Can anyone solve my error and make the long multiline equation multiline please?

Question

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}

Answer 1

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.

Answer 2

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}

---