Nonlinear analysis of the Opossum model

A model of opossums in New Zealand.  [Graphics:mmaimages/opossum_gr_1.gif]  is the opossum population, and [Graphics:mmaimages/opossum_gr_2.gif] the population infected by tuberculosis.
We'll take [Graphics:mmaimages/opossum_gr_4.gif] and [Graphics:mmaimages/opossum_gr_5.gif].

Find critical points

Critical points are where the derivatives are zero:


So the critical points are (0,0) and (8/3,2/3).

Linearize around critical points

At (0,0), the coefficient matrix of the linear system is


Similarly, at (8/3,2/3), we have


solving the linear systems

To solve the linear systems, we look for the eigenvalues and eigenvectors of the coefficient matrices.  For the first system, which is
these are


That is, the eigenvalues are -2 and 1/4, and the eigenvectors are [Graphics:mmaimages/opossum_gr_15.gif]and [Graphics:mmaimages/opossum_gr_16.gif]  A solution is therefore
Similarly, for the linearization at (8/3,2/3) we have


the eigenvalues [Graphics:mmaimages/opossum_gr_20.gif], with the second eigenvector [Graphics:mmaimages/opossum_gr_21.gif]  Solving for the real and imaginary parts of the resulting solution, we have
so that the general solution is

phase portraits

at (0,0)

At (0,0), the eigenvectors give the lines

(Let's make all of the arrows that we draw around the equilibrium points be of length 0.375:)


So that trajectories look like

at (8/3,2/3)

Here we have complex eigenvectors, so it doesn't make sense to talk about graphing them.  We know that we're going to get an inward spiral, however, because of the sines and cosines and the negative exponential multiplying them.  Let's go ahead and graph a solution of this:


The red point is the equilibrium point, and the black point is the initial condition.

the phase portrait

Now let's put this all together.  The red dots are the equilibrium points, and we have the two representative trajectories shown.


Cool!  Because we can, let's put a couple of trajectories found from the nonlinear system on top of these.


Note that the linear solution isn't quite right the farther away we get from the equilibrium points--this can be seen at the point (8/3,2/3).  However, it still gives a good picture of what's going on here.

Converted by Mathematica      March 27, 2002   (with substantial subsequent editing)