答案1
您可以使用mathtools 包中的aligned
内部功能dcases
。这是图片对齐的方式。
\documentclass[a4paper]{article}
\usepackage{mathtools}
\begin{document}
\begin{equation}
\begin{dcases}\left.
\begin{aligned}
\frac{dG(t)}{dt} &= G_{in} - \sigma_2G - a\left(c+\frac{mI}{n+I}\right)G + b \\
\frac{dI(t)}{dt} &= \frac{\sigma_1G^2}{\alpha_1^2+G^2} - d_iI(t)
\end{aligned} \right \}, \quad t \ne k_\tau, & \\
\left.
\begin{aligned}
G(t^+) &= G(t) \\
I(t^+) &= I(t) + \sigma
\end{aligned} \right \}, \quad t = k_\tau, &
\end{dcases}
\end{equation}
\end{document}
更新:
您还可以将所有等式按=
符号对齐,如下所示:
\documentclass[a4paper]{article}
\usepackage{mathtools,calc}
\newlength{\lhs}
\settowidth{\lhs}{ $\frac{dG(t)}{dt}$ }
\newcommand{\lhsbox}[1]{\makebox[\lhs][r]{$\displaystyle#1$}}
\begin{document}
\begin{equation}
\begin{dcases}\left.
\begin{aligned}
\lhsbox{\frac{dG(t)}{dt}} &= G_{in} - \sigma_2G - a\left(c+\frac{mI}{n+I}\right)G + b \\
\lhsbox{\frac{dI(t)}{dt}} &= \frac{\sigma_1G^2}{\alpha_1^2+G^2} - d_iI(t)
\end{aligned} \right \}, \quad t \ne k_\tau, & \\
\left.
\begin{aligned}
\lhsbox{G(t^+)} &= G(t) \\
\lhsbox{I(t^+)} &= I(t) + \sigma
\end{aligned} \right \}, \quad t = k_\tau, &
\end{dcases}
\end{equation}
\end{document}
答案2
这是一个采用嵌套array
环境的解决方案。
\documentclass{article}
\usepackage{array,booktabs}
\newcolumntype{L}{>{\displaystyle}l}
\begin{document}
\[
\renewcommand\arraystretch{1.33}
\setlength\arraycolsep{0pt}
\left\{
\begin{array}{L}
\left.
\begin{array}{L}
a = ldfjdl;kfsja ;kfja;ksdljfa\\
b = asjfl;a ;adjfl;asj ;adjfals
\end{array}
\right\}
,\quad t\ne kr\,. \\ \addlinespace
\left.
\begin{array}{L}
c = jfdals;jkf\\
d = dsalfjasl;fja
\end{array}
\right\}
,\quad t=kr\,.
\end{array}
\right.
\]
\结束{文档}
答案3
以下用途eqparbox
确保\eqmakebox[<tag>][<align>]{<stuff>}
所有<stuff>
相同的都<tag>
设置在具有最大宽度的框中,并进行适当的<align>
调整。这会自动执行对齐等式左侧的过程:
\documentclass{article}
\usepackage{mathtools,eqparbox}
\begin{document}
\begin{equation}
\begin{dcases}
\left. \begin{aligned}
& \eqmakebox[LHS][r]{$\dfrac{\mathrm{d}G(t)}{\mathrm{d}t}$}
= G_{in} - \sigma_2 G - a \left(c + \frac{mI}{n + I} \right) G + b \\
& \eqmakebox[LHS][r]{$\dfrac{\mathrm{d}I(t)}{\mathrm{d}t}$}
= \frac{\sigma_1 G^2}{\alpha_1^2 + G^2} - d_i I(t)
\end{aligned} \right \} \quad t \neq k_\tau, \\
\left. \begin{aligned}
& \eqmakebox[LHS][r]{$G(t^+)$} = G(t) \\
& \eqmakebox[LHS][r]{$I(t^+)$} = I(t) + \sigma
\end{aligned} \right \} \quad t = k_\tau
\end{dcases}
\end{equation}
\end{document}