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In this section you will investigate

- Conservative Systems

- Reversible Systems

- Lyapunov Functions

- methods of determining the existence of a centre

GLOBAL PHASE PORTRAITS

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.

Definition If for the system

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

1) E(x,y) is constant on a trajectory of the system

ie |  |  | for all

(x,y)

2. E(x,y) is non-constant across trajectories.

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

Worked example (1)

Verify that is a conserved quantity for the system

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

for all (x,y)

Also E(2,3) =5, E(2,1)= -3 Therefore E(x,y) is non-constant across trajectories.

Therefore E(x,y) is a conserved quantity for the system.

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1. Finding conserved quantities

It is difficult to find a conserved quantity E(x,y), One method of doing this is to find a First Integral of the system. In the problems that you will meet E(x,y) can be found by finding a First Integral as in the example below. First integrals are conserved quantities provided they exist for all (x,y) in the plane.

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 (x,y) in the plane and hence the system is conservative.

2) Dividing gives |

This function is obviously not defined for x=0. Thus the first integral is not valid in the whole plane and the system is not conservative.

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 on the end of a spring of natural length l and elasticity constant . Let the extension in the equilibrium position be e.

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

If the particle is now extended a distance x below the equilibrium point the system is no longer in equilibrium and using Newton,s equation force=massacceleration gives

…………………(1)

Substituting then

Therefore the 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) of the system remains constant throughout the motion.

The total energy of the system is obtained by integrating equation (1)

Total energy = |

Therefore E(x,y) = is a conserved quantity

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

for all (x,y)

for all x and y

therefore E(x,y) = |  | is a conserved quantity for the system.

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!.2 Phase Portrait of a Conservative System

If we can find a conserved quantity E(x,y) for a nonlinear system then the solution curves of the nonlinear system are the same as the level curves of the conserved quantity. This result is illustrated in the example below.

Worked Example (4)

Consider the predator-prey system | ………………..1

………………2

We look for a conservative quantity by finding a first integral

Dividing 2 by 1 gives |

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

0.002y+0.0025+C

This can be written in the form E(x,y)=C where E(x,y)=

The function E(x,y) is a conserved quantity as it is constant along every solution and the system is conservative.

Below you will see two diagrams. One shows the phase plane of the system ………………..1

………………2

the other shows the curves for various values of C

Compare the two diagrams.

phase portrait of the nonlinear system | Curves of E(x,y)=C

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 E(x,y) for a nonlinear system then the solution curves of the nonlinear system are the same as the level curves of the conserved quantity.

Biologically if both species are present at the outset then this model predicts periodic fluctuations about an equilibrium population.

!.3 Stability of equilibria in a Conservative System.

If the equilibria also satisfy |  | and |  | then they are stationary points for the function E(x,y).

If E(x,y) has a strict local minimum (or maximum) at an equilibrium point (p,q) then the level curves E(x,y)=C are closed curves in the region of the equilibrium point.

Since the solutions of the nonlinear system must stay on the level curves of E the phase portrait close to the equilibrium point will consist of closed curves surrounding (p,q) representing a neutrally stable equilibrium. The solutions for x(t) and y(t) will show periodic oscillations about the equilibrium. This gives a method of determining the existence of a neutrally stable equilibrium point when the equilibrium point of the linearisation is a centre.

We can determine if E(p,q) is a minimum (or maximum) by finding the eigenvalues of the Hessian matrix

H= |  | where |  | etc

If the eigenvalues are both positive the stationary point is a minimum

If the eigenvalues are both negative the stationary point is a maximum?????

If the eigenvalues are of opposite sign the equilibrium point behaves like a saddle ???

Thus , if we have a Conservative non-linear system with a conserved quantity E(x,y) then

1. the solutions of the nonlinear system lie along the level curves of E

2. If an equilibrium point is a local maximum (or minimum) of E(x,y) then the level curves of E(x,y) are closed in the neighbourhood of the equilibrium point. This implies that the equilibrium point is neutrally stable. ?????

3. If the equilibrium point is a saddle for E(x,y) then the equilibrium point is unstable but not a repeller?????

This gives a method of finding a nonlinear centre. It will be illustrated in the example below.

Returning to Worked Example (4)

You are going to do the first part of this example yourselves

1. Find the equilibrium points

2. Find the Jacobian matrix at each equilibrium point

3. Find tr, and det for each Jacobian matrix

4. Classify the equilibrium points of the linearisations

5. Using the linearisation theorem what can you deduce about the nonlinear system?

Since the linearisation of the equilibrium point (80,50) is a centre we need to look for another method to determine whether it is stable or unstable

6)To determine whether the point (80,50) is a stationary point

=0 =0

Therefore the equilibrium point (80,50) is a stationary value of E(x,y)

7)To determine if the stationary point is a maximum or minimum

=-0.00008224

-0.00010527

=0

The eigenvalues are –0.00008228 and –0.00010527. Since these are both negative the stationary point is a maximum. Thus the equilibrium point is neutrally stable and the phase portrait will show closed curves surrounding the equilibrium point (80,50) and ocsillating about it periodically. This agrees with the phase portrait already drawn.

Worked example (5)

Investigate the behaviour of the dynamical system

1. find the equilibrium points

The equilibrium points are given by

x=0 or 1 y=0

The equilibrium points are (0,0) and (1,0)

2)Find the Jacobian matrices

2. To classify the equilibrium points of the linearisations

For tr=0 det =-1 The linearisation shows a saddle. Therefore, the phase portrait of the nonlinear system shows the same geometric features close to the point (0,0) and so the equilibrium point (0,0) is unstable.

For tr=0 det=1 The linearisation shows a centre. Therefore nothing can be deduced about the stability of the equilibrium point of the nonlinear system.

3. To look for a conserved quantity for the nonlinear system

This function E(x,y)= is defined for all (x,y) and is therefore, a conserved function.

Thus the solutions of the system are the same as the level curves E(x,y)=C

4. To see if the equilibrium point is a maximum or minimum of E(x,y)

At the point (1,0) =0 =0

Therefore (1,0) is a stationary point of E(x,y)

Exx=-1+2x=1 Exy=0 Eyy=1

The Hessian matrix H=

The eignvalues are both 1 and since these are both positive the stationary point is a minimum and the equilibrium point is neutrally stable and solutions close to this equilibrium point will be periodic.

The level curves of the function E(x,y) are shown below.

Notice that one trajectory that starts from the origin and returns to it divides the closed, periodic, trajectories from the non-periodic ones. A trajectory such as this which starts and ends at a single point is called a homoclinic orbit. This trajectory is not periodic since it takes an infinite time to return to the equilibrium point.

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1. 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 to 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 ).

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.

Worked example 6

Consider the motion of a simple pendulum

where is the angle from the downward vertical, g is the acceleration due to gravity and l is the length of the pendulum.

Let then the equation can be represented by the system

Substituting

which are identical to the original equations.

The system is therefore reversible.

2.1 Stability of equilibria in a reversible system

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

.

Consider a trajectory that starts on the positive x-axis (or y axis) near the origin. The flow swirls around the origin due to the dominant centre and eventually intersect the negative x-axis (or y axis).

Now by reversibility, there must be a corresponding trajectory obtained by reflection in the x axis (or y 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 non linear centre.

This method of proving the existence of a non linear centre is illustrated below.

Worked example 7

Investigate the dynamical system

1) to find the equilibrium points

The system has two equilibrium points (0,0) and (1,0) (See worked example 4)

2. to classify the equilibrium points of their linearisations

The linearisation at the point (0,0) is a saddle and at the point (1,0) is a centre.

(See worked example 4)

3. to prove (1,0) is a nonlinear centre and neutrally stable

First we need to prove the system reversible

Making the substitutions , and and

The equations become

which are clearly identical to the original system.

Therefore the system is reversible

Therefore the phase portrait must show closed curves round the equilibrium point (1,0) and the equilibrium point is neutrally stable. This is confirmed by the phase portrait in example 5

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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. This can be extended to three dimensions.

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

provided that along each solution of the system

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

Lyapunov Functions are illustrated in the example below

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 L(x,y) is a decreasing function and hence a Lyapunov Function for the system.

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3.1 Stability of equilibria using Lyapunov functions

As before if the point (p,q) is an isolated equilibrium point of the system and (p,q) is a stationary point of L(x,y) then the level curves of L(x,y) =C are closed curves surrounding (p,q). Since L(x,y) is a decreasing function then as t increases a non equilibrium solution of the system must move across the level curves in a direction in which L(x,y) decreases.

If the equilibrium point is also a. local minimum of L(x,y)=C then the solution of the system must tend to (p,q) as . The equilibrium point (p,q) is therefore an attractor and therefore stable. This is shown in the diagram below. Note that the solution curves cross the level curves of L.

If the equilibrium point is also a local maximum of L(x,y)=C then the solution of the system must move away from (p,q) as and the equilibrium point is a repeller and therefore unstable. This is shown in the diagram below. Again note that the solution curves cross the level curves of L.

If the equilibrium point is also a saddle point of L(x,y)=C then the phase portrait of the system also resembles a saddle with the solution curves again crossing the level curves of L as shown in the diagram.

Thus , if

is a system of equations and L(x,y) 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 (p,q) is an isolated equilibrium point

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

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

3. If (p,q) behaves is a saddle point of L then the equilibrium point is unstable but not a repeller.???????

Worked example 9

1) Verify that is a Lyapunov function for the system

2) Hence find and classify the equilibrium point of the system

1)

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

Therefore L(x,y) is a Lyapunov function for the system.

2) The equilibium points of the system are given by

=0

=0

Therefore, the equilibrium point is (0,0)

(0,0) is a stationary point for L(x,y) if and

at the point (0,0)

at the point (0,0)

Therefore (0,0) is a stationary point for L(x,y)

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

tr=20, det=64 and so both eigenvalues are positive and the stationary point is a minimum of L(x,y).

Thus the equilibrium point is an attractor and is stable.

A phase portrait of the system is shown below.

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