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

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Bingham yield graph and convex monotonicity

Statement

Let $M_{3} (R)$ be the real $3 \times 3$ matrices and $I_{3}$ the identity matrix, and let $S_{0} = {D \in M_{3} (R) : D^{T} = D, tr D = 0}$ , with Frobenius product $D : E = tr (D^{T} E)$ and norm $∣ D ∣ = (D : D)^{1 / 2}$ . Define $S O (3) : = {Q \in M_{3} (R) : Q^{T} Q = I_{3}, \det Q = 1} .$ For $D$ measured in $s^{- 1}$ , let $μ_{p} \geq 0$ be in $P a s$ and $τ_{y} \geq 0$ be in $P a$ ; define $Φ (D) = μ_{p} ∣ D ∣^{2} + τ_{y} ∣ D ∣$ , in $P a s^{- 1}$ , and

\partial Φ (D) : = {τ \in S_{0} : Φ (E) \geq Φ (D) + τ : (E - D) for every E \in S_{0}} .

Define the Bingham graph by $B (D) = {2 μ_{p} D + τ_{y} D / ∣ D ∣}$ when $D \neq 0$ , and by $B (0) = {τ \in S_{0} : ∣ τ ∣ \leq τ_{y}}$ when $D = 0$ ; its graph stresses are in $P a$ . Then $Φ$ is finite, nonnegative, and convex; $B (D) = \partial Φ (D)$ for every $D$ ; the graph is monotone in the sense that $(τ_{1} - τ_{2}) : (D_{1} - D_{2}) \geq 0$ whenever $τ_{i} \in B (D_{i})$ ; every graph value satisfies $τ : D \geq 0$ ; and, for $Q \in S O (3)$ , $B (Q D Q^{T}) = Q B (D) Q^{T}$ .

Statement references

Bingham graph, norm convention, and parameters.

Reddy, An Introduction to Continuum Mechanics, Chapter 6, §6.6.2.3.

MIT OpenCourseWare, Lecture Notes on Fluid Dynamics, §1.6.2.

Facts & Assumptions

Given: The finite-dimensional real inner-product space $S_{0}$ , the parameters $μ_{p}, τ_{y} \geq 0$ , the potential $Φ$ , and the graph $B$ in the Statement.

[A1]

Generalized Newtonian, Bingham and conformation model equations: the Bingham convention is $τ = 2 μ_{p} D + τ_{y} D / ∣ D ∣$ for $D \neq 0$ , and the zero-rate value is the full trace-free Frobenius ball of radius $τ_{y}$ .

Proof

Given: The data in the Statement and Facts & Assumptions.

1.1givenalgebra

The norm is convex by the triangle inequality and positive homogeneity. For $0 \leq s \leq 1$ , $∣ s D + (1 - s) E ∣^{2} \leq (s ∣ D ∣ + (1 - s) ∣ E ∣)^{2} \leq s ∣ D ∣^{2} + (1 - s) ∣ E ∣^{2}$ , the second inequality following from $s (1 - s) (∣ D ∣ - ∣ E ∣)^{2} \geq 0$ . Thus $Φ$ is convex; it is finite everywhere and $Φ \geq 0$ , so the supporting-inequality set defined in the Statement is well-defined.

1.2givenalgebra

Let $D \neq 0$ . For every $E \in S_{0}$ , expansion of $∣ E - D ∣^{2} \geq 0$ gives $∣ E ∣^{2} \geq ∣ D ∣^{2} + 2 D : (E - D)$ , while Cauchy–Schwarz gives $∣ E ∣ \geq ∣ D ∣ + (D / ∣ D ∣) : (E - D)$ . Multiplying by $μ_{p}$ and $τ_{y}$ and adding shows $Φ (E) \geq Φ (D) + (2 μ_{p} D + τ_{y} D / ∣ D ∣) : (E - D)$ , so the stated nonzero graph value belongs to the set defined in the Statement.

1.3givenalgebra

At $D = 0$ , the supporting inequality is $μ_{p} ∣ E ∣^{2} + τ_{y} ∣ E ∣ \geq ξ : E$ for every $E$ . If $∣ ξ ∣ \leq τ_{y}$ , Cauchy–Schwarz proves it. Conversely, if $∣ ξ ∣ > τ_{y}$ , choose $E = t ξ / ∣ ξ ∣$ ; the inequality would imply $∣ ξ ∣ \leq μ_{p} t + τ_{y}$ for every $t > 0$ , which fails for sufficiently small $t$ when $μ_{p} > 0$ , and fails for every $t > 0$ when $μ_{p} = 0$ . Hence $\partial Φ (0)$ is exactly the stated stress ball, proving $B = \partial Φ$ also at zero.

2.1givenstep 1.2algebra

If $ξ$ belongs to the set defined as $\partial Φ (D)$ with $D \neq 0$ , apply its supporting inequality to $E = D + h H$ . For $h > 0$ division by $h$ gives $ξ : H \leq (Φ (D + h H) - Φ (D)) / h$ ; for $h < 0$ the reversed inequality gives $ξ : H \geq (Φ (D + h H) - Φ (D)) / h$ . The identity $(∣ D + h H ∣ - ∣ D ∣) / h = (2 D : H + h ∣ H ∣^{2}) / (∣ D + h H ∣ + ∣ D ∣) \to (D : H) / ∣ D ∣$ and the quadratic expansion in step 1.2 show that both limits equal $(2 μ_{p} D + τ_{y} D / ∣ D ∣) : H$ . Since this holds for every $H \in S_{0}$ , nondegeneracy of the Frobenius product gives $ξ = 2 μ_{p} D + τ_{y} D / ∣ D ∣$ .

2.2givenA1step 1.3algebra

If $D \neq 0$ , then $τ : D = 2 μ_{p} ∣ D ∣^{2} + τ_{y} ∣ D ∣ \geq 0$ ; if $D = 0$ , then $τ : D = 0$ for every point of the stress ball.

3.1givenstep 1.2step 2.1step 1.3

For $τ_{i} \in \partial Φ (D_{i})$ , the defining inequality gives $Φ (D_{2}) \geq Φ (D_{1}) + τ_{1} : (D_{2} - D_{1})$ and $Φ (D_{1}) \geq Φ (D_{2}) + τ_{2} : (D_{1} - D_{2})$ . Adding and cancelling the potential values yields $(τ_{1} - τ_{2}) : (D_{1} - D_{2}) \geq 0$ .

4.1givenA1algebra∎

Conjugation by $Q \in S O (3)$ preserves symmetry, trace, Frobenius norm, and the Frobenius product. Substituting $Q D Q^{T}$ into each nonzero formula therefore gives $Q τ Q^{T}$ ; at $D = 0$ , the full ball maps onto itself because $∣ Q τ Q^{T} ∣ = ∣ τ ∣$ . This proves the stated rotation-equivariance.

Depends on

Generalized Newtonian, Bingham and conformation model equations