I've found Green's function, the problem is I've never done the integration part to find the solution for a 2-D problem or higher before, so I'd like a little guidance for setting up the integral. The integral should look like this:
$$\tag{1}u(\textbf{x}) = \int_a^b f(\textbf{x'}) G(\textbf{x},\textbf{x'}) d\textbf{x'}$$
I just don't know what that's supposed to look like when we expand from vector (or bold) notation. I know it should expand into a double integral like this:
$$\tag{2}u(x,y) = \int_0^H\int_0^L f(x',y')G(x,y,x',y') dx'dy'$$ I'm just hung up on the piecewise definition of Green's function complicating things. In my example, I had a piecewise definition along y, and it was continuous along x. To illustrate it had the form $G(x,y,x',y') =G(x,x')G(y,y') = G(x,x')\cases{G(y,y'^-) & y<y' \\ G(y,y'^+) & y>y'}$. This is my guess for what I think it should be:
$$\tag{3}\boxed{u(x,y) = \int_0^L\bigg(\int_0^y f(x',y')G(x,y,x',y'^-)dy'+\int_y^H f(x',y')G(x,y,x',y'^+)dy'\bigg)dx'}$$
Please let me know if this looks correct!
Below is my work for my specific case
If my guess if correct, then for my problem:
$$u_{xx} + u_{yy} = x(x-1)y(y-1); \ \ 0 \le x \le L, 0 \le y \le H \\ BCs \cases{u(0,y) = 0 \\ u(L,y) = 0 \\ u(x,0) = 0 \\ u(x,H) = 0}$$
where I used an eigenfunction expansion to set up my Green's function and I arrived to:
$$\tag{1}G(x,y,x',y') = \sum_{m=1}^\infty c_m G(x,x') \begin{cases} G(y,y'^-) & y<y' \\ G(y,y'^+) & y>y'\end{cases}$$ $$\tag{2}c_m = \frac{2}{m\pi \sinh(\frac{m\pi}{L} H)}\\ G(x,x') = \sin(\frac{m\pi}{L}x)\sin(\frac{m\pi}{L}x'), \ \ \ G(y,y'^-) = \sinh(\frac{m\pi}{L}(y'-H)) \sinh(\frac{m\pi}{L}y), \\ G(y,y'^+) = \sinh(\frac{m\pi}{L}y')\sinh(\frac{m\pi}{L}(y-H)) $$
$$\tag{3}\boxed{u(x,y) = \sum_{m=1}^\infty c_m \Bigg[\int_0^L\bigg(\int_0^y g(y')G(y,y'^-) + \int_y^H g(y')G(y,y'^+) dy'\bigg)h(x')G(x,x') dx'\Bigg]}$$
Where I split up $f(x,y) = g(x)h(y); g(x) = x(x-1),h(y) = y(y-1)$ to make the equation look a bit cleaner.
EDIT: I computed the results for this integral and the results were atrocious not to mention it was a highly verbose result that took a very long time to find all in addition to being incorrect and I do NOT want to go back and pick through where the mistake was made. Is this where I use Green's Theorem to sidestep the area integral?