NONLINEAR SYSTEMS

INTRODUCTION

In an autonomous nonlinear system the differential equations contain nonlinear terms such as These dynamical systems have many applications including

the simple pendulum model given by

the predator prey model given by

the competing species model given by

When a plane autonomous system is linear of the form with determinant A0, there is only one equilibrium point which is at (0,0) and a precise geometrical description of solutions is possible, both near the equilibrium point and globally once the eigen values and eigen vectors of A have been found.

However, a nonlinear autonomous system may have many equilibrium points. Given an initial point we do not know which equilibrium point (if any) the solution curve will approach. We cannot even be sure that a solution curve through a point initially close to an equilibrium point will remain close to that equilibrium point. It may well be driven to a second equilibrium point .

Consider the competition model

and its plane diagram showing several solution curves.

The equilibrium points are (0,0),(0,100),(50,0) and (20,40). Click on each equilibrium point to see a phase diagram of the region close to the equilibrium point. Can you decide which equilibrium points are stable and which are unstable?

Hint Are the vector arrows pointing towards or away from the equilibrium point?

Ans. You should find that the equilibrium points (0,0) and (20,40) are unstable and the points (0,100)and (50,0) are stable.

How can we use this knowledge to help us understand the long term behaviour of the model?

If we look at the curve labelled A. Starting with small values of x and y, initially, as the curve moves away from the unstable equilibrium (0,0) towards the equilibrium (20,40) the values of x and y increase until y reaches a maximum value. The curve now comes under the influence of the unstable equilibrium point (20,40) and is repelled by it and driven towards the stable equilibrium point (50,0). Thus, x continues to increase whilst y decreases so that as but y dies out.

The biological interpretation is that the two populations are competing for a limited resource which initially is sufficient for both populations to increase. Eventually there is not enough resource to sustain both populations so y begins to decrease and becomes extinct leaving all the resources for x which then increases until its carrying capacity of 50 is reached.

Now look at the curves labelled B, C, and D and see if you can describe what is happening on each of these curves and interpret the trends.

In this model the initial conditions are crucial. Can you see why the starting point is so important?

This example suggests that a natural first step towards understanding a nonlinear system would be to classify the equilibrium points. In this unit you will learn

- How a nonlinear system can be approximated to a linear system near an equilibrium point by a process called linearisation.

that under certain conditions solutions of the nonlinear system close to an equilibrium point have the same geometric features as the solutions of its linearisation.

that there are certain cases when no conclusions can be drawn and further analysis is necessary.

TWO DIMENSIONAL SYSTEMS

This unit consists of seven chapters

prerequisites

introduction

the technique of linearisation

local stability and the classification of equilibrium points

global stability

summary and sample test papers

LINEARISATION

In this section you will learn to

understand the technique of linearisation

linearise a nonlinear system using a change of variable

linearise a nonlinear system using Taylor’s series.

LINEARISATION

The technique of linearisation involves approximating a complicated system of equations with a simpler linear system We hope to gain insight into the behaviour of the nonlinear system through an analysis of the behaviour of its linearisation. We hope that the nonlinear system will behave locally like its linearisation, at least in a qualitative sense.

In general, the linearisation of a system of equations about an equilibrium point can be achieved by changing variables so that the equilibrium point is transformed to the origin. Points in the original system close to the equilibrium point will correspond to points close to the origin in the new system. Thus we are only concerned with values of the new variables close to zero and under certain conditions the nonlinear terms can be neglected. The equations that result are linear and are the linearisation of the original system.

In the examples that you will meet the nonlinear terms will be polynomial and are small enough to be neglected.

Consider the nonlinear system

with equilibrium point (p,q).

The transformation u=x-p, v=y-q transforms the equilibrium point (p,q) to the origin

Differentiating gives and

Substituting x=u+p and y=v+q into the original equations gives

where (u,v) and (u,v) consist only of nonlinear terms.

Then the linear system

is said to be the linearisation of the nonlinear system provided that

These last conditions ensure that the nonlinear terms (u,v) and (u,v) are negligible in comparison to u and v as the equilibrium point is approached.

In the examples that you will meet the nonlinear terms (u,v) and (u,v) will be polynomials with all terms of degree two or higher.

Let

Then

<()

= 0

Therefore if consists of nonlinear polynomial terms it can be neglected for small values of u and v.

click here for a worked example

Consider the nonlinear system

Find the equilibrium points by putting

2 x + y = 0

y + x = 0

The solutions of these equations are x = 0 or 2 y = 0 or 4

Therefore the equilibrium points are (0,0) and (2,4).

For the equilibrium point (0,0) no transformation is necessary as it is already at the origin and the equations remain as

Since the nonlinear terms are polynomials they can be neglected

Hence the linearisation equations are

For the equilibrium point (2,4) use the substitution x = u + 2, y = v + 4 to move the equilibrium point to the origin.

The equations become

Neglecting the nonlinear terms since they are polynomial gives the linearisation

click here for examples for you to work

For each of the following systems a) find the equilibrium points

b) state the transformation necessary to move the equilibrium point to the origin

c) find the linearisation of the system

1) 2)

Ans. 1) a) (0,0) (1,0)

2. For (0,0) no substitution is required

For (1,0) substitute x = u +1, y = v

For (-1,0) substitute x = u –1, y = v

c)

At (0,0) linearisation is

At (1,0) linearisation is

2. a) (1,1), (1,-1), (2,2), (2,-2)

b) For (1,1) substitute x = u + 1, y = v + 1

For(1,-1) substitute x = u + 1, y = v - 1

For (2,2) substitute x = u + 2, y = v + 2

For (2,-2) substitute x = u + 2, y = v – 2

c) At (1,1) lineaisation is

At (1,-1) linearisation is

At (2,2) linearisation is

At (2,-2) linearisation is

LINEARISATION BY TAYLOR SERIES

Consider the nonlinear system

with equilibrium point (p,q).

Any function which is differentiable can be written as a Taylor series as follows

where F(x,y) consists of nonlinear polynomial terms in (x-p) and (y-q)

Therefore =

=

If we use the change of variable u=x-p, v=y-q this will transform the equilibrium point to the origin and differentiating gives and . Since (p,q) is an equilibrium point f(p,q) = g(p,q)= 0 and for points near to the equilibrium point (x-p) and (y-q) are small and the nonlinear terms and can be neglected. Substituting in the original equations gives

where

This can be written in the form where J= is called the Jacobian matrix.

click here for a worked example.

Consider the

The equilibrium points are given by

Solving these equations gives x=0 or 2, y=0 or 4

Therefore the equilibrium points are (0,0) and (2,4)

Find the Jacobian matrix

Therefore the linearisation at the point (0,0) is

Therefore the linearisation at (2,4) is

click here for examples for you to work

For the following examples find a) the equilibrium points b) the Jacobian matrix c) the linearisation at each equilibrium point

1), 2) ,

3) 4)

5) 6)

7) 8)

hint

Ans: Ans. 1) a) (0,0) (1,0)

b)

c

) At (0,0) linearisation is

At (1,0) linearisation is

2. a) (1,1), (1,-1), (2,2), (2,-2)

b)

c) At (1,1) lineaisation is

At (1,-1) linearisation is

At (2,2) linearisation is

At (2,-2) linearisation is

3) a) (0,-1)

3. Linearisation is

2. a) (0,0)

3. Linearisation is

5. a) ,

b)

2. Linearisation at is

Linearisation at is

5. a) (

) for all integer n

3. linearisation is if n is even,

if n is odd

5. a) (1,

) for all integer n

3. Linearisation is if n is even,

if n is odd

5. a) (0,0), (1,1), (-1,-1)

3. Linearisation at (0,0) is

Linearisation at (1,1) is

Linearisation at (-1,-1) is