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Yingjie Wang
2026-04-09 20:49:30 -04:00
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@@ -289,15 +289,15 @@
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} \begin{equation}
\begin{split} \begin{gathered}
\myvec{e}_1 = \mqty(1,0,0,0,0)^T,\\ \myvec{e}_1 = \mqty(1,0,0,0,0)^T \qc
\myvec{e}_2 = \mqty(0,0,-\xi_3,0,\xi_1)^T,\\ \myvec{e}_2 = \mqty(0,0,-\xi_3,0,\xi_1)^T \qc
\myvec{e}_3 = \mqty(0,0,-\xi_2,\xi_1,0)^T,\\ \myvec{e}_3 = \mqty(0,0,-\xi_2,\xi_1,0)^T,\\
\myvec{e}_4 = \mqty(0,1,\xi_1,\xi_2,\xi_3)^T,\\ \myvec{e}_4 = \mqty(0,1,\xi_1,\xi_2,\xi_3)^T \qc
\myvec{e}_5 = \mqty(0,-1,\xi_1,\xi_2,\xi_3)^T. \myvec{e}_5 = \mqty(0,-1,\xi_1,\xi_2,\xi_3)^T.
\end{split} \end{gathered}
\end{equation} \end{equation}
\end{frame} \end{frame}