I'm confused about the nonlinear dispersion relation like $$\omega(k) ∝ k^2$$.

Does this kind of wave have multiple frequencies?

I have always considered the linear dispersion case, $$\omega(k) ∝ k$$, so I cannot imagine how the wave having nonlinear dispersion relation has its form.

• This just means that different frequencies travel a different speeds. A dispersion relation is the relation between the space and time periods of a wave. If a a cosine wave will stay in place, the each dot a long the wave will remain in the same position and no oscillation will occur. In this case, $\omega = 0$ although the cosine has a certain $k$. When all the waves travel at the same speed, there is a linear relation between $\omega$ and $k$. But if high spacial frequencies travel faster than low spacial frequencies (or vice versa) you'll get a non-linear dispersion relation May 6, 2020 at 11:49
• No, you didn't get it. A dispersion relation gives a relation for cosine waves in the medium. This is the definition of the dispersion relation. May 6, 2020 at 21:32

Does this kind of wave have multiple frequencies?

For the dispersion relation to be useful (as opposed to just being true) then yes.

We find the dispersion relation of a wave by putting in a trial solution of the form $$e^{i(kx-\omega t)}$$ and then working out the conditions on $$\omega = \omega(k)$$ for the trial plane-wave waveform to be a solution of the dynamics.

The reason this is useful is that, if the governing dynamics are linear, then we can write the initial condition $$f(x,0) = f_0(x)$$ as Fourier transform, $$f(x,0) = \int_{-\infty}^\infty \tilde{f_0}(k) e^{ikx} \mathrm dk,$$ i.e. as a superposition of a continuum of plane-wave components $$e^{ikx}$$ with different spatial frequencies $$k$$, known as a wavepacket, and then evolve each of those components in time independently, using the dispersion relation we found above, $$f(x,t) = \int_{-\infty}^\infty \tilde{f_0}(k) e^{i(kx-\omega(k)t)} \mathrm dk,$$ with each spatial frequency $$k$$ present in the wavepacket's bandwidth contributing with a phase that evolves at a different temporal frequency $$\omega(k)$$.

I'm not sure what you mean by "multiple frequencies."

Do you mean does it have different (angular) frequencies at different $$k$$? Then, yes.

Or, do you mean, does it have multiple different angular frequencies, as opposed to just one, at a given fixed $$k$$? Different question; insufficient information to answer.

Anyway, the first little mental calculation you should always do when given a 1D (as opposed to vector) dispersion relation is figure out the phase speed and the group speed.

Phase speed is $$v_{\rm ph} = \omega/\lambda$$, and group speed is $$v_{\rm gr} = d\omega/dk$$. In this case, writing $$\omega = ak^2$$ for some $$a$$, $$v_{\rm ph} = ak \propto 1/\lambda,$$and, also, $$v_{\rm gr} = 2 v_{\rm ph} = 2ak \propto 1/\lambda.$$ So that tells you that these waves are kinda the opposite (in a way) of surface water waves, where long-wavelength waves move faster than short-wavelength waves, which is why if there's a storm far off in the ocean, the first waves to break on the shore are the long-period waves, and the wave period gradually gets shorter over the subsequent days. If you throw a stone into a pool of water, the spreading wave trains for waves with a wavelength more than about 5 cm are surface gravity waves that are dispersive, with the phase speed (the wave crests) moving twice as fast as the group speed (the overall speed of a train of waves, corresponding to the speed of energy being transported by the waves; if you watch closely, a wave crest will appear at the back, move through the train, and then disappear as it reaches the front of the wave train). Meaning, the waves disperse, which is the origin of the term "dispersion relation" in the first place.

In your example case however, unlike surface gravity waves, (1) shorter wavelength waves move faster than long-wavelength waves, and (2) in a group, a wave crest will appear at the front, move backwards through the bunch, and disappear at the back. It probably also has other odd properties that I haven't thought through.

Hopefully I didn't make a mistake in there but if I did I'm sure someone will correct me. :)

This is in addition to Emilio's answer. It was too long for a comment.

From the dispersion relation you can construct a 'wave equation'. Take the expression $$(\omega-\omega(k))f(x,t)=0$$, replace $$\omega$$ by $$-i\frac d{dt}$$ and every $$k$$ in $$\omega(k)$$ by $$i\frac d{dx}$$. Then $$\omega(k)=ck$$ gives you (after squaring) $$\frac{d^2f}{dt^2}=c^2\frac{d^2f}{dx^2}.$$ And $$\omega(k)=\lambda k^2$$ gives $$\frac{d^2f}{dt^2}=\lambda\frac{d^4f}{dx^4}.$$ And $$\omega(k)=c\sqrt{k^2+\frac{m^2c^2}{\hbar^2}}$$ gives $$\frac{d^2f}{dt^2}=c^2\frac{d^2f}{dx^2}-\frac{m^2c^4}{\hbar^2}f.$$ You can apply each of these wave equations to the expression for $$f(x,t)$$ in Emilio's answer to show that they are correct.

This is my non-calculus answer, since I don't know your background, so no offense if this is too simple.

A "dispersion relation" is called that, precisely because it describes how waves disperse.

Look at one wave, "wave 1", with a wavenumber $$k_1=1/{\rm m}$$, and another wave, "wave 2", with a wavenumber $$k_2=2/{\rm m}$$. That is, the second wave is a wave with half the wavelength of the first wave.

The speed of each wave (to be formal: the phase speed) is $$v_{\rm ph} = \omega/k$$. For your dispersion relation, write $$\omega = a k^2$$ for some $$a$$. Then for your dispersion relation, the wave (phase) speed is $$v_{\rm ph} = \frac{\omega}{k} = \frac{ak^2}{k} = ak.$$ Then, for wave 1, $$v_{\rm ph}^{(1)} = ak_1$$, and for wave 2, $$v_{\rm ph}^{(2)} = ak_2$$, so $$v_{\rm ph}^{(1)}:v_{\rm ph}^{(2)} = k_1:k_2 = 1/2.$$ So the second wave, with the higher wavenumber (and shorter wavelength) moves half the speed of the first wave.

So, the waves separate from each other, or in other words, they disperse.

If you have a linear dispersion relation, the waves don't disperse, so "linear dispersion relation" is actually a bit of an oxymoron, but everybody says it anyway, because this is all understood.

Also, this is just a thumbnail-sketch, because of course if you think about it some more, the waves in the example I gave can't really separate from each other, because a monochromatic wave doesn't have a beginning and an end - it's everywhere. But then, that takes us into Fourier analysis, group speed, wave packets, blah blah blah.

So that's the very simple answer. Again, depends on what level you're asking the question.