Results

The `xmaple` worksheet , or the postscript version.
The graphics page.

The only steady state solution is for x = 0, y = 0, z = 0, w = 0. (else refered to as (0,0,0,0).)

The Jacobian matrix for the model has the following form after being evaluated at (0,0,0,0).
 0 1 0 0 -1 -c1 0 3 0 0 -3/10 0 0 3/10

The eigenvalues are:

at (0,0,0,0), the following stability can be found:

1. If > 0 and c1 > 0, then (0,0,0,0) is a saddle.
2. If > 0 and c1 < 0, then (0,0,0,0) is a source.
3. If < 0 and c1 > 0, then (0,0,0,0) is a sink.
4. If < 0 and c1 < 0, then (0,0,0,0) is a source.
5. If = 0 and c1 = 0, we can say nothing about the stability of (0,0,0,0). Bifurcation theory is needed to analyse the system in this case.

Phase Portraits and Bifurcation diagram

`xrk` was used, along with `xmaple`, to discover the model's behavior. A bifurcation diagram is necessary to find the behavior of the point (0,0,0,0) for = 0 and c1 = 0. Please see the `xmaple` worksheet or the postscript version of the main work for this results section. Additional work is on the graphics page. Warning! Lots of large graphics! May take a while to load!

Relation to the van der Pol model

The van der Pol oscillator is similar to the linear pendulum equation

``` x" = -ax' -bx ```

(where the right hand side of the equation is the sum of two forces: the restoring force bx and the linear frictional force ax'), except that the frictional term is nonlinear in the van der Pol model. The van der Pol equation is

``` x" + ( + x2)*x' + x = 0 ```
where is constant.

The behavior of the x-y system (when plotted), with positive c1 and , acts much like the van der Pol forced oscillation model with < 0. (see case on graphics page)

`xmaple` worksheet on the van der Pol oscillator, and the postscript version