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Rankine--Hugoniot alone does not give uniqueness

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )) for the analytic prerequisites used below.

Let $f (u) = \frac{1}{2} u^{2}$ and let the Riemann data be $u_{L} = 0$ , $u_{R} = 1$ . Then both of the following are weak solutions of the same Cauchy problem: the expansion shock from The expansion shock is weak but not entropic and the rarefaction fan from The Burgers rarefaction Riemann solution, $u_{s h o c k} (t, x) = {\begin{cases} 0, & x < t / 2, \\ 1, & x > t / 2, \end{cases} u_{f a n} (t, x) = {\begin{cases} 0, & x \leq 0, \\ x / t, & 0 < x < t, \\ 1, & x \geq t . \end{cases}$ Only the fan is a Kruzhkov entropy solution (The Riemann solver for a strictly convex flux); the shock violates the entropy inequality. Thus the Rankine--Hugoniot condition and the weak formulation do not by themselves determine the solution, and an entropy selection is indispensable (Kruzhkov entropy solutions, Uniqueness, comparison and order preservation of entropy solutions).

Facts & Assumptions

Given: Countable Choice, the flux $f (u) = \frac{1}{2} u^{2}$ , the Riemann datum $u_{0} = 1_{(0, \infty)}$ , the expansion-shock profile $u_{s h o c k}$ and the rarefaction fan $u_{f a n}$ displayed in the statement.

[F1]

The expansion shock $u_{s h o c k}$ is a distributional weak solution with datum $u_{0}$ : it has left state $u^{-} = 0$ , right state $u^{+} = 1$ , speed $s = \frac{1}{2}$ satisfying Rankine--Hugoniot, and it fails the Kruzhkov entropy inequality (for $k = \frac{1}{2}$ the production coefficient is $\frac{1}{4} > 0$ ) (The expansion shock is weak but not entropic).

[F2]

The rarefaction fan $u_{f a n}$ is the unique Kruzhkov entropy solution of the same Riemann problem: it is a weak solution, satisfies all Kruzhkov inequalities, and attains the datum in the strong local $L^{1}$ sense (The Burgers rarefaction Riemann solution, The Riemann solver for a strictly convex flux, The self-similar Riemann problem).

[F3]

The weak formulation admits every distributional weak solution, while the entropy class is unique: two bounded Kruzhkov entropy solutions with the same datum agree almost everywhere (Uniqueness, comparison and order preservation of entropy solutions, Kruzhkov entropy solutions).

Proof

technique · direct

1.1F1F2

Two weak solutions of the same problem. By [F1] the expansion shock is a distributional weak solution with datum $u_{0}$ ; by [F2] the rarefaction fan is also a distributional weak solution with the same datum. Both are bounded and piecewise smooth.

2.1F1F2step 1.1

They differ on a set of positive measure. On the open region ${(t, x) : t > 0, t / 2 < x < t}$ the shock takes the value $1$ , while the fan takes the value $x / t < 1$ ; the region has positive Lebesgue measure, so the two classes differ.

3.1F3step 1.1step 2.1∎

Only the fan is entropic, and the entropy class is unique. The shock fails the Kruzhkov entropy inequality by [F1], so it is not a Kruzhkov entropy solution; the fan is the unique Kruzhkov entropy solution of these data by [F2], and any two bounded Kruzhkov entropy solutions with the same datum coincide almost everywhere by [F3]. Therefore Rankine--Hugoniot and the weak formulation alone determine neither the value of the solution nor its uniqueness, while the entropy condition selects the rarefaction fan and restores uniqueness in the entropy class.

Depends on

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Sources

G. A. Chechkin and A. Yu. Goritsky (translated by B. Andreianov), “S. N. Kruzhkov’s lectures on first-order quasilinear PDEs,” in Analytical and Numerical Aspects of PDEs, de Gruyter 2009, complete lecture-notes text (standard reference, not scraped)
Victor Ivrii, Partial Differential Equations, University of Toronto, current complete 415-page PDF (standard reference, not scraped)