The equation of motion of a general anharmonic oscillator includes a position-dependent force that can be expanded in a Taylor series as


I am considering a case with force $F$ higher than the critical amplitude, which means I'm in the Duffing regime.

The Duffing equation is


where the third order term is the Duffing perturbation to the simple harmonic oscillator. This destroys the symmetry around the resonance of, say, amplitude plotted against frequency.

I get that we drop the $k_0$ term with a translation to the equilibrium point. However, I don't get why the second order term disappears. The books I've read (in particular Cleland's Foundations of Nanomechanics and Radons' Nonlinear Dynamics of Nanosystems) say it's because the harmonic potential is quadratic, hence symmetrical, and hence the sign of $x$ doesn't matter, but that isn't very clear to me; it's not making click. It's probably something very basic but maybe someone can explain it to me simply and briefly?

I've also read that even-ordered nonlinearities only mix signals to DC and twice the drive frequency, as opposed to odd-ordered ones which mix back to the operating frequency band, but I don't quite understand the use of the word "mix" in that context.


1 Answer 1


If you want to generalize a potential to a class that's broader than the simple $\frac12 k_2 x^2$, it is tempting as a first step to include a small perturbation of the form $\frac13k_3x^3$. Unfortunately, this drastically changes the structure of the potential, because it becomes unbounded from below.

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Thus, you might get a slightly perturbed behaviour from a harmonic oscillator at low amplitudes, but if you drive it hard enough, then the system will cross some hill and then just run away. This behaviour is simply not what one is trying to model, which is the back-and-forth motion about a potential minimum. It is indeed possible (or I don't see why it wouldn't) to do a proper analysis of this case in the limit of small enough oscillations that the system never sees the hill, while the harmonic oscillation is still perturbed. However, the existence of the hill and its other side makes the general solution very complicated and quite different to what you really want.

For a real potential, of course, there will be a term in $x^4$ that kicks in way before the system sees the hill. Thus, after the harmonic oscillator, the next class of models that really capture the type of back-and-forth behaviour one is trying to model, for all amplitudes and all driving forces, is of the form $$V(x)=\frac12 k_2 x^2+\frac13 k_3 x^3+\frac14 k_4 x^4.$$ This succeeds in fixing the problems with a pure $x^3$ perturbation, but in solving those problems we have made the model quite a bit more complicated than we were really hoping for.

(For instance, we now have two dimensional constants, the lengths $k_2/k_3$ and $k_3/k_4$, instead of just one, and the interplay between those two will affect the behaviour. For example, if $k_3^2>\frac92k_2k_4$ the potential develops a second dip, which you don't really want.)

Thus, I think the useful classification is

The Duffing potential, $V(x)=\frac12 k_2 x^2+\frac14 k_4 x^4$, is the simplest potential which generalizes the harmonic oscillator to the anharmonic case while avoiding runaway solutions for all starting positions.

That said, if you start generalizing this model, your first step should be to put the cubic term back in. Even then, the requirement that $k_3^2$ be bounded by $k_4$ means that the cubic term must be a perturbation on top of the quartic behaviour, instead of the other way around.

There is an additional physical reason why odd-order perturbations to the potential are not very common, and it is due to the type of symmetry - or lack thereof - your system has. Specifically, even-order terms in the potential are symmetric around the equilibrium point, but including odd-order terms will make for a lopsided potential that's not symmetric any more.

For all oscillating systems, if you drive them hard enough they will become anharmonic. However, for odd-order nonlinearities to come into play, the system itself needs to be asymmetric, and that need not be the case. Indeed, depending on the setting, it can be quite hard to make a system that's asymmetric enough for those effects to be noticeable.

A good example of this is in nonlinear optics, where the relevant perturbation expansion is that of the polarization's response to the driving electric field: $$P=\chi E+\chi^{(2)}E^2+\chi^{(3)}E^3+\cdots,$$ ignoring the vector/tensor character of these quantities. It generally requires pretty high intensities to drive materials into those regimes, so using $\chi^{(2)}$ nonlinearities seems a lot better than using $\chi^{(3)}$ ones. However, for most materials, the $\chi^{(2)}$ susceptibility is zero because of symmetry considerations: the medium needs to be asymmetric on the scale of a few atoms. This is possible, for example, in non-centrosymmetric crystals, but without such a material you can't do any three-wave mixing process, which rules out second-harmonic generation, sum- and difference-frequency generation, optical parametric amplification, and a host of other toolkit essentials.

So, the bottom line on this is: the symmetry of your system matters. You can only get odd-order perturbations to the potential if your system itself is asymmetric. Even orders, on the other hand, you get for free by simply driving it hard enough.

Regarding the last part of your question, "mixing" does have a specific meaning in this context. If you have a harmonic driving that is also being driven harmonically, $$m\ddot x+m\gamma\dot x+m\omega_0^2 x=F\cos(\omega t),$$ then the solution will also oscillate at the driving frequency $\omega$, with possibly a phase delay which is not really important.

However, when you include a cubic or quartic term perturbatively, the first step is to treat those terms as external forces with the position given by the unperturbed solution, say $x(t)=x_0\sin(\omega t)$. This means that you have a harmonic system that's got an additional driving of the form $$k_3 x_0^3 \sin^3(\omega t)\quad\text{or}\quad k_4 x_0^4 \sin^4(\omega t).$$ These are hard to deal with in that specific form, but they become quite a bit easier to handle if you use the appropriate trigonometric identities to reduce them to a finite sum of harmonic drivings. Thus, the forces above reduce to $$ \begin{cases} k_3 x_0^3 \sin^3(\omega t) = \frac{3\sin(\omega t)-\sin(3\omega t)}{4}\\ k_4 x_0^4 \sin^4(\omega t)=\frac{3-4\cos(2\omega t)+\cos(4\omega t)}{8}. \end{cases} $$ This is then a harmonic driving on a harmonic system, and that we can solve. (It's only the first term in a perturbation series, but that's another story.) The mixing you read about is the fact that the final solution will contain terms that oscillate at the original frequency, but it may also contain other harmonics.

As you can see from the above, even harmonics lead to a DC perturbation plus a $2\omega$ contribution. There will also be additional terms at higher harmonics, but those are typically harder to detect. An odd-power perturbation, on the other hand, will result only in odd harmonics, including a component at the driving frequency ("operating frequency band").

In terms of the time dependence of the solution $x(t)$, this is probably not that important. However, there are large classes of systems which are easier to interact with on the frequency domain, and then these harmonics are the best way to physically understand the solutions. For example, if you're driving some slightly nonlinear electric oscillator, you're probably doing so at RF or microwave frequencies, and then it's very easy to study the output of the system via the Fourier transform capacity of an oscilloscope. Once you're there, detecting peaks at $2\omega$ or $3\omega$ can be relatively easy, and it speaks directly as to the type of nonlinearity that's present in the system.

  • $\begingroup$ The last part of your question, and of this answer, make a good Q&A pair on their own. Do consider splitting your question in two and asking that part separately. $\endgroup$ Nov 5, 2013 at 22:19
  • 1
    $\begingroup$ thank you so much for the revision and the extensive and thorough response! $\endgroup$
    – user148585
    Nov 6, 2013 at 15:43

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