我想以以下方式呈现一个等式:
然而,利用 amsmath 包,这是我能得到的最好的结果:
\begin{equation}
\begin{aligned}
\frac{\partial}{\partial t}\left[\rho\left(e+\frac{V^2}{2}\right)\right]+\nabla \cdot\left[\rho\left(e+\frac{V^2}{2} \vec{V}\right)\right] &\\
= & \rho \dot{q}+\frac{\partial}{\partial x}\left(k \frac{\partial T}{\partial x}\right)+\frac{\partial}{\partial y}\left(k \frac{\partial T}{\partial y}\right) \\
\quad+ & \frac{\partial}{\partial z}\left(k \frac{\partial T}{\partial z}\right)-\frac{\partial(u p)}{\partial x}-\frac{\partial(v p)}{\partial y}-\frac{\partial(w p)}{\partial z}+\frac{\partial\left(u \tau_{\mathrm{xx}}\right)}{\partial x} \\
\quad+ & \frac{\partial\left(u \tau_{\mathrm{yx}}\right)}{\partial y}+\frac{\partial\left(u \tau_{\mathrm{zx}}\right)}{\partial z}+\frac{\partial\left(v \tau_{\mathrm{xy}}\right)}{\partial x}+\frac{\partial\left(v \tau_{\mathrm{yy}}\right)}{\partial y}+\frac{\partial\left(v \tau_{\mathrm{zy}}\right)}{\partial z} \\
\quad+ & \frac{\partial\left(w \tau_{\mathrm{xz}}\right)}{\partial x}+\frac{\partial\left(w \tau_{\mathrm{yz}}\right)}{\partial y}+\frac{\partial\left(w \tau_{\mathrm{zz}}\right)}{\partial z}+\rho \vec{f} \cdot \vec{V}
\label{eq:energyeq}
\end{aligned}
\end{equation}
我想保留方程式标签,尽管它垂直居中于方程式上方,而不是位于最后一行下方。如有任何建议,我将不胜感激。
答案1
我会使用split
并垂直对齐所有 + 号。本地命令可以轻松完成这项工作,并\pder
大大简化输入。
\documentclass{article}
\usepackage{amsmath}
\newcommand{\pder}[2][]{\frac{\partial#1}{\partial#2}}
\begin{document}
\begin{equation}\label{eq:energyeq}
\newcommand{\locsp}{\hphantom{{}=\rho\dot{q}}}
\begin{split}
\pder{t} & \left[\rho\left(e+\frac{V^2}{2}\right)\right]
+\nabla \cdot\left[\rho\left(e+\frac{V^2}{2} \vec{V}\right)\right] \\
&= \rho \dot{q}+\pder{x}\left(k \pder[T]{x}\right)+\pder{y}\left(k \pder[T]{y}\right)
\\
&\locsp
+ \pder{z}\left(k \pder[T]{z}\right)
- \pder[(up)]{x}-\pder[(vp)]{y}
- \pder[(wp)]{z}+\pder[(u\tau_{\mathrm{xx}})]{x}
\\
&\locsp
+ \pder[(u\tau_{\mathrm{yx}})]{y}
+ \pder[(u\tau_{\mathrm{zx}})]{z}
+ \pder[(v\tau_{\mathrm{xy}})]{x}
+ \pder[(v\tau_{\mathrm{yy}})]{y}
+ \pder[(v\tau_{\mathrm{zy}})]{z}
\\
&\locsp
+ \pder[(w\tau_{\mathrm{xz}})]{x}
+ \pder[(w\tau_{\mathrm{yz}})]{y}
+ \pder[(w\tau_{\mathrm{zz}})]{z}
+ \rho \vec{f} \cdot \vec{V}
\end{split}
\end{equation}
\end{document}
答案2
...只是改变了环境内的对齐方式aligned
:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation}
\begin{aligned}
\frac{\partial}{\partial t}&\left[ \rho \left( e + \frac{V^2}{2} \right) \right] + \nabla \cdot \left[ \rho \left( e + \frac{V^2}{2} \vec{V} \right) \right] \\
&= \rho \dot{q} + \frac{\partial}{\partial x} \left (k \frac{\partial T}{\partial x} \right) + \frac{\partial}{\partial y} \left( k \frac{\partial T}{\partial y} \right) \\
&\phantom{=} + \frac{\partial}{\partial z} \left (k \frac{\partial T}{\partial z} \right) - \frac{\partial (u p)}{\partial x} - \frac{\partial (v p)}{\partial y} - \frac{\partial (w p)}{\partial z} + \frac{\partial \left( u \tau_{\mathrm{xx}} \right)}{\partial x} \\
&\phantom{=} + \frac{\partial \left( u \tau_{\mathrm{yx}} \right)}{\partial y} + \frac{\partial \left( u \tau_{\mathrm{zx}} \right)}{\partial z} + \frac{\partial \left( v \tau_{\mathrm{xy}} \right)}{\partial x} + \frac{\partial \left( v \tau_{\mathrm{yy}} \right)}{\partial y} + \frac{\partial \left( v \tau_{\mathrm{zy}} \right)}{\partial z} \\
&\phantom{=} + \frac{\partial\left(w \tau_{\mathrm{xz}}\right)}{\partial x} + \frac{\partial \left( w \tau_{\mathrm{yz}} \right)}{\partial y} + \frac{\partial \left( w \tau_{\mathrm{zz}} \right)}{\partial z} + \rho \vec{f} \cdot \vec{V}
\end{aligned}
\end{equation}
\end{document}
答案3
除了将第一行的对齐点向左移动之外,您可能还想重新排列 RHS 上的加法项,这样就可以立即清楚地看到 17 个项中的 15 个可以排列成 5 组,每组 3 个项。这种变化的副作用是,对于整个表达式,您只需 4 行(而不是 5 行)即可。
\documentclass{article} % or some other suitable document class
\usepackage{amsmath} % for 'aligned' environment
\usepackage{mleftright} % optional (for more compact display of parenth. groups)
\mleftright
\begin{document}
\begin{equation} \label{eq:energyeq}
\begin{aligned}
\frac{\partial}{\partial t}
&\left[\rho\left(e+\frac{V^2}{2}\right)\right]
+\nabla \cdot\left[\rho\left(e+\frac{V^2}{2} \vec{V}\right)\right] \\
&= \rho \dot{q}+ \rho\vec{f}\cdot\vec{V}
+\frac{\partial}{\partial x}\left(k \frac{\partial T}{\partial x}\right)
+\frac{\partial}{\partial y}\left(k \frac{\partial T}{\partial y}\right)
+\frac{\partial}{\partial z}\left(k \frac{\partial T}{\partial z}\right)\\
&\quad -\frac{\partial(u p)}{\partial x}
-\frac{\partial(v p)}{\partial y}
-\frac{\partial(w p)}{\partial z}
+\frac{\partial(u \tau_{\mathrm{xx}})}{\partial x}
+\frac{\partial(u \tau_{\mathrm{yx}})}{\partial y}
+\frac{\partial(u \tau_{\mathrm{zx}})}{\partial z}\\
&\quad +\frac{\partial(v \tau_{\mathrm{xy}})}{\partial x}
+\frac{\partial(v \tau_{\mathrm{yy}})}{\partial y}
+\frac{\partial(v \tau_{\mathrm{zy}})}{\partial z}
+\frac{\partial(w \tau_{\mathrm{xz}})}{\partial x}
+\frac{\partial(w \tau_{\mathrm{yz}})}{\partial y}
+\frac{\partial(w \tau_{\mathrm{zz}})}{\partial z} \,.
\end{aligned}
\end{equation}
\end{document}