I'm trying to compute the solution obtained via Green's Function to test it. It's for Laplace's equation for a semi-infinite slab defined for $y>0$. Here is a snippet from the document which I have been following along below showing the problem and it's solution:
I tried a couple different solutions which satisfy Laplace's Equation and plugged in for $f(\zeta)$ but they didn't work, so I'm assuming that I need to test a specific function to satisfy this problem. What should I plug in to test for this problem?
Thus far I've tried $u(x,y) = y^2 - x^2$. I tried this on the finite domain $x\in [-5,5], y\in [0,10]$ and I plugged in for $f(x,0) = f(x) = -x^2$ and I tried to get the solution for the point (0.1,0.1) to get:
$$u(0.1,0.1) = \frac{1}{\pi}\int_{-5}^5 \frac{0.1}{(0.1-\zeta)^2+0.1^2}\zeta^2d\zeta$$ and plugging this into wolframalpha to obtain a solution, I get 0.318. I was expecting $u(x,y) = 0.1^2 - 0.1^2 = 0$.
EDIT: Modifying the integral to be on an infinite domain, we instead have:
$$u(0.1,0.1) = \frac{1}{\pi}\int_{-\infty}^{\infty} \frac{0.1}{(0.1-\zeta)^2+0.1^2}\zeta^2d\zeta$$ And this integral does not converge.
Finding the indefinite integral we obtain: $$\frac{1}{\pi}\int\frac{y}{(x-\zeta)^2+y^2} = \frac{1}{4\pi}y(x\ln(x^2-2x\zeta+2\zeta^2)+2\zeta)$$. Plugging this into the expression F(b) - F(a) where F is the antiderivative, we get: $$u(0.1,0.1)=\frac{1}{4\pi}(\lim_{\zeta\rightarrow\infty}.2\zeta+.01\ln(.1^2-.2\zeta+2\zeta^2)-\lim_{\zeta\rightarrow-\infty}.2\zeta+.01\ln(.1^2-.2\zeta+2\zeta^2)) = \frac{1}{4\pi}((\infty + \infty)-(-\infty-\infty)) = \infty$$ Which means the integral does not converge.
So the question still remains, what function $f(\zeta)$ can I plug in to obtain the solution $u(x,y)$ for this problem?