PropositionStatement: 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.

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A nonzero leading defect makes the embedded difference asymptotically match the low-order defect

Statement

Fix one initial value problem and initial state. Let $D_{p} (h) : = \hat{Y} (h) - y (t + h)$ be the low member's one-step defect and $D_{p + 1} (h) : = Y (h) - y (t + h)$ the high member's defect, using the same exact initial state and step start. Suppose, as $h ↓ 0$ , $D_{p} (h) = a h^{p + 1} + O (h^{p + 2}), D_{p + 1} (h) = O (h^{p + 2})$ for a fixed vector $a \in R^{d}$ . Here $O (h^{r})$ for a vector means a remainder whose Euclidean norm is at most $C h^{r}$ for all sufficiently small positive $h$ . Then $\hat{Y} (h) - Y (h) = a h^{p + 1} + O (h^{p + 2})$ . If $a \neq 0$ , the estimator satisfies $E (h) / ∥ D_{p} (h) ∥_{2} \to 1$ . If $a = 0$ , these hypotheses do not imply that this ratio tends to one or that the estimator is nonzero; the Euler–Heun pair on $y' = t^{2}$ , $y (0) = 0$ , has ratio $3 / 2$ as computed below.

Facts & Assumptions

Given: The two outputs and Euclidean estimator in Embedded Runge–Kutta pair and difference estimator, with the two displayed defect expansions.

[F1]

Embedded Runge–Kutta pair and difference estimator: $E (h) = ∥ \hat{Y} (h) - Y (h) ∥_{2}$ .

[F2]

Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation: the Euclidean norm is continuous and satisfies $∣ ∥ u ∥_{2} - ∥ v ∥_{2} ∣ \leq ∥ u - v ∥_{2}$ .

[F3]

Euler–Heun embedded Runge–Kutta pair: Heun and Euler use the two shared stages $K_{1} = f (t, y)$ and $K_{2} = f (t + h, y + h K_{1})$ , with outputs $Y_{E} = y + h K_{1}$ and $Y_{H} = y + \frac{h}{2} (K_{1} + K_{2})$ .

Proof

1.1givenF1algebra

Subtracting the two defect identities cancels the common exact value: $\hat{Y} - Y = D_{p} - D_{p + 1} = a h^{p + 1} + O (h^{p + 2})$ . Thus there is a vector remainder $r (h) = O (h^{p + 2})$ with $\hat{Y} - Y = h^{p + 1} (a + h r_{1} (h))$ , where $r_{1}$ is bounded for all sufficiently small positive $h$ .

2.1F1F2F3step 1.1algebra∎

By [F2], $∣ E (h) / h^{p + 1} - ∥ a ∥_{2} ∣ \leq h ∥ r_{1} (h) ∥_{2} \to 0$ . If $a \neq 0$ , the same argument gives $∥ D_{p} (h) ∥_{2} / h^{p + 1} \to ∥ a ∥_{2} > 0$ ; dividing the two positive limits proves $E (h) / ∥ D_{p} (h) ∥_{2} \to 1$ . If $a = 0$ , the expansions give only $D_{p} = O (h^{p + 2})$ and $E = O (h^{p + 2})$ , with no lower bound. To verify that the ratio-one conclusion can fail for an actual embedded Runge–Kutta pair, use the Heun–Euler pair [F3] on $y' = t^{2}$ , $y (0) = 0$ : $K_{1} = 0$ , $K_{2} = h^{2}$ , $Y_{E} = 0$ , $Y_{H} = h^{3} / 2$ , and the exact endpoint is $h^{3} / 3$ . Thus $D_{1} = - h^{3} / 3$ , the estimator is $h^{3} / 2$ , and their norm ratio is $3 / 2$ , while the $h^{2}$ coefficient is zero. Conversely, for the zero field with zero initial state, both outputs and the estimator vanish identically, so no positive lower bound for the estimator follows either.

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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)