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Riemann solvers for hyperbolic systems — Examples

1 · Prerequisites

2 · Summary

These examples exercise the local scope and its failure boundaries. The acoustic system gives an exact two-wave composition, and the ideal-gas contact has equal pressure and velocity but different densities. The Sod data have a unique pressure root bracketed by their initial pressures, with a left rarefaction and right shock. Two exact velocity choices demonstrate the vacuum-threshold equality and strict-vacuum cases. The Burgers counterexample shows that an exact Roe flux-jump identity and weak Rankine–Hugoniot balance do not imply the convex entropy inequality.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

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 $u_{0} (x) = - 1$ for $x < 0$ and $u_{0} (x) = 1$ for $x > 0$ , and the convex entropy pair below.

[F1]

Roe linearization and Roe numerical flux: the Roe jump property is the identity $\tilde{a} (u_{L}, u_{R}) (u_{R} - u_{L}) = f (u_{R}) - f (u_{L})$ , 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 $[q] - s [η] \leq 0$ .

[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 $L^{1}$ initial trace.

[A1]

For Burgers' flux, $η (u) = u^{2} / 2$ and $q (u) = u^{3} / 3$ satisfy $η^{''} = 1 > 0$ and $q^{'} = η^{'} f^{'}$ .

Counterexample

1.1givenF1F3F4algebra

For Burgers' flux $f (u) = u^{2} / 2$ with $u_{L} = - 1, u_{R} = 1$ , the Roe scalar $\tilde{a} (u_{L}, u_{R}) = (u_{L} + u_{R}) / 2$ is continuous at each diagonal pair, has equal-state value $\tilde{a} (u, u) = u = f^{'} (u)$ , and satisfies the jump identity for every pair because $f (u_{R}) - f (u_{L}) = \frac{1}{2} (u_{L} + u_{R}) (u_{R} - u_{L})$ . At $(- 1, 1)$ the speed is zero and $f (1) - f (- 1) = 0 = s (u_{R} - u_{L})$ , 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 $[q] - s [η] = (1 / 3 - (- 1 / 3)) - 0 \cdot (1 / 2 - 1 / 2) = 2 / 3 > 0$ . 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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A contact discontinuity with equal velocity and pressure

Example

Fix $γ > 1$ , pressure $p_{0} > 0$ , velocity $u_{0} \in R$ , and distinct positive densities $ρ_{L}, ρ_{R}$ . Set $E_{K} = p_{0} / (γ - 1) + ρ_{K} u_{0}^{2} / 2$ and $U_{K} = (ρ_{K}, ρ_{K} u_{0}, E_{K})$ for $K = L, R$ .

Facts & Assumptions

Given: Two admissible ideal-gas states with common pressure and velocity but different densities.

[F1]

Admissible states and pressure for one-dimensional ideal-gas Euler: $F (U) = (ρ u, ρ u^{2} + p, u (E + p))$ and $E = p / (γ - 1) + ρ u^{2} / 2$ .

[F2]

Eigenstructure of the admissible ideal-gas Euler system: the middle eigenvalue is $λ_{0} = u$ and its field is linearly degenerate.

[F3]

Rankine–Hugoniot jumps and Lax shocks: a linearly-degenerate contact has equal family speeds on its two sides and is not a strict Lax shock.

Verification

1.1givenF1algebra

Both states are admissible because $ρ_{K} > 0$ and their internal energy is $p_{0} / (γ - 1) > 0$ . For either state, $F (U_{K}) - u_{0} U_{K} = (0, p_{0}, u_{0} p_{0})$ ; subtracting the two identities gives $F (U_{R}) - F (U_{L}) = u_{0} (U_{R} - U_{L})$ , the Rankine–Hugoniot equality at speed $u_{0}$ .

2.1step 1.1F2F3algebra∎

By [F2] the speed $u_{0}$ is the middle characteristic speed on both sides, so the Rankine–Hugoniot jump from step 1.1 is a linearly-degenerate contact. Density, energy, and momentum may jump while pressure and velocity remain continuous; [F3] says this contact is not a strict Lax shock because its two middle characteristic speeds are equal.

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Euler vacuum detection at and above the pressure-root threshold

Example

Let $γ = 7 / 5$ , $ρ_{L} = ρ_{R} = 1$ , and $p_{L} = p_{R} = 1$ . Then $c_{L} = c_{R} = c = \sqrt{7 / 5}$ , and the vacuum threshold is $u_{R} - u_{L} \geq 2 (c_{L} + c_{R}) / (γ - 1) = 10 c$ .

Facts & Assumptions

Given: The two positive-density, positive-pressure initial states and the vacuum criterion.

[F1]

Admissible states and pressure for one-dimensional ideal-gas Euler: positive density and pressure give admissible states and $c = \sqrt{γ p / ρ}$ .

[F2]

Ideal-gas rarefaction and shock pressure response: $Φ (0 +) = u_{R} - u_{L} - 2 (c_{L} + c_{R}) / (γ - 1)$ .

[F3]

The ideal-gas star-pressure root exists uniquely exactly below vacuum threshold: $Φ (0 +) \geq 0$ means no positive star-pressure root.

[F4]

Finite-budget safeguarded Newton–bisection for Euler star pressure: the procedure returns vacuum when $Φ (0 +) \geq 0$ .

Verification

1.1givenF1F2F3F4algebra

In the equality case take $u_{L} = - 5 c, u_{R} = 5 c$ . Then $u_{R} - u_{L} = 10 c$ , so [F2] gives $Φ (0 +) = 0$ . All thermodynamic states remain admissible by [F1]; [F3] gives no positive pressure root and [F4] returns vacuum.

2.1step 1.1F1F2F3F4algebra∎

In a strict case take $u_{L} = - 6, u_{R} = 6$ . Then $u_{R} - u_{L} = 12 > 10 \sqrt{7 / 5} = 10 c$ , since $144 > 140$ . Thus $Φ (0 +) > 0$ and the same conclusions follow from [F3]–[F4]. These cases show the threshold equality and strict-vacuum branches without using an approximate numerical pressure.

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Sod tube: a unique star-pressure root between the initial pressures

Example

For $γ = 7 / 5$ , take left primitive data $(ρ_{L}, u_{L}, p_{L}) = (1, 0, 1)$ and right data $(ρ_{R}, u_{R}, p_{R}) = (1 / 8, 0, 1 / 10)$ . Set $E_{K} = p_{K} / (γ - 1) + ρ_{K} u_{K}^{2} / 2$ and $U_{K} = (ρ_{K}, ρ_{K} u_{K}, E_{K})$ for $K = L, R$ ; both conserved states lie in $G_{γ}$ .

Facts & Assumptions

Given: The Sod initial states, the ideal-gas pressure response, its root/vacuum criterion, and the finite-budget bracket solver.

[F1]

Admissible states and pressure for one-dimensional ideal-gas Euler: $ρ > 0, p > 0$ is equivalent to the declared admissible-state inequalities for the ideal-gas model.

[F2]

Ideal-gas rarefaction and shock pressure response: $Φ (p) = f_{L} (p) + f_{R} (p) + u_{R} - u_{L}$ with rarefaction response below $p_{K}$ and shock response above it.

[F3]

The ideal-gas star-pressure root exists uniquely exactly below vacuum threshold: $Φ (0 +) < 0$ gives a unique positive root; $Φ$ is strictly increasing.

[F4]

The safeguarded Euler pressure solver preserves its root bracket and terminates: for a budget $N$ with $(3 / 4)^{N} P \leq τ$ , the safeguarded procedure returns a bracket of width at most $τ$ containing that root.

Verification

1.1givenF1F2F3algebra

The densities and pressures are positive, so [F1] gives admissible states. The sound speeds are $c_{L} = \sqrt{7 / 5}$ and $c_{R} = \sqrt{28 / 25}$ . Since both velocities are zero, $Φ (0 +) = - 2 (c_{L} + c_{R}) / (γ - 1) < 0$ . By [F3] there is one positive star pressure $p_{*}$ .

2.1step 1.1F2F3algebra

At $p = 1 / 10 = p_{R}$ , $f_{R} (p) = 0$ while $p < p_{L}$ makes the left rarefaction response strictly negative; hence $Φ (1 / 10) < 0$ . At $p = 1 = p_{L}$ , $f_{L} (p) = 0$ while $p > p_{R}$ makes the right shock response strictly positive; hence $Φ (1) > 0$ . Strict increase gives $1 / 10 < p_{*} < 1$ , so the left wave is a rarefaction and the right wave is a shock.

3.1step 2.1F2F3F4algebra∎

The contact between the two acoustic waves has pressure $p_{*}$ and common velocity $u_{*} = u_{L} - f_{L} (p_{*}) = u_{R} + f_{R} (p_{*})$ . Equality follows by rearranging $Φ (p_{*}) = 0$ . The explicit theorem bracket $[0, P]$ and [F4] give a finite-budget enclosure of any prescribed positive width when the budget meets its bound. A smaller budget may also succeed; if the procedure returns budget-exhausted, its interval still contains $p_{*}$ .

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An acoustic system resolved into two characteristic waves

Example

Consider $U_{t} + A U_{x} = 0$ with $A = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$ , base state $\overset{ˉ}{U} = (0, 0)^{T}$ , and Riemann data $U_{L} = (0, 0)^{T}$ , $U_{R} = (- ε, 0)^{T}$ . The eigenvalues are $- 1, 1$ , so the system is strictly hyperbolic and both fields are linearly degenerate. Use the constant fields $r_{-} = (1, - 1)^{T}$ and $r_{+} = (1, 1)^{T}$ . Choose $0 < ε < δ$ and $ε < 2 η$ , where $δ, η > 0$ are the local solver radii at $\overset{ˉ}{U}$ for these fields.

Facts & Assumptions

Given: The acoustic matrix, the two constant states chosen within the local solver radii, and the local solver for linearly-degenerate fields.

[F1]

Strictly hyperbolic conservation systems and characteristic families: characteristic waves have eigenvectors of the flux Jacobian, and a linear constant-coefficient system has constant eigenpairs.

[F2]

Local small-data Riemann solver near a strictly hyperbolic state: sufficiently close data are joined by the ordered local family waves; a linearly-degenerate family produces a contact.

[F3]

Local characteristic rarefaction curve and Mixed local wave curve for a genuinely nonlinear or linearly degenerate family: for a linearly-degenerate field, $Ψ_{i} (σ; U) = R_{i} (σ; U)$ , where $\partial_{σ} R_{i} = r_{i} (R_{i})$ and $R_{i} (0; U) = U$ .

Verification

1.1givenF1algebra

Direct multiplication gives eigenpairs $(- 1, r_{-} = (1, - 1)^{T})$ and $(1, r_{+} = (1, 1)^{T})$ . Since the system is linear, its two characteristic fields are linearly degenerate. The jump is $U_{R} - U_{L} = (- ε, 0)^{T} = - \frac{ε}{2} r_{-} - \frac{ε}{2} r_{+}$ .

2.1givenstep 1.1F2F3algebra

Since the chosen fields are constant, the initial-value equations in [F3] give $Ψ_{\pm} (σ; U) = U + σ r_{\pm}$ . Thus the coefficients $σ_{-} = σ_{+} = - ε / 2$ in step 1.1 are the solver's strengths for this normalization. They satisfy $∣ σ_{\pm} ∣ < η$ , and both data states lie in the prescribed radius- $δ$ neighborhood. After the first wave the intermediate state is $U_{M} = U_{L} - \frac{ε}{2} r_{-} = (- \frac{ε}{2}, \frac{ε}{2})^{T}$ ; after the second, $U_{M} - \frac{ε}{2} r_{+} = U_{R}$ . By uniqueness in [F2], these are the solver's ordered contacts. For the first jump $A (U_{M} - U_{L}) = (- 1) (U_{M} - U_{L})$ , and for the second $A (U_{R} - U_{M}) = (U_{R} - U_{M})$ , verifying Rankine–Hugoniot at the ordered speeds $- 1$ and $1$ .

3.1step 2.1F2∎

The profile equal to $U_{L}$ for $x / t < - 1$ , to $U_{M}$ for $- 1 < x / t < 1$ , and to $U_{R}$ for $x / t > 1$ is therefore the two-contact local Riemann solution.

Diagram

The horizontal axis is $x$ and the vertical axis is $t$ . Each ray from the origin has slope $d x / d t = λ$ ; the two constant wedges carry $U_{L}, U_{M}, U_{R}$ . This picture records the wave ordering only; the algebraic Rankine–Hugoniot checks above establish the solution.

Sources

Randall J. LeVeque, Numerical Methods for Conservation Laws