\documentclass{beamer}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{mathtools}
\begin{frame}
\begin{subequations}
\begin{align}
\mathcal{L}&=\left(\mathcal{T}-\mathcal{V}\right)+\sum_{i=1}^{n}\lambda_{i}L\left(\psi(\zeta_{i})-\frac{L_{i}}{L}\chi_{i}(1)\right) \nonumber \\
\mathcal{\bar{L}}&=\left(\frac{\beta^4}{2}\int_{0}^{1}\,\hspace{-0.05in} \left( \psi(\eta)\right)^2 \diff\eta + \sum_{i=1}^{n} \frac{\gamma_{i}}{2}\, \int_{0}^{1} \hspace{-0.05in} \left( \chi_{i}(\xi)\right)^2 \diff\xi \right)-\nonumber \\ &\quad \left(\frac{1}{2}\, \int_{0}^{1} \hspace{-0.05in} \left(\frac{\partial^2\psi(\eta)}{\partial \eta^2}\right)^2 \diff\eta +\sum_{i=1}^{n}\frac{\alpha_{i}}{2}\, \int_{0}^{1}\left(\frac{\partial \chi_{i}(\xi)}{\partial \xi}\right)^2 \diff\xi \right)+\nonumber \\ &\quad \sum_{i=1}^{n}\lambda_{i}^{*}\left(\psi(\zeta_{i})-\frac{L_{i}}{L}\chi_{i}(1)\right)
\end{align}
\end{subequations}
\end{frame}
答案1
完成并稍微简化了代码后,我建议使用中等大小的分数作为nccmath
分数系数,\hspace{-0.05in}
用简单的\negthinspace
(\!
)替换,并将大多数替换\left( ... \right)
为\biggl( ... \biggr)
。还改进了几个指数的位置。
amsmath
无关:加载时无需加载mathtools
。
\documentclass{beamer}
\usepackage{amssymb}
\usepackage{nccmath, mathtools}
\newcommand*{\diff}{\mathop{}\!\mathsf{d}}
\begin{document}
\begin{frame}
\begin{subequations}
\begin{align}
\mathcal{L}&=(\mathcal{T}-\mathcal{V})+\sum_{i=1}^{n}\lambda_{i}L\biggl(\psi(\zeta_{i})-\frac{L_{i}}{L}\chi_{i}(1)\biggr) \nonumber \\
\mathcal{\bar{L}}&=\biggl(\mfrac{\beta^4}{2}\!\int_{0}^{1}\! \bigl( \psi(\eta)\bigr)^2 \diff\eta + \sum_{i=1}^{n} \mfrac{\gamma_{i}}{2}\int_{0}^{1}\! \bigl( \chi_{i}(\xi)\bigr)^2 \diff\xi \biggr)-\nonumber \\
&\phantom{{}={}} \biggl(\mfrac{1}{2}\! \int_{0}^{1}\! \biggl(\frac{\partial^2\psi(\eta)}{\partial \eta^2}\biggr)^{\!2} \diff\eta +\sum_{i=1}^{n}\mfrac{\alpha_{i}}{2}\! \int_{0}^{1}\!\biggl(\frac{\partial \chi_{i}(\xi)}{\partial \xi}\biggr)^{\!\!2} \diff\xi \biggr)+\nonumber \\
&\phantom{{}={}} \sum_{i=1}^{n}\lambda_{i}^{*}\Bigl(\psi(\zeta_{i})-\frac{L_{i}}{L}\chi_{i}(1)\Bigr)
\end{align}
\end{subequations}
\end{frame}
\end{document}
答案2
您的代码无法编译。缺少 \begin{document} 和 \end{document},并且\diff
未定义(使用\newcommand
或通过使用定义它的包)。
以下 MWE:
\documentclass{beamer}
\usepackage{mathtools, amssymb}
\newcommand\diff{\,\mathrm{d}}
\begin{document}
\begin{frame}
\begin{subequations}
\begin{align}
\mathcal{L}
& = \bigl(\mathcal{T}-\mathcal{V}\bigr) +
\sum_{i=1}^{n}\lambda_{i}L\Biggl(\psi(\zeta_{i}) - \frac{L_{i}}{L}\chi_{i}(1)\Biggr) \notag \\
\mathcal{\bar{L}}
& = \Biggl(\frac{\beta^4}{2}\int_{0}^{1}
\bigl( \psi(\eta)\bigr)^2 \diff\eta +
\sum_{i=1}^{n} \frac{\gamma_{i}}{2}
\int_{0}^{1} \bigl(\chi_{i}(\xi)\bigr)^2 \diff\xi\Biggr) -
\notag \\
& \qquad\Biggl(\frac{1}{2} \int_{0}^{1}
\biggl(\frac{\partial^2\psi(\eta)}{\partial\eta^2}\biggr)^2 \diff\eta + \sum_{i=1}^{n}\frac{\alpha_{i}}{2}
\int_{0}^{1}\biggl(\frac{\partial \chi_{i}(\xi)}{\partial \xi}\biggr)^2 \diff\xi \Biggr) + \notag \\
& \qquad \sum_{i=1}^{n}\lambda_{i}^{*}\Biggl(\psi\bigl(\zeta_{i}\bigr) -
\frac{L_{i}}{L}\chi_{i}(1)\Biggr)
\end{align}
\end{subequations}
\end{frame}
\end{document}
给出:
在上面的 MWE 中,我清理了代码中的所有杂乱内容,并用 和 替换\left(
。\Biggl(
此外\right)
,Biggr)
我在方程式中以适当的步幅引入\bigl
和\biggl
,以及\bigr
和\biggr)
。我还定义了\diff
命令。
这能解决你的问题吗?