# Periodic Green's function in $\mathbb{R}^3$

A while ago I asked about the Green's function on the surface of a cylinder in $\mathbb{R}^3$, and it turned out to be very easy to compute by conformally mapping the cylinder to the half-plane. Now I'm interested in the Green's function of the Laplacian on the three-dimensional cylinder $(x, y, \cos\theta,\sin\theta)$ in $\mathbb{R}^4$; by symmetry and developability it is enough to find the Green's function $G(x,y,\theta)$ at the origin of the three-dimensional slab $\mathbb{R}^2 \times [-\pi,\pi]$, with periodic boundary conditions in $\theta$.

I thought this would be a simple calculation, but after several attempts I still have not gotten anywhere. One first approach is to use the reflection principle: $\frac{1}{\sqrt{x^2+y^2+\theta^2}}$ is (up to a constant factor) the Green's function for free space, so $$G(x,y,\theta) \propto \sum_{i=-\infty}^{\infty} \frac{1}{\sqrt{x^2+y^2+(\theta+2\pi i)^2}}$$ satisfies the periodicity condition and is harmonic everywhere it is defined; unfortunately $G(0,0,\theta)$ diverges for every $\theta$.

A second attempt was to start from $\Delta G = \delta(x)\delta(y)\delta(\theta)$ and take the Fourier transform in $\theta$; this leads to the modified Bessel equation and $$G(x,y,\theta) \propto \sum_{i=1}^{\infty} \cos(i \theta) K_0\left(i\sqrt{x^2+y^2}\right)$$ where $K_0$ is the modified Bessel function of the second kind. Unfortunately, applying Poisson summation takes me right back to the divergent $G$ above.

Where am I going wrong? How can I compute the Green's function on this domain?

UPDATE: After more carefully doing the above separation of variables I get that the Green's function must be of the form $$G(x,y,\theta) = \frac{\log \sqrt{x^2+y^2}}{4\pi^2} + \sum_{i=1}^{\infty} \left(\cos(i\theta)\left[k_i I_0\left(i \sqrt{x^2+y^2}\right) - \frac{1}{2\pi^2}K_0\left(i \sqrt{x^2+y^2}\right)\right]\right)$$ where the $k_i$ are arbitrary constants; since the $I_0$ terms are harmonic everywhere we might as well set $k_i=0$ to get $$G(x,y,\theta) = \frac{\log \sqrt{x^2+y^2}}{4\pi^2} - \frac{1}{2\pi^2} \sum_{i=1}^{\infty} \cos(i\theta)K_0\left(i \sqrt{x^2+y^2}\right)$$ which does seem to converge, painfully slowly, as $\sqrt{x^2+y^2}\to 0$. I suppose the terms can be combined into the even more dubious $$G(x,y,\theta) = - \frac{1}{2\pi^2} \sum_{i=1}^{\infty} \cos(i\theta)\left[\log \sqrt{x^2+y^2} + K_0\left(i \sqrt{x^2+y^2}\right)\right].$$ But looking at series expansions of the bracketed terms etc. hasn't yet yielded fruit.

I'm still hoping for a nice closed-form formula.

UPDATE 2: As Words points out in the comments, there is no hope for a closed form unless special cases like $\sum_{i=1}^{\infty} K_0(i)$ have a closed form. Even if a closed form is not possible, I would be happy even with a series that converges at a reasonable rate at any point away from the origin (including near $x=y=0$).

But I haven't given up yet on transforming $\sum K_0$ into something more reasonable. For instance, by Poisson summation \begin{align*} \sum_{i=1}^{\infty} K_0(i) &= \sum_{i=1}^{\infty} \begin{cases}\log i + K_0(i), & i\leq 1\\K_0(i), & i > 1\end{cases}\\ &= -\lim_{i\to 0} \frac{\log i+K_0(i)}{2} + \frac{\pi-2}{2} + \sum_{i=1}^{\infty}\left(\frac{\pi}{\sqrt{(2\pi i)^2+1}} - \frac{\textrm{Si}(2\pi i)}{\pi i}\right)\\ &= \frac{\pi+\gamma-\log 2 -2}{2} + \sum_{i=1}^{\infty}\left(\frac{\pi}{\sqrt{(2\pi i)^2+1}} - \frac{\textrm{Si}(2\pi i)}{\pi i}\right). \end{align*} Of course I don't have reason to believe at the moment that a series involving the sine integral is somehow better than one involving the modified Bessel function, but it's not obvious to me that there aren't additional or different transformations possible to tame the series.

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You may want to observe that this kind of sum would also diverge in the planar case. Yet $G$ exists. Therefore, there is a harmonic summand somewhere, which makes the sum convergent. Same here. Also consider the physical interpretation. You placed a unit source in the slab. The flow cannot escape thorough the boundary. So it essentially dissipates in two dimensions only. Its gradient should decay like $1/r$ to keep the flux constant. Therefore, $G$ has logarithmic behavior, it's negative at infinity. – user147263 Jun 6 '14 at 6:01
@wordsthatendinGRY Yes, interesting point about the scaling. – user7530 Jun 6 '14 at 6:18
Looks better now. Notice that circular integrals, $\int_{-\pi}^{\pi} G(x,y,\theta)$ must be the fundamental solution in the plane (which it is). The additional terms in the series refocus the singular source from the line $x=y=0$ into the point $x=y=\theta=0$. I was going through essentially the same steps, but decided to begin with the 2D case (where the end result is already known). That may have been a mistake, because in the 2D case I got $-\frac{1}{4\pi}|x| + \sum_{n=1}^\infty \frac{1}{n} \sinh(n|x|)\cos ny$, which has no prayer of converging... – user147263 Jun 6 '14 at 23:52
Ugh, how dumb of me to keep the growing exponential term. The 2D solution from your other question is $$G(x,\theta) = -\frac{|x|}{4\pi} + \frac{1}{2\pi}\sum_{n=1}^\infty \frac{1}{n}e^{-n|x|}\cos n\theta$$ Which magically happens to have the explicit formula $-\frac1{4\pi} \log|2\cosh (x+i\theta) - 2|$. – user147263 Jun 7 '14 at 1:23
But in 3D, we are just not so lucky as to have an explicit formula. Take $y=\theta=0$ for simplicity: what is $\sum_{n=1}^\infty K_0(nx)$? Maple does not know. Take $x=1$; what is $\sum_{n=1}^\infty K_0(n)$? Can be calculated with high precision, but the Inverse Symbolic Calculator does not recognize the result. This is hopeless. – user147263 Jun 7 '14 at 1:41

For $\sqrt{x^2+y^2} \gg 0$: $$G(x,y,\theta) = \frac{\log \sqrt{x^2+y^2}}{4\pi^2} - \frac{1}{2\pi^2}\sum_{i=1}^{\infty} \cos(i\theta) K_0\left(i \sqrt{x^2+y^2}\right).$$ For $\sqrt{x^2+y^2} \approx 0$ the above converges extremely slowly (and is undefined at $x^2+y^2=0$) and the following (derived by Poisson summation from the above) can be used instead: $$G(x,y,\theta) = \frac{\log 2-\gamma}{4\pi^2} - \frac{1}{2\pi}\sum_{i=-\infty}^{\infty} \left[\frac{\pi}{2\sqrt{(\theta+2\pi i)^2+x^2+y^2}} - \frac{\operatorname{Si}(\theta+2\pi i)}{\theta+2\pi i}\right].$$
In the above, $K_0$ is the modified Bessel function of the second kind, $\gamma$ is the Euler-Mascheroni constant, and $\operatorname{Si}$ is the sine integral.