ExampleConstruction: 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.

ARS(2,2,2) amplification on a two-cell reaction-diffusion grid

Example

Consider the two-cell no-flux graph diffusion matrix $L = (\begin{array}{cc} - 1 & 1 \\ 1 & - 1 \end{array})$ and the linear reaction-diffusion semidiscretization $y^{'} = ρ y + D L y$ with reaction rate $ρ = 1$ , diffusion coefficient $D = 10$ . Treat reaction explicitly and diffusion implicitly in The ARS(2,2,2) additive tableau and its three stages. The constant mode has amplification $221 / 200 > 1$ , matching its reaction growth. With step $h = 1 / 10$ , the nonconstant mode has amplification $R (1 / 10, - 2) = \frac{661 / 200 - (11 / 5) \sqrt{2}}{11 - 6 \sqrt{2}},$ which lies strictly between $0$ and $9 / 80$ . This is a two-mode scalar calculation for a finite grid; it makes no nonnormal-matrix or PDE stability assertion.

Facts & Assumptions

Given: The matrix $L$ , reaction rate, diffusion coefficient, step size, and the ARS(2,2,2) tableau.

[F1]

The ARS(2,2,2) additive tableau and its three stages: the explicit component is evaluated in the explicit tableau and the diffusion component is solved in the diagonally implicit tableau.

[F2]

The ARS(2,2,2) bivariate scalar stability function: for scalar eigenvalues $z_{a}, z_{b}$ with $1 - γ z_{b} \neq 0$ , the ARS amplification is the stated rational function $R (z_{a}, z_{b})$ .

[F3]

The ARS(2,2,2) coefficients satisfy the coupled second-order identities: $γ = 1 - 1 / \sqrt{2}$ , $1 - 2 γ = \sqrt{2} - 1$ , and $1 / 2 - γ^{2} = \sqrt{2} - 1$ .

[F4]

Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes: matrix multiplication is evaluated entrywise by row-by-column multiplication.

[F5]

Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ and Squaring is monotone on the nonnegatives: $\sqrt{2}$ is positive, and $a^{2} < b^{2}$ for $0 \leq a < b$ .

Proof

1.1F4algebra

Direct multiplication gives $L v_{0} = 0$ for $v_{0} = (1, 1)^{T}$ and $L v_{1} = - 2 v_{1}$ for $v_{1} = (1, - 1)^{T}$ . These vectors are linearly independent, so they form a basis of $R^{2}$ and diagonalize both $a (y) = y$ and $b (y) = 10 L y$ .

2.1F1F2F3step 1.1algebra

On each span $R v_{j}$ , every stage equation in [F1] remains in that span. The implicit denominators are $1$ on $v_{0}$ and $1 + 20 γ h > 0$ on $v_{1}$ , so the stage solves are unique there. Hence [F2] applies with $z_{a} = h$ and $z_{b} = 0$ or $z_{b} = - 20 h$ , respectively.

3.1F2step 2.1algebra

For $v_{0}$ and $h = 1 / 10$ , $z_{b} = 0$ , so [F2] gives $R (1 / 10, 0) = 1 + \frac{1}{10} + \frac{1}{200} = \frac{221}{200} > 1.$ This mode grows under the positive reaction term.

4.1F2F3F5step 2.1algebra∎

For $v_{1}$ , $z_{b} = - 2$ . Substituting [F3] into [F2] gives the displayed fraction, with denominator $(1 + 2 γ)^{2} = (3 - \sqrt{2})^{2} = 11 - 6 \sqrt{2}$ . By [F5], $7 / 5 < \sqrt{2} < 3 / 2$ : squaring the positive endpoints gives $49 / 25 < 2 < 9 / 4$ . Thus the numerator is greater than $1 / 200$ and less than $9 / 40$ , while the denominator is greater than $2$ . Consequently $0 < R (1 / 10, - 2) < 9 / 80 < 1$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

22 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

Lorenzo Pareschi and Giovanni Russo, Implicit-explicit Runge-Kutta schemes and applications to hyperbolic systems with relaxation (standard reference, not scraped)