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Properties of the explicit entropy-conservative flux

Statement

The flux in Explicit entropy-conservative two-point Euler flux is well-defined for every pair in G×G, symmetric in its two states, and consistent in the sense that Fjec(U,U)=Fj(U). It satisfies

(v(UR)−v(UL))TFjec(UL,UR)=ψj(UR)−ψj(UL).

The flux is locally Lipschitz on G×G.

Facts & Assumptions

Given: d≥1 and two states UL,UR∈G.

[F1]

Explicit entropy-conservative two-point Euler flux: with Uˉ=(UL+UR)/2, Δv=v(UR)−v(UL), and Δψj=ψj(UR)−ψj(UL), use Fj(UL) when Δv=0 and otherwise use Fj(Uˉ)+(Δψj−ΔvTFj(Uˉ))Δv/∥Δv∥2.

[F2]

Strict convexity and entropy variables for the ideal-gas Euler flux: DUv=DU2η is positive definite on G, ∇Uψj=(DUv)TFj, and ψj=vTFj−qj.

Proof

1.1F1F2F3algebra

If UL=UR, then Δv=0 and [F1] gives Fjec(UL,UR)=Fj(UL). If UL≠UR, convexity [F3] keeps the segment U(t)=UL+t(UR−UL) in G, and ΔvT(UR−UL)=∫01(UR−UL)TDU2η(U(t))(UR−UL) dt>0 by [F2]. Hence Δv≠0, so the denominator in [F1] is nonzero; also Uˉ∈G. This proves well-definedness and consistency.

1.2F1algebra

Exchanging UL and UR leaves Uˉ fixed and changes the signs of both Δv and Δψj. Thus the scalar correction coefficient in [F1] changes sign while its vector factor changes sign, leaving the flux unchanged. The diagonal branch is also symmetric.

1.3F2algebra

Fix a state X∈G and write UR=X+h/2, UL=X−h/2 for small h. The entropy compatibility in [F2] implies DUψj(X)=Fj(X)TDUv(X). Symmetric Taylor expansions therefore give Δv=DUv(X)h+O(∣h∣3) and Δψj=Fj(X)TDUv(X)h+O(∣h∣3), so N:=Δψj−ΔvTFj(X)=O(∣h∣3). Positive definiteness of DUv(X) implies ∣Δv∣≥c∣h∣ in a sufficiently small neighborhood of h=0.

2.1F1step 1.1algebra

When Δv≠0, dotting [F1] with Δv cancels its denominator and gives ΔvTFj(Uˉ)+Δψj−ΔvTFj(Uˉ)=Δψj. When Δv=0, step 1.1 gives UL=UR, so both sides are zero.

3.1F1F2step 1.1step 1.3algebra∎

On that neighborhood, the correction NΔv/∣Δv∣2 is O(∣h∣2) and tends to zero on the diagonal. Differentiating the symmetric Taylor remainders gives ∂hN=O(∣h∣2), ∂XN=O(∣h∣3), ∂hΔv=O(1), and ∂XΔv=O(∣h∣). The quotient rule then shows that the correction derivatives are bounded and tend to zero as h→0; its diagonal extension is C1. Away from the diagonal the denominator is nonzero by step 1.1 and the formula is smooth. Thus the flux is locally Lipschitz everywhere on G×G.

Depends on

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Sources