对齐连续对齐环境并保留方程编号

Question

我有以下乳胶代码：

\documentclass[12pt,letterpaper]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
\begin{subequations}
\begin{align}
\Phi_1^l(r)&=-(2l+1)\frac{q}{2la}\left(\frac{r}{a}\right)^l\\
\Phi_1^0(r)\bigg|_{q@0}&=\frac{q}{r}\\
\Phi_1^0(r)\bigg|_{q@2a}&=0\\
E_1^l(r)&=-(2l+1)\frac{q}{2ar}\left(\frac{r}{a}\right)^l\\
E_1^0(r)\bigg|_{q@0}&=-\frac{q}{r^2}\\
E_1^0(r)\bigg|_{q@2a}&=0
\end{align}
\end{subequations}
\begin{subequations}
\begin{align}
\Phi_2^l(r)&=-(2l+1)\frac{q}{2la}\left(\frac{2a-r}{a}\right)^l\\
\Phi_2^0(r)\bigg|_{q@0}&=\frac{q}{a}\\
\Phi_2^0(r)\bigg|_{q@2a}&=-\frac{q}{a}+\frac{q}{2a-r}\\
E_2^l(r)&=(1+2l)\frac{q}{2a(2a-r)}\left(\frac{2a-r}{a}\right)^l\\
E_2^0(r)\bigg|_{q@0}&=0\\
E_2^0(r)\bigg|_{q@2a}&=\frac{q}{(2a-r)^2}
\end{align}
\end{subequations}
\end{document}

我想做的是对齐两个对齐环境，同时保持子方程编号不变。我知道已经回答过有关对齐两个不同对齐环境的问题，但它们似乎对我的方程编号没有帮助，因为对齐环境位于子方程环境内。

Answer 1

这是软件包的解决方案nccmath。您只需将两个align和 subequations环境插入到 中fleqn，这是一个设计用于临时键入方程式的环境，作为fleqn的选项amsmath。此环境可以采用可选参数 - 方程式开始的左边距的距离。该值是通过反复试验选择的。

但我发现结果不太好。在我看来，在标志上对齐=并不是教条。我建议使用相同的方法，但改为在左侧对齐。

以下是两种布局的图示，其中包含以下代码：

\documentclass[12pt,letterpaper]{article}
\usepackage{amsmath, nccmath}
\usepackage{amssymb}

\begin{document}

\begin{fleqn}[7em]
  \begin{subequations}
    \begin{align}
      \Phi_1^l(r) & =-(2l+1)\frac{q}{2la}\left(\frac{r}{a}\right)^l \\
      \Phi_1^0(r)\bigg|_{q@0} & =\frac{q}{r} \\
      \Phi_1^0(r)\bigg|_{q@2a} & =0 \\
      E_1^l(r) & =-(2l+1)\frac{q}{2ar}\left(\frac{r}{a}\right)^l \\
      E_1^0(r)\bigg|_{q@0} & =-\frac{q}{r^2} \\
      E_1^0(r)\bigg|_{q@2a} & =0 %
    \end{align}
  \end{subequations}
  \begin{subequations}
    \begin{align}
      \Phi_2^l(r) & =-(2l+1)\frac{q}{2la}\left(\frac{2a-r}{a}\right)^l \\
      \Phi_2^0(r)\bigg|_{q@0} & =\frac{q}{a} \\
      \Phi_2^0(r)\bigg|_{q@2a} & =-\frac{q}{a}+\frac{q}{2a-r} \\
      E_2^l(r) & =(1+2l)\frac{q}{2a(2a-r)}\left(\frac{2a-r}{a}\right)^l \\
      E_2^0(r)\bigg|_{q@0} & =0 \\
      E_2^0(r)\bigg|_{q@2a} & =\frac{q}{(2a-r)^2}
    \end{align}
  \end{subequations}
\end{fleqn}

\begin{fleqn}[7em]
  \begin{subequations}
    \begin{align}
       & \Phi_1^l(r)=-(2l+1)\frac{q}{2la}\left(\frac{r}{a}\right)^l \\
       & \Phi_1^0(r)\bigg|_{q@0}=\frac{q}{r} \\
       & \Phi_1^0(r)\bigg|_{q@2a}=0 \\
       & E_1^l(r)=-(2l+1)\frac{q}{2ar}\left(\frac{r}{a}\right)^l \\
       & E_1^0(r)\bigg|_{q@0}=-\frac{q}{r^2} \\
       & E_1^0(r)\bigg|_{q@2a}=0 %
    \end{align}
  \end{subequations}
  \begin{subequations}
    \begin{align}
       & \Phi_2^l(r)=-(2l+1)\frac{q}{2la}\left(\frac{2a-r}{a}\right)^l \\
       & \Phi_2^0(r)\bigg|_{q@0}=\frac{q}{a} \\
       & \Phi_2^0(r)\bigg|_{q@2a}=-\frac{q}{a}+\frac{q}{2a-r} \\
       & E_2^l(r)=(1+2l)\frac{q}{2a(2a-r)}\left(\frac{2a-r}{a}\right)^l \\
       & E_2^0(r)\bigg|_{q@0}=0 \\
       & E_2^0(r)\bigg|_{q@2a}=\frac{q}{(2a-r)^2}
    \end{align}
  \end{subequations}
\end{fleqn}

\end{document} 

Answer 2

希望你不要有太多这样的对齐，

\documentclass[12pt,letterpaper]{article}
\usepackage{amsmath}
\usepackage{amssymb}

\makeatletter
\newcommand{\stepsubequations}{%
  \ifmeasuring@
  \else
    \stepcounter{parentequation}\setcounter{equation}{0}%
    \xdef\theparentequation{\arabic{parentequation}}%
  \fi
}
\makeatother

\begin{document}

\begin{subequations}
\begin{align}
\Phi_1^l(r)&=-(2l+1)\frac{q}{2la}\left(\frac{r}{a}\right)^l\\
\Phi_1^0(r)\Big|_{q@0}&=\frac{q}{r}\\
\Phi_1^0(r)\Big|_{q@2a}&=0\\
E_1^l(r)&=-(2l+1)\frac{q}{2ar}\left(\frac{r}{a}\right)^l\\
E_1^0(r)\Big|_{q@0}&=-\frac{q}{r^2}\\
E_1^0(r)\Big|_{q@2a}&=0
\\% <---- change here the value
\stepsubequations
\Phi_2^l(r)&=-(2l+1)\frac{q}{2la}\left(\frac{2a-r}{a}\right)^l\\
\Phi_2^0(r)\Big|_{q@0}&=\frac{q}{a}\\
\Phi_2^0(r)\Big|_{q@2a}&=-\frac{q}{a}+\frac{q}{2a-r}\\
E_2^l(r)&=(1+2l)\frac{q}{2a(2a-r)}\left(\frac{2a-r}{a}\right)^l\\
E_2^0(r)\Big|_{q@0}&=0\\
E_2^0(r)\Big|_{q@2a}&=\frac{q}{(2a-r)^2}
\end{align}
\end{subequations}

\begin{equation}
Check
\end{equation}

\end{document}

最后一个等式用于检查一切是否顺利。

左对齐：

\documentclass[12pt,letterpaper]{article}
\usepackage{amsmath}
\usepackage{amssymb}

\makeatletter
\newcommand{\stepsubequations}{%
  \ifmeasuring@
  \else
    \stepcounter{parentequation}\setcounter{equation}{0}%
    \xdef\theparentequation{\arabic{parentequation}}%
  \fi
}
\makeatother

\begin{document}

\begin{subequations}
\begin{align}
&\Phi_1^l(r)=-(2l+1)\frac{q}{2la}\left(\frac{r}{a}\right)^l\\
&\Phi_1^0(r)\Big|_{q@0}=\frac{q}{r}\\
&\Phi_1^0(r)\Big|_{q@2a}=0\\
&E_1^l(r)=-(2l+1)\frac{q}{2ar}\left(\frac{r}{a}\right)^l\\
&E_1^0(r)\Big|_{q@0}=-\frac{q}{r^2}\\
&E_1^0(r)\Big|_{q@2a}=0
\\% <---- change here the value
\stepsubequations
&\Phi_2^l(r)=-(2l+1)\frac{q}{2la}\left(\frac{2a-r}{a}\right)^l\\
&\Phi_2^0(r)\Big|_{q@0}=\frac{q}{a}\\
&\Phi_2^0(r)\Big|_{q@2a}=-\frac{q}{a}+\frac{q}{2a-r}\\
&E_2^l(r)=(1+2l)\frac{q}{2a(2a-r)}\left(\frac{2a-r}{a}\right)^l\\
&E_2^0(r)\Big|_{q@0}=0\\
&E_2^0(r)\Big|_{q@2a}=\frac{q}{(2a-r)^2}
\end{align}
\end{subequations}

\begin{equation}
Check
\end{equation}

\end{document}