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I'm trying to determine the electrostatic potential caused by a specified charge density function: $$ \rho_c(\vec{r}) =\begin{cases} 1 & \vec{r} \in V\\ 0 & \text{otherwise} \end{cases} $$

The solution using Green's function is then: $$ \phi(\vec{r}) = \iiint_{V} \frac{\rho_c(\vec{r}')}{4 \pi \epsilon_0 \|\vec{r}-\vec{r}'\|} d \vec{r}' $$

I can understand how this integral can be easily evaluated so long as $\vec{r} \not\in V$ since there is no singularity in the integral, however I don't know how to resolve the integral if $\vec{r} \in V$, since this appears to cause the integral to become infinite/undefined.

Suppose for a concrete example that $V$ is an axis-aligned rectangular prism with opposing corners at $(x_0,y_0,z_0)$ and $(x_1,y_1,z_1)$. How do I find $\phi(\vec{r})$ inside of this box, and how do I generalize that procedure to any arbitrary $V$?

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  • $\begingroup$ The singularity is integrable... $\endgroup$
    – Valter Moretti
    Commented Aug 31, 2021 at 11:46
  • $\begingroup$ I'd imagine it is since I can see one alternative solution to be to use the Green's function method to produce boundary conditions outside of $V$ for a Poisson PDE solver, however I don't know what mathematical tools I need to use to evaluate this integral "directly". Suppose I took the 1D case with $\rho_c(x) = 1$ for $x \in [-1, 1]$. I think this produces the integral $\phi(0) = \int_{-1}^1 \frac{1}{4\pi \epsilon_0 |x|} dx$, which doesn't converge, but using the "Poisson with BC's" approach gives $\phi(0) = \frac{6 \pi + \log(3)}{4 \pi \epsilon_0}$. $\endgroup$
    – helloworld922
    Commented Aug 31, 2021 at 12:38
  • $\begingroup$ In 1D there is no Green function, but in 2D there is and it is proportional to $\ln \sqrt{x^2+ y^2}$ which is locally integrable as well in $dxdy$ $\endgroup$
    – Valter Moretti
    Commented Aug 31, 2021 at 13:17

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Here's a very simple example: what is the improper integral $\int_0^1\frac{dt}{\sqrt{t}}$? It is $\frac{1}{2}$ even though $\frac{1}{\sqrt{t}}\to \infty$ as $t\to 0^+$. The reason is that although the function is unbounded, it is not "bad enough" for the purposes off integration. This is what @Valter Moretti means by "the singularity is integrable". It goes off to $\infty$ slowly enough that it can be integrated over to yield a finite result. On the other hand, $\int_0^1\frac{dt}{t}=\infty$.

So, for integration purposes, mere unboundedness of the function alone is not enough to deduce anything about the finiteness of the integral. In $3$ dimensions, \begin{align} \int_{\|\mathbf{r}'\|\leq 1}\frac{dV'}{\|\mathbf{r}'\|}=\int_0^1\frac{1}{r'}4\pi r'^2\,dr'=4\pi\int_0^1r'\,dr' \end{align} is certainly finite. And in general, in $\Bbb{R}^n$, \begin{align} \int_{\|\xi\|\leq 1}\frac{1}{\|\xi\|^p}\,d^n\xi&=\int_0^1\frac{1}{r^p}A_{n-1}r^{n-1}\,dr=A_{n-1}\int_0^1\frac{dr}{r^{p+1-n}} \end{align} so this is finite if and only if $p+1-n<1$, if and only if $p<n$ (here $A_{n-1}=\frac{2\pi^{n/2}}{\Gamma(n/2)}$ is the surface area of the unit sphere $S^{n-1}\subset\Bbb{R}^n$).

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  • $\begingroup$ hmm, so the 1D case is still unbounded? Or am I miss-understanding the second criteria since I can still directly solve the 1D Poisson's equation for $\rho_c = 1$ for $x \in [-1,1]$? $\endgroup$
    – helloworld922
    Commented Aug 31, 2021 at 12:48
  • $\begingroup$ @helloworld922 in the 1-D case it is not integrable as I mentioned: $\int_0^1\frac{dt}{t}=\infty$. In 3-dimensions there's no issues. I even gave the condition in $n$-dimensions for when such inverse-powers of distance yield finite integrals. (though for $1D$ solving Poisson's equation is trivial, you just integrate $u''(x)=\rho(x)$ twice (I'm ignoring the various constants)) $\endgroup$
    – peek-a-boo
    Commented Aug 31, 2021 at 12:49
  • $\begingroup$ I suppose a follow-up question is why can I solve Poisson's equation "directly" by applying appropriate BC's at $x=\pm 2$ and find $\phi(0)$? $\endgroup$
    – helloworld922
    Commented Aug 31, 2021 at 12:51
  • $\begingroup$ what do you mean? $\endgroup$
    – peek-a-boo
    Commented Aug 31, 2021 at 12:52
  • $\begingroup$ $\iint \frac{d^2 \phi(x)}{dx^2} dx dx = \phi(x)$, $\iint h(1-x) h(1+x) dx dx = \frac{1}{2} (2 c_0 x + (x - 1)^2 h(x-1) - (x+1)^2 h(x+1))+c_1$, then I just need to apply appropriate boundary conditions to find the constants of integration? $\endgroup$
    – helloworld922
    Commented Aug 31, 2021 at 13:03

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