2

I have one main equation and a few initial/boundary conditions I want to display. The first should have label 3.3 only and the conditions ideally have labels 3.3a, 3.3b, etc.

The first eq is also somewhat unique in that I want it displayed in a table-like format as it spans 3 lines and has a pattern to it. The align environment messes up the spacing so I opted for alignat:

\usepackage[tbtags]{amsmath}
\newcommand{\s}{\hspace{0.12cm}}

The 2D form of Equation \eqref{eq:pathenum1} is thus

\vspace{-0.4cm}
\begin{subequations}\label{eq:pathenum2}
\begin{alignat}{3}
    \xi_2(x,y,t+1) \s &=\s \xi_2(x-1,y+1,t) \s &+& \s \xi_2(x,y+1,t) \s &+& \s \xi_2(x+1,y+1,t) \nonumber\\
                      &+\s \xi_2(x-1,y,t) \s &+& \s \xi_2(x,y,t) \s &+& \s \xi_2(x+1,y,t) \tag{\label{eq:pathenum2}}\\
                      &+\s \xi_2(x-1,y-1,t) \s &+& \s \xi_2(x,y-1,t) \s &+& \s \xi_2(x+1,y-1,t), \nonumber
\end{alignat}

and the 3D form is displayed in Appendix Equation \eqref{appeq:pathenum3}. Initial and boundary
conditions for the 2D (and for any $D$) case are also the same as in 1D as the theory behind them
still applies and is carried forward:

\vspace{-0.45cm}
\begin{align}
    &\xi_2(0,0,1) = 1, \label{eq:pathenum2a}\\
    &\xi_2(x,y,t) = 0 \quad \mathrm{for} \quad \forall t<0, \; \forall x,y, \label{eq:pathenum2b}\\
    &\xi_2(x,y,t) = 0 \quad \mathrm{for} \quad \forall|x|,|y| > t. \label{eq:pathenum2c}
\end{align}
\end{subequations}

As you can see the numbering doesn't quite work for my main eq -- I've managed this for one-line equations elsewhere in my work but how do I fix it for this case?

1

Since you're using tbtags, the number should align to the bottom row.

You need \tag{\ref{<label>}}, not \label inside \tag.

Also the spacings you add can be removed by using the proper markup, namely &&+ instead of &+&.

Avoid blank lines before displays (which are the reasons for you to add negative spacing). A blank line can follow a display only if a new paragraph starts.

I also removed \mathrm{for} as the word is already implied in the “forall” symbols.

\documentclass{article}
\usepackage[tbtags]{amsmath}

\begin{document}
The 2D form of Equation \eqref{eq:pathenum1} is thus
\begin{subequations}\label{eq:pathenum2}
\begin{alignat*}{3}
\xi_2(x,y,t+1) &= \xi_2(x-1,y+1,t) &&+ \xi_2(x,y+1,t) &&+ \xi_2(x+1,y+1,t) \\
               &+ \xi_2(x-1,y,t)   &&+ \xi_2(x,y,t)   &&+ \xi_2(x+1,y,t) \\
               &+ \xi_2(x-1,y-1,t) &&+ \xi_2(x,y-1,t) &&+ \xi_2(x+1,y-1,t),
  \tag{\ref{eq:pathenum2}}
\end{alignat*}
and the 3D form is displayed in Appendix Equation \eqref{appeq:pathenum3}. Initial and boundary
conditions for the 2D (and for any $D$) case are also the same as in 1D as the theory behind them
still applies and is carried forward:
\begin{align}
    &\xi_2(0,0,1) = 1, \label{eq:pathenum2a}\\
    &\xi_2(x,y,t) = 0 \qquad \forall t<0, \; \forall x,y, \label{eq:pathenum2b}\\
    &\xi_2(x,y,t) = 0 \qquad \forall|x|,|y| > t. \label{eq:pathenum2c}
\end{align}
\end{subequations}

\end{document}

enter image description here

| improve this answer | |
0

You can simply decrease the equation counter after the first equation:

\documentclass{article}

\usepackage[tbtags]{amsmath}
\newcommand{\s}{\hspace{0.12cm}}
\usepackage[noabbrev]{cleveref}

\begin{document}

The 2D form of Equation \eqref{eq:pathenum1} is thus

\vspace{-0.4cm}
\begin{equation}\label{eq:pathenum2}
\begin{alignedat}{3}
    \xi_2(x,y,t+1) \s &=\s \xi_2(x-1,y+1,t) \s &+& \s \xi_2(x,y+1,t) \s &+& \s \xi_2(x+1,y+1,t) \\
                      &+\s \xi_2(x-1,y,t) \s &+& \s \xi_2(x,y,t) \s &+& \s \xi_2(x+1,y,t) \\
                      &+\s \xi_2(x-1,y-1,t) \s &+& \s \xi_2(x,y-1,t) \s &+& \s \xi_2(x+1,y-1,t),
\end{alignedat}
\end{equation}
%
and the 3D form is displayed in Appendix Equation \eqref{appeq:pathenum3}. Initial and boundary
conditions for the 2D (and for any $D$) case are also the same as in 1D as the theory behind them
still applies and is carried forward in \cref{eq:pathenum2a,eq:pathenum2b,eq:pathenum2c}:

\vspace{-0.45cm}\addtocounter{equation}{-1}
\begin{subequations}
\begin{align}
    &\xi_2(0,0,1) = 1, \label{eq:pathenum2a}\\
    &\xi_2(x,y,t) = 0 \quad \mathrm{for} \quad \forall t<0, \; \forall x,y, \label{eq:pathenum2b}\\
    &\xi_2(x,y,t) = 0 \quad \mathrm{for} \quad \forall|x|,|y| > t. \label{eq:pathenum2c}
\end{align}
\end{subequations}

\end{document}

enter image description here

| improve this answer | |

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.