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Yingjie Wang
2026-04-09 20:48:20 -04:00
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it's very clear that the principal symbol matrix is already symmetric, and thus the system is symmetrically hyperbolic, and well-posed in the $L^2$ sense. it's very clear that the principal symbol matrix is already symmetric, and thus the system is symmetrically hyperbolic, and well-posed in the $L^2$ sense.
The eigenvalues are $\lambda_1 = \lambda_2 = \lambda_3 = 0$, $\lambda_4 = -1$ and $\lambda_5 = 1$. The eigenvectors are The eigenvalues are $\lambda_1 = \lambda_2 = \lambda_3 = 0$, $\lambda_4 = -1$ and $\lambda_5 = 1$. The eigenvectors are
\begin{equation} {
\myvec{e}_1 = \mqty(1\\0\\0\\0\\0), % \fontsize{small}
\myvec{e}_2 = \mqty(0\\0\\-\xi_3\\0\\\xi_1), \begin{equation}
\myvec{e}_3 = \mqty(0\\0\\-\xi_2\\\xi_1\\0), \myvec{e}_1 = \mqty(1,0,0,0,0)^T,
\myvec{e}_4 = \mqty(0\\1\\\xi_1\\\xi_2\\\xi_3), \myvec{e}_2 = \mqty(0,0,-\xi_3,0,\xi_1)^T,
\myvec{e}_5 = \mqty(0\\-1\\\xi_1\\\xi_2\\\xi_3). \myvec{e}_3 = \mqty(0,0,-\xi_2,\xi_1,0)^T,
\end{equation} \myvec{e}_4 = \mqty(0,1,\xi_1,\xi_2,\xi_3)^T,
\myvec{e}_5 = \mqty(0,-1,\xi_1,\xi_2,\xi_3)^T.
\end{equation}
}
\end{frame} \end{frame}
\begin{frame}{Existing first order formulations} \begin{frame}{Existing first order formulations}