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Compressible viscous flow and entropy-stable viscous fluxes — Examples

1 · Prerequisites

Compressible viscous flow and entropy-stable viscous fluxes

2 · Summary

These examples compute a smooth periodic manufactured forcing, verify the continuous entropy budget for heated Couette flow, evaluate an exact three-node SBP inviscid-versus-viscous entropy rate, and exhibit the entropy-variable blowup along a positive-state sequence approaching vacuum. The wall calculation is a continuous balance, and the finite-node calculation gives an exact arithmetic check of the semidiscrete identity.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Entropy variables leave compact positive-state tubes near vacuum

Example

Take $R = 1$ , $γ = 2$ , $c_{v} = 1$ , and $s_{0} = 0$ in one dimension. For each integer $n \geq 1$ , let $U_{n} = (ρ_{n}, m_{n}, E_{n}) = (1 / n, 0, 1 / n)$ . Every state is admissible and has $p_{n} = 1 / n$ , $T_{n} = 1$ , and $s_{n} = \log n$ . Its density entropy variable is $v_{ρ, n} = 2 - \log n$ , which tends to $- \infty$ as the positive density and pressure tend to zero. The sequence approaches the excluded vacuum state $(0, 0, 0)$ , so no compact set $C ⋐ G$ contains all $U_{n}$ .

Verification

Given: The model and entropy-variable formulas with $R = 1$ , $γ = 2$ , $c_{v} = 1$ , $s_{0} = 0$ , and $U_{n} = (1 / n, 0, 1 / n)$ .

[F1] Ideal-gas compressible Navier–Stokes–Fourier model: $G = {ρ > 0, E > ∣ m ∣^{2} / (2 ρ)}$ , $p = (γ - 1) (E - ∣ m ∣^{2} / (2 ρ))$ , $T = p / (R ρ)$ , and $s = c_{v} \log (p / ρ^{γ}) + s_{0}$ .

[F2] Strict convexity and entropy variables for the ideal-gas Euler flux: $v_{ρ} = c_{v} γ - s - β ∣ u ∣^{2} / 2$ , $v_{E} = - β$ , and $β = 1 / T$ .

1.1F1givenalgebra

For $U_{n}$ , $ρ_{n} > 0$ , $ε_{n} = E_{n} = 1 / n > ∣ 0 ∣^{2} / (2 ρ_{n})$ , so $U_{n} \in G$ . The formulas [F1] give $p_{n} = 1 / n$ , $T_{n} = 1$ , and $s_{n} = \log ((1 / n) / (1 / n)^{2}) = \log n$ .

2.1F2step 1.1algebra

Here $u_{n} = 0$ and $β_{n} = 1$ . Substitution into [F2] gives $v_{ρ, n} = 2 - \log n$ and $v_{E, n} = - 1$ , hence $∣ v_{ρ, n} ∣ \to \infty$ .

3.1F1step 1.1step 2.1algebra∎

The states satisfy $U_{n} \to (0, 0, 0)$ , which is outside $G$ because its density is zero. Every compact subset of Euclidean state space is closed, so a compact $C ⋐ G$ containing every $U_{n}$ would also contain their limit, a contradiction.

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A periodic manufactured compressible viscous state

Example

On the one-dimensional periodic domain $T_{2 π}$ , fix $R > 0$ , $γ > 1$ , $μ > 0$ , $ζ \geq 0$ , and $κ > 0$ . Let $ρ (x) = 1$ , $u (x) = 0$ , $T (x) = 2 + \sin x$ , and $E (x) = c_{v} T (x)$ . With forcing $S (x) = (0, R \cos x, κ \sin x)$ in the conservative balance $\partial_{t} U + \partial_{x} F_{1} = \partial_{x} F_{1}^{v i s c} + S$ , these fields form a smooth positive stationary solution. Its pointwise physical entropy production is $κ \cos^{2} x / T (x)^{2}$ , and the forcing contribution closes the periodic mathematical-entropy balance exactly.

Verification

Given: The one-dimensional ideal-gas model, the displayed periodic state, and the forcing $S$ in the balance law.

[F1] Ideal-gas compressible Navier–Stokes–Fourier model: $p = (γ - 1) (E - m^{2} / (2 ρ))$ , $T = p / (R ρ)$ , $F_{1} = (ρ u, ρ u^{2} + p, (E + p) u)$ , and $F_{1}^{v i s c} = (0, τ, u τ + κ \partial_{x} T)$ .

[F2] Smooth ideal-gas Navier–Stokes–Fourier entropy identity: for an unforced smooth positive field, $η_{t} + \partial_{x} (η u + κ T_{x} / T) = - τ u_{x} / T - κ T_{x}^{2} / T^{2}$ .

1.1F1givenalgebra

Since $T (x) \geq 1$ , $ρ = 1$ , $u = 0$ , $m = 0$ , and $E = c_{v} T > 0$ , the state lies in $G$ . The equation of state gives $p = (γ - 1) c_{v} T = R T$ . The stress is zero because $u_{x} = 0$ , so $F_{1} = (0, R T, 0)$ and $F_{1}^{v i s c} = (0, 0, κ \cos x)$ .

2.1F1step 1.1algebra

The state is stationary, so $\partial_{t} U = 0$ . Direct differentiation gives $\partial_{x} F_{1} = (0, R \cos x, 0)$ and $\partial_{x} F_{1}^{v i s c} = (0, 0, - κ \sin x)$ . Therefore $\partial_{x} F_{1}^{v i s c} + S = \partial_{x} F_{1}$ , proving the forced balance componentwise.

3.1F1F2step 1.1step 2.1algebra∎

The physical production is $τ u_{x} / T + κ T_{x}^{2} / T^{2} = κ \cos^{2} x / T^{2}$ . For a source $S$ , multiplying the conservative equation by $v$ adds $v^{T} S$ to the right side of the mathematical-entropy identity [F2]; here $v^{T} S = - κ \sin x / T$ . The left-side entropy flux derivative is $\partial_{x} (κ \cos x / T) = - κ \sin x / T - κ \cos^{2} x / T^{2}$ , which equals minus the production plus $v^{T} S$ . Periodicity removes endpoint contributions.

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Couette flow with the thermal wall entropy balance

Example

Let $x$ be periodic, $0 < y < L$ , and fix $p_{0} > 0$ , $T_{w} > 0$ , $a \neq 0$ , $μ > 0$ , and $κ > 0$ . Define

$u (x, y) = (a y, 0), T (y) = T_{w} + \frac{μ a^{2}}{2 κ} y (L - y), ρ (y) = \frac{p_{0}}{R T (y)},$

and $m = ρ u$ , $E = p_{0} / (γ - 1) + ρ a^{2} y^{2} / 2$ . This is a smooth positive steady solution with constant pressure $p_{0}$ , no-slip wall velocities $u (x, 0) = (0, 0)$ and $u (x, L) = (a L, 0)$ , and wall temperature $T_{w}$ at both walls. Per unit periodic length, the outward physical-entropy flux through the two thermal walls equals

${[- κ \frac{T^{'}}{T}]}_{0}^{L} = \int_{0}^{L} (\frac{μ a^{2}}{T} + \frac{κ (T^{'})^{2}}{T^{2}}) d y > 0.$

This is a continuous wall-balance example; it specifies no discrete wall flux.

Verification

Given: The two-dimensional ideal-gas Navier–Stokes–Fourier model, $Ω$ periodic in $x$ and equal to $(0, L)$ in $y$ , and the displayed fields.

[F1] Ideal-gas compressible Navier–Stokes–Fourier model: $p = (γ - 1) (E - ∣ m ∣^{2} / (2 ρ))$ , $τ = 2 μ D^{0} + ζ (tr D) I$ , and $F_{j}^{v i s c} = (0, τ_{j \cdot}, u \cdot τ_{j \cdot} + κ \partial_{j} T)$ .

[F2] Smooth ideal-gas Navier–Stokes–Fourier entropy identity: physical entropy $ρ s = - η$ satisfies $\partial_{t} (ρ s) + \nabla \cdot (ρ s u - κ \nabla T / T) = τ : \nabla u / T + κ ∣ \nabla T ∣^{2} / T^{2}$ .

1.1F1givenalgebra

The profile has $T (y) \geq T_{w} > 0$ , so $ρ > 0$ . Its internal-energy density is $E - ∣ m ∣^{2} / (2 ρ) = p_{0} / (γ - 1)$ , hence [F1] gives $p = p_{0}$ . The velocity gradient has only $\partial_{y} u_{1} = a$ , so $\nabla \cdot u = 0$ , $τ_{12} = τ_{21} = μ a$ , and all entries of $τ$ are constant.

2.1F1step 1.1algebra

The mass divergence is zero because the field is independent of $x$ and $u_{2} = 0$ . The momentum divergence is zero because $p$ is constant, $τ$ is constant, and every $y$ -velocity is zero. For total energy, the Euler flux has zero divergence; the only nonzero viscous energy flux is $F_{2, E}^{v i s c} = u \cdot τ_{2 \cdot} + κ T^{'} = μ a^{2} y + κ T^{'}$ . Since $T^{''} = - μ a^{2} / κ$ , its derivative is zero. Thus all conservative equations hold.

3.1F1F2step 1.1step 2.1algebra∎

The stress production is $τ : \nabla u = μ a^{2}$ . The physical entropy law [F2], no normal velocity at either wall, and periodicity in $x$ give the boundary-flux identity after integration in $y$ . Directly, $(κ T^{'} / T)^{'} = κ T^{''} / T - κ (T^{'})^{2} / T^{2} = - μ a^{2} / T - κ (T^{'})^{2} / T^{2}$ , so integrating its negative proves the equality. The integrand is strictly positive because $a \neq 0$ , $μ > 0$ , and $T > 0$ .

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An exact three-node inviscid-versus-viscous SBP entropy budget

Example

Take $d = 1$ , $R = 1$ , $γ = 2$ , $c_{v} = 1$ , $s_{0} = 0$ , and three periodic nodes with $H = I_{3}$ and

$D = \frac{1}{2} (\begin{array}{ccc} 0 & 1 & - 1 \\ - 1 & 0 & 1 \\ 1 & - 1 & 0 \end{array}) .$

Use states $U_{1} = (1, 0, 1)$ , $U_{2} = (1, 0, 2)$ , and $U_{3} = (1, 0, 1)$ . The Euler-only entropy-conservative operator has zero total mathematical-entropy rate at this state; adding the model viscous term gives the exact rate $- κ / 8$ .

Verification

Given: The three-node data above, $κ > 0$ , the SBP flux-differencing formula, and the entropy-variable viscous block.

[F1] Periodic diagonal-norm SBP compressible viscous scheme: the Euler term is $2 D F^{e c} (U_{i}, U_{ℓ})$ , the viscous term is $D (K_{11} (U_{i}) (D v)_{i})$ , and the periodic grid satisfies $D^{T} = - D$ and $D 1 = 0$ .

[F2] Properties of the explicit entropy-conservative flux: $F^{e c} (U_{i}, U_{ℓ}) = F^{e c} (U_{ℓ}, U_{i})$ and $(v_{ℓ} - v_{i})^{T} F^{e c} (U_{i}, U_{ℓ}) = ψ (U_{ℓ}) - ψ (U_{i})$ .

[F3] Positive-semidefinite entropy-variable factorization of viscous flux: the one-dimensional nodal viscous entropy production is $2 μ T ∣ G^{0} ∣^{2} + ζ T (tr G)^{2} + κ T^{2} (D v_{E})^{2}$ .

[F4] Strict convexity and entropy variables for the ideal-gas Euler flux: $v = \nabla_{U} η$ , $v_{m} = u / T$ , and $v_{E} = - 1 / T$ .

1.1F1givenalgebra

The displayed $D$ is skew-symmetric, and each row sums to zero, so with $H = I_{3}$ it satisfies [F1]. The states have $ρ_{i} = 1$ , $p_{i} = E_{i} > 0$ , and $T_{i} = E_{i}$ , hence are admissible.

2.1F1F3F4step 1.1algebra

Formula [F4] gives $v_{E} = (- 1, - 1 / 2, - 1)$ and $v_{m} = (0, 0, 0)$ . Multiplication by $D$ gives $D v_{E} = (1 / 4, 0, - 1 / 4)$ and $D v_{m} = 0$ . Since $u_{i} = 0$ , the strain array $G$ in [F3] vanishes at every node.

2.2F1F2F4step 1.1algebra

By [F4] and the chain rule, the Euler-only operator has entropy derivative $- 2 \sum_{i, ℓ} D_{i ℓ} v_{i}^{T} F^{e c} (U_{i}, U_{ℓ})$ . Skew-symmetry and [F2] rewrite this as $- \sum_{i, ℓ} D_{i ℓ} (ψ (U_{i}) - ψ (U_{ℓ})) = 0$ , since both row and column sums of $D$ vanish.

3.1F1F3F4step 1.1step 2.1algebra∎

For the viscous term, [F4] and matrix skew-symmetry give the entropy derivative $v^{T} D W = - (D v)^{T} W = - \sum_{i} (D v_{i})^{T} K_{11} (U_{i}) (D v_{i})$ for the nodal flux array $W_{i} = K_{11} (U_{i}) (D v)_{i}$ . The zero strain and $T = (1, 2, 1)$ from step 1.1 reduce [F3] to $- κ \sum_{i} T_{i}^{2} (D v_{E, i})^{2} = - κ (1 / 16 + 0 + 1 / 16) = - κ / 8$ . Thus adding viscosity changes the Euler-only rate from zero to the stated negative value.

Compressible viscous flow and entropy-stable viscous fluxes — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Entropy variables leave compact positive-state tubes near vacuum

Example

Verification

A periodic manufactured compressible viscous state

Example

Verification

Couette flow with the thermal wall entropy balance

Example

Verification

An exact three-node inviscid-versus-viscous SBP entropy budget

Example

Verification

Sources