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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Objective generalized-Newtonian dissipation

Statement

Let $u$ be the $C^{1}$ Eulerian velocity induced by a $C^{2}$ incompressible motion in Euclidean three-space, and set $L = \nabla u$ , $D = (L + L^{T}) / 2$ , and $\dot{γ} = (2 D : D)^{1 / 2}$ . Here incompressibility means $div u = tr L = 0$ . For any finite continuous viscosity $η : [0, \infty) \to [0, \infty)$ and the stress law $τ = 2 η (\dot{γ}) D$ , one has $τ : D \geq 0$ . Under a $C^{2}$ rigid change of observer $x^{*} = Q (t) x + c (t)$ , where $Q (t) \in S O (3)$ , the transformed rate of strain is $D^{*} = Q D Q^{T}$ , so the transformed stress is $τ^{*} = Q τ Q^{T}$ .

For the Carreau formula $η_{C} = η_{\infty} + (η_{0} - η_{\infty}) (1 + (λ \dot{γ})^{2})^{(n - 1) / 2}$ , assume $η_{0} \geq η_{\infty} \geq 0$ , $λ \geq 0$ , and $n > 0$ . It is finite and nonnegative for every finite $\dot{γ}$ . If $λ > 0$ and $η_{0} > η_{\infty}$ , it decreases to $η_{\infty}$ for $0 < n < 1$ , is the constant $η_{0}$ for $n = 1$ , and increases without bound for $n > 1$ ; if $λ = 0$ or $η_{0} = η_{\infty}$ , it is constant. For the regularized power formula $η_{R} = m (ε^{2} + {\dot{γ}}^{2})^{(n - 1) / 2}$ , assume $m > 0$ , $ε > 0$ , and $n > 0$ ; it is finite and strictly positive for every finite $\dot{γ}$ . For the unregularized power stress $τ = 2 m {\dot{γ}}^{n - 1} D$ at $\dot{γ} > 0$ , with $m > 0, n > 0$ , the extension $τ = 0$ at $D = 0$ is continuous, even though the viscosity diverges there when $0 < n < 1$ .

The dimensions are $[D] = [\dot{γ}] = [ε] = s^{- 1}$ , $[η] = [η_{0}] = [η_{\infty}] = [η_{R}] = [η_{U}] = P a s$ , $[λ] = s$ , $[m] = P a s^{n}$ , and $[τ] = P a$ ; $n$ is dimensionless.

Statement references

constitutive formulas and units.

rigid-observer transformation of the velocity gradient and rate of strain.

Reddy, An Introduction to Continuum Mechanics, Chapter 6.

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

Facts & Assumptions

Given: The incompressible smooth-motion kinematics, the finite nonnegative continuous viscosity, and the constitutive formulas and parameter ranges in the Statement.

[A1]

Generalized Newtonian, Bingham and conformation model equations: $D = (L + L^{T}) / 2$ , $\dot{γ} = (2 D : D)^{1 / 2}$ , $τ = 2 η (\dot{γ}) D$ , and the displayed Carreau and regularized or unregularized power parameter conventions.

[F1]

Rigid observer calculus and objective strain rate: for $Q, c \in C^{2}$ , $Q (t) \in S O (3)$ , and corresponding points under $x^{*} = Q x + c$ , the transformed fields satisfy $L^{*} = Q L Q^{T} + Q^{'} Q^{T}$ and $D^{*} = Q D Q^{T}$ .

[F2]

Real powers for positive bases, with the zero-base positive-exponent convention: for $a > 0$ and $r \in R$ , $a^{r} = \exp (r \log_{n a t} a)$ , where $\log_{n a t}$ is the natural logarithm.

[F3]

Continuity and derivatives of positive-base real powers: for $r \in R$ , $x \mapsto x^{r}$ is continuous and differentiable on $(0, \infty)$ , with derivative $r x^{r - 1}$ .

[F4]

The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents: for positive bases, $a^{r + s} = a^{r} a^{s}$ , $(a b)^{r} = a^{r} b^{r}$ , $(a / b)^{r} = a^{r} / b^{r}$ , and $(a^{r})^{s} = a^{r s}$ .

[F5]

The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ : on a positive closed interval, the difference of a continuous function's endpoint values equals its derivative at an interior point times the interval length, when the function is differentiable in the interior.

[F6]

The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ : $\exp (x) > 0$ and $\exp (- x) = 1 / \exp (x)$ for every $x \in R$ .

Proof

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

1.1givenA1algebra

If $D \neq 0$ , then $τ : D = 2 η (\dot{γ}) (D : D) = η (\dot{γ}) {\dot{γ}}^{2} \geq 0$ ; if $D = 0$ , then $τ = 0$ and the contraction is $0$ .

1.2givenA1F2F6algebra

For finite $\dot{γ}$ , the Carreau base $1 + (λ \dot{γ})^{2}$ is finite and at least $1$ . By [F2], its real power is the exponential of a finite real number, hence is finite and positive by [F6]; the parameter inequalities then make both summands nonnegative.

1.3givenA1F2F3F4F5F6algebrachoose

Since $ε^{2} + {\dot{γ}}^{2} > 0$ and is finite, [F2] and [F6] show that its real power is finite and positive; multiplication by $m > 0$ proves the regularized viscosity claim. For $\dot{γ} > 0$ , the unregularized law has $∣ τ ∣ = 2 m {\dot{γ}}^{n - 1} ∣ D ∣ = \sqrt{2} m {\dot{γ}}^{n}$ . For $n > 0$ , [F2], [F3], [F5] and positivity from [F6] make $x \mapsto x^{n}$ strictly increasing on $(0, \infty)$ ; given $ρ > 0$ , choose $δ = ρ^{1 / n} > 0$ . If $0 < \dot{γ} < δ$ , then ${\dot{γ}}^{n} < δ^{n} = ρ$ by [F4]. Thus ${\dot{γ}}^{n} \to 0$ as $\dot{γ} ↓ 0$ , so the zero extension of $τ$ is continuous. If $0 < n < 1$ , put $p = 1 - n > 0$ . By [F2] and [F6], ${\dot{γ}}^{- p} = 1 / {\dot{γ}}^{p}$ . For any $M > 0$ , choose $δ = (m / M)^{1 / p} > 0$ . When $0 < \dot{γ} < δ$ , monotonicity gives ${\dot{γ}}^{p} < δ^{p} = m / M$ , so $η_{U} = m {\dot{γ}}^{- p} > M$ . Hence $η_{U} \to + \infty$ as $\dot{γ} ↓ 0$ .

2.1givenA1F2F3F4F5F6step 1.2algebra

If $λ > 0$ , the base $y = 1 + (λ \dot{γ})^{2}$ increases from $1$ to $+ \infty$ as $\dot{γ}$ runs from $0$ to $+ \infty$ . Put $q = (n - 1) / 2$ . If $q > 0$ , [F2], [F3], [F5] and [F6] show $y \mapsto y^{q}$ is strictly increasing; for every $M > 0$ , $y > M^{1 / q}$ implies $y^{q} > M$ by strict monotonicity and [F4], so $y^{q} \to + \infty$ . If $q < 0$ , put $p = - q > 0$ ; [F2] and [F6] give $y^{q} = 1 / y^{p}$ , and for every $ρ > 0$ , $y > ρ^{- 1 / p}$ implies $0 < y^{q} < ρ$ , so $y^{q} \to 0$ . If $q = 0$ , [F2] and [F6] give $y^{0} > 0$ , while [F4] gives $y^{0} = y^{0} y^{0}$ , hence $y^{0} = 1$ . Consequently the Carreau factor decreases to $0$ , is constantly $1$ , or increases without bound for $0 < n < 1$ , $n = 1$ , or $n > 1$ , respectively. Since $η_{0} - η_{\infty} > 0$ , adding $η_{\infty}$ gives the stated limits and monotonicities. If $λ = 0$ or $η_{0} = η_{\infty}$ , direct substitution gives a constant.

3.1F1A1algebra∎

By [F1], $D^{*} = Q D Q^{T}$ . Orthogonality gives $(Q D Q^{T})^{T} (Q D Q^{T}) = Q D^{T} D Q^{T}$ , so cyclicity of the finite matrix trace yields $D^{*} : D^{*} = tr (Q D^{T} D Q^{T}) = tr (D^{T} D Q^{T} Q) = D : D$ . Thus ${\dot{γ}}^{*} = \dot{γ}$ . Consequently $τ^{*} = 2 η ({\dot{γ}}^{*}) D^{*} = Q (2 η (\dot{γ}) D) Q^{T} = Q τ Q^{T}$ ; the zero-rate extension obeys the same identity.

Objective generalized-Newtonian dissipation

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources