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方程线分解

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方程线分解 
\documentclass[12pt,a4paper]{article}
\usepackage{amsmath}\usepackage{amssymb}
\textwidth=16.5cm  \oddsidemargin=-0.10cm \evensidemargin=-0.10cm  \topmargin=-1.0cm \textheight=24.5cm

\newcommand{\piRsquare}{\pi r^2}        

\title{The small amplitude expansion: The class of theoritical considered}
\author{Md. X }     
\date{January 26, 2013}                 
\begin{document} \baselineskip=18pt
\section{Introduction}  
  \begin{align}
    \left(\nabla p_1\right){}^2 g_2-\frac{1}{3} \text{Cos}[2 \tau ] \left(\nabla p_1\right){}^2 g_2-\frac{1}{24} \lambda  \text{Cos}[2 \tau ] g_2 p_1^4-\frac{1}{24} \lambda  \text{Cos}[4 \tau ] g_2 p_1^4+\frac{19}{72} g_2^3 p_1^4-\frac{1}{9} \text{Cos}[2 \tau ] g_2^3 p_1^4+\frac{5}{72} \text{Cos}[4 \tau ] g_2^3 p_1^4 \\
    & \qquad -\frac{5}{8} g_2 g_3 p_1^4-\frac{1}{2} \text{Cos}[2 \tau ] g_2 g_3 p_1^4+\frac{1}{8} \text{Cos}[4 \tau ] g_2 g_3 p_1^4+\frac{3}{8} g_4 p_1^4+\frac{1}{2} \text{Cos}[2 \tau ] g_4 p_1^4+\frac{1}{8} \text{Cos}[4 \tau ] g_4 p_1^4-\frac{5}{2} \text{Cos}[\tau ] g_2^2 p_1^2 p_2+\frac{1}{2} \text{Cos}[3 \tau ] g_2^2 p_1^2 p_2 \\
    & \qquad +\frac{9}{4} \text{Cos}[\tau ] g_3 p_1^2 p_2+\frac{3}{4} \text{Cos}[3 \tau ] g_3 p_1^2 p_2+\frac{1}{2} g_2 p_2^2+\frac{1}{2} \text{Cos}[2 \tau ] g_2 p_2^2+g_2 p_1 p_3+\text{Cos}[2 \tau ] g_2 p_1 p_3-\frac{5}{6} \text{Sin}[\tau ] g_2^2 p_1^2 q_2+\frac{1}{2} \text{Sin}[3 \tau ] g_2^2 p_1^2 q_2 \\
    & \qquad +\frac{3}{4} \text{Sin}[\tau ] g_3 p_1^2 q_2+\frac{3}{4} \text{Sin}[3 \tau ] g_3 p_1^2 q_2+\text{Sin}[2 \tau ] g_2 p_2 q_2+\frac{1}{2} g_2 q_2^2-\frac{1}{2} \text{Cos}[2 \tau ] g_2 q_2^2+\text{Sin}[2 \tau ] g_2 p_1 q_3+g_2 p_1 \text{$\Delta $p}_1\\
    & \qquad -\frac{1}{3} \text{Cos}[2 \tau ] g_2 p_1 \text{$\Delta $p}_1-\text{Cos}[\tau ] \text{$\Delta $p}_2-\text{Sin}[\tau ] \text{$\Delta $q}_2-\frac{2}{3} \text{Cos}[2 \tau ] g_2 p_1^2 \omega _2-\text{Cos}[\tau ] p_2 \omega _2-\text{Sin}[\tau ] q_2 \omega _2-\text{Cos}[\tau ] p_1 \omega _3+\phi _4[\tau ]+\phi _4''[\tau ] =0
    \end{align}
\end{document}

我尝试拆分方程，但失败了。我可以拆分其他方程，但这不适用于多个方程。我使用过 & \qquad，但不起作用。有人能帮助我吗？

答案1

&对于较长的机械生成的术语，尝试插入和对齐结构是否有意义尚不清楚\\。要真正获得人类可读的输出，您需要将其完全分解，为子术语命名等，另一种方法是让整个过程顺其自然：

在此处输入图片描述

\documentclass{article}
\usepackage{amsmath}
\begin{document}

\begin{flushleft}\linespread{2}\selectfont
$\displaystyle
(\nabla p_1){}^2 g_2-\frac{1}{3} \text{Cos}[2 \tau ] \left(\nabla p_1\right){}^2 g_2-\frac{1}{24} \lambda  \text{Cos}[2 \tau ] g_2 p_1^4-\frac{1}{24} \lambda  \text{Cos}[4 \tau ] g_2 p_1^4+\frac{19}{72} g_2^3 p_1^4-\frac{1}{9} \text{Cos}[2 \tau ] g_2^3 p_1^4+\frac{5}{72} \text{Cos}[4 \tau ] g_2^3 p_1^4 
 -\frac{5}{8} g_2 g_3 p_1^4-\frac{1}{2} \text{Cos}[2 \tau ] g_2 g_3 p_1^4+\frac{1}{8} \text{Cos}[4 \tau ] g_2 g_3 p_1^4+\frac{3}{8} g_4 p_1^4+\frac{1}{2} \text{Cos}[2 \tau ] g_4 p_1^4+\frac{1}{8} \text{Cos}[4 \tau ] g_4 p_1^4-\frac{5}{2} \text{Cos}[\tau ] g_2^2 p_1^2 p_2+\frac{1}{2} \text{Cos}[3 \tau ] g_2^2 p_1^2 p_2 
+\frac{9}{4} \text{Cos}[\tau ] g_3 p_1^2 p_2+\frac{3}{4} \text{Cos}[3 \tau ] g_3 p_1^2 p_2+\frac{1}{2} g_2 p_2^2+\frac{1}{2} \text{Cos}[2 \tau ] g_2 p_2^2+g_2 p_1 p_3+\text{Cos}[2 \tau ] g_2 p_1 p_3-\frac{5}{6} \text{Sin}[\tau ] g_2^2 p_1^2 q_2+\frac{1}{2} \text{Sin}[3 \tau ] g_2^2 p_1^2 q_2 
 +\frac{3}{4} \text{Sin}[\tau ] g_3 p_1^2 q_2+\frac{3}{4} \text{Sin}[3 \tau ] g_3 p_1^2 q_2+\text{Sin}[2 \tau ] g_2 p_2 q_2+\frac{1}{2} g_2 q_2^2-\frac{1}{2} \text{Cos}[2 \tau ] g_2 q_2^2+\text{Sin}[2 \tau ] g_2 p_1 q_3+g_2 p_1 \text{$\Delta $p}_1
-\frac{1}{3} \text{Cos}[2 \tau ] g_2 p_1 \text{$\Delta $p}_1-\text{Cos}[\tau ] \text{$\Delta $p}_2-\text{Sin}[\tau ] \text{$\Delta $q}_2-\frac{2}{3} \text{Cos}[2 \tau ] g_2 p_1^2 \omega _2-\text{Cos}[\tau ] p_2 \omega _2-\text{Sin}[\tau ] q_2 \omega _2-\text{Cos}[\tau ] p_1 \omega _3+\phi _4[\tau ]+\phi _4''[\tau ] =0
\hfill\qquad\refstepcounter{equation}(\theequation)
$
\end{flushleft}

\end{document}

答案2

这个等式中有一些非常长的行。您需要&在第一行前面进行对齐，这样您才能看到以下行：

\documentclass{article}
\usepackage[margin=1in,landscape]{geometry}
\usepackage{amsmath,amssymb}
\pagestyle{empty}
\begin{document}

\begin{align}
\left(\nabla p_1\right){}^2 & g_2-\frac{1}{3} \text{Cos}[2 \tau ] \left(\nabla p_1\right){}^2 g_2-\frac{1}{24} \lambda  \text{Cos}[2 \tau ] g_2 p_1^4-\frac{1}{24} \lambda  \text{Cos}[4 \tau ] g_2 p_1^4+\frac{19}{72} g_2^3 p_1^4-\frac{1}{9} \text{Cos}[2 \tau ] g_2^3 p_1^4+\frac{5}{72} \text{Cos}[4 \tau ] g_2^3 p_1^4 \\
& \qquad -\frac{5}{8} g_2 g_3 p_1^4-\frac{1}{2} \text{Cos}[2 \tau ] g_2 g_3 p_1^4+\frac{1}{8} \text{Cos}[4 \tau ] g_2 g_3 p_1^4+\frac{3}{8} g_4 p_1^4+\frac{1}{2} \text{Cos}[2 \tau ] g_4 p_1^4+\frac{1}{8} \text{Cos}[4 \tau ] g_4 p_1^4-\frac{5}{2} \text{Cos}[\tau ] g_2^2 p_1^2 p_2+\frac{1}{2} \text{Cos}[3 \tau ] g_2^2 p_1^2 p_2 \\
& \qquad +\frac{9}{4} \text{Cos}[\tau ] g_3 p_1^2 p_2+\frac{3}{4} \text{Cos}[3 \tau ] g_3 p_1^2 p_2+\frac{1}{2} g_2 p_2^2+\frac{1}{2} \text{Cos}[2 \tau ] g_2 p_2^2+g_2 p_1 p_3+\text{Cos}[2 \tau ] g_2 p_1 p_3-\frac{5}{6} \text{Sin}[\tau ] g_2^2 p_1^2 q_2+\frac{1}{2} \text{Sin}[3 \tau ] g_2^2 p_1^2 q_2 \\
& \qquad +\frac{3}{4} \text{Sin}[\tau ] g_3 p_1^2 q_2+\frac{3}{4} \text{Sin}[3 \tau ] g_3 p_1^2 q_2+\text{Sin}[2 \tau ] g_2 p_2 q_2+\frac{1}{2} g_2 q_2^2-\frac{1}{2} \text{Cos}[2 \tau ] g_2 q_2^2+\text{Sin}[2 \tau ] g_2 p_1 q_3+g_2 p_1 \text{$\Delta $p}_1\\
& \qquad -\frac{1}{3} \text{Cos}[2 \tau ] g_2 p_1 \text{$\Delta $p}_1-\text{Cos}[\tau ] \text{$\Delta $p}_2-\text{Sin}[\tau ] \text{$\Delta $q}_2-\frac{2}{3} \text{Cos}[2 \tau ] g_2 p_1^2 \omega _2-\text{Cos}[\tau ] p_2 \omega _2-\text{Sin}[\tau ] q_2 \omega _2-\text{Cos}[\tau ] p_1 \omega _3+\phi _4[\tau ]+\phi _4''[\tau ] =0
\end{align}

\end{document}

在此处输入图片描述

此外，我建议将\text{Cos}和分别更改\text{Sin}为\cos和\sin。

最后，这是一个非常长且难以理解的公式。作为读者，我会发现它很难解析。

答案3

Mathematica 的自动输出不能直接插入，需要一些手动工作。

例如，你会发现无用的\left和\right，还有无用的{}和错误的 \text{$\Delta$p}_2$。

我将长等式在加号和减号处分开，并用\begin{multline}和将其包围\end{multline}；然后我\\在可能合适的位置插入以免线条过长。

此外，我\text{Cos}还\text{Sin}使用适当的命令进行了更改。

\documentclass{article}
\usepackage[pass,showframe]{geometry} % just for the example (it shows the margins)
\usepackage{amsmath,amssymb}
\DeclareMathOperator\Cos{Cos}
\DeclareMathOperator\Sin{Sin}

\begin{document}

\begin{multline}
(\nabla p_1)^2 g_2
-\frac{1}{3} \Cos[2\tau] (\nabla p_1)^2 g_2
-\frac{1}{24} \lambda \Cos[2\tau] g_2 p_1^4
-\frac{1}{24} \lambda \Cos[4\tau] g_2 p_1^4
\\
+\frac{19}{72} g_2^3 p_1^4
-\frac{1}{9} \Cos[2\tau] g_2^3 p_1^4
+\frac{5}{72} \Cos[4 \tau ] g_2^3 p_1^4
-\frac{5}{8} g_2 g_3 p_1^4
\\
-\frac{1}{2} \Cos[2\tau] g_2 g_3 p_1^4
+\frac{1}{8} \Cos[4\tau] g_2 g_3 p_1^4
+\frac{3}{8} g_4 p_1^4
+\frac{1}{2} \Cos[2\tau] g_4 p_1^4
\\
+\frac{1}{8} \Cos[4\tau] g_4 p_1^4
-\frac{5}{2} \Cos[\tau] g_2^2 p_1^2 p_2
+\frac{1}{2} \Cos[3 \tau ] g_2^2 p_1^2 p_2
+\frac{9}{4} \Cos[\tau] g_3 p_1^2 p_2
\\
+\frac{3}{4} \Cos[3\tau] g_3 p_1^2 p_2
+\frac{1}{2} g_2 p_2^2
+\frac{1}{2} \Cos[2\tau] g_2 p_2^2
+g_2 p_1 p_3
\\
+\Cos[2 \tau ] g_2 p_1 p_3
-\frac{5}{6} \Sin[\tau] g_2^2 p_1^2 q_2
+\frac{1}{2} \Sin[3\tau] g_2^2 p_1^2 q_2
+\frac{3}{4} \Sin[\tau] g_3 p_1^2 q_2
\\
+\frac{3}{4} \Sin[3\tau] g_3 p_1^2 q_2
+\Sin[2 \tau ] g_2 p_2 q_2
+\frac{1}{2} g_2 q_2^2
-\frac{1}{2} \Cos[2\tau] g_2 q_2^2
\\
+\Sin[2\tau] g_2 p_1 q_3
+g_2 p_1 \Delta p_1
-\frac{1}{3} \Cos[2\tau] g_2 p_1 \Delta p_1
-\Cos[\tau ] \Delta p_2
\\
-\Sin[\tau ] \Delta q_2
-\frac{2}{3} \Cos[2\tau] g_2 p_1^2 \omega _2
-\Cos[\tau ] p_2 \omega _2
-\Sin[\tau ] q_2 \omega _2
\\
-\Cos[\tau ] p_1 \omega _3
+\phi _4[\tau ]
+\phi _4''[\tau ] =0
\end{multline}

\end{document}

在此处输入图片描述

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