From b3e1a8e379b138540773ef976287d3ff6197d238 Mon Sep 17 00:00:00 2001
From: Yingjie Wang <phywyj@gmail.com>
Date: Thu, 9 Apr 2026 20:49:30 -0400
Subject: [PATCH] update: auto commit

---
 report.tex | 18 +++++++++---------
 1 file changed, 9 insertions(+), 9 deletions(-)

diff --git a/report.tex b/report.tex
index 786cd91..1c1caff 100644
--- a/report.tex
+++ b/report.tex
@@ -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.
 
     The eigenvalues are $\lambda_1 = \lambda_2 = \lambda_3 = 0$, $\lambda_4 = -1$ and $\lambda_5 = 1$. The eigenvectors are
-      \begin{equation}
-        \begin{split}
-          \myvec{e}_1 = \mqty(1,0,0,0,0)^T,\\
-          \myvec{e}_2 = \mqty(0,0,-\xi_3,0,\xi_1)^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}_5 = \mqty(0,-1,\xi_1,\xi_2,\xi_3)^T.
-        \end{split}
-      \end{equation}
+    \begin{equation}
+      \begin{gathered}
+        \myvec{e}_1 = \mqty(1,0,0,0,0)^T \qc
+        \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}_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.
+      \end{gathered}
+    \end{equation}
     
   \end{frame}