LemmaStatement: 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.

A vector field along an ODE solution has an explicit derivative bound

Statement

Let $d \geq 1$ , let $V \subseteq R \times R^{d}$ be open, let $f : V \to R^{d}$ have continuous first partial derivatives, and let $y : I \to R^{d}$ solve $y^{'} = f (t, y)$ on an open interval $I$ . Fix $a < b$ with $[a, b] \subset I$ such that $(s, y (s)) \in V$ for every $s \in [a, b]$ . Suppose finite constants $F_{j}, M_{t, i}, M_{i j} \geq 0$ satisfy for all $s \in [a, b]$

$∣ f_{j} (s, y (s)) ∣ \leq F_{j}, ∣ \partial_{t} f_{i} (s, y (s)) ∣ \leq M_{t, i}, ∣ \partial_{z_{j}} f_{i} (s, y (s)) ∣ \leq M_{i j} (1 \leq i, j \leq d) .$

Define $g_{i} (s) : = f_{i} (s, y (s))$ and $B_{i} : = M_{t, i} + \sum_{j = 1}^{d} M_{i j} F_{j}$ . Then $g_{i}$ is continuous on $[a, b]$ , differentiable on $(a, b)$ , and

$g_{i}^{'} (s) = \partial_{t} f_{i} (s, y (s)) + \sum_{j = 1}^{d} \partial_{z_{j}} f_{i} (s, y (s)) f_{j} (s, y (s)), ∣ g_{i}^{'} (s) ∣ \leq B_{i} .$

Facts & Assumptions

Given: The open ODE domain, continuous first partial derivatives, solution $y^{'} = f (t, y)$ on $[a, b]$ , and the displayed componentwise bounds.

[F1]

First-order systems, initial value problems, and solutions on intervals: an IVP solution is differentiable, its graph lies in the open domain, and it satisfies $y^{'} (s) = f (s, y (s))$ .

[F2]

If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative: continuous first partial derivatives give total differentiability with derivative matrix equal to the Jacobian.

[F3]

Total differentiability gives a local $O (∥ h ∥_{2})$ increment bound and therefore continuity: total differentiability implies continuity.

[F4]

The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ : the derivative of a composition is the composition of the total derivatives.

Proof

1.1givenF1F2F3

By [F2]–[F3], $f$ is continuous on $V$ . Since $y^{'} = f (s, y (s))$ , the derivative of $y$ is continuous on $I$ ; hence $y$ is continuously differentiable and the path $q (s) : = (s, y (s))$ has derivative $q^{'} (s) = (1, f (s, y (s)))$ .

2.1F1F2F3F4algebra∎

For each $i$ , the component map $f_{i}$ is totally differentiable, so [F4] applied to $g_{i} = f_{i} \circ q$ gives the displayed derivative formula. The bounds give $∣ g_{i}^{'} (s) ∣ \leq M_{t, i} + \sum_{j = 1}^{d} M_{i j} F_{j} = B_{i}$ for every $s \in (a, b)$ . Continuity of $g_{i}$ on $[a, b]$ follows from continuity of $f$ and $y$ .

Depends on

Used by

Forward Euler has uniform local order one and a one-step Lipschitz bound Theorem

Dependency tree · two levels

18 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, Section III.2 (standard reference, not scraped)
MIT OpenCourseWare, 18.330 Introduction to Numerical Analysis (Spring 2012), Chapter 5 (standard reference, not scraped)