Is energy conserved in a Van der Pol oscillator? 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.
 A: $\def\half{{\textstyle {1 \over 2}}}$
Multiply your first equation by $x'(t)$ and integrate between $t_1$ and $t_2$:
$$x' x'' = \mu\,(1 - x^2)\,{x'}^2 - x\,x'$$
$$\eqalign{
&\half\,{x'}^2(t_2) - \half\,{x'}^2(t_1) = {}\cr
&\qquad -\half\,x^2(t_2) + \half\,x^2(t_1) +
\mu\!\int_{t_1}^{t_2}\! [1 - x^2(t')]\,{x'}^2(t')\>dt'.\cr}$$
Reordering
$$\eqalign{
&\Bigl[\half\,{x'}^2(t_2) + \half x^2(t_2)\Bigr] - 
\Bigl[\half\,{x'}^2(t_1) + \half\,x^2(t_1)\Bigr] = {}\cr 
&\qquad\qquad \mu\!\int_{t_1}^{t_2}\![1 - x^2(t')]\,{x'}^2(t')\>dt'.\cr}$$
At left side you read variation of kinetic energy plus potential energy between $t_1$ and $t_2$. At right side, were it not for the $x^2$, you'd have energy "creation": a positive term, showing that energy increases so that oscillation is enhanced.
But there is the $x^2$ term, which has the opposite sign, showing a dissipation of energy, i.e. a damped oscillation. It's negligible for small $x$ whereas it becomes important for large $x$. You may then expect that for some intermediate oscillation amplitude dissipation and creation might balance, thus giving rise to a sustained (stable) oscillation.  This the limit cycle.
A better understanding is possible trying an explicit form for $x(t)$:
$$x(t) = a\,\sin \omega t.$$
Although you can't expect this is the right form of the limit cycle, it may give a semiquantitative idea of what will happen. I leave for you to compute the integral and to show that for $a>2$ dissipation prevails whereas for $a<2$ there is a net energy creation. A balance is obtained for $a=2$.
Therefore if you start from rest or in general from an amplitude $<2$ the oscillation will grow larger, until balance is reached. The opposite happens if you start from $a>2$: amplitude decreases until the limit cycle is reached from above. This shows stability of the limit cycle.
