Integration of Poisson's Equation

[mastrofoffi]

TL;DR Summary

I have a doubt about integration boundaries in Poisson's equation. Specifically, my 'intuition'(which is clearly wrong) would lead me to integrate in opposite order to the exact one.

I have a gaussian charge distribution, in gaussian units
$$
\rho(\mathbf r) = q\frac{\alpha^3}{\pi^{3/2}}\exp( -\alpha^2 r^2 )
$$
and I want to solve Poisson's equation to find the electrostatic potential
$$
\nabla^2 \psi(\mathbf r) = -4\pi\rho(\mathbf r).
$$
Since the charge distribution has spherical symmetry I move to polar coordinates where Poisson's equation reads
$$
\frac{ \text{d}^2 }{ \text{d} r^2} r\psi(r) = -4\pi r \rho(r).
$$

Now what I would do in order to solve the equation is to integrate in 2 steps, so the first integration would give me a radial expression for the electric field, and integrating the electric field i would obtain the potential. If i want the potential at some point $r$, I would expect the integration on the density $\rho$ to be taken from the origin to $r$, and the subsequent integration from $\infty$ to $r$, so in maths:
$$
\psi(r) = -4\pi\frac{1}{r}\int_{\infty}^{r}\text{d} s \int_{0}^{s} \text{d} t \rho(t)t
$$
This is clearly wrong, since I get a $1-e^{-...}$ term from the first integral, returning an $\infty$ when integrated again. It turns out the correct integration is the other way around:
$$
\psi(r) = -4\pi\frac{1}{r}\int_{0}^{r}\text{d}s \int_{\infty}^{s} \text{d} t\rho(t) t
$$
I know this is correct since it agrees with what I expected(I am using this to develop the decomposition of electrostatic energy in Ewald summation), but I can't understand what's the reasoning behind the 'choice' of the integration boundaries. I am able solve the equation by other means (i.e. use gauss's theorem to write $\psi(r) = \int_{\infty}^{r}\text{d}s E(s) = \int_{\infty}^{r}\text{d}s \frac{1}{s^2} \int_{\Omega(s)} \text{d}\mathbf t \rho(\mathbf t)$ where $\Omega(s)$ is a sphere of radius $s$ in $\mathbb R^3$ so I'm not asking about how to get the solution, but only how to setup the integrals correctly and why, both from a mathematical and from a more 'practical' physical perspective.

Thank you.

 

[rude man]

Summary: I have a doubt about integration boundaries in Poisson's equation. Specifically, my 'intuition'(which is clearly wrong) would lead me to integrate in opposite order to the exact one.Since the charge distribution has spherical symmetry I move to polar coordinates where Poisson's equation reads

Did you mean spherical coordinates I hope?
Anyway, your equation looks wrong. There should be 2 terms on the lhs.

 

[mastrofoffi]

Did you mean spherical coordinates I hope?
Anyway, your equation looks wrong. There should be 2 terms on the lhs.

In italian we use the equivalent of 'polar' and 'spherical' interchangeably, I'm sorry if it is not correct in english, I'll keep it in mind next time, but yes indeed I meant spherical.

I already know "my" equation for $\psi$(the first one) is wrong, that's the point of the topic.
Yet I don't think I understand why do you say there should be 2 terms; expand please. I'm using a reference zero potential at infinity, if that's what you mean(can't think of anything else), and I know the second equation for $\psi$ gives me the same result as the indirect solution via gauss' theorem, therefore it is either correct, or a mathematical accident which implies I'm missing something very basic.
Thank you for your time.

 

[maajdl]

See this for spherical coordinates on mathworld:

And from wiki for cylindrical:

In any case, the problem you mention with the 1−e−... would not occur because of the expression for the Laplacian.

Where did you get your expression for the Laplacian?

 

[rude man]

I'm sorry, you were right, you have the correct equation. When expanded it becomes two terms on the left-hand side:

$d^2V/dr^2+(2/r)dV/dr = −ρ/ϵ$

in mks units (Ihave forgotten cgs if that's what you were using).

I will repond again when and if I can be of further help.

 

[rude man]

I believe you should keep your equation as is.
Then you can integrate rV twice:

$d(rV)/dr = (-1/\epsilon) \int\rho(r) \, dr + C_1 = - (1/\epsilon) f(r) + C_1$
$rV = (-1/\epsilon) \int f(r) \, dr + C_1r + C_2$

then divide lhs & rhs by r to get V(r)

so you need to know V(r) at some value of r in addition to r = infinity, for example at r = a.

 

[mastrofoffi]

See this for spherical coordinates on mathworld:

View attachment 246993

And from wiki for cylindrical:

View attachment 246994

In any case, the problem you mention with the 1−e−... would not occur because of the expression for the Laplacian.

Where did you get your expression for the Laplacian?

The radial component of the spherical laplacian that you posted is equal to that of my formula.
I'm writing from my mobile now and it's a bit weird to write latex so I won't post a proof here, but you can easily write both of these expressions as
$$
\frac2r\frac{\partial}{\partial r} + \frac{\partial^2}{\partial r^2}
$$
which means they are identical. (Notice I already multiplied by $r$ on both sides of my Poisson's eq.)

I'll check the other answers tomorrow, and maybe try to explain better what my question is if needed.

 

[vanhees71]

I'm not sure, what you are after, but the standard solution is in terms of the Coulomb potential (using your Gaussian units)
$$\phi(\vec{r})=\int_{\mathbb{R}^3} \mathrm{d}^3 r' \frac{\rho(\vec{r}')}{|\vec{r}-\vec{r}'|}.$$
This is of course too complicated to use for your case.

Here it's easier to start from the differential equation and use spherical coordinates. Due to the spherical symmetry of your charge distribution, i.e., the fact that there's no vector and thus no special direction involved, you can conclude that also the potential must be spherically symmetric, i.e.,
$$\phi(\vec{r})=\phi(r).$$
Then, using the Laplacian in spherical coordinates, you get
$$\Delta \phi=\frac{1}{r} \partial_r^2 (r \phi)=-A \exp(-\alpha r^2),$$
where $A=4 \pi q \alpha^3/\pi^{3/2}$.
Then you first get
$$\partial_r^2 (r \phi)=-A r \exp(-\alpha r^2),$$
which is easy to integrate
$$\partial_r (r \phi) = \frac{A}{2 \alpha} \exp(-\alpha r^2) +C.$$
The next integral leads to an error function,
$$r \phi=\frac{A \sqrt{\pi}}{4 \alpha^{3/2}} [\text{erf}(r \sqrt{\alpha})-1] + C r,$$
where
$$\text{erf}(x)=\frac{2}{\sqrt{\pi}} \int_0^x \mathrm{d} y \exp(-y^2).$$
Finally you get
$$\phi(r)=\frac{A \sqrt{\pi}}{4 \alpha^{3/2} r} \text{erf}(r \sqrt{\alpha}).$$
This is of course only a particular integral. You can add an arbitrary solution of the homogenous equation, which reads
$$\phi_0(r)=\frac{C_1}{r}+C_2.$$
Since there's no point charge in the origin, you have of course $C_1=0$. And usually one puts the constant $C_2$ such that $\phi(r) \rightarrow 0$ for $r \rightarrow \infty$. Thus also $C_2=0$, and the above is the unique solution.

The electric field is given by the gradient, which is easy to calculate:
$$\vec{E}=-\vec{\nabla} \phi(r)=-\frac{\vec{r}}{r} \phi'(r).$$

 

[mastrofoffi]

Thank you @vanhees71 and @rude man, I agree with your results, but that was not the point.
A more straightforward way to ask my question will probably be the following:
given that I do know how to obtain the potential in different ways(green functions, indefinite integration+solving boundary value problem as you two did, gauss' theorem) , I only want to understand how could I do this same thing via definite integration as shown in the last equation written in OP; see pic below for proof that the 2nd equation I gave for $\psi$ is correct(don't care about the italian stuff, its a screen I took from my personal notes and the steps should be quite easy to follow).
More precisely then, the question is: is it possible to somehow 'infer' the correct integration boundaries in order to solve Poisson's equation as of OP?

 

[rude man]

My only additional comments:

(1) you cannot "infer" two boundary conditions for a second-order ODE. They must be given One you have already assumed: V(∞) = 0. The second must be V(r=a).

(2) your original equation was wrong after all; it's dimensionally incorrect. It should be
(1/r) d/dr {d(rV)/dr} = -ρ(r)/ε (in SI units).

But you can still proceed along the lines I have indicated, carrying the 1/r term you omitted in your first post:
d/dr d(rV)/dr) = -rρ(r)/ε
and double-integrating rV as before.