Question: 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.

Please help me out.

Answer: 
> 
> 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. :)