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: 
It seems you're getting very messed up with some things. One by one.

You ask wether $f(x)=\sin (x^2-t^2)$ is a travelling wave. The answer is "no, it isn't", with the usual definition of wave.

The reason is that that function does not satisfy the wave equation. It's as easy as that: it doesn't satisfy the wave equation → it is not a wave. Nothing else.

Second point, you seem to be worried about the fact that it LOOKS LIKE a solution, but you plugged that into the equation and it was not satisfied, so it might be similar to a solution, but it is not. If it were a solution, the wave equation would have been satisfied.

And why is it? That's because the wave equation, writen as

$\frac{\partial^2 f(x,t)}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 f(x,t)}{\partial t^2}$

is a **linear** equation. "Linear" means that "if X and Y are solutions", then any linear combination of them $aX + bY$, with $a,b$ real numbers, is also a solution. The equaiton is linear because it is made of derivatives, which are linear (sum of derivatives = derivative of the sum). You can show it explicitly.

Yep, it doesn't say anything about multiplication, that's because multiplication ISN'T a solution.

a sinus of a multiplication isn't translated as a sum either, $\sin(AB)\neq sin(A) +sin(B)$. Since they are NOT the same, $\sin(AB)$ is not a linear combination of basic solutions, so it's not a solution either.

In sum

* A function is only a wave if it satisfies the wave equation.
* The basic solution of the wave equation is a plain wave.
* Since the wave equation is linear, the most general solution to the wave equation is a linear combination of basic solutions.
* Any solution can be thus written as a linear combination of them.
* Any function can remind you the basic solution, but no matter how similar they are, what matters is if they are solutions or not.

And, regarding your specific question:

* Your function does not satisfy the wave equation, so it is not a wave.
* That's because it is not a linear combination of basic solutions.

One last thing, I've just seen your comment

> 
> So, can I conclude in general that $f[(x−vt)(x+vt)]$ is not a traveling wave?
> 
> 
> 

What you should have learnt so far with this post is... "let's check it". Spoiler: it isn't. As I said, this is not a linear combination of the basic solution $\sin(x\pm vt)$, but if you are not sure, plug it again in the wave equation!

Substitute in the wave equation and you will see that it does not satisfy the wave equation. Anytime you ask yourself if something is a wave, just check if it satisfies that?

$f$ is an applicant to be a wave. The requirement is to satisfy that equation. If that equation is not satisfied, $f$ is discarded. (discarded as a wave, but it can be a solution of other different equation).
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)