Neumann problem for the Laplacian with Dirac delta functions

Question

I have encountered a problem while dealing with the adjoint method in potential flow that is also described, in a similar fashion, in (eq. 39) of this paper. The problem is essentially this: $$\begin{cases} \Delta f= 0 & B \\ \partial_n f=\delta(x-x_0) & \partial B^+ \\ \partial_n f=-\delta(x-x_0) & \partial B^-. \\ \end{cases}$$

$B$ is the domain exterior to a unit disc in $\mathbb{R}^2$, $\delta$ is the Dirac delta function, $x_0$ is the point $(1,0)$, and $\partial B^\pm$ are the half-circles corresponding to the non-negative (resp. negative) values of the Cartesian coordinate $y$. I can't really make sense of this and I'm inclined to think that the problem is somehow ill-defined or has a trivial solution, but I don't know how to proceed. Any clues?

Answer 1

It occurs to me that perhaps one can split the problem in two

$$\begin{cases} \Delta f^+= 0 & B \\ \partial_n f^+=\delta(x-x_0) & \partial B \\ \end{cases}$$

and

$$\begin{cases} \Delta f^-= 0 & B \\ \partial_n f^-=-\delta(x-x_0) & \partial B \\ \end{cases}$$

and use Poisson formula $$ f(x)=-\frac{1}{\pi} \int_{\partial B} \partial_n f(y) \log\vert x-y\vert dy $$ for the Neumann problem to each case and then add the solutions. For each of the problems separately, the formal solution would be $$ f^\pm=\mp \frac{1}{\pi}\log\vert x-x_0\vert $$ so, apparently, the total solution $f^++f^-=0$. One has of course to be careful when dealing with this delta functions, and, besides, each of the subproblems is probably not well defined as $$ \int_{\partial B} \partial_n f(y) dy \ne 0 $$ but I guess this could be arranged by adding or subtracting 1 from the Dirac delta in the Neumann b.c. for each subproblem, which would still yield $f^++f^-=0$. I am aware that the argument is a bit sloppy, but this is the first thing that came to mind.

Alternatively, one can split the singularity as follows $$\begin{cases} \Delta f= 0 & B \\ \partial_n f =\delta(x-x_\epsilon)-\delta(x-x_{-\epsilon}) & \partial B \\ \end{cases}$$ where $x_{\pm\epsilon}=(\cos\epsilon,\pm\sin\epsilon)$ (with $\epsilon <<1)$ and use again Poisson formula to solve the problem with the same (zero) result.