我正在使用 amsmath 包。我可以分解括号外的方程,但我也想分解这两个大括号内的方程。以下是代码:
\begin{equation}
\begin{aligned}
\displaystyle\frac{c_{i,j,m}^{n+1}-c_{i,j,m}^n}{k}=\frac{D_C}{2}\bigg\lbrace \frac{c_{i+1,j,m}^{n}-2 c_{i,j,m}^{n}+c_{i-1,j,m}^{n}}{h^2}+\frac{c_{i+1,j,m}^n-c_{i-1,j,m}^n}{2 r h}
+\frac{c_{i,j-1,m}^n-2 c_{i,j,m}^n+c_{i,j+1,m}^n}{(r l)^2}+\frac{c_{i,j,m-1}^n-2 c_{i,j,m}^n+c_{i,j,m+1}^n}{(r q)^2}\bigg\rbrace \\
+\bigg\lbrace\frac{ V_{IPR}\left(\frac{d_{i,j,m}^n}{d_{i,j,m}^n+K_{IP}}\right)^3\left(\frac{c_{i,j,m}^n}{c_{i,j,m}^n+K_{A}}\right)^3\left(\frac{K_{In}}{K_{In}+c_{i,j,m}^n}\right)^3 (c_E-c_{i,j,m}^n)-V_{SP}\frac{{c_{i,j,m}^n}^2}{K_{SP}^2+{c_{i,j,m}^n}^2}+ V_{Le}(c_E-c_{i,j,m}^n)}{F_C}\bigg\rbrace,
\end{aligned}
\end{equation}
答案1
非常难。我可以提出以下建议:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation}
\begin{split}
\frac{c_{i,j,m}^{n+1}-c_{i,j,m}^n}{k}=\frac{D_C}{2}
& \begin{aligned}[t]
\biggl\lbrace & \frac{c_{i+1,j,m}^{n}-2 c_{i,j,m}^{n}+c_{i-1,j,m}^{n}}{h^2} \\
&+\frac{c_{i+1,j,m}^n-c_{i-1,j,m}^n}{2 r h} \\
&+\frac{c_{i,j-1,m}^n-2 c_{i,j,m}^n+c_{i,j+1,m}^n}{(r l)^2} \\
&+\frac{c_{i,j,m-1}^n-2 c_{i,j,m}^n+c_{i,j,m+1}^n}{(r q)^2}\biggr\rbrace
\end{aligned}
\\
{}+\frac{1}{F_C}
& \begin{aligned}[t]
\biggl\lbrace & V_{IPR}\biggl(\frac{d_{i,j,m}^n}{d_{i,j,m}^n+K_{IP}}\biggr)^3
\biggl(\frac{c_{i,j,m}^n}{c_{i,j,m}^n+K_{A}}\biggr)^3\\
&\hphantom{V_{IPR}}\makebox[0pt][r]{${}\cdot{}$}
\biggl(\frac{K_{In}}{K_{In}+c_{i,j,m}^n}\biggr)^3
(c_E-c_{i,j,m}^n) \\
&-V_{SP}\frac{{c_{i,j,m}^n}^2}{K_{SP}^2+{c_{i,j,m}^n}^2}
+ V_{Le}(c_E-c_{i,j,m}^n)\bigg\rbrace,
\end{aligned}
\end{split}
\end{equation}
\end{document}
答案2
使用包multlined
中定义的环境mathtools
并使用包\mfrac
中定义的nccmath
:
\documentclass{article}
\usepackage{geometry}
\usepackage{nccmath, mathtools}
\usepackage{lipsum}% for dummy text
\begin{document}
\lipsum[66]
\begin{equation}%\medmath{
\begin{multlined}[0.9\linewidth]
\mfrac{c_{i,j,m}^{n+1}-c_{i,j,m}^n}{k} =
\mfrac{D_C}{2} \bigg\lbrace \mfrac{c_{i+1,j,m}^{n}
- 2 c_{i,j,m}^{n}+c_{i-1,j,m}^{n}}{h^2}
+ \mfrac{c_{i+1,j,m}^n-c_{i-1,j,m}^n}{2 r h} \\
+ \mfrac{c_{i,j-1,m}^n-2 c_{i,j,m}^n+c_{i,j+1,m}^n}{(r l)^2}
+ \mfrac{c_{i,j,m-1}^n-2 c_{i,j,m}^n+c_{i,j,m+1}^n}{(r q)^2}\biggr\rbrace \\
+ \mfrac{1}{F_C} \biggl\lbrace
V_{IPR}\biggl(\mfrac{d_{i,j,m}^n}{d_{i,j,m}^n+K_{IP}}\biggr)^3\biggl(\mfrac{c_{i,j,m}^n}{c_{i,j,m}^n
+ K_{A}}\biggr)^3\biggl(\mfrac{K_{In}}{K_{In}+c_{i,j,m}^n}\biggr)^3 (c_E-\mathrlap{c_{i,j,m}^n)} \\
- V_{SP}\mfrac{{c_{i,j,m}^n}^2}{K_{SP}^2+{c_{i,j,m}^n}^2}
+ V_{Le}(c_E-c_{i,j,m}^n)\biggr\rbrace,
\end{multlined}%}
\end{equation}
\end{document}
答案3
如果您加载几何图形以获得更合理的边距和fleqn
环境,则可以使用4 行nccmath
来完成:alignat
\documentclass{article}
\usepackage{nccmath, mathtools}
\usepackage[showframe]{geometry}
\begin{document}
\mbox{}
\begin{fleqn}
\begin{equation}
\begin{alignedat}{2}
\frac{c_{i,j,m}^{n+1}-c_{i,j,m}^n}{k} & =\frac{D_C}{2}& & \bigg\lbrace \frac{c_{i+1,j,m}^{n}-2 c_{i,j,m}^{n}+c_{i-1,j,m}^{n}}{h^2}+\frac{c_{i+1,j,m}^n-c_{i-1,j,m}^n}{2 r h} \\
& & & {} + \frac{c_{i,j-1,m}^n-2 c_{i,j,m}^n+c_{i,j+1,m}^n}{(r l)^2}+\frac{c_{i,j,m-1}^n-2 c_{i,j,m}^n+c_{i,j,m+1}^n}{(r q)^2}\biggr\rbrace \\
& {} +\frac{1}{F_C} & & \biggl\lbrace V_{IPR}\biggl(\frac{d_{i,j,m}^n}{d_{i,j,m}^n+K_{IP}}\biggr)^3\biggl(\frac{c_{i,j,m}^n}{c_{i,j,m}^n
+K_{A}}\biggr)^3\biggl(\frac{K_{In}}{K_{In}+c_{i,j,m}^n}\biggr)^3 (c_E-\mathrlap{c_{i,j,m}^n)}
\\%
& & & {}-V_{SP}\frac{{c_{i,j,m}^n}^2}{K_{SP}^2+{c_{i,j,m}^n}^2}+ V_{Le}(c_E-c_{i,j,m}^n)\biggr\rbrace,
\end{alignedat}
\end{equation}
\end{fleqn}
\end{document}
答案4
这可能有点离题,但我不相信读者会从这么长的方程式中受益。我宁愿把它分开。(一些括号大小是从其他答案中改编而来的。)
\documentclass{article}
\usepackage[fleqn]{amsmath}
\begin{document}
\begin{equation}
\frac{c_{i,j,m}^{n+1}-c_{i,j,m}^n}{k}=\frac{D_C}{2}c_D
+\frac{1}{F_C} c_F\;,
\end{equation}
where
\begin{subequations}
\begin{align}
c_D&= \biggl\lbrace \frac{c_{i+1,j,m}^{n}-2 c_{i,j,m}^{n}+c_{i-1,j,m}^{n}}{h^2}
+\frac{c_{i+1,j,m}^n-c_{i-1,j,m}^n}{2 r h} \\
&\hphantom{= \biggl\lbrace}
+\frac{c_{i,j-1,m}^n-2 c_{i,j,m}^n+c_{i,j+1,m}^n}{(r \ell)^2}
+\frac{c_{i,j,m-1}^n-2 c_{i,j,m}^n+c_{i,j,m+1}^n}{(r q)^2}\biggr\rbrace
\;, \notag\\
c_F&=\biggl\lbrace V_{IPR}\,\biggl(\frac{d_{i,j,m}^n}{d_{i,j,m}^n+K_{IP}}
\frac{c_{i,j,m}^n}{c_{i,j,m}^n+K_{A}}
\frac{K_{In}}{K_{In}+c_{i,j,m}^n}\biggr)^3(c_E-c_{i,j,m}^n)\notag\\
&\hphantom{= \biggl\lbrace}
-V_{SP}\frac{{c_{i,j,m}^n}^2}{K_{SP}^2+{c_{i,j,m}^n}^2}
+ V_{Le}\,(c_E-c_{i,j,m}^n)\bigg\rbrace \;.
\end{align}
\end{subequations}
\end{document}
然后你就可以去掉这些花括号了。
\documentclass{article}
\usepackage[fleqn]{amsmath}
\begin{document}
\begin{equation}
\frac{c_{i,j,m}^{n+1}-c_{i,j,m}^n}{k}=\frac{D_C}{2}c_D
+\frac{1}{F_C} c_F\;,
\end{equation}
where
\begin{subequations}
\begin{align}
c_D&= \frac{c_{i+1,j,m}^{n}-2 c_{i,j,m}^{n}+c_{i-1,j,m}^{n}}{h^2}
+\frac{c_{i+1,j,m}^n-c_{i-1,j,m}^n}{2 r h} \notag\\
&\hphantom{= }
+\frac{c_{i,j-1,m}^n-2 c_{i,j,m}^n+c_{i,j+1,m}^n}{(r \ell)^2}
+\frac{c_{i,j,m-1}^n-2 c_{i,j,m}^n+c_{i,j,m+1}^n}{(r q)^2}
\;, \\
c_F&= V_{IPR}\,\biggl(\frac{d_{i,j,m}^n}{d_{i,j,m}^n+K_{IP}}
\frac{c_{i,j,m}^n}{c_{i,j,m}^n+K_{A}}
\frac{K_{In}}{K_{In}+c_{i,j,m}^n}\biggr)^3(c_E-c_{i,j,m}^n)\notag\\
&\hphantom{= }
-V_{SP}\frac{{c_{i,j,m}^n}^2}{K_{SP}^2+{c_{i,j,m}^n}^2}
+ V_{Le}\,(c_E-c_{i,j,m}^n) \;.
\end{align}
\end{subequations}
\end{document}
我发现这更易读。额外的好处是这些表达式现在都有标签,因此您以后可以以一种不令人困惑的方式引用它们。