Applied Dynamical Systems
Study guide for
APM2614
Prof A Batubenge-Tshidibi
Department of Mathematical Sciences
University of South Africa, Pretoria
© 2015 University of South Africa
All rights reserved
Printed and published by the
University of South Africa
Muckleneuk, Pretoria
APM2614/001/2016–2017
60157445
iii
APM2614/1
Contents
Page
Contents
iii
Preface
v
CHAPTER 1 Overview
1
CHAPTER 2 Linear Autonomous Dynamical Systems
7
CHAPTER 3 Linear Control Theory
45
CHAPTER 4 Autonomous Nonlinear Systems
70
CHAPTER 5 Advanced Topics
108
APPENDIX
116
iv
v
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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.
N.T. BISHOP
Revised by S.A. DE SWARDT
F.E.S. BULLOCK
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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 one do (other than give up)? One could use
(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 one wants a solution to a specific problem, but is not very helpful for obtaining a general understanding
of the behaviour of the solutions. For this we turn to
(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, with a parachute, jumps out of an aeroplane. The equation describing her motion is
k
dv
= g − v 1.9 , v > 0
dt
m
(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 5ms −1 . What is the least possible value for the ratio k/m?
2
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
v=V
v=0
dv
=
g − mk v 1.9
t=T
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, leads NOWHERE. So we resort to
Solution 2 (Dynamical systems approach)
The equation (1) has a so–called fixed point (i.e. a point where
v = v∗ ≡
mg
k
dv
= 0) at
dt
1/1.9
.
(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 diagram below:
0
v*
v
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
or, in the case of a system of differential equations,
d x1
dt
d x2
dt
= x1 − x1 x2
= x2 −
x1 x2
.
2
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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
1
x(n + 1) = (x(n)2 − x(n))
2
are autonomous, but
dx
= x2 − t
dt
and
1
x(n + 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.
(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.
4
(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 by examples. Note that sometimes stable points are called attractors, or attracting points, and
unstable points are called repellors, or repelling points.
Examples
1.
dx
= x 2 − x.
dt
We solve the equation x∗2 − x∗ = 0 to get the fixed points. This gives x∗ = 0 or x∗ = 1. If x < 0 then d x/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:
0
1
(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
= x 2.
dt
Solving the equation x∗2 = 0 gives x∗ = 0 as the only fixed point.
We see that d x/dt > 0 if x < 0 or if x > 0, so we get the phase line below:
0
This system is neither stable nor unstable. The reasons why will be discussed later – we included this example to
show some of the subtleties that can arise in dynamical systems theory.
1
3. x(n + 1) = − x(n) + 6.
2
To find the fixed point(s) we substitute x∗ for x(n) and x(n + 1) in the difference equation to get x∗ = − 12 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(n)
5
*
4
*
*
*
*
3
2*
1
0
n
1
2
3
4
5
(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.
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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
Let x1 = y, x2 =
dy
,
dt
and x3 =
d2 y
.
dt 2
d3 y
dy
+
3
dt
dt
2
+ e y = 3.
Then we get an equivalent first order system of equations
dx3
dt
dx2
dt
dx1
dt
=
1
3 − e x1 − 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 (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
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 on pages 96 and 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)
6
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.
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CHAPTER 2
LINEAR AUTONOMOUS
DYNAMICAL SYSTEMS
Linear Autonomous Dynamical systems are described by the equations
P
x = 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 vary with time. We consider in section 1 the simplest
such 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, 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
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∗ =
Let
f
.
1−a
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
= a X (t) + ax∗ + f
= a X (t) ,
(3a)
(3b)
(4)
8
since x∗ = − af .
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) = a X (n) + x∗ (a − 1) + f
= a X (n) ,
since x∗ =
f
.
1−a
Equation (2) therefore becomes
Ẋ = a X or X (n + 1) = a X (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
Ẋ (t)
dt =
a dt
X (t)
n X (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 = a X 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) = a X (0)
X (2) = a X (1) = a (a X (0)) = a 2 X (0)
X (3) = a X (2) = a a 2 X (0) = a 3 X (0)
..
.
clearly yielding
X (n) = a n X (0) .
(6b)
Using (4), (3a) and (3b), the solutions of the original equation (2) are therefore:
x(t) = −
f
f
+ eat x(0) +
a
a
or
(7)
x(n) =
f
f
+ a n x(0) −
.
1−a
1−a
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[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 a n → 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.
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
1
X (n + 1) + 4 = (X (n) + 4) + 2
2
clearly yielding
1
X (n + 1) = X (n)
2
10
with solution
1
2
X (n) =
n
X (0).
Transforming back to the original variable x gives
1
2
n
x(n) =
1
2
n
=
(x(0) − 4) + 4
(−2) + 4 = −
n−1
1
2
+ 4.
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:
x(n)
5
4
*
*
*
*
3
4
5
*
3
2*
1
0
n
1
2
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.
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.
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Example 3
Consider the discrete system:
2
x(n + 1) = − x(n),
3
This is easy to solve. The solution is x(n) = 3(−2/3)n .
x(0) = 3.
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
Making the substitution
1
x∗ = .
2
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
x (t) −
1 −2t
1
= x (0) −
e ,
2
2
which, together with the initial condition x (0) = −1, leads to the solution
x (t) =
=
Example 5
Consider the system
1 −2t
1
+ −1 −
e
2
2
1 3 −2t
− e .
2 2
3
dx
= x − 3,
dt
2
x(0) = 0.
The solution is
3
x(t) = −2e 2 t + 2.
12
2. Two Dimensional Systems
We now move on to the two–dimensional case, i.e. in equation (1) x and f are 2–vectors and A is a 2 × 2 matrix.
The theory will be described for the continuous case (P
x = Ax + f), and the corresponding equations for the discrete
case will be summarised in section 5. The dynamical system
P
x = Ax + f
(9)
has a fixed point provided P
x = 0, so the fixed point is given by
Ax + f = 0.
Therefore
x∗ = −A−1 f.
(10)
X(t) = x(t) − x∗
(11)
P = AX.
X
(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 what it is.]
Substituting (13) into (12) gives
Kλeλt = AKeλt .
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 will not assume any prior knowledge
of eigenvalue theory, and where necessary, standard results will simply be stated.
Equation (14) may be rewritten:
(A − λI)K = 0
where I is the identity matrix
1 0
0 1
(15)
. We require that K should be non–trivial (i.e. K = 0), which implies
that
det(A − λI) = 0.
(16)
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[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
b
c
d −λ
i.e.
=0
(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
det(A). The solution of (19) is
λ1 =
σ+
σ−
σ2 − 4
, λ2 =
2
= ad − bc is
σ2 − 4
.
2
(20)
In general, the procedure for obtaining the solution of equation (12) is as follows: First, we use (15) to find the
Kx
eigenvector K(=
):
Ky
(A − λI)K = 0
(15)
Now, because det(A − λI) = 0, (15) leads to two equations for K x , K y that are multiples of each other:
a K x + bK y = 0
ca K x + cbK y = 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 .
For any choice of X(0) we can find c1 and c2 such that
X(0) = c1 K1 + c2 K2 .
(21)
14
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
(B) σ 2 − 4
(C) σ 2 − 4
> 0: there are two distinct real solutions for λ.
< 0: there are two distinct complex solutions for λ.
= 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
P = AX with A =
, then X
1 3
5 3
= x + 3y
= 5x + 3y.
(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
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 a zero
determinant means, for a 2 × 2 matrix, that one row is a multiple of the other. Let, for example
K1 =
a
b
.
Then (23) becomes
−5a + 3b = 0
5a − 3b = 0
...(A)
...(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 = 53 c1 and our vector K 1 would be given by
K1 = c1
1
5
3
.
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This however is just a constant multiple of the previous solution of K 1 :
c1
1
5
3
1
= c1
5
5
3
!
Similarly for K2 , equation (15) is
3 3
5 5
K2 = 0
so that
K2 = c2
1
−1
.
Thus from (13), and using the principle of superposition for linear systems, the general solution of (22) is:
X=
x
y
= c1
3
5
e6t + c2
1
−1
e−2t .
(24)
[Reminder: this means
x = 3c1 e6t + c2 e−2t
y = 5c1 e6t − c2 e−2t ]
Note that X → ∞ as t → ∞.
The above solution is the general solution because it has two arbitrary constants, which 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.
Figure 1: Saddle point
Example 2
Consider the system:
dx
= 2x + 8y,
dt
dy
= 4y,
dt
(25)
16
so that clearly
A=
2 8
0 4
.
Then (19) becomes
λ2 − 6λ + 8 = 0.
a
b
Solving this equation we get λ1 = 2, λ2 = 4. Let K1 =
be the eigenvector corresponding to the eigenvalue
λ1 = 2. Then, from equation (15) we need to solve
(A − 2I) K1 = 0,
i.e.
0 8
0 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
be the eigenvector corresponding to the eigenvalue λ2 = 4. Then, from (A − λ2 I) K1 = 0
d
(see equation (15)), we get
−2 8
c
0
=
0 0
d
0
Next, let K2 =
i.e.
−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
K2 = c2
4
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 → ∞.
17
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Figure 2: Unstable node
Example 3
Consider the system:
dx
dt
= −x − 2y + 4
dy
= x − 4y + 2.
dt
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
=
dt
−1 −2
1 −4
X.
Equation (19) becomes
λ2 + 5λ + 6 = 0,
with solutions λ1 = −2, λ2 = −3. Substituting into equation (15) we get
1 −2
1
2
The solutions are K1 = c1
2
1
, K2 = c2
X=
2 −2
1 −1
K1 = 0,
1
1
x − x∗
y − y∗
K2 = 0.
. Then
= c1
2
1
e−2t + c2
1
1
e−3t ,
(27)
18
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 → ∞.
Figure 3: Stable node
Case (B)
In this case equation (19) has two distinct complex solutions for λ. Writing β =
λ1 =
σ
σ
+ iβ, λ2 = − iβ.
2
2
4
−σ 2 /2, the solutions are
(29)
Note that λ1 and λ2 are complex conjugates of each other.
Example 1
Consider the system
dx
dt
= 6x − y,
dy
dt
= 5x + 4y.
Then
A=
6 −1
5
4
,
so that equation (19) becomes
λ2 − 10λ + 29 = 0.
(30)
19
APM2614/1
The roots λ1 , λ2 are
λ1 = 5 + 2i, λ2 = 5 − 2i
(31)
Thus equation (15) for K1 becomes
−1
(1 − 2i)
5
(−1 − 2i)
K1 = 0.
(32)
[Note that, 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
.
(33)
K1 = c1
1 − 2i
Similarly, we find the other eigenvector to be
K2 = c2
1
1 + 2i
(34)
.
Note that if the matrix A is real the vector K1 is the complex conjugate of K2 . It is simple to prove (but it won’t be
done here) that this is always the case. Thus the solution of (30) can be written
X=
x
y
1
1 − 2i
= c1
e(5+2i)t + c2
1
1 + 2i
e(5−2i)t .
(35)
It would, however, 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:
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)
Let
C1 = c1 + c2 and C2 = c1 i − c2 i.
(37)
Then
X=
x
y
=
C1
+C2
cos 2t
cos 2t + 2 sin 2t
e5t
sin 2t
−2 cos 2t − sin 2t
e5t .
(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 ). Using the notation K1 to mean the complex conjugate of K1 , so
K1 = K2 , and defining:
1
B1 = Re(K1 ) = (K1 + K1 )
2
20
1
B2 = Im(K1 ) = (K1 − K1 ),
2
(where the constant c1 multiplying K1 is taken as 1), then
(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 manipulation, and
is omitted.
N.B. – In the above formula the eigenvector K1 must be worked out for the eigenvalue λ1 which has positive imaginary part.
In example 1, K1 is given by (33), so that from
1
1 − 2i
1
1
=
0
−2
+i
,
we have that
B1 = Re (K1 ) =
1
1
, B2 = Im (K1 ) =
1
1
cos 2t −
Also σ /2 = 5 and β = 2. Thus the solution is
X = e5t C1
+e5t C2
0
−2
0
−2
cos 2t +
0
−2
.
sin 2t
1
1
sin 2t
which, as expected, is the same as (38).
The solution curves are shown in Figure 4. The fixed point at (0,0) is called an unstable focus.
21
APM2614/1
Figure 4: Unstable focus
Example 2
Consider the system
d
dt
x
y
=
The fixed point (x∗ , y∗ ) is found by solving
x
y
+
=−
2
1
−1
2
1
− 2 −1
x∗
y∗
−1
2
1
− 2 −1
2
1
(41)
.
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 K 1 =
−i
− 12
2
−i
a
b
0
0
=
i.e.
−ia + 2b = 0 ...(1)
1
− a − ib = 0 ...(2)
2
From (1), choose a = 1 (say), then b =
i
2
so that
K1 =
=
1
i
2
1
0
+i
0
1
2
,
a
b
is found from solving
22
and therefore
1
0
B1 =
0
and B2 =
.
1
2
Hence
X =
x − x∗
y − y∗
+C2
1
0
= e−t C1
0
1
2
cos t +
1
0
cos t−
sin t
0
1
2
sin t
.
Thus
x = 2 + e−t (C1 cos t + C2 sin t)
1
1
y = e−t ( C1 sin t + C2 cos t).
2
2
(42)
The solution curves are shown in Figure 5. The fixed point at (0, 0) is a stable focus.
Figure 5: Stable focus
Example 3
Consider the system
P=
X
2
8
−1 −2
X.
Equation (19) is:
λ2 + 4 = 0
so that
λ1 = 2i, λ2 = −2i.
The equation for K1 is:
8
(2 − 2i)
−1
(−2 − 2i)
K1 = 0
(43)
23
APM2614/1
which has a solution
2 + 2i
−1
K1 =
2
−1
=
+i
2
0
so that
B1 =
2
−1
B2 =
2
0
and
.
Since σ = 0 the solution is
X = C1
2 cos 2t − 2 sin 2t
− cos 2t
2 cos 2t + 2 sin 2t
− sin 2t
+ C2
.
(44)
The solution curves are shown in Figure 6. The fixed point (0,0) is called a centre.
Figure 6: Centre
Case (C)
Here we have “Equal roots”, so σ 2 − 4 = 0. This is a special case which, for the purposes of this module, is not
very important. So we just briefly summarise the solutions. There are really two sub–cases:
C − 1:
Here
which, for X =
x (t)
y (t)
P=
X
a 0
0 a
=
a 0
0 a
X
(45)
, can be written as
ẋ (t)
ẏ (t)
x (t)
y (t)
24
i.e.
ẋ (t) = ax (t) and ẏ = ay (t) .
These two equations clearly have the solutions
x (t) = c1 eat and y (t) = c2 eat
so that
X=
C − 2:
All other cases in which σ 2 − 4
x
y
= c1
1
0
eat + c2
0
1
eat .
(46)
= 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
P=
X
3 −18
2 −9
X.
The eigenvalue equation is
λ2 + 6λ + 9 = 0
or
(λ + 3)2 = 0,
so
λ = −3.
Then the equation for K is
6 −18
2 −6
K=0
(52)
25
which has a solution K =
3
1
APM2614/1
. Then from (51), the equation for P is
6 −18
2 −6
i.e.
3
1
P=
2a − 6b = 1.
Now choose (say) b = 0, then a =
1
2
X = c1
so that P =
3
1
1
2
0
e−3t + c2
. Thus the general solution of (52) is
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 first–order 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 pure 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:
26
For stability we require
> 0 and σ < 0.
4. Higher Dimensional Systems
We now investigate the general case for continuous systems
P
x = Ax + f
(54)
where the dimensionality is taken as m. As before, assuming A−1 exists,
P
x = 0 at x∗ = −A−1 f.
(55)
X = x − x∗
(56)
P = AX.
X
(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).
Then a general solution of (54) is:
m
x=
i=1
(ci Ki eλi t ) − A−1 f.
(58)
(59)
For a large system, carrying out the procedure to obtain (59) from (54) is at least tedious if not impractical. So,
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 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 it is 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
Thus det(A − λI) is


−4 1
1

P =
X
 1 5 −1  X.
0 1 −3


−4 − λ
1
1


det 
1
5−λ
−1  ,
0
1
−3 − λ
(61)
27
APM2614/1
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 3–dimensional generalisation of a saddle point.
Example 2
Consider the system




−1
1 −2
−1




P
x =  1 −1
0  x+  3  .
1
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
P = AX, and the eigenvalues are found from 0 = det(A − λI):
Then putting X = x − x∗ gives X


−1 − λ
1
−2


det 
1
−1 − λ
0

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.
(63)
28
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 terms e±i t 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
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.
There is a fixed point x∗ given by solving
(65)
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)
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 again use
eigenvalue theory. 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
m
X(0) =
ci Ki
(69)
i=1
where the constants ci are unique. Putting (69) into (68) gives
m
X(n) =
i=1
ci An Ki =
m
i =1
ci λin Ki =
m
λin 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, by examples.
Example 1
29
APM2614/1
Consider the system
x(n + 1) = x(n) + 2y(n),
y(n + 1) = −x(n) + 4y(n)
(71)
with
x(0) = 1,
y(0) = 2.
Let
X (n) =
x (n)
y (n)
.
Then equation (71) can be written
1 2
−1 4
X (n + 1) =
with
X (0) =
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
, λ2 = 3, K2 =
.
λ1 = 2, K1 =
1
1
We must now express X(0) in terms of K1 and K2 , i.e. 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
X(n) = c1 K1 λn1 + c2 K2 λn2 = −1
2
1
2n + 3
1
1
3n .
Thus
x(n) = −2n+1 + 3n+1
y(n) = −2n + 3n+1 .
(72)
Example 2
Consider the system
X (n + 1) =
0.5 −0.25
1
0.5
X (n)
with
X (0) =
1
2
.
The eigenvalue equation is
λ2 − λ + 0.5 = 0
(73)
30
so that
λ1 = 0.5 + 0.5i,
λ2 = 0.5 − 0.5i.
The eigenvectors are
K1 =
i
2
K2 =
−i
2
,
.
Thus the general solution is
i
2
X(n) = c1 (0.5 + 0.5i)n
The constants c1 , c2 are determined from X(0) =
+ c2 (0.5 − 0.5i)n
1
2
.
=
i (c1 − c2 )
2 (c1 + c2 )
−i
2
.
We get
1
2
which are solved to give
c1 = 0.5 − 0.5i
c2 = 0.5 + 0.5i.
Therefore
X(n) = (0.5 − 0.5i)(0.5 + 0.5i)n
+ (0.5 + 0.5i) (0.5 − 0.5i)n
i
2
−i
2
.
(74)
There is one important point to make about equation (74). The second term is the complex conjugate of the first
term, and 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, i.e.
z = rei θ
where
r
θ
a 2 + b2
b
= tan−1 ,
a
=
31
APM2614/1
we write:
1
√ e−i π/4 ,
2
1 iπ /4
√ e .
2
(0.5 − 0.5i) =
(0.5 + 0.5i) =
Then (74) becomes
X(n) =
1
√
2
n+1
1
+ √
2
n+1
1
√
2
n+1
=
1
√
2
n+1
=
1
√
2
n−1
=
ei π/2
2
eiπ/4(n−1)
e−iπ /2
2
eiπ /4(1−n)
eiπ/4(n−1)+iπ/2 + ei π/4(1−n)−i π/2
2ei π/4(n−1) + 2eiπ /4(1−n)
2 cos(π/2 + (n − 1)π/4)
4 cos((n − 1)π/4)
− sin((n − 1)π/4)
2 cos((n − 1)π/4)
(75)
,
where the second last line of (75) was derived from
cos t =
eit + e−i t
.
2
Example 3
Consider the system
x (n + 1) =
2 1
0 2
2
−1
x (n) +
Then the fixed point is given by
2 1
0 2
x∗ =
−2
−1
x∗ +
Therefore
1 1
0 1
−2
−1
x∗ +
and hence
x∗ =
1
1
.
= 0,
.
Let
X(n) = x(n) − x∗ .
Then
X (n + 1) =
2 1
0 2
The eigenvalue equation is
(λ − 2)2 = 0.
X (n) .
.
(76)
32
There is therefore a repeated root at λ1 = 2. As in the continuous system case, we describe the procedure for
1
].
dealing with this situation, but do not prove it. Let K1 be the single eigenvector, [so that here K1 = c1
0
Then define P by
(A − λI)P = K1 .
[Here P = c2
0
1
(76)
.] Then the general solution is
X(n) = c1 λn1 K1 + c2 (λn1 P + nλ1n−1 K1 )
(77)
so that, for this example,
X(n) = c1 2n
1
0
+ c2 2n
0
1
+ n2n−1
1
0
1
0
+
.
Thus
x(n) = c1 2n
1
0
+ c2 2n
0
1
+ n2n−1
1
1
.
(78)
For discrete systems, the fixed points can be classified as follows:
(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) λ1 , λ2 real and both greater than 1. The behaviour is analogous to an unstable node.
(c) λ1 , λ2 real and 0 < λ1 , λ2 < 1. The behaviour is analogous to a stable node.
(d) λ1 , λ2 real and 0 < λ1 < 1 < λ2 . The behaviour is analogous to a saddle point.
(e) One, or both λ1 , λ2 negative. There is no analogy with the continuous situation. The solution keeps jumping
from one side of the fixed point to the other (as in 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, or 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
33
APM2614/1
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”, which is proportional to its size;
this “hitting strength” reduces the other force:
dx
= −α y,
dt
dy
= −βx
dt
(79)
where α and β are constants which 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
λ1 = − aβ, λ2 = + aβ.
(80)
The eigenvector equations are then
√
αβ −α
√
αβ
−β
and
√
− αβ
−α
√
−β
− αβ
These equations give
K1 = c1
√
α
√
β
K 1x
K 1y
K 2x
K 2y
, K2 = c2
=0
= 0.
√
− α
√
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:
34
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:
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). Of
course, 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 depend 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
− αβt
e
e αβt ,
+ c2
X (t) = c1 √
√
β
β
the values of the constants c1 and c2 in terms of the initial conditions x (0), y (0):
x (0)
y (0)
=
√
α (c1 − c2 )
√
β (c1 + c2 )
yielding
c1 =
c2 =
√
√
β + y (0) α
√
2 αβ
√
√
y (0) α − x (0) β
.
√
2 αβ
x (0)
,
35
APM2614/1
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
=
,
t→∞
y (t)
0
that is
c1
√
α
√
β
lim e
t→∞
√
− αβt
+ c2
√
− α
√
β
lim e
t→∞
√
αβt
=
0
0
.
This clearly gives
c2 = 0
yielding the desired relationship
y (0) = x (0) β/α.
We can therefore clearly summarize the outcome of the battle in terms of the initial conditions as:
(a) If y(0) > x(0)
β
α
then x is totally eliminated.
(b) If y(0) < x(0)
β
α
then y is totally eliminated.
√
(c) In the case that 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

0
 S1 0 · · · · · · · · ·

..

0
.
 0 S2

x (n + 1) =  .
..
..
..
.
.
 .. 0
.

.
.
.
 .
..
..
..
.
. ..
 .
0
0 0
0
Sm−1
(83)






 x (n) .





(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
36
eigenvalue” means the eigenvalue of greatest absolute magnitude. It can be shown [Luenberger, 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 repeated root). Then if K1 is the corresponding eigenvector, it is also real. Thus the
solution is of the form
x(n) = c1 K1 λn1 +
m
i =2
= λn1 c1 K1 +
m
i=2
Now because λ1 > |λi | then as n → ∞, the
ci Ki λin
m
ci Ki
i=2
λi
λ1
n
.
(85)
term tends to zero. Thus, as n → ∞,
x(n) → λn1 c1 K1 .
(86)
Thus, for large n, the population distribution amongst 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 = 32 . 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 12
1
2
0

5
3

−2 0  K1 = 0.
2/3 −2
(87)
37
Let
APM2614/1

K 11


K1 =  K 12  .
K 13

Suppose that K 13 = 1, then the last row of (87)
2
K 1 − 2K 13 = 0,
3 2
gives
K 12 = 3,
so that from the middle row
1
K 1 − 2K 13 = 0,
2 2
we have
K 11 = 12.
Thus
and, for large n,
(iii) Safety of buildings


12


K1 =  3  .
1
(88)
 

x1
12

 

 x2  ∝  3  .
x3
1
(89)

Mechanical structures (bridges, buildings, etc.) are built from materials having 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 = MR
x(t), where M is the mass matrix. Thus
R
x(t) = −M−1 Ax = −Bx
(90)
where


1 −1
0


A = a  −1
2 −1  ,
0 −1
2


1 0 0


M = m  0 1 0 .
0 0 2
(91)
38
(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 (R
x) rather than
first order (P
x). There are two different, but ultimately equivalent, ways of handling equation (90).
(a) We look for solutions of the form
x = Ke±i ωt .
(92)
Equation (92) is inspired by the solution of ẍ + ω2 x = 0, x = e±iωt . Substituting (92) into (90) gives
−ω2 Ke±i ωt = −Be±i ωt K.
Therefore
ω2 K = BK.
(93)
Thus ω2 is an eigenvalue, and K the corresponding eigenvector, of the matrix B. Now B = M−1 A, thus


1 −1
0
a 

B =  −1
2 −1  ,
m
1
0 − 12
so that the eigenvalue equation is
0 = det B − w 2 I
a
m
=
− w2
− ma
0
− ma
2a
− w2
m
a
− 2m
0
− ma
a
− w2
m
a
2a
a2
a
− w2
− w2
− w2 −
m
m
m
2m 2
a
a a
+
−
− w2
m
m m
a 3
1
− (1 − λ)
=
(1 − λ) (2 − λ) (1 − λ) −
m
2
a 3
1
=
(1 − λ) λ2 − 3λ + 2 − − 1
m
2
1
a 3
=
(1 − λ) λ2 − 3λ −
m
2
=
where λ =
m 2
w .
a
39
APM2614/1
Hence we solve
1
2
(1 − λ) λ2 − 3λ −
=0
(94)
to find the solutions
λ1 = 1,
√
7)/2,
√
= (3 − 7)/2,
λ2 = (3 +
λ3
so that
ω1 =
a
m
1/2
a
m
a
= 0.42
m
ω2 = 1.68
ω3
,
1/2
1/2
,
.
(95)
Thus the lowest value of ω is ω3 , and therefore the lowest natural frequency (frequency ν = ω/2π) is
ν = 0.067
a
m
1/2
.
(96)
Problems occur when the lowest frequency is small, meaning in practice when ν ≤ 10 (cycles/second).
Thus one requires
m
10 2
= 22277.
(97)
≥
a
0.067
The ratio elastic coefficient: 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
 
 
 


x1
y4
ẋ1
y1
 .   .   .   . 
 ..  =  ..  ,  ..  =  ..  .
 
 
 


y3
x3
y6
x3
(98)
Then the system (90) is P
y = Cy where
C=
0 I
−B 0
.
(99)
It can be shown that, provided B is invertible (which it is in this case), then the eigenvalues of C are
±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.
40
EXERCISES
1. For 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)
dy
dx
= 5x + y − 1,
= −2x + 3y − 2.
dt
dt
(c)
dx
dy
= −2x + y + 4,
= −2x + 4y + 6.
dt
dt
(d) X(n + 1) =
(e) X(n + 1) =
−1 2
−7 4
1 2
1
2
X(n).
X(n) +
1
2
−4
−1
.
2. Find the general solution of the following systems, and state the location and nature of the singular point.
(a)
dx
dy
= 3x + 3y − 4,
= −x − 3y + 5.
dt
dt
(b)
dX
=
dt
−2 3
−3 5
X.
dy
dx
= 6x − y − 5,
= 5x + 2y − 3.
dt
dt


−1 −1 0
dX  3

=  4 − 32 3  X.
(d)
dt
1
1
− 12
8
4
(c)
3. For the following systems, find the solution that satisfies the given initial condition; state the location and
nature of the singular point.
(a)
dX
=
dt
−1 −2
3
4
(b)
dX
=
dt
4 −5
5 −4
(c)
dx
= −x + y, with x(0) = 2.
dt
X+
3
3
subject to X(0) =
X subject to X(0) =
dy
= x + 2y + z − 5, with y(0) = 3.
dt
dz
= 3y − z − 1, with z(0) = 4.
dt
6
2
.
3
4
.
41
APM2614/1
4. Find the general solution of the following discrete dynamical systems, and then find the particular solution
for the given initial condition. State the nature of the singular point.
(a) X(n + 1) =
−2 12
5
(b) X(n + 1) =
2 −5
1 −2
X(n), with X (0) =
X(n), with X (0) =
2
2
5
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.
(d) X(n + 1) =
with X(0) =
13
25
−9 −17
2
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 assumptions, and considering only those females less than 39
years old gives the following model:

 


x1 (k + 1)
0 α 1−α
x1 (k)

 


0   x2 (k) 
 x2 (k + 1)  =  1 0
x3 (k + 1)
x3 (k)
0 1
0
where 0 ≤ α ≤ 1.
(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 (that is,
the population at year k + 1 would be α times the population at year k). The population distribution, however,
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 itself, from rural to
42
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 be easily 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
0 ≤ β ≤ min
r(k)
u(k)
.
α
α
,
1−γ γ
is necessary and sufficient in order to guarantee that both urban and rural populations remain nonnegative
given any nonnegative initial conditions.
43
APM2614/1
7. Find the general solution of the following systems, and state the location and nature of the singular point.
(a) X (n + 1) =
7 −1
5
3
X (n)
(b) X (n + 1) =
6 1
−2 4
X (n) −
(c) X (n + 1) =
5 −5
5 −3
X (n)
1
2
(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 the following systems.
(a)
(b)
(c)
(d)

2 −1 2


X (n + 1) =  −1
2 0  X (n)
−1
0 2


3
5 1


X (n + 1) =  −5 −5 4  X (n)
0
0 3


6 −4 0


X (n + 1) =  1
1 2  X (n)
0
2 6


2 0
0


X (n + 1) =  2 3 −1  X (n)

0 1
1
9. Find the general solution of the following systems, and state the location and nature of the singular point.
(b)
dx
=z
dt

(c)

(a)
dX 
=
dt
dX 
=
dt
dy
dz
= −z
=y
dt
dt



−1 1
0
0



1 2
1  X+  −4 
0 3 −1
−2

1 1 4

0 2 0 X
1 1 1
44
10. Write the following as first–order systems 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)
(c)
d2 x
dt 2
+ 4 dx
+3=0
dt
(d)
d2 x
dt 2
− 5 dx
+6=0
dt
45
APM2614/1
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 (for example, turning a tap
on or off to control the level of water in a tank, or a heater on or off to control the temperature in a room). In this
chapter we consider some aspects of linear dynamical systems subject to linear controls.
Mathematically such systems are described by the equation
P
x(t) = Ax(t) + Bu(t)
for continuous systems, and
x(k + 1) = Ax(k) + Bu(k)
in the discrete case. 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, the results being 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).
46
Our aim in this chapter is to study two important properties of the above control systems:
(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
The question here is “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 to represent such systems by
certain diagrams. 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 having the
value of the incoming line. The delay (d) is the basic dynamic component of discrete–time systems. If x(k) comes
in to 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 goes in to the integrator then x goes out. The way in which these
diagrams are put together to form linear control systems is best described by examples.
Example 1 (Pure Integrator)
47
APM2614/1
The equation for this simple system is
ẋ(t) = u(t).
The output x(t) is the integral of the input function u(t). The diagram is therefore:
Figure 2
Example 2
Consider the control system
ẋ1
ẋ2
=
3 0
1 2
x1
x2
+
1
0
u(t)
y (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
Figure 4
is
48
Putting these two diagrams together and noting 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)
is
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, providing a force of u 1 (t) per unit mass, and the brake, which produces a retarding
force of u 2 (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 = u 1 − u 2 .
In matrix notation:
ẋ1
ẋ2
=
0 1
0 0
x1
x2
+
0
0
1 −1
u1
u2
.
49
APM2614/1
The dynamic diagram for this system is:
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) + L θ̈(t) = g sin θ(t)
where g is the gravitational constant.
We also have
x(t) = u(t) + L sin θ (t).
2
If we assume that the stick is nearly at rest close to the vertical position, then we can put θ̇ = 0, cos θ = 1 and
sin θ = θ, and the above equations can be written
ẍ(t) =
g
g[x(t) − u(t)].
L
50
Put
v(t) = ẋ(t),
then the matrix equation for the system is
ẋ (t)
v̇ (t)
The dynamic diagram for the system is

0
= g
L
1
0


x (t)
v (t)

0
+  g  u(t).
−
L

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
m L 2 θ̈ = −mgL sin θ
or
θ̈ +
g
sin θ = 0.
L
Define
x1 = θ
x2 = θ̇.
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Then we get
ẋ1 = x2
ẋ2 = −
g
sin x1 .
L
The output y is the angle θ so
y = x1 .
If the angle θ is small we can put sin x1 = x1 and the matrix equations for the system are then


0 1
ẋ1
 x1
=  g
−
0
ẋ2
x2
L
y =
1 0
x1
x2
.
The dynamic diagram is
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
52
of proportionality λ).
Let V1 , V2 denote the volume of water in each tank. Then
d V1
dt
d V2
dt
= u − y1 + λ
= u − y2 + λ
V2
V1
,
−
A2
A1
V2
V1
,
−
A1
A2
V1
A1
V2
y2 = µ .
A2
y1 = µ
Thus
dV1
dt
dV2
dt
V2
V1
+λ
+ u,
A1
A2
V2
V1
− (λ + µ)
+ u.
= λ
A1
A2
= −(λ + µ)
The final combined outflow is
y=µ
V2
V1
.
+
A1
A2
The system equations in matrix form are therefore
V̇1
V̇2

λ+µ
 − A1

= 


λ
A1
y =
µ µ
A1 A2
−

λ
A2 



λ+µ 
V1
V2
V1
V2
+
1
1
u
A2
.
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 this section we give a precise definition
for discrete systems and show how to determine if 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.
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Here are some examples to illustrate this definition.
Example 8
Consider the system with dynamic diagram
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 if 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 will also need the following lemma. We omit
the proof – if you are interested, the proof is on page 278 of Luenberger.
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, ..., A N −1 B]
54
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
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, giving finally
x(N ) = A N −1 Bu(0) + A N −2 Bu(1) + ... + Bu(N − 1).
The expression on the right is simply a linear combination of the columns of the matrix
K = B, AB, A2 B, ..., A N −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.
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The following examples show how to apply this important theorem.
Example 10
Consider the control system in example 8. We have
A=
a 0
1 b
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
M=
0 0
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 dynamic diagram
Figure 14
and the equations for this modified system are
x1 (k + 1) = ax1 (k) + u(k)
x2 (k + 1) = x1 (k) + bx2 (k).
Then
A=
a 0
1 b
and
B=
1
0
.
56
The controllability matrix M is therefore
1 a
0 1
M = [B, AB] =
.
Therefore rank M = 2, and hence the system is completely controllable.
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, and 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
and n = 2. Then
A =
−2
2
1 −1
B =
1
0
,
1 −2
0
1
M = [B, AB] =
.
Clearly rank M = 2 = n, and so the system is completely controllable. Let us find a control sequence which
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
x(0) =
0
0
x(2) =
1
2
and
.
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)
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since x (0) = 0.
This gives
1
2
−2
1
=
1
0
u(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, and so 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
P
x(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
K=
t1
0
T
e−At BBT e−A t dt
58
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
P
x(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 ,
t1
x(t1 ) =
eA(t1 −t) Bu(t)dt
(1)
0
Then, using the definition of eA(t1 −t) ,
x(t1 ) =
t1
0
I + A(t1 − t) +
A2
(t1 − t)2 ... Bu(t)dt
2!
or
x(t1 ) = B
t1
0
u(t)dt + AB
t1
+A2 B
0
t1
0
(t1 − t)u(t)dt
(t1 − t)2
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 which transfers the zero vector to a given vector x1 . By Lemma 2
for any t1 > 0 the matrix
K=
t1
T
e−At BBT e−A t dt
0
has an inverse K . Now, given x1 , select any t1 and let
−1
T
u(t) = BT e−A t K−1 e−At1 x1 .
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We verify that this input transfers the state from zero to x1 at time t1 by substituting for u in equation (1):
x(t1 ) =
t1
0
T
eA(t1 −t) BBT e−A t K−1 e−At1 x1 dt
= 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 P
x in the equations and the delay symbol by the integrator symbol in the dynamic diagrams.
Let us explicitly calculate the control in the continuous–time version of example 12.
Example 14
The system is 1–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 us find a control u which transfers 0 to 1 at time
t = 1. The matrix K in this example is
K =
t1
0
1
T
e−At B B T e−A t dt =
Therefore
K −1 =
Hence
0
e−t (1)(1)e−t dt =
1 − e−2
.
2
2
.
1 − e−2
T
u(t) = B T e−A t K −1 e−At1 x1 =
We can check that this control does work. We have x(0) = 0 and
2e−(1+t)
.
1 − e−2
dx
2e−(1+t)
.
= x +u = x +
dt
1 − e−2
Solving this 1st order linear differential equation for x(t) gives
x(t) =
and we see x(1) = 1.
e et − e−t
e2 − 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.
60
Example 15
The matrix form of the system equations is
ẋ
v̇
Therefore

0 1

= g
0
L

x
v


0
+  g  u.
−
L


0 1

A= g
0
L
and

0
B =  g .
−
L

The controllability matrix M is then

g 
0 −
L .
M= g
−
0
L
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 simultaneously balanced? 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:
ẋ1 = v 1
g
v̇ 1 =
(x1 − u)
L1
ẋ2 = v 2
g
v̇ 2 =
(x2 − u).
L2
In matrix form:






ẋ1
v̇ 1
ẋ2
v̇ 2


0
  g
  L
= 1
  0
 
0
1
0
0
0
0
0
0
g
L2

0


0 


1 

0
x1
v1
x2
v2
Calculating the controllability matrix M in the usual way gives

0
− Lg1
0

2

g
0
− Lg 2
 − L1
1
M=
 0
0
− Lg2


2
− Lg2
0
− Lg 2
2



0
 

  − Lg 
+
1  u.
  0 
 

− Lg2
2
− Lg 2
1
0
2
− Lg 2
2
0




.



If L 1 = L 2 then rank M = 4 and the system is completely controllable, ie the sticks can be simultaneously
balanced. However, if L 1 = L 2 then rank M = 2, so the system is not completely 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
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of the stick! I do find it quite surprising that two sticks of unequal length can be simultaneously balanced – 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 us investigate under what conditions this system is completely
controllable. In this case
1
B=
1
and

λ
λ+µ
+
−

A1
A2 


.
AB = 



λ+µ 
λ
−
A1
A2
Hence the controllability matrix for this system is

λ+µ
λ
 1 − A1 + A2

M = [B, AB] = 


λ
λ+µ
1
−
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+µ λ+µ
λ
λ
−
+
−
A1
A2
A1
A2
(2λ + µ) (A2 − A1 )
.
A 1 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 .
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. The definition is
62
Definition
The discrete–time system
x(k + 1) = Ax(k) + Bu(k)
y(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.
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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) = Ax(k) + Bu(k)
y(k) = Cx(k) + Du(k)
is completely observable if and only if the pn × n observability matrix


C


 CA





2
S =  CA

 .

 ..



n−1
CA
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=
a 1
0 b
C=
0 1
.
a 1
0 b
=
and
For CA we get
CA =
0 1
0 b
.
Therefore the observability matrix is
S=
0 1
0 b
.
Since rank S = 1 the system is not completely observable.
Example 20
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).
64
The matrix A is the same as in example 19 and for C we get
C=
1 0
.
a 1
0 b
=
Therefore
CA =
1 0
a 1
,
and the observability matrix S is
S=
1 0
a 1
.
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
P
x = 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

has rank n.




S=



C
CA
CA2
..
.
CAn−1









Example 21
Under what conditions is the system of water tanks in example 7 completely observable? In this case
C=
µ µ
A1 A2
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and the observability matrix is
C
CA

µ
A1


=

 µ (λ (A1 − A2 ) − µA2 )
A21 A2
The determinant of the observability matrix is
det
C
CA
=
µ
A2



.

µ (λ (A2 − A1 ) − µA1 ) 
A1 A22
µ2 (A2 − A1 ) (µ + 2λ)
.
A21 A22
Of course λ > 0 and µ > 0 so the system is completely observable if and only if A1 = A2 .
7. Feedback
Control systems can be classified into two types: open loop and closed loop or feedback control systems. In an
open loop system the control input follows a pre–set 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 and closed loop systems are provided by robots at traffic intersections which
change at fixed intervals of time, and those which measure traffic flow in some way and change accordingly. Two
advantages of feedback control are:
(ii) Feedback control is often simpler than open loop control. For example, one could design a complicated open
loop control for a traffic robot which 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, and it is this aspect of feedback that we will discuss in this section.
Recall from Chapter 2 that the dynamical system P
x = 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
P
x = 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
where C is a constant m × n matrix. Substituting for u in the system equation we get
P
x = (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.
66
Theorem 5 (Eigenvalue placement theorem)
Suppose the control system
P
x = 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.
A very simple example to illustrate this theorem is
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 = C x
has the solution
x(t) = keCt .
If C < 0 then eCt → 0 as t → ∞, so the system with feedback u = C x 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
A =
1
0
g
L
0
B =
,
− Lg
.
c1 c2
.
Suppose that
C=
Then
A + BC =
g
L
0
1
(1 − c1 ) − Lg c2
.
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
. For this value of C the closed loop system will
example, to make both roots −1, choose c1 = 1 + Lg and c2 = 2L
g
be stable.
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Exercises for Chapter 3
1. Investigate the controllability of the system
P
x = Ax + Bu
if u(t) is a scalar, and
(a) A =

−5 1
0 4
0 1 0

(b) A =  0 0 1
0 0 0

3 3 6

(c) A =  1 1 2
2 2 4
2. Consider the system
,
1
.
1
 
0
 
B =  0 .
1
 
0
 
B =  0 .
B=


,


,
1
P
x = Ax + Bu
y = Cx
where
A =
2 −5
−4
0
B =
1
−1
and
C=
1 1
.
Is this system
(i) completely controllable?
(ii) completely observable?
3. For the discrete–time system
x(k + 1) = Ax(k) + bu(k)
where
A=
2 1
0 2
let
b1 =
0
1
,
b2 =
1
0
.
68
(a) For each b, determine if 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
P (t)
w
P
y (t)
=
A11 A12
A21 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 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.
(a) Determine whether each system is controllable, 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, observable.
(c) The systems are now connected in a feedback configuration as shown in the figure below, to produce S4 .
Determine whether S4 is controllable, observable.
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7. Two pendulums are connected by a spring as shown in the diagram below:
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 θ i2 and higher powers can be
neglected) the equations of motion are
m θ̈ 1 = −mgθ 1 − u − ka 2 (θ 1 − θ 2 )
m θ̈ 2 = −mgθ 2 + u + ka 2 (θ 1 − θ 2 )
Take
x1 = θ 1 + θ 2
x2 = θ 1 − θ 2
x3 = ẋ1
x4 = ẋ2
as state variables. Is the system completely controllable?
70
CHAPTER 4
AUTONOMOUS NONLINEAR
SYSTEMS
The introduction of nonlinearity into a system introduces a rich variety of behaviours that are qualitatively different
from the linear case. For example,
z n+1 = z n2 + c
where z n 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.
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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(x 2 − 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).
4.2 LINEARISATION ABOUT SINGULAR POINTS
For clarity, we start in one dimension:
dx
= g(x)
dt
and suppose that x = x∗ is a fixed point, i.e. g(x∗ ) = 0. Then by Taylor’s theorem, for x near x∗ :
(1)
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.
72
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)
dx
= ax(k − x)
dt
(5)
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
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:
unstable
At x = k we have g (k) = −ak < 0: asymptotically stable
2.
x(n + 1) = x(n)2 − 3x(n) + 4
(6)
This equation has fixed points given by solving
x∗ = x∗2 − 3x∗ + 4.
Thus
x∗2 − 4x∗ + 4 = 0
so that
x∗ = 2.
d
Furthermore, g (x) = x 2 − 3x + 4 so that g (x) = dx
(x 2 − 3x + 4) = 2x − 3, and at x = x∗ = 2 we have g (2) = 1.
Thus linearisation does not say whether x∗ = 2 is a stable or unstable fixed point.
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)
(7)
dt
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where x, g are vectors. For example, the system
dx
= x 2 + y 2 − 6,
dt
dy
= x2 − y
dt
(8)
would have g1 = x 2 + y 2 − 6, g2 = x 2 − y. Then the Jacobian of the system is


∂g1 ∂g1
 ∂x
∂y 


.
A=


 ∂g2 ∂g2 
∂x
∂y
(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
(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 ( , 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)
74
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 = x 2 + y 2 − 6
ẏ = x 2 − y
(14)
The critical points are found by solving ẋ = 0 and ẏ = 0 simultaneously:
ẋ = 0 ⇒ x 2 + y 2 − 6 = 0
(1)
ẏ = 0 ⇒ x 2 − y = 0
(2)
From (2) we have x 2 = 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
√
√
2, 2
and P2 = − 2, 2 .
P1 =
The Jacobian matrix is
A=
and so at P1
A=
Thus
√
4
2 2
√
2 2 −1
.
√
= −10 2 < 0
√
= 2 2−1
σ
so, since
2x 2y
2x −1
< 0, the critical point is locally a saddle point.
At P2
A=
from which
σ
√
−2 2
4
√
−2 2 −1
,
√
= 10 2
√
= −2 2 − 1 < 0
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and
σ2 − 4
<0
so that P2 is locally a stable focus.
2.
ẍ + x − x 3 = 0
(15)
Writing the equation as a first order system with y = ẋ:
ẋ = y ẏ = x 3 − x.
(16)
Solving ẋ = 0, ẏ = 0 simultaneously we get:
ẋ = 0 ⇒ y = 0
ẏ = 0 ⇒ x 3 − 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
A=
0
1
2x − 1 0
2
.
Thus
A1 =
A2 = A3 =
0 1
−1 0
0 1
2 0
,
.
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 = x 2 − 2y + 3,
y = x y − 7y.
(i)
(ii)
The equation (ii) is 0 = y(x −8), and is satisfied by y = 0 or x = 8. Substituting y = 0 into (i) gives x 2 −x +3 = 0,
which has no real solution. Substituting x = 8 in (i) gives y = 29 12 . Thus the only singular point is
P1 = (x∗ , y∗ ) = (8, 29.5).
The Jacobian of the system is
A=
2x
−2
y x −7
,
76
thus
A1 =
16 −2
29.5
1
at the singular point P1 . The eigenvalue equation has solutions that are complex and |λ| > 1.
Thus the singular point P1 is unstable.
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
=
=
dx
dx/dt
g2 (x, y)
(18)
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 d x/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 d x/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.
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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 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:
ẋ =
ẏ =
r1
x(P1 − x − α 12 y)
P1
r2
y(P2 − y − α 21 x)
P2
(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 )α 12 x y and
(−r2 /P2 )α 21 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) or P1 − x − α 12 y = 0 (ii)
ẏ = 0: y = 0 (iii) or P2 − y − α 21 x = 0 (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 , α 12 , α 21 . 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
78
which it can evolve. The cases are presented diagrammatically as:
Figure 1
We will investigate case II here, i.e. the case where, as can clearly be seen from the sketch, P1 > P2 /α 21 and
P2 > P1 /α 12 . The other cases are left to you to do as exercises.
The Jacobian is:
Thus at (0, 0) we get

2r1 x
r1 y
 r1 − P1 − α 12 P1

J=


−r2 α 21 y
P2
J=
−α 12r1 x
P1



.

2r2 y r2 α 21 x 
r2 −
−
P2
P2
r1 0
0 r2
so that
= r1r2 > 0
σ
σ2 − 4
= r1 + r2
= (r1 + r2 )2 − 4r1r2 = (r1 − r2 )2 > 0.
The origin is therefore always an unstable node.
(23)
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At the singular point (0, P2 )
0
r1 (1 − α 12 P2 /P1 )
−r2 α 21
−r2
J=
.
(24)
In this case, case II, P2 > P1 /α 12 so the first term in the Jacobian is negative. Thus, by setting
β = 1 − α 12
P2
P1
and keeping in mind that β < 0 (from P2 > P1 /α 12 ), we have
= −r1r2 β > 0
σ
= r1 β − r2 < 0
and
σ2 − 4
= (r1 β − r2 )2 + 4r1r2 β
= r12 β 2 − 2r1r2 β + r22 + 4r1r2 β
= r12 β 2 + 2r1r2 β + r22
= (r1 β + r2 )2
> 0.
The singular point (0, P2 ) is therefore clearly a stable node.
Similarly, at the point (P1 , 0),
−r1 α 12
−r1
0 r2 (1 − α 21 P1 /P2 )
J=
(25)
and P1 > P2 /α 21 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:
At Q,

Thus
−r1 x∗
P1


J=
 −r α y
2 21 ∗
P2
r1r2 x∗ y∗
(1 − α 21 α 12 ),
P1 P2
r1 x∗ r2 y∗
.
= −
+
P1
P2
=
σ
−α 12r1 x∗ 

P1

.
−r2 y∗ 
P2
(26)
80
At Q, both x∗ , y∗ are positive, so σ is clearly negative, and the sign of
In case II, P2 > P1 /α 12 and P1 > P2 /α 21 , thus
P2 >
will depend on the sign of (1 − α 21 α 12 ).
P1
P2
>
α 12
α 12 α 21
(27)
so that clearly
α 12 α 21 > 1.
Thus
< 0 and Q is a saddle point.
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
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trajectories. In this case:
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.
Example
The above may be clearer if instead of using arbitrary constants r1 , P1 , α 12 , r2 , P2 , α 21 , we let the constants have
definite numerical values:
r1 = 0.1
P1 = 10000
α 12 = 1
r2 = 0.2
P2 = 15000
α 21 = 2.
The condition defining the case being considered here, case II, is P2 > P1 /α 12 and P1 > P2 /α 21 , which is clearly
satisfied. The fixed points, and the values of the Jacobian at the fixed points, are:
1. (0, 0). From (23), J =
0.1 0
0 0.2
Clearly this is an unstable node.
.
82
2. (0, P2 ). From (24),
0.1 (1 − 1.5)
0
−0.2 × 2
−0.2
J =
−0.05
0
−0.4 −0.2
=
Thus
= +0.01, σ = −0.25; thus σ 2 > 4
3. (P1 , 0). From (25), J =
Thus
−0.1
−0.1
0
−0.0666
.
and the fixed point is a stable node.
.
= 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)
J=
Thus
−0.05
−0.05
−0.13333 −0.06666
.
= −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) = x(n)(1 − a + by(n))
r
y(n + 1) = y(n)(1 + r − y(n) − cx(n))
P
(28)
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 = x(1 − a + by)
ry
− cx)
y = y(1 + r −
P
(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
x =
1−
,y= .
c
bP
b
(30)
83
APM2614/1
The Jacobian is
J=
1 − a + by
bx
−cy
1 + r − 2rPy − cx
.
(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. b P > 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.
(i) x = 0, y = 0, J =
1−a
0
0
1+r
(34)
.
The eigenvalues are 1–a and 1 + r, so it is unstable and analogous to a saddle point.
(ii) x = 0, y = P, J =
1 − a + bP
0
−c P
1−r
(35)
.
Since b P > a, the diagonal elements of J, and therefore in this case its eigenvalues, are λ1 = 1 − a + b P > 1
and λ2 = 1 − r with |λ2 | = |1 − r| < 1. Thus the singular point is again unstable and analogous to a saddle
point.


r
a
1
b 1−
c
bP 

a
a
r
.
1−
,y = ,J=
(36)
(iii) x =


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
, and note that 0 < d < 2;
bP
a
e = ar 1 −
and note that e > 0.
bP
d =
Then the Jacobian can be rewritten as
so that



J=

1
−
ca
b
be
ac
1−d
= 1−d +e
σ
= 2 − d.





84
The eigenvalues of J are (from equation (20) in Chapter 2):
√
2 − d ± 4 − 4d + d 2 − 4 + 4d − e
2
√
2 − d ± d2 − e
,
=
2
which is real for d 2 − e ≥ 0. If d 2 − e < 0, the eigenvalues are complex and are given by
√
2 − d ± e − d 2i
.
2
For the real case d 2 − e ≥ 0 we have
√
√
λ1 = 1 + 12 −d + d 2 − e < 1 + 12 −d + d 2 = 1.
√
Thus 0 < λ1 < 1. Also, λ2 = 1 + 12 −d − d 2 − e < 1, and also λ2 > 1 + 12 (−d − d) = 1 − d > −1.
Thus −1 < λ2 < 1.
For the complex case d 2 − e < 0,
|λ| =
=
so that, for stability we need |λ| < 1, hence,
d
1−
2
2
+
√
e − d2
2
e
1−d + ,
4
e
− d < 0.
4
In terms of the parameters a, b, r, an P, this means
b P < a + 4.
From equation (32) we now conclude that the condition
a < bP < a + 4
will ensure that this fixed point is asymptotically stable.
85
APM2614/1
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:
1. (0, 0), J =
0.5 0
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.
2. (0, P), J =
1.5
0
−10 0.5
.
The eigenvalues are 1.5 and 0.5, so the fixed point is unstable and analogous to a saddle point.
3. (25, 500), J =
1 0.025
−5 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
dx
= −βx y,
= βx y − γ y
dt
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 − βx y
dy
dt
= βx y − γ y.
(38)
86
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 − βx y
dy
dt
= βx y − γ y.
(39)
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.
The isocline diagram looks like:
The Jacobian is:
J=
The singular points are therefore:
1. (0, 0), J =
α 0
0 −γ
α − 2αx − β y − 2α y −2αy − βx − 2αx
βy
βx − γ
.
(40)
.
It is clearly a saddle point.
2. (1, 0), J =
−α −2α − β
0
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 − βx y = 0.
87
APM2614/1
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=
Thus
−0.0079 −0.217
0.011
0
.
= 0.002, σ = −0.008, so that σ 2 < 4 . The fixed point is therefore a stable focus.
The phase plane diagram is therefore:
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.
88
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
, with x∗ ∈
.
(2) V (x) is real-valued and continuous and has continuous first partial derivatives.
(3) V (x) has a unique minimum at x∗ , i.e. V (x∗ ) < V (x) for all x ∈
(4) The value of V decreases along any trajectory in
, x = x∗ .
.
(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)
,
1 + x1 (k)2 + x2 (k)2
x1 (k)
1 + x1 (k)2 + x2 (k)2
(42)
which has x = (0, 0) as an equilibrium point. Let us define the function
V (x) = x12 + x22 .
(43)
89
APM2614/1
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) 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)
x1 (k)
=
+
2
2
1 + x1 (k) + x2 (k)
1 + x1 (k)2 + x2 (k)2
=
=
x2 (k)2 + x1 (k)2
1 + x1 (k)2 + x2 (k)2
V (x (k))
1 + x1 (k)2 + x2 (k)2
2
2
2
≤ V (x (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))
(45)
dt
together with a given equilibrium point x∗ . It is assumed that g is continuous. The requirement (41-4), 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
V̇ (x) ≡
d
(V (x (t))) ≤ 0
dt
(46)
with equality if and only if x = x∗ .
By the chain rule, this is
∂ V dx1
d
∂ V dx2
∂ V dxn
+
+···+
≤ 0.
(V (x (t))) =
dt
∂x1 dt
∂ x2 dt
∂ xn dt
(47)
Now apply the system equation (45):
∂V
∂V
∂V
d
g1 (x (t)) +
g2 (x(t)) + · · · +
gn (x (t)) ≤ 0
(V (x (t))) =
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 : → R, with the domain ⊆ Rn ] such that
(i) V (x) is continuous and has continuous first partial derivatives;
(ii) x∗ ∈
;
(iii) V (x) has a unique minimum at x∗ ;
90
(iv)
is either Rn or bounded, and if it is bounded then V (x) is constant on the boundary of
;
(v) V satisfies (48) [or equivalently (49)] with equality if and only if x = x∗ .
Then for any trajectory that starts in
(x(t = 0) ∈
),
lim x(t) = x∗ .
t→∞
[Alternatively, x∗ is asymptotically stable with
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
d V (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
cannot leave . 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
for all x ∈
(52)
, 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 : → R, with the domain ⊆ Rn ] such that
(i) V (x) is continuous;
(ii) x∗ ∈
;
(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∗ .
91
Then there exists a domain
, with x∗ ∈
⊆
APM2614/1
, such that for any trajectory that starts in
,
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
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) = x12 + x22 is a Liapunov function for the system (42),
x1 (k + 1) =
x2 (k + 1) =
x2 (k)
,
1 + x1 (k)2 + x2 (k)2
x1 (k)
.
1 + x1 (k)2 + x2 (k)2
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
J=
0 1
1 0
.
(53)
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 = −x 3 − x y 2 + y
(54)
ẏ = −y 3 − x 2 y − x
V = x 2 + y2
(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(−x 3 − x y 2 + y) + 2y(−y 3 − x 2 y − x)
= −2x 4 − 2x 2 y 2 + 2x y − 2y 4 − 2x 2 y 2 − 2x y
= −2(x 4 + 2x 2 y 2 + y 4 )
= −2(x 2 + y 2 )2 .
(56)
2
Since −2 x 2 + 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 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)
92
so that
= 1, σ = 0 and the fixed point may be stable or unstable.
If a Liapunov function is defined over a large region , 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 ]. Therefore, the region over which the Liapunov
function is defined, delineates a region 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) = x12 + 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.
From Newton’s laws of motion we have:
M R θ̈ = −Mg sin θ − Mk θ̇.
(59)
93
APM2614/1
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
V (θ, ω) =
1
M R 2 ω2 + Mg R(1 − cos θ).
2
(61)
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 R 2 ωω̇ + Mg R sin θ.θ̇
g
kω2
= M R 2 − ω sin θ −
R
R
2
= −k M Rω ≤ 0.
+ Mg R ω sin θ
(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.
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
94
ẋr = R,
ẏr = yr = 0.
(63)
The fact that the velocity of the dog is D means that
ẋd2 + ẏd2 = D 2 .
(64)
The velocity vector of the dog always pointing toward the rabbit means that:
ẋd = −k(xd − xr )
(65)
(66)
ẏd = −kyd .
The constant k is found by calculating ẋd2 + ẏd2 from (65) and (66), and then equating it to D 2 from (64). We
find
D
(67)
k=
1
(xd − xr )2 + yd2 2
so that (65) and (66) become:
ẋd
= −
ẏd
= −
(xd − xr ) D
(xd − xr )2 + yd2
yd D
(xd − xr )2 + yd2
(68)
.
The system is more meaningful when expressed in relative coordinates, representing the difference in position
of the rabbit and dog. Define
(69)
x = xd − xr , y = yd
then
ẋd = ẋ + R, ẏd = ẏ
and (68) becomes:
ẋ = −
xD
x2
+
y2
− R,
ẏ =
(70)
−y D
x 2 + y2
.
(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) = x 2 + y.
(72)
Now
d
V (x(t), y(t))
dt
xD
2y (−y D)
= 2x −
−R +
x 2 + y2
x 2 + y2
V̇ (x, y) =
= −2D x 2 + y 2 − 2Rx.
(73)
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APM2614/1
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 − x 2 − y 2 ), ẏ = −x + y(1 − x 2 − y 2 )
(74)
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 x 2 + 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 − 2.2x 2 , 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 > 0 such that if |x(0) − G| < 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
96
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 s denote the region where V (x) < s. Assume that s is bounded and that V (x) ≤ 0
[or V̇ (x) ≤ 0 in continuous time]. Let S be the set of points within s 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 s
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
( s ) of attraction.
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 − x 2 − y 2 ], ẏ = −x + y[1 − x 2 − y 2 ].
Define
V (x, y) = (1 − x 2 − y 2 )2 .
(77)
Then
V̇ (x, y) = −2(1 − x 2 − y 2 )(2x ẋ + 2y ẏ)
= −4(x 2 + y 2 )(1 − x 2 − y 2 )2 .
(78)
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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 1
be the region where V (x) < 1 [then 0 ∈
/ 1 ]. 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 1 tends towards the limit cycle consisting of the
unit circle.
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
(81)
|d − a1 | < |b − a1 |.
[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 .]
98
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)
Thus the 2-cycle is an attractor if
|g (a1 )g (a2 )| < 1.
(85)
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 − 2.2x 2 .
(89)
with
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Thus (88) is:
∴
0 = g(3.2x − 2.2x 2 ) − x
0 = 3.2(3.2x − 2.2x 2 ) − 2.2(3.2x − 2.2x 2 )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
(−121x 2 + 231x − 105) = 0.
125
(91)
Thus using the quadratic formula
√
21 ± 21
x=
= 0.746 or 1.163.
22
Thus the 2-cycle is, as stated earlier (0.746; 1.163). To investigate stability note that
(92)
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 − 2x y
ẏ = 2x y − y
(b) ẋ = y − x 2 + 2
ẏ = x 2 − x y
(c) ẋ = −3x + y 2 + 2
ẏ = x 2 − y 2
(d) ẋ = −2x y
ẏ = y − x + x y − y 3
100
(e) ẋ = x 2 − y 2 − 1
ẏ = 2y
(f) ẋ = 2x − y 2
ẏ = −y + x y
(g) ẋ = x y − 3y − 4
ẏ = y 2 − x 2
(h) ẋ = x(1 − x 2 − 2y 2 )
ẏ = y(3 − x 2 − 3y 2 )
(i) ẋ = −2x + y + 10
y
.
ẏ = 2x − y − 15 y+5
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 − x 3 ) − x 2 = 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).
7. For the system
ẋ(t) = sin[x(t) + y(t)]
ẏ(t) = e x(t) − 1
determine all the equilibrium points, and classify each equilibrium point as asymptotically stable or unstable.
8. The van der Pol equation. The equation
ẍ + [x 2 − 1]ẋ + x = 0
arises in the study of vacuum tubes. Show that if
it is unstable.
< 0, the origin is asymptotically stable and that for
>0
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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 + bx y −
ẏ = −cx y + dy −
where
1
and
(a) When
2
2
1x
2 y,
are positive constants.
< d, show that there is a new critical point in the first quadrant that is a centre.
(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
k
β
ẍ = − 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.
[Hint:
d
y|y|
dy
= 2|y|.]
11. The plane autonomous system
y
x−x
1+y
y
x −y+β
ẏ = −
1+y
ẋ = 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
102
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.
13. Use phase plane analysis to analyze the solutions to the dynamical system
2
a(n + 1) = [2 − a(n) − b(n)]a(n)
3
2
b(n + 1) = [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
A=
4
1
.
15. No Marginal Stability. Consider the system
ẋ1 = x1 a 2 − x12 − x22 + x2 a 2 + x12 + x22
ẋ2 = −x1 a 2 + x12 + x22 + x2 a 2 − x12 − x22
and the function V (x) = x12 + 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 = −x 3
(b) ẋ = −x 3 − y 2
ẏ = x y − y 3
(c) ẋ = y(1 − x)
ẏ = −x(1 − y)
V = −x − log(1 − x) − y − log(1 − y)
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17. The van der Pol Equation. The equation
ẍ + [x 2 − 1]ẋ + x = 0
arises in the study of vacuum tubes. Show that if
trajectories.
> 0 there is a limit cycle that is approached by all other
[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.]
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 − e x1 (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 ) =
b 4
1
1
b2
1
x1 + (x1 + cx2 )4 − x14 + x22 + bx2 x3 + x32 .
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?
104
(b) Show that V̇ has the form
V̇ (x1 , x2 , x3 ) =
γ x22
3
x1 + cx2
2
2
3
+ x12
4
and determine the constant γ .
(c) Is V always a Liapunov function if b and c satisfy the conditions of part (a)?
(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:
m
dθ
d 2θ
= ω2 m sin θ cos θ − mg sin θ − β
.
2
dt
dt
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(a) If ω2 < g/ , show that (θ = 0, dθ/dt = 0) is a stable fixed point and is the only critical point with
−π < θ < π.
(b) If ω2 > g/ , 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 = yx 2 − x
ẏ = −x 3 − 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:
d I1
dt
= (0.5 − 0.01I2 − 0.0005I1 )I1
d I2
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:
d I1
dt
d I2
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.
106
30. The competing species model for species x, y is:
dx
dt
dy
dt
x
0, 1
−
y
10000 10000
y
0, 2x
= 0.2y 1 −
−
15000 15000
= 0.1x 1 −
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 ) = x12 + x22 is a Liapunov function at the origin for each of the following systems:
(a) ẋ1 = −x2 − x13 ,
ẋ2 = x1 − x23 ;
(b) ẋ1 = −x13 + x2 sin x1 ,
(c) ẋ1 = −x1 − 2x22 ,
(d) ẋ1 = −x1 sin2 x1 ,
(e) ẋ1 = −(1 − x2 )x1 ,
ẋ2 = −x2 − x12 x2 − x1 sin x1 ;
ẋ2 = 2x1 x2 − x23 ;
ẋ2 = −x2 − x25 ;
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 − x12 − x22 )(x12 + x22 + 1)
ẋ2 = −x1 − x2 (1 − x12 − x22 )(x12 + x22 + 1);
(b) ẋ1 = x2 , ẋ2 = −x2 + x23 − x15 .
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
x12
x22
+
< 1.
a2
b2
34. Given the system
ẋ1 = x2 , ẋ2 = x2 − x13
show that a positive definite function of the form
V (x1 , x2 ) = ax14 + bx12 + cx1 x2 + dx22
can be chosen such that V̇ (x1 , x2 ) is also positive definite. Hence deduce that the origin is unstable.
35. Show that the origin of the system
ẋ1 = x22 − x12 , ẋ2 = 2x1 x2
is unstable by using
V (x1 , x2 ) = 3x1 x22 − x13 .
107
36. Show that the fixed point at the origin of the system
ẋ1 = x14 , ẋ2 = 2x12 x22 − x22
is unstable by using the function
V (x1 , x2 ) = αx1 + βx2
for a suitable choice of the constants α and β.
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108
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:
• bifurcation
• chaos
• 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 ν e (the escape
velocity), it will fall back to earth; if its velocity exceeds ν e , it will break free of the earth’s gravitational force and
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go into outer space.
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 m 1 of equilibrium
values, but for another value of the parameter we have a different number of equilibrium values, say m 2 , 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 + x 2 ).
dt
(1)
√
x∗ = −1 and, for a ≥ 1, x∗ = ± a − 1.
(2)
First, we find all the fixed points:
Next, we determine stability of the fixed points. To do this we need to know g (x) :
g (x) = (1 − a + x 2 ) + (1 + x)2x = 1 − a + 2x + 3x 2 .
(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.
110
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.
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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) + r x(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 = r x − y − x z
ż = x y − 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 = 83 , 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” each made up of spiralling
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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 = 83 , 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
(10)
z(n + 1) = z(n)2
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)
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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
114
pictures suggest that M is a connected set, a fact that has been confirmed mathematically.
Julia sets for c at various points in the Mandelbrot set.
Exercises
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 + 1) = x(n) + 0.5 1 −
x (n)
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
θ n+1 = θ 2n , rn+1 = rn2
with θ ∈ [0, 2π). If θ 2n ≥ 2π, say θ 2n = ψ n + k(2π) with ψ n ∈ [0, 2π) and k an integer, then we put
θ n+1 = ψ n . 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.
3. Consider the dynamic system
dx
= 4x − x 2 − b.
dt
Determine the fixed points and their nature and draw a bifurcation diagram.
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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?
(b) State the dynamic equation that is used for generating the Mandelbrot set, and then define the Mandelbrot set.
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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
A=
1 3
5 4
and
4
2
B=
then
[A, B] =
1 3 4
5 4 2
.
This notation is extended in an obvious way to more than two matrices.
A
denotes the ( p + q) × m matrix obtained by
B
placing A above B. Obviously, this notation can also be extended to more than two matrices.
2. Rank of a Matrix
Similarly if A is an p × m matrix and B is a q × m matrix then
The set of all column m–vectors









k1
k2
k3
..
.
km









is denoted by Rm . (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
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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
is
1 2
0 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
C=
1
t2
, D=
t3
3
t sin t
118
then
d
C=
dt
0
2t
while
t1
0
and
t1
0



Ddt = 




1dt 
t1
0
 
=
 
1 3
t1

t
t 2 dt
31

t1


Cdt = 


t1
0
,





 1
4
3t1
  4 t1
0
 
=
 
t1

1 2
t 1 − cos t1
sin t dt
21
0
t1
t 3 dt
0
t1
0
3t 3
1 cos t
d
D=
dt
,
tdt
0
3dt
4. Exponential Function
The exponential function e x , 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
This follows easily from the definition of eAt . We have
d At
e
dt
d
A2 t 2 A3 t 3
I + At +
+
+ ...
dt
2!
3!
A3 t 2 A4 t 3
= A + A2 t +
+
+ ...
2!
3!
A2 t 2
A2 t 2
+ . . . or I + At +
+ ... A
= A I + At +
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) .
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5. Solution of the Equation P
x = Ax + Bu.
Multiply both sides of the equation P
x = Ax + Bu on the left by e−At . This gives
e−At P
x(t) = e−At Ax(t) + e−At Bu(t)
or
x(t) − e−At Ax(t) = e−At Bu(t).
e−At P
But
d −At
x(t) − e−At Ax(t)
e x(t) = e−At P
dt
so we get
d −At
e x(t) = e−At Bu(t).
dt
Therefore
t
e−At x(t) =
Multiply both sides of this equation on the left by e
x(t) = eAt
At
e−As Bu(s)ds.
a
to get
t
a
t
e−As Bu(s)ds =
eA(t−s) Bu(s)ds.
a
If we assume that x(0) = 0 then we can take a = 0 and we get
t
x(t) =
0
eA(t−s) Bu(s)ds.