A model of opossums in New Zealand. is the opossum population, and the population infected by tuberculosis.
We'll take and .
Critical points are where the derivatives are zero:
So the critical points are (0,0) and (8/3,2/3).
At (0,0), the coefficient matrix of the linear system is
Similarly, at (8/3,2/3), we have
To solve the linear systems, we look for the eigenvalues and eigenvectors of the coefficient matrices. For the first system, which is
That is, the eigenvalues are -2 and 1/4, and the eigenvectors are and A solution is therefore
Similarly, for the linearization at (8/3,2/3) we have
the eigenvalues , with the second eigenvector Solving for the real and imaginary parts of the resulting solution, we have
so that the general solution is
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
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.
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.