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Latex 方程定位问题

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Latex 方程定位问题 
\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{xxx }       
\date{January 26, 2013}                 
\begin{document} \baselineskip=18pt
\section{Introduction}


\section{Reconstruction of the article equation(15) }
Given, $$\phi_1= p_1cos(\tau+\alpha)$$
$$\nabla\phi_1= -p_1sin(\tau+\alpha) \nabla \alpha+\nabla p_1cos(\tau+\alpha)$$
$$\Delta\phi_1= -p_1 \nabla \alpha \cos(\tau+\alpha) \nabla \alpha-p_1 sin(\tau+\alpha) \Delta\alpha -\nabla \alpha \sin(\tau+\alpha)\nabla p_1-\nabla p_1 \nabla \alpha sin(\tau+ \alpha)+\Delta p_1cos(\tau+\alpha)$$
$$\Delta\phi_1= \cos(\tau+\alpha) [-p_1 \Delta \alpha + \Delta p_1]- sin(\tau+\alpha) [p_1\Delta \alpha+2\nabla \alpha \nabla p_1]$$
 Differentiating $\phi_1$,
$$\phi_1= p_1cos(\tau+\alpha)$$
$$\dot\phi_1= -p_1 sin(\tau+\alpha)$$
$$\ddot\phi_1= -p_1cos(\tau+\alpha)$$
Again,
$$\phi_2 = p_2\cos(\tau + \alpha) + q_2\sin(\tau + \alpha) + \frac{g_2}{6}p_1^2[\cos(2\tau + 2\alpha) - 3] $$
$\omega_1=0$ for the bounding conditions.
$$\dot\phi_2 = -p_2\sin(\tau + \alpha) + q_2\cos(\tau + \alpha) + \frac{g_2}{6}p_1^2[-2 \sin(2\tau + 2\alpha) ] $$
$$\ddot\phi_2 = -p_2\cos(\tau + \alpha)- q_2\sin(\tau + \alpha) - \frac{4g_2}{6}p_1^2[\cos(2\tau + 2\alpha) ] $$
Putting these values in equation,
\begin{align*}
 &\ddot\phi_3+\phi_3+2g_2\phi_1\phi_2+g_3\phi_1^3-\ddot\phi_1-\Delta\phi_1
   +\omega_1\ddot\phi_2+\omega_2\ddot\phi_1 =0
\\[\bigskipamount]
&\begin{aligned}
 &\ddot\phi_3+\phi_3+2g_2p_1 \cos(\tau+\alpha)
  [p_2\cos(\tau + \alpha)  + q_2\sin(\tau + \alpha) +
   \frac{g_2}{6}p_1^2[\cos(2\tau + 2\alpha) - 3]] \\
  &\quad{}+g_3p^3_1\cos^3(\tau+\alpha)
    +p_1\cos(\tau+\alpha)-\cos(\tau+\alpha) [-p_1 \Delta \alpha+\Delta p_1] \\
  &\quad {}-\sin(\tau+\alpha) [p_1\Delta \alpha+2\nabla \alpha \nabla p_1]
    +\omega_2p_1\cos(\tau+\alpha)=0 \\
\end{aligned}
\\[\bigskipamount]
&\begin{aligned}
 &\ddot\phi_3+\phi_3+g_2p_1 p_2 [1+\cos2(\tau+\alpha)]
  + g_2p_1 q_2\sin2(\tau+\alpha) \\
 &\quad{} + \frac{2 g_2^2 p_1^3}{6} \cos(\tau +\alpha)[2\cos^2(\tau+\alpha)-4] \\
 &\quad{} + g_3p_1^3[\frac{1}{4}(3\cos(\tau +\alpha)+\cos3(\tau+\alpha))]
          + p_1\cos(\tau+\alpha) \\
 &\quad{} -\cos(\tau+\alpha)[\Delta p_1-p_1 \Delta\alpha]
          +\sin(\tau+\alpha)[p_1\Delta \alpha+2 \nabla p_1 \nabla \alpha]
           \omega_2p_1\cos(\tau+\alpha) =0
\end{aligned}
\\[\bigskipamount]
&\begin{aligned}
 & \ddot\phi_3+\phi_3+g_2p_1 p_2 + g_2p_1 p_2  cos2(\tau+\alpha)] + g_2p_1 q_2sin2(\tau+\alpha)  \\
 &\quad{} + \frac{2 g_2^2 p_1^3}{3} cos^3(\tau +\alpha)-\frac{8 g_2^2 p_1^3}{6} cos(\tau +\alpha)+g_3p_1^3[\frac{1}{4}(3cos(\tau +\alpha)+cos3(\tau+\alpha))] \\
 &\quad{} +p_1cos(\tau+\alpha)-cos(\tau+\alpha)[\Delta p_1-p_1 \Delta \alpha]+sin(\tau+\alpha)[p_1\Delta \alpha+2 \nabla p_1 \nabla \alpha]+\omega_2p_1cos(\tau+\alpha)=0
\end{aligned}
\\[\bigskipamount]
&\begin{aligned}
 & \ddot\phi_3+\phi_3+g_2p_1 p_2 + g_2p_1 p_2  cos2(\tau+\alpha)] + g_2p_1 q_2sin2(\tau+\alpha)  \\
 &\quad{} + \frac{2 g_2^2 p_1^3}{3}[\frac{1}{4}(3cos(\tau +\alpha)+cos3(\tau+\alpha))] -\frac{8 g_2^2 p_1^3}{6} cos(\tau +\alpha)+g_3p_1^3[\frac{1}{4}(3cos(\tau +\alpha)+cos3(\tau+\alpha))] \\
 &\quad{} +p_1cos(\tau+\alpha)-cos(\tau+\alpha)[\Delta p_1-p_1 \Delta\ alpha]+sin(\tau+\alpha)[p_1\Delta \alpha+2 \nabla p_1 \nabla \alpha]+\omega_2p_1cos(\tau+\alpha)=0 \\
\end{aligned}
\\[\bigskipamount]
&\begin{aligned}
 & \ddot\phi_3+\phi_3+sin(\tau+\alpha)[p_1\Delta \alpha+2 \nabla p_1 \nabla \alpha] -cox(\tau+ \alpha)[\nabla p_1\\
 &\quad{} -p_1\nabla \alpha+\frac{5}{6}g_2^2p_1^3- \frac{3}{4}g_3p_1^3-p_1 +\omega_2 p_1] \\
 &\quad{} +\frac{ p_1^3}{12}(2g_2^2+3g_3)cos3(\tau + \alpha)+g_2 p_1 [p_2+ p_2 cos2(\tau +\alpha)+q_2 sin2(\tau+\alpha)] =0\\
\end{aligned}
\\[\bigskipamount]
&\begin{aligned}
 & \ddot\phi_3+\phi_3+sin(\tau+\alpha)[p_1\Delta \alpha+2 \nabla p_1 \nabla \alpha] -cox(\tau+ \alpha)[\Delta p_1-p_1\nabla \alpha \\
 &\quad{} + \lambda p_1^3-p_1+\omega_2 p_1] +\frac{p_1^3}{12}(2g_2^2+3g_3)cos3(\tau + \alpha) \\
 &\quad{} +g_2 p_1 [p_2+ p_2 cos2(\tau +\alpha)+q_2 sin2(\tau+\alpha)]=0
\end{aligned}
\end{align*}
\end{document}

答案1

这是一个建议；我只排版前三个方程，用它们作为其余方程的模型。

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{align*}
&\ddot\phi_3+\phi_3+2g_2\phi_1\phi_2+g_3\phi_1^3-\ddot\phi_1-\Delta\phi_1 
   +\omega_1\ddot\phi_2+\omega_2\ddot\phi_1 =0
\\[\bigskipamount]
&\begin{aligned}
 &\ddot\phi_3+\phi_3+2g_2p_1 \cos(\tau+\alpha)
  [p_2\cos(\tau + \alpha)  + q_2\sin(\tau + \alpha) + 
   \frac{g_2}{6}p_1^2[\cos(2\tau + 2\alpha) - 3]] \\
  &\quad{}+g_3p^3_1\cos^3(\tau+\alpha) -p_1\cos(\tau+\alpha)
    -p_1\cos(\tau+\alpha)-\cos(\tau+\alpha) [-p_1 \Delta \alpha-\Delta p_1] \\
  &\quad {}-\sin(\tau+\alpha) [p_1\Delta \alpha+2\nabla \alpha \nabla p_1] 
    +\omega_2p_1\cos(\tau+\alpha)=0 \\
\end{aligned}
\\[\bigskipamount]
&\begin{aligned}
 &\ddot\phi_3+\phi_3+g_2p_1 p_2 [1+\cos2(\tau+\alpha)] 
  + g_2p_1 p_2\sin2(\tau+\alpha) \\
 &\quad{} + \frac{25g_2^2 p_1^3}{6} \cos(\tau +\alpha)[2\cos(\tau+\alpha-4)] \\
 &\quad{} + g_3p_1^3[\frac{1}{4}(3\cos(\tau +\alpha)+\cos3(\tau+\alpha))] 
          + p_1\cos(\tau+\alpha) \\
 &\quad{} -\cos(\tau+\alpha)[\Delta p_1-p_1 \Delta\alpha]
          +\sin(\tau+\alpha)[p_1\Delta \alpha+2 \nabla p_1 \nabla \alpha]
           \omega_2p_1\cos(\tau+\alpha) =0
\end{aligned}
\end{align*}
\end{document}

在此处输入图片描述

始终使用\cos和\sin来实现功能；您可以通过更改\tau+\alpha为类似或之类的东西来节省空间\hat{\tau}。

我怀疑是否有人真的能读懂这么大的方程式，尤其是当它们都显示在同一屏幕上时。最好将每个方程式视为单独的显示，并在其间添加一些解释性文字。

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