Neumann condition for Poisson equation Solving $ \nabla^2u = 1 $ for spherically symmetric u in the region $r < a,  a > 0$, with the following conditions at r = a (separately) 
(a) $u = 0$
(b) $\nabla u \cdot n = 0 $ where n is the outward normal of the region.
So generally $ u(r) = 1/6 r^2 + Ar^{-1} + B $ where A,B are arbitrary. For (a) I said A = 0 to remove singularity at r=0 and then applying the condition to get $u(r) = 1/6 (r^2 - a^2) $. However I'm not sure how to approach (b). I thought that the normal is the unit radial vector so $\nabla u\cdot n = u'(r)$, but then unless we leave the singularity at r=0 we get $(1/3)a = 0 $ which is clearly wrong. Leaving the singularity gives $A = -(1/3)a^3 $ which would make sense since then the solution is unique up to a constant as required, but why is it then ok for the singularity to persist?

Note that a necessary condition for the existence of a solution to the Neumann problem is that 
$$\begin{align}\int_{r\le a} (1)\,dV&=\int_{r\le a}\nabla^2 u(r)\,dV\\\\
&=\oint_{r=a}\nabla u(r)\cdot \hat n\,dS \tag 1
\end{align}$$
The left-hand side of $(1)$ is simply $\frac{4\pi a^3}{3}$ while if $u'(a)=0$, the right-hand side of $(1)$ is zero.  Therefore, there exists no solution to the Neumann problem in this case. 
