10

Following Problem: I have a PDE and two boundary conditions what i actually want is:

PDE (1)

BC1 (1a)

BC2 (1b)

and what i have is:

PDE (1)

BC1 (2a)

BC2 (2b)

is there something which provides a functionallity like this?

MWE:

\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}
  • The easiest way is to remove \nonumber in the main equation and use \addtocounter{equation}{-1} just before \begin{subequations} – user31729 Dec 30 '14 at 10:50
  • 1
    utf8x option to inputenc is outdated, the standalone option has no effect to article – user31729 Dec 30 '14 at 10:56
  • In fact you have PDEs, not ODEs... – user31729 Dec 30 '14 at 11:06
  • \nonumber was a failure because of copy and paste; yes it's PDE...too focused on the Latex-Problem. i'll edit it. – user69453 Dec 30 '14 at 12:18
8

Use \tag and the fact that \label at the start of a subequations environment refers to the current main equation number.

\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}

I removed all the superfluous \left and \right. Also \text{s} has been changed to \subs for better readability and input (note that _{\text{s,0}} is wrong and should be _{\text{s},0}).

The last equation needs equation, not being in the align.

Important. Don't forget ~ in cases such as Eq.~\eqref{eq:litdiff}.

enter image description here

7

A \addtocounter{equation}{-1} will do. However, I don't get the nonumber directive in a equation you refer to later, so I deleted it:

\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} 

enter image description here

  • Personally, I think this one is a little cleaner than the OP's preferred answer. And it seems less "hack"-y to me. – grfrazee Aug 11 '15 at 19:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.