以下问题:我有一个 PDE 和两个边界条件,我真正想要的是:
偏微分方程 (1)
BC1(1a)
BC2(1b)
我拥有的是:
偏微分方程 (1)
BC1(2a)
BC2(2b)
有什么东西可以提供这样的功能吗?
梅威瑟:
\documentclass[standalone]{article}
\usepackage[utf8x]{inputenc}
\usepackage{amsmath}
\begin{document}
\begin{equation}
\nonumber \frac{\partial}{\partial t}\left(\widetilde{c}_{\text{s}}\left(r,z,t\right)+c_{\text{s,0}}\right)=\frac{D_{\text{s}}}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial \left(\widetilde{c}_{\text{s}}\left(r,z,t\right)+c_{\text{s,0}}\right)}{\partial r}\right)
\label{eq:litdiff}
\end{equation}
The boundary conditions of Eq. \eqref{eq:litdiff} are given by
\begin{subequations}
\begin{align}
D_{\text{s}}\frac{\partial}{\partial r} \widetilde{c}_{\text{s}}(0,z,t)&=0\label{eq:diffusionpartialdiff_boundaries0}\\
D_{\text{s}}\frac{\partial}{\partial r} \widetilde{c}_{\text{s}}(R_{\text{s}},z,t) &= -j(z,t)\label{eq:diffusionpartialdiff_boundariesRs}
\intertext{and the initial condition is given by}
\widetilde{c}_{\text{s}}(r,z,0) = 0 \quad r\in \left[0;R_{\text{s}}\right]. \label{eq:diffusionpartialdiff_init}
\end{align}
\end{subequations}
\end{document}
答案1
使用并且在环境开始时指的是当前主方程编号的\tag
事实。\label
subequations
\documentclass{article}
\usepackage{amsmath}
\newcommand{\subs}{\textup{s}}
\begin{document}
\begin{subequations}\label{eq:litdiff}
\begin{equation}
\frac{\partial}{\partial t}(\widetilde{c}_{\subs}(r,z,t)+c_{\subs,0})=
\frac{D_{\subs}}{r^{2}}\frac{\partial}{\partial r}
\left(
r^{2}\frac{\partial (\widetilde{c}_{\subs}(r,z,t)+c_{\subs,0})}{\partial r}
\right)
\tag{\ref{eq:litdiff}}
\end{equation}
The boundary conditions of Eq.~\eqref{eq:litdiff} are given by
\begin{align}
D_{\subs}\frac{\partial}{\partial r} \widetilde{c}_{\subs}(0,z,t)
&=0\label{eq:diffusionpartialdiff_boundaries0}\\
D_{\subs}\frac{\partial}{\partial r} \widetilde{c}_{\subs}(R_{\subs},z,t)
&= -j(z,t)\label{eq:diffusionpartialdiff_boundariesRs}
\end{align}
and the initial condition is given by
\begin{equation}
\widetilde{c}_{\subs}(r,z,0) = 0 \quad r\in [0;R_{\subs}].
\label{eq:diffusionpartialdiff_init}
\end{equation}
\end{subequations}
\end{document}
我删除了所有多余的\left
和\right
。 也\text{s}
已更改为 ,\subs
以提高可读性和输入效果(请注意 是_{\text{s,0}}
错误的,应该是_{\text{s},0}
)。
最后一个等式需要equation
,不在align
。
重要的。~
在诸如此类的情况下不要忘记Eq.~\eqref{eq:litdiff}
。
答案2
A\addtocounter{equation}{-1}
可以。但是,我不明白nonumber
你稍后提到的公式中的指令,所以我删除了它:
\documentclass{article}
\usepackage[utf8x]{inputenc}
\usepackage{amsmath}
\begin{document}
\begin{equation}
\frac{\partial}{\partial t}\left(\widetilde{c}_{\text{s}}\left(r,z,t\right)+c_{\text{s,0}}\right)=\frac{D_{\text{s}}}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial \left(\widetilde{c}_{\text{s}}\left(r,z,t\right)+c_{\text{s,0}}\right)}{\partial r}\right)
\label{eq:litdiff}
\end{equation}
The boundary conditions of Eq. \eqref{eq:litdiff} are given by\addtocounter{equation}{-1}
\begin{subequations}
\begin{align}
D_{\text{s}}\frac{\partial}{\partial r} \widetilde{c}_{\text{s}}(0,z,t)&=0\label{eq:diffusionpartialdiff_boundaries0}\\
D_{\text{s}}\frac{\partial}{\partial r} \widetilde{c}_{\text{s}}(R_{\text{s}},z,t) &= -j(z,t)\label{eq:diffusionpartialdiff_boundariesRs}
\intertext{and the initial condition is given by}
\widetilde{c}_{\text{s}}(r,z,0) = 0 \quad r\in \left[0;R_{\text{s}}\right]. \label{eq:diffusionpartialdiff_init}
\end{align}
\end{subequations}
From condition \eqref{eq:diffusionpartialdiff_boundaries0}, we see that…
\end{document}