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Recently I faced several moral issues while I was solving Poisson's equation in a simple case of point charge in $\overline{r}=\overline{0}$. They concern Fourie transform and improper integral calculation.

Here is the deal. I have $\Delta\phi(\overline{r})=4\pi e\delta(\overline{r})$.
Let's take Fourier transform as $F(\overline{p})=\int{f(\overline{r})\exp(-i\overline{r}\cdot\overline{p})} \, d\overline{r}$ and the inverse transform as

$$f(\overline{r})=\frac{1}{(2\pi)^3}\int{F(\overline{p})\exp(i\overline{r}\cdot\overline{p})} \, d\overline{p}.$$

So I can rewrite my equation as

$$\frac{1}{(2\pi)^3}\Delta \left[\int{\Phi(\overline{p})\exp(i\overline{r}\cdot\overline{p})} \, d\overline{p}\right]=\frac{4\pi e}{(2\pi)^3}\left[\int{\exp(i\overline{r}\cdot\overline{p})} \, d\overline{p}\right]$$

or $$\int{[-p^2\Phi(\overline{p})]\exp(i\overline{r}\cdot\overline{p})d\overline{p}}=\int{[4\pi e]\exp(i\overline{r}\cdot\overline{p})d\overline{p}}.$$

This gives me algebraical equation for $\Phi(\overline{p})$:
$\Phi(\overline{p})=-\frac{4\pi e}{p^2}$.
After that I can find the potential as
$\phi(\overline{r})=-\frac{4\pi e}{(2\pi)^3}\int{\frac{\exp(i\overline{r}\cdot\overline{p})}{p^2}d\overline{p}}=-\frac{e}{2\pi^2}\int{\frac{\exp(i\overline{r}\cdot\overline{p})}{p^2}d\overline{p}}$.
And here comes my first confusion: as we don't use the inverse transform the choise of coefficients in straight and invers Fourie transforms results the solution. If we take $F(\overline{p})=\frac{1}{(2\pi)^3}\int{f(\overline{r})\exp(-i\overline{r}\cdot\overline{p})}d\overline{r}$ and $f(\overline{r})=\int{F(\overline{p})\exp(i\overline{r}\cdot\overline{p})}d\overline{p}$, it gives us
$\phi(\overline{r})=-4\pi e\int{\frac{\exp(i\overline{r}\cdot\overline{p})}{p^2}d\overline{p}}$.
What am I doing wrong?

Let's consider, that everything is OK with the above. Now I have to calculate the integral $\int{\frac{\exp(i\overline{r}\cdot\overline{p})}{p^2}d\overline{p}}$. First of all I pass on to spherical coordinate system:
$\int\limits_{0}^{2\pi}d\phi\int\limits_{0}^{+\infty}dp\int\limits_0^\pi\exp(ipr\cos\theta)\sin\theta d\theta=\\ |t=\cos\theta|= -2\pi\int\limits_{0}^{+\infty}dp\int\limits_1^{-1}\exp(iprt)dt= \frac{2\pi}{ir}\int\limits_{0}^{+\infty}\frac{[\exp(ipr)-\exp(-ipr)]dp}{p}$
Then a little trick:
$\int\limits_{0}^{+\infty}\frac{[\exp(ipr)-\exp(-ipr)]dp}{p}= \int\limits_{0}^{+\infty}\frac{\exp(ipr)dp}{p}-\int\limits_{0}^{+\infty}\frac{\exp(-ipr)dp}{p}=\\ |p\rightarrow-p\ in\ second\ integral|= \int\limits_{-\infty}^{+\infty}\frac{\exp(ipr)dp}{p}$.
So we have $\phi(\overline{r})=-\frac{e}{i\pi r}\int\limits_{-\infty}^{+\infty}\frac{\exp(ipr)dp}{p}$ and last thing to do is to calculate $\int\limits_{-\infty}^{+\infty}\frac{\exp(ipr)dp}{p}$.
To do this I extend integration curve to the top complex half-plane with curve $C_\infty$, which goes counter-clockwise. It gives me

$$\int\limits_{-\infty}^{-\epsilon}[\cdots]+\int\limits_{C_\epsilon}[\cdots]+ \int\limits_{+\epsilon}^{+\infty}[\cdots]+\int\limits_{C_\infty}[\cdots]$$

which is equal to $0$ as $\int\limits_{C_\epsilon}[...]$ goes over the pole $p=0$ ($p=\epsilon\exp(i\alpha)$, $\alpha=[\pi;0]$) and we have no poles in closed curve.
$\int\limits_{C_\infty}[...]=0$ so $\int\limits_{-\infty}^{-\epsilon}[...]+ \int\limits_{+\epsilon}^{+\infty}[...]=-\int\limits_{C_\epsilon}[...]$.
With $\epsilon\rightarrow0$ we have the sought-for integral in the left part of the last equation. So the goal is to find
$\int\limits_{C_\epsilon}\frac{\exp(ipr)dp}{p}$.
And there is my second problem. I suppose, that I should somehow bring it to Cauchy's integral formula, but I just can't do it. My flair tells me that it is just half the result, that we have if we take the full circle instead of half of it, but I would like to find something stricter.

P.S. I expect someone to advice me using Green's function, but I prefer to find out, what is wrong with my solution. Basically because I hardly understand the essense of Green's function :)

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migrated from mathoverflow.net Aug 12 '13 at 21:06

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2  
Moral issues with a math problem makes it sound like you think some types of math are sinful. :P And using a Green's function to solve this problem would be fruitless, as the solution is the Green's function. –  Muphrid Aug 12 '13 at 23:09

1 Answer 1

  1. When you change the normalization for the Fourier transform, the constant changes when you apply the rule for differentiation, and that's why the constant is missing.

  2. When you take the Fourier transform of the constant function as in your second part, you are expecting to get again the dirac delta right? You can't rigorously write the integral like (the first rewrite where you take the Fourier transform of the delta function and try to apply the inverse), because you aren't going to get a function back that way. It looks like you tried anyway, and there are going to be convergence issues. In this case, you need to study Fourier transforms on the space of distributions, and that you can find for instance here: Fourier Transform of Dirac Delta Function.

At the end of the day, you can still take the Fourier transform of both sides as you did (kind of), but don't expect to use the normal inverse Fourier integral to recover a non-function!

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