Consider a rectangular area, defined by the region $x=0,x=a,y=0,y=b$. Now, there is a potential $\phi(x,y)$ defined in this region, which satisfies, $\nabla^2 \phi=0$, and the following boundary conditions: $\phi(0,y)=\phi(x,0)=\phi(a,y)=0$ and $\phi(x,b)=Ax/a$.
My Argument What I am really confused about, is whether a boundary value problem is solvable, if it boundary values themselves contradict each other. For e.g: At $(a,b)$ in the above problem, the potential is both A and 0. As such, even if I consider an infinite series (of sines and sinhs), it does not converge to a single value near the corner $(a,b)$ as the $\phi(x,y)$ itself is not single valued 'everywhere'.
Can someone help me out with this conceptual difficulty ? Is there a formal solution to such 'contradicting boundary' value problems?