A moment closure based on a projection on the boundary of the realizability domain: 1D case

Teddy Pichard

doi:10.3934/krm.2020045

Article Contents

2020, Volume 13, Issue 6: 1243-1280. Doi: 10.3934/krm.2020045

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A moment closure based on a projection on the boundary of the realizability domain: 1D case

Teddy Pichard

CMAP, École Polytechnique, CNRS UMR7641, Institut Polytechnique de Paris, Palaiseau, France

Received: November 30, 2019

Revised: July 31, 2020

Early access: September 2020

Published: November 2020

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Abstract

This work aims to develop and test a projection technique for the construction of closing equations of moment systems. One possibility to define such a closure consists in reconstructing an underlying kinetic distribution from a vector of moments, then expressing the closure based on this reconstructed function.

Exploiting the geometry of the realizability domain, i.e. the set of moments of positive distribution function, we decompose any realizable vectors into two parts, one corresponding to the moments of a chosen equilibrium function, and one obtain by a projection onto the boundary of the realizability domain in the direction of equilibrium function. A realizable closure of both of these parts are computed with standard techniques providing a realizable closure for the full system. This technique is tested for the reduction of a radiative transfer equation in slab geometry.

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Mathematics Subject Classification: 35L40, 47B35, 35B09.

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Figure 1. Schematic representation of a ray starting at a point $ \mathbf{V}\in\mathcal{R}_{\mathbf{b}} $ directed by $ -\mathbf{V}_{eq} $ and crossing $ \partial\mathcal{R}_{\mathbf{b}} $ in $ \mathbf{W}(\bar{x}) $

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Figure 2. Moments of order 0 (left) and 1 (right) obtained with $ P_7 $, $ K_7 $, $ \Pi_7 $ and reference solution for the simple beam test case

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Figure 3. Discrete $ l^1 $ (top left), $ l^2 $ (top right) and $ l^\infty $ (bottom) errors on the moment of order 0 compared to a reference solution for the $ P_N $, $ K_N $ and $ \Pi_N $ as a function of $ N $ for the simple beam test case

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Figure 4. Moments of order 0 (left) and 1 (right) obtained with $ P_N $, $ K_N $, $ \Pi_N $ for $ N = 2 $ (first line), $ N = 3 $ (second line), $ N = 6 $ (third line), $ N = 7 $ (fourth line) and reference solution for the double beam test case

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Figure 5. Discrete $ l^1 $ (top left), $ l^2 $ (top right) and $ l^\infty $ (bottom) errors on the moment of order 0 compared to a reference solution for the $ P_N $, $ K_N $ and $ \Pi_N $ as a function of $ N $ for the double beam test case

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Figure 6. Representation of the measures representing the vector $ \mathbf{f}(x = L/2) $ with $ P_N $, $ K_N $ and $ \Pi_N $ models with $ N = 2 $ (top left), $ 3 $ (top right), $ 4 $ (middle left), $ 5 $ (middle right), $ 6 $ (bottom left), $ 7 $ (bottom right) for the double beam test case

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Figure 7. Moments of order 0 (left) and 1 (right) obtained with $ P_N $, $ K_N $, $ \Pi_N $ for $ N = 2 $ (first line), $ N = 3 $ (second line), $ N = 6 $ (third line), $ N = 7 $ (fourth line) and a reference $ P_{24} $ solution for the point source test case

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Figure 8. Discrete $ l^1 $ (top left), $ l^2 $ (top right) and $ l^\infty $ (bottom) errors on the moment of order 0 compared to a reference $ P_{24} $ simulation for the $ P_N $, $ K_N $ and $ \Pi_N $ as a function of $ N $ for the point source test case

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Figure 9. Representation of the measures representing the vector $ \mathbf{f}(x = L/2) $ with $ P_N $, $ K_N $ and $ \Pi_N $ models for $ N = 2 $ (top left), $ 3 $ (top right), $ 4 $ (middle left), $ 5 $ (middle right), $ 6 $ (bottom left), $ 7 $ (bottom right) for the point source test case

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Figure 10. Moments of order 0 (left) and 1 (right) obtained with $ P_N $, $ K_N $, $ \Pi_N $ for $ N = 2 $ (first line), $ N = 3 $ (second line), $ N = 6 $ (third line), $ N = 7 $ (fourth line) and those of the analytical solution for the Riemann problem

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Figure 11. Representation of the measures representing the vector $ \mathbf{f}(x = L/2) $ with $ P_N $, $ K_N $ and $ \Pi_N $ models for $ N = 2 $ (top left), $ 3 $ (top right), $ 4 $ (middle left), $ 5 $ (middle right), $ 6 $ (bottom left), $ 7 $ (bottom right) for the Riemann problem

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