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Conditional admissibility of the specified non-Newtonian branches

Physical scope

Given selected stated law and balances, derive objective nonnegative mechanical dissipation; conformation entropy requires explicit isothermal Helmholtz realization/thermostat.

Statement

Work in the smooth incompressible branch of the adopted nonrelativistic continuum model. Let $T > 0$ , let $J \subseteq R$ be an open interval containing $[0, T]$ , let $Ω \subseteq R^{3}$ be open, and let $u \in C^{2} (J \times Ω; R^{3})$ be the Eulerian velocity induced by a $C^{2}$ material motion with orientation-preserving $C^{2}$ diffeomorphic time slices, and satisfy $div u = 0$ . At a spatial point let $L = \nabla u$ , $D = (L + L^{T}) / 2$ , $tr D = 0$ , and use the Frobenius contraction $S : T = tr (S^{T} T)$ with norm $∣ S ∣ = (S : S)^{1 / 2}$ . Let $M_{3} (R)$ denote the real $3 \times 3$ matrices, and define ${S y m}_{0} (3) = {S \in M_{3} (R) : S = S^{T}, tr S = 0}$ and ${S y m}^{+} (3) = {C \in M_{3} (R) : C = C^{T}, z^{T} C z > 0 for every z \neq 0}$ . A constitutive response is objective when under every rigid observer change $x^{*} = Q (t) x + c (t)$ , with $Q \in C^{2} ([0, T], S O (3))$ and $c \in C^{2} ([0, T], R^{3})$ , its second-order tensor value at corresponding events transforms as $S^{*} = Q S Q^{T}$ . For the generalized Newtonian branch, let $\dot{γ} = (2 D : D)^{1 / 2}$ , let $η : [0, \infty) \to [0, \infty)$ be finite and continuous, and set $τ = 2 η (\dot{γ}) D$ . Define $I_{3}$ as the $3 \times 3$ identity matrix; in SI units, $L, D, \dot{γ}$ have units $s^{- 1}$ , $η, η_{s}, μ_{p}$ have units $P a s$ , and $τ, τ_{y}, G$ have units Pa. For the Bingham branch, let $μ_{p} \geq 0$ in $P a s$ and $τ_{y} \geq 0$ in $P a$ , and set $B (D) = {2 μ_{p} D + τ_{y} D / ∣ D ∣}$ for $D \neq 0$ and $B (0) = {S \in {S y m}_{0} (3) : ∣ S ∣ \leq τ_{y}}$ . For the conformation branch, take a $C^{1}$ trajectory $x : [0, T] \to Ω$ with $x^{'} (t) = u (t, x (t))$ . Set $L (t) = \nabla u (t, x (t))$ , $D (t) = (L (t) + L (t)^{T}) / 2$ , and let the dimensionless $C \in C^{1} ([0, T]; M_{3} (R))$ be symmetric, so $C (t) = C (t)^{T}$ for all $t$ , and satisfy

C^{'} = L C + C L^{T} - (C - I_{3}) / λ, C (0) = C_{0} \in {S y m}^{+} (3),

where $G > 0$ is a modulus in Pa, $λ > 0$ is a relaxation time in seconds, and $η_{s} \geq 0$ is a solvent viscosity in Pa s. Define $τ_{p} = G (C - I_{3})$ , $τ_{s} = 2 η_{s} D$ , and $w (C) = \frac{G}{2} (tr C - \log \det C - 3)$ in J m $^{- 3}$ . Then the generalized Newtonian and Bingham branches have objective stress responses and nonnegative instantaneous mechanical power $τ : D \geq 0$ for every allowed stress. On every forward time $t \in [0, T]$ , the conformation solution remains positive definite, its constitutive response is objective, and

(τ_{s} + τ_{p}) : D - \frac{d}{d t} w (C) = 2 η_{s} D : D + \frac{G}{2 λ} (tr C + tr C^{- 1} - 6) \geq 0.

Thus the selected smooth branches have nonnegative mechanical or stored-energy dissipation under their stated model hypotheses. This is a conditional deduction, not a claim of empirical material applicability or existence of a fluid partial-differential-equation solution.

For an entropy-production interpretation of the conformation inequality, additionally adopt the following separate isothermal realization along the trajectory. Fix $T_{0} > 0$ in K and let $e_{v}, s_{v} \in C^{1} ([0, T]; R)$ be internal-energy and entropy densities per current volume, in $J m^{- 3}$ and $J K^{- 1} m^{- 3}$ , respectively. Let $h \in C^{0} ([0, T]; R)$ be the signed heat-removal rate per current volume, in $W m^{- 3}$ , positive for heat passed to an ideal external thermostat at the same temperature $T_{0}$ . Assume this is the only heat exchange, carrying entropy out at rate $h / T_{0}$ , and adopt the energy balance $e_{v}^{'} = (τ_{s} + τ_{p}) : D - h$ and the Helmholtz free-energy identification $w (C) = e_{v} - T_{0} s_{v}$ . Define the internal entropy-production density by $π_{s} : = s_{v}^{'} + h / T_{0}$ , in $W K^{- 1} m^{- 3}$ . Derivatives at the time endpoints are one-sided. Then

T_{0} π_{s} = (τ_{s} + τ_{p}) : D - \frac{d}{d t} w (C) \geq 0.

These additional thermodynamic assumptions are not inferred from the continuum balances or the stored-energy identity alone.

Statement references

J. N. Reddy, An Introduction to Continuum Mechanics, 2nd ed., §§5.4.2–5.4.3, equations (5.4.11), (5.4.20), and (5.4.23)–(5.4.30): energy, entropy exchange, and Helmholtz conventions

J. N. Reddy, An Introduction to Continuum Mechanics, 2nd ed., Chapter 6, equation (6.5.15) and §§6.6.1–6.6.2.3, PDF pp. 25–27 (printed pp. 245–247)

Chiang C. Mei, MIT 1.63J/2.21J Lecture Notes on Fluid Dynamics, §1.6.2, equations (1.6.13)–(1.6.18), complete 4-page PDF

John W. Barrett and Sébastien Boyaval, Existence and Approximation of a (Regularized) Oldroyd-B Model, arXiv:0907.4066v2, §1.1 and §2.3, PDF pp. 2–11

S. Boyaval, T. Lelièvre, and C. Mangoubi, Free-energy-dissipative schemes for the Oldroyd-B model, ESAIM: M2AN 43 (2009), §§2.1.1–2.1.2, PDF pp. 4–6

David Tong, Fluid Mechanics, Part I, §§1.1.1–1.1.3, PDF pp. 6–9

Facts & Assumptions

Given: The adopted smooth continuum branch and the three constitutive cases in the Statement.

[F1]

Restricted non-Newtonian constitutive examples: a material motion is a $C^{2}$ map with orientation-preserving $C^{2}$ diffeomorphic time slices, its Eulerian velocity is $u (t, χ (t, a)) = \partial_{t} χ (t, a)$ , and $S O (3) = {Q : Q^{T} Q = I_{3}, \det Q = 1}$ ; under a rigid observer change with $Q (t) \in S O (3)$ , objective tensor responses transform as $S^{*} = Q S Q^{T}$ .

[F2]

Restricted non-Newtonian constitutive examples: the generalized Newtonian branch defines $\dot{γ} = (2 D : D)^{1 / 2}$ and $τ = 2 η (\dot{γ}) D$ with finite continuous $η : [0, \infty) \to [0, \infty)$ .

[F3]

Restricted non-Newtonian constitutive examples: the selected parameter $τ_{y}$ is the Frobenius radius of the Bingham stress ball at rest, and the graph is set-valued there.

[F4]

Restricted non-Newtonian constitutive examples: the conformation branch adopts the displayed upper-convected equation, polymer and solvent stresses, and stored-energy density $w (C)$ on a smooth incompressible trajectory.

[F5]

Objective generalized-Newtonian dissipation: for the incompressible generalized Newtonian law with finite continuous $η : [0, \infty) \to [0, \infty)$ , $τ : D \geq 0$ ; under the stated $C^{2}$ rigid observer change the rate and stress transform by orthogonal conjugation.

[F6]

Bingham yield graph and convex monotonicity: the trace-free Bingham graph is the subdifferential of $Φ (D) = μ_{p} ∣ D ∣^{2} + τ_{y} ∣ D ∣$ ; its values are monotone, every value satisfies $τ : D \geq 0$ , and $B (Q D Q^{T}) = Q B (D) Q^{T}$ .

[F7]

Objective conformation positivity and stored-energy dissipation: for the stated smooth forward trajectory, continuous matrix field $L$ , positive-definite initial $C_{0}$ , $G, λ > 0$ , and $η_{s} \geq 0$ , the solution remains positive definite and satisfies the exact displayed stored-energy dissipation identity.

[F8]

Objective conformation positivity and stored-energy dissipation: under the stated rigid observer change, the conformation equation is covariant, the conformation and both stresses transform by orthogonal conjugation, and the stored-energy density is invariant.

[A1]

For the optional isothermal realization only, adopt the Statement's $T_{0} > 0$ , densities $e_{v}, s_{v}$ , heat-removal rate $h$ , energy balance $e_{v}^{'} = (τ_{s} + τ_{p}) : D - h$ , Helmholtz identification $w (C) = e_{v} - T_{0} s_{v}$ , and entropy-exchange convention $π_{s} = s_{v}^{'} + h / T_{0}$ . These are additional physical assumptions and a definition, not consequences of [F1]–[F8].

Proof

1.1F1F2F5algebra

In the generalized Newtonian branch, ${\dot{γ}}^{2} = 2 D : D$ and $η (\dot{γ}) \geq 0$ by [F2], so $τ : D = 2 η (\dot{γ}) D : D = η (\dot{γ}) {\dot{γ}}^{2} \geq 0$ . Under a rigid observer change, [F5] gives $D^{*} = Q D Q^{T}$ and $τ^{*} = Q τ Q^{T}$ ; orthogonal conjugation preserves the Frobenius product, hence ${\dot{γ}}^{*} = \dot{γ}$ . By the definition in [F1], the response is objective.

1.2F3F5F6algebra

In the Bingham branch, when $D \neq 0$ , each allowed stress is $τ = 2 μ_{p} D + τ_{y} D / ∣ D ∣$ , so $τ : D = 2 μ_{p} ∣ D ∣^{2} + τ_{y} ∣ D ∣ \geq 0$ . When $D = 0$ , every allowed stress has $τ : D = 0$ ; the graph value there is nonempty since it contains $0$ , and for $D \neq 0$ the graph value is a singleton. The kinematic identity $D^{*} = Q D Q^{T}$ in [F5] and the graph equivariance in [F6] give $B (D^{*}) = Q B (D) Q^{T}$ , so the graph response is objective, including at zero rate.

1.3F4F7F8algebra

For the conformation branch, [F7] supplies positive definiteness and the exact identity in the Statement under its forward-time and regularity hypotheses; [F8] supplies objectivity. Set $A = C - I_{3}$ . Since $C$ is positive definite, it is invertible and $C^{- 1}$ is positive definite: for each nonzero $z$ , $z^{T} C^{- 1} z = (C^{- 1} z)^{T} C (C^{- 1} z) > 0$ . Matrix multiplication gives $A C^{- 1} A = C + C^{- 1} - 2 I_{3}$ , so $tr C + tr C^{- 1} - 6 = tr (A^{T} C^{- 1} A) = \sum_{j = 1}^{3} a_{j}^{T} C^{- 1} a_{j} \geq 0$ , where $a_{j}$ are the columns of $A$ . Since $η_{s} \geq 0$ , the displayed stored-energy dissipation is nonnegative.

2.1givenA1step 1.1step 1.2step 1.3algebra∎

Steps 1.1–1.3 establish the conditional nonnegative mechanical or stored-energy dissipation and objectivity. Under the additional assumptions [A1], constancy of $T_{0}$ gives $\frac{d}{d t} w (C) = e_{v}^{'} - T_{0} s_{v}^{'} = (τ_{s} + τ_{p}) : D - h - T_{0} s_{v}^{'}$ . Rearrangement yields $(τ_{s} + τ_{p}) : D - \frac{d}{d t} w (C) = T_{0} (s_{v}^{'} + h / T_{0}) = T_{0} π_{s} \geq 0$ by step 1.3. Since $T_{0} > 0$ , $π_{s} \geq 0$ ; no entropy conclusion is used or asserted without [A1].

Conditional admissibility of the specified non-Newtonian branches

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources