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A High-Order Discontinuous Galerkin Method for One-Fluid Two-Temperature Euler Non-equilibrium Hydrodynamics

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Abstract

In this work, we present a high-order discontinuous Galerkin (DG) method for solving the one-fluid two-temperature Euler equations for non-equilibrium hydrodynamics. In order to achieve optimal order of accuracy as well as suppress potential numerical oscillations behind strong shocks, special jump terms are applied in the DG spatial discretization for the nonconservative equation of electronic internal energy. Moreover, inspired by the solution procedure of Riemann problem, we develop a new HLLC (Harten–Lax–van Leer Contact) approximate Riemann solver for the one-fluid two-temperature Euler equations and use it as a building block for the high-order discontinuous Galerkin method. Several key features of the proposed HLLC approximate Riemann solver are analyzed. Finally, we design typical test cases to numerically verify and demonstrate the performance of the proposed method.

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Data Availability

Data sets generated during the current study are available from the corresponding author on reasonable request.

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Acknowledgements

The author would like to thank Zhifang Du for helpful discussion. This work is supported by the National Natural Science Foundation of China No.12171046 and No. 12031001.

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  1. Institute of Applied Physics and Computational Mathematics, Beijing, 100094, People’s Republic of China

    Jian Cheng

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  1. Jian Cheng
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Appendix A

Appendix A

In the appendix, we give the eigenvalues and eigenvectors for the matrixes \(\textbf{A}_x(\textbf{U})\) and \(\textbf{A}_y(\textbf{U})\), which are obtained by rewritten (16) into a quasi-linear form as follows

$$\begin{aligned} \frac{\partial \textbf{U}}{\partial t} + \textbf{A}_x(\textbf{U})\frac{\partial \textbf{U}}{\partial x} + \textbf{A}_y(\textbf{U})\frac{\partial \textbf{U}}{\partial y}=0. \end{aligned}$$
(68)

The five eigenvalues of \(\textbf{A}_x(\textbf{U})\) are

$$\begin{aligned} \lambda _1 = u+c, \quad \lambda _2=\lambda _3=\lambda _4=u,\quad \lambda _5 = u-c. \end{aligned}$$
(69)

The corresponding right eigenvectors of \(\textbf{A}_x(\textbf{U})\) are given as

$$\begin{aligned} \mathbf {R_1}= \begin{pmatrix} 1 \\ u+c\\ v\\ H+uc\\ \gamma _\textrm{e} e_\textrm{e}\\ \end{pmatrix}, \mathbf {R_2}= \begin{pmatrix} 1 \\ u \\ v \\ \frac{q^2}{2} \\ 0\\ \end{pmatrix}, \mathbf {R_3}= \begin{pmatrix} 0\\ 0\\ 1\\ v\\ 0\\ \end{pmatrix}, \mathbf {R_4}= \begin{pmatrix} 0\\ 0\\ 0\\ \gamma _\textrm{i}-\gamma _\textrm{e}\\ \gamma _\textrm{i}-1\\ \end{pmatrix}, \mathbf {R_5}= \begin{pmatrix} 1 \\ u-c \\ v \\ H-uc\\ \gamma _\textrm{e} e_\textrm{e} \end{pmatrix},\nonumber \\ \end{aligned}$$
(70)

where \(H = \frac{E+p}{\rho }\) and \(q^2=u^2+v^2\). The left eigenvectors of \(\textbf{A}_x(\textbf{U})\) are given as

$$\begin{aligned} \mathbf {L_1}= & {} \begin{pmatrix} \frac{(\gamma _\textrm{i}-1)q^2-2uc}{4c^2}, &{} \frac{-(\gamma _\textrm{i}-1)u+c}{2c^2}, &{} \frac{-(\gamma _\textrm{i}-1)v}{2c^2}, &{} \frac{(\gamma _\textrm{i}-1)}{2c^2}, &{} \frac{(\gamma _\textrm{e}-\gamma _\textrm{i})}{2c^2} \\ \end{pmatrix},\nonumber \\ \mathbf {L_2}= & {} \begin{pmatrix} \frac{-(\gamma _\textrm{i}-1)q^2+2c^2}{2c^2}, &{} \frac{(\gamma _\textrm{i}-1)u}{c^2}, &{} \frac{(\gamma _\textrm{i}-1)v}{c^2}, &{} \frac{-(\gamma _\textrm{i}-1)}{c^2}, &{} \frac{(\gamma _\textrm{i}-\gamma _\textrm{e})}{c^2} \\ \end{pmatrix},\nonumber \\ \mathbf {L_3}= & {} \begin{pmatrix} -v, &{} 0, &{} 1, &{} 0, &{} 0\\ \end{pmatrix},\nonumber \\ \mathbf {L_4}= & {} \begin{pmatrix} \frac{-\gamma _\textrm{e}e_\textrm{e}q^2}{2c^2}, &{} \frac{\gamma _\textrm{e}e_\textrm{e}u}{c^2}, &{} \frac{\gamma _\textrm{e}e_\textrm{e}v}{c^2}, &{} \frac{-\gamma _\textrm{e}e_\textrm{e}}{c^2}, &{} \frac{\gamma _\textrm{e}e_\textrm{e}+\gamma _\textrm{i}e_\textrm{i}}{c^2} \\ \end{pmatrix},\nonumber \\ \mathbf {L_5}= & {} \begin{pmatrix} \frac{(\gamma _\textrm{i}-1)q^2+2uc}{4c^2}, &{} \frac{-(\gamma _\textrm{i}-1)u-c}{2c^2}, &{} \frac{-(\gamma _\textrm{i}-1)v}{2c^2}, &{} \frac{(\gamma _\textrm{i}-1)}{2c^2}, &{} \frac{(\gamma _\textrm{e}-\gamma _\textrm{i})}{2c^2} \\ \end{pmatrix}. \end{aligned}$$
(71)

Similarly, the five eigenvalues of \(\textbf{A}_y(\textbf{U})\) are

$$\begin{aligned} \lambda _1 = v+c, \quad \lambda _2=\lambda _3=\lambda _4=u,\quad \lambda _5 = v-c. \end{aligned}$$
(72)

The corresponding right eigenvectors of \(\textbf{A}_y(\textbf{U})\) are given as

$$\begin{aligned} \mathbf {R_1}= \begin{pmatrix} 1 \\ u\\ v+c\\ H+vc\\ \gamma _\textrm{e} e_\textrm{e}\\ \end{pmatrix}, \mathbf {R_2}= \begin{pmatrix} 1 \\ u \\ v \\ \frac{q^2}{2} \\ 0\\ \end{pmatrix}, \mathbf {R_3}= \begin{pmatrix} 0\\ 1\\ 0\\ u\\ 0\\ \end{pmatrix}, \mathbf {R_4}= \begin{pmatrix} 0\\ 0\\ 0\\ \gamma _\textrm{i}-\gamma _\textrm{e}\\ \gamma _\textrm{i}-1\\ \end{pmatrix}, \mathbf {R_5}= \begin{pmatrix} 1 \\ u \\ v-c \\ H-vc\\ \gamma _\textrm{e} e_\textrm{e} \end{pmatrix},\nonumber \\ \end{aligned}$$
(73)

and the left eigenvectors of \(\textbf{A}_y(\textbf{U})\) are given as

$$\begin{aligned} \begin{aligned} \mathbf {L_1}&= \begin{pmatrix} \frac{(\gamma _\textrm{i}-1)q^2-2vc}{4c^2}, &{} \frac{-(\gamma _\textrm{i}-1)u}{2c^2}, &{} \frac{-(\gamma _\textrm{i}-1)v+c}{2c^2}, &{} \frac{(\gamma _\textrm{i}-1)}{2c^2}, &{} \frac{(\gamma _\textrm{e}-\gamma _\textrm{i})}{2c^2} \\ \end{pmatrix},\\ \mathbf {L_2}&= \begin{pmatrix} \frac{-(\gamma _\textrm{i}-1)q^2+2c^2}{2c^2}, &{} \frac{(\gamma _\textrm{i}-1)u}{c^2}, &{} \frac{(\gamma _\textrm{i}-1)v}{c^2}, &{} \frac{-(\gamma _\textrm{i}-1)}{c^2}, &{} \frac{(\gamma _\textrm{i}-\gamma _\textrm{e})}{c^2} \\ \end{pmatrix},\\ \mathbf {L_3}&= \begin{pmatrix} -u, &{} 1, &{} 0, &{} 0, &{} 0\\ \end{pmatrix},\\ \mathbf {L_4}&= \begin{pmatrix} \frac{-\gamma _\textrm{e}e_\textrm{e}q^2}{2c^2}, &{} \frac{\gamma _\textrm{e}e_\textrm{e}u}{c^2}, &{} \frac{\gamma _\textrm{e}e_\textrm{e}v}{c^2}, &{} \frac{-\gamma _\textrm{e}e_\textrm{e}}{c^2}, &{} \frac{\gamma _\textrm{e}e_\textrm{e}+\gamma _\textrm{i}e_\textrm{i}}{c^2} \\ \end{pmatrix},\\ \mathbf {L_5}&= \begin{pmatrix} \frac{(\gamma _\textrm{i}-1)q^2+2vc}{4c^2}, &{} \frac{-(\gamma _\textrm{i}-1)u}{2c^2}, &{} \frac{-(\gamma _\textrm{i}-1)v-c}{2c^2}, &{} \frac{(\gamma _\textrm{i}-1)}{2c^2}, &{} \frac{(\gamma _\textrm{e}-\gamma _\textrm{i})}{2c^2} \\ \end{pmatrix}. \end{aligned} \end{aligned}$$
(74)

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Cheng, J. A High-Order Discontinuous Galerkin Method for One-Fluid Two-Temperature Euler Non-equilibrium Hydrodynamics. J Sci Comput 100, 82 (2024). https://doi.org/10.1007/s10915-024-02640-z

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