# is this function a traveling wave? [closed]

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?

## closed as off-topic by Mitchell, Jon Custer, Emilio Pisanty, Cosmas Zachos, valerioFeb 9 '18 at 0:06

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• Have you tried checking if it satisfies the wave equation? – knzhou Jan 28 '18 at 18:24
• Oh! It doesn't satisfy the wave equation but what yo do you think about my professor question, is it right? – Sama Jan 28 '18 at 18:41
• They messed up, but it doesn't matter that much! False things are false. – knzhou Jan 28 '18 at 18:43
• So, can I conclude in general that $f[(x-vt)(x+vt)]$ is not a traveling wave? – Sama Jan 28 '18 at 18:49
• Sorry, I don't understand your point, could you explain me more, please? – Sama Jan 28 '18 at 18:54

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.

• 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).

• I completely agree with your view of this issue! – freecharly Jan 29 '18 at 19:48

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:

• So the multiplication of two solutions is not a solution but what about $\sin[(x-t)(x+t)]$? – Sama Jan 28 '18 at 18:45
• Depends on what you mean by "traveling wave." Here's what your function looks like: sandbox.open.wolframcloud.com/app/objects/…. The peaks of the function clearly travel away from the origin as time passes, but they travel slower and slower the further away they get. In addition, new peaks are created faster and faster as time passes. You might want to ask your professor what, exactly, he means by "traveling wave." – probably_someone Jan 28 '18 at 18:52
• The reason it depends so much on definition is because $\sin(x^2-t^2)$ probably is an oscillatory solution to some really odd potential, and sometimes we call those solutions "waves" even if they don't follow the wave equation. – probably_someone Jan 28 '18 at 18:53
• @probably_someone I cannot open your link. – Farcher Jan 28 '18 at 19:13
• @Farcher Link produced the animation that is now in my answer. – probably_someone Jan 28 '18 at 19:29

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.

• Is $f(x,t)=0$ a traveling wave? – probably_someone Jan 28 '18 at 23:54
• @probably_someone - $f(x,t)=0$ is an equation, not a function. – freecharly Jan 29 '18 at 0:02
• Alright then, is the function that returns $0$ for all $x$ and $t$ a traveling wave? – probably_someone Jan 29 '18 at 0:02
• @probably_someone- This is a trivial function. Like any constant in x and t, it also fulfills the linear wave equation. But it is not a a wave solution – freecharly Jan 29 '18 at 0:57
• Ah, I see where the semantic difference is. This whole time, we (OP included) have been conflating the questions "Is X a traveling wave?" and "Is X a solution of the wave equation?" when the two questions may have different answers. – probably_someone Jan 29 '18 at 1:47