The Van der Pol Oscillator is governed by a 2nd order ODE with nonlinear damping. The 'position' of the oscillator is the solution to
$$x''(t) = \mu (1 - x^2(t)) x'(t) - x(t)$$
Here $\mu$ controls how much nonlinear damping there is. When $\mu = 0$ the ODE reduces to
$$x''(t) = - x(t)$$
which is simply the equation of the simple harmonic oscillator ('a spring') with spring constant $k = 1$.
It's well-known that the Van der Pol oscillator has a non-trivial limit cycle for most initial conditions in the $x - x'$ plane. However, this is actually surprising to me!
I am a little surprised because the damped harmonic oscillator governed by
$$x''(t) = - \mu x'(t) - x(t)$$
(with $\mu > 0$, I think)
doesn't have a limit cycle. Instead, the oscillator is damped until $x, x' = 0, 0$ in the limit as $t \rightarrow \infty$. In elementary physics textbooks people like to explain this as the oscillator losing energy due to friction.
This seems to imply that in the limit as $t \rightarrow \infty$, the damped harmonic oscillator 'loses all its energy'.
On the other hand, since the Van der Pol oscillator has a non-trivial limit cycle, with non-zero speed, does that mean that the oscillator has some non-zero amount of (kinetic?) energy in its steady state?
How is the Van der Pol oscillator different from the damped harmonic oscillator?
How can/should I quantify the energy in these oscillators?
Context: I've been playing around with simulations of oscillators in using numpy/scipy.