The Neumann numerical boundary condition for transport equations

Jean-François Coulombel; Frédéric Lagoutière

doi:10.3934/krm.2020001

Article Contents

2020, Volume 13, Issue 1: 1-32. Doi: 10.3934/krm.2020001

This issue Previous Article Next Article A kinetic approach of the bi-temperature Euler model

The Neumann numerical boundary condition for transport equations

Jean-François Coulombel^1, and
Frédéric Lagoutière^2, ,

1.
Institut de Mathématiques de Toulouse; UMR5219, Université de Toulouse; CNRS, Université Paul Sabatier, F-31062 Toulouse Cedex 9, France
2.
Université de Lyon, Université Claude Bernard Lyon 1, Institut Camille Jordan (CNRS UMR5208), 43 boulevard du 11 novembre 1918, F-69622 Villeurbanne Cedex, France

^* Corresponding author: Frédéric Lagoutière
^* Corresponding author: Frédéric Lagoutière

Received: October 2018

Revised: June 2019

Published: February 2020

Both authors are supported by the ANR project BoND, ANR-13-BS01-0009, and by the ANR project NABUCO, ANR-17-CE40-0025

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Abstract

In this article, we show that prescribing homogeneous Neumann type numerical boundary conditions at an outflow boundary yields a convergent discretization in $ \ell^\infty $ for transport equations. We show in particular that the Neumann numerical boundary condition is a stable, local, and absorbing numerical boundary condition for discretized transport equations. Our main result is proved for explicit two time level numerical approximations of transport operators with arbitrarily wide stencils. The proof is based on the energy method and bypasses any normal mode analysis.

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Mathematics Subject Classification: Primary: 65M12, 65M06, 65M20.

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Figure 1. The mesh on $ \mathbb{R}^+ \times (0,L) $ in blue, and the "ghost cells" in red ($ r = p = 2 $ here)

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Figure 2. Top: updating iteratively the ghost values at the outflow boundary ($ r = p = k_b = 2 $). Bottom: updating the numerical approximation in the interior

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Figure 3. Initial condition $ u^{0,2} $ with $ 40 $ cells in $ (0, 1) $

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Figure 4. Numerical and exact solutions at time $ t = 0.2625 $ with initial condition $ u^{0,2} $ ($ 40 $ cells)

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Figure 5. Numerical and exact solutions at time $ t = 0.5075 $ with initial condition $ u^{0,2} $ ($ 40 $ cells)

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Table 1. Error for datum $ u^{0,1} $. The result presented in this paper guarantees that the order is at least $ 3/2 $ when $ k_b = 2 $, and at least $ 1/2 $ when $ k_b = 1 $. The "order" that is indicated in bracket is a "local" order computed, for each line $ k $, as $ -\log(e_k/e_{k-1})/\log(2) $

Number of cells $ J $	Measured error with $ k_b = 2 $	Measured error with $ k_b = 1 $
10	$ 2.53 \times 10^{-3} $	$ 8.34 \times 10^{-3} $
20	$ 8.28 \times 10^{-4} $ (order 1.61)	$ 4.92 \times 10^{-3} $ (order 0.76)
40	$ 2.31 \times 10^{-4} $ (order 1.84)	$ 2.63 \times 10^{-3} $ (order 0.90)
80	$ 6.09 \times 10^{-5} $ (order 1.93)	$ 1.40 \times 10^{-3} $ (order 0.91)
160	$ 1.56 \times 10{-5} $ order 1.96)	$ 7.21 \times 10^{-4} $ (order 0.96)
320	$ 3.97 \times 10^{-6} $ (order 1.97)	$ 3.66 \times 10^{-4} $ (order 0.98)
640	$ 1.00 \times 10^{-6} $ (order 1.99)	$ 1.84 \times 10^{-4} $ (order 0.99)
1280	$ 2.52 \times 10^{-7} $ (order 1.99)	$ 9.24 \times 10^{-5} $ (order 0.99)

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Table 2. Error for datum $ u^{0,2} $. The result presented in this paper guarantees that the order is at least $ 3/2 $ when $ k_b = 2 $, and at least $ 1/2 $ when $ k_b = 1 $. Concerning the order, see Table 1

Number of cells $ J $	Measured error with $ k_b = 2 $	Measured error with $ k_b = 1 $
10	$ 2.80 \times 10^{-3} $	$ 1.03 \times 10^{-2} $
20	$ 8.25 \times 10^{-4} $ (order 1.76)	$ 5.78 \times 10^{-3} $ (order 0.83)
40	$ 2.53 \times 10^{-4} $ (order 1.71)	$ 3.09 \times 10^{-3} $ (order 0.91)
80	$ 7.81 \times 10^{-5} $ (order 1.69)	$ 1.62 \times 10^{-3} $ (order 0.93)
160	$ 2.36 \times 10^{-5} $ (order 1.73)	$ 8.29 \times 10^{-4} $ (order 0.97)
320	$ 7.11 \times 10^{-6} $ (order 1.73)	$ 4.19 \times 10^{-4} $ (order 0.98)
640	$ 2.14 \times 10^{-6} $ (order 1.74)	$ 2.11 \times 10^{-4} $ (order 0.99)
1280	$ 6.43 \times 10^{-7} $ (order 1.73)	$ 1.06 \times 10^{-4} $ (order 0.99)

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Table 3. Error for datum $ u^{0,3} $. The result presented in this paper does not apply to the case $ k_b = 2 $, and guarantees that the convergence order is at least at least $ 1/2 $ when $ k_b = 1 $. Concerning the order, see Table 1

Number of cells $ J $	Measured error with $ k_b = 2 $	Measured error with $ k_b = 1 $
10	$ 2.84 \times 10^{-3} $	$ 1.08 \times 10^{-2} $
20	$ 9.18 \times 10^{-4} $ (order 1.63)	$ 6.00 \times 10^{-3} $ (order 0.85)
40	$ 3.02 \times 10^{-4} $ (order 1.60)	$ 3.20 \times 10^{-3} $ (order 0.91)
80	$ 9.76 \times 10^{-5} $ (order 1.63)	$ 1.67 \times 10^{-3} $ (order 0.93)
160	$ 3.08 \times 10^{-5} $ (order 1.66)	$ 8.56 \times 10^{-4} $ (order 0.97)
320	$ 9.72 \times 10^{-6} $ (order 1.66)	$ 4.32 \times 10^{-4} $ (order 0.98)
640	$ 3.08 \times 10^{-6} $ (order 1.66)	$ 2.17 \times 10^{-4} $ (order 0.99)
1280	$ 9.71 \times 10^{-7} $ (order 1.67)	$ 1.09 \times 10^{-4} $ (order 0.99)

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Table 4. Spectral radii and $ l^2 $ induced norms of the linear operator associated with the scheme

$ J $	Spectral radius, $ k_b = 1 $	$ l^2 $ norm, $ k_b = 1 $	Spectral radius, $ k_b = 2 $	$ l^2 $ norm, $ k_b = 2 $
20	0.7100	0.9999	0.7098	1.0035
80	0.74300	0.9999	0.7513	1.0035
320	0.9208	0.9999	0.9212	1.0035
1280	0.9817	0.9999	0.9805	1.0035

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