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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.

Depends on

Used by

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Dependency tree · two levels

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Sources

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