如何对齐方程式的这些行以防止第二行超出页面？我尝试了其他帖子中建议的方法

Question

以下是此部分的代码：

\begin{equation}
\begin{split}
\frac{\partial \varphi}{\partial z} & =  \frac{1}{2} \frac{\rho_{obj}a^{2}}{3M}(\xi_{0}(\xi_{0}^{2}-1) - \xi_{core}(\xi_{0}^{2}-1)) \cdot \Bigg\{\Bigg[\Bigg(3\eta \bigg(\frac{3\xi^2 - 1}{4}\ln\bigg(\frac{\xi + 1}{\xi - 1}\bigg)\Bigg) - \frac{3\xi}{2}\bigg) \cdot 
\\
& \phantom{ {}= } \frac{1}{2}\Bigg(\frac{z+a}{\sqrt{x^2 + y^2 + (z+a)^2}} - \frac{z-a}{\sqrt{x^2 + y^2 + (z-a)^2}}\Bigg)\Bigg] + \Bigg[\Bigg(-\frac{1}{\xi^2 - 1} - \frac{1}{2}(3\eta^2-1)\frac{6\xi\ln\bigg(\frac{\xi + 1}{\xi - 1}\bigg) - \frac{2(3\xi^2 - 1)}{\xi^2 - 1}}{4} - \frac{3}{2}\Bigg) \cdot
\\
& \phantom{ {}= } \frac{1}{2}\Bigg(\frac{z+a}{\sqrt{x^2 + y^2 + (z+a)^2}} + \frac{z-a}{\sqrt{x^2 + y^2 + (z-a)^2}}\Bigg)\Bigg]\Bigg\}.
\label{eqn:eq83}
\end{split}
\end{equation}  

Answer 1

我将使用包multlined提供的环境mathtools，它是 的扩展amsmath。

\documentclass{article}
\usepackage{mathtools}

\begin{document}
\begin{equation}
\begin{split}
\frac{\partial \varphi}{\partial z} & =  \frac{1}{2} \frac{\rho_{obj}a^{2}}{3M}(\xi_{0}(\xi_{0}^{2}-1) - \xi_{core}(\xi_{0}^{2}-1)) \cdot \Bigg\{\Bigg[\Bigg(3\eta \bigg(\frac{3\xi^2 - 1}{4}\ln\bigg(\frac{\xi + 1}{\xi - 1}\bigg)\Bigg) - \frac{3\xi}{2}\bigg) \cdot 
\\
& 
\begin{multlined}
  \phantom{ {}= } \frac{1}{2}\Bigg(\frac{z+a}{\sqrt{x^2 + y^2 + (z+a)^2}} 
  - \frac{z-a}{\sqrt{x^2 + y^2 + (z-a)^2}}\Bigg)\Bigg] \\
  + \Bigg[\Bigg(-\frac{1}{\xi^2 - 1}
  - \frac{1}{2}(3\eta^2-1)\frac{6\xi\ln\bigg(\frac{\xi + 1}{\xi - 1}\bigg) 
  - \frac{2(3\xi^2 - 1)}{\xi^2 - 1}}{4} - \frac{3}{2}\Bigg) \cdot
\end{multlined}
\\
& \phantom{ {}= } \frac{1}{2}\Bigg(\frac{z+a}{\sqrt{x^2 + y^2 + (z+a)^2}} + \frac{z-a}{\sqrt{x^2 + y^2 + (z-a)^2}}\Bigg)\Bigg]\Bigg\}.
\label{eqn:eq83}
\end{split}
\end{equation}  
\end{document}

Answer 2

我会将其分成五行并删除一层的支撑。

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\begin{equation}
\label{eqn:eq83}
\begin{split}
\frac{\partial \varphi}{\partial z}
&= \frac{1}{2}\frac{\rho_{obj}a^{2}}{3M}(\xi_{0}(\xi_{0}^{2}-1)
   - \xi_{\mathrm{core}}(\xi_{0}^{2}-1)) \\
&\qquad \cdot \biggl[
   \begin{aligned}[t]
   &\biggl(3\eta \biggl(\frac{3\xi^2 - 1}{4}\ln\biggl(\frac{\xi + 1}{\xi - 1}\biggr)\biggr)
    - \frac{3\xi}{2}\biggr)\\
   & \cdot \frac{1}{2}
     \biggl(\frac{z+a}{\sqrt{x^2+y^2+(z+a)^2}}-\frac{z-a}{\sqrt{x^2+y^2+(z-a)^2}}\biggr)
   \\
   &+\bigg(-\frac{1}{\xi^2 - 1} - \frac{1}{2}(3\eta^2-1)
       \frac{6\xi\ln\bigl(\frac{\xi + 1}{\xi - 1}\bigr) - \frac{2(3\xi^2 - 1)}{\xi^2 - 1}}{4}
    - \frac{3}{2}\biggr)
   \\
   &\cdot\frac{1}{2}\biggl(\frac{z+a}{\sqrt{x^2+y^2+(z+a)^2}}+\frac{z-a}{\sqrt{x^2+y^2+(z-a)^2}}\biggr)
   \biggr].
   \end{aligned}
\end{split}
\end{equation}

\end{document}

Answer 3

最近发布的“autobreak”软件包可能对你有帮助，下面是用法：

    \documentclass{book}
\usepackage{amsmath,autobreak}
\begin{document}

\begin{align}
\everyafterautobreak{\quad}
\begin{autobreak}
\frac{\partial \varphi}{\partial z}
= \frac{1}{2} \frac{\rho_{obj}a^{2}}{3M}(\xi_{0}(\xi_{0}^{2}-1) - \xi_{core}(\xi_{0}^{2}-1))
\cdot \Bigg\{\Bigg[\Bigg(3\eta \bigg(\frac{3\xi^2 - 1}{4}\ln\bigg(\frac{\xi + 1}{\xi - 1}\bigg)\Bigg) - \frac{3\xi}{2}\bigg)
\cdot \frac{1}{2}\Bigg(\frac{z+a}{\sqrt{x^2 + y^2 + (z+a)^2}} - \frac{z-a}{\sqrt{x^2 + y^2 + (z-a)^2}}\Bigg)\Bigg]
+ \Bigg[\Bigg(-\frac{1}{\xi^2 - 1} - \frac{1}{2}(3\eta^2-1)\frac{6\xi\ln\bigg(\frac{\xi + 1}{\xi - 1}\bigg) - \frac{2(3\xi^2 - 1)}{\xi^2 - 1}}{4} - \frac{3}{2}\Bigg)
\cdot \frac{1}{2}\Bigg(\frac{z+a}{\sqrt{x^2 + y^2 + (z+a)^2}} + \frac{z-a}{\sqrt{x^2 + y^2 + (z-a)^2}}\Bigg)\Bigg]\Bigg\}.
\end{autobreak}\label{eqn:eq83}
\end{align}  
\end{document}