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A nonuniform insulated temperature profile relaxes to its mean

Physical scope

An exact initial-profile example inside the constant-capacity, constant-conductivity insulated rectangular model; it is a conditional calculation, not an observation.

Statement

Assume Countable Choice as in Constant-conductivity heat evolution on an insulated rectangle. Let $L_{i} > 0$ be lengths in metres and $Ω = \prod_{i = 1}^{3} (0, L_{i})$ . Let $c > 0$ be a constant volumetric heat capacity in $J m^{- 3} K^{- 1}$ , $k > 0$ a constant scalar conductivity in $W m^{- 1} K^{- 1}$ , and $u_{⋆}$ an arbitrary reference energy density in $J m^{- 3}$ , with $u (θ) = u_{⋆} + c θ$ as in Constant-conductivity heat evolution on an insulated rectangle. Let $T_{b}$ and $ϵ$ be temperatures in kelvins with $T_{b} > ϵ / 30 > 0$ . For $r \in [0, 1]$ set $h (r) = r^{2} (1 - r)^{2} - 1 / 30$ and prescribe $g (x) = T_{b} + ϵ h (x_{1} / L_{1}) (x \in \overline{Ω}) .$ Use its even $2 L_{i}$ -periodic extension in each coordinate as the initial datum on $R^{3}$ . The unique insulated heat solution has constant spatial mean $T_{b}$ and satisfies $\int_{Ω} ∣ T (x, t) - T_{b} ∣^{2} d x \leq \frac{ϵ^{2} ∣ Ω ∣}{2100} e^{- 2 (k / c) t / C_{Ω}}, C_{Ω} = \frac{3}{2} \max_{i} L_{i}^{2} .$ This gives an exactly computed initial variance and the conditional relaxation estimate of the model.

Facts & Assumptions

Given: Countable Choice; lengths $L_{i}$ in metres; $c, k, u_{⋆}, T_{b}, ϵ$ with the units and hypotheses in the Statement; the profile $h (r) = r^{2} (1 - r)^{2} - 1 / 30$ ; and the field $g (x) = T_{b} + ϵ h (x_{1} / L_{1})$ .

[A1]

Countable Choice is the explicit hypothesis used by the heat-evolution supplier The Axiom of Countable Choice ( ${A C}_{ω}$ ).

[F1]

Constant-conductivity heat evolution on an insulated rectangle: the even-periodic compatible initial field has the unique Gaussian-convolution heat solution with positive bounds and homogeneous Neumann faces.

[F2]

Exponential relaxation of the constant-conductivity Neumann heat model: for this solution, the mean is conserved and $E (t) \leq e^{- 2 (k / c) t / C_{Ω}} E (0)$ .

Example

The function $h$ is dimensionless, and $g$ has units of kelvins. The extension uses even reflection at each face and period $2 L_{i}$ ; the initial field is nonconstant but compatible with insulated walls.

Verification

1.1givenalgebra

Write $q (r) = r^{2} (1 - r)^{2}$ . Then $q^{'} (r) = 2 r (1 - r) (1 - 2 r)$ , so $q^{'} (0) = q^{'} (1) = 0$ , and $q^{''} (0) = q^{''} (1) = 2$ . Also $0 \leq q \leq 1 / 16$ on $[0, 1]$ . The constant and first two derivatives therefore match across the even-reflection and periodic seams; the extension is $C^{2}$ , with bounded uniformly continuous derivatives. Since $h = q - 1 / 30$ , its values lie in $[- 1 / 30, 7 / 240]$ , so $g \geq T_{b} - ϵ / 30 > 0$ .

1.2givenalgebra

Expand $q (r) = r^{2} - 2 r^{3} + r^{4}$ , so $\int_{0}^{1} q (r) d r = 1 / 3 - 1 / 2 + 1 / 5 = 1 / 30$ . Also $q (r)^{2} = r^{4} - 4 r^{5} + 6 r^{6} - 4 r^{7} + r^{8}$ , whence $\int_{0}^{1} q (r)^{2} d r = \frac{1}{5} - \frac{4}{6} + \frac{6}{7} - \frac{4}{8} + \frac{1}{9} = \frac{1}{630} .$ Therefore $\int_{0}^{1} h (r) d r = 0$ and $\int_{0}^{1} h (r)^{2} d r = \int_{0}^{1} (q - 1 / 30)^{2} d r = \frac{1}{630} - \frac{1}{900} = \frac{1}{2100} .$ Integrating over the other two coordinates now gives $\overline{g} = T_{b}$ and $E (0) = ϵ^{2} L_{2} L_{3} L_{1} / 2100 = ϵ^{2} ∣ Ω ∣ / 2100$ .

2.1A1F1F2step 1.1step 1.2∎

The profile satisfies every initial-data hypothesis of [F1]; its vanishing normal derivative follows from $q^{'} (0) = q^{'} (1) = 0$ . Apply [F1] for existence and [F2] with the computed $E (0)$ to obtain the stated conserved mean and decay bound. This is a deduction within the adopted continuum model.

A nonuniform insulated temperature profile relaxes to its mean

Statement

Facts & Assumptions

Example

Verification

Depends on

Dependency tree · 2 levels

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