 Solution

Let $r=\Delta \tau /h^2$ and impose periodic indices modulo $J$. With ${\mathbf{\Phi }}^n=({\Phi }_0^n,\dots ,{\Phi }_{J-1}^n{)}^T$, the linear system iswhere $S$ shifts one grid cell cyclically. For $J\ge 3$, $A$ has diagonal $1+2r$, adjacent and corner entries $-r$, and all other entries zero. It is a cyclic tridiagonal matrix, rather than a nonperiodic tridiagonal one.

The periodic Laplacian annihilates the constant vector, so its steady equation requires $\boxed{\sum _j{\rho }_j=0}$. A strictly positive gravitational density violates this condition: summing the update gives ${\bar{\Phi }}^{n+1}={\bar{\Phi }}^n-4\pi G\Delta \tau \bar{\rho }$, an indefinitely drifting mean. Periodic gravitational calculations normally solve for $\rho -\bar{\rho }$ and fix the arbitrary mean gravitational potential. The per-step matrix $A$ itself remains invertible; the issue is existence of a periodic steady solution of Poisson's equation.