Finite-budget Parareal correction sweeps

Input: an admissible propagator pair from Fine and coarse propagators on a common macro grid, initial state $x_0$, number of macro intervals $N$, integer sweep budget $K\ge 0$, and optionally a wall-clock budget $\tau \ge 0$ and certificate data $(\mathcal{C},c,Q,V)$. Here $c\in \mathcal{C}$ is a supplied certificate, $Q$ is a declared predicate on completed rows, and $V(W,c)$ is a supplied deterministic verifier that terminates with a Boolean for every completed row $W$ and satisfies $V(W,c)=\mathrm{t}\mathrm{r}\mathrm{u}\mathrm{e}\Rightarrow Q(W)$. Without certificate data, take $V$ to return false. Every map call is required to terminate with a state or declared failure. Each map call either returns one state or a declared failure status. The output contains every fully completed row of the table $U_n^k$ for $0\le k\le K$, the last completed endpoint when one exists, and a status. If initialization never completes, return an empty completed-row table and no endpoint.

Initialize $U_0^0=x_0$ and, in increasing order of $n$, set $U_{n+1}^0=G_n(U_n^0)$. Store the values $G_n(U_n^0)$. For each $k=0,\dots ,K-1$, set $U_0^{k+1}=x_0$. Evaluate $F_n(U_n^k)$ independently for all $n$; the stored $G_n(U_n^k)$ values are reused from the preceding row. Then, in increasing order of $n$, compute $G_n(U_n^{k+1})$ and set $U_{n+1}^{k+1}=G_n(U_n^{k+1})+F_n(U_n^k)-G_n(U_n^k)\mathrm{.}$ The fine calls in a sweep can run in parallel, while the corrected row is computed sequentially.

If a map call fails, return the last fully completed row and status propagation-failure; a prefix of the failed row may be reported separately but is not a completed iterate. If the time budget expires at a call boundary, return the last fully completed row and status budget-exhausted. When initialization and all $K$ requested sweeps finish, call $V$ on the final row and $c$ (or use false when no certificate was supplied). Return status converged if this returns true, certifying the declared predicate $Q$; otherwise return status budget-exhausted. Completion of the finite budget alone is not a convergence certificate.

For the exact-arithmetic recurrence, an update computed with inexact fine/coarse values or an inexact correction solve can be written as the displayed right-hand side plus a combined defect ${\eta }_n^{k+1}$. A residual tolerance alone does not bound that defect in forward error.