I want to calculate Green's Function to solve $\triangle u = f(x,\ y)$, using Laplace Transforms. My plan was to tailor boundary conditions to the problem which simplify the computation.
Because the Laplacian is self-adjoint, my associated Green's Function equation can immediately be written as $G_{xx} + G_{yy} = \delta(x - x_G,\ y - y_G)$.
Laplace Transforming $y$ to $s$ and $G$ to $\hat G$, I get $\hat G_{xx} + s^2 \hat G - sG(x,\ 0) - G_y(x,\ 0) = e^{-y_G s} \delta(x - x_G)$. Thus to simplify things, I'll go back and include $u(x,\ 0) = 0,\ u_y(x,\ 0) = 0$ in the problem statement, so that the terms with $G(x,\ 0)$ and $G_y(x,\ 0)$ drop out. Actually, I'm not $100$% sure if this is correct. I'm waiting on the results of some other questions to understand/play around with this more. Regardless, let's assume I prescribe some boundary conditions on $u$ such that $G(x,\ 0)$ and $G_y(x,\ 0)$ drop out. Now I have the ODE $\hat G_{xx} + s^2 \hat G = e^{-y_G s} \delta(x - x_G)$.
Laplace Transforming $x$ to $r$ and $\hat G$ to $\hat{\hat G}$, I get $r^2 \hat{\hat G} - r \hat G(0,\ s) - \hat G_x(0,\ s) + s^2 \hat{\hat G} = e^{-y_G s}e^{-x_G r}$. Again to simplify things, I'll go back and include $u(0,\ y) = 0,\ u_x(0,\ y) = 0$ in the problem statement. Now I have $\hat{\hat G} = e^{-y_G s}e^{-x_G r}\frac{1}{r^2 + s^2}$.
Inverting from $r$ back to $x$, by applying the shifting theorem, I think I should get $\hat G = e^{-y_G s}\frac{\sin(s(x - x_G))}{s}H[x - x_G]$, where $H[\ ]$ is the Heaviside step function.
Inverting from $s$ back to $y$, the first factor is another shift, and the last factor is constant with respect to $s$, so the challenge is to find the inverse transform of $\frac{\sin(s(x - x_G))}{s}$. This does not seem easy to find directly (using tables or software; I don't know how to do it with Complex Analysis), so I break the sine function into exponentials and use $\frac{e^{\_\_\_s}}{s} \iff H[y + \_\_\_]$ for each exponential. After some algebra, I arrive at $$G = \frac{H[y - y_G - xi + x_G i]H[x - x_G]H[y - y_G]i - H[y - y_G + xi - x_G i]H[x - x_G]H[y - y_G]i}{2}$$
The Boundary Value Problem has been refined into $\triangle u = f(x,\ y),\ u(x,\ 0) = 0,\ u_y(x,\ 0) = 0,\ u(0,\ y) = 0,\ u_x(0,\ y) = 0$. Is this the correct Green's Function for this BVP? If so, then how can I proceed to clean it up into a real, and more useful, expression and use it to solve the BVP for some particular forcing function? If not, then what went wrong? If the $i$'s cannot be eliminated, is this an indication that I should use the Fourier Transform instead of the Laplace Transform for one or both steps? Would a better choice of boundary conditions produce a more palatable Green's Function?