Take the 2-minute tour ×
TeX - LaTeX Stack Exchange is a question and answer site for users of TeX, LaTeX, ConTeXt, and related typesetting systems. It's 100% free, no registration required.
\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}
share|improve this question
3  
I do not understand the question; do you ask how to do in technically or how to do it best typographically? –  malin Feb 5 '13 at 9:53
    
@ malin, run the tex file I wrote, then you will understand. –  Complex Guy Feb 5 '13 at 9:57
5  
@Forhad: You'll probably get better answers if you make it as easy as possible for others to help you. That mainly means two things: a) provide a compilable minimal document that shows the issue (i.e. starting from \documentclass, so people can compile the document straight away, without having to fill in the blanks first) b) explain your problem in more detail. Ideally, people wouldn't even need to compile your code first: Attach a screenshot of the ouput and explain why you're not happy with the result. –  Jake Feb 5 '13 at 10:03
    
@Forhad Your input seems to line up correctly for me, in the sense that all of the & work to create alignment. However, the text is far too wide to fit into a standard page width (article class with no alterations). I suspect you want & on lines 3 onward, as with the settings as given all of these lines comes 'before' the alignment point. –  Joseph Wright Feb 5 '13 at 10:06
    
I added the starting , you can run now. –  Complex Guy Feb 5 '13 at 10:24
add comment

1 Answer

up vote 7 down vote accepted

Here's a proposal; I typeset only the first three equations, use them as a model for the remaining ones.

\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}

enter image description here

Always use \cos and \sin for the functions; you can save space by changing \tau+\alpha into something like \hat{\tau} or whatever.

I doubt that anybody can really read such big equations, particularly if they are altogether in the same display. It's probably better to treat each one as a separate display with some explanatory text in between.

share|improve this answer
    
It worked. Thanks –  Complex Guy Feb 5 '13 at 11:41
    
Your suggestion was worked but now it showing little error, can you check please? –  Complex Guy Feb 5 '13 at 16:07
    
@Forhad I get no error, only warnings that some lines are too long; split them. You still have wrong cos and sin –  egreg Feb 5 '13 at 16:14
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.