 Solution

Write $s={\mu }_x+{\mu }_y$. The update can be rearranged asWhen $s\le 1$, the right side is a convex combination of old values: its coefficients are nonnegative and sum to one. Assume all old values lie in $[U_{\min },U_{\max }]$. A new maximum $v>U_{\max }$ is interior because boundary values are zero and $U_{\max }\ge 0$. At its node the left side is at least $(1+s)v-sv=v$, while the right side is at most $U_{\max }$, a contradiction. A new minimum $v<U_{\min }$ similarly makes the left side at most $v$ and the right side at least $U_{\min }$, again a contradiction.

The initial values satisfy the bounds by construction; mathematical induction provesThe new-time matrix is a strictly diagonally dominant matrix with positive diagonal and nonpositive off-diagonal entries, so the linear update is uniquely solvable. This Crank-Nicolson maximum principle under a step restriction concerns preservation of value bounds, which is a stronger requirement than scalar A-stability.