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方程式格式（对齐）

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方程式格式（对齐）

根据 问题我已经写入文件，但出现了错误。您能检查一下吗？

\documentclass{article}
\usepackage{amsmath}

\begin{document}
\begin{align}
&\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]]+g_3p^3_1\cos^3(\tau+\alpha)-\cos(\tau+\alpha) [-p_1 (\nabla \alpha)^2+\Delta p_1]\\
&-\sin(\tau+\alpha) [p_1\Delta \alpha+2\nabla \alpha \nabla p_1]\qquad+\omega_2p_1\cos(\tau+\alpha)=0
\end{aligned}\nonumber\\
&\begin{aligned}
& \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))]\\
  & -\cos(\tau+\alpha)\left[\Delta p_1--p_1 (\nabla \alpha)^2\right]  \\
  & +\sin(\tau+\alpha)\left[p_1\Delta \alpha+2 \nabla p_1 \nabla \alpha\right]\\
 & + \omega_2p_1\cos(\tau+\alpha) =0
\end{aligned}\nonumber\\
&\begin{aligned}
& \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))]\\
 &  -\cos(\tau+\alpha)[\Delta p_1--p_1 (\nabla \alpha)^2] \\
 & +\sin(\tau+\alpha)\left[p_1\Delta \alpha+2 \nabla p_1 \nabla \alpha\right]+\omega_2p_1\cos(\tau+\alpha)=0
\end{aligned}\\
&\begin{aligned}
 & \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}[\frac{1}{4}(3\cos(\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))] \\
 &  -\cos(\tau+\alpha)[\Delta p_1-p_1 (\nabla \alpha)^2] \\
 &+\sin(\tau+\alpha)[p_1\Delta \alpha+2 \nabla p_1 \nabla \alpha]+\omega_2p_1\cos(\tau+\alpha)=0
\end{aligned} \\
&\begin{aligned}
    &\ddot\phi_3&+\phi_3+\sin(\tau+\alpha)[p_1\Delta \alpha+2 \nabla p_1 \nabla \alpha] \\
    & -\cos(\tau+ \alpha)\left[\Delta p_1 -p_1 (\nabla \alpha)^2+\frac{5}{6}g_2^2p_1^3- \frac{3}{4}g_3p_1^3-p_1 +\omega_2 p_1\right] \\
    & +\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}\nonumber\\
&\begin{aligned}\label{eq10}
   & \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 (\nabla \alpha)^2 + \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{aligned}
\end{align}
\end{document}

答案1

align您可能读过一些有关和环境的内容aligned。对齐非常简单。&告诉公式在哪里对齐。

\documentclass{article}
\usepackage{amsmath}

\begin{document}
\begin{align}
    &\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]]+g_3p^3_1\cos^3(\tau+\alpha) \\
                &-\cos(\tau+\alpha) [-p_1 (\nabla \alpha)^2+\Delta p_1] \\
                &-\sin(\tau+\alpha) [p_1\Delta \alpha+2\nabla \alpha \nabla p_1]+\omega_2p_1\cos(\tau+\alpha)=0
     \end{aligned} \nonumber \\
    &\begin{aligned}
     \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))] \\
                &-\cos(\tau+\alpha)\left[\Delta p_1--p_1 (\nabla \alpha)^2\right]  \\
                &+\sin(\tau+\alpha)\left[p_1\Delta \alpha+2 \nabla p_1 \nabla \alpha\right]\\
                &+\omega_2p_1\cos(\tau+\alpha)=0
     \end{aligned} \nonumber \\
    &\begin{aligned}
     \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))] \\
                &-\cos(\tau+\alpha)[\Delta p_1--p_1 (\nabla \alpha)^2] \\
                &+\sin(\tau+\alpha)\left[p_1\Delta \alpha+2 \nabla p_1 \nabla \alpha\right]+\omega_2p_1\cos(\tau+\alpha)=0
     \end{aligned} \\
    &\begin{aligned}
     \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}[\frac{1}{4}(3\cos(\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))] \\
                &-\cos(\tau+\alpha)[\Delta p_1-p_1 (\nabla \alpha)^2]+\sin(\tau+\alpha)[p_1\Delta \alpha+2 \nabla p_1 \nabla \alpha] \\
                &+\omega_2p_1\cos(\tau+\alpha)=0
     \end{aligned} \\
    &\begin{aligned}
     \ddot\phi_3&+\phi_3+\sin(\tau+\alpha)[p_1\Delta \alpha+2 \nabla p_1 \nabla \alpha] \\
                &-\cos(\tau+ \alpha)\left[\Delta p_1 -p_1 (\nabla \alpha)^2+\frac{5}{6}g_2^2p_1^3- \frac{3}{4}g_3p_1^3-p_1 +\omega_2 p_1\right] \\
                &+\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} \nonumber \\
    &\begin{aligned} \label{eq10}
     \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 (\nabla \alpha)^2 + \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{aligned}
\end{align}
\end{document}

您的示例（垂直方向）大于一页。您可能需要将其拆分。例如两次：

\begin{align}
&\begin{aligned}
\end{aligned}\nonumber\\
&\begin{aligned}
\end{aligned}
\end{align}
%
\begin{align}
&\begin{aligned}
\end{aligned}\\
&\begin{aligned}
\end{aligned}
\end{align}

编辑

正如卡尔科勒\allowdisplaybreaks[1]，如果将其放入你的序言中，那么所有这些公式的对齐都是可能的。

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