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Conservation and entropy dissipation of the periodic SBP scheme

Statement

Let Ui(t)∈G, i∈I, be any differentiable solution of Periodic diagonal-norm SBP compressible viscous scheme. Then every component of the weighted conservative state ∑iHiiUi(t) is constant, and

ddt∑i∈IHiiη(Ui(t))=−∑i∈IHii∑j,k=1d(Djvi)TKjk(Ui)(Dkvi)≤0.

The exact entropy production is the finite nodal sum displayed above.

Facts & Assumptions

Given: The finite periodic node set, diagonal matrix H, derivative matrices Dj, and semidiscrete equation of Periodic diagonal-norm SBP compressible viscous scheme, with all nodal states remaining in G.

[F1]

Periodic diagonal-norm SBP compressible viscous scheme: H is positive diagonal, HDj+DjTH=0, Dj1=0, and the semidiscrete residual is the sum of the symmetric two-point Euler flux differencing term and the divergence of Kjk(Ui)(Dkv)i.

[F2]

Properties of the explicit entropy-conservative flux: Fjec(Ui,Uℓ)=Fjec(Uℓ,Ui) and (vℓ−vi)TFjec(Ui,Uℓ)=ψj(Uℓ)−ψj(Ui).

[F3]

Positive-semidefinite entropy-variable factorization of viscous flux: the full spatial-state block matrix K(U) is symmetric positive semidefinite.

Proof

1.1F1algebra

For each direction set Qj=HDj. Then [F1] gives QjT=−Qj and Qj1=HDj1=0. Skew-symmetry also gives 1TQj=−(Qj1)T=0.

2.1F1F2step 1.1algebra

Multiply the semidiscrete equation by Hii and sum in i. For the Euler term, each component of ∑i,ℓ(Qj)iℓFjec(Ui,Uℓ) is zero: swapping i and ℓ changes the sign of Qj by step 1.1 and leaves the flux unchanged by [F2]. The viscous term sums to 1THDjW=1TQjW=0 for each nodal flux array W. Thus the weighted sum of each conservative component has derivative zero.

2.2F1F2step 1.1algebra

By the chain rule, the Euler contribution to the entropy derivative is −2∑j,i,ℓ(Qj)iℓviTFjec(Ui,Uℓ). Skew-symmetry and flux symmetry [F2], after swapping the two indices in one copy of the sum, rewrite this as −∑j,i,ℓ(Qj)iℓ(vi−vℓ)TFjec(Ui,Uℓ). The entropy-potential identity [F2] changes the summand to (Qj)iℓ(ψj(Ui)−ψj(Uℓ)). Both its row sum and column sum vanish by step 1.1, so the total Euler contribution is zero.

3.1F1F3algebra∎

For each viscous direction, vTHDjW=−(Djv)THW by DjTH=−HDj. Applying this to Wi=∑kKjk(Ui)(Dkv)i shows that the remaining entropy derivative is exactly the negative sum in the Statement. Each nodal quadratic form is nonnegative by [F3], and each weight Hii is positive; hence the derivative is nonpositive.

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