ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generated

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 longer run exposes age-cutoff outflow error

Example

Let $μ = β = 0$ , initial density $n_{0} (a) = e^{- a}$ , cutoff $A = 4$ , and $Δ a = Δ t = 1$ . Compare runs to horizons $T = 1, 2, 3$ using the upwind age scheme.

Verification

Given: The finite-volume shift update and cutoff-tail bound.

[F1] Linear age-structured renewal transport problem: with zero mortality and fertility, the exact characteristic solution is $n (t, a) = e^{- (a - t)} 1_{a \geq t}$ .

[F2] Finite-volume upwind scheme for age renewal transport: when $λ = 1$ and $μ = 0$ , each cell mass shifts one cell to the right per step and the last-cell flux is discarded as outflow.

[F3] Age-cutoff survival, outflow and infinite-age tail bounds: for $A > T$ , $\sup_{0 \leq t \leq T} ∥ n (t) - n^{A} (t) ∥_{1} \leq 2 e^{β_{\infty} T} \int_{A - T}^{\infty} n_{0} (a) d a$ , separately from discretization; cumulative outflow is bounded by the same initial tail without the exponential prefactor.

1.1F1algebra

The exact solution in [F1] has total mass $\int_{t}^{\infty} e^{- (a - t)} d a = 1$ at every $t$ . The exact initial cell masses are $M_{j}^{0} = e^{- (j - 1)} - e^{- j}$ for $1 \leq j \leq 4$ .

2.1F2step 1.1algebra

Since $λ = 1$ and $μ = 0$ , [F2] gives $M_{1}^{k + 1} = 0$ and $M_{j}^{k + 1} = M_{j - 1}^{k}$ for $2 \leq j \leq 4$ . After $k$ steps, the retained total is $\sum_{j = 1}^{4 - k} M_{j}^{0} = 1 - e^{- (4 - k)}$ , so the error from total mass $1$ is $e^{- (4 - k)}$ .

3.1F3step 2.1algebra∎

For $T = k = 1, 2, 3$ , the errors are respectively $e^{- 3}, e^{- 2}, e^{- 1}$ . The cumulative outflow is $\sum_{r = 0}^{k - 1} M_{4 - r}^{0} = e^{- (4 - k)} - e^{- 4}$ , and the initial mass omitted above $A$ is $e^{- 4}$ . Their sum equals the exact tail $\int_{4 - k}^{\infty} e^{- a} d a = e^{- (4 - k)}$ in [F3]. The larger error at the longer horizon is the age-cutoff tail, not a time-discretization defect.

Depends on

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Sources

Benoît Perthame, Introduction to Structured Equations in Biology, full lecture notes (standard reference, not scraped)