Is it possible to make this physicists' argument rigorous using distributions?

Question

We are interested in the motion of an incompressible, irrotational fluid through the infinite channel $\{z:-a<\mathrm{Im}(z)<a\}\subset\mathbb C$ induced by a single point source of strength $Q$ at $z=0$. The method of images would have us reflect the point source in both boundaries to create image sources, and then reflect those image sources, and so on, until we have an image source at each point $z=2kai,k\in\mathbb Z$; which then "implies", by superposing all of these, that the complex velocity "potential" looks like $$w(z)``="\frac{Q}{2\pi}\lim_{n\to\infty}\sum_{k=-n}^n\log(z-2kai),$$which we then are able to differentiate termwise to obtain a correct (and convergent) expression for the velocity. As a pure mathematician by training, two scary things immediately jump out to me:

1. The logarithms here are multivalued, as this wouldn't be a valid potential otherwise- we need to be able to differentiate this potential everywhere away from the sources to obtain a smooth velocity field. So without defining branches, I am finding it very hard to think about the imaginary part of the sumand.
2. The real part of the sumand is $O(\log|k|)$, and so the sumand shoots off to infinity, let alone the sum itself. So this potential as it stands doesn't look very sensible- and that's even before we've tried to justify interchanging the summation and differentiation to obtain a velocity!

My gut instinct is far more concerned by 2) than by 1)- I somewhat trust the physicists' justification for thinking about a multivalued logarithm here, and I can easily imagine some "patchwork" reasoning to make sense of that aspect of the sum. After all, the potential itself doesn't matter much- the velocity is the only thing of interest, and each point away from the sources will have some small enough neighbourhood in which I can make sure the logarithms are well-defined, after which I can probably patch up all the little pieces to give me a smooth velocity field.

My hope for 2) is that the "correct" notion of a velocity potential should always be given in terms of distribution theory. From the little that I know about distributions, I could imagine that the series expression for $w$ somehow converges in distribution to a well-defined distribution of some description, and that then the "nice" behaviour of distributions interacting with differentiation makes the termwise differentiation of that sum fine. I vaguely remember having seen similar ideas show up in the context of Fourier series, so it doesn't seem too far-fetched to me that this approach might work. However, I have yet to take a formal distribution theory course (I know a few definitions and not much more), and so I wouldn't know where to start trying to formalise any of this into a convincing, rigorous argument. Is it possible to do so? And if so, what would this formalisation look like?

Answer 1

When your source is a delta-function then you already inside distribution land, but the resolution to your questions have little to do with this. Once you have gotten to the potential for one source (the function $\log(z)$) then all of the subsequent steps, the application of the method of images, can be formally justified in normal function land. The TLDR answer to the question about what is the "correct" velocity potential is simply that there isn't one "correct" potential.

The potential, when including the first $n$ pairs of symmetric images, can be written $$w_n(z) = \frac{Q}{2\pi}\left[ \log(z) + 2\log(2^na^nn!) + \log\prod_{k=1}^n\left(1 + \frac{z^2}{4a^2k^2}\right)\right].$$ The last product converges to $\sinh(\pi z/2a)/(\pi z/2a)$ as $n\to \infty$ (see e.g. this), while the middle term diverges. The potential is however only defined up to a constant so we are free to choose the zero point. This is often glossed over in usual treatments as the next step is anyway to take the derivative of it to find the flow and then this "divergent constant" is differentiated away. To do it more rigorously, we can define the potential to be zero at some point $z_*$ and then our potential is $$\tilde{w}_n(z) \equiv w_n(z) - w_n(z_*) = \frac{Q}{2\pi}\left[ \log(z/z_*) + \log\prod_{k=1}^n\left(1 + \frac{z^2}{4a^2k^2}\right)/\left(1 + \frac{z_*^2}{4a^2k^2}\right)\right],$$ which is free of infinities as $n\to\infty$ and gives us $$\tilde{w}(z) \equiv \lim_{n\to\infty}\tilde{w}_n(z) = \frac{Q}{2\pi}\log\left(\sinh\left(\frac{z\pi}{2a}\right) / \sinh\left(\frac{z_*\pi}{2a}\right)\right).$$ The derivative of this gives the final result, without having to do term-by-term differentiation, $$v = \frac{d\tilde{w}(z)}{dz} = \frac{Q}{4a}\coth\left(\frac{z\pi}{2a}\right).$$ Note that this is also independent of the arbitrary zero-point (as it better be). If you want to apply term-by-term differentiation instead then I suggest you first compute $$\frac{d\tilde{w}_n(z)}{dz} = \frac{Q}{2\pi}\left[ \frac{1}{z} + \sum_{k=1}^n \frac{2z}{4a^2k^2 + z^2} \right]$$ and then try to show that this series converges uniformly inside the channel. The relevant theorem that then guarantees term-by-term differentiation is listed here. If we call the sum above $f_n(z)$ then a simplest approach is to just estimate $|f(z) - f_n(z)| = \sum_{k=n+1}\frac{2|z|}{|4a^2k^2 + z^2|}$ on a disc $|z| \leq R$ and show that we can make it arbitrarily small independent of $z$. For example if we put $k_* = \lfloor R/2a \rfloor$ then a straight forward estimation gives us $|f(z) - f_n(z)| \leq \frac{k_*}{a}\sum_{k=n-k_*+1}\frac{1}{k^2}$ which is enough to get uniform convergence on any compact subset of $\mathbb{C}$.

Finally, lets get back to your question about what exactly is the logarithm above? There is indeed some ambiguity for the potential with regards to what branch you choose. But just as with the zero-point above you are free to define this yourself and again this choice does not affect any of the physics. The difference between branches is just a constant so it's like the "infinite constant" above - any ambiguity gets differentiated away. Also note that none of the images $\log(z + 4ika)$ gets close to the branch-cut as we move around the channel so issues of moving across a branch-cut only applies for the main source ($\log(z)$). If you really wanted to avoid getting close to the branch-cut at all you could also restrict the analysis to $\Re z > 0$ and then extend to $\Re z < 0$ using symmetry of the problem: $v(z) = -v(-z)$, though it's not really needed.