TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-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.

Heun's quadratic dense output has local order two

Statement

Under the smoothness and bounded-derivative hypotheses of Local orders and leading estimator term for the Euler–Heun pair, for a fixed initial value and all sufficiently small $h > 0$ , $\sup_{0 \leq θ \leq 1} ∥ P_{n} (θ) - y (t_{n} + θ h) ∥_{2} \leq C h^{3}$ for a constant $C$ independent of $h$ and $θ$ . Thus this continuous extension has local order two uniformly throughout the accepted step, with a defect separate from the endpoint defect.

Facts & Assumptions

Given: The Heun stage and dense-output formulas, the smooth vector field and bounded derivatives from Local orders and leading estimator term for the Euler–Heun pair, and $0 \leq θ \leq 1$ .

[F1]

Quadratic continuous extension for Heun's method and Euler–Heun embedded Runge–Kutta pair: $P_{n} (θ) = y_{n} + h (θ K_{1} + \frac{1}{2} θ^{2} (K_{2} - K_{1}))$ , and $Y_{H} - Y_{E} = \frac{h}{2} (K_{2} - K_{1})$ .

[F2]

Local orders and leading estimator term for the Euler–Heun pair: uniformly for $0 \leq θ \leq 1$ , $y (t_{n} + θ h) = y_{n} + θ h K_{1} + \frac{1}{2} θ^{2} h^{2} g + O (h^{3})$ and $Y_{H} - Y_{E} = \frac{1}{2} h^{2} g + O (h^{3})$ . Together with [F1], the latter gives $K_{2} - K_{1} = h g + O (h^{2})$ .

Proof

1.1F1F2algebra

From [F1] and [F2], $\frac{h}{2} (K_{2} - K_{1}) = \frac{1}{2} h^{2} g + O (h^{3})$ ; since $h > 0$ , division by $h / 2$ gives $K_{2} - K_{1} = h g + O (h^{2})$ . Substituting this into the defining formula for $P_{n}$ in [F1], and using $0 \leq θ^{2} \leq 1$ , gives $P_{n} (θ) = y_{n} + θ h K_{1} + \frac{1}{2} θ^{2} h^{2} g + O (h^{3})$ uniformly in $θ$ .

2.1F2step 1.1algebra∎

The uniform exact-path expansion in [F2] has the same first two terms as step 1.1, with an $O (h^{3})$ remainder independent of $θ$ . Subtracting it from the expansion of $P_{n} (θ)$ proves the asserted uniform estimate.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Ernst Hairer, course notes, Chapter III: Équations Différentielles Ordinaires, Exercise 10 (Dense output), printed p. 85 (standard reference, not scraped)
Ernst Hairer, Ordinary Differential Equations, lecture notes, Exercise 10 (standard reference, not scraped)