I was reading a research paper and I observed the conversion of 2-D integral of Green's Function into its finite form. It used the free space Green's function to solve the following Poisson's Equation: $$\nabla^2 \chi(\textbf{x}) = \alpha(\textbf{x}); \ \ \ \textbf{x} = (x,y)$$. We have the equation for the solution via the free space Green's function as: $$\chi(\textbf{x}) = \frac{1}{2\pi}\int_D \ln(r)\alpha(\textbf{x}')d\textbf{x}' \ \ \ \ \text{where: } \textbf{x}' = (x',y'),r = \sqrt{(x-x')^2+(y-y')^2}$$. According to the paper, the discrete form of this equation (which I need help unpacking and I'll box it to emphasize its importance) is
$$\boxed{\chi(\textbf{x}_{i.j}) = \frac{1}{2\pi} \sum_p \sum_q \alpha(\textbf{x}_{p,q}) \times \Big(\ln(|p-i,q-j|)+\ln(\delta s)\Big)\delta s^2} \\ \text{where } \delta s = \delta x = \delta y, \text{ and } \textbf{x}_{p,q} = (x_p,y_q) = (p \delta s, q\delta s)$$.
So far what I can assume based on my own knowledge is couple of things.
- First, given from the paper that $\delta s = \delta x = \delta y$, I'm assuming that the term $\delta s^2$ is the finite form of $d\textbf{x}' = dxdy$ which is coming from the Riemann sum equivalent of the integrals.
- Second, that and $\ln(|p-i,q-j|)$ with the use of a comma looks odd to me, but I'm guessing it would be written out as $\ln(\sqrt{(p-i)^2+(q-j)^2})$.
- and Third, that the finite form of r is $\sqrt{(p\Delta x - i\Delta x)^2 + (q\Delta y - j\Delta y)^2} = \sqrt{\big((p-i)^2+(q-j)^2\big)\delta s^2}$ and taking the natural log gives: $\frac{1}{2}\Big(\ln\big((p-i)^2 + (q-j)^2\big)+2\ln(\delta s)\Big)$. Redistributing the $\frac{1}{2}$ gives $\ln(r(\textbf{x}_{p,q})) = \ln\big(\sqrt{(p-i)^2 + (q-j)^2}\big)+\ln(\delta s) = \ln(|p-i,q-j|)+\ln(\delta s)$.
What's confusing me is that this looks like some vectorized form hence the cross product, however neither are vectors, both are magnitudes. $\alpha$ is actually defined as $\alpha = \nabla \cdot \textbf{V}$ meaning it's a scalar field representing divergence, and we're using r as the scalar distance between $\textbf{x}$ and $\textbf{x}'$ making it a scalar field as well, so unless $\times$ means something else in the finite difference scheme or whatever context might give it a different meaning, I don't understand what this operator is supposed to mean. Any thoughts?