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}
答案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}
答案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}
答案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}