方程式数量:1、1a、1b

方程式数量:1、1a、1b

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。\labelsubequations

\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\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}

在此处输入图片描述

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} 

在此处输入图片描述

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