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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.

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Sources

Eitan Tadmor, Entropy Stable Schemes, Chapter 18, Handbook of Numerical Methods for Hyperbolic Problems: Basic and Fundamental Issues, 2016, complete chapter (standard reference, not scraped)
Pablo Fernandez, Ngoc-Cuong Nguyen, and Jaime Peraire, Entropy-stable hybridized discontinuous Galerkin methods for the compressible Euler and Navier-Stokes equations, complete author preprint (standard reference, not scraped)