LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-21

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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 solution whose graph approaches a compact interior region at a finite endpoint extends past that endpoint

Statement

Let $x : (α, β) \to R^{n}$ solve a Picard-Lindelof ODE and suppose $β < \infty$ . If there are $t_{j} ↑ β$ for which $(t_{j}, x (t_{j}))$ lies in one compact subset $K$ of the open ODE domain, then $x$ extends to a solution beyond $β$ . A solution whose graph has a sequence approaching a compact interior endpoint state extends past that endpoint. The reflected statement holds at a finite left endpoint.

Facts & Assumptions

Given: The solution, finite endpoint, compact set, and sequence in the Statement.

[L1]

A bounded sequence in $R^{n}$ , $n \geq 1$ , has a convergent subsequence (For $n \geq 1$ every bounded sequence in $R^{n}$ has a convergent subsequence).

[L2]

For an integrable vector-valued function on $[a, b]$ with $a \leq b$ , ${∥ \int_{a}^{b} f ∥}_{2} \leq \int_{a}^{b} ∥ f ∥_{2}$ (For $a \leq b$ and $f : [a, b] \to R^{m}$ integrable when $a < b$ , $∥ \int_{a}^{b} f ∥_{2} \leq \int_{a}^{b} ∥ f ∥_{2}$ ; for $a < b$ , $∥ f ∥_{2}$ is integrable).

[L3]

Picard-Lindelöf gives a unique local solution through each point of the open ODE domain (Picard-Lindelöf local existence and uniqueness for first-order systems).

[L4]

Locally unique solutions agreeing at a common endpoint glue to a solution on the union interval (Locally unique ODE solutions agree on overlaps and glue across a common endpoint).

[L5]

A subset of Euclidean space is compact if and only if it is closed and bounded (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

Proof

technique · direct

1.1

Compactness makes the sequence of graph points $(t_{j}, x (t_{j}))$ bounded, so [L1] in $R^{n + 1}$ gives a subsequence converging to $(β, p)$ ; [L5] makes $K$ closed, hence $(β, p) \in K$ and lies in the interior of the ODE domain.

givenL1L5

2.1

Choose a compact cylinder about $(β, p)$ inside the ODE domain and let $M$ bound the field there; for large $j$ , $x (t_{j})$ lies within half its state radius and $M (β - t_{j})$ is smaller than the remaining half, so a first-exit argument using [L2] keeps the whole tail in that cylinder and gives $x (t) \to p$ ; [L3] starts a solution at $(β, p)$ and [L4] glues it to $x$ past $β$ , with $M = 0$ immediate.

step 1.1L2L3L4∎

Depends on

Used by

At a finite maximal time an ODE solution leaves every compact subset of the domain Corollary

Dependency tree · two levels

83 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

Gerald Teschl, Ordinary Differential Equations and Dynamical Systems, Ch. 2 (standard reference, not scraped)
Jiri Lebl, Basic Analysis I, Section 6.3 (standard reference, not scraped)