ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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Static cosine relaxation

Example

Use the fixed-background box, lapse $N$ , constants and material model of Constructed supported static heat evolution. Let $\overline{ϑ} > 0$ and $δ \in R$ satisfy $∣ δ ∣ < \overline{ϑ}$ , and take the initial redshifted temperature

ϑ_{0} (x) = \overline{ϑ} + δ \cos ⁣ (\frac{π x^{1}}{L_{1}}), x \in \overline{Ω} .

It has a positive even-reflected periodic extension, and the corresponding exact solution and proper-temperature field are

ϑ (t_{K}, x) = \overline{ϑ} + δ \exp ⁣ (- \frac{A_{0} π^{2} t_{K}}{C_{V} L_{1}^{2}}) \cos ⁣ (\frac{π x^{1}}{L_{1}}), T (t_{K}, x) = \frac{ϑ (t_{K}, x)}{N (x)} .

The proper temperature tends to $\overline{ϑ} / N (x)$ , which varies spatially when the lapse is nonconstant, while the redshifted temperature $N T$ tends to the constant $\overline{ϑ}$ .

Facts & Assumptions

Given: The box, lapse, heat coefficients, state function and solution class of Constructed supported static heat evolution, with $L_{1}, C_{V}, A_{0} > 0$ .

[A1]

$\overline{ϑ} > 0$ and $∣ δ ∣ < \overline{ϑ}$ .

[F1]

Constructed supported static heat evolution: for positive even-reflected periodic initial data, $ϑ = N T$ solves $C_{V} \partial_{t_{K}} ϑ = A_{0} Δ_{x} ϑ$ , satisfies zero normal derivative on each face, and is the unique solution in the stated continuous/classical class.

Verification

Given: The explicit initial profile and the constants in [A1].

1.1givenA1algebra

Put $r = A_{0} π^{2} / (C_{V} L_{1}^{2}) > 0$ . The formula ${\tilde{ϑ}}_{0} (x) = \overline{ϑ} + δ \cos (π x^{1} / L_{1})$ on $R^{3}$ is smooth, even in each coordinate, and $2 L_{i}$ -periodic in each coordinate; $∣ δ ∣ < \overline{ϑ}$ makes it positive. Its average on $Ω$ is $\overline{ϑ}$ , since $\int_{0}^{L_{1}} \cos (π x^{1} / L_{1}) d x^{1} = {[\frac{L_{1}}{π} \sin (π x^{1} / L_{1})]}_{0}^{L_{1}} = 0.$ Differentiating the displayed time-dependent profile gives $\partial_{t_{K}} ϑ = - r δ e^{- r t_{K}} \cos (π x^{1} / L_{1})$ and $Δ_{x} ϑ = - (π^{2} / L_{1}^{2}) δ e^{- r t_{K}} \cos (π x^{1} / L_{1})$ . Thus $C_{V} \partial_{t_{K}} ϑ = A_{0} Δ_{x} ϑ$ , and its initial trace is exactly $ϑ_{0}$ .

2.1step 1.1F1A1algebra∎

The bound $∣ δ e^{- r t_{K}} \cos (π x^{1} / L_{1}) ∣ \leq ∣ δ ∣ < \overline{ϑ}$ proves positivity. Direct differentiation gives $\partial_{x^{1}} ϑ = - (π δ / L_{1}) e^{- r t_{K}} \sin (π x^{1} / L_{1})$ , which vanishes at $x^{1} = 0, L_{1}$ ; the other two normal derivatives vanish because the profile is independent of $x^{2}, x^{3}$ . Hence it satisfies every boundary condition of [F1], and uniqueness identifies it with the theorem's heat solution. Using $\cos^{2} z = (1 + \cos (2 z)) / 2$ and integrating the cosine term over $[0, L_{1}]$ gives $\int_{0}^{L_{1}} \cos^{2} (π x^{1} / L_{1}) d x^{1} = L_{1} / 2$ . Therefore $∥ ϑ (t_{K}, \cdot) - \overline{ϑ} ∥_{L^{2} (Ω)} = ∣ δ ∣ \sqrt{∣ Ω ∣ / 2} e^{- r t_{K}} .$ Dividing by the fixed positive lapse gives $T = ϑ / N$ .

Depends on

Constructed supported static heat evolution

Static cosine relaxation

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources