Laplace equation in a rectangle, Dirichlet to Neumann map Consider the problem
$$\Delta u = 0,(x,y) \in \Omega=(0,1)^2 \\
u(0,y)=u(1,y)=1 \\
u(x,0)=u(x,1)=0.$$
This problem can be solved exactly. The solution is
$$u(x,y)=4\sum_{\text{odd } n} \frac{1}{n\pi} \sin(n\pi x) \left ( 1 - \frac{\sinh(n\pi y)+\sinh(n\pi(1-y))}{\sinh(n\pi)} \right ).$$
Note that the boundary conditions on the left and right edges are not strictly speaking satisfied; these are only satisfied in a weak sense. Essentially the right way to think about this is to round the corners out into quarter-circles of a small radius $r$, make the boundary value go smoothly (but rapidly) from $0$ to $1$ around each corner, and then take the limit $r \to 0$. In the limit $r \to 0$ there is a Gibbs phenomenon appearing which spoils the boundary conditions.
I am trying to analytically compute or approximate $\frac{\partial u}{\partial y}(x,1)$. Unfortunately, it appears that the convergence properties of this sum are too bad to allow term-by-term differentiation with respect to $y$ (which is not a huge surprise, because the boundary conditions are singular at the corners). How can this be circumvented to extract the boundary derivative data?

The derivative series is just fine for $0 < y < 1$:
$$
    u_{y}(x,y) = 4\sum_{\mbox{odd $n$}}\sin(n\pi x)\frac{\cosh(n\pi(1-y))-\cosh(n\pi y)}{\sinh(n\pi)}
$$
For $\delta < y < 1-\delta$ the fraction in the sum is bounded by $C e^{-n\pi\delta}$.
