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Heat and viscous entropy specialization — Examples

1 · Prerequisites

2 · Summary

The example substitutes one positive cosine mode into the supported static heat theorem. Direct differentiation verifies its decay rate, reflected Neumann faces, positive proper temperature, and exact exponential L2 relaxation. The lapse enters through T=ϑ/N; the proper-temperature limit varies spatially when the lapse is nonconstant, while the redshifted temperature NT approaches a spatial constant.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-adaptedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Static cosine relaxation

Example

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

ϑ0(x)=ϑ‾+δcos⁡ ⁣(πx1L1),x∈Ω‾.

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

ϑ(tK,x)=ϑ‾+δexp⁡ ⁣(−A0π2tKCVL12)cos⁡ ⁣(πx1L1),T(tK,x)=ϑ(tK,x)N(x).

The proper temperature tends to ϑ‾/N(x), which varies spatially when the lapse is nonconstant, while the redshifted temperature NT tends to the constant ϑ‾.

Facts & Assumptions

Given: The box, lapse, heat coefficients, state function and solution class of Constructed supported static heat evolution, with L1,CV,A0>0.

[A1]

ϑ‾>0 and ∣δ∣<ϑ‾.

[F1]

Constructed supported static heat evolution: for positive even-reflected periodic initial data, ϑ=NT solves CV∂tKϑ=A0Δxϑ, 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=A0π2/(CVL12)>0. The formula ϑ~0(x)=ϑ‾+δcos⁡(πx1/L1) on R3 is smooth, even in each coordinate, and 2Li-periodic in each coordinate; ∣δ∣<ϑ‾ makes it positive. Its average on Ω is ϑ‾, since ∫0L1cos⁡(πx1/L1) dx1=[L1πsin⁡(πx1/L1)]0L1=0. Differentiating the displayed time-dependent profile gives ∂tKϑ=−rδe−rtKcos⁡(πx1/L1) and Δxϑ=−(π2/L12)δe−rtKcos⁡(πx1/L1). Thus CV∂tKϑ=A0Δxϑ, and its initial trace is exactly ϑ0.

2.1step 1.1F1A1algebra∎

The bound ∣δe−rtKcos⁡(πx1/L1)∣≤∣δ∣<ϑ‾ proves positivity. Direct differentiation gives ∂x1ϑ=−(πδ/L1)e−rtKsin⁡(πx1/L1), which vanishes at x1=0,L1; the other two normal derivatives vanish because the profile is independent of x2,x3. Hence it satisfies every boundary condition of [F1], and uniqueness identifies it with the theorem's heat solution. Using cos⁡2z=(1+cos⁡(2z))/2 and integrating the cosine term over [0,L1] gives ∫0L1cos⁡2(πx1/L1) dx1=L1/2. Therefore ∥ϑ(tK,⋅)−ϑ‾∥L2(Ω)=∣δ∣∣Ω∣/2 e−rtK. Dividing by the fixed positive lapse gives T=ϑ/N.

Sources