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Non-dimensionalization of a wave equation with point source

  1. Aug 2, 2010 #1
    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.

    [itex]\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}[/itex]

    The term with the (3D) delta function throws me off. The delta function has units of [itex][L]^{-3}[/itex]. 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.
    Last edited: Aug 2, 2010
  2. jcsd
  3. Aug 5, 2010 #2
    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.
  4. Aug 5, 2010 #3
    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.
  5. Aug 5, 2010 #4
    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 [itex][L]^{-3}[/itex].

    If it is unclear [itex]\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}[/itex]. [itex]\vec{r}_s[/itex] 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 [itex]x = L x^*[/itex]) 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.
    Last edited: Aug 5, 2010
  6. Aug 5, 2010 #5
    Oh ok maybe I get what you mean... probably the property you need is


    and for the 3-dimensional delta function


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

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

    with [tex]B=4\pi A L^8[/tex] and [tex]C=\omega L/c[/tex] dimensionless constants.
    I hope I haven't made mistakes and that it helps.
    Last edited: Aug 5, 2010
  7. Aug 5, 2010 #6
    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.
    Last edited: Aug 5, 2010
  8. Aug 5, 2010 #7
    Oh I forgot... edited!
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