# Confusion regarding a basic boundary value problem

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?

• If there is a discontinuity in $t0$, the limit when $t$ in less than $t0$ and $t$ greater than $t0$ is different and so the function has no definite value in $t0$. You could say that it is multivalued in $t0$. You could have a boundary value problem with $\Phi$ equal to zero if $x <0$ and $\Phi$ equal to 1 of $x>0$. The Fourier series converge toward the half value, but not uniformly. – Vincent Fraticelli Feb 12 at 9:00