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A Roe jump property does not imply entropy admissibility

Statement refuted

For a scalar conservation law, the Roe exact flux-jump identity by itself guarantees entropy admissibility.

Facts & Assumptions

Given: Burgers' scalar conservation law, its Roe matrix, the Riemann step data u0(x)=−1 for x<0 and u0(x)=1 for x>0, and the convex entropy pair below.

[F1]

Roe linearization and Roe numerical flux: the Roe jump property is the identity a~(uL,uR)(uR−uL)=f(uR)−f(uL), which does not itself impose an entropy inequality.

[F2]

Convex entropy admissibility for a scalar conservation law: for an entropy pair (η,q), a moving jump of speed s is entropy-admissible only if [q]−s[η]≤0.

[F3]

Rankine–Hugoniot jumps and Lax shocks: a Rankine–Hugoniot jump of speed s is defined by [f]=s[u].

[F4]

A moving jump is a weak solution exactly under Rankine–Hugoniot: a Rankine–Hugoniot jump with the corresponding Riemann step data is a distributional weak solution and has the required local L1 initial trace.

[A1]

For Burgers' flux, η(u)=u2/2 and q(u)=u3/3 satisfy η′′=1>0 and q′=η′f′.

Counterexample

1.1givenF1F3F4algebra

For Burgers' flux f(u)=u2/2 with uL=−1,uR=1, the Roe scalar a~(uL,uR)=(uL+uR)/2 is continuous at each diagonal pair, has equal-state value a~(u,u)=u=f′(u), and satisfies the jump identity for every pair because f(uR)−f(uL)=12(uL+uR)(uR−uL). At (−1,1) the speed is zero and f(1)−f(−1)=0=s(uR−uL), so the stationary jump satisfies the Roe identity [F1] and, by [F4], is a weak solution with the Riemann data in the Given clause.

2.1givenA1F2step 1.1algebra∎

For [A1], the entropy production coefficient is [q]−s[η]=(1/3−(−1/3))−0⋅(1/2−1/2)=2/3>0. This violates the necessary entropy inequality [F2]. Thus the Roe jump property and weak Rankine–Hugoniot balance do not guarantee entropy admissibility.

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