格式化方程式

Question

我有以下等式

\documentclass[12pt]{article}
    \usepackage[a4paper]{geometry}
    \usepackage{mathtools}
\begin{document}
\begin{equation*}
u(x) = \left\{
\begin{multlined}
      \shoveleft{u_0(x)}\\
      \shoveright{\text{for all } x \in \mathcal{D}_0}\\
      u_{i}(x) + \underline{u}_{0} - \underline{u}_{i} + 1 + \varepsilon_i +
      \sum_{\mathclap{ \{j: j \geq 1 \text{ \& }  j \leq i-1 \}}}
      (\underline{u}_{j} - \overline{u}_{j} + 1 + \varepsilon_{j}),\\
      \shoveright{\text{for all }  x \in \mathcal{D}_{i},\ i \in \Gamma,\ i \geq 1}\\
      u_{i}(x) + \underline{u}_{0} - \overline{u}_{j} - 1 - \varepsilon_i -
      \sum_{\mathclap{\{j: j \leq -1 \text{ \& } j\geq i+1\}}} (\underline{u}_{j} - \overline{u}_{j} - 1 - \varepsilon_j),  \\
      \shoveright{\text{for all }  x \in \mathcal{D}_{i}, i \in \Gamma, i \leq -1}\\
\end{multlined}
\right.
\end{equation*}
\end{document} 

结果还可以，但我不太满意。有什么线索可以更好地呈现它吗？

Answer 1

我有两个建议：

\documentclass[12pt]{article}
\usepackage[a4paper]{geometry}
\usepackage{mathtools}

\usepackage{lipsum}

\begin{document}

\lipsum[1][1-3]
\begin{equation*}
u(x) = \left\{
\begin{aligned}
&\makebox[0.8\displaywidth][s]{$u_0(x)$\hfill for all $x \in \mathcal{D}_0$}
\\[2ex]
&\begin{multlined}[c][0.8\displaywidth]
   u_{i}(x) + \underline{u}_{0} - \underline{u}_{i} + 1 + \varepsilon_i +
   \sum_{1\leq j \leq i-1}
     (\underline{u}_{j} - \overline{u}_{j} + 1 + \varepsilon_{j}),
 \\
   \text{for all $x \in \mathcal{D}_{i}$, $i \in \Gamma$, $i \geq 1$}
 \end{multlined}
\\[2ex]
&\begin{multlined}[c][0.8\displaywidth]
   u_{i}(x) + \underline{u}_{0} - \overline{u}_{j} - 1 - \varepsilon_i -
   \sum_{i+1\leq j\leq -1}
     (\underline{u}_{j} - \overline{u}_{j} - 1 - \varepsilon_j),
  \\
  \text{for all $x \in \mathcal{D}_{i}$, $i \in \Gamma$, $i \leq -1$}
 \end{multlined}
\end{aligned}
\right.
\end{equation*}
\lipsum[1][1-3]
\begin{equation*}
u(x) = 
\begin{cases}
u_0(x) & \text{for all $x \in \mathcal{D}_0$}
\\[2ex]
\begin{aligned}[b]
  u_{i}(x)
  &+ \underline{u}_{0} - \underline{u}_{i} + 1 + \varepsilon_i
  \\
  &+ \sum_{1\leq j \leq i-1}
       (\underline{u}_{j} - \overline{u}_{j} + 1 + \varepsilon_{j}),
\end{aligned}
& \text{for all $x \in \mathcal{D}_{i}$, $i \in \Gamma$, $i \geq 1$}
\\[4ex]
\begin{aligned}[b]
  u_{i}(x)
  &+ \underline{u}_{0} - \overline{u}_{j} - 1 - \varepsilon_i \\
  &- \sum_{i+1\leq j\leq -1}
       (\underline{u}_{j} - \overline{u}_{j} - 1 - \varepsilon_j),
\end{aligned}
& \text{for all $x \in \mathcal{D}_{i}$, $i \in \Gamma$, $i \leq -1$}
\end{cases}
\end{equation*}

\end{document}