如何在漂亮的视图中写出 3 个长等式？

Question

我有 3 个长方程式，我想在漂亮的视图中写出它们。它们现在看起来很丑，有什么帮助吗？还有一个问题，我\left[在第三个方程式的第一行使用了，然后\\在下一行使用了，当我使用\right]它时会出现错误并且不显示第二个括号，我该如何修复它？

输出如下，

\documentclass[a4paper,14pt]{extreport}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{breqn}
\usepackage{flexisym}    % I don't know what is this package, could you tell me what is it?
\newcommand{\p}{\partial}
\begin{document}
\begin{align}
&\frac{\p}{\p \theta}\left(\bar{h}_0^3 \frac{\p\bar{p}_3}
{\p\theta}\right)+\left(\frac{D}{L}\right)^2 \frac{\p}{\p \bar{z}} 
\left(\bar{h}_0^3 \frac{\p \bar{p}_3}{\p \bar{z}}\right)+\frac{\p}{\p 
\theta}\left(3\bar{h}_0^2 \cos\theta \frac{\p\bar{p}_{1}}{\p \theta}\right)+ 
\notag \\ 
&\left(\frac{D}{L}\right)^2 \frac{\p}{\p \bar{z}} \left(\bar{h}_0^2 
\cos\theta \frac{\p \bar{p}_{1}}{\p \bar{z}}\right)+\frac{\p}{\p 
\theta}\left(3\bar{h}_0\cos\theta^2 \frac{\p\bar{p}_{0}}{\p \theta}\right) 
+\notag\\
&\left(\frac{D}{L}\right)^2 \frac{\p}{\p \bar{z}} \left(\bar{h}_0 
\cos\theta^2 \frac{\p \bar{p}_{0}}{\p \bar{z}}\right) =  0 \label{eq:p3}\\
&\frac{\p}{\p \theta}\left(\bar{h}_0^3 \frac{\p\bar{p}_4}{\p \theta}\right) 
+ \left(\frac{D}{L}\right)^2 \frac{\p}{\p \bar{z}} \left(\bar{h}_0^3 
\frac{\p \bar{p}_4}{\p \bar{z}}\right)+\frac{\p}{\p 
\theta}\left(3\bar{h}_0^2 \sin\theta \frac{\p\bar{p}_{2}}{\p \theta}\right)+ 
\notag \\ 
&\left(\frac{D}{L}\right)^2 \frac{\p}{\p \bar{z}} \left(\bar{h}_0^2 
\sin\theta \frac{\p \bar{p}_{2}}{\p \bar{z}}\right)+\frac{\p}{\p 
\theta}\left(3\bar{h}_0\sin\theta^2 \frac{\p\bar{p}_{0}}{\p \theta}\right) 
+\notag\\
&\left(\frac{D}{L}\right)^2 \frac{\p}{\p \bar{z}} \left(\bar{h}_0 
\sin\theta^2 \frac{\p \bar{p}_{0}}{\p \bar{z}}\right) =  0  \label{eq:p4}\\
&\frac{\p}{\p \theta}\left(\bar{h}_0^3 \frac{\p\bar{p}_5}{\p \theta}\right) 
+ \left(\frac{D}{L}\right)^2 \frac{\p}{\p \bar{z}} \left(\bar{h}_0^3 
\frac{\p \bar{p}_5}{\p \bar{z}}\right) + \frac{\p}{\p \theta}\left[   
3\bar{h}_0^2 \left(\cos\theta \frac{\p\bar{p}_{2}}{\p \theta} +\notag \\
&\sin\theta \frac{\p\bar{p}_{1}}{\p \theta}\right)\right]  + \left(\frac{D}
{L}\right)^2 \frac{\p}{\p \bar{z}} \left[ \bar{h}_0^2\left( \cos\theta 
\frac{\p \bar{p}_{2}}{\p \bar{z}} + \sin\theta \frac{\p \bar{p}_{1}}{\p 
\bar{z}}\right)\right] +\notag \\
&\frac{\p}{\p \theta} \left(\bar{h}_0 \cos\theta\: \sin\theta \:\frac{\p 
\bar{p}_{0}}{\p \theta}\right) +
\left(\frac{D}{L}\right)^2 \frac{\p}{\p \bar{z}} \left(\bar{h}_0 
\cos\theta\: \sin\theta \:\frac{\p \bar{p}_{0}}{\p \bar{z}}\right) =  
0\label{eq:p5}
\end{align}
\end{document}

Answer 1

您使用align但未指定任何实际对齐点，也不能在一条线上使用左/右对齐，但在几乎所有情况下，最好还是使用固定大小。我使用了，\bigl在\bigr这里您可能更喜欢稍大一点\Bigl或\biggl……

我还增加了一行以避免分裂内部术语（删除时不再产生错误\left/\right但分裂看起来不太好）。

您可以将其标记为对齐中的三个编号行，而不是使用\notag，而左侧则进一步拆分（例如，aligned但我在这里没有这样做。

\documentclass[a4paper,14pt]{extreport}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}

\begin{document}
\let\p\partial

\begin{align}
\frac{\p}{\p \theta}\bigl(\bar{h}_0^3 \frac{\p\bar{p}_3}
{\p\theta}\bigr)+\bigl(\frac{D}{L}\bigr)^2 \frac{\p}{\p \bar{z}} 
\bigl(\bar{h}_0^3 \frac{\p \bar{p}_3}{\p \bar{z}}\bigr)+\frac{\p}{\p 
\theta}\bigl(3\bar{h}_0^2 \cos\theta \frac{\p\bar{p}_{1}}{\p \theta}\bigr)+{}\notag \\ 
\bigl(\frac{D}{L}\bigr)^2 \frac{\p}{\p \bar{z}} \bigl(\bar{h}_0^2 
\cos\theta \frac{\p \bar{p}_{1}}{\p \bar{z}}\bigr)+\frac{\p}{\p 
\theta}\bigl(3\bar{h}_0\cos\theta^2 \frac{\p\bar{p}_{0}}{\p \theta}\bigr) 
+{}\notag\\
\bigl(\frac{D}{L}\bigr)^2 \frac{\p}{\p \bar{z}} \bigl(\bar{h}_0 
\cos\theta^2 \frac{\p \bar{p}_{0}}{\p \bar{z}}\bigr) &=  0 \label{eq:p3}\\
\frac{\p}{\p \theta}\bigl(\bar{h}_0^3 \frac{\p\bar{p}_4}{\p \theta}\bigr) 
+ \bigl(\frac{D}{L}\bigr)^2 \frac{\p}{\p \bar{z}} \bigl(\bar{h}_0^3 
\frac{\p \bar{p}_4}{\p \bar{z}}\bigr)+\frac{\p}{\p 
\theta}\bigl(3\bar{h}_0^2 \sin\theta \frac{\p\bar{p}_{2}}{\p \theta}\bigr)+{}
\notag \\ 
\bigl(\frac{D}{L}\bigr)^2 \frac{\p}{\p \bar{z}} \bigl(\bar{h}_0^2 
\sin\theta \frac{\p \bar{p}_{2}}{\p \bar{z}}\bigr)+\frac{\p}{\p 
\theta}\bigl(3\bar{h}_0\sin\theta^2 \frac{\p\bar{p}_{0}}{\p \theta}\bigr) 
+{}\notag\\
\bigl(\frac{D}{L}\bigr)^2 \frac{\p}{\p \bar{z}} \bigl(\bar{h}_0 
\sin\theta^2 \frac{\p \bar{p}_{0}}{\p \bar{z}}\bigr) &=  0  \label{eq:p4}\\
\frac{\p}{\p \theta}\bigl(\bar{h}_0^3 \frac{\p\bar{p}_5}{\p \theta}\bigr) 
+ \bigl(\frac{D}{L}\bigr)^2 \frac{\p}{\p \bar{z}} \bigl(\bar{h}_0^3 
\frac{\p \bar{p}_5}{\p \bar{z}}\bigr) + {}\notag\\
\frac{\p}{\p \theta}\bigl[   
3\bar{h}_0^2 \bigl(\cos\theta \frac{\p\bar{p}_{2}}{\p \theta} +%{}\notag \\
\sin\theta \frac{\p\bar{p}_{1}}{\p \theta}\bigr)\bigr]  +{}\notag\\
 \bigl(\frac{D}
{L}\bigr)^2 \frac{\p}{\p \bar{z}} \bigl[ \bar{h}_0^2\bigl( \cos\theta 
\frac{\p \bar{p}_{2}}{\p \bar{z}} + \sin\theta \frac{\p \bar{p}_{1}}{\p 
\bar{z}}\bigr)\bigr] +{}\notag \\
\frac{\p}{\p \theta} \bigl(\bar{h}_0 \cos\theta\: \sin\theta \:\frac{\p 
\bar{p}_{0}}{\p \theta}\bigr) +
\bigl(\frac{D}{L}\bigr)^2 \frac{\p}{\p \bar{z}} \bigl(\bar{h}_0 
\cos\theta\: \sin\theta \:\frac{\p \bar{p}_{0}}{\p \bar{z}}\bigr) &
=   0\label{eq:p5}
\end{align}
\end{document}

Answer 2

为了直观地强调每个方程中的六个加法项之和必须等于0，我将这个条件放在开始每个公式，而不是把它藏在最后。

我还会将材料分组，使每行有两个加法项；这样，公式的内部结构就变得更加明显：三个项分别涉及和的偏导数\theta。\bar{z}对于第三个方程，这种方法需要使用额外的行；我相信可读性方面的收益值得这种改变。

为了增加可读性，我还会在方程式之间添加一些垂直空白。

\documentclass[a4paper,14pt]{extreport}
\usepackage{amsmath} % for "align" environment
\newcommand{\p}{\partial}
\begin{document}
\begin{align}
0 &= \frac{\p}{\p \theta}\biggl(\bar{h}_0^3 
  \frac{\p\bar{p}_3}{\p\theta}\biggr) 
  +\biggl(\frac{D}{L}\biggr)^{\!\!2} \frac{\p}{\p \bar{z}} 
  \biggl(\bar{h}_0^3 \frac{\p \bar{p}_3}{\p \bar{z}}\biggr) 
  \notag \\
&\quad +\frac{\p}{\p\theta}\biggl(3\bar{h}_0^2 \cos\theta \,
  \frac{\p\bar{p}_{1}}{\p \theta}\biggr)  
  +\biggl(\frac{D}{L}\biggr)^{\!\!2} \frac{\p}{\p \bar{z}} 
  \biggl(\bar{h}_0^2 \cos\theta \,\frac{\p \bar{p}_{1}}{\p \bar{z}}\biggr)
  \notag\\
&\quad +\frac{\p}{\p \theta}\biggl(3\bar{h}_0\cos\theta^2 \,
  \frac{\p\bar{p}_{0}}{\p \theta}\biggr) 
  +\biggl(\frac{D}{L}\biggr)^{\!\!2} \frac{\p}{\p \bar{z}} 
  \biggl(\bar{h}_0 \cos\theta^2 \,
  \frac{\p \bar{p}_{0}}{\p \bar{z}}\biggr) 
  \label{eq:p3}\\[2ex]
0 &= \frac{\p}{\p \theta}\biggl(\bar{h}_0^3 
  \frac{\p\bar{p}_4}{\p \theta}\biggr) 
  + \biggl(\frac{D}{L}\biggr)^{\!\!2} \frac{\p}{\p \bar{z}} 
  \biggl(\bar{h}_0^3 \frac{\p \bar{p}_4}{\p \bar{z}}\biggr)
  \notag \\ 
&\quad+\frac{\p}{\p\theta} \biggl(3\bar{h}_0^2 \sin\theta \,
  \frac{\p\bar{p}_2}{\p \theta}\biggr) 
  +\biggl(\frac{D}{L}\biggr)^{\!\!2} \frac{\p}{\p \bar{z}} 
  \biggl(\bar{h}_0^2 \sin\theta \,
  \frac{\p \bar{p}_2}{\p \bar{z}}\biggr)
  \notag\\
&\quad+\frac{\p}{\p \theta}\biggl(3\bar{h}_0\sin\theta^2 \,
  \frac{\p\bar{p}_{0}}{\p \theta}\biggr)
  +\biggl(\frac{D}{L}\biggr)^{\!\!2} \frac{\p}{\p \bar{z}} 
  \biggl(\bar{h}_0 \sin\theta^2 \,\frac{\p \bar{p}_{0}}{\p \bar{z}}\biggr)  
  \label{eq:p4}\\[2ex]
0 &= \frac{\p}{\p \theta}\biggl(\bar{h}_0^3 
  \frac{\p\bar{p}_5}{\p \theta}\biggr) 
  + \biggl(\frac{D}{L}\biggr)^{\!\!2} \frac{\p}{\p \bar{z}} 
  \biggl(\bar{h}_0^3 \frac{\p \bar{p}_5}{\p \bar{z}}\biggr) 
  \notag \\
&\quad + \frac{\p}{\p \theta}\biggl[3\bar{h}_0^2 
  \biggl(\cos\theta \,\frac{\p\bar{p}_2}{\p \theta} 
  +\sin\theta \,\frac{\p\bar{p}_{1}}{\p \theta}\biggr)\biggr]  
  \notag \\
&\qquad 
  + \biggl(\frac{D}{L}\biggr)^{\!\!2} \frac{\p}{\p \bar{z}} 
  \biggl[ \bar{h}_0^2\biggl( \cos\theta \,\frac{\p \bar{p}_2}{\p \bar{z}} 
  + \sin\theta \,\frac{\p \bar{p}_{1}}{\p \bar{z}}\biggr)\biggr] 
  \notag \\
&\quad+\frac{\p}{\p \theta} \biggl(\bar{h}_0 \cos\theta \sin\theta \,
  \frac{\p \bar{p}_{0}}{\p \theta}\biggr) 
  + \biggl(\frac{D}{L}\biggr)^{\!\!2} \frac{\p}{\p \bar{z}} 
  \biggl(\bar{h}_0 \cos\theta \sin\theta \,
  \frac{\p \bar{p}_{0}}{\p \bar{z}}\biggr) 
  \label{eq:p5}
\end{align}
\end{document}

Answer 3

这是 Bernard 解决方案的变体。考虑到空间限制，我认为对齐特定术语没有多大意义。因此我的解决方案看起来很像 Mico 的解决方案。

我同意 Mico 的评论，左右\Bigg(lr)变得太大，看起来不均匀。这是一个调整

\documentclass[a4paper,14pt]{extreport}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
%\usepackage[showframe]{geometry}
\usepackage{mathtools}
\usepackage{amssymb}
\usepackage{esdiff}
\newcommand\DLtwo{\biggl(\frac{D}{L}\biggr)^2}

\begin{document}

\begin{align}
  \MoveEqLeft \diffp{}{\theta}\biggl(\bar{h}_0^3
    \diffp{\bar{p}_3}{\theta}\biggr) + \DLtwo \diffp{}{{\bar z}}
  \biggl(\bar{h}_0^3
    \diffp{\bar{p}_3}{{\bar{z}}}\biggr)+\diffp{}{\theta}\biggl(3\bar{h}_0^2
    \cos\theta \diffp{\bar{p}_{1}}{\theta}\biggr) \notag
  \\
  & + \DLtwo \diffp{}{{\bar{z}}} \biggl(\bar{h}_0^2 \cos\theta
    \diffp{\bar{p}_{1}}{{\bar{z}}}\biggr)+\diffp{}{\theta}\biggl(3\bar{h}_0\cos\theta^2
    \diffp{\bar{p}_{0}}{\theta}\biggr) \notag
  \\
  & + \DLtwo \diffp{}{{\bar{z}}} \biggl(\bar{h}_0 \cos\theta^2
    \diffp{\bar{p}_{0}}{{\bar{z}}}\biggr) = 0 \label{eq:p3}
  \\[2ex]
%%
  \MoveEqLeft \diffp{}{\theta}\biggl(\bar{h}_0^3
    \diffp{\bar{p}_4}{\theta}\biggr) + \DLtwo \diffp{}{{\bar{z}}}
  \biggl(\bar{h}_0^3
    \diffp{\bar{p}_4}{{\bar{z}}}\biggr)
  +\diffp{}{\theta}\biggl(3\bar{h}_0^2
    \sin\theta \diffp{\bar{p}_{2}}{\theta}\biggr) \notag
  \\
  & + \DLtwo \diffp{}{{\bar{z}}} \biggl(\bar{h}_0^2 \sin\theta
    \frac{\bar{p}_{2}}{\bar{z}}\biggr)+\diffp{}{\theta}\biggl(3\bar{h}_0\sin\theta^2
    \diffp{\bar{p}_{0}}{\theta}\biggr) \notag
  \\
  & + \DLtwo \diffp{}{{\bar{z}}} \biggl(\bar{h}_0\sin\theta^2
    \frac{\bar{p}_{0}}{\bar{z}}\biggr) = 0 \label{eq:p4}
  \\[2ex]
%
  \MoveEqLeft \diffp{}{\theta}\biggl(\bar{h}_0^3
    \diffp{\bar{p}_5}{\theta}\biggr) + \DLtwo \diffp{}{{\bar{z}}}
  \biggl(\bar{h}_0^3 \diffp{\bar{p}_5}{{\bar{z}}}\biggr) 
  \notag
  \\
  &+
  \diffp{}{\theta}\biggl[3\bar{h}_0^2 \biggl(\cos\theta
  \diffp{\bar{p}_{2}}{\theta} + \sin\theta
  \diffp{\bar{p}_{1}}{\theta}\biggr)\biggr] \notag
  \\
  & + \DLtwo \diffp{}{{\bar{z}}} \biggl[
    \bar{h}_0^2 \biggl( \cos\theta \diffp{\bar{p}_{2}}{{\bar{z}}} +
    \sin\theta \diffp{\bar{p}_{1}}{{\bar{z}}}\biggr)\biggr]
    \notag
    \\
    &+\diffp{}{\theta} \biggl(\bar{h}_0 \cos\theta\: \sin\theta
      \:\diffp{\bar{p}_{0}}{\theta}\biggr)
    \notag
  \\
  & + \DLtwo \diffp{}{{\bar{z}}} \biggl(\bar{h}_0 \cos\theta\:
    \sin\theta \:\diffp{\bar{p}_{0}}{{\bar{z}}}\biggr) =
  0\label{eq:p5}
\end{align}

\end{document} 

Answer 4

我想建议将分布的 (D/L) ²项和最外层偏导数分解出来。如果我这样做，从 Mico 的代码开始，我可以在两行上拟合第一个方程，我认为这会使结构更加清晰。例如，第二行上没有 3 的因子这一事实现在非常明显（您确定除了内部微分的变化之外，受 ∂/∂θ 约束的项不应该与受 ∂/∂z̅ 约束的项相同吗？）并且没有 ∂p̅ ₁ /∂■ 也是如此（我看到它出现在第三个方程中，所以它可能应该是那样的）。

我还引入了一个\pd宏来减少输入，并改进了源和排版格式的水平对齐。比我更了解环境的人可能不需要做太多事情align就能做到这一点。\phantom

\documentclass[a4paper,14pt]{extreport}
\usepackage{amsmath} % for "align" environment
\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}}
\begin{document}
\begin{align}
  0 &=\quad\phantom{\biggl(\frac{D}{L}\biggr)^{\!\!2}}
  \pd{}{\theta}
  \biggl(
        \bar{h}_0^3                \pd{\bar{p}_3}{\theta}
     + 3\bar{h}_0^2 \cos\theta   \,\pd{\bar{p}_1}{\theta}
     + 3\bar{h}_0   \cos\theta^2 \,\pd{\bar{p}_0}{\theta}
  \biggr)
\notag\\&\quad
+ \biggl(\frac{D}{L}\biggr)^{\!\!2}
  \pd{}{\bar{z}}
  \biggl(
                   \bar{h}_0^3                \pd{\bar{p}_3}{\bar{z}}
     + \phantom{3} \bar{h}_0^2 \cos\theta   \,\pd{\bar{p}_1}{\bar{z}}
     + \phantom{3} \bar{h}_0   \cos\theta^2 \,\pd{\bar{p}_0}{\bar{z}}
  \biggr)
\label{eq:p3}
\end{align}
\end{document}