Problem: On the upper right quarter plane $D = {(x,y):0 < x < a}, 0 < y < b; \partial D = C = (x,y):(x=0,y),(x,y=0),(x=a,y),(x,y=b)$, we have the BVP: $$\frac{\partial^2 u(x,y)}{\partial x^2} + \frac{\partial^2 u(x,y)}{\partial y^2} = F = 0 \in \ \ D \\ u = f \in C $$ $$BCs\cases{u(0,y) = y^2, \\ u(a,y) = y^2-a^2, \\ u(x,0) = -x^2, \\ u(x,b) = b^2 - x^2}$$, Use Green's Function to find the expression for the solution $u(x,y)$.
From re-arranging and plugging into the 2-D variants of Green's second identity, we obtain $$u(x,y) = \int\int_DGFdA+\int_Cf\nabla G \cdot \hat{n}dS$$ Where F is the function$\in D$ on the RHS of our PDE where we replace $\xi$ and $\eta$ so we have $F=0$ and that simplifies the problem to: $$\boxed{u(x,y) = \int_C f\nabla G \cdot \hat{n}dS}$$
We know that Green's Function for a $\Delta$ operator is $G(x,y,\xi,\eta) = -\frac{1}{2\pi}\ln(\sqrt{(x-\xi)^2+(y-\eta)^2}) + h$ Where we know for h: $\nabla^2h = 0, \ \ (\xi,\eta)\in D$ and $h = -\frac{1}{2\pi}\ln(\sqrt{(x-\xi)^2+(y-\eta)^2}), \ (\xi,\eta) \in C$
We have 3 mirror image points for the upper right plane so we have the points $(x,-y),(-x,y),(-x,-y)$ which we use to find h. Using these points we get $$\boxed{G = \frac{1}{2\pi}\ln(\sqrt{(\xi-x)^2+(\eta-y)^2})-\frac{1}{2\pi}\ln(\sqrt{(\xi-x)^2+(\eta+y)^2})- \frac{1}{2\pi}\ln(\sqrt{(\xi+x)^2+(\eta-y)^2})+\frac{1}{2\pi}\ln(\sqrt{(\xi+x)^2+(\eta+y)^2})}$$
And now we can formulate the integrand which gives: $$f\nabla G \cdot \hat{n}= (\frac{\partial G}{\partial \xi} + \frac{\partial G}{\partial \eta}) \\ = \frac{1}{2\pi}\bigg(\Big(\frac{\xi-x}{(\xi-x)^2+(\eta-y)^2}-\frac{\xi-x}{(\xi-x)^2+(\eta+y)^2}-\frac{\xi+x}{(\xi+x)^2+(\eta-y)^2}+\frac{\xi+x}{(\xi+x)^2+(\eta+y)^2}\Big)+\Big(\frac{\eta-y}{(\xi-x)^2+(\eta-y)^2}-\frac{y+\eta}{(\xi-x)^2+(\eta+y)^2}-\frac{\eta-y}{(\xi+x)^2+(\eta-y)^2}+\frac{y+\eta}{(\xi+x)^2+(\eta+y)^2}\Big)\bigg)$$
Where $\cdot \hat{n}$ goes away since the gradient vector is normal to the surface at all points on the boundary. Also, we're integrating on the boundary where either $\xi = 0$ or $\eta = 0$, so we can simplify the derivatives down to:
$$\frac{\partial G}{\partial \xi}_{\xi = 0} = \frac{1}{2\pi}\frac{-8xy\eta}{(x^2+(\eta-y)^2)(x^2+(\eta+y)^2)} = -\frac{1}{\pi}\frac{4xy\eta}{(x^2+(\eta-y)^2)(x^2+(\eta+y)^2)}$$
$$\frac{\partial G}{\partial \eta}_{\eta = 0} = \frac{1}{2\pi}\frac{8xy\xi}{(y^2+(x-\xi)^2)(y^2+(x+\xi)^2)} = \frac{1}{\pi}\frac{4xy\xi}{(y^2+(x-\xi)^2)(y^2+(x+\xi)^2)}$$
I'm assuming that for the derivatives, we also need to plug in for the other end of the boundaries at $\xi = a$ and $\eta = b$??
And then lastly we plug this part of the integrand and all 4 Boundary Conditions into the expression for the solution:
This part below is where I know I'm messing up
I'm assuming we want to do a piecewise integration along $0\le \xi \le x$ to $x\le \xi \le a$ and $0\le \eta \le y$ to $y\le \eta \le b$. That's how I saw it done in 1-D for a string with BCs y(0) = y(L) = 0 so I'm applying guesswork here. That example also used a different Green's Function for each interval so I think I DO need to create another set of Green's Functions for the corresponding boundaries at $\xi = a$ and $\eta = b$, but for now this is what I have below. I'm also assuming that for the BCs we can't plug in f(x,b) or f(a,y) or can we?? Below is what I have obtained from my assumptions:
$$u(x,y) = -\frac{1}{\pi}\bigg(\int_0^{y} \frac{4xy\eta}{(x^2+(\eta-y)^2)(x^2+(\eta+y)^2)}f(0,\eta)d\eta + \int_{y}^b \frac{4xy\eta}{(x^2+(\eta-y)^2)(x^2+(\eta+y)^2)}f(0,b)d\eta \bigg)+\frac{1}{\pi}\bigg(\int_0^{x} \frac{4xy\xi}{(y^2+(x-\xi)^2)(y^2+(x+\xi)^2)}f(\xi,0)d\xi + \int_{x}^a \frac{4xy\xi}{(y^2+(x-\xi)^2)(y^2+(x+\xi)^2)}f(a,0)d\xi \bigg)$$ $$=\boxed{-\frac{1}{\pi}\bigg(\int_0^{y} \frac{4xy\eta^3}{(x^2+(\eta-y)^2)(x^2+(\eta+y)^2)}d\eta + \int_{y}^b \frac{4xy\eta b^2}{(x^2+(\eta-y)^2)(x^2+(\eta+y)^2)}d\eta \bigg)+\frac{1}{\pi}\bigg(\int_0^{x} \frac{4xy\xi^3}{(y^2+(x-\xi)^2)(y^2+(x+\xi)^2)}d\xi + \int_{x}^a \frac{4xy\xi a^2}{(y^2+(x-\xi)^2)(y^2+(x+\xi)^2)}d\xi \bigg)}$$
Again, I'm pretty sure I need to go back and make another set of Green's functions for the other 2 boundaries, but I'd like some input first.
Below is a picture of this exact same problem I took from HERE but the main difference is that it used a plane $x>0$, $y>0$. I simply made it finite in the positive and x and y directions.
