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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Two Euler quadrature nodes rescaled to a positive margin

Example

With weights w1=w2=1/2, cell mean Uˉ=(1,0,1), and candidate conserved states V1=(1,4,1) and V2=(1,−4,1), apply the limiter to K7/5,1/2. It returns θ∗=1/4, giving states (1,1,1) and (1,−1,1) with positive pressure and the original cell mean.

Facts & Assumptions

Given: The weights, mean, nodes, and set K7/5,1/2. Compute the maximal scaling and verify the mean and positive margin.

[F1]

Maximal convex scaling toward an admissible cell mean: the limiter chooses the largest θ∈[0,1] keeping each Uˉ+θ(Vj−Uˉ) in K.

[F2]

Positive-margin admissible states for ideal-gas Euler: Kγ,ε requires ρ≥ε and E−m2/(2ρ)≥ε.

[F3]

Admissible states and pressure for one-dimensional ideal-gas Euler: p=(γ−1)(E−m2/(2ρ)).

[F4]

Convex scaling preserves the cell mean and positive-margin node states: the maximal scaling exists, preserves the weighted mean, and returns nodes in the positive-margin set.

Verification

1.1givenF2algebra

The weighted mean is (V1+V2)/2=(1,0,1). Its density is 1 and its internal-energy density is 1, so it belongs to K7/5,1/2 by [F2]. The scaled states are (1,4θ,1) and (1,−4θ,1), each with internal energy 1−8θ2. Hence they lie in K7/5,1/2 exactly when 1−8θ2≥1/2, or 0≤θ≤1/4.

2.1givenF1F2F3F4algebra∎

The largest feasible parameter is θ∗=1/4 by [F1]. The outputs are (1,1,1) and (1,−1,1); each has internal-energy density 1/2 and pressure (2/5)(1/2)=1/5>0 by [F3]. Their weighted mean is (1,0,1), also as guaranteed by [F4].

Depends on

Used by

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