我在子方程模式中具有以下方程组,我想将其中一些长表达式拆分为 2-3 行:
\documentclass[12pt]{report}
\usepackage{amsmath}
\begin{document}
\begin{subequations}
\begin{align}
\frac{\mathrm{d} h}{\mathrm{d} t} &= -\frac{3}{2} \frac{J_2\mu R^2}{r^3} \sin^2i \sin 2u \\
\frac{\mathrm{d} e}{\mathrm{d} t} &= \frac{3}{2} \frac{J_2\mu R^2}{h r^3} \left \{\frac{h^2}{\mu r}\sin \theta \left ( 3\sin^2i \sin 2u-1 \right ) -\sin 2u \sin^2i \left [ \left ( 2+e\cos \theta\right ) \cos\theta +e \right ]\right \} \\
\frac{\mathrm{d} \theta}{\mathrm{d} t} &=\frac{h}{r^2} +\frac{3}{2} \frac{J_2\mu R^2}{e h r^3} \left \{\frac{h^2}{\mu r}\cos \theta \left ( 3\sin^2i \sin 2u-1 \right ) + \left ( 2+e\cos \theta \right ) \sin 2u \sin^2i \sin \theta \right \} \\
\frac{\mathrm{d} \Omega}{\mathrm{d} t} &= -3 \frac{J_2\mu R^2}{h r^3} \sin^2i \cos i \\
\frac{\mathrm{d} i}{\mathrm{d} t} &= -\frac{3}{4} \frac{J_2\mu R^2}{h r^3} \sin 2u \sin 2i \\
\frac{\mathrm{d} \omega}{\mathrm{d} t} &= \frac{3}{2} \frac{J_2\mu R^2}{e h r^3} \left \{\frac{h^2}{\mu r}\cos \theta \left ( 1-3\sin^2 i \sin^2 u \right )-\left ( 2+e\cos \theta\right )\sin 2u \sin^2i \sin \theta +2e\cos^2 i \sin^2 u\right \}
\end{align}
\label{J2_gauss}
\end{subequations}
\end{document}
我跟着这个帖子但每次我进入拆分模式时,都会遇到有关匹配括号的错误。
答案1
目前无法引入换行符的原因在于您过度使用了\left
and \right
:这些指令必须成对出现,并且不允许跨行拆分。我建议您 (a) 删除所有\left
and\right
指令 — 10 个实例中有 7 个是不需要的,并且 (b) 使用\Bigl
and调整 3 对花括号的大小。然后就可以直接引入合适的换行符;不要忘记同时\Bigr
引入三个指令。\notag
\documentclass{article}
\usepackage[margin=1in]{geometry}
\usepackage{amsmath}
\counterwithin{equation}{section}
\DeclareMathOperator{\dee}{d\mskip-1.5mu}
\newcommand\diff[2]{\frac{\dee #1}{\dee #2}}
\begin{document}
\setcounter{section}{4} % just for this example
\setcounter{equation}{12}
\begin{subequations}
\begin{align}
\diff{h}{t}
&= -\frac{3}{2}\frac{J_2\mu R^2}{r^3}\sin^2 i\sin 2u \\
\diff{e}{t}
&= \frac{3}{2} \frac{J_2\mu R^2}{h r^3} \Bigl\{ \frac{h^2}{\mu r}\sin\theta (3\sin^2i\sin 2u-1) \notag\\
&\qquad -\sin 2u\sin^2i [ (2+e\cos\theta)\cos\theta +e ] \Bigr\} \\
\diff{\theta}{t}
&=\frac{h}{r^2} +\frac{3}{2} \frac{J_2\mu R^2}{e h r^3}
\Bigl\{ \frac{h^2}{\mu r}\cos\theta (3\sin^2i\sin 2u-1) \notag\\
&\qquad +(2+e\cos\theta)\sin 2u\sin^2i\sin\theta \Bigr\} \\
\diff{\Omega}{t}
&= -3 \frac{J_2\mu R^2}{h r^3}\sin^2i\cos i \\
\diff{i}{t}
&= -\frac{3}{4} \frac{J_2\mu R^2}{h r^3}\sin 2u\sin 2i \\
\diff{\omega}{t}
&= \frac{3}{2} \frac{J_2\mu R^2}{e h r^3}
\Bigl\{\frac{h^2}{\mu r}\cos\theta (1-3\sin^2 i\sin^2 u) \notag\\
&\qquad -(2+e\cos\theta)\sin 2u\sin^2i\sin\theta +2e\cos^2 i\sin^2 u \Bigr\}
\end{align}
\label{J2_gauss}
\end{subequations}
\end{document}