ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge 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 fast-decay test problem is rejected at the minimum step

Example

Run the Euler–Heun controller on the scalar fast-decay equation $y^{'} = - 1000 y$ , $y (0) = 1$ , with $T = 0.004$ , $h_{\min} = 0.002$ , $h_{\max} = h_{0} = 0.004$ , $a t o l = 10^{- 4}$ , $r t o l = 0$ , and attempt budget $N = 2$ , $p = 1$ , safety $σ = 0.9$ , and scale factors $γ_{\min} = 0.2$ , $γ_{\max} = 5$ . The computed pair difference rejects both attempted steps, so the procedure stops at its minimum-step safeguard. This scalar fast-decay equation is used as a stiffness illustration: the calculation verifies repeated rejection of this explicit trial only, and gives no general stiffness diagnosis.

Facts & Assumptions

Given: The scalar field $f (y) = - 1000 y$ , starting value $1$ , trial sizes $0.004$ and $0.002$ , and controller parameters in the Example.

[F1]

Euler–Heun embedded Runge–Kutta pair: $K_{1} = f (y_{n})$ , $K_{2} = f (y_{n} + h K_{1})$ , $Y_{E} = y_{n} + h K_{1}$ , and $Y_{H} = y_{n} + \frac{h}{2} (K_{1} + K_{2})$ .

[F2]

Finite-budget adaptive Runge–Kutta controller: reject if $Q > 1$ , scale by $\max (γ_{\min}, \min (γ_{\max}, σ Q^{- 1 / (p + 1)}))$ , clamp the next proposal to the step bounds, and return minimum-step-rejected for a rejection at $h_{\min}$ .

[F3]

The bounded adaptive Runge–Kutta controller terminates: this finite-budget controller reaches a terminal status after finitely many attempts.

[F4]

Local orders and leading estimator term for the Euler–Heun pair: the Euler–Heun pair has low order $p = 1$ .

Verification

1.1givenF1F2F4algebra

At $h = 0.004$ , $K_{1} = - 1000$ , the Euler stage value is $- 3$ , and $K_{2} = 3000$ . Thus $Y_{E} = - 3$ , $Y_{H} = 5$ , the difference magnitude is $8$ , and $Q = 8 / 10^{- 4} = 80000 > 1$ . The proposed scale is clamped to $0.2$ , so the next step is $\max (0.002, 0.2 \cdot 0.004) = 0.002$ .

2.1F1F2F3F4step 1.1algebra∎

At $h = 0.002$ , $K_{1} = - 1000$ , the Euler stage value is $- 1$ , and $K_{2} = 1000$ . Thus $Y_{E} = - 1$ , $Y_{H} = 1$ , the difference magnitude is $2$ , and $Q = 20000 > 1$ . This rejection occurs at $h_{\min}$ , so [F2] returns minimum-step-rejected; finite termination is also covered by [F3].

Depends on

Used by

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Sources

Ernst Hairer, course notes, Chapter III: Équations Différentielles Ordinaires, §III.2 (standard reference, not scraped)
Tobin A. Driscoll and Richard J. Braun, Fundamentals of Numerical Computation, Chapter 6 (standard reference, not scraped)