CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-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 stable forward Euler step can still make a concentration negative

Statement refuted

For every linear finite conservative PDS, if each eigenvalue $λ$ of its system matrix satisfies $∣ 1 + h λ ∣ \leq 1$ , then the forward Euler step preserves nonnegative compartment values.

Witness

Use state domain $Ω = R^{2}$ and two compartments with $p_{21} = d_{12} = y_{1}$ and all other transfers zero. The exact system is $y_{1}^{'} = - y_{1}$ , $y_{2}^{'} = y_{1}$ . From the nonnegative initial state $(1, 0)$ , take the forward Euler step $h = 3 / 2$ . The step is in the absolute-stability interval for the negative scalar eigenvalue $- 1$ , yet its first component is negative.

Facts & Assumptions

Given: The stated two-compartment transfer system, initial state $(1, 0)$ , and $h = 3 / 2$ .

[F1]

Finite conservative production destruction systems: paired internal rates $p_{i j} = d_{j i}$ give a finite conservative PDS.

[F2]

Orders and scalar stability functions of Euler, midpoint, and classical RK4: forward Euler has stability function $R_{E} (z) = 1 + z$ and has uniform local order at least one under the stated smooth compact-tube hypotheses.

[F3]

Forward Euler has the closed unit disk centered at −1 as its absolute-stability region: on $y^{'} = - ρ y$ , forward Euler is absolutely stable exactly when $0 < h ρ \leq 2$ .

Refutation

1.1givenF1algebra

The rates satisfy $p_{21} = d_{12} = y_{1} \geq 0$ and all other pair rates vanish, so [F1] gives $y_{1}^{'} = - y_{1}$ and $y_{2}^{'} = y_{1}$ with conserved sum $y_{1} + y_{2}$ .

2.1givenstep 1.1algebra

Forward Euler gives $y^{1} = y^{0} + h F (y^{0}) = (1, 0) + (3 / 2) (- 1, 1) = (- 1 / 2, 3 / 2)$ , so the first concentration is negative even though the sum remains $1$ .

3.1givenF2F3step 1.1algebra

The system matrix is lower triangular with eigenvalues $- 1$ and $0$ . For $λ = - 1$ , [F3] gives absolute stability because $0 < h ρ = 3 / 2 < 2$ ; for $λ = 0$ , [F2] gives multiplier $R_{E} (0) = 1$ . Thus $∣ 1 + h λ ∣ \leq 1$ for both eigenvalues even though step 2.1 has a negative component.

4.1F2step 2.1step 3.1∎

The vector field is linear and smooth, so the compact-tube hypotheses of [F2] hold locally and forward Euler has first-order local consistency. Steps 2.1 and 3.1 show that neither this accuracy order nor scalar absolute stability implies positivity preservation.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 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.