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 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.