[Introduction | Background | Results | Conclusions]

Results

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

Steady States

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

Steady State Stability

The Jacobian matrix for the model has the following form after being evaluated at (0,0,0,0).

Jacobian matrix entries
0	1	0	0
-1	-c₁	0	3
0	0		-3/10
0	0	3/10

The eigenvalues are:

lambda +/- (3/10)*i, -1/2*c1 +/- 1/2*(root(c1^2 -4))

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

If > 0 and c₁ > 0, then (0,0,0,0) is a saddle.
If > 0 and c₁ < 0, then (0,0,0,0) is a source.
If < 0 and c₁ > 0, then (0,0,0,0) is a sink.
If < 0 and c₁ < 0, then (0,0,0,0) is a source.
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 lambda

= 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 (alpha) is constant.

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

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

Michelle Vadeboncoeur <mrvahs@wpi.edu>
Forest Lee-Elkin <yusuf@wpi.edu>