I came across this problem in the study of surface waves in an oscillating cylindrical vessel of liquid.

There are various eigenmodes described using Bessel functions, and energy transfer can happen between them through nonlinear effects (usually to higher order modes or modes with a frequency that is close to a multiple of the base frequency).

This is a very complicated system to explore nonlinearity in, so I would like to know if there are simpler systems that could be used as an analogy.

  • $\begingroup$ I am not really sure about your specific problem, but it does seem to be similar to the idea of parametric resonances - where exciting just one degree of freedom can excite a different degree of freedom. In case this is close to what you're looking for, a pretty simple system to look at this is in a spring pendulum - a pendulum bob swinging from an elastic string. This shows a parametric resonance at a particular relation between the frequencies of the spring mode and the oscillatory mode. $\endgroup$
    – ShKol
    Dec 23, 2022 at 14:56
  • $\begingroup$ This is pretty well outlined in in several articles but this: iopscience.iop.org/article/10.1088/1361-6404/aaf146 is a nice one according to me. $\endgroup$
    – ShKol
    Dec 23, 2022 at 14:56
  • 3
    $\begingroup$ Please add some details or reference to your question $\endgroup$
    – basics
    Dec 23, 2022 at 14:58
  • $\begingroup$ To narrow down the answers: Can the model of a simple harmonic oscillator be tweaked to add a nonlinearity such that energy under certain circumstances leaks from one oscillation frequency to another? Say the oscillator is forced at its resonant frequency, but energy through (nonlinearities) ends up at other frequencies too. $\endgroup$
    – Chillpadde
    Dec 23, 2022 at 18:56
  • $\begingroup$ Also, I'd prefer if the system could be purely mechanical. $\endgroup$
    – Chillpadde
    Dec 24, 2022 at 18:03

1 Answer 1


I think a very nice and also relevant example is that of second harmonic generation (SHG). I will summarize it briefly, but you can find many resources online.

Maxwell's wave equation in a medium is (setting $c=1$) $$\nabla^2 E - \frac{\partial^2 E}{\partial t^2} = \frac{\partial^2 P}{\partial t^2},$$ where P is the polarization of the medium. In linear optics one uses the ansatz $P(\omega) = \chi(\omega)E(\omega)$, where $\chi$ is called the linear response function. More generally, one can expand $P$ in a Mac Laurin series, but here we will truncate at the second order nonlinearity: $$P = \chi E + 2dE^2.$$ The parameter d is a material property, and by symmetry considerations can only be non-zero if the material does not posses inversion symmetry (just as a side info, but not important here). Inserting this expansion into the wave equation gives $$\nabla^2 E - n^2\frac{\partial^2 E}{\partial t^2} = -S,$$ where $S = -\frac{\partial^2}{\partial t^2}(2dE^2) =: \frac{\partial^2}{\partial t^2}P_\mathrm{NL}$ is called the source term and $n$ is the refractive index.

This can be solved exactly with coupled wave theory, but let's just look at the problem in the Born approximation to get better intuition. Let's assume our zeroth order solution is just a plane wave: $E_0(r,t) = \cos(\omega t - kr)$. Then the second order part of the polarization is $$P_\mathrm{NL} = 2dE^2 \approx 2dE_0^2 = d(1 + \cos(2\omega t - 2kr)).$$

If you insert this back into the wave equation, you see easily that this gives rise to an electric field oscillating at $2\omega$. In a photon picture you can think of two photons of frequency $\omega$ coming from the "incoming" beam $E_0$ combining into one photon of frequency $2\omega$ (energy conservation is not violated) of the "outgoing" beam $E_1$, where $E_1$ is the first-order Born approximation. So you excite the system with frequency $\omega$, but the 2nd order nonlinearity of the medium shifts that into the second harmonic.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.