I want to find the solution to the equation: $$ x \partial_y f - y \partial_x f = \delta^{(2)}(x,y) \,, $$ where $f : \mathbb{R}^2 \to \mathbb{R}$ is a scalar function, and $\delta^{(2)}(x,y)$ is the 2-dimensional Dirac Delta function centered at the origin of $\mathbb{R}^2$.
I know that the solution of the homogeneous equation $\partial_y f - y \partial_x f = 0$ is: $$ G(x,y) = h(x^2 + y^2) \,, $$ for any single-argument function $h$.So shouldn't $G(x,y)$ for a certain $h$ be the solution to the equation above, like what happens for the Green function of the Laplacian in 3D?
However, I'm not sure how to prove that. I can require $f$ to be a solution of the first equation in a weak sense: $$ \int_B dx dy \, g(x,y) \left( x \partial_y f - y \partial_x f \right) = g(0,0) \,, $$ where the integral is over a compact region $B$ that includes the origin. I can then write the left-hand side as: $$ \int_B dx dy \, g(x,y) \epsilon^{ij} x_i \partial_j f \,, $$ where $\epsilon^{12}=-\epsilon^{21}=1$ and $\epsilon^{11}=\epsilon^{22}=0$, and I can integrate by part the derivative: $$ \int_B dx dy \, g(x,y) \epsilon^{ij} x_i \partial_j f = \int_B dx dy \partial_j \left(g(x,y) \epsilon^{ij} x_i \, f \right) - \int_B dx dy \, f \epsilon^{ij} x_i \partial_j g(x,y) $$ $$ = \oint_{\partial B} d\sigma_j \epsilon^{ij} x_i \, f \, g(x,y) - \int_B dx dy \, f \epsilon^{ij} x_i \partial_j g(x,y) $$ where $d\sigma_j$ is the normal to the boundary curve $\partial B$.
This is where I stop... taking $g(x,y) = 1$ I can prove that the first of the two integral is zero when $f(x,y) = G(x,y)$, but then of course the second integral is zero too. Is it possible to show that the above expression is equal to $g(0,0)$, or perhaps it's just not true that $G(x,y)$ weakly solves the first equation?
The 2D curl operator is $\partial_{\theta}$ in polar coordinates, so since the 2D Dirac delta distribution $\delta^2(x,y)$ does not depend on $\theta={\rm atan2}(x,y)$, the Greens function is morally $G(x,y)=\theta\delta^2(x,y)$, which is not mathematically well-defined.
A better strategy is to not use a Greens function and instead to Fourier expand a general inhomogeneous source term in a Fourier series $$ \partial_{\theta}f(r,\theta)~=~\sum_{n\in\mathbb{Z}}c_n(r)e^{in\theta},$$ which has a formal (possibly multivalued) solution $$ f(r,\theta)~=~\theta c_0(r)~+~\sum_{n\in\mathbb{Z}\backslash\{0\}}\frac{c_n(r)}{in}e^{in\theta} ~+~h(r). $$