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Poisson equation with a dirac delta source.

  1. May 23, 2015 #1
    Consider:

    ##\nabla^{2} V(\vec{r})= \delta(\vec{r})##

    By taking the Fourier transform, the differential equation dissapears. Then by transforming that expression back I find something like ##V(r) \sim \frac{1}{r}##.

    I seem to have lost the homogeneous solutions in this process. Where does this happen?
     
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  3. May 23, 2015 #2

    Orodruin

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    Taking the Fourier transform you are essentially saying that your domain is infinite and that your function (##V##) is square integrable. The homogeneous solution is not square integrable.
     
  4. May 23, 2015 #3
    Thanks. Maybe you could get this one as well since you already bothered to answer so I don't have to make another post:

    What's the difference in the evolution of ##f## for ##t>0## between saying

    1) ##\partial_{x}^{2} f - \frac{1}{c^{2}} \partial_{t}^{2}f = -\delta(x) \delta(t)## where nothing is happening before ##t=0##

    2) ##\partial_{x}^{2} f - \frac{1}{c^{2}} \partial_{t}^{2}f = 0 ## and to impose a delta peak as initial condition?
     
  5. May 23, 2015 #4

    Orodruin

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    Essentially nothing, but depending a bit on which initial condition you put a delta in (the wave equation is second order in t and so needs two initial conditions).
     
  6. May 23, 2015 #5
    I was talking about the condition on ##f##, the derivative condition is zero everywhere.

    Is it correct to say that the first expression is more general. It can encompass the state of the system at ##t<0##. For example, if an oscillatory motion already exists before this delta peak, this expression can still be in agreement with it.

    The second one is only a description for ##t>0## of a special case of the above description (namely, when everything is zero before the delta peak).
     
  7. May 23, 2015 #6

    Orodruin

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    Well, if you have initial conditions, you are encoding anything that happened before that time in those conditions. So for solving it for future ##t## it does not really matter. But the domain of the other problem is larger, yes.
     
  8. May 29, 2015 #7

    vanhees71

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    Well, it's all correct with your calculation of the Green function of the Laplace operator. The solution is not square integrable, because it's a distribution and not a function since the source is already a distribution. The Green's function in fact is
    $$G(\vec{x})=-\frac{1}{4\pi |\vec{x}|} \; \Leftrightarrow \; \Delta G(\vec{x})=\delta^{(3)}(\vec{x}).$$
     
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