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 $\omega(k)$.
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.