From c84562a69166a5880f97fc165133a646fd58f382 Mon Sep 17 00:00:00 2001 From: Yingjie Wang Date: Fri, 10 Apr 2026 03:02:21 -0400 Subject: [PATCH] update: auto commit --- report.tex | 11 ++++++++++- 1 file changed, 10 insertions(+), 1 deletion(-) diff --git a/report.tex b/report.tex index 0a66eac..87953b1 100644 --- a/report.tex +++ b/report.tex @@ -907,7 +907,16 @@ $} Then what about adding all possible terms with undetermined coefficients, and solve the coefficients by the conditions of hyperbolicity? - Yes, I thought about it. We need a function that takes the principal symbol matrix as input, and outputs the conditions on the coefficients for the principal symbol matrix to be diagonalizable with real eigenvalues. I do wrote such a function, but it cannot be simple. When apply to our case, it will run forever. + Yes, I thought about it. We need a function that takes the principal symbol matrix with lots of undetermined coefficients as input, and outputs the conditions on the coefficients for the principal symbol matrix to be diagonalizable with real eigenvalues. I do wrote such a function, but it cannot be simple. When apply to our case, it will run forever. + \end{frame} + + \begin{frame}{Appendix}{Why can't we fully automate the workflow using Mathematica?} + More details: we cannot just compute the eigenvalues and eigenvectors, then check if the eigenvectors form a complete basis. Think about this simple matrix: + \begin{equation*} + \mqty(1 & a \\ 0 & 1), + \end{equation*} + if we ask Mathematica to compute the eigenvectors, it will give us $\myvec{v}_1 = \mqty(1 \\ 0)$ and $\myvec{v}_2 = \mqty(0 \\ 0)$, because unless $a=0$, this matrix have a Jordan block. So the output of Mathematica is correct except on a zero-measure set. If we simply check if the eigenvectors are linearly independent, we will miss the condition $a=0$ and get false. + \end{frame} \end{document} \ No newline at end of file