2.2
LOCAL STABILITY
In this section you will

investigate the geometric behaviour of nonlinear systems and their linearisations
close to their equilibrium points.

learn under what conditions a nonlinear system behaves locally like its
linearisation

learn how to classify the equilibrium points
2.2.1 Linearisation theorem
The idea of linearisation is to use a linear system to approximate the behaviour of
solutions of a nonlinear system near an equilibrium point We are going to investigate the
equilibrium points of the linearisation and see what, if anything, we can deduce about the
equilibrium points of the nonlinear system. Can we deduce the stability of their
equilibrium points? Can we say anything about the geometric behaviour of solutions
close to the equilibrium points?
Consider the nonlinear system
x  f ( x , y )
y  g ( x, y)
with equilibrium point
bp, qg. We
have already seen that the system may be
approximated in the neighbourhood of the equilibrium point by the linear system
u O L
uO
L

J
M
v P
vP
N
QM
N
Q
where J is the Jacobian matrix
f
L
M
x
M

g
M
M
Nx
f
y
g
y
4
O
P
P
P
P
Q
Look at the tr- det diagram for linear systems. The boundary lines for stability are
trace  0 and det  0 . Since linearisation is an approximation it is when the trace and
determinant of the Jacobian matrix lie on the boundary lines between stability and nonstability and the system is, therefore, very sensitive to small changes that there are likely
to be differences in the long term behaviour of a nonlinear system and its linearisation.
You already know that if det  0 the linear system is non simple and without further
investigation nothing can be deduced about the behaviour of the system close to an
equilibrium point. Likewise a nonlinear system is not a simple system if its linearisation
is non simple and we are, therefore, unable to classify its equilibrium points from its
linearisation. The stability under the condition trace  0 needs careful investigation.
Investigation 1
Consider again the competing species model
x  0.2 x  0.004 x 2  0.003xy  f ( x, y)
y  01
. y  0.001y 2  0.003xy  g( x, y)
b gb gb g b g
with equilibrium points 0,0 , 50,0 , 0,100 and 20,40
Here is its phase plane diagram.
5
You are going to compare the phase portraits of the nonlinear system close to its
equilibrium points with those of its linearisations.
You will need to use the material on the web for this investigation.
This investigation suggests that the behaviour of a nonlinear system close to an
equilibrium point mimics the behaviour of its linearisation. The extent to which this is
true is summarised in the following important theorem, proofs of which were published
independently by the American mathematician Philip Hartman (1964) and the Soviet
mathematician David Grobman (1965).
2.2.2
Hartman Grobman linearisation theorem
If the linearisation matrix J has no zero or purely imaginary eigenvalues, then the phase
portrait for the nonlinear system near an equilibrium point is similar to the phase portrait
of its linearisation.
This theorem tells us that:

If a linear system has a zero eigenvalue we can deduce nothing about the
nonlinear system. A zero eigenvalue corresponds to the line det  0 in the tr-det
diagram and the system is therefore, non simple.

If the linear system has purely imaginary eigenvalues we can deduce nothing
about the corresponding non linear system. For purely imaginary eigenvalues the
linearisation is a centre. Hence, if the linearisation is a centre we are unable to
learn anything about the non linear system from its linearisation.

If the linearisation shows a stable node or stable focus then all solutions of the
linearised system approach (0,0) as t   . Hence for the non linear system
solutions which start near the equilibrium point approach it as t   and the
equilibrium point is an attractor and stable.
6

If the linearisation shows an unstable node or an unstable focus all solutions
move away from (0,0) and are unbounded. Hence for the nonlinear system all
points which start close to the equilibrium point move away from it and the
equilibrium point is a repeller and unstable.

If the linearisation shows a saddle then some solutions approach (0,0) as t  
and others move away from it. Hence the non linear system shows similar
behaviour close to an equilibrium point and the equilibrium point is a nonlinear
saddle and unstable.
Explanation of the results obtained in investigation 1 in terms of the linearisation
theorem.
First we need to calculate the linearisations at each equilibrium point.
To calculate the Jacobian matrices
Let
0.2 x  0.004 x 2  0.003xy  f ( x, y)
01
. y  0.001y 2  0.003xy  g( x, y)
f
g
f
g
 0.2 .008 x  0.003 y ,
 0.003 y ,
 0.003x ,
 01
.  0.002 y  0.003x
x
x
y
y
Therefore
J ( 0,0) 
J ( 50,0) 
0.2 0 I
F
G
H0 01. J
K, J
0.2
F
G
H0
( 0 ,100 )
IJ, J
0.05K

015
.
7
( 20 , 40 )
01
.
F
G
H0.3

IJ,
01
.K
0.08
F
G
.
H012
0
IJ
0.06K
0.06

b g
tre
Jb gj  0,dete
Jb gj  0, tr e
Jb gj 4 dete
Jb gj  0
At the point 50,0
2
50, 0
50, 0
50, 0
50, 0
b g
so the eigenvalues are real and negative and the linearisation at the point 50,0
is a stable node and by the linearisation theorem the nonlinear system mimics
this behaviour close to the equilibrium point and, therefore, the nonlinear
equilibrium point is an attractor and stable;

bg
trc
J h
 0,dete
Jb gj  0, tr e
Jb gj 4 dete
Jb gj  0
At the point 0,0
2
0,0
0,0
0,0
0, 0
bg
so the eigenvalues are real and positive and the linearisation at the point 0,0 is
an unstable node and by the linearisation theorem the nonlinear system mimics
this behaviour close to the equilibrium point and, therefore, the nonlinear
equilibrium point is a repeller and unstable.

b g
At the point 20,40
e j
det Jb20,40g  0
so the eigenvalues are real with one positive and one negative and the
linearisation at the point (20,40) is a saddle point and by the linearisation
theorem the nonlinear system mimics this behaviour close to the equilibrium
point and, therefore, the nonlinear equilibrium point is a nonlinear saddle and
unstable.
b g
tre
Jb gj  0, dete
Jb gj  0, tr e
Jb gj 4 dete
Jb gj  0
0,100gis a
so the eigenvalues are both negative and the linearisation at the point b

At the point 0,100
2
0,100
0,100
0,100
0,100
stable improper node and by the linearisation theorem the nonlinear system mimics
this behaviour close to the equilibrium point and, therefore, the nonlinear equilibrium
point is an attractor and stable.
8

Worked Example 1
Here is a worked example based on the linearisation theorem. There is a parallel example
for you written for the web for those who prefer to use Maple.
Consider the nonlinear system
x  x  y  x 2
y   x  y
(1) Find the equilibrium points of the system and classify its linearisation.
(2) What does this tell you about the stability of the nonlinear system?
(3) Verify your answer to (2) by looking at the phase plane diagram of the nonlinear
system near its equilibrium points.
Find the equilibrium points
x  x  y  x 2  0
y   x  y  0
Therefore x  0 or 2, y = 0 or 2
bg bg
The equilibrium points are 0,0 and 2,2
Find the Jacobian matrix.
Let
bg
gb
x, yg
 x  y
f x, y  x  y  x 2
f
f
g
g
 1  2 x,
 1,
 1,
1
x
y
x
y
9
J
1  2 x 1O
L
M
N1 1P
Q
bg
Stability at the point 0,0
1 1O
L
M
1 1P
N
Q
tr bg
J  2  0, detbg
J  2  0, tr bg
J  4detbg
J  4  0
J
2
The eigenvalues are imaginary with positive real parts and the linearisation is an unstable
focus. By the linearisation theorem the nonlinear system mimics this behaviour and the
bg
point 0,0 is a repeller and unstable.
bg
Stability at the point 2,2
3 1O
L
M
1 1P
N
Q
tr bg
J  2  0,detbg
J  2  0, tr bg
J  4 detbg
J  12  0
J
2
The eigenvalues are real and of opposite signs so the linearisation is an unstable saddle.
By the linearisation theorem the nonlinear system mimics this behaviour and the point
b2,2gis a nonlinear saddle and unstable.
These results can be verified by examining the phase plane diagram of the nonlinear
system. You will find the program for this at the end of the Maple version of the worked
example.
2.2.3
Limitations of the Hartman Grobman linearisation theorem.
Note that the Hartman-Grobman theorem has two serious limitations. First it tells us
nothing when the linearisation is non simple or a centre. Secondly, when the theorem
does apply it provides information only about the behaviour of solutions near the
10
equilibria. In order to make global predictions we need more information. These
limitations will be illustrated in the following investigations.
Investigation 2
To investigate solutions of a nonlinear system when the linearisation is a centre.
You will need to use the material on the web for this investigation. You are going to plot
the phase diagrams of the nonlinear system close to its equilibrium point and the phase
diagram of its linearisation and compare them.
You will see that the phase diagram for the linearisation shows a centre where as the
phase diagram for the nonlinear system shows an unstable equilibrium point with the
trajectories spiralling away from its equilibrium point.
Thus if the equilibrium point of the linearisation is a centre further analysis is necessary
to determine the stability of the equilibrium point of the nonlinear system.
Investigation 3.
To look at the global phase portrait of a nonlinear system.
You will need to use the material on the web for this investigation. You are going to plot
the phase diagrams of the nonlinear system both close to its equilibrium point and also
over a wider range of values of x and y and compare them.
You will see that locally the phase portrait shows that the equilibrium point is a stable
attractor where as the global portrait shows the existence of a closed trajectory. This is an
example of a limit cycle which will be discussed in the next unit.
2.2.4
Summary
(1) If the linearisation is a stable node or stable focus the equilibrium point of the
nonlinear system is an attractor and stable.
11
(2) If the linearisation is an unstable node or unstable focus the equilibrium point of the
nonlinear system is a repeller and unstable
(3) If the linearisation shows a saddle then the nonlinear system is a nonlinear saddle and
is unstable
(4) If the linearisation shows a centre it is not yet possible to determine the nature of the
equilibrium point of the nonlinear system.
(5) If the linearisation is a non simple system the nonlinear system is also classed as non
simple and we are unable to classify the system from its linearisation.

Worked Example 2
Here is another worked example for you to look at. There is a parallel example on the
web for those who prefer to use Maple.
Find the equilibria of
x  3x  y 2  2
y  x 2  y 2
and determine their stability
To find the equilibria
x  2 xy =0
y  y  x  xy  y 3 =0
b gb gb g
Therefore the equilibria are 0,0 , 0,1 , 0,1
Find the Jacobian matrix
Let
f ( x, y)  3x  y 2  2
g( x, y)  x 2  y 2
f
 2 y
x
f
 2 x
y
g
 1  y
x
12
g
 1  3y 2  x
y
2 y
L
M
1  y
N
J
O
1  3y P
Q
2 x
2
bg
Stability at the point 0,0
0 0O
L
M
1 1P
N
Q
tre
Jb gj  1,dete
Jb gj  0 ,
J ( 0,0) 
0, 0
0, 0
so the linearisation is nonsimple and so nothing can be deduced about the equilibrium
point of the nonlinear system from its linearisation. The phase portrait (which you will
plot at the end of this example) suggests the equilibrium point is unstable.
bg
Stability at the point 0,1
2 0 O
L
M
N0 2P
Q
tre
Jb gj  4,dete
Jb gj  4, tr e
Jb gj 4 dete
Jb gj  0
J ( 0,1) 
2
0,1
0 ,1
0,1
0,1
so the linearisation is a stable node and by the linearisation theorem the equilibrium point
of the nonlinear system is an attractor and stable.
b g
Stability at the point 0, 1
2
0I
F
G
H2 2J
K
tre
Jb gj  0,dete
Jb gj  4
J ( 0, 1) 
0 ,1
0 ,1
so the linearisation is a saddle and by the linearisation theorem the equilibrium point of
the nonlinear system is a nonlinear saddle and unstable.
13
Phase portrait
Plot the phase diagram for the system. You will find the program for this at the end of the
parallel Maple example. Examine it carefully and check that it confirms the above
analysis.
2.2.5
Examples 1
Use the Linearisation Theorem to classify where possible the equilibrium points of the
following systems.
(a) x  y 2  3x  2
y  x 2  y 2
(b) x  y
y   x  x 3
(c) x  sin x  y
y  y
b g
(d)
x  x  y  e x
y  x  y  y 2
(e)
x  y
y   1  x 2  x 4 y  x
(f)
x  3y  xy  4
y  y  x
c
2
h
2
14