Consider the two dimensional laplacian $$\Delta=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}$$
Set $x=r \cos \theta$ , $y =r \sin \theta$ , then we have $$\Delta=\frac{\partial^2}{\partial r^2}+ \frac1r \frac{\partial }{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2} \,\,\,\,\,\,\,\,\,\, (1)$$
First, to deduce the above equation , we need to know $\frac{\partial \theta}{\partial y}$ . However , the function $\theta(x,y)$ is not even continuous with respect to $y$ $$\lim_{y \to 0^+} \theta(1,y)=0 \neq 2\pi=\lim_{y \to 0^-} \theta(1,y)$$
Second , if we want a function $u$ in the unit ball which has $\Delta = 0$ ; then $u_{xx}+u_{yy}|_{(x,y)=(0,0)}$ has to be zero . However , we can not use polar system to show this , since $\theta(x,y)$ can only defind in $R^2-\{0\}$ .
My question :
Why we can use polar system to solve laplace equation in the open ball ? I can not see why the equation $(1)$ is valid when $(r,\theta)$ is not continuous with respect to $(x,y)$ .