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 chemical-dosing model under a constant admissible dose

Example

Let $x (t)$ be the concentration of a decaying reagent after dosing, with $\dot{x} = - x + u$ , $x (0) = 0$ , $0 \leq u \leq 1$ , and $0 \leq t \leq 1$ . Use zero running and terminal costs to focus on the state trajectory. For the admissible constant dose $u (t) = 1$ , the exact concentration is $x (t) = 1 - e^{- t}$ . A one-cell explicit-midpoint transcription from zero returns the stage value $X_{2} = 1 / 2$ and endpoint approximation $x_{1} = 1 / 2$ , whereas the exact endpoint is $1 - e^{- 1}$ . The controller's embedded difference alone is not an error certificate.

Facts & Assumptions

Given: The linear concentration model, constant input, one-cell mesh, and explicit midpoint tableau $a_{21} = 1 / 2$ , $b_{1} = 0$ , $b_{2} = 1$ , $c_{1} = 0$ , $c_{2} = 1 / 2$ .

[F1]

The unique state solves its Volterra integral equation for bounded measurable linear dynamics and an $L^{2}$ input (The bounded measurable linear ODE has a unique affine state map).

[F2]

The one-step stage and endpoint formulas for an explicit RK tableau are those of Explicit Runge–Kutta tableaus, stage maps, and failure status.

[F3]

Piecewise-constant controls are defined interval by interval on the mesh (Time meshes and piecewise-constant control parameterization).

[F4]

An embedded adaptive RK estimate is a local pair difference; controller termination does not certify a global exact error (Finite-budget adaptive Runge–Kutta controller, The bounded adaptive Runge–Kutta controller terminates).

[A1]

Exact real arithmetic is used for the displayed values.

Verification

1.1F1givenalgebra

The exact integral equation is $x (t) = \int_{0}^{t} (- x (s) + 1) d s$ . The function $x (t) = 1 - e^{- t}$ has $x (0) = 0$ and derivative $e^{- t} = - (1 - e^{- t}) + 1$ , so it solves the equation; uniqueness follows from [F1].

2.1F2F3A1step 1.1algebra

On the one-cell mesh, the explicit midpoint first slope is $K_{1} = - x_{0} + u = 1$ . The second stage is $X_{2} = x_{0} + \frac{1}{2} K_{1} = \frac{1}{2}$ , so $K_{2} = - X_{2} + u = \frac{1}{2}$ . The output is $x_{1} = x_{0} + K_{2} = \frac{1}{2}$ by [F2]; the exact endpoint is $1 - e^{- 1}$ by step 1.1.

3.1F4step 2.1algebra∎

The exponential series gives $e = 1 + 1 + \sum_{k \geq 2} 1 / k! > 2$ , so $1 - e^{- 1} > 1 / 2$ and the midpoint endpoint differs from the exact endpoint. The procedure returns the stated value, not a bound on the true error, by [F4].

Depends on

Used by

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Dependency tree · two levels

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Sources

S. Gros and M. Diehl, Numerical Optimal Control, draft course manuscript, Chapter 13 §§13.1–13.3 (standard reference, not scraped)
J. Monnier, Optimal Control of ODEs: The Linear-Quadratic (LQ) Case, course notes, Chapter 2 §§2.3–2.4 (standard reference, not scraped)