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I know this question is old but I found an error that I wanted to clear up in @DrHansGruber 's answer. It took me a few hours to spot the issue and I wanted to post the solution to save anyone else from working through this. First, as has been stated by Hans and Peter, there are infinitely many solutions to the 1D Poisson equation with Neumann BCs at both ...


Problem solved: I just have to include the solution of the homogeneous equation into the solution of the differential equation. The solution to the differential equation is $$V_H(r) = -\frac{K}{r} - \frac{r+1}{r} \exp(-2r).$$ By requiring that $\lim_{r \to 0} V_H(r)$ is real we get $K=-1$ and $$V_H(r) = \frac{1}{r} - \frac{r+1}{r} \exp(-2r).$$


Maybe this isn't a helpful response but the reason this happens for the matrix form of Laplacians is because this actually happens for the true infinite-dimensional Laplacians in some settings. In recangular coordinates, $N$-D Laplacians act exactly like their discretized counterparts in this way. For example, let $\Delta_3$ by the Laplacian on $[0,1]^3$ ...


This is my attempt at providing some intuition. Everything I state might be obvious, moreover it doesn't have much to do with physics, so this could be a non-answer. I will ignore boundary conditions. Imagine we have a 2D grid with function values $u_{ij}$ for $i=1\ldots m, j=1\ldots n$. Let the $x$ axis point downward, the $y$ axis to the right. N.b. $z$ ...

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