I have a 2-D Poisson Equation with homogeneous boundary conditions, and I used an eigenfunction along x, and a Fourier sine series expands the solution to $u(x,y) = \sum_{n=1}^\infty a_n(y)\sin(\frac{n\pi}{L}x)$. Plugging this into our Poisson Equation, I get the following Green's function:
$$G(\textbf{x},\textbf{x}_0) = \sum_{n=1}^\infty\frac{2\sin(\frac{n\pi x_0}{L})\sin(\frac{n\pi x}{L})}{n\pi\sinh(\frac{n\pi H}{L})} \cases{\sinh(\frac{n\pi(y_0-H)}{L})\sinh(\frac{n\pi y}{L}) & $y<y_0$\\ \sinh(\frac{n\pi y_0}{L})\sinh(\frac{n\pi (y-H)}{L}) & $y>y_0$}$$
And I know using Green's function, the solution is of the form:
$$\tag{1}\boxed{u(\textbf{x}) = \int f(\textbf{x}_0)G(\textbf{x},\textbf{x}_0)dA_0}$$.
Expanding out the integral:
$$\tag{2}u(x,y) = \int_0^L\int_0^H f(x,y) G(x,y,x_0,y_0) dy_0 dx_0$$
and plugging in for our Green's function (which is piecewise so it has a different definition at different parts of the domain of integration), I get:
$$\tag{3}u(x,y) = \int_0^L \Bigg[\int_0^y\bigg(\frac{2}{\pi}\sum_{n=1}^\infty \frac{\sin(\frac{n\pi x_0}{L})\sin(\frac{n\pi x}{L})}{n \sinh(\frac{n\pi H}{L})}f(x,y)\sinh(\frac{n\pi(y_0-H)}{L})\sinh(\frac{n\pi y}{L}) \bigg)dy_0 \\ + \int_y^H \bigg(\frac{2}{\pi}\sum_{n=1}^\infty \frac{\sin(\frac{n\pi x_0}{L})\sin(\frac{n\pi x}{L})}{n \sinh(\frac{n\pi H}{L})}f(x,y)\sinh(\frac{n\pi y_0}{L})\sinh(\frac{n\pi (y-H)}{L}) \bigg)dy_0 \Bigg]dx_0$$
Where I assumed $0 < y_0 < H$. Reorganizing and separating the x and y dependent functions, we get:
$$\tag{4}u(x,y) = \int_0^L \frac{2}{\pi}\sum_{n=1}^\infty \frac{\sin(\frac{n\pi x_0}{L})\sin(\frac{n\pi x}{L})}{n \sinh(\frac{n\pi H}{L})}\Bigg[\int_0^y\bigg(f(x,y)\sinh(\frac{n\pi(y_0-H)}{L})\sinh(\frac{n\pi y}{L}) \bigg)dy_0 \\ + \int_y^H \bigg(f(x,y)\sinh(\frac{n\pi y_0}{L})\sinh(\frac{n\pi (y-H)}{L}) \bigg)dy_0 \Bigg]dx_0$$.
And swapping places between the sum and the integral and some more reorganzing:
$$\tag{5}u(x,y) = \frac{2}{\pi}\sum_{n=1}^\infty \frac{\sin(\frac{n\pi x}{L})}{n \sinh(\frac{n\pi H}{L})} \int_0^L \sin(\frac{n\pi x_0}{L}) \Bigg[\sinh(\frac{n\pi y}{L})\int_0^y\bigg(f(x,y)\sinh(\frac{n\pi(y_0-H)}{L}) \bigg)dy_0 \\ + \sinh(\frac{n\pi (y-H)}{L})\int_y^H \bigg(f(x,y)\sinh(\frac{n\pi y_0}{L}) \bigg)dy_0 \Bigg]dx_0$$.
The forcing term for PDE is $f(x,y) = \sin(\pi x)\sin(\pi y)$
$$\tag{6}u(x,y) = \frac{2}{\pi}\sum_{n=1}^\infty \frac{\sin(\frac{n\pi x}{L})}{n \sinh(\frac{n\pi H}{L})} \int_0^L \sin(\pi x_0) \sin(\frac{n\pi x_0}{L})\Bigg[\sinh(\frac{n\pi y}{L})\int_0^y\bigg(\sin(\pi y_0)\sinh(\frac{n\pi(y_0-H)}{L}) \bigg)dy_0 \\ + \sinh(\frac{n\pi (y-H)}{L})\int_y^H \bigg(\sin(\pi y_0)\sinh(\frac{n\pi y_0}{L}) \bigg)dy_0 \Bigg]dx_0$$.
Is this correct? Actually, I mostly care if (3) is correct, so if someone could verify at least that, that should be fine.
I figured out what I did wrong; the integrands were in the incorrect places and this was done because I misunderstood the inequalities. When we integrated along $y_0$ from 0 to 1, that is to say $$0\le y_0 \le 1$$ and $$0<y<1$$, so for some point $y=y_0$ on the number line, to the left of the point, $y>y_0$ and $y<y_0$ to the right. So, the integral for (3) should have been:
$$\tag{3}u(x,y) = \int_0^L \Bigg[\int_0^y\bigg(\frac{2}{\pi}\sum_{n=1}^\infty \frac{\sin(\frac{n\pi x_0}{L})\sin(\frac{n\pi x}{L})}{n \sinh(\frac{n\pi H}{L})}f(x,y)\sinh(\frac{n\pi y_0}{L})\sinh(\frac{n\pi (y-H)}{L})) \bigg)dy_0 \\ + \int_y^H \bigg(\frac{2}{\pi}\sum_{n=1}^\infty \frac{\sin(\frac{n\pi x_0}{L})\sin(\frac{n\pi x}{L})}{n \sinh(\frac{n\pi H}{L})}f(x,y)\sinh(\frac{n\pi(y_0-H)}{L})\sinh(\frac{n\pi y}{L} \bigg)dy_0 \Bigg]dx_0$$
and from there, it can be simplified further to:
$$u(x,y) = \frac{2}{\pi}\sum_{n=1}^\infty \frac{\sin(\frac{n\pi x}{L})}{n \sinh(\frac{n\pi H}{L})} \Bigg[\int_0^L \bigg(\sinh(\frac{n\pi (y-H)}{L}))\int_0^y \sinh(\frac{n\pi y_0}{L}) dy_0 + \\ \sinh(\frac{n\pi y}{H})\int_y^1 \sinh(\frac{n\pi(y_0-H)}{L}) dy_0 \bigg)\sin(n\pi x_0)dx_0\Bigg]$$