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\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{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 $\phi_1$,
$$\phi_1= p_1cos(\tau+\alpha)$$
$$\dot\phi_1= -p_1 sin(\tau+\alpha)$$
$$\ddot\phi_1= -p_1cos(\tau+\alpha)$$
$$\phi_2 = p_2\cos(\tau + \alpha) + q_2\sin(\tau + \alpha) + \frac{g_2}{6}p_1^2[\cos(2\tau + 2\alpha) - 3] $$
$\omega_1=0$ 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,
   +\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]] \\
    +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 \\
 &\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
 & \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
 & \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 \\
 & \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\\
 & \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
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closed as too localized by Andrew Swann, lockstep, bloodworks, ienissei, Seamus Feb 6 '13 at 9:38

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Please, reduce the code to a minimal working example (MWE), you may even find the error yourself then. Also, don't use $$ ...$$ (tex.stackexchange.com/q/503). For several consecutive equations, use gather or align. – Torbjørn T. Feb 5 '13 at 13:28
Is the problem solved? – Complex Guy Feb 5 '13 at 13:32
You can't use \\ between \left[ and \right]. That is atleast one problem I saw. – xfoo Feb 5 '13 at 13:34
I don't think there is a major problem in my tex file. Just look for a while please. – Complex Guy Feb 5 '13 at 13:35
@All: Please don't downvote below a score of -1, even if the question in it's current form needs some improvement. A score of -1 is enough to show that the question needs work, anything below that is of no use. Also, if you downvote, please leave a comment explaining why you did so. – Joseph Wright Feb 5 '13 at 19:23

You need to add \right. and \left. when you want to have a parenthesis that spans several lines:

&-Acosx- Bsinx -9Ccos3x-9Dsin3x-4Ecos2x-4Fsin2x+Acosx+ Bsinx +Ccos3x+Dsin3x\\
 &\quad{}+Ecos2x+Fsin2x+G+(p_1\Delta\alpha+2\nabla\alpha\nabla p_1)\sin(\tau+\alpha)-\left[\Delta p_1+\omega_2p_1+\lambda p_1^3+\right.\\
&\left.\quad{}-p_1(\nabla\alpha)^2\right]\cos(\tau+\alpha)+\frac{1}{12}p_1^3(2g_2^2+3g_3)\cos(3\tau+3\alpha)+g_2p_1\left[q_2\sin(2\tau+2\alpha)+p_2\cos(2\tau+2\alpha)+p_2\right] =0
share|improve this answer
This doesn't necessarily lead to equal sized parenthesis, as the scaling only takes into account the symbols on the current line. May be better to use \biggl and \biggr (or one of the similar commands) instead of \left and \right. – Torbjørn T. Feb 5 '13 at 14:05
To be honest I didn't look at the output after "fixing" the compilation error. Now that I have, I have to say the size of his parenthesis is probably the least of his problems, the equation is not just reaching into the margins, it's actually going on well beyond the end of the page for quite a number of equations. Lot of work to be done there. – myrtille Feb 5 '13 at 14:27

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