ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A nonlinear Parareal iterate can leave its local Lipschitz tube

Example

Let $D = [0, 1]$ , $F (x) = x^{2}$ , and $G (x) = x$ . Both maps send $D$ into itself. On $D$ , $G$ is $1$ -Lipschitz and $F - G$ is $1$ -Lipschitz. With $x_{0} = 1 / 4$ and two macro intervals, the first corrected row contains $U_{2}^{1} = - 1 / 8$ , outside $D$ . The tube condition in the binomial error theorem must therefore be checked for corrected iterates; invariance of the individual maps on $D$ is insufficient.

Facts & Assumptions

Given: The finite Parareal recurrence and the stated maps on $D = [0, 1]$ .

[F1]

Finite-budget Parareal correction sweeps: $U_{n + 1}^{k + 1} = G (U_{n}^{k + 1}) + F (U_{n}^{k}) - G (U_{n}^{k})$ , with $U_{n + 1}^{0} = G (U_{n}^{0})$ .

[F2]

A conditional binomial error bound for Parareal: the estimate requires all fine and Parareal iterates under consideration to belong to one common set where the two Lipschitz inequalities hold.

[A1]

$N = 2$ , $x_{0} = 1 / 4$ , $F (x) = x^{2}$ , $G (x) = x$ , and $D = [0, 1]$ .

Verification

1.1givenA1algebra

For $x, y \in D$ , $∣ G (x) - G (y) ∣ = ∣ x - y ∣$ . Also $(F - G) (x) - (F - G) (y) = (x - y) (x + y - 1)$ , and $∣ x + y - 1 ∣ \leq 1$ on $D$ , so $F - G$ is $1$ -Lipschitz there. Both maps take $D$ into $D$ .

2.1A1F1step 1.1algebra

Coarse initialization gives $U_{1}^{0} = U_{2}^{0} = 1 / 4$ , while the first corrected node is $U_{1}^{1} = F (1 / 4) = 1 / 16$ . Applying [F1] again, $U_{2}^{1} = G (1 / 16) + F (1 / 4) - G (1 / 4) = 1 / 16 + 1 / 16 - 1 / 4 = - 1 / 8$ .

3.1givenF2step 2.1∎

Since $- 1 / 8 \notin D$ , the first corrected table already leaves the set on which the Lipschitz bounds were verified. Thus [F2] cannot be invoked for a sweep whose comparisons use that value; no divergence conclusion follows.

Depends on

Used by

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Sources

Martin J. Gander, Time Parallel Time Integration, Chapter 2: Methods Based on Multiple Shooting (standard reference, not scraped)
Martin J. Gander, 50 Years of Time Parallel Time Integration (standard reference, not scraped)