Why does $G = \log|(x,y) - (ξ,η)| $ satisfy the laplace equation for all $(x,y) \ne (ξ,η)$
Could someone explain why this is true?
I thought $\frac{\partial G}{\partial x} = \frac{1}{(x,y) - (ξ,η)}$ and $\frac{\partial G}{\partial y} = \frac{1}{(x,y) - (ξ,η)}$
and so $\frac{\partial^2 G}{\partial x^2}+ \frac{\partial^2 G}{\partial y^2} = \frac{-1}{[(x,y) - (ξ,η)]^2} + \frac{-1}{([(x,y) - (ξ,η)]^2}$
Your calculation of the partial derivatives is wrong. The expression $(x,y) - (\xi,\eta)$ is a vector and doesn't make sense in a denominator. Instead you should write $$G(x,y) = \log \sqrt{(x-\xi)^2 + (y - \eta)^2} = \frac 12 \log \left( (x-\xi)^2 + (y - \eta)^2\right)$$ and work from there. For instance, $$\frac{\partial G}{\partial x}(x,y) = \frac{x-\xi}{(x-\xi)^2 + (y - \eta)^2}$$ so that $$ \frac{\partial^2 G}{\partial x ^2}(x,y) = \frac{ -(x-\xi)^2 + (y-\eta)^2}{(x-\xi)^2 + (y - \eta)^2}.$$