Question: Is the function $\psi(x,t)=\sin(x^2-t^2)$ a traveling wave?

It can be written as $\sin[(x-t)(x+t)] $ and I know that $\sin(x\pm t)$ are traveling waves but is $\sin[(x-t)(x+t)]$ a traveling wave? My professor asked us to check out whether or not the multiplication of two traveling waves is a traveling wave, I rememeber from my course of "oscilations and waves" (I'm taking optics) that the sum and rest of two solutions of a wave equation is also a solution but I don't remember anything about the multiplication, so I don't believe the multiplication of two solutions is a solution. But even if the multiplication was a solution, my professor's question is wrong, isn't it? Because if the multiplication was a solution, then $\sin(x-t)\sin(x+t)$ would be the solution and not $\sin[(x-t)(x+t)]$ or what do you think?

Answer: 
Regarding the question, "Is the product of two traveling waves a traveling wave?" the answer is no. Take two traveling pulses: $f(x,t)=1$ on $[t,t+1]$ and is $0$ elsewhere, and $g(x,t)=1$ on $[t+4,t+5]$ and is $0$ elsewhere. At any time, there are no points where both $f$ and $g$ are nonzero, so their product is $0$, which is not a traveling wave.

EDIT:

Here's the animation I was talking about in the comments:

[![enter image description here](https://i.stack.imgur.com/v2OcN.gif)](https://i.stack.imgur.com/v2OcN.gif)
This function is not the solution of the wave equation which has general wave solutions of the form $$ y=f(x-vt)+ g(x+vt)$$ where $f$ and $g$ are arbitrary functions with $f(x-vt)$ travelling with constant shape and velocity $v$ in positive x-direction, and $g(x+vt)$ in negative x-direction. 

The function $$y=\sin [(x-t)(x+t)]$$ obviously has zeros that also travel with velocities $\pm 1m/s$ along the positive and negative x-axis. But the shape of the function changes strongly during propagation. Thus I would not consider this a "travelling wave". 

The question whether the product of two solutions of the wave equation is also a solution is completely different to the question regarding this mathematical function. If you consider solutions of the wave equation of the form $$y=A(\omega)\exp i[k(\omega)x-\omega t]$$ which is just a Fourier decomposition of the arbitrary function $f$ and $g$ considered above, then the product of two such solutions is again a solution of the wave equation: $$A\_1 A\_2\exp i[(k\_1+k\_2)x-(\omega\_1 +\omega\_2) t]$$ with a the wave velocity $$v=\frac {\omega\_1 +\omega\_2} {k\_1+k\_2}$$ which, due to $\omega =v k$, is the same as the velocities $$v=v\_1=\frac{\omega\_1}{k\_1}=\frac{\omega\_2}{k\_2}=v\_2$$ of the multiplied waves.

**NOTE:** This multiplicative superposition of waves with the same wave velocities $v$ but different frequencies $\omega$, which are solutions of the the classical wave equation, is used in the signal transmission by *amplitude modulated radio waves*.