Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)
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.

Existence and uniqueness of transport along a piecewise-smooth path

Statement

Let n≥1 and let 0=t0<t1<⋯<tm=1 be a finite partition. For each j, let Aj:[tj−1,tj]→Mn(R) be continuous. There is a unique continuous function U:[0,1]→Mn(R) that is C1 on each partition interval, satisfies U′(t)=−Aj(t)U(t) in its interior, and has U(0)=In. Derivatives at interval endpoints are the corresponding one-sided derivatives.

Facts & Assumptions

Given: A positive integer n, a finite partition of [0,1], one continuous matrix coefficient on each closed subinterval, and the initial matrix In.

[F1]

Linear matrix ODEs have unique global solutions on a fixed interval: a continuous matrix coefficient on a compact interval gives a unique global solution to the linear matrix initial-value problem.

Proof

technique · direct
1.1givenF1construct

Apply [F1] on [t0,t1] with initial matrix In to obtain the unique C1 solution on the first interval.

2.1givenF1step 1.1construct

For each j=2,…,m, apply [F1] on [tj−1,tj] with initial value equal to the terminal value already constructed at tj−1. Adjacent solutions agree at their shared endpoint by construction, so the finite family pastes to a continuous piecewise-C1 solution on [0,1].

3.1givenF1step 1.1step 2.1induction∎

Any other such function has initial value In on the first interval, so it agrees there by [F1]. If it agrees through tj−1, then it has the same initial value as the constructed solution on the jth interval, so [F1] makes them agree there. Induction through the finite partition proves uniqueness on all of [0,1].

Depends on

Dependency tree · 2 levels

Sources