Non-dimensionalization of a wave equation with point source

[LightningStrike]

I've been trying to non-dimensionalize a wave equation with a moving point source, but the peculiar properties of the delta function have confused me. How does one non-dimensionalize an equation with a delta function?

For example, the equation I'm looking at is something like the one below.

$\nabla^2 P + 4 \pi A \delta(\vec{r} - \vec{r}_s) \sin(\omega t) = \frac{1}{c} \frac{\partial^2 P}{\partial t^2}$

The term with the (3D) delta function throws me off. The delta function has units of $[L]^{-3}$. By changing the inside of the delta function to non-dimensionalized vectors, the dimension of the delta function is lost, and the equation is no longer dimensionally consistent.

Any hints?

By the way, I know the solution for the unbounded case, but I'm interested in a certain geometry.

 

[LightningStrike]

Does anyone know how to non-dimensionalize this? It'd really help with the research I'm doing. I'm convinced that I'm simply missing something obvious.

 

[Petr Mugver]

A delta function has the dimension of the inverse of it's argument. In your equation the dimension of A must be [m]/[P] and you need c^2 instead of c on the right-hand side.

 

[LightningStrike]

Yes, c instead of c^2 is a typo.

And I understand that a delta function has the dimension of the inverse of its argument. As I've said, the 3D delta function has units of $[L]^{-3}$.

If it is unclear $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$. $\vec{r}_s$ is the location of the sound source.

The problem is that when non-dimensionalizing this equation if I substitute in the definition of my dimensionless variables (i.e. something to the effect of $x = L x^*$) in the delta function, the argument of the delta function still has units. So there is no dimensionless parameter that is multiplied by delta function and sine function. I want a dimensionless parameter to multiply the delta function and sine function so I can solve the non-dimensionalized problem. How to resolve this problem is unclear to me.

I've rewritten the paragraph about a few times because it sounded convoluted. Let me know if you want another explanation.

 

[Petr Mugver]

Oh ok maybe I get what you mean... probably the property you need is

$$\delta(ax)=\frac{\delta(x)}{|a|}$$

and for the 3-dimensional delta function

$$\delta^3(a\mathbf{x})=\frac{\delta^3(\mathbf{x})}{|a|^3}$$

so that, if you put $$\mathbf{x}'=L\mathbf{x}$$ and $$t'=\omega t$$ your equation becomes

$$\nabla'^2P+B\delta^3(\mathbf{x}'-\mathbf{x}'_s)\sin(t')=C^2\frac{\partial^2P}{\partial t'^2}$$

with $$B=4\pi A L^8$$ and $$C=\omega L/c$$ dimensionless constants.
I hope I haven't made mistakes and that it helps.

 

[LightningStrike]

Yes, that seems to be precisely what I needed. I was not aware of that property. Thank you.

You neglected the sine function, and I think you meant x = L x', but I can figure the remainder out.

 

[Petr Mugver]

You neglected the sine function, and I think you meant x = L x', but I can figure the remainder out.

Oh I forgot... edited!