Discrete Poisson equation

Remarks

Example: let $\Omega = \set{(0, 0)}$, $h = 1$, $g = 0$ on the four neighbours and $f(0, 0) = -4$. The equation reads $0 + 0 + 0 + 0 - 4u(0, 0) = -4$, so $u(0, 0) = 1$ is the solution.

Non-example: the function with $u(0,0) = 0$ and $u = 0$ on $\partial\Omega$ satisfies the boundary condition but gives $0 \ne -4$ on the left, so it is not a solution.

For a finite $\Omega$, the problem has exactly one solution for all $f$ and $g$; this follows from the discrete maximum principle. Ordering the unknowns gives a linear system $A \mathbf{u} = \mathbf{b}$, where $A$ is sparse, symmetric and negative definite. On a rectangle the matrix has a block tridiagonal form with $-4$ on the diagonal, before scaling by $h^{-2}$. The same definition works in $\Z^d$ with $2d$ neighbours and $-2d$ in place of $-4$.