Consider any Hamiltonian system, not necessarily conservative. For any bounded regionin phase space, letbe the area ofand letbe the region of phase space obtained by taking a regionat timeand letting it evolve through a timecontains pointssuch thatinandwhereIf the system is conservative thenwhereis the flow of the system.

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An autonomous Hamiltonian system has a Hamiltonian function that is independent of time, This function takes a constant value along any phase curve. We can show this by differentiatingwith respect to

If the curve is a phase curve then Hamilton's equations are satisfied:

and

hence

Conversely if the system is conservative then the Hamiltonian is autonomous, since

Henceand the Hamiltonian is autonomous.

Sinceis conserved,a constant, defines a curve in phase space. It is enough, when drawing the phase diagram for a Hamiltonian system, to draw a selection of curvesfor various constantsThe differencefor any two phase curves will be constant throughout the phase plane.

Once the phase curve is drawn we can find the speed of the system on a phase curve by finding

If necessary a velocity field can be drawn using the velocity vector field.

]]>separation of variables gives

Integration now gives

Example: The equation of radioactive decay,is an autonomous system. Separation of variables and integration gives

Rearrangement gives

The original equation implies that the probability of decay does not depend when the decay takes place. This is a vital property, used in radioactive dating.

In general all the solutions of an autonomous system are relation. If two solution areand withthenmust satisfy

The above discussion can be extended to higher order systems in the natural way.

]]>{jatex options:inline}\dot{y}=G(x,y){/jatex}

This system may be very difficult or impossible to solve, but we can find the nature of the solution about a point {jatex options:inline}(x_0,y_0){/jatex} by linearising the system about the point {jatex options:inline}(x_0,y_0){/jatex}

{jatex options:inline}\dot{x}=\frac{\partial F}{\partial x}|_{(x_0, y_0)}(x-x_0) +\frac{\partial F}{\partial y}|_{(x_0, y_0)}(y-y_0) + higher \: order \: terms{/jatex}

{jatex options:inline}\dot{y}=\frac{\partial G}{\partial x}|_{(x_0, y_0)}(x-x_0) +\frac{\partial G}{\partial y}|_{(x_0, y_0)}(y-y_0) + higher \: order \: terms{/jatex}

We can define {jatex options:inline}x'=x-x_0, \: y'=y-y_0{/jatex} and the system becomes, in matrix form:

{jatex options:inline}\begin{pmatrix}\dot{x}'\\ \dot{y}'\end{pmatrix} = \left( \begin{array}{cc} \frac{\partial F}{\partial x}|_{(x_0, y_0)} & \frac{\partial F}{\partial y}|_{(x_0, y_0)} \\ \frac{\partial G}{\partial x}|_{(x_0, y_0)} & \frac{\partial G}{\partial y}|_{(x_0, y_0)} \end{array} \right) \begin{pmatrix}x'\\ y'\end{pmatrix} {/jatex}

We are especially interested in the point(s) {jatex options:inline}(x_0,y_0){/jatex} that satisfy {jatex options:inline}\dot{x}=\dot{y}=0{/jatex}. These are called critical points.

The linearisation matrix {jatex options:inline}\left( \begin{array}{cc} \frac{\partial F}{\partial x}|_{(x_0, y_0)} & \frac{\partial F}{\partial y}|_{(x_0, y_0)} \\ \frac{\partial G}{\partial x}|_{(x_0, y_0)} & \frac{\partial G}{\partial y}|_{(x_0, y_0)} \end{array} \right){/jatex} has eigenavlues {jatex options:inline}\lambda_1 , \: \lambda_2{/jatex} and corresponding eigenvectors {jatex options:inline}\mathbf{v}_1, \: \mathbf{v}_2{/jatex}.

If {jatex options:inline}\lambda_1 \: \lambda_2 \lt 0{/jatex} then the point {jatex options:inline}(x_0, y_0){/jatex} is stable. Any initial point on the eigenline (a line at starting at the point {jatex options:inline}(x_0, y_0){/jatex} in the direction of an eigenvector) will tend to the point {jatex options:inline}(x_0, y_0){/jatex} along that eigenline. Any initial point not on an eigenline will tend to the eigenline will the eigenline with the corresponding most negative eigenvalue, then towards the point {jatex options:inline}(x_0, y_0){/jatex}. This is a stable node.

If {jatex options:inline}\lambda_1 \: \lambda_2 \gt 0{/jatex} then the point {jatex options:inline}(x_0, y_0){/jatex} is unstable. Any initial point on the eigenline will tend to move away from the point {jatex options:inline}(x_0, y_0){/jatex} along that eigenline. Any initial point not on an eigenline will tend to the eigenline will the eigenline with the corresponding most positive eigenvalue, then away from the point {jatex options:inline}(x_0, y_0){/jatex}. This is an unstable node.

If {jatex options:inline}\lambda_1 \gt 0, \: \lambda_2 \,t 0{/jatex} then the point {jatex options:inline}(x_0, y_0){/jatex} is unstable. Any initial point not on the eigenline corresponding to {jatex options:inline}\lambda_2{/jatex} will tend to the point {jatex options:inline}(x_0, y_0){/jatex}.Any other initial point on the eigenline corresponding to {jatex options:inline}\lambda_1{/jatex} will tend to move away from the point {jatex options:inline}(x_0, y_0){/jatex} along that eigenline, or will tend to that eigenline away from the point {jatex options:inline}(x_0, y_0){/jatex}.This is a saddle point.

If the eigenvalues are complex {jatex options:inline}\lambda_1=a+bi, \: \lambda_2 =a-bi{/jatex} then if {jatex options:inline}a \lt 0{/jatex} the point {jatex options:inline}(x_0, y_0){/jatex} is stable (a stable spiral) and unstable (an unstable spiral)if {jatex options:inline}a \gt 0{/jatex}. If {jatex options:inline}a = 0{/jatex} then any point will move in a circle about the point {jatex options:inline}(x_0, y_0){/jatex}. The point {jatex options:inline}(x_0, y_0){/jatex} is a centre.

If there is only one eigenvalue {jatex options:inline}\lambda{/jatex} and only one eigenvector {jatex options:inline}\mathbf{v}{/jatex} then the point {jatex options:inline}(x_0,y_0){/jatex} is a stable degenerate node if {jatex options:inline}\lambda \lt 0{/jatex} and an unstable degenerate node if {jatex options:inline}\lambda \gt 0{/jatex}.

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This means that the nature of the fixed point is determined completely by the trace and determinant of the linearisation matrixas long asThis is illustrated below.

]]>These eigenvalues are

The eigenvectors corresponding to these eigenvectors will be complex when solved using

We can however transform coordinates. When we transform to polar coordinates, the system becomes much simpler. If the eigenvalues arethen define

Similarlyand differentiation gives

Hence

We obtain the simple equationsand

We can solve these to obtainand

Elimination ofbetween these two equations gives

Ifandorandthenincreases exponentially asincreases. Ifandhave the same sign thendecreases exponentially asincreases.

]]>finding the fixed points amounts to solving the simultaneous equationsIn general there may be none, one or many fixed points, each of which can be investigated individually to ascertain it's nature.

Example: Find the fixed points of the system

Putting the second of these equal to 0 and factorising gives

From this we deduceorSubstitutinginto the first, which we set equal to 0, gives

Henceandare fixed points of the system.

Substitutinginto the first and setting it equal to 0 gives

Henceandare fixed points of the system

Example: Determine the fixed points of the system

Putand the system becomes

Putting the second equal to 0 givesand substituting this into the second, which we set equal to 0 gives

The fixed point areand

]]>Example: Find a perturbation expansion for the solution to

Substituteto obtain

Grouping together coefficients offor eachgives

Then

Then

(1) can be solved exactly to giveand

The Taylor expansion of the first coincides with the perturbation expansion ifbut the second root isand is not present in the unperturbed equationso cannot be given by the perturbed equation. We can find a perturbation expansion for the other root by puttingto obtain the equationand assuming an expansion

Example: Find a perturbation expansion for the root nearof the equationup to the term in

Assume

Grouping powers ofgives

Then

]]>Example: Find a perturbation expansion for the solution to

Substituteto obtain

Grouping together coefficients offor eachgives

Then

Then

(1) can be solved exactly to giveand

The Taylor expansion of the first coincides with the perturbation expansion ifbut the second root isand is not present in the unperturbed equationso cannot be given by the perturbed equation. We can find a perturbation expansion for the other root by puttingto obtain the equationand assuming an expansion

Example: Find a perturbation expansion for the root nearof the equationup to the term in

Assume

Grouping powers ofgives

Then

]]>One example is shown above,This motion has three fixed points, at These points may be classified as stable, or unstable.

The functionmay be expanded in a Taylor series about a fixed pointto givewith solution

whereIfthentends tofor any particle with initial position sinceand iftends tosince

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