The units of the d'Alembertian are distance$^{-2}$. It should be the case that the inhomogeneous wave equation describing $$\square u = f$$ should have matching units on both sides. My understanding is that $u$ is dimensionless (this may be wrong). Thus the units of the forcing function should be distance$^{-2}$. Is this right? If so, what can it possibly mean?

Another way to approach this problem is to think of d'Alembert's formula for the inhomogeneous wave equation.

Say we have initial conditions $u(x,0)= g(x)$ and $u_t(x,0) = 0$. Let the forcing function be $f(x,t)$. Say for the sake of argument that $g(x) = \cos(kx)$, where $x$ has units of distance and $k$ has units of inverse distance (the wave number). So $u(x,0)$ is dimensionless and thus $u(x,t)$ for all $t$ is dimensionless. We have no initial velocity so we can neglect that integral in d'Alembert's formula. We are left with:

$$2u(x,t) - g(x+ct) + g(x-ct) = \frac{1}{c}\int_0^t \int_{x-c(t-s)}^{x+c(t-s)} f(\xi,s) d\xi ds.$$

If the units of $f$ are distance$^{-2}$, the RHS is not dimensionless but has dimensions time$^2$/distance$^{2}$.

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    $\begingroup$ Your statement, u is dimensionless, is definitely wrong. You have not stated what u is physically. It could be pressure, velocity potential, or some other state variable. Whatever it is, it has units. If the equation is scaled to make it dimensionless then f has to scaled too. $\endgroup$ – ggcg Jan 28 '19 at 17:22
  • $\begingroup$ OK, but I think the question remains. Say $u$ has units of pressure. So the units on the left hand side are Pascals/distance^2. Then $f$ has to have the same units. I can't see how to make the integral in d'Alembert's formula have those units. It has units Pascals*time^2/distance^2. Right? $\endgroup$ – Wapiti Jan 29 '19 at 20:19

Then your integral is not defined consistently relative to the diff eq. Your measure needs to scale.

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  • $\begingroup$ Can you expand on this? The answer is to gnomic to be useful. $\endgroup$ – Wapiti Feb 4 '19 at 3:40
  • $\begingroup$ I think I meant this to be a response to your comment not an answer. $\endgroup$ – ggcg Feb 8 '19 at 11:52

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