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First-order HLL updates are averages of adjacent half-cell fan averages

Statement

Use the first-order HLL update of First-order finite-volume update using supplied HLL speeds. At a face with left and right states $U_{L}, U_{R}$ , supplied speeds $S_{L} < 0 < S_{R}$ , and physical flux values $F_{L} = F (U_{L}), F_{R} = F (U_{R})$ , define $U_{*} : = \frac{S_{R} U_{R} - S_{L} U_{L} - (F_{R} - F_{L})}{S_{R} - S_{L}} .$ If the one-sided CFL bounds $2 ν (- S_{L}) \leq 1$ and $2 ν S_{R} \leq 1$ hold, the averages of the HLL fan over the adjacent left and right half-cells are, respectively, $A^{-} = (1 - 2 ν (- S_{L})) U_{L} + 2 ν (- S_{L}) U_{*}, A^{+} = (1 - 2 ν S_{R}) U_{R} + 2 ν S_{R} U_{*} .$ For every cell $i$ , the updated mean satisfies $U_{i}^{+} = \frac{1}{2} (A_{i - 1 / 2}^{+} + A_{i + 1 / 2}^{-}),$ where each face uses its own supplied speeds and intermediate state.

Facts & Assumptions

Given: A periodic uniform mesh, the HLL middle branch at each face, and the stated CFL bounds. Derive the half-cell averages and the finite-volume decomposition.

[F1]

HLL two-wave approximate flux: when $S_{L} < 0 < S_{R}$ , the HLL flux is $(S_{R} F_{L} - S_{L} F_{R} + S_{L} S_{R} (U_{R} - U_{L})) / (S_{R} - S_{L})$ .

[F2]

The HLL flux follows from the conservative two-wave model: the displayed $U_{*}$ and the assigned model flux $F_{*} = F_{L} + S_{L} (U_{*} - U_{L})$ satisfy both conservative model balances; $F_{*}$ is the middle-branch HLL flux and need not equal $F (U_{*})$ .

[F3]

HLL two-wave approximate flux: in the middle branch, the approximate HLL model has three constant regions separated by $S_{L}, S_{R}$ and one intermediate state.

[F4]

First-order finite-volume update using supplied HLL speeds: $U_{i}^{+} = U_{i} - ν ({\hat{F}}_{i + 1 / 2} - {\hat{F}}_{i - 1 / 2})$ with shared HLL face fluxes.

Proof

1.1givenF1F2algebra

By [F2], the displayed $U_{*}$ is the intermediate state of the conservative HLL fan. Write $D = S_{R} - S_{L} > 0$ . Substitution gives $U_{*} - U_{L} = (S_{R} (U_{R} - U_{L}) - (F_{R} - F_{L})) / D$ and $U_{*} - U_{R} = (S_{L} (U_{R} - U_{L}) - (F_{R} - F_{L})) / D$ . Therefore $F_{L} + S_{L} (U_{*} - U_{L}) = F_{R} + S_{R} (U_{*} - U_{R})$ , and expanding either side yields exactly [F1].

2.1givenF2F3F4step 1.1construct

By [F3] and [F2], in coordinates with the face at $x = 0$ the HLL profile is $U_{L}$ for $x / t < S_{L}$ , $U_{*}$ for $S_{L} < x / t < S_{R}$ , and $U_{R}$ for $x / t > S_{R}$ . On $[- h / 2, 0]$ at time $Δ t$ , the two portions have lengths $h / 2 + S_{L} Δ t$ and $- S_{L} Δ t$ ; dividing by $h / 2$ gives $A^{-}$ . The CFL condition makes both lengths nonnegative. On $[0, h / 2]$ , the portions have lengths $S_{R} Δ t$ and $h / 2 - S_{R} Δ t$ ; dividing by $h / 2$ gives $A^{+}$ , again with nonnegative lengths.

3.1step 1.1step 2.1F4algebra∎

At the right face of cell $i$ , step 1.1 gives ${\hat{F}}_{i + 1 / 2} = F (U_{i}) + S_{L, i + 1 / 2} (U_{*, i + 1 / 2} - U_{i})$ ; at its left face it gives ${\hat{F}}_{i - 1 / 2} = F (U_{i}) + S_{R, i - 1 / 2} (U_{*, i - 1 / 2} - U_{i})$ . Substitution in [F4] yields $U_{i}^{+} = (1 - ν a_{R} - ν a_{L}) U_{i} + ν a_{R} U_{*, i + 1 / 2} + ν a_{L} U_{*, i - 1 / 2}$ , where $a_{R} = - S_{L, i + 1 / 2}$ and $a_{L} = S_{R, i - 1 / 2}$ . The half-sum $\frac{1}{2} (A_{i - 1 / 2}^{+} + A_{i + 1 / 2}^{-})$ has those same three coefficients, proving the identity.

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Sources

D. I. Ketcheson, R. J. LeVeque, and M. J. del Razo, Riemann Problems and Jupyter Solutions (standard reference, not scraped)
D. I. Ketcheson, R. J. LeVeque, and M. J. del Razo, Riemann Problems and Jupyter Solutions (standard reference, not scraped)