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The Burgers rarefaction Riemann solution

Statement

Assume Countable Choice (The Axiom of Countable Choice (ACω)) for the analytic prerequisites used below.

Let f(u)=12u2 and uL<uR. Then the entropy solution of the Riemann problem (The self-similar Riemann problem) is the centred rarefaction u(t,x)={uL,x≤uLt,x/t,uLt<x<uRt,uR,x≥uRt. The middle branch satisfies ut+uux=0, the outer branches are constant, and the values match continuously across the rays x=uLt and x=uRt. For every t>0 the profile is locally Lipschitz in (t,x) (on the fan ux=1/t); thus the chain rule gives zero distributional production for every convex entropy pair on R×(0,∞). The solution attains the Riemann data in the strong local L1 sense as t↓0. For uL=0, uR=1, the fan is u=x/t on 0<x<t (The Riemann solver for a strictly convex flux, Kruzhkov entropy solutions).

Facts & Assumptions

Given: Countable Choice, the flux f(u)=12u2, states uL<uR, the centred rarefaction profile u of the statement, and a test function φ∈Cc∞(ΠT).

[F1]

The strictly convex Riemann solver: for f∈C2 strictly convex with uL<uR, the unique Kruzhkov entropy solution of the Riemann problem is the centred rarefaction uL for x/t≤f′(uL), (f′)−1(x/t) for f′(uL)<x/t<f′(uR), and uR for x/t≥f′(uR); it satisfies the weak conservation law, all Kruzhkov entropy inequalities and the strong local L1 initial trace (The Riemann solver for a strictly convex flux, Kruzhkov entropy solutions, The self-similar Riemann problem).

[F2]

For f(u)=12u2 one has f′(u)=u and f′′≡1>0, so f is strictly convex with (f′)−1(ξ)=ξ for all ξ (directly, the Jensen gap for a,b is λ(1−λ)(a−b)2/2, positive for a≠b and 0<λ<1).

[F3]

Calculus on the self-similar profile: the chain rule computes ut=−ξU′(ξ)/t and ux=U′(ξ)/t for u(t,x)=U(x/t); a continuous piecewise C1 profile with equal traces across an interface produces no interface term in the weak or entropy residual, since the traces of u, f(u), and of η(u), q(u) for continuous pairs coincide from both sides (The chain rule, in one line from Carathéodory: if g is differentiable at c and f is differentiable at g(c), then f∘g is differentiable at c with (f∘g)′(c)=f′(g(c)) g′(c)).

Proof

technique · direct
1.1F1F2

Specialisation of the solver. By [F2], f′(u)=u and (f′)−1(ξ)=ξ; the three branches of the strictly convex Riemann solver [F1] read uL for x/t≤uL, x/t for uL<x/t<uR, and uR for x/t≥uR, which is exactly the displayed centred rarefaction. Hence by [F1] it is the unique Kruzhkov entropy solution of the Riemann problem, satisfies the weak conservation law, all Kruzhkov entropy inequalities, and attains the Riemann datum in the strong local L1 sense.

1.2F2F3

Direct check of the middle branch and the interfaces. On the fan, u(t,x)=x/t, so ut=−x/t2, ux=1/t and ut+uux=−x/t2+(x/t)(1/t)=0 by the chain rule [F3]; on the two outer regions u is constant, so both derivatives vanish there. At x=uLt the three-branch formula gives uL from both the first and middle branches, and at x=uRt it gives uR from both the middle and last branches; the traces of u and of f(u) therefore agree across the rays, so by [F3] no interface terms arise in the weak residual and the profile is a distributional weak solution.

2.1F1F3step 1.1

Entropy production and regularity. For any convex C2 pair (η,q) with q′=η′f′, the chain rule gives ∂tη(u)+∂xq(u)=η′(u)(ut+f′(u)ux)=η′(u)(ut+uux) on each smooth branch; this vanishes on the fan by step 1.2 and on the outer branches because u is constant. Across the rays the traces of η(u) and q(u) agree because u is continuous there, so no interface measure arises: the entropy production is identically 0 on R×(0,∞) for every convex C2 pair. (For the non-smooth Kruzhkov pairs, the entropy inequalities are supplied by the solver [F1].) On the fan ux=1/t, so the profile is locally Lipschitz on every compact subset of the open strip t>0; no uniform Lipschitz bound as t↓0 is claimed.

3.1F1F3step 1.1step 2.1∎

Initial trace and the special case. The discrepancy from the initial step is supported between min⁡{0,uL}t and max⁡{0,uR}t, and is bounded by uR−uL. Thus ∫K∣u(t)−u0∣≤(uR−uL)(max⁡{0,uR}−min⁡{0,uL})t→0. When uL=0, uR=1, the fan is x/t on 0<x<t, with outer states 0 and 1. For nonsmooth convex pairs, smooth convex approximation and uniform convergence of the integral fluxes pass the zero-production identity of step 2.1 to the limit; thus production is zero, not merely nonpositive.

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