I have not seen a worked out solution to a 2-D BVP before, so I went over one of the exercises in Richard Haberman's Applied Partial Differential Eautions and picked a random example that I thought might be a good one to start with to attempt applying Green's function to. The following examples is 8.6 which reads:
$$\nabla^2 u = e^{2y}\sin(x) \\ \cases{u(0,y) = 0 & u(x,0) = 0 \\ u(\pi,y) = 0 & u(x,H) = f(x)} $$
I know the process for how to solve this using method of separation of variables and eigenfunction expansions and that our solution using those methods typically gives us an infinite sine series and some coefficient that we find using orthogonality of sines. How would we solve this same problem using Green's function instead? Or at the very least, I'd like to see the process for obtaining Green's function, $G(x,y,x',y')$, for this problem.
Some things I know coming out of the gate:
$$\tag{1} \text{L}G(x,y) = \delta(x'-x,y'-y)\text{ in }D$$ where L is the linear differential operator, and where I'm guessing our D would be the bounded domain $D = [0,\pi],[0,\text{H}]$ $$\tag{2} G = 0 \text{ on } C$$ Where C is the curve that bounds D and circulates in a positive direction. My intuition is that Green's Theorem will come up here should this condition come up in the process of obtaining $G(x,y,x',y')$.
I've also seen that for a Laplacian differential operator, $G(\boldsymbol{r},\boldsymbol{r'}) = \frac{1}{2\pi}\ln(|\boldsymbol{r}-\boldsymbol{r'}|)$ where I'm assuming $\boldsymbol{r}-\boldsymbol{r'} = <x \hat{i} + y\hat{j}>-<x' \hat{i} + y' \hat{j}> = <(x-x')\hat{i}+(y-y')\hat{j}>$ and the magnitude of this vector would give: $|\boldsymbol{r}-\boldsymbol{r'}| = \sqrt{(x-x')^2+(y-y')^2}$. However, I don't know if this Green function satisfies all PDEs that use a Laplacian or the one in our BVP example for that matter.