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.

  • $\begingroup$ I've never heard about this type of oscillator, so this is a really interesting topic. The limit cycle isn't really surprising, since it's remember that non-linearity usually introduces limit cycles to oscillatory-like systems (you may want to check chaos theory). Indeed, you can see that near the origin the system is unstable ($x=dx/dt=0$), and from from the origin it's damped. As for the non-trivial limit cycle, the oscillator indeed has a non-zero kinetic energy in its steady state since it has a non-zero speed, and you can actually quantify it by the usual $E_K=\frac{1}{2}m \frac{dx}{dt}$ $\endgroup$
    – Charlie
    Mar 24, 2019 at 23:52
  • $\begingroup$ Since the damped harmonic oscillator has a linear damp term, it's characterized by the value of the factor of that term, resulting in either overdamping (no oscillation), critical damping (quickly return to steady state), or underdamping (oscillations with decaying exponential amplitude). As such, it has no limit cycles. On the other hand, the Van der Pol oscillator can have two possible states depending on the factor in the damping term: either unstable equilibrium (near the origin) or damped oscillations (which is a clear difference from the linear case). $\endgroup$
    – Charlie
    Mar 24, 2019 at 23:56
  • $\begingroup$ I think the energy of the oscillator should be quantified in the usual way, with the kinetic and potential terms in function of time. Note that you can actually quantify the energy lost at a given time by the taking the difference between the initial energy and the sum of kinetic-potential energies for a given time $t$. $\endgroup$
    – Charlie
    Mar 24, 2019 at 23:58
  • $\begingroup$ The original van der Pol oscillator describes a triode whose anode is loaded with a parallel LCR while the grid is magnetically coupled (transformer) to the resonator's inductor. There is no issue of energy conservation, all dissipation is in the resistive load R and is fed (replenished, if you wish) by the effectively negative resistance of the triode facing the transformer. $\endgroup$
    – hyportnex
    Mar 25, 2019 at 1:00

1 Answer 1


$\def\half{{\textstyle {1 \over 2}}}$ 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.

  • $\begingroup$ This is really interesting; thank you! So another way to interpret the RHS integral in your equation (2) is to say that once the system has evolved to the (stable) limit cycle, I should be able to measure the RHS integral for any choice of $t_1, t_2$ and find that it is always equal to zero. In contrast if we were to derive an analogous equation for damped harmonic motion, we'd find that the RHS integral always evaluates to a strictly negative quantity as long as $x' \neq 0$. $\endgroup$
    – overseas
    Mar 26, 2019 at 16:03
  • 1
    $\begingroup$ @overseas I should be able to measure the RHS integral for any choice of t1,t2 and find that it is always equal to zero Not quite. I apologize for being rather inaccurate. Cancellation only occurs if $t_2=t_1+T$, $T$ being the period of limit cycle. $\endgroup$
    – Elio Fabri
    Mar 26, 2019 at 16:19

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