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In case it matters, I'm using win edit as editor.

\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{xyz }       
\date{January 26, 2013}                 
\begin{document} \baselineskip=18pt
\section{Reconstruction of the article equation(15)}
Given, $$\phi_1= p_1\cos(\tau+\alpha)$$
$$\nabla\phi_1= -p_1\sin(\tau+\alpha) \nabla \alpha+\nabla p_1cos(\tau+\alpha)$$
$$\Delta\phi_1= -p_1 \nabla \alpha \cos(\tau+\alpha) \nabla \alpha-p_1 \sin(\tau+\alpha) \Delta\alpha -\nabla \alpha \sin(\tau+\alpha)\nabla p_1-\nabla p_1 \nabla \alpha \sin(\tau+ \alpha)+\Delta p_1\cos(\tau+\alpha)$$
$$\Delta\phi_1= \cos(\tau+\alpha) [-p_1 \Delta \alpha + \Delta p_1]- \sin(\tau+\alpha) [p_1\Delta \alpha+2\nabla \alpha \nabla p_1]$$
 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 \\
\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]] +g_3p^3_1\cos^3(\tau+\alpha)-\cos(\tau+\alpha) [-p_1 \Delta \alpha+\Delta p_1] -\sin(\tau+\alpha) [p_1\Delta \alpha+2\nabla \alpha \nabla p_1] +\omega_2p_1\cos(\tau+\alpha)=0
\end{align}
\begin{align}
\ddot\phi_3+\phi_3+g_2p_1 p_2 [1+\cos2(\tau+\alpha)] + g_2p_1 q_2\sin2(\tau+\alpha)  + \frac{2 g_2^2 p_1^3}{6} \cos(\tau +\alpha)[2\cos^2(\tau+\alpha)-4]
 + g_3p_1^3[\frac{1}{4}(3\cos(\tau +\alpha)+\cos3(\tau+\alpha))]+ p_1\cos(\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_1\cos(\tau+\alpha) =0
\end{align}
\begin{align}
\ddot\phi_3+\phi_3+g_2p_1 p_2 + g_2p_1 p_2  \cos2(\tau+\alpha)] + g_2p_1 q_2\sin2(\tau+\alpha) + \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}(3\cos(\tau +\alpha)+\cos3(\tau+\alpha))]+p_1\cos(\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_1\cos(\tau+\alpha)=0
\end{align}
\begin{align}
 \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) + \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}(3\cos(\tau +\alpha)+\cos3(\tau+\alpha))]  +p_1\cos(\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_1\cos(\tau+\alpha)=0
\end{align}
\begin{align}
 \ddot\phi_3+\phi_3+sin(\tau+\alpha)[p_1\Delta \alpha+2 \nabla p_1 \nabla \alpha] -\cos(\tau+ \alpha)[\Delata p_1 - p_1\Delata\alpha+\frac{5}{6}g_2^2p_1^3- \frac{3}{4}g_3p_1^3-p_1 +\omega_2 p_1] +\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{align}
\begin{align}
\ddot\phi_3+\phi_3+sin(\tau+\alpha)[p_1\Delta \alpha+2 \nabla p_1 \nabla \alpha] -\cos(\tau+ \alpha)[\Delta p_1-p_1\Delta \alpha \\ + \lambda p_1^3-p_1+\omega_2 p_1] +\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{align}

\end{document}
share|improve this question
    
Not exactly a MWE... Simply look at the error log and you'll see that there are typos in your code (\Delata instead of \Delta). –  Jubobs Mar 1 '13 at 8:53
    
Mr.Downvoter please mention why? give him benefit of doubt and help him with links to minimal working example (MWE) that illustrates your problem. It will be much easier for us to reproduce your situation and find out what the issue is when we see compilable code, starting with \documentclass{...} and ending with \end{document}. –  texenthusiast Mar 1 '13 at 8:57
    
Aobut the error: you wrote \Delata instead of \Delta. About multiline: you have examples in the amslatex manual (online here for example). See section 3.3 or 3.4 –  JLDiaz Mar 1 '13 at 8:57
    
@Forhad Try Buildingup or hacking down approach step by step each section of the .tex document and compile to diagnose the error,later start developing from simple example. –  texenthusiast Mar 1 '13 at 9:00
    
Some tips for your next paper: (1) use the geometry-package for setting the page margins; (2) If you want to set lengths manually (baselineskip for example), use the LaTeX \setlength{\<command name>}{<length>} – the way you did could cause errors in combination with other packages; (3) use a 'real' environment like align, gather, etc. instead of $$ –  ralfix Mar 1 '13 at 9:21
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2 Answers

up vote 4 down vote accepted

As mentioned in the comments, look out for typos in macro names (in your case \delata instead of \delta). You can use the split environment to introduce line breaks in a long equation. Here is a working (if not very pleasing to the eye) example with one of you long equations:

\documentclass[12pt,a4paper]{article}
\usepackage{amsmath}
\usepackage{amssymb}

\begin{document}

\begin{equation}
\begin{split}
    & \ddot\phi_3+\phi_3+sin(\tau+\alpha)[p_1\Delta \alpha+2 \nabla p_1 \nabla \alpha] \\
    & \qquad -\cos(\tau+ \alpha)[\Delta p_1 - p_1\Delta\alpha+\frac{5}{6}g_2^2p_1^3- \frac{3}{4}g_3p_1^3-p_1 +\omega_2 p_1] 
    +\frac{ p_1^3}{12}(2g_2^2+3g_3)\cos3(\tau + \alpha) \\
    & \qquad +g_2 p_1 [p_2+ p_2 cos2(\tau +\alpha)+q_2 \sin2(\tau+\alpha)] =0
\end{split}
\end{equation}

\end{document}

Marc Van Dongen (LaTeX & Friends) recommends to put a linebreak:

  • before a + or - sign, and a \qquad on the next line,
  • before a × sign, and a \qquad on the next line.
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Your code suffers from a number of problems (in addition to having grossly overlong lines):

  • Typos such as \Delata in several instances, which throw immediate errors. (There are probably further math errors in the code, as opposed to TeX syntax errors.)
  • Omission of \ before sin and cos in quite a few cases
  • Use of Plain-TeX $$ directives instead of something that's more robust/stable
  • The equations in the align group don't seem to feature any alignment points
  • Long strings of math manipulations without connecting phrases or explanations what's going on. Unless your intended readership already knows exactly what's going on, you should consider providing a bit more "connecting tissue" to signal to the readers where the argument is going.

The following code does the following:

  • Consistent use of align and align* environments
  • The long group of numbered aligned equations is "aligned" as follows: the first term of each equation sticks out to the left by a small amount, and subsequent lines are aligned on the first + or - operators
  • I've replaced the [ and ] brackets with \bigl[ and \bigr] directives
  • Minor stuff: (i) Setting \allowdisplaybreaks allows LaTeX to insert a page break where necessary; (ii) Some \phantom{-} instructions to get better alignment of some of the equations' right-hand sides; (iii) Fix syntax errors and insert missing \ before "sin" and "cos"; (iv) use of the geometry package in lieu of a slew of TeX primitives to set up the page geometry; (v) use of the setspace package in lieu of fiddling with the low-level \baselineskip parameter; (vi) suggestions for "connective tissue" to help the reader(s) along.

There are almost certainly better ways to choose some of the line breaks, but the code below will hopefully serve as a good jumping-off point.

\documentclass[12pt,a4paper]{article}
\usepackage{amsmath}
\allowdisplaybreaks
\usepackage{amssymb}

%\textwidth=16.5cm  \oddsidemargin=-0.10cm \evensidemargin=-0.10cm  \topmargin=-1.0cm \textheight=24.5cm
\usepackage[margin=2cm]{geometry}

\newcommand{\piRsquare}{\pi r^2}        
\usepackage{setspace}
\setstretch{1.15}

\begin{document} 
%\section{Reconstruction of the article equation(15)}
Given:
\begin{align*}
\phi_1 &= p_1\cos(\tau+\alpha)\\
\nabla\phi_1 &= -p_1\sin(\tau+\alpha) \nabla \alpha+\nabla p_1\cos(\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\\
&\qquad{}-\nabla p_1 \nabla \alpha \sin(\tau+ \alpha)+\Delta p_1\cos(\tau+\alpha)\\
\Delta\phi_1&= \cos(\tau+\alpha) \bigl[-p_1 \Delta \alpha + \Delta p_1\bigr]- 
    \sin(\tau+\alpha) \bigl[p_1\Delta \alpha+2\nabla \alpha \nabla p_1\bigr]
\end{align*}

Differentiating $\phi_1$ [with respect to $\tau$]:
\begin{align*}
\phi_1     &= \phantom{-}p_1\cos(\tau+\alpha)\\
\dot\phi_1 &= -p_1 \sin(\tau+\alpha)\\
\ddot\phi_1&= -p_1\cos(\tau+\alpha)
\end{align*}

Again,
\begin{align*}
\phi_2       &= \phantom{-}p_2\cos(\tau + \alpha) + 
  q_2\sin(\tau + \alpha) + 
  \frac{g_2}{6}p_1^2\bigl[\cos(2\tau + 2\alpha) - 3\bigr]
\intertext{$\omega_1=0$ for the bounding conditions.}
\dot\phi_2   &= -p_2\sin(\tau + \alpha) + q_2\cos(\tau + \alpha) + 
    \frac{g_2}{6}p_1^2\bigl[-2 \sin(2\tau + 2\alpha) \bigr] \\
\ddot\phi_2  &= -p_2\cos(\tau + \alpha)- q_2\sin(\tau + \alpha) - 
    \frac{4g_2}{6}p_1^2\bigl[\cos(2\tau + 2\alpha) \bigr] 
\end{align*}

Putting these values in equation [xx] [leads to the following first-order conditions (??)]:
\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\\
\ddot\phi_3{}+{}&\phi_3+2g_2p_1 \cos(\tau+\alpha)\bigl[p_2\cos(\tau + \alpha) + 
    q_2\sin(\tau + \alpha)\bigr] \notag\\
{}+{}&\frac{g_2}{6}p_1^2\bigl[\cos(2\tau + 2\alpha) - 3\bigr]\bigr]  +g_3p^3_1\cos^3 
    (\tau+\alpha)-\cos(\tau+\alpha) \bigl[-p_1 \Delta \alpha+\Delta p_1\bigr] \notag\\
{}-{}&\sin(\tau+\alpha) \bigl[p_1\Delta \alpha+2\nabla \alpha \nabla p_1\bigr] 
    +\omega_2p_1\cos(\tau+\alpha)=0\\
\ddot\phi_3{}+{}&+\phi_3+g_2p_1 p_2 \bigl[1+\cos2(\tau+\alpha)\bigr] + 
    g_2p_1 q_2\sin2(\tau+\alpha)  \notag\\
{}+{}&\frac{2 g_2^2 p_1^3}{6} \cos(\tau +\alpha)\bigl[2\cos^2(\tau+\alpha)-4\bigr]
    + g_3p_1^3\bigl[\frac{1}{4}(3\cos(\tau +\alpha)+\cos3(\tau+\alpha))\bigr] \notag\\
{}+{}&p_1\cos(\tau+\alpha) -\cos(\tau+\alpha)\bigl[\Delta p_1-p_1 \Delta\alpha\bigr] 
    +\sin(\tau+\alpha)\bigl[p_1\Delta \alpha+2 \nabla p_1 \nabla \alpha\bigr]\notag\\
{}+{}&\omega_2p_1\cos(\tau+\alpha) =0\\
\ddot\phi_3{}+{}&\phi_3+g_2p_1 p_2 + g_2p_1 p_2  \cos2(\tau+\alpha)\bigr] + 
    g_2p_1 q_2\sin2(\tau+\alpha) \notag\\
{}+{}&\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)\notag\\
{}+{}&g_3p_1^3\bigl[\frac{1}{4}(3\cos(\tau +\alpha) +\cos3(\tau+\alpha))\bigr]+
    p_1\cos(\tau+\alpha)\notag\\
{}-{}&\cos(\tau+\alpha)\bigl[\Delta p_1-p_1 \Delta \alpha\bigr] +\sin(\tau+\alpha)
    \bigl[p_1\Delta \alpha+2 \nabla p_1 \nabla \alpha\bigr]\notag\\
{}+{}&\omega_2p_1\cos(\tau+\alpha)=0\\
\ddot\phi_3{}+{}&\phi_3+g_2p_1 p_2 + g_2p_1 p_2  \cos2(\tau+\alpha)\bigr] + 
    g_2p_1 q_2\sin2(\tau+\alpha) \notag\\
{}+{}&\frac{2 g_2^2 p_1^3}{3}\bigl[\frac{1}{4}(3\cos(\tau +\alpha)+\cos3
    (\tau+\alpha))\bigr] -\frac{8 g_2^2 p_1^3}{6} \cos(\tau +\alpha)\notag\\
{}+{}&g_3p_1^3\bigl[\frac{1}{4}(3\cos(\tau +\alpha)+ \cos3(\tau+\alpha))\bigr]  
    +p_1\cos(\tau+\alpha)\notag\\ 
{}-{}&\cos(\tau+\alpha)\bigl[ \Delta p_1 -p_1 \Delta \alpha\bigr]
+\sin(\tau+\alpha)\bigl[p_1\Delta \alpha+2 \nabla p_1 \nabla \alpha\bigr]\notag\\
{}+{}&\omega_2p_1\cos(\tau+\alpha)=0\\
\ddot\phi_3{}+{}&\phi_3+\sin(\tau+\alpha)\bigl[p_1\Delta \alpha+2 \nabla p_1 
    \nabla \alpha\bigr] 
-\cos(\tau+ \alpha)\bigl[\Delta p_1 - p_1\Delta\alpha\notag\\
{}+{}&\frac{5}{6}g_2^2p_1^3 - \frac{3}{4}g_3p_1^3-p_1 
+\omega_2 p_1\bigr] +\frac{ p_1^3}{12}(2g_2^2+3g_3)\cos3(\tau + \alpha)\notag\\
{}+{}&g_2 p_1 \bigl[p_2+ p_2 \cos2(\tau +\alpha)+q_2 \sin2(\tau+\alpha)\bigr] =0\\
\ddot\phi_3{}+{}&\phi_3+\sin(\tau+\alpha)\bigl[p_1\Delta \alpha+2 \nabla p_1 
    \nabla \alpha\bigr] \notag\\
{}-{}&\cos(\tau+ \alpha)\bigl[\Delta p_1-p_1\Delta \alpha + \lambda p_1^3-p_1+
    \omega_2 p_1\bigr] \notag\\
{}+{}&\frac{p_1^3}{12}(2g_2^2+3g_3)\cos3(\tau + \alpha)+g_2 p_1 \bigl[p_2+ p_2 
    \cos2(\tau +\alpha)\notag\\
{}+{}&q_2 \sin2(\tau+\alpha)\bigr]=0
\end{align}

\end{document}

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