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如何在带分数的子方程模式下拆分和对齐方程

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如何在带分数的子方程模式下拆分和对齐方程

如何分裂这个方程组？

\begin{subequations}
\begin{align}
    u_0&=\frac{c_{1,1} (c_{1,0}^4 (I_{0,1}-2 (I_{-1,0}-I_{-1,1}+I_{0,0}))+c_{1,0}^2 (c_{0,0}^2 (2 I_{-1,1}+I_{0,1})-c_{1,1}^2 (I_{-1,1}+2 I_{0,1}))+2 c_{1,1}^4 (I_{-1,0}+I_{0,0})-c_{0,0}^2 c_{1,1}^2 (I_{-1,1}+2 I_{0,1}))}{2 (c_{1,0}^2-c_{1,1}^2) (3 c_{1,1} c_{0,0}^2-2 (c_{1,0}^2+c_{1,1}^2) c_{0,0}+3 c_{1,0}^2 c_{1,1})}\\
    u_1&= \frac{c_{1,0}^4 (c_{1,1} (2 I_{-1,0}+I_{-1,1}+2 I_{0,0}-I_{0,1})-2 c_{0,0} I_{-1,1})+c_{1,1} c_{1,0}^2 (c_{0,0}^2 (I_{-1,1}-I_{0,1})+2 c_{1,1} c_{0,0} I_{0,1}-2 c_{1,1}^2 (I_{-1,0}+I_{0,0}))}{(c_{1,0}^2-c_{1,1}^2) (3 c_{1,1} c_{0,0}^2-2 (c_{1,0}^2+c_{1,1}^2) c_{0,0}+3 c_{1,0}^2 c_{1,1})}\\
    u_2&= \frac{c_{1,1}^2 (c_{1,0}^2 (c_{1,1} (I_{-1,0}-I_{-1,1}+I_{0,0}+I_{0,1})-c_{0,0} I_{0,1})-c_{1,1} (c_{0,0}^2 (I_{-1,1}-I_{0,1})-c_{0,0} c_{1,1} I_{-1,1}+c_{1,1}^2 (I_{-1,0}+I_{0,0})))}{(c_{1,0}^2-c_{1,1}^2) (3 c_{1,1} c_{0,0}^2-2 (c_{1,0}^2+c_{1,1}^2) c_{0,0}+3 c_{1,0}^2 c_{1,1})}
\end{align}
\end{subequations}

在此处输入图片描述 我知道如何像这样在单方程环境中进行拆分

u_0=\frac{\begin{split}
&c_{1,1} (c_{1,0}^4 (-2 I_{-1,1}-2 I_{0,1}+I_{-1,0}+2 I_{0,0})+2 c_{0,0}^4 (I_{-1,1}+I_{0,1})\\
&-\left(c_{1,0}^2+c_{1,1}^2\right) c_{0,0}^2 (2 I_{-1,0}+I_{0,0})+c_{1,0}^2 c_{1,1}^2 (I_{-1,0}+2 I_{0,0}))
\end{split}}
{\begin{split}
&2 \left(c_{0,0}^2-c_{1,0}^2\right) \left(2 c_{1,1} c_{0,0}^2-3 \left(c_{1,0}^2+c_{1,1}^2\right) c_{0,0}+2 c_{1,0}^2 c_{1,1}\right)
\end{split}}
\end{equation}

在此处输入图片描述

但在子方程环境中不起作用。提前致谢。

答案1

我有两个主要建议。首先，充分利用2三个子方程的分母项相同（第一个子方程中最多一个因子）这一事实。我们称这个公分母为V。这允许你写

V^{-1}(<long numerator term>)

代替

\frac{<long numerator term>}{V}

其次，根据需要，使用aligned环境来分成<long numerator term>两行或三行。

第三个小建议是将“外”括号从 改为(...)。[...]这将大大提高可读性。

在此处输入图片描述

\documentclass{article}
\usepackage{geometry}
\usepackage{amsmath} % for 'subequations' and 'aligned' environments
\newcommand\mybox[1]{\begin{aligned}[t] #1 \end{aligned}}
\begin{document}
\noindent
Put $V=(c_{1,0}^2-c_{1,1}^2) (3 c_{1,1} c_{0,0}^2-2 (c_{1,0}^2+c_{1,1}^2) c_{0,0}+3 c_{1,0}^2 c_{1,1})$. Then
\begin{subequations}
\begin{align}
    u_0&=\tfrac{1}{2}V^{-1} \mybox{%
     &\bigl\{c_{1,1} \bigl[ c_{1,0}^4 [I_{0,1}-2 (I_{-1,0}-I_{-1,1}+I_{0,0})]\\
     &\quad +c_{1,0}^2 [c_{0,0}^2 (2 I_{-1,1}+I_{0,1})-c_{1,1}^2 (I_{-1,1}+2 I_{0,1})] \\
     &\quad +2 c_{1,1}^4 (I_{-1,0}+I_{0,0})-c_{0,0}^2 c_{1,1}^2 (I_{-1,1}+2 I_{0,1})\bigr] \bigr\}} \\[2\jot]
u_1&= \hphantom{\tfrac{1}{2}} V^{-1}\mybox{%
     &\bigl\{c_{1,0}^4 [c_{1,1} (2 I_{-1,0}+I_{-1,1}+2 I_{0,0}-I_{0,1})-2 c_{0,0} I_{-1,1}]\\
     &\quad +c_{1,1} c_{1,0}^2 [c_{0,0}^2 (I_{-1,1}-I_{0,1})+2 c_{1,1} c_{0,0} I_{0,1}-2 c_{1,1}^2 (I_{-1,0}+I_{0,0})] \bigr\} }\\[2\jot]
u_2&= \hphantom{\tfrac{1}{2}}V^{-1}\mybox{%
     &\bigl\{ c_{1,1}^2 \bigl[ c_{1,0}^2 [c_{1,1} (I_{-1,0}-I_{-1,1}+I_{0,0}+I_{0,1})-c_{0,0} I_{0,1}]\\
     &\quad -c_{1,1} [c_{0,0}^2 (I_{-1,1}-I_{0,1})-c_{0,0} c_{1,1} I_{-1,1}+c_{1,1}^2 (I_{-1,0}+I_{0,0})]\bigr] \bigr\} }
\end{align}
\end{subequations}
\end{document}

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