Poisson equation with a dirac delta source.

[Coffee_]

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?

 

[Orodruin]

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.

 

[Coffee_]

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?

 

[Orodruin]

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).

 

[Coffee_]

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).

 

[Orodruin]

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.

 

[vanhees71]

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
$$G(\vec{x})=-\frac{1}{4\pi |\vec{x}|} \; \Leftrightarrow \; \Delta G(\vec{x})=\delta^{(3)}(\vec{x}).$$