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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 $p_{*}$ .

Depends on

Used by

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Sources

Randall J. LeVeque, Numerical Methods for Conservation Laws (standard reference, not scraped)