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Constant-conductivity heat evolution on an insulated rectangle

Physical scope

Existence and uniqueness for the three-dimensional rigid Fourier heat model with constant positive volumetric heat capacity and scalar conductivity, homogeneous Neumann walls, and even-periodic uniformly regular positive initial data; no general rough-data or arbitrary-domain PDE claim is made.

Statement

Assume Countable Choice. Let $L_{i} > 0$ be lengths measured in metres, set $Ω = \prod_{i = 1}^{3} (0, L_{i})$ , and take the open temperature interval $T = (0, \infty)$ with temperatures measured in kelvins. Fix a constant volumetric heat capacity $c > 0$ in $J m^{- 3} K^{- 1}$ , a constant scalar conductivity $k > 0$ in $W m^{- 1} K^{- 1}$ , and an arbitrary reference energy density $u_{⋆}$ in $J m^{- 3}$ . Define $u : T \to R$ by $u (θ) = u_{⋆} + c θ$ , so $u^{'} (θ) = c$ , and define $K : \overline{Ω} \times T \to R^{3 \times 3}$ by $K (x, θ) = k I_{3}$ as in Fourier heat-flux constitutive model. Put $a = k / c$ in $m^{2} s^{- 1}$ .

Let $g : R^{3} \to T$ be $C^{2}$ and satisfy, for every $i \in {1, 2, 3}$ and $x \in R^{3}$ , $g (x_{1}, \dots, - x_{i}, \dots, x_{3}) = g (x_{1}, \dots, x_{i}, \dots, x_{3}), g (x_{1}, \dots, x_{i} + 2 L_{i}, \dots, x_{3}) = g (x_{1}, \dots, x_{i}, \dots, x_{3}) .$ Assume every coordinate derivative $D^{α} g$ with $∣ α ∣ \leq 2$ is bounded and uniformly continuous, and that $0 < T_{\min} \leq g (y) \leq T_{\max} < \infty$ for all $y$ , with these bounds measured in kelvins.

For $x \in \overline{Ω}$ and $t > 0$ define $T (x, t) = \int_{R^{3}} Γ (x - y, a t) g (y) d y, Γ (z, θ) = (4 π θ)^{- 3 / 2} e^{- ∣ z ∣^{2} / (4 θ)},$ and set $T (x, 0) = g (x)$ . Here $y$ is measured in metres, $θ = a t$ in the kernel has units $m^{2}$ , and the normalized kernel $Γ (\cdot, a t)$ has units $m^{- 3}$ , so the integral has units of kelvins. Also $\nabla T = (\partial_{x_{1}} T, \partial_{x_{2}} T, \partial_{x_{3}} T)$ , $Δ T : = \sum_{i = 1}^{3} \partial_{x_{i}}^{2} T$ , and on a face with outward unit normal $n$ , $\partial_{n} T : = n \cdot \nabla T$ . For every finite $τ > 0$ , this is the unique function $T$ continuous on $\overline{Ω} \times [0, τ]$ whose first time derivative for $t > 0$ and spatial derivatives through order two extend continuously to $\overline{Ω} \times [0, τ]$ , solving $c \partial_{t} T = k Δ T$ on $Ω \times (0, τ]$ , with $\partial_{n} T = 0$ on each face and the given initial value. The time derivative at $t = 0$ is interpreted from the right. The field satisfies $T_{\min} \leq T (x, t) \leq T_{\max}$ for all $t \geq 0$ , its spatial derivatives through order two are continuous to $t = 0$ , and its right time derivative there is $a Δ g$ .

On $I = (0, \infty)$ , set $j_{Q} (x, t) = - k \nabla T (x, t)$ . Then $(T, u, j_{Q}, K)$ is a field record of Rectangular local-equilibrium field data satisfying Local energy balance for an insulated continuum body and Fourier heat-flux constitutive model, so the solution is global in time for this restricted physical continuum model.

Facts & Assumptions

Given: Countable Choice and the geometric, caloric, conductivity, and initial-profile data specified in the Statement.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): for every sequence of nonempty sets there is a function selecting one element from each set. Countable Choice is the only choice principle assumed here.

[A2]

The stated energy-density map is $u (θ) = u_{⋆} + c θ$ on $T = (0, \infty)$ , so its heat capacity is the constant $u^{'} (θ) = c > 0$ on the full temperature range.

[A3]

Local energy balance for an insulated continuum body: $\partial_{t} u + div j_{Q} = 0$ and $j_{Q} \cdot n = 0$ on each insulated face.

[A4]

Fourier heat-flux constitutive model: $j_{Q} = - K \nabla T$ ; the constant scalar case is $K = k I_{3}$ .

[F1]

The heat Cauchy problem for bounded uniformly continuous data: assuming Countable Choice, convolution of a bounded uniformly continuous $g : R^{3} \to R$ with the Gaussian heat kernel gives a bounded function, smooth for positive time, solving $U_{θ} = Δ U$ , and converging locally uniformly to $g$ as $θ ↓ 0$ .

[F2]

Normalisation, parabolic scaling, heat equation and derivative bounds for the heat kernel: $Γ (z, θ) > 0$ , has unit integral, and its derivatives have Gaussian tail bounds; $Γ (\cdot, θ)$ is an $L^{1}$ approximate identity as $θ ↓ 0$ .

[F3]

Spatial and time derivatives pass through heat convolution for positive time: for bounded data the heat convolution is smooth for positive time, derivatives pass through its integral, and $U_{θ} = Δ U$ .

[F4]

$L^{1}$ approximate identities converge uniformly on compacta for bounded continuous functions: convolution by an $L^{1}$ approximate identity converges uniformly on every compact set for bounded continuous data.

[F5]

Rectangular local-equilibrium field data: $T, u, c, j_{Q}, K$ have the domains, regularity classes, Euclidean derivatives, and SI units stated in its Definition.

[F6]

Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation: translation preserves Lebesgue measure.

[F7]

A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not: under Countable Choice, an invertible linear map changes Lebesgue measure by the absolute determinant of its matrix.

[F8]

Newton–Leibniz needs only continuity on $[a, b]$ , differentiability on $(a, b)$ , and a Riemann-integrable extension of the interior derivative: a continuous function on a closed interval with an integrable interior derivative has endpoint increment equal to the derivative integral.

[F9]

The heat kernel on $R^{n}$ and its causal extension: for $θ > 0$ , $Γ (z, θ) = (4 π θ)^{- 3 / 2} \exp (- ∣ z ∣^{2} / (4 θ))$ on $R^{3}$ .

[F10]

Fubini's theorem for L^1 functions on a sigma-finite product: absolutely integrable functions on sigma-finite product measure spaces have equal iterated and product integrals.

[F11]

$C^{k}$ maps and multi-index derivative notation in Euclidean space: $C^{k}$ means that every coordinate derivative through order $k$ exists and is continuous on the open domain.

Proof

1.1A1A2A3A4F1givenalgebra

The assumptions on $g$ make it bounded and uniformly continuous, so [F1] applies. Write $U (x, θ) = Γ_{θ} * g (x)$ and set $T (x, t) = U (x, a t)$ . The chain rule gives $\partial_{t} T = a U_{θ} (x, a t) = a Δ U (x, a t) = a Δ T$ for $t > 0$ ; [F1] also gives the formula, smoothness, and local uniform initial convergence. Since $k = a c$ , this is $c \partial_{t} T = k Δ T$ . With $u^{'} (T) = c$ , $u_{t} = c T_{t}$ , and $j_{Q} = - k \nabla T$ from [A3, A4], the same equality gives $\partial_{t} u + div j_{Q} = c \partial_{t} T - k Δ T = 0$ .

1.2F1F2F5F6F7F9A4givenalgebra

Positivity and unit mass in [F2] imply $T_{\min} \int Γ \leq T (x, t) \leq T_{\max} \int Γ$ , hence $T_{\min} \leq T \leq T_{\max}$ . Translation by $2 L_{i}$ in the convolution variable preserves $g$ and Lebesgue measure by [F6], so $T$ is $2 L_{i}$ -periodic in $x_{i}$ . Reflection $y_{i} \mapsto - y_{i}$ preserves Lebesgue measure by [F7] because its matrix has determinant of absolute value one; evenness of the kernel [F9] and of $g$ then makes $T$ even in each coordinate. For $t > 0$ , evenness gives $\partial_{x_{i}} T = 0$ on the face $x_{i} = 0$ ; periodicity gives $\partial_{x_{i}} T = \partial_{x_{i}} T$ at opposite faces $x_{i} = - L_{i}, L_{i}$ , while evenness makes these two values negatives, so both vanish. The same boundary derivative holds at $t = 0$ for $g$ , and [A4] gives $j_{Q} \cdot n = 0$ on every face.

1.3F2F3F4F8F10F11givenalgebra

By [F11], the stated $C^{2}$ hypothesis gives continuous derivatives through spatial order two. For $∣ α ∣ \leq 2$ and $t > 0$ , differentiating under the integral and then integrating by parts in each coordinate gives $D^{α} T (x, t) = \int Γ (x - y, a t) D^{α} g (y) d y$ . The products of kernel derivatives with bounded derivatives of $g$ are absolutely integrable by the Gaussian bound [F2], so [F10] licenses the iterated integrals; on each finite coordinate interval Newton–Leibniz [F8] gives the integration-by-parts identity, and the boundary terms vanish as the interval expands because of the Gaussian tails. Each $D^{α} g$ is bounded and continuous, so [F4] makes this expression converge uniformly on $\overline{Ω}$ to $D^{α} g$ as $t ↓ 0$ . Thus the spatial derivatives through order two extend continuously to the initial time.

2.1F1F3F4F8step 1.1step 1.3

For $t > 0$ , the heat equation and the same integration-by-parts argument give $\partial_{t} T (x, t) = a \int Γ (x - y, a t) Δ g (y) d y$ . By [F4] this converges uniformly on $\overline{Ω}$ to $a Δ g (x)$ . For $0 < ε < t$ , [F8] gives $T (x, t) - T (x, ε) = \int_{ε}^{t} \partial_{s} T (x, s) d s$ . Letting $ε ↓ 0$ is valid uniformly in $x$ by step 1.1 and the uniform limit just established; consequently $T (x, t) - g (x) = \int_{0}^{t} \partial_{s} T (x, s) d s$ , and dividing by $t$ proves the right derivative $\partial_{t} T (x, 0) = a Δ g (x)$ .

3.1F5F8F10step 1.2step 1.3step 2.1algebra∎

For uniqueness, let $T_{1}, T_{2}$ be two solutions in the stated regularity class and set $w = T_{1} - T_{2}$ . Then $w_{t} = a Δ w$ , $\partial_{n} w = 0$ , and $w (\cdot, 0) = 0$ . At each fixed time, $w$ and its spatial derivatives through order two are continuous on the compact rectangle, so the integrands $w \partial_{i}^{2} w$ and $∣ \partial_{i} w ∣^{2}$ are bounded and hence $L^{1}$ for the finite Lebesgue product measure. Apply [F10] successively in the three coordinate factors to identify the rectangle integrals with the iterated one-dimensional integrals. The product rule and [F8] then give $\int_{0}^{L_{i}} w \partial_{i}^{2} w = [w \partial_{i} w]_{0}^{L_{i}} - \int_{0}^{L_{i}} ∣ \partial_{i} w ∣^{2}$ . The boundary terms vanish by the homogeneous Neumann condition, so $E^{'} (t) = 2 a \int_{Ω} w Δ w = - 2 a \int_{Ω} ∣ \nabla w ∣^{2} \leq 0$ . Thus $E (0) = 0$ and $E (t) \geq 0$ imply $E (t) = 0$ ; continuity of $w$ gives $w = 0$ everywhere. The formula therefore gives the unique solution in the stated class.

Depends on

Dependency tree · 2 levels

Sources

John K. Hunter, Notes on Partial Differential Equations, §5.1