t1 > 0 the matrix
Z t1
T
K=
e−At BBT e−A t dt
0
67
has an inverse K−1 . Now, given x1 , select any t1 and let
T
u(t) = BT e−A t K−1 e−At1 x1 .
We verify that this input transfers the state from zero to x1 at time t1 by substituting for u in
equation (1), as follows:
Z t1
x(t1 )
=
T
eA(t1 −t) BBT e−A t K−1 e−At1 x1 dt
0
= eAt1 KK−1 e−At1 x1 = x1
Since x1 was chosen arbitrarily it follows from the definition that the system is completely controllable.
Note that in the above proof we select any t1 > 0, so in fact the state can be transferred to x1 in
an arbitrarily short period of time.
All the examples used to illustrate controllability in discrete systems can also be used for continuous
systems – simply replace x(k + 1) by ẋ in the equations and the delay symbol by the integrator
symbol in the dynamic diagrams. Let’s explicitly calculate the control in the continuous-time version of example 12.
Example 14
The system is one-dimensional with equation
ẋ = x + u.
As before, A = B = 1 and the controllability matrix M is
M = [1].
Since rank M = 1 this system is completely controllable. Let’s find a control u that transfers 0 to
1 at time t = 1. The matrix K in this example is
Z t1
K=
e
−At
T −AT t
BB e
Z 1
dt =
0
e−t (1)(1)e−t dt =
0
Therefore
K −1 =
2
.
1 − e−2
Hence
T
u(t) = B T e−A t K −1 e−At1 x1 =
2e−(1+t)
.
1 − e−2
We can check whether this control works. We have x(0) = 0 and
2e−(1+t)
dx
=x+u=x+
.
dt
1 − e−2
1 − e−2
.
2
68
CHAPTER 3. LINEAR CONTROL THEORY
Solving this first-order linear differential equation for x(t) gives
x(t) =
e (et − e−t )
e2 − 1
and we see x(1) = 1.
We can now apply Theorem 2 to Example 5 on stick balancing. We know that it is possible to
balance a stick on one’s hand so the system should be completely controllable. Let’s verify this.
69
Example 15
The matrix form of the system equations is
#
#
"
"
0
0 1
x
ẋ
g
g u.
=
+
0
−
v̇
v
L
L
Therefore
"
0
g
L
1
0
"
0
g
−
L
#
A=
and
B=
#
.
The controllability matrix M is then

0
M= g
−
L
−
g 
L .
0
Since rank M = 2 the system is controllable, as we expected but you may be surprised by the
following extension of the problem: Can two sticks (with masses concentrated at their tops) placed
side by side a little distance apart on one hand be balanced simultaneously? Let the suffixes 1 and
2 distinguish the various variables for the two sticks. The same control is applied to both sticks,
and there is no interference between the sticks, so the system equations will be the same as for one
stick but repeated twice follows:
In matrix form:

 
0
ẋ1
 v̇1   Lg

  1
 ẋ2  =  0
0
v̇2
ẋ1
=
v̇1
=
ẋ2
=
v̇2
=
1
0
0
0
v1
g
(x1 − u)
L1
v2
g
(x2 − u).
L2
0
0
0
g
L2


 
0
0
x1
g 

 
0 
  v1  +  − L 1  u
1   x2   0 
0
− Lg2
v2
Calculating the controllability matrix M in the usual way gives


2
0
− Lg1
0
− Lg 2
1 

g2
 −g

0
0
−
2


L1
M =  L1
2 .
g
g
 0

−
0
−
L2
L22 

g
g2
− L2
0
− L2
0
2
If L1 ̸= L2 then rank M = 4 and the system is completely controllable, that is the sticks can be
balanced simultaneously. However, if L1 = L2 then rank M = 2, so the system is not completely
70
CHAPTER 3. LINEAR CONTROL THEORY
controllable and the sticks cannot be balanced. It is not surprising that two sticks of equal length
cannot be balanced, since if they could it would imply (because the sticks move independently)
that the control used to balance a stick did not depend on the state of the stick. I do find it quite
surprising that two sticks of unequal length can be balanced simultaneously - in fact if any finite
number of sticks, no two of equal length, are placed side by side on a (big enough) hand, in theory
they can all be balanced simultaneously. This is because the form of the controllability matrix
remains the same as that of the above M, its dimension just gets larger. Note that the mass of a
stick plays no role in the analysis presented here.
Example 16
Consider the water tanks system in Example 7. Let’s investigate under what conditions this system
is completely controllable. In this case
1
B=
1
and

−


AB = 


λ
λ+µ
+
A1
A2 

.
λ
λ+µ 
−
A1
A2
Hence the controllability matrix for this system is



M = [B, AB] = 

1
−
1

λ+µ
λ
+
A1
A2 

.
λ
λ+µ 
−
A1
A2
It is important to note that the values A1 and A2 are unknowns so that the rank of the controllability matrix will clearly depend on these values. Another thing to remember (See paragraph 2 of
the appendix):
If M is a n × n matrix then rank M =n if and only if det M ̸= 0.
For the above matrix we have
det M =
=
λ
λ+µ λ+µ
λ
−
+
−
A1
A2
A1
A2
(2λ + µ) (A2 − A1 )
.
A1 A2
Since λ and µ are positive, it follows that
det M = 0 ⇔ A1 = A2 .
Therefore the rank of M is 2 if and only if A1 ̸= A2 .
71
Thus the system is completely controllable except when the tanks have the same cross-sectional
areas. For example, if A1 ̸= A2 and the tanks are both empty to start with, then both tanks can
be filled to equal levels, even if one is much larger than the other.
5. Observability for discrete-time systems
The definition and results for observability are very similar to those for controllability.
72
CHAPTER 3. LINEAR CONTROL THEORY
Definition
The discrete–time system
x(k + 1)
y(k)
=
Ax(k) + Bu(k)
= Cx(k) + Du(k)
is completely observable if there is a positive integer N such that knowledge of the outputs y(0), y(1), y(2), ..., y(N −
1) when all inputs u(k) are zero is sufficient to determine the value of the initial state x(0).
(As usual, A is n × n, B is n × m, C is p × n, and D is p × m.)
Note that once the initial state x(0) is known, all subsequent states can be determined.
Observability is important because inputs are often determined by observation of outputs. If the
outputs do not give full information about the state vector, then it may not be possible to formulate
a suitable control scheme. Thus in general we should have the ability to detect what the system is
doing (observability) and to change the system behaviour (controllability).
The following two examples serve to illustrate the above definition.
Example 17
Consider the system with dynamic diagram shown in figure 15 below:
Figure 15
The system has the single output y which is not connected at all to x1 , and therefore there is no
way to infer the value of x1 from y. So this system is not completely observable.
Example 18
Suppose we have a system with the same dynamic diagram as in the previous example, except that
the single output is taken from x1 instead of x2 . The system is now completely observable. This
is easily shown as follows: Suppose we know y(0) and y(1), that is, x1 (0) and x1 (1). From Figure
15 we see that
x1 (1) = ax1 (0) + x2 (0).
But we know x1 (0) and x1 (1), so we can solve for x2 (0). Hence if we know y(0) and y(1) we can
determine x(0) and therefore the system is completely observable.
73
The following theorem is analogous to Theorem 1. We will omit the proof since it is similar to that
of Theorem 1.
Theorem 3
The system
x(k + 1)
=
y(k)
Ax(k) + Bu(k)
= Cx(k) + Du(k)
is completely observable if and only if the pn × n observability matrix


C
 CA

2

S =  CA
 ..
 .






CAn−1
has rank n.
Let us apply this theorem to examples 17 and 18:
Example 19
The system equations for the dynamic diagram in figure 15 are:
x1 (k + 1)
=
ax1 (k) + x2 (k)
x2 (k + 1)
=
bx2 (k)
y(k)
= x2 (k)
Then
a
0
1
b
0
1
a
0
1
b
0
0
1
b
A=
and
C=
.
For CA we get
CA =
0
1
=
Therefore the observability matrix is
S=
.
Since rank S = 1 the system is not completely observable.
Example 20
0 b
.
74
CHAPTER 3. LINEAR CONTROL THEORY
If we modify figure 15 as described in example 18 then the system equations are
x1 (k + 1)
=
ax1 (k) + x2 (k)
x2 (k + 1)
=
bx2 (k)
y(k)
=
x1 (k).
The matrix A is the same as in example 19 and for C we get
C= 1 0 .
Therefore
CA =
1
0
a
0
1
b
1
a
0
1
=
a 1
,
and the observability matrix S is
S=
.
Hence rank S = 2 and the system is completely observable – as we concluded earlier.
6. Observability for Continuous–Time Systems
For continuous–time systems the results are analogous to those for discrete–time systems. We begin
with the definition.
Definition
A system
ẋ = Ax(t) + Bu(t)
y(t)
= Cx(t) + Du(t)
is completely observable if there is a t1 > 0 such that knowledge of y(t), for all t, 0 ≤ t ≤ t1 , when
u(t) = 0 for all t, is sufficient to determine x(0).
The following theorem (we omit the proof) gives a condition for complete observability for continuous systems:
Theorem 4
The system in the above definition is completely observable if and only if the observability matrix


C
 CA



 CA2

S=

 ..

 .

CAn−1
has rank n.
Example 21
75
Under what conditions is the system of water tanks in example 7 completely observable? In this
case
µ µ
C=
A1 A2
and the observability matrix is

C
CA
µ
A1


=
 µ (λ (A1 − A2 ) − µA2 )
µ
A2



.
µ (λ (A2 − A1 ) − µA1 ) 
A21 A2
A1 A22
The determinant of the observability matrix is
µ2 (A2 − A1 ) (µ + 2λ)
C
det
=
.
CA
A21 A22
Of course λ > 0 and µ > 0 so the system is completely observable if and only if A1 ̸= A2 .
7. Feedback
There are two types of control systems, namely: open-loop and closed-loop or feedback control systems. In an open-loop system the control input follows a preset pattern without taking account of
the output or state variables, whereas in a feedback control system the control input is obtained
by the output being fed back in some way to the input. Simple illustrations of open-loop and
closed-loop systems are provided by robots at traffic intersections that change at fixed intervals of
time, and those that measure traffic flow in some way and change accordingly. The following are
two advantages of feedback control:
(ii) Feedback control is often simpler than open-loop control. For example, one could design a
complicated open-loop control for a traffic robot that varied the green and red cycle lengths
according to the expected traffic flow at various hours of the day, but feedback control would
be simpler and more effective.
(ii) Feedback control can rapidly adjust to changes in the state of the system. This means that
feedback control can improve the stability of the system. We discuss this aspect of feedback
in this section.
In Chapter 2 you learned that the dynamical system ẋ = Ax is stable at a fixed point if and only
if the real part of the eigenvalues of A are negative. Now consider the control system
ẋ = Ax + Bu.
Suppose that we apply linear feedback to this system, which means each control variable is a linear
combination of the state variables. Then
u = Cx
76
CHAPTER 3. LINEAR CONTROL THEORY
where C is a constant m × n matrix. Substituting for u in the system equation we get
ẋ = (A + BC)x.
Therefore if we can find a matrix C such that the eigenvalues of A + BC have negative real part
then the system with feedback will be stable. The following theorem (whose proof we omit) shows
that if the system without feedback is completely controllable, then we can find such a feedback
matrix C.
Theorem 5 (Eigenvalue placement theorem)
Suppose the control system
ẋ = Ax + Bu
is completely controllable and λ1 , λ2 , ..., λn are n complex numbers such that any which are not
purely real occur in conjugate pairs. Then there exists a real matrix C such that the eigenvalues
of A + BC are the complex numbers λ1 , λ2 , ..., λn . In particular, we can choose C such that all
the eigenvalues have negative real part.
The followin very simple example to illustrates this theorem
Example 22
Consider the one-dimensional system
ẋ(t) = u(t).
This is certainly completely controllable. In this case A = 0, B = 1, and A + BC = C. The
eigenvalue of C is C, and we can choose C real and C < 0. Then
ẋ = Cx
has the solution
x(t) = keCt .
If C < 0 then eCt → 0 as t → ∞, so the system with feedback u = Cx is stable.
Example 23 (Stick balancing again)
Since the stick balancing system in examples 5 and 15 is completely controllable we must be able to
find a feedback matrix C such that the closed-loop system is stable. Let’s verify this. The matrices
A and B are
0 1
,
A =
g
0
L 0
B =
.
− Lg
Suppose that
C=
Then
A + BC =
c1
c2
.
0
1
g
g
(1
−
c
)
−
1
L
L c2
.
77
As explained in Chapter 2, page 30, the eigenvalues of A + BC are found by solving the equation
det((A + BC) − λI) = 0. This is a quadratic equation:
λ2 +
g
g
c2 λ − (1 − c1 ) = 0.
L
L
We can choose the solutions of this equation to be whatever we like by taking appropriate values
for c1 and c2 . For example, to make both roots −1, choose c1 = 1 + Lg and c2 = 2L
g . For this value
of C the closed loop system will be stable.
Exercises for Chapter 3
1. Investigate the controllability of the system
ẋ = Ax + Bu
if u(t) is a scalar, and
−5 1
1
(a) A =
,
B=
.
0 4
1




0
0 1 0
B =  0 .
(b) A =  0 0 1  ,
1
0 0 0




0
3 3 6
B =  0 .
(c) A =  1 1 2  ,
1
2 2 4
2. Consider the system
ẋ =
Ax + Bu
y
Cx
=
where
A
2
−4
1
−1
1
=
B =
−5
0
and
C=
Is this system
(i) completely controllable?
(ii) completely observable?
1
.
78
CHAPTER 3. LINEAR CONTROL THEORY
3. For the discrete–time system
x(k + 1) = Ax(k) + bu(k)
where
A=
2
0
1
2
0
1
1
0
let
b1
=
b2
=
,
.
(a) For each b, determine whether the system is controllable.
(b) For each b that results in a completely controllable system, find the shortest input
sequence that drives the state to zero if
2
x(0) =
.
1
4. Show that feedback does not destroy complete controllability. Specifically, show that if the
system
x(k + 1) = Ax(k) + Bu(k)
is completely controllable, then the system
x(k + 1) = Ax(k) + BCx(k) + Bu(k)
is also completely controllable.
(Hint: Rather than checking the rank condition, apply the original definition of complete
controllability.)
5. Suppose that the partitioned system
ẇ (t)
A11
=
ẏ (t)
A21
A12
A22
w (t)
y (t)
with output y(t) is completely observable. Show that the combination of A11 as system and
A21 as output matrix is a completely observable pair.
(Hint: You may find it simpler to prove complete controllability of the transposed combination.)
6. Consider the following two dynamic systems:
S1
S2
ẋ1 = x2 + u
ẋ2 = −2x1 − 3x2
y = αx1 + x2
ẋ3 = x3 + w
z = x3
S1 has state (x1 , x2 ), control u and output y. S2 has state x3 , control w and output z.
79
(a) Determine whether each system is controllable and observable. (Note: α is a parameter.)
(b) These two systems are connected in series, with w = y. The resulting system is called
S3 . Determine whether it is controllable and observable.
(c) The systems are now connected in a feedback configuration to produce S4 as shown in
the following figure:
Determine whether S4 is controllable and observable.
7. Two pendulums are connected by a spring as shown in the following diagram:
The masses of the rods, which have unit length, can be neglected. Equal and opposite control
forces u(t) are applied to the particles, which have mass m. For small oscillations (so that θi2
and higher powers can be neglected) the equations of motion are
mθ̈1
=
−mgθ1 − u − ka2 (θ1 − θ2 )
mθ̈2
=
−mgθ2 + u + ka2 (θ1 − θ2 )
Take
x1
= θ1 + θ2
x2
= θ1 − θ2
x3
= ẋ1
x4
= ẋ2
as state variables. Is the system completely controllable?
SOLUTIONS
1. Let M = [B AB] =
1
1
−4
4
(a) ∴ rank M = 2, hence the system is controllable.
80
CHAPTER 3. LINEAR CONTROL THEORY
(b) Let M = [B AB A2 B]

0
 0
1

1
0  , Since rank(M ) = 3
0
0
1
0

0
after swap r1 and r3  0
1
0
1
0

1
0 .
0
Hence the system is controllable.
(c) Let M = [B AB A2 B]

0
 0
1

48
16  , rank(M ) = 2 < 3.
32
6
2
4
Hence the system is not controllable.
2.
(i) Let M = [B AB].
1
7
, rank(M ) = 2, hence the system is controllable.
−1 −4
(ii) Let S =
C
CA
=
1
−2
1
−5
1
0
2
0
1
0
2
0
3. (a) (i) Let M = [b1 Ab1 ] =
.
(ii) Let M = [b2 Ab2 ] =
0
1
transfer in one step.
(b) (i) We get for b1 =
0
(ii) We get for b2 =
1
transfer in one step.
=
1
0
1
−3
rank(S)=2, hence the system is observable.
, rank(M ) = 2, hence the system is not controllable
, rank(M ) = 1, hence the system is not controllable.
, u(0) = 2 and u(1) = −3, hence it is not possible to do the
u(0) = 1 and u(1) = 0, hence it is not possible to do the
4. Refer to Chapter 3, section 3.3, Theorem 1.
5. Refer to Chapter 3, section 3,3, Theorem 2.
81
6. (a) System of S2 is trivial that it is both observable and controllable.
C
α
1
Let S =
=
, rank (S) = 2, hence it is observable.
CA
−2 α − 3
1
0
Let M = [B AB] =
, rank (M ) = 2, hence it is controllable.
−2 −2
(b) Refer to Example 15 in Chapter 3.
(c) Refer to Example 15 in Chapter 3.
7. Refer to Example 6, Figure 3.10.
Chapter 4
AUTONOMOUS NON–LINEAR
SYSTEMS
The introduction of nonlinearity into a system introduces a rich variety of behaviours that are
qualitatively different from the linear case. For example,
zn+1 = zn2 + c
where zn and c are complex, leads to the amazing richness and beauty of the Mandelbrot set (which
is discussed further in Chapter 5). However, we start by using the results of linear theory, and
we will see that in most cases the behaviour of a nonlinear system in the neighbourhood of a fixed
point can be found from linear theory. The crucial phrase in the previous sentence is “in the
neighbourhood of a fixed point”: far from the fixed point the behaviour is dominated by nonlinear
effects and linear approximations are not valid.
4.1 STABILITY
At this stage, we need to be rather more precise about the concept of stability. A linear system
has, in general, only one fixed point, and the whole space is affected by the stability (or not) of the
fixed point. In nonlinear systems there may be several fixed points, and the stability (or not) about
each fixed point is in general a concept that is valid only locally. First, a trajectory means a path
x(t) or x(n) representing a solution of the system. The diagrams in Chapter 2 showing the solution
behaviour near a fixed point are diagrams of trajectories. A fixed point x∗ is asymptotically stable
iff there exists a neighbourhood N of x∗ such that every trajectory that starts in N approaches x∗
as time tends to infinity:
Continuous case: lim x(t) = x∗ .
t→∞
Discrete case: lim x(n) = x∗ .
n→∞
Comment: The neighbourhood N represents the idea that the stability is local. While all trajectories in the space may not tend to x∗ , there is a region around x∗ , i.e. N , in which the trajectories
do tend to x∗ .
Sometimes it is also useful to refer to the domain of asymptotic stability of a fixed point x∗ : this
is simply the largest neighbourhood N for which the definition of asymptotically stable holds.
82
83
Example. In linearised theory, a stable node and a stable focus (figures 3 and 5 of Chapter 2) are
both asymptotically stable.
The next definition that we need is that of an unstable fixed point. Unfortunately the definition
is a bit contorted, so study it carefully: A fixed point x∗ is unstable in a neighbourhood N of x∗
iff there exists a sequence xi of points, with limi→∞ xi = x∗ , such that every trajectory, with an xi
as starting point, contains points that are outside N.
Example. In linearized theory, an unstable node, an unstable focus and a saddle point (figures 1,
2 and 4 of Chapter 2) are all unstable.
Finally, we define marginal stability. A fixed point x∗ is stable iff it is not unstable. Of course it
is easy to see that
asymptotic stability ⇒ stability,
but the converse is not true. A fixed point x∗ is defined to be marginally stable iff it is neither
unstable nor asymptotically stable.
Example. In linearised theory, a centre (figure 6 of Chapter 2) is marginally stable.
The examples given above are all for linear systems and the domain of stability is the whole space.
A simple example illustrating a finite domain of stability is:
dx
= x(x2 − 1).
dt
The singular points are at x = 0, x = −1, x = +1 and the phase diagram for the system is:
Thus x = −1 and x = +1 are unstable, and x = 0 is stable with domain of stability (−1, 1).
84
CHAPTER 4. AUTONOMOUS NON–LINEAR SYSTEMS
4.2 LINEARISATION ABOUT SINGULAR POINTS
For clarity, we start in one dimension:
dx
= g(x)
(1)
dt
and suppose that x = x∗ is a fixed point, i.e. g(x∗ ) = 0. Then by Taylor’s theorem, for x near x∗ :
dx
≈ g(x∗ ) + (x − x∗ )g ′ (x∗ ).
dt
Thus
dx
≈ (x − x∗ )g ′ (x∗ ) .
(2)
dt
In the language of the first part of Chapter 2, g ′ (x∗ ) is equivalent to the constant a, and it is clear
that the fixed point x = x∗ is
asymptotically stable for g ′ (x∗ ) < 0 and unstable for g ′ (x∗ ) > 0.
For discrete systems in one dimension:
x(n + 1) = g(x(n))
(3)
and at a fixed point x = x∗ we have g(x∗ ) = x∗ . Then applying Taylor’s theorem
x(n + 1) ≈ x∗ + (x(n) − x∗ )g ′ (x∗ ).
Thus
x (n + 1) − x∗
≈ g ′ (x∗ ),
x (n) − x∗
(4)
and the system is asymptotically stable if |g ′ (x∗ )| < 1 and unstable if |g ′ (x∗ )| > 1.
Of course it may happen that g ′ (x∗ ) is 0 (continuous case) or 1 (discrete case), and then linearisation cannot say whether x = x∗ is stable or unstable.
Examples
1. The logistic equation of population growth is given by
dx
= ax(k − x)
dt
(5)
where a and k are positive constants.
There are two fixed points, at x = 0 and x = k.
Then g(x) = ax (k − x) so that g ′ (x∗ ) = ak − 2ax∗ , and therefore:
At x = 0 we have g ′ (0) = ak > 0:
′
At x = k we have g (k) = −ak < 0:
unstable
asymptotically stable
85
2.
x(n + 1) = x(n)2 − 3x(n) + 4
(6)
This equation has fixed points given by solving
x∗ = x2∗ − 3x∗ + 4.
Thus
x2∗ − 4x∗ + 4 = 0
so that
x∗ = 2.
′
d
(x2 − 3x + 4) = 2x − 3, and at x = x∗ = 2
Furthermore, g (x) = x − 3x + 4 so that g (x) = dx
′
we have g (2) = 1. Thus linearisation does not say whether x∗ = 2 is a stable or unstable fixed point.
2
In order to describe linearisation in higher dimensions it is first necessary to introduce some notation.
In the continuous case the system is written as
dx
= g(x)
dt
(7)
where x, g are vectors. For example, the system
dx
= x2 + y 2 − 6,
dt
dy
= x2 − y
dt
would have g1 = x2 + y 2 − 6, g2 = x2 − y. Then the Jacobian of the system is


∂g1 ∂g1
 ∂x
∂y 


A=
.
 ∂g
∂g2 
2
∂x
∂y
(8)
(9)
For the above example we have
A=
2x 2y
2x −1
.
(10)
The procedure for analysing the behaviour about the fixed points of a nonlinear system is:
1. Find the fixed point(s).
2. For each fixed point evaluate the Jacobian.
It can be shown that the behaviour of the nonlinear system in the neighbourhood of the fixed point
is similar to that of the linear system
dx
= Ax
(11)
dt
in the neighbourhood of the origin, provided the linear system is not on the boundary between
two qualitatively different types of behaviour. For a system of two equations, and referring to
the diagram on page 41 of Chapter 2, this means that the singular point behaves as shown in the
diagram UNLESS
86
CHAPTER 4. AUTONOMOUS NON–LINEAR SYSTEMS
(a) ∆ = 0
or
(b) σ = 0, ∆ > 0
or
(c) σ 2 = 4∆.
For a system of three or more equations, there is no short–cut to finding the behaviour of the linear
system, and one has to calculate the eigenvalues. If the real parts of all the eigenvalues are < 0,
then the nonlinear system is asymptotically stable in a neighbourhood of the singular point. If
the real part of at least one eigenvalue is > 0, then the singular point is unstable, at least in some
neighbourhood. If at least one eigenvalue has real part exactly 0 and the others have real part
≤ 0, then linear theory cannot say anything about the behaviour of the singular point.
The above can be formulated as a precise theorem and proven. However the proof involves (ϵ, δ)
techniques in real analysis, and is omitted.
In the discrete case the dynamic system is written:
x(n + 1) = g(x(n))
(12)
and the Jacobian matrix A is defined as in the continuous case. A fixed point x = x∗ is defined by
the condition
x∗ = g(x∗ ).
(13)
If the eigenvalues of A are denoted by λi , then the fixed point is locally (i.e. in a neighbourhood of
x∗ ) asymptotically stable if |λi | < 1; locally unstable if there exists an eigenvalue with |λi | > 1; and
linear theory cannot say whether or not the fixed point is stable if the largest value of |λi | is exactly 1.
Examples
1.
ẋ =
x2 + y 2 − 6
ẏ =
x2 − y
14
The critical points are found by solving ẋ = 0 and ẏ = 0 simultaneously:
ẋ =
0 ⇒ x2 + y 2 − 6 = 0
(1)
ẏ
0 ⇒ x2 − y = 0
(2)
=
From (2) we have x2 = y which, substituted in (1) yields
y + y 2 − 6 = 0 ⇒ (y − 2) (y + 3) = 0,
i.e.
y = 2 or
y = −3.
We only consider real solutions of x and y and thus
√ ignore the option y = −3, since this would give
complex values for x. From y = 2 we get x = ± 2 and our two critical points are therefore given
by
√ √ P1 =
2, 2
and P2 = − 2, 2 .
87
The Jacobian matrix is
A=
and so at P1
2x 2y
2x −1
A=
√
2√2
2 2
△ =
√
−10 2 < 0
√
2 2−1
4
−1
.
Thus
σ
=
so, since △ < 0, the critical point is locally a saddle point.
At P2
A=
√
−2√2
−2 2
△ =
σ =
√
10 2
√
−2 2 − 1 < 0
4
−1
,
from which
and
σ 2 − 4△ < 0
so that P2 is locally a stable focus.
2.
ẍ + x − x3 = 0
(15)
Writing the equation as a first order system with y = ẋ:
ẋ = y ẏ = x3 − x.
Solving ẋ = 0, ẏ = 0 simultaneously we get:
ẋ = 0 ⇒ y = 0
ẏ = 0 ⇒ x3 − x = x (x − 1) (x + 1) = 0
yielding the critical points P1 = (0, 0), P2 = (1, 0) and P3 = (−1, 0). The Jacobian matrix is
0
1
A=
.
2x2 − 1 0
Thus
0
−1
0
2
A1 =
A2 = A3 =
1
,
0
1
.
0
(16)
88
CHAPTER 4. AUTONOMOUS NON–LINEAR SYSTEMS
Now, det(A2 ) = det(A3 ) = −2 < 0, and therefore both P2 and P3 are saddle points. The matrix
A1 has ∆ = 1, σ = 0. It is therefore a focus, but since σ is on the borderline between stable and
unstable, it is not clear whether the behaviour near P1 is a stable focus, an unstable focus, or a
centre.
3.
x(n + 1) =
x(n)2 − 2y(n) + 3
y(n + 1) =
x(n)y(n) − 7y(n)
(17)
To find the singular points, we need to find values (x∗ , y∗ ) that satisfy
x =
x2 − 2y + 3,
(i)
y
xy − 7y.
(ii)
=
The equation (ii) is 0 = y(x − 8), and is satisfied by y = 0 or x = 8. Substituting y = 0 into (i)
gives x2 − x + 3 = 0, which has no real solution. Substituting x = 8 in (i) gives y = 29 21 . Thus the
only singular point is
P1 = (x∗ , y∗ ) = (8, 29.5).
The Jacobian of the system is
A=
thus
A1 =
−2
x−7
16 −2
29.5
1
2x
y
,
at the singular point P1 . The eigenvalue equation has solutions that are complex and |λ| > 1.
Thus the singular point P1 is unstable.
89
Phase Plane Analysis
This is usually applied to a system of two continuous equations, and is a means of obtaining the
overall behaviour of the system from the nature of the fixed points and other elementary considerations. It is more difficult to apply the methods to discrete systems, and the cases to be considered
here will all be continuous.
It is useful to start by defining some terms. A trajectory is a path in the (x, y) plane of a solution
of the system. The diagrams in Chapter 2 showing the solution behaviour near a fixed point are
diagrams of trajectories. An important fact about trajectories is that at any non–singular point
on the phase plane
dy/dt
g1 (x, y)
dy
=
=
(18)
dx
dx/dt
g2 (x, y)
has a definite value. Thus, at non–singular points the direction of a trajectory is unique, and
trajectories do not cross each other.
The x-isocline and y-isocline are curves in the (x, y) plane on which dx/dt = 0 and dy/dt = 0,
respectively. They are obtained by solving the (algebraic) equations
g1 (x, y) = 0, g2 (x, y) = 0.
(19)
The importance of the isoclines is that:
(a) singular points are at the intersection of an x–isocline and a y–isocline;
(b) they divide the (x, y) plane into regions in which the direction of a trajectory is known to
within 90o ;
(c) on the x–isocline dx/dt = 0, so trajectories crossing the x–isocline must be parallel to the
y–axis at the crossing point; similarly, trajectories crossing the y–isocline must be parallel to
the x–axis at the crossing point.
Phase plane analysis usually proceeds as follows:
(a) Calculate and draw the isoclines.
(b) Find the location of the fixed points.
(c) Determine the nature of each fixed point by calculating its Jacobian matrix.
(d) Sketch the direction of the trajectories in each region, bounded by isoclines, of the (x, y)
plane.
(e) Attempt to sketch trajectories throughout the whole (x, y) plane.
These techniques are illustrated in some application models that are discussed in the next section.
4.3 SOME APPLICATIONS
Application 1: Lotka–Volterra competition model
The model describes a competitive interaction between two species that are competing for the
food, water and other resources of an ecosystem. The use of these resources by one species inhibits
90
CHAPTER 4. AUTONOMOUS NON–LINEAR SYSTEMS
the ability of the other species to use them, thereby reducing the growth rate. The model aims
to determine under what conditions two competing species can coexist. Let x and y denote the
populations of the two species, then the model is represented by the dynamical system:
r1
x(P1 − x − α12 y)
P1
r2
y(P2 − y − α21 x)
P2
ẋ =
ẏ =
20
Note that in the absence of one of the species, say y = 0, the equation for x is ẋ = (r1 /P1 )x(P1 − x),
which is the standard logistic equation for population growth; it has a stable fixed point at x = P1 .
Similarly, if x = 0 then y is described by the logistic equation. The effect of competition is represented by the terms (−r1 /P1 )α12 xy and (−r2 /P2 )α21 yx.
In this model it does not make sense to talk about negative populations, so we are only interested
in solutions with x, y ≥ 0.
The isoclines are:
ẋ = 0: x = 0 (i)
ẏ = 0: y = 0 (iii)
or
or
P1 − x − α12 y = 0
P2 − y − α21 x = 0
(ii)
(iv)
(21)
The fixed points are therefore:
(0, 0) , (0, P2 ) , (P1 , 0) Q
(22)
where Q is at the intersection of (ii) and (iv). There are four separate cases to consider, depending
on the values of the constants P1 , P2 , α12 , α21 . The constants r1 , r2 do not appear in the isocline
equations and so do not affect the fixed points of the system: they affect the rate at which the
dynamic system evolves, but not the end–point(s) to which it can evolve. The cases are presented
diagrammatically as:
91
Figure 1
We will investigate case II here, i.e. the case where, as can clearly be seen from the sketch,
P1 > P2 /α21 and P2 > P1 /α12 . The other cases are left to you to do as exercises.
The Jacobian is:

r1 −


J=

2r1 x
r1 y
− α12
P1
P1
−r2 α21 y
P2
−α12 r1 x
P1



.
2r2 y r2 α21 x 
r2 −
−
P2
P2
(23)
Thus at (0, 0) we get
r1
0
J=
0
r2
so that
△ =
σ
r1 r2 > 0
=
r1 + r2
2
σ 2 − 4△ =
2
(r1 + r2 ) − 4r1 r2 = (r1 − r2 ) > 0.
The origin is therefore always an unstable node.
At the singular point (0, P2 )
J=
r1 (1 − α12 P2 /P1 )
0
−r2 α21
−r2
.
(24)
92
CHAPTER 4. AUTONOMOUS NON–LINEAR SYSTEMS
In this case, case II, P2 > P1 /α12 so the first term in the Jacobian is negative. Thus, by setting
β = 1 − α12
P2
P1
and keeping in mind that β < 0 (from P2 > P1 /α12 ), we have
△
= −r1 r2 β > 0
σ
= r1 β − r2 < 0
and
σ 2 − 4△ =
2
(r1 β − r2 ) + 4r1 r2 β
=
r12 β 2 − 2r1 r2 β + r22 + 4r1 r2 β
=
r12 β 2 + 2r1 r2 β + r22
=
(r1 β + r2 )
>
0.
2
The singular point (0, P2 ) is therefore clearly a stable node.
Similarly, at the point (P1 , 0),
J=
−r1
0
−r1 α12
r2 (1 − α21 P1 /P2 )
(25)
and P1 > P2 /α21 so the last term in the Jacobian is negative and the singular point can also shown
to be a stable node.
In order to deal with the singular point Q, we could solve (ii) and (iv), substitute into the Jacobian
and then evaluate σ and ∆. However, the algebra involved in such a direct approach is rather
lengthy (and therefore it is easy to make a mistake). It is better to approach the algebra with some
cunning. At the point Q equations (ii) and (iv) apply, so it is permissible to use them to simplify
the diagonal terms of the Jacobian. The result is:
93
At Q,
−r1 x∗
P1



J=
 −r2 α21 y∗
P2

−α12 r1 x∗

P1

.
−r2 y∗ 
P2
(26)
Thus
∆
σ
r1 r2 x∗ y∗
(1 − α21 α12 ),
P 1 P2
r1 x∗
r2 y∗
= −
+
.
P1
P2
=
At Q, both x∗ , y∗ are positive, so σ is clearly negative, and the sign of ∆ will depend on the sign
of (1 − α21 α12 ). In case II, P2 > P1 /α12 and P1 > P2 /α21 , thus
P2 >
P2
P1
>
α12
α12 α21
so that clearly
α12 α21 > 1.
Thus ∆ < 0 and Q is a saddle point.
(27)
94
CHAPTER 4. AUTONOMOUS NON–LINEAR SYSTEMS
The phase plane diagram is therefore:
In the above diagram we show:
(a) the direction of the trajectories on each isocline;
(b) the range of directions of the trajectories in each region bounded by isoclines;
(c) the singular points and information about their nature.
Note from the sketch that a singular point occurs only where an x–isocline intersects with a y–
isocline. Since the y–axis is an x–isocline and since the trajectories crossing the x–isocline are
parallel to the y–axis, the y–axis itself will be a trajectory. A similar argument follows for the
x–axis. The direction of the trajectories are away from the unstable singular points, towards the
stable points. Note also that the trajectories crossing the other x–isocline are crossing it parallel
to the y–axis and the trajectories crossing the y–isocline are crossing it parallel to the x–axis.
From a diagram like that above there is usually only one way to sketch the global, qualitative
behaviour of the trajectories. In this case:
95
Conclusion: Depending on the initial values of x and y, one or other of the species dies out, and
the other approaches the stable value given by the logistic equation for a single population.
The other cases I, III and IV are left as exercises. Here we shall simply state that stable coexistence
is possible only in case I.
96
CHAPTER 4. AUTONOMOUS NON–LINEAR SYSTEMS
Example
The above may be clearer if instead of using arbitrary constants r1 , P1 , α12 , r2 , P2 , α21 , we let the
constants have definite numerical values:
r1
=
0.1
P1
=
10000
α12
=
1
r2
=
0.2
P2
=
15000
α21
=
2.
The condition defining the case being considered here, case II, is P2 > P1 /α12 and P1 > P2 /α21 ,
which is clearly satisfied. The fixed points, and the values of the Jacobian at the fixed points, are:
0.1 0
1. (0, 0). From (23), J =
.
0 0.2
Clearly this is an unstable node.
2. (0, P2 ). From (24),
J
0.1 (1 − 1.5)
−0.2 × 2
−0.05
0
−0.4 −0.2
=
=
0
−0.2
.
Thus ∆ = +0.01, σ = −0.25; thus σ 2 > 4∆ and the fixed point is a stable node.
−0.1
−0.1
3. (P1 , 0). From (25), J =
.
0
−0.0666
Thus ∆ = 0.00666, σ = −0.1666; thus σ 2 > 4∆ and the fixed point is a stable node.
4. Solving (21), (ii) and (iv) we obtain Q : (5000, 5000). Thus from (26)
−0.05
−0.05
J=
.
−0.13333 −0.06666
Thus ∆ = −0.0033333, σ = −0.116666; thus the fixed point is a saddle point.
Application 2: A discrete predator–prey model
The model is represented by the dynamical system
x(n + 1) =
y(n + 1) =
x(n)(1 − a + by(n))
r
y(n)(1 + r − y(n) − cx(n))
P
28
97
where x(n) is the number of predators and y(n) is the number of prey. If x(n) = 0, then y(n) obeys
the usual logistic equation. For x(n) > 0, the term cx(n) reduces the growth rate of the prey, so
allowing for the fact that they are being eaten by the predators. In the absence of any prey, the
growth rate of the predators is negative, because they have nothing to eat. The term by(n) models
the fact that the growth rate of the predators depends on the size of their food supply.
For discrete systems we analyse the behaviour near the fixed points, but it is not possible to apply
the other phase plane techniques that were described for continuous systems. The fixed points are
obtained by solving simultaneously:
x(1 − a + by)
ry
− cx)
y(1 + r −
P
x=
y=
(i)
(ii)
29
(i) has solutions: x = 0 or y = a/b; substituting these solutions into (ii) leads to the fixed points:
x=
0, y = 0
x=
0, y = P
r
a a
1−
,y= .
c
bP
b
x=
The Jacobian is
J=
1 − a + by
−cy
bx
1 + r − 2ry
P − cx
30
.
(31)
At this stage we should say a little about the values of the constants a, b, c, r, P . They are all
> 0, and the conditions that we will impose are
1. bP > a
(32)
What does this mean? If the prey population is at its (logistic) equilibrium value P , and a
small number of predators are introduced, then the predators would have enough to eat and
their population would grow. It is clear that if this condition is not satisfied then there is no
possibility of a stable situation involving both predators and prey.
2. r < 2
(33)
This is required so that the model consisting of prey only is stable at y = P.
We now calculate the Jacobian, and estimate its eigenvalues, at the three critical points.
1−a
0
(i) x = 0, y = 0, J =
.
0
1+r
The eigenvalues are 1–a and 1 + r, so it is unstable and analogous to a saddle point.
1 − a + bP
0
(ii) x = 0, y = P , J =
.
−cP
1−r
(34)
(35)
Since bP > a, the diagonal elements of J, and therefore in this case its eigenvalues, are
λ1 = 1 − a + bP > 1 and λ2 = 1 − r with |λ2 | = |1 − r| < 1. Thus the singular point is again
unstable and analogous to a saddle point.
98
CHAPTER 4. AUTONOMOUS NON–LINEAR SYSTEMS
r
a 
1
−

c
bP 
a a
r
.
1−
,y= ,J=
(iii) x =
(36)


c
bP
b
ca
ra
−
1−
b
bP
It can be shown that both eigenvalues of J are less than 1, although the algebra is rather
intricate. For simplicity, let
ra
d =
, and note that 0 < d < 2;
bP
a and note that e > 0.
e = ar 1 −
bP

1
b
Then the Jacobian can be rewritten as


J=

be
ac
1

−
ca
b
△
=
1−d+e
σ
=
2 − d.
1−d



so that
The eigenvalues of J are (from equation (20) in Chapter 2):
√
2 − d ± 4 − 4d + d2 − 4 + 4d − e
2
√
2
2−d± d −e
,
=
2
which is real for d2 − e ≥ 0. If d2 − e < 0, the eigenvalues are complex and are given by
√
2 − d ± e − d2 i
.
2
For the real case d2 − e ≥ 0 we have
√ √
λ1 = 1 + 12 −d + d2 − e < 1 + 12 −d + d2 = 1.
√
Thus 0 < λ1 < 1. Also, λ2 = 1 + 12 −d − d2 − e < 1, and also λ2 > 1 + 12 (−d − d) =
1 − d > −1. Thus −1 < λ2 < 1.
For the complex case d2 − e < 0,
v
u
2
u
d
|λ| = t 1 −
+
2
r
e
=
1−d+ ,
4
√
e − d2
2
!
99
so that, for stability we need |λ| < 1, hence,
e
− d < 0.
4
In terms of the parameters a, b, r, an P , this means
bP < a + 4.
From equation (32) we now conclude that the condition
a < bP < a + 4
will ensure that this fixed point is asymptotically stable.
Example
We let the constants a, b, c, P , r have definite numerical values:
a
=
0.5
b =
0.001
c =
0.01
P
=
1000
r
=
0.5.
The conditions (32) and (33) are clearly satisfied. The fixed points and corresponding Jacobians
are:
0.5 0
1. (0, 0), J =
.
0 1.5
The eigenvalues are 0.5 and 1.5, so the fixed point has one eigenvalue greater than 1, and is
therefore unstable and analogous to a saddle point.
1.5
0
2. (0, P ), J =
.
−10 0.5
The eigenvalues are 1.5 and 0.5, so the fixed point is unstable and analogous to a saddle point.
100
CHAPTER 4. AUTONOMOUS NON–LINEAR SYSTEMS
3. (25, 500), J =
1
−5
0.025
0.75
.
Thus the eigenvalue equation is:
λ2 − 1.75λ + 0.875 = 0
which has solutions
λ = 0.875 ± 0.331i.
√
Thus |λ| = 0.8752 + 0.3312 ≈ 0.875; this is less than 1 so the fixed point is stable, and
analogous to a stable focus, because of the imaginary part in λ.
Application 3: A model of the AIDS epidemic
The basic idea is to divide the population into two groups: healthy but susceptible to the disease,
x(t), and sick, y(t). Then the basic assumption is that the disease spreads at a rate proportional
to xy. A simple model would be:
dy
= βxy − γy
dt
dx
= −βxy,
dt
(37)
where γ represents the death rate, and β is the infection rate. The system (37) is useful for a disease
with a short time–scale, but in the case of AIDS we also need to take into account population growth.
We get
dx
=
dt
αx − δ(x + y)2 − βxy
dy
=
dt
βxy − γy.
38
When y = 0, this system reduces to the logistic equation. For simplicity, we will scale the population
to have a value of 1 at the logistic population value x = P = α/δ; this implies α = δ. Thus x and
y are given as fractions of P . To translate into numbers of people, we would multiply by P . For
example P ≈ 50 million for South Africa, and P ≈ 6 × 109 for the world. Then the system is
dx
=
dt
αx − α(x + y)2 − βxy
dy
=
dt
βxy − γy.
In some places the algebra can be rather heavy, and we will sometimes use numerical values for
the constants α, β, γ:
α
=
0.02
β
=
0.5
γ
=
0.2.
39
101
The isocline diagram looks like:
The Jacobian is:
J=
α − 2αx − βy − 2αy
βy
−2αy − βx − 2αx
βx − γ
.
(40)
The singular points are therefore:
1. (0, 0), J =
α
0
0
−γ
.
It is clearly a saddle point.
2. (1, 0), J =
−α
0
−2α − β
β−γ
.
If β < γ then J has two negative eigenvalues and the fixed point is a stable node. If β > γ,
which is the case considered here, then ∆ < 0 and the fixed point is a saddle point.
3. To find the fixed point, we substitute x = γ/β into
αx − α(x + y)2 − βxy = 0.
For the case considered here, this gives
y 2 + 10.8y − 0.24 = 0.
The positive solution is y = 0.022. So the fixed point is (x∗ , y∗ ) = (0.4, 0.022). Thus J is
J=
−0.0079
0.011
−0.217
0
.
Thus ∆ = 0.002, σ = −0.008, so that σ 2 < 4∆. The fixed point is therefore a stable focus.
The phase plane diagram is therefore:
102
CHAPTER 4. AUTONOMOUS NON–LINEAR SYSTEMS
Discussion
The values of α, β, γ have been chosen as follows. The constant α = 0.02 represents a
population that grows at 2% per year, if resources are available. Putting β = 0.5 means that
the number of infected people increases by about 50% every year. If γ = 0.2 then about 20%
of infected people die every year. In this case the stable value of the population is reduced to
(x∗ + y∗ ), i.e. 0.437 of what it could be in the absence of AIDS. The crucial factor is γ/β, and
the goal of public health policy must be to increase γ/β. According to this model if public
education etc. can decrease β so that γ/β > 1, then the AIDS epidemic will fade away. This
can be understood as saying that if the average infective, before he dies, passes the disease
on to less than, on average, one other person, then the number of infectives will diminish to
zero.
Of course this model is rather crude, but it does show the power of applied dynamical systems
theory to make predictions about important matters in everyday life.
4.4 LIAPUNOV STABILITY THEORY
The method of determining stability by linearisation about a fixed point is sometimes referred to
as Liapunov’s first method. This section is about Liapunov ’s second method, also called Liapunov ’s
direct method. It works explicitly with the nonlinear system and has the advantage of being useful
in marginal situations; and also that it can be applied beyond a small region near the fixed point
– it may even determine global stability.
The basic idea is to seek a function that continually decreases towards a minimum as the system
evolves (this idea comes from energy in classical mechanics). Suppose that x∗ is an equilibrium
point of a given dynamical system. Then V (x) is a Liapunov function if:
(1) V (x) is defined in a domain Ω, with x∗ ∈ Ω.
(2) V (x) is real-valued and continuous and has continuous first partial derivatives.
103
(3) V (x) has a unique minimum at x∗ , i.e. V (x∗ ) < V (x) for all x ∈ Ω, x ̸= x∗ .
(4) The value of V decreases along any trajectory in Ω.
(41)
The key requirement is (4). Later it will be formulated more precisely, and in somewhat different
ways for the discrete and continuous cases. We will also show that if a Liapunov function exists,
then the equilibrium point x∗ is asymptotically stable.
The idea behind a Liapunov function is shown in the above figure. The function V (x) has a minimum of zero at x = x∗ . A trajectory in the (x1 , x2 ) plane is shown: it is such that, when projected
into the surface V (x), its direction must be downhill.
One important point to remember about a Liapunov function is that it is only a function of the
coordinates (x1 , x2 , ..., xn ). It is not defined by motion of the system. Rather, evolution of the
system means that the coordinates (x1 , x2 , ..., xn ) change, and in this way V (x) changes. Also,
note that Liapunov functions are not unique.
Example
Consider the system
x1 (k + 1) =
x2 (k + 1) =
x2 (k)
2
1 + x1 (k) + x2 (k)
x1 (k)
2
,
2
2
1 + x1 (k) + x2 (k)
42
which has x = (0, 0) as an equilibrium point. Let us define the function
V (x) = x21 + x22 .
(43)
This V is defined everywhere, is real-valued and continuous, and has a unique minimum at the
equilibrium point. To determine how V changes along a trajectory, consider any two vectors, x(k)
104
CHAPTER 4. AUTONOMOUS NON–LINEAR SYSTEMS
and x(k + 1), related by the system equations. We find
V (x (k + 1))
=
=
x1 (k + 1)2 + x2 (k + 1)2
"
#2
x2 (k)
2
2
=
=
2
1 + x1 (k) + x2 (k)
"
+
#2
x1 (k)
2
1 + x1 (k) + x2 (k)
2
2
x2 (k) + x1 (k)
2
1 + x1 (k) + x2 (k)
2
2
V (x (k))
h
i2 ≤ V (x (k)) .
2
2
1 + x1 (k) + x2 (k)
(44)
Thus the function V obeys condition (4), and therefore V is a Liapunov function.
Liapunov theorem for continuous systems
We now focus on a continuous system
dx
(t) = g(x(t))
dt
(45)
together with a given equilibrium point x∗ . It is assumed that g is continuous. The requirement (414), that along any trajectory V is decreasing, is expressed as follows. Suppose x(t) is a trajectory.
Then V (x(t)) is a function of time and represents the values of V along the trajectory. The
requirement (41-4) is then
d
V̇ (x) ≡
(V (x (t))) ≤ 0
(46)
dt
with equality if and only if x = x∗ .
By the chain rule, this is
∂V dx1
∂V dx2
∂V dxn
d
(V (x (t))) =
+
+···+
≤ 0.
dt
∂x1 dt
∂x2 dt
∂xn dt
(47)
Now apply the system equation (45):
d
∂V
∂V
∂V
(V (x (t))) =
g1 (x (t)) +
g2 (x(t)) + · · · +
gn (x (t)) ≤ 0
dt
∂x1
∂x2
∂xn
(48)
[Note: For those of you also taking MAT215 = APM212, this is more conveniently written as
d
(V (x(t))) = (∇V ) · g ≤ 0
dt
(49)
where ∇ is the gradient operator.]
Theorem Suppose that the dynamical system (45) has an equilibrium point x∗ , that the function
g(x) in (45) is continuous, and that there exists a function V (x)[V : Ω → R, with the domain
Ω ⊆ Rn ] such that
105
(i) V (x) is continuous and has continuous first partial derivatives;
(ii) x∗ ∈ Ω;
(iii) V (x) has a unique minimum at x∗ ;
(iv) Ω is either Rn or bounded, and if it is bounded then V (x) is constant on the boundary of Ω;
(v) V satisfies (48) [or equivalently (49)] with equality if and only if x = x∗ .
Then for any trajectory that starts in Ω(x(t = 0) ∈ Ω),
lim x(t) = x∗ .
t→∞
[Alternatively, x∗ is asymptotically stable with Ω contained in the domain of stability.]
Sketch of proof. From (v), V (x(t)) is monotonically decreasing with t; further, by (iii) V is
bounded below by V (x∗ ). Thus V∞ ≡ limt→∞ V (x(t)) exists. We now observe that V∞ = V (x∗ ),
since only at x = x∗ does dV (x(t))/dt = 0. Thus limt→∞ V (x(t)) = V (x∗ ). Since V (x) has a
unique minimum at x = x∗ , this implies
lim x(t) = x∗ .
t→∞
□
Note that condition (iv) above was not used explicitly, but it is necessary to ensure that a trajectory that starts in Ω cannot leave Ω. Note also that condition (v) can be weakened to V̇ ≤ 0 with
equality only at x = x∗ and at isolated points on any trajectory; the proof is somewhat intricate
and is omitted.
Liapunov theorem for discrete systems
The dynamical system is:
x(n + 1) = g(x(n))
(50)
and x = x∗ is an equilibrium point. The requirement (41-4) is expressed as follows. If at t = n
the value of the Liapunov function is V (x(n)), then at the next time instant V = V (x(n + 1)) =
V (g(x(n))). Define
∆V (x) = V (g(x)) − V (x)
(51)
then the condition that V is decreasing can be expressed as
∆V (x) ≤ 0
(52)
for all x ∈ Ω, with equality if and only if x = x∗ .
Theorem Suppose that the dynamical system (50) has an equilibrium point x∗ , that the function
g(x) in (50) is continuous, and that there is a function V (x)[V : Ω → R, with the domain Ω ⊆ Rn ]
such that
(i) V (x) is continuous;
106
CHAPTER 4. AUTONOMOUS NON–LINEAR SYSTEMS
(ii) x∗ ∈ Ω;
(iii) V (x) has a unique minimum at x = x∗ ;
(iv) ∆V , defined by (51), satisfies (50), with equality if and only if x = x∗ .
Then there exists a domain Ω′ , with x∗ ∈ Ω′ ⊆ Ω, such that for any trajectory that starts in Ω′ ,
lim x(n) = x∗
n→∞
[i.e. x∗ is asymptotically stable].
Proof. Omitted. The basic idea is the same as for the continuous case, but because a trajectory
can jump out of Ω the proof of the theorem is rather complicated.
Liapunov theory can also be used to show that a fixed point is unstable. In the above theorems, if
the condition (v) V̇ ≤ 0 (continuous case) or (iv) ∆V ≤ 0 (discrete case) is replaced by (v) V̇ ≥ 0
(continuous case) or (iv) ∆V ≥ 0 (discrete case), then the fixed point x = x∗ is unstable.
Examples
1. We have already shown that V (x) = x21 + x22 is a Liapunov function for the system (42),
x1 (k + 1)
=
x2 (k + 1)
=
x2 (k)
2
2,
2
2.
1 + x1 (k) + x2 (k)
x1 (k)
1 + x1 (k) + x2 (k)
Thus the equilibrium point (0, 0) is asymptotically stable. Further, this conclusion could not
have been obtained by analysing the Jacobian at (0,0), which is
0 1
J=
.
(53)
1 0
The eigenvalues are ±1, which is a borderline case, so linear theory provides no information
about whether or not the equilibrium point is stable.
2. ẋ = −x3 − xy 2 + y
3
(54)
2
ẏ = −y − x y − x
V = x2 + y 2
(55)
Clearly the system (54) has a fixed point at (0, 0), and V (x, y) trivially satisfies the conditions
(i) to (iv) of the continuous Liapunov theorem. In this case (48) gives
2x(−x3 − xy 2 + y) + 2y(−y 3 − x2 y − x)
=
−2x4 − 2x2 y 2 + 2xy − 2y 4 − 2x2 y 2 − 2xy
=
−2(x4 + 2x2 y 2 + y 4 )
=
−2(x2 + y 2 )2 .
56
107
2
Since −2 x2 + y 2 ≤ 0 with equality only at x = y = 0, the condition (v) is satisfied.
Thus the fixed point (0, 0) is asymptotically stable. In this case Ω is the whole space, so all
trajectories tend towards (0, 0). Note that linear analysis is of no help. At (0, 0):
J=
0
−1
1
0
(57)
so that ∆ = 1, σ = 0 and the fixed point may be stable or unstable.
If a Liapunov function is defined over a large region Ω, we can say more than if it is defined
only in a small region. In fact, if the initial point x(0) has V (x(0)) = q, then the subsequent
trajectory never goes outside the region V (x) ≤ q [provided this region is contained in Ω].
Therefore, the region Ω over which the Liapunov function is defined, delineates a region Ω in
which system behaviour is related to the equilibrium point. This information about extent
of stability cannot be obtained by linearization. For this reason a Liapunov analysis often
follows a linearization analysis of a fixed point even if stability has been established.
Suppose a Liapunov function can be found that does not change along any trajectory: V̇ (x) ≡
0, or ∆V (x) ≡ 0. Thus V is constant along any trajectory and V is said to be a constant of
motion. We then know that any trajectory must lie on a contour of the function V . In an
n-dimensional space a contour defines an (n − 1)-dimensional surface.
Example. Consider the system
ẋ1 = x2 , ẋ2 = −x1 .
(58)
Let V (x) = x21 + x22 . Then V̇ (x) = 2x1 x2 + 2x2 (−x1 ) = 0. Thus V is a constant of the motion and
the system trajectories are circles centred at the origin.
In practice, we are given the equations describing a system and have to try to construct a suitable
Liapunov function. This can be extremely difficult unless we use the context, or original motivation,
of the system equations. This idea is illustrated in two examples.
1. A simple swinging pendulum has an equilibrium point when hanging straight down. Further,
it is intuitively clear that, assuming there is friction at the bearing, then the system will
approach the equilibrium point; i.e. it is stable. We assume that the pendulum is of length R
and has point mass M concentrated at the end. The position of the pendulum at any time is
described by the angle θ with the vertical. We assume that the frictional force is proportional
to the speed of the pendulum.
108
CHAPTER 4. AUTONOMOUS NON–LINEAR SYSTEMS
From Newton’s laws of motion we have:
M Rθ̈ = −M g sin θ − M k θ̇.
(59)
Expressing (59) as a first-order system:
θ̇ = ω, ω̇ = −
g
k
sin θ − ω.
R
R
(60)
There is a fixed point at (θ = 0, ω = 0).
The question now is: How do we find a Liapunov function for the system (60)? We want
something that we expect will decrease with time. The physical situation is that energy is
being dissipated by friction, and thus we try the mechanical energy of the system. [It was this
type of situation that led to the development of the Liapunov theorem.] V = kinetic energy
+ potential energy, so
1
V (θ, ω) = M R2 ω 2 + M gR(1 − cos θ).
(61)
2
It is clear that, without the motivation of energy, it would be difficult to guess the form (61)
from (60).
We now show that (61) is a Liapunov function. It is continuous and so are its first partial
derivatives, and it has a unique minimum at (θ = 0, ω = 0). Note that the range of θ can be
taken as −π < θ ≤ π. We compute V̇ :
V̇ (θ, ω) =M R2 ω ω̇ + M gR sin θ.θ̇
g
kω 2
2
=M R − ω sin θ −
+ M gR ω sin θ
R
R
= − kM Rω 2 ≤ 0.
62
Thus V̇ ≤ 0 with equality whenever ω = 0. This includes the fixed point (θ = 0, ω = 0) as
well as points that are isolated on any trajectory. Thus the fixed point is asymptotically stable.
109
2. Suppose a dog is chasing a rabbit. The rabbit runs a straight course along the x-axis at
constant velocity R. The dog runs at a constant velocity D, but in such a way as to always
point directly toward the rabbit. Let xr , yr and xd , yd denote the x and y coordinates of the
rabbit and dog respectively. Then
ẋr = R, ẏr = yr = 0.
(63)
The fact that the velocity of the dog is D means that
ẋ2d + ẏd2 = D2 .
(64)
The velocity vector of the dog always pointing toward the rabbit means that:
ẋd =
−k(xd − xr )
(65)
ẏd =
−kyd .
66
The constant k is found by calculating ẋ2d + ẏd2 from (65) and (66), and then equating it to
D2 from (64). We find
D
(67)
k=
12
2
(xd − xr ) + yd2
so that (65) and (66) become:
ẋd =
(xd − xr ) D
−q
2
(xd − xr ) + yd2
ẏd =
−q
yd D
.
68
2
(xd − xr ) + yd2
The system is more meaningful when expressed in relative coordinates, representing the difference in position of the rabbit and dog. Define
x = xd − xr , y = yd
(69)
110
CHAPTER 4. AUTONOMOUS NON–LINEAR SYSTEMS
then
ẋd = ẋ + R, ẏd = ẏ
(70)
and (68) becomes:
ẋ = − p
xD
x2 + y 2
− R, ẏ = p
−yD
x2 + y 2
.
(71)
We examine the system (71). Will the dog catch the rabbit? This is equivalent to asking
whether a trajectory of (71) with arbitrary initial condition x(0), y(0) will eventually go to
the origin, where the relative coordinates are zero. Since (71) is undefined at the origin and
the origin is therefore not an equilibrium point, we cannot apply Liapunov’s theorem to this
problem. However, it is still useful to seek a Liapunov function for (71). The dog is trying to
catch the rabbit by minimising the distance between them. It is therefore natural to try the
distance (or distance squared) from the origin as a Liapunov function:
V (x, y) = x2 + y.
(72)
Now
V̇ (x, y) =
=
=
d
V (x(t), y(t))
dt
!
2y (−yD)
xD
−R + p
2x − p
2
2
x +y
x2 + y 2
p
−2D x2 + y 2 − 2Rx.
73
If D > R it easily follows that V̇ (x, y) < 0 everywhere except at x = y = 0 where V̇ (x, y) = 0.
Thus V (x, y) decreases along any trajectory if D > R. Hence if the dog runs faster than the
rabbit, he will catch it.
(Note that conditions (i) to (iv) are also satisfied. Thus V (x, y) is a Liapunov function.)
Attractors
So far we have talked about equilibrium and stability in terms of one isolated point. While this is
(usually) the case for linear systems, non-linearity permits more interesting possibilities. We start
by generalising the concept of an equilibrium point to that of an invariant set:
A set G is an invariant set (for a dynamical system) if whenever a point x on a system trajectory
is in G, the trajectory remains in G.
An equilibrium point is perhaps the simplest example of an invariant set. Also, if a system has
several equilibrium points then the union of these points is an invariant set. Here are some more
substantial examples.
1. ẋ = y + x(1 − x2 − y 2 ), ẏ = −x + y(1 − x2 − y 2 )
(74)
111
The origin is an equilibrium point, and it can easily be shown that the system has no other
equilibrium points. However, once the system is on the unit circle x2 + y 2 = 1, it will stay
there. To see this, observe that at any point on the unit circle the direction of a trajectory
is given by ẏ/ẋ = −x/y, which is the direction of a tangent to the unit circle. Thus the unit
circle is an invariant set of the system (74).
2. x(n + 1) = 3.2x(n) − 2.2x(n)2
(75)
The system has equilibrium points at x = 0 and 1. Writing (75) as x(n + 1) = g(x(n)), we
see that g(x) = 3.2x − 2.2x2 , so that g ′ (x) = 3.2 − 4.4x. Thus |g ′ (x)| > 1 at both equilibrium
points and they are therefore unstable. However, if we calculate the evolution of this system
on a computer we find that it settles down to a stable 2-cycle:
x (n) = . . . , 0.746; 1.163; 0.746; 1.163; 0.746; . . . .
(76)
Thus the system has an invariant set G = {0.746; 1.163}.
Having generalised the concept of an equilibrium point to that of an invariant set, we now generalise
the concept of stability. Although the phrase “asymptotically stable” is sometimes used to describe
an invariant set, it is more usual to talk about an attractor.
An invariant set G is an attractor if there exists ϵ > 0 such that if |x(0)−G| < ϵ then limt→∞ |x(t)−
G| = 0 (continuous case), or limn→∞ |x(n) − G| = 0 (discrete case). Note that |x − G| means the
greatest lower bound of the distance between x and every point in G, i.e. the shortest distance
between x and G.
In general, invariant sets and attractors may have an extremely complicated structure. However,
here we will restrict attention to cases where the attractor is a closed curve, i.e. analogous to a
circle, and is then called a limit cycle; or consists of a finite number (n) of points and is called an
n-cycle. Examples 1 and 2 above illustrated these two cases.
We now investigate conditions under which an invariant set is an attractor.
First we state a general result, which is an extension of the Liapunov theory of the previous section.
Theorem Consider the dynamical system (50) [or (45)]. Let V (x) be a scalar function with continuous first partial derivatives and let s > 0. Let Ωs denote the region where V (x) < s. Assume
that Ωs is bounded and that ∆V (x) ≤ 0 [or V̇ (x) ≤ 0 in continuous time]. Let S be the set of
points within Ωs where ∆V (x) = 0 [or V̇ (x) = 0], and let G be the largest invariant set within S i.e.
the union of all the invariant sets within S. Then every trajectory in Ωs tends to G as time increases.
Proof Omitted.
Comment. Not only does the theorem show that G is an attractor, but it also gives a lower
estimate of the domain (Ωs ) of attraction.
112
CHAPTER 4. AUTONOMOUS NON–LINEAR SYSTEMS
A special case of the above result is the Poincaré-Bendixson theorem which applies to 2-dimensional
continuous systems: If a region R of the phase plane contains no fixed points, and trajectories enter
but never leave R, then R contains a limit cycle. Again the proof is omitted.
This implies that the only attractors in 2-dimensional continuous systems are fixed points and limit
cycles. Attractors with a more complex structure are found in higher dimensional continuous systems, and in discrete systems.
Example. Consider (74) again –
ẋ = y + x[1 − x2 − y 2 ], ẏ = −x + y[1 − x2 − y 2 ].
Define
V (x, y) = (1 − x2 − y 2 )2 .
(77)
Then
V̇ (x, y) =
−2(1 − x2 − y 2 )(2xẋ + 2y ẏ)
=
−4(x2 + y 2 )(1 − x2 − y 2 )2 .
78
The function V (x, y) has the graph shown:
V̇ (x, y) ≤ 0 everywhere, with equality at x = y = 0 and on the unit circle. V has a maximum at
x = y = 0 and therefore by a simple linearization argument, as in the previous chapter, this is an
unstable equilibrium point. Let Ω1 be the region where V (x) < 1 [then 0 ∈
/ Ω1 ]. In applying the
theorem the set S is the unit circle. We have already seen that the unit circle is an invariant set;
thus every trajectory in Ω1 tends towards the limit cycle consisting of the unit circle.
113
For an n-cycle in a discrete system we can use linear methods to determine stability. We illustrate
this for a 2-cycle in a 1-dimensional system
x(n + 1) = g(x(n))
(79)
and suppose that (a1 , a2 ) form a 2-cycle:
a1 = g(a2 ), a2 = g(a1 ).
(80)
Let b be near a1 . Let c = g(b), so c is near a2 , and let d = g(c), so that d is near a1 . Then the
2-cycle is an attractor if d is nearer to a1 than b, i.e. if
|d − a1 | < |b − a1 |.
(81)
[Technically we should also check that if e = g(d), then e is nearer to a2 than c is. However, it
turns out that this is not necessary because the condition (80) is symmetric in a1 , a2 .]
114
CHAPTER 4. AUTONOMOUS NON–LINEAR SYSTEMS
Using Taylor’s theorem,
c =g(b) = g((b − a1 ) + a1 ) = g(a1 ) + (b − a1 )g ′ (a1 )
=a2 + (b − a1 )g ′ (a1 ).
82
Similarly,
d = g(c) = a1 + (c − a2 )g ′ (a2 ).
(83)
Substituting for (c − a2 ) from (82) into (83) gives
d =a1 + (b − a1 )g ′ (a1 )g ′ (a2 )
∴
(d − a1 )
=g ′ (a1 ) g ′ (a2 ) .
(b − a1 )
84
|g ′ (a1 )g ′ (a2 )| < 1.
(85)
Thus the 2-cycle is an attractor if
We will not prove the condition for an n-cycle to be an attractor, but it is the natural generalisation
of (85):
|g ′ (a1 )g ′ (a2 ) · · · g ′ (an )| < 1.
(86)
Example. Consider (75) again:
x(n + 1) = 3.2x(n) − 2.2x(n)2 .
We will show algebraically that the system has a 2-cycle, and that it is an attractor. Having a cycle
(a1 , a2 ) means:
a2 = g(a1 ), a1 = g(a2 ) ∴ a1 = g(g(a1 )).
(87)
We therefore solve
0 = g(g(x)) − x
(88)
g(x) = 3.2x − 2.2x2 .
(89)
with
115
Thus (88) is:
0 =g(3.2x − 2.2x2 ) − x
∴
0 =3.2(3.2x − 2.2x2 ) − 2.2(3.2x − 2.2x2 )2 − x.
90
Now x = g(g(x)) is also solved by solutions of x = g(x), so we already know that x = 0 and x = 1
are solutions of (90). Thus we can divide (90) by x(x − 1). Doing all the algebra we get
11
(−121x2 + 231x − 105) = 0.
125
(91)
Thus using the quadratic formula
√
21 ± 21
x=
= 0.746 or 1.163.
22
(92)
Thus the 2-cycle is, as stated earlier (0.746; 1.163). To investigate stability note that
g ′ (x) = 3.2 − 4.4x
(93)
|g ′ (0.746).g ′ (1.163)| = 0.158
(94)
Thus
which is less than 1 and the 2-cycle is therefore an attractor.
Comment. The dynamical system (75) represents logistic population growth, with a large growth
factor (3.2). As such it could be used to describe the population of various species of insect. The
oscillations in population size predicted by the model are observed in nature.
Exercises
1. Consider the cases I, III and IV of the Lotka–Volterra competition model. For each case, find
the nature of the singular points and sketch the phase plane diagrams with the trajectories.
2. For the AIDS epidemic model consider the case γ/β > 1. Find the nature of the singular
points and sketch the phase plane diagram with the trajectories.
3. Classify (if possible) each critical point of the given plane autonomous system as a stable
node, an unstable node, a stable spiral point, an unstable spiral point, or a saddle point.
(a) ẋ = 1 − 2xy
ẏ = 2xy − y
Will be solve in details, the rest follow the same. We solve the critical point by letting ẋ = ẏ = 0.
1 − 2xy = 0
· · · · · · (1)
2xy − y = 0
· · · · · · (2)
116
CHAPTER 4. AUTONOMOUS NON–LINEAR SYSTEMS
from (1), 1 = 2xy, subt. in (2),
we get 1 − y = 0, ⇒ y = 1, now subt y = 1 in 1 = 2xy,
we get x = 21 then P1 =
1
2; 1
now we form the Jacobian Matrix, of the form:

J =
∂
∂x (1 − 2xy)
∂
∂y (1 − 2xy)
∂
∂x (2xy − y)
∂
∂y (2xy − y)

−2y
−2x
J =




2x − 1
2y
Now, Subt. P1 = ( 12 , 1) in the Jacobian, we get

−2
−1
2
0
J =

 , it takes us back to chapter 1 and chapter 2.

−2
−1
2
0

 , σ = −2, ∆ = 2
J =
λ2 + 2λ + 2,
λ1 = 1 − i
λ2 = −1 + i
by 2.3 of chapter 2 bullet number 5. The fixed point is a stable focus or stable spiral point.
(b) ẋ = y − x2 + 2
ẏ = x2 − xy
(c) ẋ = −3x + y 2 + 2
ẏ = x2 − y 2
(d) ẋ = −2xy
ẏ = y − x + xy − y 3
(e) ẋ = x2 − y 2 − 1
ẏ = 2y
(f) ẋ = 2x − y 2
ẏ = −y + xy
(g) ẋ = xy − 3y − 4
ẏ = y 2 − x2
(h) ẋ = x(1 − x2 − 2y 2 )
ẏ = y(3 − x2 − 3y 2 )
(i) ẋ = −2x + y + 10
y
ẏ = 2x − y − 15 y+5
.
117
4. Classify (if possible) each critical point of the given second–order differential equation as a
stable node, an unstable node, a stable spiral point, an unstable spiral point, or a saddle.
(a) θ̈ = (cos θ − 0.5) sin θ, |θ| < π.
(b) ẍ + ẋ(1 − x3 ) − x2 = 0.
5. Consider the nonlinear dynamical system
a(n + 1)
=
[1.8 − 0.8a(n) − 0.2b(n)]a(n)
b(n + 1)
=
[1.8 − 0.8b(n) − 0.2a(n)]b(n).
(a) Find the equilibrium vectors (or points) for this system, and for each equilibrium vector,
determine if it is asymptotically stable or unstable.
(b) Pick a point (a(0), b(0)) close to each equilibrium vector and compute (a(1), b(1)), (a(2),
b(2)) and (a(3), b(3)) using this nonlinear system to see if (a(k), b(k)) seems to agree
with your results in part (a).
6. Repeat Problem 5 for the nonlinear dynamical system
a(n + 1)
=
[1.4 − 0.4a(n) − 1.2b(n)]a(n)
b(n + 1)
=
[1.4 − 0.4b(n) − 1.2a(n)]b(n).
ẋ(t)
=
ẏ(t)
= ex(t) − 1
7. For the system
sin[x(t) + y(t)]
determine all the equilibrium points, and classify each equilibrium point as asymptotically
stable or unstable.
8. The van der Pol equation. The equation
ẍ + ϵ[x2 − 1]ẋ + x = 0
arises in the study of vacuum tubes. Show that if ϵ < 0, the origin is asymptotically stable
and that for ϵ > 0 it is unstable.
9. In many fishery science models, the rate at which a species is caught is assumed to be directly
proportional to its abundance. If both predator and prey are being exploited in this manner,
the Lotka-Volterra differential equations take the form
ẋ =
−ax + bxy − ϵ1 x
ẏ
−cxy + dy − ϵ2 y,
=
where ϵ1 and ϵ2 are positive constants.
(a) When ϵ2 < d, show that there is a new critical point in the first quadrant that is a centre.
118
CHAPTER 4. AUTONOMOUS NON–LINEAR SYSTEMS
(b) Volterra’s principle states that a moderate amount of exploitation increases the average number of prey and decreases the average number of predators. Is this fisheries
model consistent with Volterra’s principle?
10. Suppose that the damping force of free, damped motion is proportional to the square of the
velocity, and the differential equation is
ẍ = −
β
k
ẋ|ẋ| − x.
m
m
(a) Write the second-order differential equation as a plane autonomous system and find all
critical points.
(b) The system is called overdamped when (0,0) is a stable node and is called underdamped
when (0,0) is a stable spiral point. Physical considerations suggest that (0,0) must be an
asymptotically stable critical point. Show that the system is necessarily underdamped.
d
y|y| = 2|y|.]
[Hint: dy
11. The plane autonomous system
y
x−x
1+y
y
x−y+β
= −
1+y
ẋ =
ẏ
α
arises in a model for the growth of microorganisms in a chemostat, a simple laboratory device
in which a nutrient from a supply source flows into a growth chamber. In the system, x denotes
the concentration of the microorganisms in the growth chamber, y denotes the concentration
of nutrients, and α and β are positive constants that can be adjusted by the experimenter.
Show that when α > 1 and β(α − 1) > 1, the system has a unique critical point (x∗ , y∗ ) in
the first quadrant, and demonstrate that this critical point is a stable node.
12. Consider the dynamical system
a(n + 1)
=
[1.4 − 0, 4a(n) − 1.2(n)]a(n)
b(n + 1)
=
[1.4 − 0.4b(n) − 1.2a(n)]b(n).
Use phase plane analysis in the first quadrant to analyze the solutions to this system. Do this
by constructing the lines
a = 0 and 0.4 − 0.4a = 1.2b = 0
and drawing arrows to the left or right according to the solution to an appropriate inequality,
and then constructing the lines
b = 0 and 0.4 − 0.4b − 1.2a = 0
and drawing arrows up or down. Finally, combine the two constructions to get a phase plane
analysis of the behaviour of solutions.
119
13. Use phase plane analysis to analyze the solutions to the dynamical system
a(n + 1)
b(n + 1)
2
[2 − a(n) − b(n)]a(n)
3
2
= [2 − b(n) − a(n)]b(n)
3
=
in the first quadrant.
14. Use phase plane analysis to analyze the solutions to the dynamical system
a(n + 1) − a(n)
=
1.2[1 − b(n)]a(n)
b(n + 1) − b(n)
=
0.5[1 − 5b(n) + a(n)]b(n)
for the predator-prey model. Notice that the solutions oscillate about the equilibrium vector
4
A=
.
1
15. No Marginal Stability. Consider the system
ẋ1 = x1 a2 − x21 − x22 + x2 a2 + x21 + x22
ẋ2 = −x1 a2 + x21 + x22 + x2 a2 − x21 − x22
and the function V (x) = x21 + x22 . Show that the system is stable for a = 0 and unstable for
a ̸= 0. Thus, the transition from (asymptotic) stability to instability does not pass through a
point of marginal stability.
16. Prove that the origin is asymptotically stable for each of the systems below using Liapunov’s
direct method. [In parts (a) and (b) find a suitable Liapunov function. In part (c) try the
suggested function.]
(a) ẋ = y
ẏ = −x3
(b) ẋ = −x3 − y 2
ẏ = xy − y 3
(c) ẋ = y(1 − x)
ẏ = −x(1 − y)
V = −x − log(1 − x) − y − log(1 − y)
17. The van der Pol Equation. The equation
ẍ + ϵ[x2 − 1]ẋ + x = 0
arises in the study of vacuum tubes. Show that if ϵ > 0 there is a limit cycle that is approached
by all other trajectories.
[Hint: The origin is unstable, and Liapunov’s method will show that trajectories a long way
from the origin move towards the origin. Now use the Poincaré-Bendixson theorem.]
120
CHAPTER 4. AUTONOMOUS NON–LINEAR SYSTEMS
18. Consider the dynamical system
A(n + 1) = 3, 2A(n) − 3, 2A2 (n).
The numbers a = 0 and a = 0.6875 are equilibrium values for this dynamical system.
(a) Write down the equation, f (f (a)) − a = 0, factor out the terms a and a − 0.6875, then
use the quadratic formula to find the 2-cycle for this equation.
(b) Show that this 2-cycle is an attractor.
19. Consider the dynamical system
A(n + 1) = 3.2A(n) − 1.6A2 (n).
(a) Approximate the numbers a1 and a2 that form a 2-cycle for this dynamical system, by
letting ao = 1 and computing A(1), A(2), and so forth, until you repeat the same two
numbers. Round off each A(k) value to 3 decimal places.
(b) Show that this 2-cycle is an attractor by computing
f ′ (a1 )f ′ (a2 ).
20. Determine whether the origin is a stable equilibrium point for each of the following systems:
(a) x1 (k + 1) = 2x1 (k) + x2 (k)2
x2 (k + 1) = x1 (k) + x2 (k)
(b) x1 (k + 1) = 1 − ex1 (k)x2 (k)
x2 (k + 1) = x1 (k) + 2x2 (k).
21. Given the system
ẋ1
= x2
ẋ2
= x3
ẋ3
= −(x1 + cx2 )3 − bx3
consider the function
V (x1 , x2 , x3 ) =
1
1
b2
1
b 4
x1 + (x1 + cx2 )4 − x41 + x22 + bx2 x3 + x23 .
4
4c
4c
2
2
(a) What condition on b > 0, c > 0 insures that V (x1 , x2 , x3 ) > 0 for (x1 , x2 , x3 ) ̸= 0?
(b) Show that V̇ has the form
V̇ (x1 , x2 , x3 ) = γx22
"
3
x1 + cx2
2
2
3
+ x21
4
#
and determine the constant γ.
(c) Is V always a Liapunov function if b and c satisfy the conditions of part (a)?
121
(d) Is the origin asymptotically stable under these conditions?
22. Consider the dynamical system
A(n + 1) = 3.1A(n) − 3.1A2 (n).
(a) Show that a1 = 0.76456652 and a2 = 0.558014125 form a 2-cycle by showing that, if
A(0) = a1 , then A(1) = a2 , and A(2) = a1 .
(b) Show that this 2-cycle is stable by showing that
|f ′ (a1 )f ′ (a2 )| < 1.
23. Consider the dynamical system
A(n + 1) = 3.3A(n) − 3.3A2 (n).
(a) Find a 2-cycle for this dynamical system.
(b) Show that this 2-cycle is stable by showing that
|f ′ (a1 )f ′ (a2 )| < 1.
24. Consider the dynamical system
A(n + 1) = 3.6A(n) − 3.6A2 (n).
(a) Find a 2-cycle for this system.
(b) Show that this 2-cycle is unstable by showing that
|f ′ (a1 )f ′ (a2 )| > 1.
25. Consider the dynamical system
A(n + 1) = 3.5A(n) − 2.5A2 (n).
(a) Show that a1 = 0.701237896 is one point in a 4-cycle by finding a2 = f (a1 ), a3 = f (a2 ),
and a4 = f (a3 ), and showing that a1 = f (a4 ).
(b) Show that this 4-cycle is stable by showing that
|f ′ (a1 )f ′ (a2 )f ′ (a3 )f ′ (a4 )| < 1.
26. The rod of a pendulum is attached to a horizontally movable joint and rotates at a constant
angular speed ω. The resulting differential equation for θ (the angle that the pendulum makes
with the vertical) is:
mℓ
d2 θ
dθ
= ω 2 mℓ sin θ cos θ − mg sin θ − β
.
dt2
dt
(a) If ω 2 < g/ℓ, show that (θ = 0, dθ/dt = 0) is a stable fixed point and is the only critical
point with −π < θ < π.
122
CHAPTER 4. AUTONOMOUS NON–LINEAR SYSTEMS
(b) If ω 2 > g/ℓ, show that (θ = 0, dθ/dt = 0) is unstable and that there are two additional
stable critical points with −π < θ < π.
27.(a) State and prove Liapunov’s theorem for a continuous-time dynamical system.
(b) By finding a suitable Liapunov function show that the origin is a stable fixed point for the
dynamical system:
ẋ
= yx2 − x
ẏ
= −x3 − y
What is the domain of stability?
28.(a) A farmer’s crops are eaten by one species of insect, I1 , who are in turn eaten by another
species, I2 . The populations are described by the dynamical system:
dI1
dt
=
(0.5 − 0.01I2 − 0.0005I1 )I1
dI2
dt
=
(−0.5 + 0.001I1 )I2 .
Determine the location of the equilibrium point for which both I1 and I2 are non-zero.
Determine the nature of the equilibrium point.
(b) The farmer now decides to try to reduce the pest population I1 by spraying his crops with
insecticide. The effect of this is to amend the dynamical system equations to:
dI1
dt
dI2
dt
=
((0.5 − k) − 0.01I2 − 0.0005I1 )I1
=
((−0.5 − k) + 0.001I1 )I2 .
For small k, how is the value of I1 at the equilibrium point affected? Was spraying with
insecticide a good idea?
29(a) Given a nonlinear discrete dynamical system in one variable, x,
x(n + 1) = g(x(n))
that forms a 2-cycle
a1 = g(a2 ), a2 = g(a1 ),
state and prove the condition for the 2-cycle to be an attractor.
(b) Given that (0.746 and 1.163) form a 2-cycle of
x(n + 1) = 3.2x(n) − 2.2x(n)2
determine whether or not the 2-cycle is stable.
123
30. The competing species model for species x, y is:
dx
x
0, 1
= 0.1x 1 −
−
y
dt
10000 10000
dy
y
0, 2x
= 0.2y 1 −
−
dt
15000 15000
Restrict attention to x, y ≥ 0 and sketch the phase plane diagram. Find the location and
nature of all singular points. Is coexistence of the species x and y possible?
31. Show that V (x1 , x2 ) = x21 + x22 is a Liapunov function at the origin for each of the following
systems:
(a) ẋ1 = −x2 − x31 , ẋ2 = x1 − x32 ;
(b) ẋ1 = −x31 + x2 sin x1 , ẋ2 = −x2 − x21 x2 − x1 sin x1 ;
(c) ẋ1 = −x1 − 2x22 , ẋ2 = 2x1 x2 − x32 ;
(d) ẋ1 = −x1 sin2 x1 , ẋ2 = −x2 − x52 ;
(e) ẋ1 = −(1 − x2 )x1 , ẋ2 = −(1 − x1 )x2 .
Find domains of stability at the origin for each of these systems.
32. Find domains of stability for the following systems by using an appropriate Liapunov function:
(a) ẋ1 = x2 − x1 (1 − x21 − x22 )(x21 + x22 + 1)
ẋ2 = −x1 − x2 (1 − x21 − x22 )(x21 + x22 + 1);
(b) ẋ1 = x2 , ẋ2 = −x2 + x32 − x51 .
33. Use V (x1 , x2 ) = (x1 /a)2 + (x2 /b)2 to show that the system
ẋ1 = x1 (x1 − a), ẋ2 = x2 (x2 − b). a, b > 0,
has an asymptotically stable origin. Show that all trajectories tend to the origin as t → ∞ in
the region
x2
x21
+ 22 < 1.
2
a
b
34. Given the system
ẋ1 = x2 , ẋ2 = x2 − x31
show that a positive definite function of the form
V (x1 , x2 ) = ax41 + bx21 + cx1 x2 + dx22
can be chosen such that V̇ (x1 , x2 ) is also positive definite. Hence deduce that the origin is
unstable.
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CHAPTER 4. AUTONOMOUS NON–LINEAR SYSTEMS
35. Show that the origin of the system
ẋ1 = x22 − x21 , ẋ2 = 2x1 x2
is unstable by using
V (x1 , x2 ) = 3x1 x22 − x31 .
36. Show that the fixed point at the origin of the system
ẋ1 = x41 , ẋ2 = 2x21 x22 − x22
is unstable by using the function
V (x1 , x2 ) = αx1 + βx2
for a suitable choice of the constants α and β.
Chapter 5
ADVANCED TOPICS
In this final Chapter we will look at some recent developments in Dynamical Systems Theory. This
will be done mainly by example, i.e. we will write down explicit dynamical systems and investigate
certain aspects of their behaviour. In this way we will illustrate the following complex features
that can arise in nonlinear dynamical systems:
ˆ bifurcation
ˆ chaos
ˆ fractals.
Bifurcation theory
We often think that the world is continuous in the sense that a small change in input causes a small
change in the output. But this is not always the case, as exemplified by the phrase “the straw that
broke the camel’s back”. Bifurcation theory is the study of that breaking point, that is, it is the
study of the point at which the qualitative behaviour of a system changes.
Complex behaviour for nonlinear dynamical systems
While you may not be familiar with the term “bifurcation”, you are certainly familiar with the
concept. Consider a vertical rectangular metal bar. Suppose you start putting weights on the top
of the bar. At first nothing happens, but after you put enough weight on the bar, the bar bends
to one side or the other. There is some weight ω such that if the total weight is less than ω the
bar has one equilibrium value (it is straight), but if the weight exceeds ω then the bar has three
equilibrium values (bent to the left and bent to the right which are stable equilibriums, and straight
which is an unstable equilibrium). The weight ω is a bifurcation value in that as the weight on
the bar goes from less than ω to greater than ω, one equilibrium point divides (or bifurcates) into
three equilibrium values.
Another example of bifurcation is the velocity ν of a rocket. If its velocity is below a fixed velocity
νe (the escape velocity), it will fall back to earth; if its velocity exceeds νe , it will break free of the
earth’s gravitational force and go into outer space.
125
126
CHAPTER 5. ADVANCED TOPICS
The key to bifurcation theory is that we have one unknown input (or parameter) in our dynamical
system (the weight on the rod, the velocity of the rocket). For some values of the parameter
we have one number m1 of equilibrium values, but for another value of the parameter we have a
different number of equilibrium values, say m2 , that is, the qualitative behaviour of the system we
are studying changes as the parameter changes.
We look at an example of bifurcation involving a 1–dimensional dynamic system with 1 parameter.
We will see that the situation is described in a 1 + 1 = 2–dimensional diagram. In general the
bifurcation properties of a d–dimensional dynamic system with p parameters are described in a
(d + p)–dimensional space.
Example
Consider
dx
= g(x) = (1 + x)(1 − a + x2 ).
dt
First, we find all the fixed points:
√
x∗ = −1 and, for a ≥ 1, x∗ = ± a − 1.
(1)
(2)
Next, we determine stability of the fixed points. To do this we need to know g ′ (x) :
g ′ (x) = (1 − a + x2 ) + (1 + x)2x = 1 − a + 2x + 3x2 .
(3)
Fixed points:
(i) x∗ = −1, g ′ (x∗ ) = 1 − a − 2 + 3 = 2− a. Stable for a > 2, unstable for a < 2.
√
√
√
(ii) x∗ = + a − 1, g ′ (x∗ ) = 1 − a + 2 a − 1 + 3(a − 1) = 2(a − 1) + 2 a − 1. Thus g ′ (x∗ ) > 0
for a > 1 and the fixed point is unstable.
√
(iii) x∗ = − a − 1,
√
√
√
g ′ (x∗ ) = 2(a − 1) − 2 a − 1 = 2 a − 1 a − 1 − 1 . This is negative (thus stable) for
1 < a < 2, and positive (thus unstable) for a > 2.
127
We can therefore draw the bifurcation diagram:
In this case there are points of bifurcation at a = 1 and at a = 2.
Chaos
From a mathematical point of view there are three ingredients necessary for a dynamical system to
exhibit chaotic behaviour. We will not give formal definitions of these ingredients, but the essential
idea is to have a combination of attraction and repulsion. More precisely we need:
1. An attractor G (as defined in the previous Chapter).
2. The attractor must be transitive. This means that if x(0) is close to G then in the future
x(t), or x(n), will get arbitrarily close to every point of G. For example, if G is a stable limit
cycle, or a stable n–cycle, then it is transitive. But if G consists of two distinct fixed points
then it is not transitive.
3. The crucial condition for chaotic behaviour is that the system should have sensitive dependence
on the initial values. That is, for every point x(0) there are points y(0) arbitrarily close to
x(0), such that at some times in the future x(t) and y(t) (or x(n) and y(n)) are not close to
each other. An example of such a system is
x(n + 1) = 2x(n)
(4)
x(n) = 2n x(0).
(5)
|x(n) − y(n)| = |2n xo − 2n y0 | = 2n |x0 − y0 | → ∞
(6)
which has solution
Given initial values, say x0 , y0 , we find
no matter how close the initial values x0 and y0 are to each other. Thus this system has
sensitive dependence on initial values. But notice that it does not have an attractor G, and
it is not chaotic.
128
CHAPTER 5. ADVANCED TOPICS
The following two examples illustrate the concept of chaos in dynamical systems theory.
1. We have already mentioned the logistic equation of population growth. In discrete form it
can be written as
x(n + 1) = x(n) + rx(n)(1 − x(n)/N ).
(7)
The dynamic system (7) has fixed points at 0 and N . We have seen that the fixed point at
0 is unstable (for r > 0), and that the fixed point at N is stable for 0 < r < 2. At the end
of the last Chapter we saw that for r > 2 the fixed point at N becomes unstable, and instead
the system develops stable n−cycles. [We showed that if r = 2.2 there is a stable 2–cycle.]
Now what happens as r is further increased? First, it is convenient to re–write (7) as
x(n + 1) = λx(n)(1 − x(n)).
(8)
This is simply a re–scaling and is brought about by writing x(n) = ay(n) for suitable a. It
turns out that λ = 1 + r. In terms of λ the system does not have a√stable (non–zero) fixed
point for λ ≥ 3; instead
√ it may have stable n−cycles. For 3 < λ < 6 + 1 there is a stable
2–cycle, but for λ > 6 + 1 there is no stable 2–cycle; instead there may be a stable 4–cycle.
As λ increases we find that the order n of the n−cycle must become larger and larger to
ensure stability – n−cycles with low n exist, but they are unstable. Finally, at λ = 3.57,
there are n−cycles (with n arbitrarily large) but they are all unstable.
For 3.57 < λ < 4, the dynamic system (8) is chaotic.
2. Consider thermal convection of a horizontal layer of fluid heated from below: the warm fluid
may rise owing to its buoyancy and circulate in cylindrical rolls. Under certain conditions
these cells are a series of parallel rotating cylindrical rolls; see the figure on the next page.
Lorenz used the continuity equation and Navier–Stokes equations from fluid dynamics, together with the heat conduction equation to describe the behaviour of one of these rolls. A
series of approximations and simplifications lead to the Lorenz equations
ẋ =σ(y − x)
ẏ =rx − y − xz
ż =xy − bz.
(9)
The term x represents the rate of rotation of the cylinder, z represents the deviation from
a linear vertical temperature gradient, y corresponds to the difference in temperature at
opposite sides of the cylinder and σ, b and r are constants. The non–linearity in the second
and third equations results from the non–linearity of the equations of flow.
With σ = 10, b = 38 , r = 28 (values often chosen for study) the trajectories are concentrated
onto an attractor of a highly complex form. This Lorenz attractor consists of two “discs”
129
each made up of spiralling trajectories. Certain trajectories leave each of the discs almost
perpendicularly and flow into the other disc. If a trajectory x(t) is computed, the following
behaviour is typical. As t increases, x(t) circles around one of the discs a number of times
and then “flips” over to the other disc. After a few loops round this second disc, it flips back
to the original disc. This pattern continues, with an apparently random number of circuits
before leaving each disc.
The above figure shows a view of the Lorenz attractor for σ = 10, b = 38 , r = 28. Note the
spiralling round the two discs and the “jumps” from one disc to the other. Points that are initially
close together soon have completely different patterns of residence in the two discs of the attractor.
Thus the system exhibits sensitive dependence on initial values, and is chaotic.
The Lorenz equations (9) describe a part of the Physics of the Earth’s atmosphere. Since the
Lorenz system is chaotic it follows that so is the Earth’s atmosphere. This leads to an important
practical conclusion: precise long–term weather prediction is impossible.
Fractals
Consider the dynamic system
z(n + 1) = z(n)2
(10)
where z is complex, z = x + iy. The system could also be written as
x(n + 1) = x(n)2 − y(n)2 , y(n + 1) = 2x(n)y(n),
but the behaviour is clearer in terms of complex z. The system (10) has fixed points at
z∗ = 0, z∗ = 1
(11)
and it is straightforward to check that z∗ = 0 is stable and z∗ = 1 is unstable. There is also
an invariant set at |z| = 1, because if |z(n)| = 1 then |z(n + 1)| = 1; this invariant set is not
an attractor, but is unstable. It forms the boundary of the domain of stability of the fixed point
z∗ = 0. If |z(0)| < 1 then z(n) → 0; if |z(0)| > 1 then z(n) → ∞.
We now amend the dynamic system (10) to
z(n + 1) = z(n)2 + c,
(12)
130
CHAPTER 5. ADVANCED TOPICS
where c is a constant complex number, and again we are interested in the domain of stability. The
question can be formulated in two different ways.
(a) For fixed c, find the domain of stability, i.e. find those points z(0) such that |z(n)| is bounded
for all n > 0. The boundary of this domain of stability is called the Julia set. If c = 0,
the Julia set is the unit circle |z| = 1, and for small c one would expect that the Julia set
would be a deformation of the unit circle. In is indeed close to the unit circle, but whereas
the unit circle is a smooth curve, the Julia set for non–zero c is not smooth. It has a very
complex structure, and a magnification of any region has approximately the same shape as
the original. It is a fractal. The definition of a fractal is beyond the scope of this course,
but the idea is illustrated in figure on page 13.
(b) For fixed z(0) (usually taken as z(0) = 0), find those values of c for which |z(n)| is bounded
for all n > 0. This is the Mandelbrot set M . It is a bifurcation diagram in the sense that as
you cross the boundary of M you go from bounded to unbounded solutions, or vice versa.
Often pictures of the Mandelbrot set, and of Julia sets, are given in colour. From their
popularity it is clear that these pictures are having an influence on modern art and design.
The definition of the Julia and Mandelbrot sets is a yes/no issue – either a point is in the
set, or it isn’t. So how is colour introduced into what is essentially a black/white picture?
In practice diagrams of the Julia and Mandelbrot sets are calculated by computer. Starting
with z(0), we apply (12) repeatedly to find z(1), z(2), ..., z(N ), where N is some (previously
decided) maximum. If |z(n)| ≤ a, n = 0, ..., N , where a is a constant usually taken as 2, then
the trajectory starting at z(0) is regarded as bounded, so that c ∈ M , or z(0) ∈ interior of
the Julia set. Otherwise the trajectory may be regarded as escaping to infinity. In this case,
let k be the first iterate such that |z(k)| > a. The larger the value of k, the closer z(0) is to
the boundary. It is this value of k that is used to determine the colour in diagrams of the
Mandelbrot and Julia sets.
Pictures of the Mandelbrot set suggest that it has a highly complicated form – see the figure on
the next page. It has certain obvious features: a main cardioid to which a series of prominent
circular “buds” are attached. Each of these buds is surrounded by further buds, and so on.
However, this is not all. In addition, fine, branched “hairs” grow outwards from the buds, and
these hairs carry miniature copies of the entire Mandelbrot set along their length. It is easy
to miss these hairs in computer pictures. However, accurate pictures suggest that M is a
connected set, a fact that has been confirmed mathematically.
131
Julia sets for c at various points in the Mandelbrot set.
Exercises
Only problem 3 will be solved in details, the rest follow the same.
1. A simple fish harvesting model uses the logistic equation,
amended by assuming that a proportion b of the population is harvested in each time period:
x (n)
x(n + 1) = x(n) + 0.5 1 −
x(n) − bx(n)
N
where N is the population in the absence of harvesting. Determine the fixed points and their
nature (i.e. stable or unstable) and draw a bifurcation diagram.
2. A discrete 2–dimensional dynamic system is described in terms of polar coordinates (r, θ) :
1
θn+1 = θn2 , rn+1 = rn2
with θ ∈ [0, 2π). If θn2 ≥ 2π, say θn2 = ψn + k(2π) with ψn ∈ [0, 2π) and k an integer, then we
put θn+1 = ψn . Show that the unit circle r = 1 is an attractor. Show also that the system is
not transitive, thereby implying that it is not chaotic.
132
CHAPTER 5. ADVANCED TOPICS
3. Consider the dynamic system
dx
= 4x − x2 − b.
dt
Determine the fixed points and their nature and draw a bifurcation diagram.
Solution:
We find the fixed point by solving,
4x − x2 − b = 0
−x2 + 4x − b = 0
x2 − 4x + b = 0
thus gives
x1 = 2 +
x2 = 2 −
√
√
4−b
4−b
Now,
g(x) = 4x − x2 − b
g ′ (x) = 4 − 2x
(i) for
x1
g ′ (x1 )
=2+
√
4−b
= 4 − 2(2 +
√
4 − b)
√
= −2 4 − b the fixed is stable for ∞ < b < 4
(ii) for
x2
g ′ (x2 )
=2−
√
4−b
= 4 − 2(2 −
√
4 − b)
√
=4−4+2 4−b
√
= 2 4 − b, the fixed is stable for ∞ < b < 4
(iii) Graph refer to example an 5.1.
4(a) A model of harvesting a fixed amount b of a population (that would otherwise obey the logistic
equation) is:
A(n + 1) = 1.8A(n) − 0.8A(n)2 − b.
For b > 0 find the fixed points of the system, and then draw the bifurcation diagram. What
limit is imposed on the amount of harvesting?
133
(b) State the dynamic equation that is used for generating the Mandelbrot set, and then define
the Mandelbrot set.
Chapter 6
APPENDIX
1. Notation
If A is an n × p matrix and B is an n × q matrix then [A, B] denotes the n × (p + q) matrix obtained
by placing B alongside and to the right of A. For example, if
1 3
A=
5 4
and
B=
then
[A, B] =
1
5
4
2
3
4
4
2
.
This notation is extended in an obvious way to more than two matrices.
A
Similarly if A is an p × m matrix and B is a q × m matrix then
denotes the (p + q) × m
B
matrix obtained by placing A above B. Obviously, this notation can also be extended to more
than two matrices.
134
APPENDIX
135
2. Rank of a Matrix
The set of all column m–vectors

k1
 k2

 k3

 ..
 .







km
m
is denoted by R . (The vectors can also be written in row form, but in this chapter we often think
of the columns of a matrix as vectors, so for us it is more convenient to use column vectors.) As
explained in module MAT103, we can multiply these vectors by scalars and add them, so forming
linear combinations of vectors. Thus if
x1 , x2 , x3 , ..., xn
are scalars and
a1 , a2 , a3 , a4 , ..., an
are n column m–vectors, we can form the linear combination
x1 a1 + x2 a2 + x3 a3 + ... + xn an .
If a set of x′ s, not all of them zero, can be found such that
x1 a1 + x2 a2 + x3 a3 + ... + xn an = 0
then the set of vectors {a1 , a2 , a3 , ..., an } is said to be linearly dependent. Otherwise we say they
are linearly independent.
Our interest in the concept of linear independence lies in the following fact:
If S is a subset of Rn then every vector in Rn can be expressed as a linear combination of vectors
in S if and only if S contains a subset of n linearly independent vectors.
Now suppose we have a n × m matrix A. The columns of A can be thought of as vectors in Rn .
The rank of A is defined as the maximum number of linearly independent columns of A. (It can
be shown that this is also equal to the maximum number of linearly independent rows of A, where
each row is thought of as a vector in Rm .) In the proof of Theorem 1 we use the following result:
If A is an n × m matrix, then every vector in Rn can be expressed as a linear combination of
columns of A if and only if rank A = n.
It is important to be able to calculate the rank of a matrix. The rank can be calculated by first
reducing the matrix to row–echelon form (this process is explained in the module MAT103) and
then counting the number of non–zero rows in the row–echelon matrix. This number will be the
rank of the matrix. For example, the row–echelon form of the matrix
1 2
A=
4 8
136
APPENDIX
is
1
0
2
0
and therefore rank A = 1.
Another result which is sometimes useful is:
If A is an n × n matrix then rank A = n if and only if det A ̸= 0.
3. Derivatives and Integrals of Matrices
The calculation of derivatives and integrals of matrices is very simple – you merely differentiate or
integrate each entry in the matrix using the calculus rules you learned in MAT101 and MAT102.
For example, if
3
1
t
3
C=
, D=
t2
t sin t
then
d
C=
dt
0
2t
d
D=
dt
,
3t3
1
0
cos t

t1

1 3
t
3 1


,
while
 R t1
Z t1
0
Cdt = 
0
1dt
R t1 2
t dt
0

=

and
Z t1
0
 R t1 3
t dt
0
Ddt = 
R t1
tdt
0
R t1
3dt

 1
t41
 4
=

R t1
1 2
sin t dt
t
0
2 1
0
3t1
1 − cos t1


.

4. Exponential Function
The exponential function ex , where x ∈ R, can be written as a power series:
ex = 1 + x +
x2
x3
xk
+
+ ... +
+ ...
2!
3!
k!
which converges for all x ∈ R. If A is a square matrix, we define
eA = I + A +
A2
A3
A4
Ak
+
+
+ ... +
+ ...
2!
3!
4!
k!
It can be shown that this series converges for any square matrix A. If A is an n × n matrix then
eA will also be an n × n matrix. In this module you will never have to calculate the value of eA , all
you need to know is the above definition, and the following results:
If t ∈ R and A is a constant matrix, then
d At
e = AeAt = eAt A.
dt
APPENDIX
137
This follows easily from the definition of eAt . We have
d At
A2 t2
A3 t3
d
e
I + At +
+
+ ...
=
dt
dt
2!
3!
A4 t3
A3 t2
+
+ ...
= A + A2 t +
2!
3! A2 t2
A2 t2
= A I + At +
+ . . . or I + At +
+ ... A
2!
2!
= AeAt or eAt A.
It can also be shown that
(i) If A is a square matrix then eA e−A = e−A eA = I, and therefore the inverse of eA is e−A .
(ii) If s, t ∈ R then eAt eAs = eAs eAt = eA(s+t) .
5. Solution of the Equation ẋ = Ax + Bu.
Multiply both sides of the equation ẋ = Ax + Bu on the left by e−At . This gives
e−At ẋ(t) = e−At Ax(t) + e−At Bu(t)
or
e−At ẋ(t) − e−At Ax(t) = e−At Bu(t).
But
d −At
e
x(t) = e−At ẋ(t) − e−At Ax(t)
dt
so we get
d −At
e
x(t) = e−At Bu(t).
dt
Therefore
e
−At
Z t
x(t) =
e−As Bu(s)ds.
a
Multiply both sides of this equation on the left by eAt to get
Z t
Z t
At
−As
x(t) = e
e
Bu(s)ds =
eA(t−s) Bu(s)ds.
a
a
If we assume that x(0) = 0 then we can take a = 0 and we get
Z t
x(t) =
0
eA(t−s) Bu(s)ds.