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An additive error ledger for inexact Parareal sweeps

Statement

Use the notation and Lipschitz hypotheses of A conditional binomial error bound for Parareal, with A=max⁡{1,a}. Let computed iterates Vnk∈D satisfy V0k=x0, ∥Vn0−xn∥≤M, and, for each requested correction, ∥Vn+1k+1−[Gn(Vnk+1)+(Fn−Gn)(Vnk)]∥≤η for a constant η≥0. Define Pnk=M(nk)Anbk and a nonnegative forcing array by R0k=0, Rn0=0, and Rn+1k+1=ARnk+1+bRnk+η. Then ∥Vnk−xn∥≤Pnk+Rnk for every completed node and sweep.

The base envelope M includes coarse-initialization error relative to the fine trajectory. The per-update defect η must bound the actual combined effect of fine/coarse evaluation errors and correction-solve error; a residual tolerance alone supplies no forward-error bound.

Facts & Assumptions

Given: The Parareal Lipschitz hypotheses, the stated computed recurrence defect, and the definitions of Pnk and Rnk.

[F1]

A conditional binomial error bound for Parareal: with A=max⁡{1,a}, the exact-arithmetic error is bounded by Pnk=M(nk)Anbk.

[A1]

Every computed comparison point lies in the common set on which the two Lipschitz bounds hold, and the combined update defect has norm at most η.

Proof

1.1givenA1algebra

Set enk=∥Vnk−xn∥. Subtracting xn+1=Fn(xn) from the computed recurrence, splitting off the update defect, and applying the two Lipschitz inequalities gives en+1k+1≤aenk+1+benk+η≤Aenk+1+benk+η.

1.2givenF1A1algebra

At node 0, e0k=0, P0k=0 for k>0, and R0k=0; at k=0, the base coarse-error assumption gives en0≤M≤Pn0+Rn0.

2.1F1A1step 1.1step 1.2inductionalgebra∎

Assume enk≤Pnk+Rnk at node n for every completed sweep. For k≥0, step 1.1 gives en+1k+1≤A(Pnk+1+Rnk+1)+b(Pnk+Rnk)+η. The P terms satisfy APnk+1+bPnk=MAnbk+1(A(nk+1)+(nk))≤MAn+1bk+1((nk+1)+(nk))=Pn+1k+1 because A≥1 and Pascal's identity holds. The remaining terms equal Rn+1k+1 by definition. Therefore the bound holds at node n+1; induction proves it for all nodes and sweeps.

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