In the phase portrait, you see that the solutions near the equilibrium (31.25,12.03) spiral around it. Can we explain this with the complex eigenvalues you found?

Consider the general form of the solution of  from the video:

The λ‘s are complex numbers and the solution contains factors . For exponential functions of complex numbers, we can use Euler’s formula

Take the first eigenvalue λ1, which is complex of the form λ1=a+bi. Then

( gives a similar expression.) The solution  is now an expression with the complex i in it many times. Of course,  denotes a vector with (shifted) population sizes, so it should be real. In this course, we will not go into the details of how you can rewrite the solution for  to a real expression, but just state the result:

When the eigenvalues of the Jacobian matrix are complex numbers: λ=a±bi,
the solutions for  contain factors  and .

Looking at these factors, we see that each component of  has oscillating parts: either , or  or both. So all terms oscillate with an angular frequency of b radians per day, where b is the imaginary part of the eigenvalues. The graphs of the solutions versus time oscillate up and down. In the phase portrait the trajectories will circle the origin.

Each component of  also contains a factor . If a<0 that factor decreases in time, so the oscillations will be damped and in phase space the trajectories will spiral in towards the origin. If a>0, the exponential factor will increase in time, making the oscillations larger and larger, and in the phase plane the trajectories will spiral outwards.

This type of behavior around an equilibrium point makes it a spiral point. When the real part a of the eigenvalue is negative, so a<0, all solutions that start near the origin will end up in the origin, so that spiral point is a stable equilibrium point. When a>0, solutions that start near the origin will spiral away from the origin, so then the origin is called an unstable spiral point.

An equilibrium point  is called a stable spiral point if the Jacobian matrix  has two complex eigenvalues  with negative real parts: a<0. A typical sketch of the solutions near a stable spiral point in the phase plane is given by

An equilibrium point  is called an unstable spiral point if the Jacobian matrix  has two complex eigenvalues  with positive real parts: a>0. A phase portrait of an unstable spiral point is given in the following graph.

When a=0, the solution does not spiral inwards nor outwards, so the solution will be periodic, and in phase space the trajectory is a closed curve. Even though that curve is most often an ellips, the equilibrium point is called a circle point.

An equilibrium point  is called a circle point if the Jacobian matrix  has two complex eigenvalues with zero real parts: .