LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passverified 2026-10-23 (gpt-6.1-sol)

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Newtonian/Fourier dissipation and exact coefficient conditions

Statement

Assume ${A C}_{ω}$ : every sequence of nonempty sets admits a choice function. At a fixed state, let $μ, λ \in R$ , $K \in R^{3 \times 3}$ , $T > 0$ , $g \in R^{3}$ , and let $D \in R^{3 \times 3}$ be symmetric. Put $θ : = tr D = \sum_{i = 1}^{3} D_{i i}, D_{0} : = D - \frac{θ}{3} I, ζ : = λ + \frac{2 μ}{3}, sym K : = \frac{K + K^{T}}{2},$ where $I$ is the $3 \times 3$ identity and $∣ D_{0} ∣^{2} : = \sum_{i, j = 1}^{3} (D_{0})_{i j}^{2}$ . Use the Newtonian stress expression $τ : = 2 μ D + λ θ I$ and the Fourier flux expression $q : = - K g$ , and define the quadratic residual $Q (D, g, T) : = \frac{τ : D}{T} - \frac{q \cdot g}{T^{2}}, τ : D : = \sum_{i, j = 1}^{3} τ_{i j} D_{i j} .$ Here $g$ represents an independently varied temperature-gradient value. The coefficient equivalence is an algebraic statement about $Q$ ; its interpretation as an entropy residual is conditional on the local state and balance hypotheses of the Gibbs identity. For these definitions the exact algebraic decomposition is $Q (D, g, T) = \frac{2 μ ∣ D_{0} ∣^{2} + ζ θ^{2}}{T} + \frac{g^{T} (sym K) g}{T^{2}} .$ Then $Q (D, g, T) \geq 0$ for every symmetric $D$ , every $g \in R^{3}$ , and every $T > 0$ if and only if $μ \geq 0, ζ \geq 0, x^{T} (sym K) x \geq 0 for every x \in R^{3} .$ In particular, $λ$ need not be nonnegative: if $μ > 0$ and $ζ = 0$ , then $λ = - 2 μ / 3$ and the coefficient conditions still hold when $sym K$ is positive semidefinite. This is an algebraic coefficient test for the conditional entropy residual; it neither derives a constitutive law nor assigns material coefficients. In the physical SI interpretation, $μ, λ, ζ$ have units $P a s$ , $D$ has units $s^{- 1}$ , $K$ has units $W m^{- 1} K^{- 1}$ , $g$ has units $K m^{- 1}$ , and $T$ has units $K$ .

Statement references

Kelly, Solid Mechanics Part III, Appendix 4.A “Isotropic Functions,” pp. 406–411

Krzysztof Wilmanski, Continuum Thermodynamics, §4.2 “Isotropy,” pp. 46–50

Rohan Abeyaratne, Continuum Mechanics, Volume II, Chapter 12 §12.2, Problem 12.8, printed pp. 358–360

Facts & Assumptions

Given: The coefficients, temperature, matrix, vectors, quadratic residual, and the countable-choice assumption in the Statement.

[F1]

Gibbs cancellation and entropy-production identity: under its displayed state, contact, smooth-motion, continuity, integral-balance, and ${A C}_{ω}$ hypotheses, the entropy residual is $τ : D / T - q \cdot \nabla T / T^{2}$ . The Gibbs interpretation retains all displayed state, balance, contact and choice hypotheses.

[F2]

Linear isotropic symmetric stress classification: if $L : Sym (3) \to Sym (3)$ is real-linear and satisfies $L (Q D Q^{T}) = Q L (D) Q^{T}$ for every $Q \in SO (3)$ and $D \in Sym (3)$ , then unique $λ, μ \in R$ give $L (D) = 2 μ D + λ (tr D) I$ ; conversely every map of this form is real-linear and satisfies that equivariance. This is a classification under those hypotheses, not a derivation of a material response law.

[F3]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): ${A C}_{ω}$ says every sequence of nonempty sets admits a choice function. It is carried because [F1]'s current local-balance supplier explicitly assumes ${A C}_{ω}$ ; no full Axiom of Choice is assumed.

[A1]

The matrices $D, τ, K$ and vector $g$ are evaluated at one fixed state, and $D$ and $g$ range independently over all stated values.

[A2]

Assume ${A C}_{ω}$ exactly as stated in the Statement; no full Axiom of Choice is assumed.

Proof

1.1A1A2F1F2F3givenalgebra

Under the conditional Gibbs identity [F1] and the countable-choice premise [A2, F3], substituting the stress and flux expressions assumed in the Statement, with $g = \nabla T$ , identifies $Q$ with the entropy residual. The stress expression is a member of the linear isotropic family classified by [F2]; that classification imposes no sign condition on its coefficients. The trace of $D_{0}$ is $θ - \frac{θ}{3} tr I = 0$ , and $D = D_{0} + \frac{θ}{3} I$ .

2.1A1step 1.1algebra

Since $D_{0} : I = tr D_{0} = 0$ and $I : I = 3$ , contraction gives $τ : D = 2 μ ∣ D_{0} ∣^{2} + (λ + \frac{2 μ}{3}) θ^{2} = 2 μ ∣ D_{0} ∣^{2} + ζ θ^{2}$ . Also $- q \cdot g = g^{T} K g = g^{T} (sym K) g$ : the scalar $g^{T} (K - K^{T}) g$ equals its transpose $g^{T} (K^{T} - K) g$ , so it is zero. Hence $Q (D, g, T) = (2 μ ∣ D_{0} ∣^{2} + ζ θ^{2}) / T + g^{T} (sym K) g / T^{2}$ .

3.1A1step 2.1algebra

If $μ \geq 0$ , $ζ \geq 0$ , and $sym K$ is positive semidefinite, every term in step 2.1 is nonnegative for all $D, g, T > 0$ . This proves sufficiency.

3.2A1step 2.1givenalgebra

For necessity, take $T = 1$ , $g = 0$ , and $D = diag (1, - 1, 0)$ . Then $θ = 0$ , $D_{0} = D$ , and $∣ D_{0} ∣^{2} = 2$ , so step 2.1 gives $Q = 4 μ \geq 0$ and hence $μ \geq 0$ .

3.3A1step 2.1givenalgebra

Take $T = 1$ , $g = 0$ , and $D = I / 3$ . Then $θ = 1$ and $D_{0} = 0$ , so $Q = ζ \geq 0$ .

3.4A1step 2.1givenalgebra

Take $T = 1$ , $D = 0$ , and arbitrary $g \in R^{3}$ . Then $Q = g^{T} (sym K) g \geq 0$ for every $g$ , which is exactly positive semidefiniteness of $sym K$ .

4.1step 3.1algebra∎

If $μ > 0$ and $ζ = 0$ , then $λ = - 2 μ / 3 < 0$ , while step 3.1 still gives nonnegative residual whenever $sym K$ is positive semidefinite. Thus nonnegativity of $λ$ is not necessary.

Remarks

The algebra in steps 2.1–3.4 is pointwise and needs $D$ and $g$ to vary independently. The identification of $Q$ with a fluid entropy residual is conditional on [F1]; experiments or material observations do not prove these mathematical coefficient conditions.

Newtonian/Fourier dissipation and exact coefficient conditions

Statement

Statement references

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources