答案1
使用aligned
具有单个可扩展括号的环境:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation}
\left.
\begin{aligned}
\nabla \times \vec{\mathbf{E}}(\vec{\mathbf{r}}) &=
-j \omega \mu_0 \vec{\mathbf{H}}(\vec{\mathbf{r}}) \\
\nabla \times \vec{\mathbf{H}}(\vec{\mathbf{r}}) &=
j \omega \epsilon \vec{\mathbf{H}}(\vec{\mathbf{r}})
\end{aligned}
\right\}
\longrightarrow
\begin{aligned}
\nabla^2 \vec{\mathbf{E}}(\vec{\mathbf{r}}) &=
-\omega^2 \mu_0 \epsilon \vec{\mathbf{E}}(\vec{\mathbf{r}}) \\
\nabla^2 \vec{\mathbf{H}}(\vec{\mathbf{r}}) &=
\omega^2 \mu_0 \epsilon \vec{\mathbf{H}}(\vec{\mathbf{r}})
\end{aligned}
\end{equation}
\end{document}
答案2
环境的另一种可能性rcases
来自mathtools
:
\documentclass{article}
\usepackage{mathtools}
\begin{document}
\begin{equation}
\begin{rcases}
\nabla\times\vec{\mathbf{E}}(\vec{\mathbf{r}}) = -j \omega\mu_0 \vec{\mathbf{H}}(\vec{\mathbf{r}}) \\
\nabla\times\vec{\mathbf{H}}(\vec{\mathbf{r}}) = j \omega\epsilon \vec{\mathbf{H}}(\vec{\mathbf{r}})
\end{rcases}
\longrightarrow
\begin{aligned}
\nabla^2 \vec{\mathbf{E}}(\vec{\mathbf{r}}) &= -\omega^2\mu_0\epsilon \vec{\mathbf{E}}(\vec{\mathbf{r}}) \\
\nabla^2 \vec{\mathbf{H}}(\vec{\mathbf{r}}) &= \omega^2\mu_0\epsilon \vec{\mathbf{H}}(\vec{\mathbf{r}})
\end{aligned}
\end{equation}
\end{document}