ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passverified 2026-10-23 (gpt-6.1-sol)

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A negative bulk coefficient yields a negative quadratic residual

Example

Assume ${A C}_{ω}$ : every sequence of nonempty sets admits a choice function. Let $μ > 0$ , $ζ < 0$ , $T > 0$ , and $θ \in R ∖ {0}$ ; set $λ = ζ - \frac{2 μ}{3}$ , $K = I_{3}$ , $D = \frac{θ}{3} I_{3}$ , and $g = 0$ , where $I_{3}$ is the $3 \times 3$ identity matrix. For $M \in R^{3 \times 3}$ define $tr M : = \sum_{i = 1}^{3} M_{i i}$ and $∣ M ∣^{2} : = \sum_{i, j = 1}^{3} M_{i j}^{2}$ ; for $K \in R^{3 \times 3}$ define $sym K : = (K + K^{T}) / 2$ . With $D_{0} = D - \frac{tr D}{3} I_{3}$ and $Q (D, g, T) : = \frac{2 μ ∣ D_{0} ∣^{2} + ζ (tr D)^{2}}{T} + \frac{g^{T} (sym K) g}{T^{2}},$ one has $D_{0} = 0$ and $tr D = θ$ , so $Q (D, g, T) = ζ θ^{2} / T < 0$ . Thus a positive shear coefficient alone does not ensure a nonnegative quadratic residual, even though this choice satisfies $x^{T} (sym K) x \geq 0$ for every $x \in R^{3}$ .

Example references

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

Facts & Assumptions

Given: The coefficient and matrix choices in the Example and the countable-choice assumption stated there.

[F1]

Newtonian/Fourier dissipation and exact coefficient conditions: under its conditional Gibbs-residual interpretation, $Q = (2 μ ∣ D_{0} ∣^{2} + ζ θ^{2}) / T + g^{T} (sym K) g / T^{2}$ , where $ζ = λ + 2 μ / 3$ .

[F2]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): ${A C}_{ω}$ is the axiom that every sequence of nonempty sets admits a choice function.

[A1]

The choices satisfy $μ > 0$ , $ζ < 0$ , $T > 0$ , and $θ \neq 0$ ; $K = I_{3}$ , $D = (θ / 3) I_{3}$ , and $g = 0$ .

Verification

1.1A1algebra

From the definitions, $tr D = θ$ , so $D_{0} = (θ / 3) I_{3} - (θ / 3) I_{3} = 0$ . Since $K = I_{3}$ , $sym K = I_{3}$ and $x^{T} (sym K) x = \sum_{i = 1}^{3} x_{i}^{2} \geq 0$ for every $x \in R^{3}$ . Also $g^{T} (sym K) g = 0$ because $g = 0$ .

2.1A1F1F2step 1.1algebra∎

By [F1] and the countable-choice premise [F2], step 1.1 gives $Q = ζ θ^{2} / T$ . Since $ζ < 0$ , $θ^{2} > 0$ , and $T > 0$ , this value is strictly negative. Therefore the stated positive-shear example verifies the claimed failure of nonnegative residual.

Depends on

Dependency tree · 2 levels

Sources

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