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A Spherical Harmonic Discontinuous Galerkin Method for Radiative Transfer Equations with Vacuum Boundary Conditions

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Abstract

In this paper, we propose and analyze a spherical harmonic discontinuous Galerkin (SH-DG) method for solving the radiative transfer equations with vacuum boundary conditions. To incorporate vacuum boundary conditions in spherical harmonic approximations, we first embed the original domain into a larger computational area of rectangular type with an extra pure absorbing layer and then establish a perturbation problem with a periodic condition at the boundary of the extended domain. Since the outflow radiative intensity at the outer boundary of the extended area can be made arbitrarily small by sufficiently increasing the magnitude of absorption in or the thickness of the absorbing layer, such a replacement of the boundary condition only causes a minimal difference between the solution of the perturbation problem and the original problem in the original domain, but will benefit the construction of the discretization scheme. Then based on the analysis of the perturbation problem and the SH-DG method for solving the radiative transfer equation with periodic boundary conditions, the well-posedness and the error estimates are derived for the approximation solution arising from the SH-DG method for solving the radiative transfer equation with vacuum boundary conditions. Numerical examples with both periodic and vacuum boundary conditions are included to validate the theoretical results.

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Acknowledgements

The authors are grateful to the reviewers for carefully reading this paper.

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Authors and Affiliations

  1. Department of Mathematics, California State University, Bakersfield, CA, 93311, USA

    Qiwei Sheng

  2. School of Mathematical Sciences, Tongji University, Shanghai, 200092, People’s Republic of China

    Cheng Wang

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Appendix

Appendix

In this section, the proofs of some technical results are given.

1.1 Proof of Theorem 6

We introduce the operator \(\mathcal {L}: \varvec{W}\rightarrow (L^2(X))^L\) by \(\mathcal {L}\varvec{u}:= \varvec{A}\cdot \nabla \varvec{u} + (\sigma _{\mathrm {a}}\varvec{I} + \sigma _{\mathrm {s}} \varvec{R})\varvec{u}\). We will show that \(\mathcal {L}\) is an isomorphism. The following results will be needed.

Lemma 9

Under conditions (2.8), there exists \(\alpha >0\), such that for all \(\varvec{u}\in \varvec{W}\), \(\Vert \mathcal {L}\varvec{u}\Vert _X\ge \alpha \Vert \varvec{u}\Vert _{\varvec{V}}\).

Proof

Let us denote \(\varvec{n}\cdot \varvec{A}=\sum _{i=1}^{d}n_i\varvec{A}^{(i)}\), where \(\varvec{n}(\mathbf {x})=(n_1,n_2,\cdots ,n_d)^\mathsf {T}\) is the unit outward normal vector at \(\mathbf {x}\in \partial X\). Since \(\varvec{u}\) is periodic, we have \(\int _{\partial X}\varvec{u}^{\mathsf {T}}\varvec{n}\cdot \varvec{A}\varvec{u}\,\mathrm {d}\mathbf {x}=0\). Note that

$$\begin{aligned} (\mathcal {L}\varvec{u},\varvec{u})_X&= \int _X \varvec{u}^{\mathsf {T}} \varvec{A}\cdot \nabla \varvec{u}\,\mathrm {d}\mathbf {x}+ \int _X \varvec{u}^{\mathsf {T}}(\sigma _{\mathrm {a}}\varvec{I} + \sigma _{\mathrm {s}} \varvec{R})\varvec{u}\,\mathrm {d}\mathbf {x}\\&= \frac{1}{2}\int _{\partial X} \varvec{u}^{\mathsf {T}}\varvec{n}\cdot \varvec{A}\varvec{u}\,\mathrm {d}\mathbf {x}+ \int _X \varvec{u}^{\mathsf {T}}(\sigma _{\mathrm {a}}\varvec{I} + \sigma _{\mathrm {s}} \varvec{R})\varvec{u}\,\mathrm {d}\mathbf {x}\\&\ge \sigma _0 \Vert \varvec{u}\Vert _X^2, \end{aligned}$$

i.e., \((\mathcal {L}\varvec{u},\varvec{u})\ge \sigma _0 \Vert \varvec{u}\Vert _X^2\), which implies \(\sup _{\varvec{v}\in (L^2(X))^L\backslash \{\varvec{0}\}}\frac{(\mathcal {L}\varvec{u},\varvec{v})_X }{\Vert \varvec{v}\Vert _X}\ge \sigma _0 \Vert \varvec{u}\Vert _X\). Therefore,

$$\begin{aligned} \sup _{\varvec{v}\in (L^2(X))^L\backslash \{\varvec{0}\}}\frac{(\mathcal {L}\varvec{u},\varvec{v})_X }{\Vert \varvec{v}\Vert _X}&\ge \sup _{\varvec{v}\in (L^2(X))^L\backslash \{\varvec{0}\}}\frac{(\varvec{A}\cdot \nabla \varvec{u},\varvec{v})_X }{\Vert \varvec{v}\Vert _X} - (\sigma _{\mathrm {a}} + \sigma _{\mathrm {s}}) \Vert \varvec{u}\Vert _X \\&\ge \Vert \varvec{A}\cdot \nabla \varvec{u}\Vert _X - \frac{\sigma _{\mathrm {a}} + \sigma _{\mathrm {s}}}{\sigma _0}\sup _{\varvec{v}\in (L^2(X))^L\backslash \{\varvec{0}\}}\frac{(\mathcal {L}\varvec{u},\varvec{v})_X }{\Vert \varvec{v}\Vert _X}. \end{aligned}$$

Hence,

$$\begin{aligned} \left( 1 + \frac{1+\sigma _{\mathrm {a}} + \sigma _{\mathrm {s}}}{\sigma _0}\right) \sup _{\varvec{v}\in (L^2(X))^L\backslash \{\varvec{0}\}}\frac{(\mathcal {L}\varvec{u},\varvec{v})_X }{\Vert \varvec{v}\Vert _X} \ge \Vert \varvec{A}\cdot \nabla \varvec{u}\Vert _X + \Vert \varvec{u}\Vert _X = \Vert \varvec{u}\Vert _{\varvec{V}}, \end{aligned}$$

from which the claim follows. \(\square \)

Lemma 10

Under the condition (2.8), for all \(\varvec{v}\in (L^2(X))^L\), if \((\mathcal {L}\varvec{u},\varvec{v})=0\) for any \(\varvec{u}\in \varvec{W}\), then \(\varvec{v}=0\).

Proof

Let \(\varvec{v}\in (L^2(X))^L\) be such that, for all \(\varvec{u}\in \varvec{W}\), \((\mathcal {L}\varvec{u},\varvec{v})_X=0\). Therefore, we have

$$\begin{aligned} (\varvec{u},-\varvec{A}\cdot \nabla \varvec{v} + (\sigma _{\mathrm {a}} + \sigma _{\mathrm {s}} \varvec{R})\varvec{v})_X + (\varvec{u},\varvec{n}\cdot \varvec{A}\varvec{v})_{\partial X}=(\mathcal {L}\varvec{u},\varvec{v})_X=0, \quad \forall \varvec{u}\in \varvec{W}, \end{aligned}$$

where \(\varvec{A}\cdot \nabla \varvec{v}\) is understood in the distribution sense. As a result,

$$\begin{aligned} -\varvec{A}\cdot \nabla \varvec{v} + (\sigma _{\mathrm {a}} + \sigma _{\mathrm {s}} \varvec{R})\varvec{v}=0, \end{aligned}$$
(7.1)

i.e., \(\varvec{v}\in \varvec{V}\). Furthermore, \((\varvec{u},\varvec{n}\cdot \varvec{A}\varvec{v})_{\partial X}=0\) for all \(\varvec{u}\in \varvec{W}\). Let \(\breve{\mathbf {k}}\) be a vector with one component equal to either 1 or \(-1\) and all other two equal to zero. Define \(\varGamma _1=\{\mathbf {x}\in \partial X: \mathbf {x}+\breve{\mathbf {k}}\in \partial X\}\) and \(\varGamma _2=\partial X\backslash \varGamma _1\). Due to the periodic condition, the following decompositions hold: \(\varvec{v}=\varvec{u}_1 + \varvec{v_1}\), where \(\varvec{u}_1\in \varvec{W}\) and \(\varvec{u}_1=\varvec{v}\) on \(\varGamma _1\); \(\varvec{v}=\varvec{u}_2 + \varvec{v_2}\), where \(\varvec{u}_2\in \varvec{W}\) and \(\varvec{u}_2=\varvec{v}\) on \(\varGamma _2\). Therefore, on \(\partial X\),

$$\begin{aligned} (\varvec{v},\varvec{n}\cdot \varvec{A}\varvec{v})_{\partial X}&= \left( \varvec{u}_1 + \varvec{v_1}, \varvec{n}\cdot \varvec{A}(\varvec{u}_2 + \varvec{v_2})\right) _{\partial X} \\&= (\varvec{u}_1,\varvec{n}\cdot \varvec{A}\varvec{u}_2)_{\partial X} + (\varvec{u}_1,\varvec{n}\cdot \varvec{A}\varvec{v}_2)_{\partial X} +(\varvec{v}_1,\varvec{n}\cdot \varvec{A}\varvec{u}_2)_{\partial X} + (\varvec{v}_1,\varvec{n}\cdot \varvec{A}\varvec{v}_2)_{\partial X} \\&= (\varvec{v}_1,\varvec{n}\cdot \varvec{A}\varvec{v}_2)_{\varGamma _1} + (\varvec{v}_1,\varvec{n}\cdot \varvec{A}\varvec{v}_2)_{\varGamma _2}\\&= 0, \end{aligned}$$

where the last equality is due to the fact that \(\varvec{v}_1=0\) on \(\varGamma _1\) and \(\varvec{v}_2=0\) on \(\varGamma _2\). Taking the inner product of (7.1) with \(\varvec{v}\) and owing to the boundary condition satisfied by \(\varvec{v}\), we have

$$\begin{aligned} 0 = (\varvec{v},-\varvec{A}\cdot \nabla \varvec{v} + (\sigma _{\mathrm {a}} + \sigma _{\mathrm {s}} \varvec{R})\varvec{v})_X = (\mathcal {L}\varvec{v},\varvec{v})_X - (\varvec{v},\varvec{n}\cdot \varvec{A}\varvec{v})_{\partial X} = (\mathcal {L}\varvec{v},\varvec{v})_X \ge \sigma _0 \Vert \varvec{v}\Vert _X, \end{aligned}$$

i.e., \(\varvec{v}=0\). \(\square \)

With these two lemmas, we are ready to prove Theorem 6.

Proof of Theorem 6

By Lemmas 9 and 10, and employing the Banach-Nečas-Babuška (BNB) Theorem (cf. Theorem 1.1 in [17] and Theorem 2.6 in [23]), the operator \(\mathcal {L}\) is an isomorphism from \(\varvec{W}\) to \((L^2(X))^L\), from which we conclude that (4.12) admits a unique solution \(\varvec{u}\in \varvec{W}\). \(\square \)

Remark 7

The injectivity of \(\mathcal {L}\) is a direct consequence of Lemmas 9. From Lemma 10, we can infer that \(\mathcal {L}\) is surjective. Therefore \(\mathcal {L}\) is bijective.

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Sheng, Q., Wang, C. A Spherical Harmonic Discontinuous Galerkin Method for Radiative Transfer Equations with Vacuum Boundary Conditions. J Sci Comput 88, 9 (2021). https://doi.org/10.1007/s10915-021-01530-y

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