The Van der Pol Oscillator:  PHY 320 class notes.

We begin with dimensionless equation of motion and use the "stability matrix" method of finding the eigenvalues of the linearization about the (0,0) fixed point.  Here's the "vdot" equation (I use the explicit time dependence - "x[t]" - because I'll be solving theequation of motion numerically later.

Classify  the fixed point

To find the motion near the (0,0) fixed point, we evaluate this at x->0, :

To classify the fixed point we need to find the eigenvalues:

The eigenvalues are complex roots for (implying a focus or spiral fixed point) and real for (implying a node fixed point).  Note if the eigenvalues are pure imaginary, consistent with this special case, which reduces to a simple harmonic oscillator.  Thus, we can classify the fixed point based on the parameter ε :

    0 < ε < 2:    Unstable focus  (Re() > 0)
    -2 < ε < 0:      Stable focus  (Re() < 0)
    ε > 2:        Unstable node ( > 0)
    ε < -2:        Stable node  ( < 0)
    ε = 2:        Unstable star (Im() = 0, Re() > 0)
    ε = -2:        Stable star   (Im() = 0, Re() < 0)
    ε = 0:        Elliptic  (Re() = 0)

Plot the phase portrait

Unstable Focus at the origin

Now consider the phase portrait for ε = 0.5,which makes both eigenvalues complex with positive real part; i.e. the linear beahvior should be an unstable focus, spiralling outward from the origin. However,we also see that the equation of motion has a damping term when x > 1 so, far from the fixed point, orbits should head toward the origin.  What happens in between cannot be predicted strictly from the linearized equations, but we can calculate it numerically.

Begin by defining the "vdot" for this special case, then solve the equations of motion numerically using the Mathematica "NDSolve" command.  Use a small initial value for and zero for .

Plotting the solution shows that our linearized prediction is correct:  the orbit spirals out from the origin, but then seems to get "caught" on a closed curve away from the fixed point:

This closed curve represents a periodic, oscillatory orbit.  What happens if we start the orbit outside this periodic orbit?

It looks like this orbit heads inward and gets caught by the same periodic orbit.  Put the 2 orbits together on the same plot:

The periodic, oscillatory orbit is called a limit cycle.  This limit cycle is an attracting, or stable, limit cycle because all orbits end up on it.  It is an intrinsically nonlinear phenomenon which occurs in many nonlinear oscillators.

Unstable Node at the origin

Now let's look at a case which has a node at the origin; e.g. ε = 3.  Here the eigenvalues are both real and positive, so we expect an unstable node flowing into a limit cycle attractor:

Since we expect a node, it's important to get the behavior near the origin;  to get the part of the orbit near the origin, intergate backwards in time:

You get the idea: Now try a few more initial conditions, then put them all togehter on one plot.  I won't save each individual plot, but you can see them all in the combined plot below.

             & nbsp;         The Van der Pol "bird" limit cycle.

You can see the node at the fixed point with its asymptotic direction as the orbits leave the origin.  But as soon as the orbits get away from the linear region, they head for the limit cycle.  I haven't shown orbits beginning outside the limit cycle, but they are also attracted to it.

We still haven't explored the situation when there are stable fixed points at the origin, i.e. when orbits are attracted toward the origin so they can't go toward an attracting limit cycle.  Try it and see what happens!