使用带有 2 个 && 的 alignat 时出现问题

Question

我无法对齐下面示例中的第二个块......

\documentclass{article}

\usepackage{amsmath}
\usepackage{amssymb}

\begin{document}
\begin{alignat}{2}\nonumber
    \frac{\partial U^\varepsilon_{p,q}}{\partial x_3} &= \frac{i\omega\Gamma_1(\kappa_1) m(x_3/\varepsilon^2)}{\varepsilon}\bigg(pU^\varepsilon_{p,q} - qU^\varepsilon_{p,q}\bigg) \\ \nonumber
    &+\frac{i\omega\Gamma_2(\kappa_1) m(x_3/\varepsilon^2)}{2\varepsilon}e^{2i\omega\bar\zeta(\kappa_1)x_3/\varepsilon}&&\bigg(pU^\varepsilon_{p+1,q}e^{ih\bar\zeta(\kappa_1)x_3 - i\omega\bar{c}_{66}\kappa_1\lambda x_3/(\bar{c}_{44}\bar{\zeta}(\kappa_1))} \\ \nonumber
    &\phantom{+} &&-qU^\varepsilon_{p,q-1}e^{-ih\bar\zeta(\kappa_1)x_3 + i\omega\bar{c}_{66}\kappa_1\lambda x_3/(\bar{c}_{44}\bar{\zeta}(\kappa_1))}\bigg) \\ \nonumber
    &+\frac{i\omega\Gamma_2(\kappa_1) m(x_3/\varepsilon^2)}{2\varepsilon}e^{-2i\omega\bar\zeta(\kappa_1)x_3/\varepsilon}&&\bigg(pU^\varepsilon_{p-1,q}e^{-ih\bar{\zeta}x_3 + i\omega\bar{c}_{66}\kappa_1\lambda x_3/(\bar{c}_{44}\bar{\zeta}(\kappa_1))} \\
    &\phantom{+} &&-qU^\varepsilon_{p,q+1}e^{ih\bar{\zeta}(\kappa_1)x_3 - i\omega\bar{c}_{66}\kappa_1\lambda x_3/(\bar{c}_{44}\bar{\zeta}(\kappa_1))}\bigg) 
\end{alignat}
\end{document}

出现 \bigg(p 的第二行与其前面的指数之间不应该有太大间隙，有人可以给我一些建议吗？谢谢。

Answer 1

您可以在第一行添加一个负空间，让 TeX 认为它更短。

由于指数中的负号，仍然会存在小的差距，但我觉得对齐更为重要。

\documentclass{article}
\usepackage{geometry}
\usepackage{amsmath}
\usepackage{amssymb}

\begin{document}

\begin{alignat}{2}
\frac{\partial U^\varepsilon_{p,q}}{\partial x_3}
&= \frac{i\omega\Gamma_1(\kappa_1) m(x_3/\varepsilon^2)}{\varepsilon}
   \bigl(pU^\varepsilon_{p,q} - qU^\varepsilon_{p,q}\bigr) \hspace{-5em}
\nonumber \\
&+ \frac{i\omega\Gamma_2(\kappa_1) m(x_3/\varepsilon^2)}{2\varepsilon}
   e^{2i\omega\bar\zeta(\kappa_1)x_3/\varepsilon}
   &&\Bigl(pU^\varepsilon_{p+1,q}e^{ih\bar\zeta(\kappa_1)x_3 -
     i\omega\bar{c}_{66}\kappa_1\lambda x_3/(\bar{c}_{44}\bar{\zeta}(\kappa_1))}
\nonumber \\
&  &&-qU^\varepsilon_{p,q-1}e^{-ih\bar\zeta(\kappa_1)x_3 +
     i\omega\bar{c}_{66}\kappa_1\lambda x_3/(\bar{c}_{44}\bar{\zeta}(\kappa_1))}\Bigr)
\nonumber \\
&+ \frac{i\omega\Gamma_2(\kappa_1) m(x_3/\varepsilon^2)}{2\varepsilon}
   e^{-2i\omega\bar\zeta(\kappa_1)x_3/\varepsilon}
   &&\Bigl(pU^\varepsilon_{p-1,q}e^{-ih\bar{\zeta}x_3 +
     i\omega\bar{c}_{66}\kappa_1\lambda x_3/(\bar{c}_{44}\bar{\zeta}(\kappa_1))}
\nonumber \\
&  &&-qU^\varepsilon_{p,q+1}e^{ih\bar{\zeta}(\kappa_1)x_3 -
     i\omega\bar{c}_{66}\kappa_1\lambda x_3/(\bar{c}_{44}\bar{\zeta}(\kappa_1))}\Bigr)
\end{alignat}

\end{document}

\Bigl请注意and的用法，\Bigr原因有二：

1. \bigg太大；
2. 应该将分隔符设置为打开和关闭原子。

在第一行\bigl似乎\bigr已经足够了。

我之所以使用geometry，是因为根据您的图片，您的文本宽度比默认值大。

Answer 2

我建议这种基于\multlined嵌套的布局aligned，使用更小的括号和geometry包，以获得更合适的边距：

\documentclass{article}
\usepackage{geometry}
\usepackage{mathtools}
\usepackage{amssymb}

\begin{document}

\begin{equation}
\begin{aligned}[b]
    \frac{\partial U^\varepsilon_{p,q}}{\partial x_3} &=\frac{i\omega\Gamma_1(\kappa_1) m(x_3/\varepsilon^2)}{\varepsilon}\mathrlap{\Bigl(pU^\varepsilon_{p,q} - qU^\varepsilon_{p,q}\Bigr)} \\
    &\begin{multlined} + \frac{i\omega\Gamma_2(\kappa_1) m(x_3/\varepsilon^2)}{2\varepsilon}e^{2i\omega\bar\zeta(\kappa_1)x_3/\varepsilon}\Bigl(pU^\varepsilon_{p+1,q}e^{ih\bar\zeta(\kappa_1)x_3 - i\omega\bar{c}_{66}\kappa_1\lambda x_3/(\bar{c}_{44}\bar{\zeta}(\kappa_1))} \\[-1.5ex]
   -qU^\varepsilon_{p,q-1}e^{-ih\bar\zeta(\kappa_1)x_3 + i\omega\bar{c}_{66}\kappa_1\lambda x_3/(\bar{c}_{44}\bar{\zeta}(\kappa_1))}\Bigr)
    \end{multlined}\\
    &\begin{multlined}[b] + \frac{i\omega\Gamma_2(\kappa_1) m(x_3/\varepsilon^2)}{2\varepsilon}e^{-2i\omega\bar\zeta(\kappa_1)x_3/\varepsilon}\Bigl(pU^\varepsilon_{p-1,q}e^{-ih\bar{\zeta}x_3 + i\omega\bar{c}_{66}\kappa_1\lambda x_3/(\bar{c}_{44}\bar{\zeta}(\kappa_1))} \\[-1.5ex]
    -qU^\varepsilon_{p,q+1}e^{ih\bar{\zeta}(\kappa_1)x_3 - i\omega\bar{c}_{66}\kappa_1\lambda x_3/(\bar{c}_{44}\bar{\zeta}(\kappa_1))}\Bigr)
    \end{multlined}
\end{aligned}
\end{equation}

\end{document}