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Entropy production in the local Fourier heat model

Physical scope

Conditional local entropy and energy balance for a sufficiently regular, fixed-composition local-equilibrium continuum field on a rectangular insulated body with a positive-semidefinite symmetric conductivity tensor.

Statement

Let $I \subseteq R$ be an open time interval, $Ω = \prod_{i = 1}^{3} (0, L_{i})$ a fixed rectangle, and $(T, u, j_{Q}, K)$ a field record from Rectangular local-equilibrium field data satisfying the adopted local-energy, Fourier, and local-entropy postulates Local energy balance for an insulated continuum body, Fourier heat-flux constitutive model, and Local-equilibrium caloric entropy density. Assume additionally that $j_{Q}$ and its first spatial derivatives are continuous on every compact subset of $\overline{Ω} \times I$ . Then, throughout $Ω \times I$ , $\partial_{t} s (T) + div ⁣ (\frac{j_{Q}}{T}) = \frac{(\nabla T)^{T} K_{s} (x, T) \nabla T}{T^{2}} \geq 0,$ where $K_{s} = (K + K^{T}) / 2$ . For every compact time interval $[t_{0}, t_{1}] \subset I$ , the model's total entropy $S (t) = \int_{Ω} s (T (x, t)) d x$ is differentiable and nondecreasing, while $\int_{Ω} u (T (x, t)) d x$ is constant. These are conditional deductions of the declared continuum model, not observations or a claim that every medium obeys Fourier's law.

Facts & Assumptions

Given: A rectangular field record on $\overline{Ω} \times I$ with $T > 0$ , $u = u (T)$ , $c (T) = u^{'} (T) > 0$ , $j_{Q} = - K (x, T) \nabla T$ , $K_{s} ⪰ 0$ , $\partial_{t} u + div j_{Q} = 0$ , and $j_{Q} \cdot n = 0$ on every face.

[F1]

Rectangular local-equilibrium field data: $T$ is $C^{1}$ in time and $C^{2}$ in space up to the rectangular faces; $u$ and its heat capacity $c = u^{'}$ are the stated local functions, and $\nabla$ and $div$ are Euclidean coordinate derivatives.

[A1]

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.

[A2]

Fourier heat-flux constitutive model: $j_{Q} = - K \nabla T$ and $z^{T} K_{s} z \geq 0$ for every $z \in R^{3}$ .

[A3]

Local-equilibrium caloric entropy density: $s^{'} (T) = c (T) / T$ for the declared local-equilibrium entropy density.

[A4]

The flux $j_{Q}$ and its first spatial derivatives are continuous on each compact time slab through the boundary, so the coordinate divergence and face traces in the proof are defined and integrable.

[F2]

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

[F3]

Riemann--Fubini on product rectangles, with lower and upper section integrals and content-zero exceptional sections: iterated Riemann integrals of continuous functions on product rectangles equal the multidimensional integral.

Proof

1.1F1A3algebra

By [A3] and the chain rule, $\partial_{t} s (T) = s^{'} (T) \partial_{t} T = (c (T) / T) \partial_{t} T = (\partial_{t} u) / T$ , since $u^{'} (T) = c (T)$ .

1.2A2F1algebra

Substituting $j_{Q} = - K \nabla T$ from [A2] makes the right side $(\nabla T)^{T} K \nabla T / T^{2}$ . For any real vector $z$ , $z^{T} K z = z^{T} K_{s} z$ because the scalar $z^{T} (K - K^{T}) z$ equals its own negative; hence [A2] and $T > 0$ make this quantity nonnegative.

1.3F2F3A1A4

For each coordinate $i$ , apply [F2] to the $i$ th component of $j_{Q} / T$ on $[0, L_{i}]$ : its integral is the endpoint difference $[(j_{Q})_{i} / T]_{x_{i} = 0}^{x_{i} = L_{i}}$ . By [F3] these one-dimensional integrals may be integrated over the other two coordinates and summed. The result is $\int_{Ω} div (j_{Q} / T) d x = 0$ because $j_{Q} \cdot n = 0$ on every face and $T > 0$ . The same calculation gives $\int_{Ω} div j_{Q} d x = 0$ .

2.1A1step 1.1algebra

The coordinate product rule and $\nabla (1 / T) = - \nabla T / T^{2}$ give $div (j_{Q} / T) = (div j_{Q}) / T - j_{Q} \cdot \nabla T / T^{2}$ . Substituting [A1] and step 1.1 yields $\partial_{t} s (T) + div (j_{Q} / T) = - j_{Q} \cdot \nabla T / T^{2}$ .

3.1F1F2A1step 1.2step 1.3step 2.1algebra∎

On $\overline{Ω} \times [t_{0}, t_{1}]$ , $\partial_{t} s (T)$ and $\partial_{t} u (T)$ are continuous. The one-dimensional fundamental theorem [F2] expresses each time difference quotient as the average of its continuous time derivative; uniform continuity on this compact set makes those quotients converge uniformly in $x$ . Therefore differentiation passes through the bounded-box integrals. Integrating the local balance in step 2.1 and using the zero face integral in step 1.3 gives $S^{'} (t) = \int_{Ω} (\nabla T)^{T} K_{s} \nabla T / T^{2} d x$ . Its integrand is nonnegative by step 1.2. Integrating [A1] and using step 1.3 gives $\frac{d}{d t} \int_{Ω} u (T) d x = 0$ .

Entropy production in the local Fourier heat model

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources