Text in word format

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

LOCAL STABILITY AND THE CLASSIFICATION OF THE EQUILIBRIUM POINTS

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

with equilibrium point (p,q).

We have already seen that the system may be approximated in the neighbourhood of the

equilibrium point by the linear system |

Where J is the Jacobian matrix |

Look at the tr- det diagram for linear systems. The boundary lines for stability are tr=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 non-stability 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 tr=0 needs careful investigation.

INVESTIGATION 1

Consider again the competing species model

with equilibrium points (0,0), (0,100), (50,0) and (20,40)

Here is its phase plane diagram.

To calculate the Jacobian matrices

| =0.2-0.008x-0.003y |  | =-0.003x |  | =-0.003y |  | =0.1-0.002y-0.003x

therefore |  |

For det >0 and tr >0 so both eigenvalues have positive real parts and the

the linearisation shows an unstable equilibrium point

tr>4det so the eigenvalues are real and the linearisation is an unstable

node.

Click on the point (0,0) on the phase plane diagram to see solutions close to the point (0,0) for both the nonlinear system and its linearisation. Compare the two diagrams. Do the linearisation and the nonlinear system both show the same geometric behaviour close to the equilibrium point?

For det >0 and tr <0 so the eigenvalues both have negative real parts and the

linearisation shows a stable equilibrium point.

tr>4det so the eigenvalues are real and the linearisation is a stable node.

Click on the point (50,0) on the phase plane diagram to see solutions close to the point (50,0) for both the nonlinear system and its linearisation.. Compare the two diagrams. Do the linearisation and the nonlinear system both show the same geometric behaviour close to the equilibrium point?

For det <0 so the eigenvalues are real and have opposite signs and the

linearisation is an unstable saddle point.

Click on the point (20,40) on the phase plane diagram to see solutions close to the point (20,40) for both the linearisation and the nonlinear system. Compare the two diagrams. Do the linearisation and the nonlinear system both show the same geometric behaviour close to the equilibrium point?

For det >0 and tr <0 so the eigenvalues both have negative real parts and the

linearisation predicts a stable equilibrium point.

tr=4det so it is a repeated eigenvalue and a closer look at the Jacobian

matrix shows that the linearisation is an improper node.

Click on the point (0,100) on the phase plane diagram to see solutions close to the point (0,100) for both the linearisation and the nonlinear system. Compare the two diagrams. Do the

linearisation and the nonlinear system both show the same geometric behaviour close to the equilibrium point?

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).

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

1)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.

2)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.

3)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 tand the equilibrium point is an attractor and stable.

4)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.

5)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.

If we take another look at INVESTIGATION 1

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 attractor and stable.

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

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.

The linearisation at the point (0,100) is a 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.

This is in full agreement with the results we obtained by comparing the phase plane diagrams.

Here is an example for you to try based on the linearisation theorem.

Example

Consider the nonlinear system

.

1. find the equilibrium point

2. find the Jacobian matrix at the equilibrium point

3. find values for tr, det, and tr-4det. What does this tell you about the eigen values?

4. classify the equilibrium point of the linearisation. What does this tell you about the stability of the nonlinear system?

5. To verify your answer to (4) click here to see a phase plane diagram of the nonlinear system near its equilibrium point.

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 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.

Consider the nonlinear system

There is only one equilibrium point (0,0)

, , ,

The Jacobian matrix is given by

tr=0 det=2 so the eigen values are imaginary with real part zero. The linearisation is therefore a centre.

Click on (0,0) to see a phase plane diagram of solutions close to the equilibrium point for the nonlinear system. The diagram shows an unstable equilibrium point with the solution curve spiralling away from the equilibrium point (0,0). In this case we were unable to predict the geometric behaviour of the nonlinear system from its linearisation.

INVESTIGATION 3

To look at the global phase portrait of a nonlinar system.

Consider the nonlinear system

There is only one equilibrium point (0,0)

The Jacobian matrix is given by |

Since det=2 and tr=2 the equilibrium point of the linearisation is a stable focus. The linearisation theorem tells us that the nonlinear system shows the same geometric behaviour close to an equilibrium point. This can be verified by clicking on phase plot to see a phase plot of the nonlinear system for the range –0.1 and

The range can now be extended to –2<x<2 and –2<y<2 . Click on phase plot to see a phase plot of the nonlinear system for this larger range. You can see that the global portrait does not resemble that of a linear stable focus but in fact shows the existence of a closed trajectory. This is an example of a limit cycle which will be discussed in the next unit.

SUMMARY

1. If the linearisation is a stable node or stable focus the equilibrium point of the nonlinear system is an attractor and stable.

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

Find the equilibria of

and determine their stability

At the equilibria =0

=0

Therefore the equilibria are (0,0), (0,-1), (0,1)

Find the Jacobian matrices

At the point (0,0) tr = 1, det = 0, so the linearisation is nonsimple and so nothing can be deduced about the equilibrium point of the nonlinear system from its linearisation.

At the point (0,-1) tr = 0, det = -4 so the linearisation is a saddle and by the linearisation theorem the equilibrium point of the nonlinear system is a nonlinear saddle and unstable

At the point (0,1) tr = -4, det = 4, tr-4 det = 0 so the linearisation is a stable node and by the linearisation theorem the equilibrium point of the nonlinear system is an attractor and stable

Look carefully at the following diagram which is a phase portrait of the system and check that it confirms the above analysis.

Click here for examples for you to do.