带括号的子方程和对齐方程组

Question

我正在寻找在子方程环境中对齐方程组的最佳方法，左侧带有括号，并带有引言。这是我目前所拥有的：

\documentclass{article}
\usepackage{amssymb}  
\usepackage{amsmath}
\usepackage{empheq}

\begin{document}

\begin{subequations}\label{pb:init} 
  \begin{empheq}[left=\empheqlbrace]{gather}
  \textnormal{find $u_1 : \Omega_1 \to \mathbb{R}$ and $u_2 : \Omega_2 \to \mathbb{R}$ such that}, \nonumber\\
  \begin{gathered}\label{eq:initOmega1}
    \begin{array}{rclll}
      -\mu_1 \Delta u_1 & = & f_1 & \text{ in } & \Omega_1, \\
      u_1 & = & 0 & \text{ on } & \Gamma_1,
    \end{array}
  \end{gathered} \\
  \begin{gathered}\label{eq:initOmega2}
    \begin{array}{rclll}
      -\mu_2 \Delta u_2 & = & f_2 & \text{ in } & \Omega_2, \\
      u_2 & = & 0 & \text{ on } & \Gamma_2,
    \end{array}
  \end{gathered}\\
  \begin{gathered}\label{eq:initCouplage}
    \begin{array}{rclll}
      u_1 & = & u_2 & \text{ on } & \Gamma, \\
      \mu_1\nabla u_1 \cdot n_1 & = & -  \mu_2 \nabla u_2 \cdot n_2 & \text{ on } & \Gamma,
    \end{array}
  \end{gathered}
  \end{empheq}
\end{subequations}

\end{document}

问题在于“=”符号没有彼此对齐，并且引导句位于中央而不是左侧。

你知道怎样做吗？

Answer 1

我不明白为什么介绍性文字应该在括号的范围内。

\documentclass{article}
\usepackage{amssymb}  
\usepackage{amsmath}
\usepackage{empheq}
\usepackage{eqparbox}

\newcommand{\matheqbox}[2]{\eqmakebox[#1][l]{$\displaystyle#2$}}

\begin{document}

\begin{subequations}\label{pb:init} 
Find $u_1\colon \Omega_1 \to \mathbb{R}$ and 
$u_2\colon \Omega_2 \to \mathbb{R}$ such that
\begin{empheq}[left=\empheqlbrace]{align}
\label{eq:initOmega1}
\begin{split}
-\mu_1 \Delta u_1 &= \matheqbox{A}{f_1} \text{ in $\Omega_1$}, \\
              u_1 &= \matheqbox{A}{0}   \text{ on $\Gamma_1$},
\end{split}
\\[1ex]
\label{eq:initOmega2}
\begin{split}
-\mu_2 \Delta u_2 & = \matheqbox{A}{f_2} \text{ in $\Omega_2$}, \\
              u_2 & = \matheqbox{A}{0}   \text{ on $\Gamma_2$},
\end{split}
\\[1ex]
\label{eq:initCouplage}
\begin{split}
                      u_1 & = \matheqbox{A}{u_2} \text{ on $\Gamma$}, \\
\mu_1\nabla u_1 \cdot n_1 & = \matheqbox{A}{-\mu_2 \nabla u_2 \cdot n_2}
                                                    \text{ on $\Gamma$},
\end{split}
\end{empheq}
\end{subequations}

\end{document}

第一个参数应该\matheqbox是任意标签（ASCII 字符），对于该技巧的每个应用来说都是不同的。

等号处没有对齐：

\documentclass{article}
\usepackage{amssymb}  
\usepackage{amsmath}
\usepackage{empheq}
\usepackage{eqparbox}

\newcommand{\matheqbox}[2]{\eqmakebox[#1][l]{$\displaystyle#2$}}

\begin{document}

\begin{subequations}\label{pb:init} 
Find $u_1\colon \Omega_1 \to \mathbb{R}$ and 
$u_2\colon \Omega_2 \to \mathbb{R}$ such that
\begin{empheq}[left=\empheqlbrace]{align}
\label{eq:initOmega1}
\begin{split}
&\matheqbox{A}{-\mu_1 \Delta u_1 = f_1} \text{ in $\Omega_1$}, \\
&\matheqbox{A}{u_1 = 0}                 \text{ on $\Gamma_1$},
\end{split}
\\[1ex]
\label{eq:initOmega2}
\begin{split}
&\matheqbox{A}{-\mu_2 \Delta u_2 = f_2} \text{ in $\Omega_2$}, \\
&\matheqbox{A}{u_2 = 0}                 \text{ on $\Gamma_2$},
\end{split}
\\[1ex]
\label{eq:initCouplage}
\begin{split}
&\matheqbox{A}{u_1 = u_2} \text{ on $\Gamma$}, \\
&\matheqbox{A}{\mu_1\nabla u_1 \cdot n_1 = -\mu_2 \nabla u_2 \cdot n_2}
                          \text{ on $\Gamma$},
\end{split}
\end{empheq}
\end{subequations}

\end{document}

Answer 2

这是一个解决方案，它使用 (a) 三个独立的aligned环境和 (b) 左侧和右侧的“数学框”来存储和显示前四个=符号左侧和右侧的材料。另请注意，我已将介绍性句子从显示的方程式组中移除，并且我主张在当前用例中使用\colon它而不是。:

\documentclass{article}
\usepackage{amssymb,amsmath,empheq}
%% code for right-hand-side and left-hand-side boxes
\newlength\mylenr
\newcommand\longblurbr{-\mu_2\nabla u_2\cdot n_2 \text{ on $\Gamma$.}}
\settowidth\mylenr{$\displaystyle\longblurbr$}
\newcommand\rbox[1]{\parbox{\mylenr}{$#1$}}
\newlength\mylenl
\newcommand\longblurbl{\mu_1\nabla u_1 \cdot n_1 }
\settowidth\mylenl{$\displaystyle\longblurbl$}
\newcommand\lbox[1]{\parbox{\mylenl}{\hfill$#1$}} 

\begin{document}
Find $u_1 \colon \Omega_1 \to \mathbb{R}$ and $u_2 \colon \Omega_2 \to \mathbb{R}$ such that
\begin{subequations}\label{pb:init} 
\begin{empheq}[left=\empheqlbrace]{gather}
    \begin{aligned}\label{eq:initOmega1}
      \lbox{-\mu_1 \Delta u_1} &= \rbox{f_1\text{ in $\Omega_1$,}} \\
      \lbox{u_1}               &= \rbox{0  \text{ on $\Gamma_1$,}}
    \end{aligned}\\[1ex]
    \begin{aligned} \label{eq:initOmega2}
      \lbox{-\mu_2 \Delta u_2} &= \rbox{f_2\text{ in $\Omega_2$,}} \\
      \lbox{u_2}               &= \rbox{0  \text{ on $\Gamma_2$,}}
    \end{aligned}\\[1ex]
    \begin{aligned}\label{eq:initCouplage}
      u_1        &= u_2 \text{ on $\Gamma$,} \\
     \longblurbl &= \longblurbr
    \end{aligned}
\end{empheq}
\end{subequations}

\end{document}