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The ideal-gas star-pressure root exists uniquely exactly below vacuum threshold

Statement

For the data of Ideal-gas rarefaction and shock pressure response, let $Φ (0 +) : = u_{R} - u_{L} - \frac{2 (c_{L} + c_{R})}{γ - 1} .$ The residual $Φ$ is $C^{1}$ and strictly increasing on $(0, \infty)$ . If $Φ (0 +) < 0$ , then $Φ$ has exactly one zero $p_{*} > 0$ . By the acoustic wave-response lemma this is the unique positive pressure at which the left and right wave curves can be joined by a contact. It lies in the explicit bracket $0 < p_{*} < P$ , where, writing $D = \max {0, u_{L} - u_{R}}$ and $K = \sqrt{A_{L}} + \sqrt{A_{R}}$ , $P = 2 \max {p_{L}, p_{R}} + \frac{6 D^{2}}{K^{2}} + p_{L} .$ If $Φ (0 +) \geq 0$ , then there is no positive matching pressure. The limiting left and right rarefaction-tail velocities are ordered with a vacuum gap, or coincide at the zero-pressure equality threshold. No other vacuum condition is asserted.

Facts & Assumptions

Given: Positive densities and pressures, $γ > 1$ , and the response functions from Ideal-gas rarefaction and shock pressure response.

[F1]

Ideal-gas rarefaction and shock pressure response: on each side the rarefaction and shock formulas define $f_{K} (p)$ and $Φ (p) = f_{L} (p) + f_{R} (p) + u_{R} - u_{L}$ for $p > 0$ .

[F2]

Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a, b]$ takes every value between $f (a)$ and $f (b)$ : a continuous function on a closed real interval takes every value between its endpoint values.

[F3]

Ideal-gas acoustic wave curves have the stated pressure response: the left and right star velocities are $u_{L} - f_{L} (p)$ and $u_{R} + f_{R} (p)$ ; they match at a positive pressure exactly when $Φ (p) = 0$ , and their star states are then joined by a contact.

Proof

1.1givenF1algebra

On $0 < p \leq p_{K}$ , differentiation gives $f_{K}^{'} (p) = \frac{c_{K}}{γ p_{K}} {(\frac{p}{p_{K}})}^{- (γ + 1) / (2 γ)} > 0$ ; on $p \geq p_{K}$ , it gives $f_{K}^{'} (p) = \frac{\sqrt{A_{K}} (p + p_{K} + 2 B_{K})}{2 (p + B_{K})^{3 / 2}} > 0$ . At $p = p_{K}$ both derivatives equal $1 / \sqrt{γ ρ_{K} p_{K}}$ , so $f_{K}$ is $C^{1}$ and strictly increasing on $(0, \infty)$ . Thus $Φ$ is also $C^{1}$ and strictly increasing.

1.2F1algebra

For $p \geq 2 p_{K}$ , we have $p - p_{K} \geq p / 2$ and $p + B_{K} \leq 3 p / 2$ , because $0 < B_{K} < p_{K} \leq p / 2$ . The shock formula therefore gives $f_{K} (p) \geq \sqrt{A_{K} p / 6}$ . The stated $P$ exceeds both $2 p_{K}$ and $6 D^{2} / K^{2}$ , so $f_{L} (P) + f_{R} (P) \geq K \sqrt{P / 6} > D$ , and consequently $Φ (P) > 0$ .

2.1step 1.1F1algebra

The rarefaction formula in [F1] tends to $- 2 c_{K} / (γ - 1)$ as $p ↓ 0$ , hence $\lim_{p ↓ 0} Φ (p) = Φ (0 +)$ . If this limit is nonnegative, strict increase makes $Φ (p) > 0$ for every $p > 0$ , so there is no positive zero.

3.1step 1.1step 2.1step 1.2F2choose

If $Φ (0 +) < 0$ , continuity at the extended endpoint $0$ gives some $ϵ \in (0, P)$ with $Φ (ϵ) < 0$ . Applying [F2] to $Φ$ on $[ϵ, P]$ gives a zero in $(ϵ, P)$ ; strict increase from step 1.1 makes it unique on $(0, \infty)$ . Thus $0 < p_{*} < P$ .

4.1step 2.1step 3.1F3algebra∎

Taking the limits in [F3] using step 2.1 gives the left tail velocity $v_{L} = u_{L} + 2 c_{L} / (γ - 1)$ and the right tail velocity $v_{R} = u_{R} - 2 c_{R} / (γ - 1)$ , so $v_{R} - v_{L} = Φ (0 +)$ . If this right-minus-left gap is positive, the tails leave a vacuum interval; at equality they meet. If it is negative, step 3.1 gives the unique positive pressure at which the two wave responses match and the contact joins their star states. This proves the stated threshold and matching interpretation.

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Sources

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