如果您能帮助我重新排列以下内容以获得更好的形式来显示这个长等式,我将不胜感激。
\documentclass[preprint,12pt,authoryear]{elsarticle}
\rho \epsilon {\bf \ddot u_{0}(v},t)=\epsilon \int_{\mathcal{R}} f(\eta_{00},\xi_{00})dA'+\\
\frac{1}{2} \epsilon^2 \int_{\mathcal{R}} \left( \big [\frac{\partial \bf f}{\partial \eta}(\eta_{00},\xi_{00}) \big]{\bf a'(v},t)+\big [\frac{\partial \bf f}{\partial \xi}(\eta_{00},\xi_{00})]k \right )dA'
答案1
您可以使用align*
环境来显示两线方程。
请注意,我放大了第二行的方括号。我还{\bf ...}
用指令替换了各种构造\mathbf{...}
。
\documentclass[preprint,12pt,authoryear]{elsarticle}
\usepackage{amsmath} % for 'align*' environment
\begin{document}
\begin{align*}
\rho \epsilon \ddot{\mathbf{u}}_{0}(\mathbf{v},t)
&=\epsilon \int_{\mathcal{R}} f(\eta_{00},\xi_{00})\,dA'\\
&\quad+\frac{1}{2} \epsilon^2 \int_{\mathcal{R}}
\biggl( \biggl[ \frac{\partial \mathbf{f}}{\partial \eta}(\eta_{00},\xi_{00}) \biggr]\mathbf{a}'(\mathbf{v},t)+
\biggl[ \frac{\partial \mathbf{f}}{\partial \xi}(\eta_{00},\xi_{00})\biggr]k \biggr) \,dA'
\end{align*}
\end{document}
答案2
如果你还想要参考,你可以使用split
提供的环境数学内部equation
:
\documentclass[preprint,12pt,authoryear]{elsarticle}
\usepackage{amsmath}
\begin{document}
\begin{equation}\label{e:myeqn}
\begin{split}
\rho \epsilon \mathbf{\ddot u_{0}(v},t)={} &\epsilon \int_{\mathcal{R}} f(\eta_{00},\xi_{00})dA'+\\
&\frac{1}{2} \epsilon^2 \int_{\mathcal{R}} \left( \bigg [\frac{\partial \mathbf{f} }{\partial \eta}(\eta_{00},\xi_{00}) \bigg]\mathbf{a'(v},t)+\bigg [\frac{\partial \mathbf{f}}{\partial \xi}(\eta_{00},\xi_{00})\bigg]k \right )dA'
\end{split}
\end{equation}
We can now refer to equation \eqref{e:myeqn}.
\end{document}
编辑以纳入 Mico 和daleif 的更正。