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: 
So going back to the real basics (i.e. before students are introduced to "the" wave equation or allowed to think that it is the only wave equation) traveling waves were traditionally defined in terms like

> 
> a moving disturbance
> 
> 
> 

and (so long as $\psi$ is interpreted as a alteration of *something* from equilibrium) then the function in front of you meets that definition. 

That said it has some strange properties and may not represent the solution to any physical system.
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*.