2. 3 global phase portraits

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2.3 GLOBAL PHASE PORTRAITS

In this section you will investigate

• Conservative Systems

• Reversible Systems

• Lyapunov Functions

• methods of determining the existence of a nonlinear centre

The linearisation theorem tells us that in many cases the solutions of a nonlinear system near an equilibrium point mimic the solutions of the linearisation of the system at that point. However, the result does not apply if the linearisation is non simple or shows a centre. Furthermore if it does apply it tells us only about local behaviour. To make further progress with the analysis of nonlinear systems we need to look at a global analysis. There are certain special systems and methods which we are going to study. These include Conservative Systems, Reversible Systems and Lyapunov Functions.

1. Conservative Systems

The concept of a conservative dynamical system is important in real world applications. Conservative systems are ones in which a quantity is conserved in the sense that it is constant on any trajectory of the system. Often the conserved quantity has a physical meaning such as the total energy in the case of a system arising in classical mechanics or spin in the case of quantum mechanics. The following definition provides a formal statement of the concept.
If for the system

there exists a function E(x,y) such that its partial derivatives are continuous and satisfies the following conditions

• is constant on a trajectory of the system i.e

Therefore

for all (x,y)

• is non-constant across trajectories.

then E(x,y) is a conserved quantity and the system is conservative.

A method for verifying that a function is a conservative quantity is illustrated in the following example. There is a parallel example on the web for those who prefer to use Maple.

• Worked Example 1

Verify that

is a conserved quantity for the system

If is a conserved quantity then it must satisfy
for all (x,y)

Also

.
Therefore is non-constant across trajectories.

Therefore is a conserved quantity for the system.

Finding Conserved Quantities using First Integrals

It is difficult to find a conserved quantity . One method of doing this is to find a First Integral of the system. First Integrals are conserved quantities provided they exist for all in the plane. In the problems that you will meet can be found by finding a First Integral as in the example below. Again there is a parallel example on the web worked in Maple.

• Worked Example 2

Find a first integral for the dynamical systems

(1) (2)

and decide whether they are conservative systems.

(1) Dividing gives

Thus is a first integral for all in the plane and hence the system is conservative.
(2) Dividing gives

This function is obviously not defined for . Thus the first integral is not valid in the whole plane and the system is not conservative.

Finding conserved quantities using the total energy.

A second method of finding a conserved quantity is to look at the total energy. In many physical systems the total energy is conserved and kinetic and potential energy are interchanged during the motion of the system. This method is illustrated by considering the motion of a particle oscillating on the end of an elastic spring.

• Worked Example 3

Consider a particle of mass m kg on the end of a spring of natural length l cm and elasticity constant . Let the extension in the equilibrium position be e cm and the height of the particle above the ground be h cm.

Since the system is in equilibrium the total force is zero and

If the particle is now extended a distance x cm below the equilibrium point the system is no longer in equilibrium and using Newton's equation

force=mass acceleration

gives

Substituting and
The second order equation can be written as a system of equations

The total energy for the system is the sum of the P.E. (potential energy) and K.E. (kinetic energy). This total energy remains constant throughout the motion.
The kinetic energy of the system
The potential energy in the spring
The potential energy of the mass

The total energy is given by

where k is a constant.
Therefore

is a conserved quantity.
To prove is a conserved quantity

If is a conserved quantity then it must satisfy

for all (x,y)

Therefore

for all .

therefore

is a conserved quantity for the system.
Examples 1

1. Find the first integrals of the following systems together with their domains of definition. Hence decide which are conservative systems.

Phase Portrait of a Conservative System

How can a conserved quantity help you to understand the phase portrait of a nonlinear system?

Investigation 1

The following pair of equations are a nonlinear system which model a predator-prey interaction.

You are going to find a conserved quantity for the system. You will then compare the global phase portrait of the nonlinear system with the diagram of the level curves of the conserved quantity. The material for this investigation is on the web.
It can be seen that the two diagrams show the same geometric features. Both show the point (80,50) is a stable equilibrium surrounded nearby by periodic orbits.
This example illustrates the fact that if we can find a conserved quantity for a nonlinear system then the solution curves of the nonlinear system are the same as the level curves of the conserved quantity.
This follows from the fact that if then is a constant with respect to time. Consider the point at time . As t increases (or decreases) the point must move so that remains a constant. Thus the point is constrained to move on the level curve of .
Since the nolinear system was a model for a predator-prey interaction the result can be

interpreted in terms of the species. If both species are present at the outset then this

model predicts periodic fluctuations of both species about an equilibrium population.

Stability of equilibria in a Conservative System.

If an equilibrium point of a conservative system also satisfies

then it is stationary point for the function .

If also the eigenvalues of the Hessian matrix

• are both positive then the stationary of is a local minimum.

• are both negative then the stationary point of is a local maximum.

• are one positive and one negative then the stationary point is a local saddle.

If the stationary point of is a local minimum or maximum then the level curves of are closed in the neighbourhood of the equilibrium point. Thus the solutions of the nonlinear system also form closed curves surrounding the equilibrium point and the equilibrium point is a nonlinear centre and neutrally stable.This gives a method for determining the existence of a nonlinear centre.

If the equilibrium point is a local saddle for then the equilibrium point is unstable and a nonlinear saddle.
Use of conservative systems for determining the existence of a nonlinear centre.

One of the limitations of the linearisation theorem is its inability to prove the existence of a nonlinear centre. If the equilibrium point of the linearisation is a centre it does not follow that the equilibrium point of the nonlinear system is also a centre and further investigation is necessary. You can now do this further analysis for a conservative system. The method is illustrated in the following example. There is a parallel example on the web for those who prefer to use Maple.

• Worked Example (4)

Consider the predator-prey system

If x represents the predator then if its prey y is present it will grow but if its prey y is absent then it has no food and it will die out. This is modelled by the first equation. If the predator x is absent then the prey will increase exponentially but if the predator x is present then there are limits on the growth of the prey. This is modelled by the second equation and together the two equations form a nonlinear system for modelling a predator prey interaction.
To determine the long term behaviour of the system it will be necessary to examine its phase portrait.
To find the equilibrium points

The equilibrium points are given by

The solutions of the equations are

The equilibrium points are therefore and .

To classify the equilibrium points of the linearisations

The Jacobian matrix is given by

At the point the Jacobian matrix is

,
The equilibrium point of the linearisation is unstable and a saddle.

By the Hartman Grobman linearisation theorem the equilibrium point of the nonlinear system is unstable and a nonlinear saddle.

Thus the point (0,0) is unstable and a nonlinear saddle.
At the point (80,50) the Jacobian matrix is

The equilibrium point of the linearisation is a centre and thus we cannot determine the stability of the point from its linearisation.

To investigate this equilibrium point further we will look for a conserved quantity for the system.

To find a conserved quantity for the nonlinear system.

Look for a conserved quantity by finding a first integral

Dividing g by f gives

Integrating gives –0.1 +0.002y=-0.0025x+0.02lnx+C

0.002y+0.0025+C

This can be written in the form where

The function is a conserved quantity as it is constant along every solution and the system is conservative.

To determine if the point is a stationary point of .

At the point

At the point .

Therefore the equilibrium point (80,50) is a stationary value of
To determine if the stationary point is a maximum or minimum

The Hessian matrix is given by

The eigenvalues of H are –0.00008224 and –0.00010527. Since these are both negative the stationary point is a maximum. Thus the equilibrium point is neutrally stable and a nonlinear centre and the phase portrait will show closed curves surrounding the equilibrium point (80,50). The graphs of and show periodic ocsillations about the equilibrium point.
Phase portrait of the nonlinear system.

Draw the phase portrait of the nonlinear system. You will find the commands for this at the end of parallel example on the web.

Study the phase portrait and check that it agrees with the results obtained in the last example.

This agrees with the phase portrait already drawn.

Examples 2

1 a) Show

is a conservative function for the system

1. Find all the equilibrium points and show that all the equilibrium points are

stationary points for .

1. Find the Hessian matrix for and its eigenvalues at each equilibrium point. Hence classify the equilibrium points.

2. Plot a phase portrait for the system and check that youre answers to part © are correct.

3. Repeat question 1 for

and the nonlinear system

3 a) Show the equilibrium points of the nonlinear system

are .

1. Using the Linearisation Theorem show that one equilibrium point is a nonlinear saddle but the other point needs further investigation.

2. By finding a First Integral find a conserved quantity for the system.

3. By finding the Hessian matrix show the unclassified equilibrium point is a minimum stationary point for the conserved quantity and is, therefore a nonlinear centre.

4. Plot a phase portrait for the system.

Note: the existence of the nonlinear saddle could also have been proved using the Hessian matrix. Check this yourself.

1. Repeat question (3) for the system

2.3.2 Reversible Systems

Many dynamical systems have time-reversed symmetry in the sense that their behaviour is independent of the direction of time. For example a film of a swinging undamped pendulum would appear be the same whether played forwards or backwards.

Definition A plane dynamical system

is said to be reversible if the equations are invariant under the change of variables , and (or , and ).
Notice that in many dynamical systems x represents the displacement and y represents a velocity i.e. . Hence changing the sign of t also changes the sign of y but not of x.
To prove a dynamical system is reversible.

A method for proving a dynamical system reversible will be illustrated in the following example.

• Worked example 5

Consider the motion of a simple pendulum modelled by the equation

where x is the angle the string makes with the downward vertical, g is the acceleration due to gravity and l is the length of the string.

x

The second order differential equation can be reduced to a nonlinear system by using the following substitution.

Let , then and the equation can be represented by the system

To prove the system reversible.
Making the substitutions

the equations become

and

which are identical to the original equations.

The system is therefore reversible.

Phase portrait of a reversible system

Reversible systems have the property that if {x(t),y(t)}is a solution of the system so is {x(-t),y(-t)}. This means that every trajectory lying above the x-axis (or right of the y axis) must have a twin obtained by reflection in the x-axis (or y-axis) differing in time direction i.e the phase portrait is symmetrical about either the x-axis or the y-axis.

Stability of equilibria in a reversible system.

How can the existence of a reversible system help in the understanding of the long term behaviour of a nonlinear system?

Consider a trajectory of a reversible system that starts on the x-axis to the right of the equilibrium point. This trajectory will swirl round the equilibrium point and eventually intersect the axis to the left of the equilibrium point.

Now by reversibility, there must be a corresponding trajectory obtained by reflection in the x-axis with the same end points but reversed in time direction.

Together the two trajectories form a closed orbit. Hence the equilibrium point is a centre.

Likewise a trajectory which starts on the y-axis above the equilibrium point will swirl round the equilibrium point and eventually intersect the y-axis at a point below the equilibrium point. By reversibility there must be a corresponding trajectory obtained by reflection in the y-axis. Together the two trajectories form a closed orbit and the equilibrium point must be a centre.

Theorem.

If a reversible system has an equilibrium point which has a linearisation with a centre, the non-linear system must have a non-linear centre.

Thus a reversible system can be useful in proving the existence of a nonlinear centre. The method is illustrated in the example below and in a parallel example using Maple.

• Worked example 6

Analyse the behaviour of the dynamical system given by

To find the equilibrium points

Therefore

therefore or

The equilibrium points are and .

To classify the equilibrium points of the linearisations.

The Jacobian matrix of the linearisation is given by

At the point (0,0)

The equilibrium point of the linearisation is unstable and a saddle. By the Hartman Grobman linearisation theorem the equilibrium point of the nonlinear system exhibits the same geometric behaviour and is therefore unstable and a nonlinear saddle.
At the point (1,0)

The equilibrium point of the linearisation is a centre. Thus the linearisation theorem breaks down and it is necessary to do a further analysis to determine the stability of the equilibrium point of the nonlinear system.

This further analysis can be done by finding a conserved quantity or by proving the nonlinear system reversible. In this example we are going to prove the system reversible (compare this with worked example 4 where we found a conserved quantity).

To prove the system is reversible

Making the substitutions

and

and
The original equations become

and

which are clearly identical to the original system.

Therefore the system is reversible

Thus by the reversibility theorem since the equilibrium point of the linearisation is a centre the equilibrium point of the nonlinear system is neutrally stable and a nonlinear centre.
Thus the equilibrium point are is unstable and a nonlinear saddle and the point (1,0) is neutrally stable and a nonlinear centre.
Phase portrait of the nonlinear system

Look at the phase portrait of the system. If necessary you will find the commands at the end of the Maple version of this example.

Examples 3.

1 Prove the following systems are reversible and use Maple to plot the phase portrait.

a)

b)

c)
2 Consider the system defined by

where f is an even and both f and g are differentiable. Show that

1. the system is invariant under time reversal symmetry ;

2. the equilibrium points cannot be nodes or foci.

3. Illustrate the results by considering the systems

3 For the system

1. find the equilibrium point, linearise the system at the equilibrium point and

show that the equilibrium point of the linearisation is a centre.

1. Prove the system is reversible and hence classify the equilibrium point of the nonlinear system.

4 Repeat question (3) for the system

1. Consider the system

1. Write the system as a pair of nonlinear differential equations.

2. Show the system has an infinity of equilibrium points at .

3. By using the Linearisation theorem and proving the system reversible show that the equilibrium points are alternately nonlinear saddles and nonlinear centres.

2.3.3. Lyapunov Functions

Lyapunov Functions are named after the Russian mathematician Alexander Lyapunov (1857-1918). For a two dimensional system a Lyapunov Function has the following definition.

Definition A function V(x,y) is a Lyapunov function for a system

provided that

1. is positive definite

i.e.

b) along each solution of the system

i.e.

for every solution (x(t),y(t)) that is not an equilibrium point with strict inequality except for a discrete set of t’s.

This definition can be extended to three dimensional systems.

A method of verifying a Lyapunov Function is illustrated in the example below. There is a parallel example in Maple on the web.

• Worked example 7

Verify that

is a Lyapunov function for the system

for all(x,y) and only 0 at the point (0,0).

Therefore is a decreasing function and hence a Lyapunov Function for the system.
Examples 4

1. Verify that

is a Lyapunov function for the system

1. Repeat question (1) for

and the system

Stability of equilibria using Lyapunov functions.

How can a Lyapunov function help in the understanding of the stability of a nonlinear system?

Investigation2

You are going to draw the phase portrait of a nonlinear system and the level curves of a Lyapunov function for that system to see if you can see a connection between them.

The material for this investigation is on the web.
This investigation shows that the solution curves of the nonlinear system all cross the level curves of the Lyapunov system.

As before if the point is an isolated equilibrium point of the system and is a stationary point of then the level curves of are closed curves surrounding .

Consider a solution curve passing through a point on the curve .

Since is a decreasing function then as t increases, providing is not an equilibrium point of the system, the solution curve must move across the level curves of in a direction in which decreases.

If, as in the investigation, the equilibrium point is also a. local minimum of then the solution of the system must tend to as . The equilibrium point is therefore an attractor and therefore stable.
If the equilibrium point is also a local maximum of then the solution of the system must move away from as and the equilibrium point is a repeller and therefore unstable.
If the equilibrium point is also a saddle point of then the phase portrait of the system also resembles a saddle with the solution curves again crossing the level curves of as shown in the diagram.
Thus , if

is a system of equations and is a Lyapunov function for the system that has continuous first and second partial derivatives. Assume that the only way for which can remain zero along a solution for any length of time is if the solution is an equilibrium. Then if is an isolated equilibrium point and

• If (p,q) is a strict local minimum of V then is an attractor and is stable.

• If (p,q) is a strict local maximum of V then is a repeller and is unstable.

• If behaves like a saddle point of V then the equilibrium point is unstable but not a repeller.

Worked example 8

Verify that

is a Lyapunov function for the system

Hence find and classify the equilibrium point of the system

To prove is a Lyapunov Function.

First prove positive definite

for all (x,y) except x=y=0

and

Therefore is positive definite.

Secondly prove

for all x and y and equals 0 only at the point (0,0).
Therefore is a Lyapunov function for the system.
To find the equilibium points of the system.

At the equilibrium points

=0

=0
Therefore, the equilibrium point is (0,0)

To prove is also a stationary point of .

(0,0) is a stationary point for if

which is 0 at the point (0,0).

which is 0 at the point (0,0).

Therefore (0,0) is a stationary point for .
To determine the nature of the stationary point.

The type of stationary point is determined by the eigenvalues of the Hessian matrix.

The Hessian matrix is given by

and so both eigenvalues are positive and the stationary point is a minimum of .

Thus the equilibrium point is an attractor and is stable.

Draw a phase portrait of the system. This can be done by using the commands at the end of the Maple version of this example.
• Examples 5

1. For each of the following systems

2. show

is a Lyapunov function for the system.

1. show is an equilibrium point of the system.

2. show is a stationary point for

3. determine the type of stationary point by looking at the eigenvalues of the Hessian matrix

4. determine the stability of the equilibrium point.

A)

B)

C)

1. Repeat question (1) for

and the following systems

a)

b)

1. Show that the system

has no closed orbits by construsting a Lyapunov function

with suitable a and b.

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