The wave equation is $\frac{\partial^2 \chi}{\partial t^2} = c^2 \frac{\partial^2 \chi}{\partial x^2}$. I'll be understanding it in terms of sound.

The wave equation is solved by many periodic functions like $\chi(x, t) = \sin(kx - \omega t)$.

However, there are valid physical phenomena which do not solve the wave equation.

Consider an infinitely long tube. The tube is full of air which is accelerating uniformly in the $+x$ direction. The displacement of that air vs time looks like $\chi(x, t) = t^2$. Plugging this function into the wave equation we get $2 \ne c^2 \times 0$.

What can be concluded from the fact that this function fails to solve the wave equation?

My guess is that its failure to solve the equation means that the case of uniformly accelerating air violates one of the assumptions made in deriving the wave equation for sound. However, I haven't been able to figure out how (I'm reading Feynman's lecture on sound and the wave equation).

I've found a few somewhat related questions and not quite been satisfied by the answers:

  • $\begingroup$ it means the solution is not a wave… certainly $\chi(x,t)=t^2$ does not describe a wave… $\endgroup$ Jan 7, 2022 at 2:55
  • 1
    $\begingroup$ It means the wave equation you are asking about doesn't describe the situation you've giving, for instance water waves in shallow water are not modeled by the wave equation, but are modeled by the Korteweg-De Vries equation. You might however get a better answer on mathematics stackexchange than here, at least if you are looking for some rigor. $\endgroup$
    – Triatticus
    Jan 7, 2022 at 3:00
  • $\begingroup$ I guess what I'm confused about is why the wave equation doesn't describe the situation I gave. The wave equation derivation for sound that I'm looking at just works from Newton's laws and a couple of other assumptions which aren't violated by the situation I gave as far as I can tell so it seems like the wave equation should for more than just waves. $\endgroup$
    – jrpear
    Jan 7, 2022 at 3:14
  • $\begingroup$ Why do we show that quantities in many physical processes need to be solutions of PDEs? One of the reasons is that is saves us the trouble of having to list millions of non-solutions and argue separately why each one is ruled out. $\endgroup$ Jan 7, 2022 at 3:39
  • $\begingroup$ Thanks for the comments, I've got much to think on. $\endgroup$
    – jrpear
    Jan 7, 2022 at 4:00

1 Answer 1


Sure enough, this is because one of the assumptions made in the derivation was violated. I missed it several times.

Feynman's derivation assumes that the force on a particular volume of air is $F_{net} = P(x + \Delta x, t) - P(x, t)$. If you have some other force causing the uniform acceleration, this assumption does not hold.

So one of many answers to "What does it mean when a function doesn't solve the wave equation?" is:

Assuming the wave equation for the domain you're interested in was derived from some universally true laws and some conditionally approximately true assumptions, a physically possible system will not satisfy the wave equation when the system violates an assumption made in the derivation.

  • $\begingroup$ Actually, a uniformly accelerating volume of air will develop a pressure (and/or density) gradient opposing the acceleration. Incidentally, this still satisfies the wave equation in the rest frame of the cylinder (the second derivative in space vanishes, and the time derivative vanishes because the gradient is static in the rest frame.) $\endgroup$
    – TLDR
    Jan 7, 2022 at 4:05
  • $\begingroup$ Oh is this a special relativity thing or a statistical consequence of the microscopic motion of the atoms? Or something else? $\endgroup$
    – jrpear
    Jan 7, 2022 at 4:10
  • $\begingroup$ @jrpear you need to use @<username> in comments for a commenter to be notified of your reply. $\endgroup$
    – antimony
    Jan 8, 2022 at 1:12
  • $\begingroup$ @jrpear The pressure gradient is caused by interactions between molecules in the air; in the rest frame of the cylinder (or with respect to an observer moving with the cylinder, which is just a plain old relativity thing, nothing 'special' about it) the acceleration is indistinguishable from a gravitational field, so you might expect there to be an air pressure gradient analogous to the gradient with altitude in Earth's atmosphere. $\endgroup$
    – TLDR
    Jan 8, 2022 at 6:52
  • $\begingroup$ @TLDR Ahhh that's what you were saying, got it. Thanks! $\endgroup$
    – jrpear
    Jan 8, 2022 at 23:45

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