A stencil computation repeatedly updates every cell of a grid from its neighbours' values at the previous timestep. Simulating T steps on N cells directly costs Theta(NT), and a line of work beginning with Ahmad et al. reduces this by composing many timesteps into one linear operator and applying it with a Fast Fourier Transform. That technique needs to know which cells will still obey the same operator when the composed step ends, and in a free-boundary problem they do not: the region governed by a given rule is determined by the solution and moves as it evolves. We study one spatial dimension, a three-point stencil with time-varying coefficients, and a computed region that is a single interval whose two endpoints move by arbitrary amounts at every step, revealed online. Let B be the horizon plus the total variation of the boundary trajectory. We give a schedule whose work is O((B+N) log T log(N+B)) and whose span is O(T log T log(N+B)), and we prove that the values it computes are exact. The best existing bound for a region that moves requires its boundary to travel at most one cell per timestep. We drop that requirement and lose nothing by it: a boundary obeying it has B <= 3T, so our bound stays near-linear on every trajectory the earlier result covers. Elsewhere, B grows only by the distance the boundary actually travels – one jump of width N costs T + 2N. The reason total variation suffices is that everything the two endpoints touch over a time window of any length lies in two intervals, one per endpoint. This cannot be relaxed: with p regions the bound degrades by a factor p, and at p = sqrt(T) there is an instance on which the work is Theta(T^{3/2}) while B + N = Theta(T). All results are machine-checked in Lean 4, apart from the classical convolution bound, which is imported as an interface.