编写 CASES 括号的左侧

Question

我想在案例括号的左侧写一些内容，如图所示。我使用的代码：

\begin{equation}
\begin{split}
& E(T,\alpha) = \\
&
    \begin{cases}
    1 & \text{$\alpha < \alpha_{vi}$} \\
    \textrm{$E_0(T)$} & \textrm{$\alpha$ = $\alpha_{vi}$}\\
    \text{$E_0(T) + (\alpha-\alpha_{vi})(E_\infty(T)-E_{0}(T))$} & \text{$\alpha_{vi} < \alpha < 1$}\\
    \text{$E_\infty(T)$} & \text{$\alpha = 1$}
    \end{cases}
\end{split}
\label{eq:chile}
\end{equation}

我想添加红色字母：

Answer 1

一种可能性是：

\documentclass{article}
\usepackage{amsmath}

\begin{document}
\begin{equation}
    \begin{split}
       & E(T,\alpha) = \\
       & 
\begin{matrix}
  \textsf{I}    \\[0.5ex]
  \textsf{II}   \\[0.5ex]
  \textsf{III}  \\[0.5ex]
  \textsf{IV}
\end{matrix}\quad
\begin{cases}
    1 & \text{$\alpha < \alpha_{vi}$} \\
    \textrm{$E_0(T)$} & \textrm{$\alpha$ = $\alpha_{vi}$}\\
    \text{$E_0(T) + (\alpha-\alpha_{vi})(E_\infty(T)-E_{0}(T))$} & \text{$\alpha_{vi} < \alpha < 1$}\\
    \text{$E_\infty(T)$} & \text{$\alpha = 1$}
\end{cases}
    \end{split}
\end{equation}
\end{document}

Answer 2

{NiceArray}的可能性是nicematrix。即使更改数组的高度、深度或行，输出仍将是正确的。

 \documentclass{article}
 \usepackage{nicematrix}

 \begin{document}

 \begin{equation}
 \begin{split}
 & E(T,\alpha) = \\
 &
 \begin{NiceArray}{>{\color{red}}llc}
 \text{I}   & 1 & \alpha < \alpha_{vi} \\
 \text{II}  & E_0(T) & \alpha$ = $\alpha_{vi}\\
 \text{III} & E_0(T) + (\alpha-\alpha_{vi})(E_\infty(T)-E_{0}(T)) & \alpha_{vi} < \alpha < 1 \\
 \text{IV}  & E_\infty(T) & \alpha = 1
 \CodeAfter
   \SubMatrix{\{}{1-2}{4-2}{.}
 \end{NiceArray}
 \end{split}
 \label{eq:chile}
 \end{equation}

 \end{document} 

您需要进行多次编译（因为nicematrix在后台使用了 PGF/Tikz）。

Answer 3

还有另一种解决方案，无需案例，只需替换为blkarray包bigdelim：

\documentclass{article}
\usepackage{amsmath}
\usepackage{blkarray}
\usepackage{bigdelim}
\usepackage[svgnames]{xcolor}

\begin{document}

\begin{equation}
 \begin{split}
 & E(T,\alpha) = \\
 &
\begin{blockarray}{>{\color{IndianRed}\mathsf{\Roman{BAenumi}}}l@{\;}c@{}ll}
 & \ldelim\{{4}{*}& 1 & \alpha < \alpha_{vi} \\
 & & E_0(T) & \alpha = \alpha_{vi} \\
 & & E_0(T) + (\alpha-\alpha_{vi})(E_\infty(T)-E_{0}(T)) & \alpha_{vi} < \alpha < 1 \\
 & & E_\infty(T) & \alpha = 1
\end{blockarray}
 \end{split}
\end{equation}

\end{document}