[Tex/LaTex] Block of equations with only some being subequations

equationsnumberingsubequations

I'm trying to make a block of equations where the first two are standalone equations, and the second two are subequations, ie

X=XXXX (1)

X=XXXX (2)

X=XXXX (3a)

=XXXX (3b)

I'm using the amsmath package, with align and subequation, but I can't figure out how to do this without making a separate block for Eqs 1-2 and one for Eqs 3a-3b. That solution is unsatisfactory, because then there is a big vertical gap between the two blocks, and I want them to appear all as one.

Here's the code I have so far, in case it's unclear.

\begin{align}
T(x) &= \frac{m}{2{k_{B}}} \left\langle \mathbf{\tilde{v}}{(\mathbf{x}_{i},t)}^{2} \right\rangle\\ 
\mathbf{u}(x) &= \left\langle \mathbf{v}_{i}(\mathbf{x}_{i},t) \right\rangle
\end{align}
\begin{subequations} 
\begin{align}
P_{xy}(x) &= \frac{1}{A} \left\langle \sum\limits_{i} {m \tilde{v}_{i,x} \tilde{v}_{i,y}} + \frac{1}{2} \sum\limits_{i} {\sum\limits_{j\ne i} {r_{ij,y} F_{ij,x}}} \right\rangle \\
\begin{split}
 &=\frac{m}{A} \Bigg\langle \sum\limits_{i}{{{{\tilde{v}}}_{i,x}}{{{\tilde{v}}}_{i,y}}} + \frac{a^3\omega _{pd}^2}{4} \\
 &\quad \times \sum\limits_{i}{\sum\limits_{j\ne i}{r_{ij,y}{\left\{ \frac{e^{(-{{r}_{ij}}/\lambda )}}{{{r}_{ij}}}\left( \frac{1}{r_{ij}}+\frac{1}{\lambda} \right) \right\}_{x}}}} \Bigg\rangle
\end{split}
\end{align}
\end{subequations}

Any suggestions would be greatly appreciated!

Best Answer

It's possible with a little manual work. With \startsubequation you fix the equation number and each \tagsubequation command will step the subcounter. Until you use \tagsubequation, the main equation number will not increase.

\documentclass{article}

\usepackage{amsmath}

\makeatletter
\newcounter{manualsubequation}
\renewcommand{\themanualsubequation}{\alph{manualsubequation}}
\newcommand{\startsubequation}{%
  \setcounter{manualsubequation}{0}%
  \refstepcounter{equation}\ltx@label{manualsubeq\theequation}%
  \xdef\labelfor@subeq{manualsubeq\theequation}%
}
\newcommand{\tagsubequation}{%
  \stepcounter{manualsubequation}%
  \tag{\ref{\labelfor@subeq}\themanualsubequation}%
}
\let\subequationlabel\ltx@label
\makeatother


\begin{document}

\begin{align}
T(x) &=
  \frac{m}{2k_{B}} \biggl\langle \mathbf{\tilde{v}}(\mathbf{x}_{i},t)^{2} \biggl\rangle
\\ 
\mathbf{u}(x) &=
  \langle \mathbf{v}_{i}(\mathbf{x}_{i},t) \rangle
\\
\startsubequation\subequationlabel{eq:3}\tagsubequation\label{eq:3a}
P_{xy}(x) &=
   \frac{1}{A} \biggl\langle \sum_{i} m \tilde{v}_{i,x} \tilde{v}_{i,y} 
  +\frac{1}{2} \sum_{i} \sum_{j\ne i} r_{ij,y} F_{ij,x} \biggr\rangle
\\
\tagsubequation\label{eq:3b}
\begin{split}
&=
   \frac{m}{A} \biggl\langle \sum_{i}\tilde{v}_{i,x}\tilde{v}_{i,y}+\frac{a^3\omega_{pd}^2}{4}
 \\
&\quad \times \sum_{i}\sum_{j\ne i}r_{ij,y}\left\{ \frac{e^{(-r_{ij}/\lambda)}}{r_{ij}}
  \left( \frac{1}{r_{ij}}+\frac{1}{\lambda} \right) \right\}_{x} \biggr\rangle
\end{split}
\end{align}

Let's see the next number
\begin{equation}
a=b
\end{equation}

Also references: \eqref{eq:3}, \eqref{eq:3a} and \eqref{eq:3b}

\end{document}

You're using too many braces, in my opinion.

Related Solutions

[Tex/LaTex] Number only some equations in an equation array

You can use \notag and/or \nonumber from amsmath (or improvised mathtools package):

\documentclass{article}
\usepackage{mathtools,amssymb}


\begin{document}
\begin{align}
  x &= y \\\nonumber
  x &= 3 \\
  y &= 10 \\ \notag
 \therefore\ x &\neq y \\
   x &= y \\\nonumber
  x &= 3 \\
  y &= 10 \\ \notag
 \therefore\ x &\neq y \\
 ? &=?
\end{align}
\end{document}

enter image description here

[Tex/LaTex] flush text on the left and centre equation of the same line (not working!)

Here are three possibilities; the first two are variants based on the flalign environment, with a different positioning of the equations w.r.t.the left margin text. The third possibility paves the texts aligned w.r.t. each other, at some distance from the equations. It's based on alignat:

\documentclass[a4paper, 12pt, default, numbered, print, index]{article}
\usepackage{mathtools}

\usepackage{eqparbox} \newcommand{\eqmathbox}2[M]{\eqmakebox[#1]{$\displaystyle#2$}}

\DeclareFontFamily{U}{mathx}{\hyphenchar\font45}
\DeclareFontShape{U}{mathx}{m}{n}{
<5><6><7><8><9><10>
<10.95><12><14.4><17.28><20.74><24.88>
mathx10
}{}
\DeclareSymbolFont{mathx}{U}{mathx}{m}{n}
\DeclareFontSubstitution{U}{mathx}{m}{n}
\DeclareMathAccent{\widebar}{0}{mathx}{"73}

\begin{document}

We can simplify our analysis by assuming the flow is inviscid ($\tau_{i,j}$ = 0), an ideal gas ($p = ρRT$), and calorically perfect ($c_p$, $c_v$ = constant). Thus, for a homogeneous, steady, uniform gas flow in a straight duct without heat conduction, the linearised governing equations can be written as:

\begin{subequations}
\begin{flalign}
\label{eq:linear_EOM_vector_mass}
 & \rlap{Mass} & &\eqmathbox{\frac{\widebar{D}\rho'}{Dt} + \bar{ρ}∇ · \boldsymbol{u}' = \mathbf{0}} & \\[0.8ex]
\label{eq:linear_EOM_vector_mom}
 & \rlap{Momentum} & &\eqmathbox{\frac{\widebar{D}\mathbf{u}'}{Dt} + \frac{1}{\bar{ρ}}
∇ p' = \mathbf{0}}
\end{flalign}
\end{subequations}

\begin{subequations}
\begin{flalign}
\label{eq:linear_EOM_vector_mass}
 & \text{Mass} & & \frac{\widebar{D}\rho'}{Dt} + \bar{ρ}∇ · \boldsymbol{u}' = \mathbf{0} &\hspace{12em} \\[0.8ex]
\label{eq:linear_EOM_vector_mom}
 & \text{Momentum} & & \frac{\widebar{D}\mathbf{u}'}{Dt} + \frac{1}{\bar{ρ}}
∇ p' = \mathbf{0}&&
\end{flalign}
\end{subequations}

\begin{subequations}
\begin{alignat}{2}
\label{eq:linear_EOM_vector_mass}
 & \text{Mass} &\hspace{3em} &\frac{\widebar{D}\rho'}{Dt} + \bar{ρ}∇ · \boldsymbol{u}' = \mathbf{0} \\[0.8ex]
\label{eq:linear_EOM_vector_mom}
 & \text{Momentum} & & \frac{\widebar{D}\mathbf{u}'}{Dt} + \frac{1}{\bar{ρ}}
∇ p' = \mathbf{0}
\end{alignat}
\end{subequations}

\end{document}

Edit: Here is how to adapt the first solution to the last version of the O.P.'s code (two variants):

\documentclass[a4paper,12pt,default,numbered,print,index]{article}
\usepackage{lipsum}
\usepackage{enumitem}
\usepackage{graphicx} % Required for the inclusion of images
\usepackage{setspace} % for use of \singlespacing and \doublespacing
\usepackage{pdfpages}
\usepackage{cite}
\usepackage[section]{placeins}
\usepackage{comment}
\usepackage{siunitx}
\usepackage{color}
\usepackage{ragged2e}
\usepackage{esvect}
\usepackage{mathtools}
\usepackage{lscape}
\usepackage{tabularx}
\usepackage{multirow}
\usepackage{array}
\usepackage{soul}
\usepackage{bm}
\usepackage{url}
\usepackage{xparse}
\usepackage{hyperref}
\newcommand{\myeqlabel}[1]{\rlap{\bfseries#1}}
\usepackage{eqparbox}
\newcommand{\eqmathbox}[2][M]{\eqmakebox[#1]{$\displaystyle#2$}}

    \DeclareFontFamily{U}{mathx}{\hyphenchar\font45}
    \DeclareFontShape{U}{mathx}{m}{n}{
    <5><6><7><8><9><10>
    <10.95><12><14.4><17.28><20.74><24.88>
    mathx10
    }{}
    \DeclareSymbolFont{mathx}{U}{mathx}{m}{n}
    \DeclareFontSubstitution{U}{mathx}{m}{n}
    \DeclareMathAccent{\widebar}{0}{mathx}{"73}


\begin{document}

We can simplify our analysis by assuming the flow is inviscid ($\tau_{i,j} = 0$), an ideal gas ($p = \rho RT$), and calorically perfect ($c_p$, $c_v$ = constant). Thus, for a homogeneous, steady, uniform gas flow in a straight duct without heat conduction, the linearised governing equations can be written as:
\begin{subequations}
\begin{flalign}
& \myeqlabel{Mass} &&
    \eqmathbox{\frac{\widebar{D}\rho'}{Dt} + \bar{\rho}\,\nabla\cdot\boldsymbol{u}'
            = \mathbf{0}} & \label{eq:linear_EOM_vector_mass}\\
& \myeqlabel{Momentum} &&
    \eqmathbox{\frac{\widebar{D}\mathbf{u}'}{Dt} + \frac{1}{\bar{\rho}} \nabla p'
            = \mathbf{0}} & \label{eq:linear_EOM_vector_mom}\\
& \myeqlabel{Energy} && 
    \eqmathbox{c_p\frac{\widebar{D}}{Dt}(\bar{\rho}\,T' + \rho'\widebar{T}) + \nabla \cdot (\boldsymbol{u'}\bar{p} + \bar{\boldsymbol{u}}p')
            = \dot{q}'\bar{\rho} + \bar{\dot{q}}\rho' }& \label{eq:linear_EOM_vector_energy}\\
& \myeqlabel{Entropy} && 
    \eqmathbox{\frac{\widebar{D}s'}{Dt}
            = \frac{\dot{q}'}{\bar{\rho}\,\bar{T}}} & \label{eq:linear_EOM_vector_entropy}\\
& \myeqlabel{Vorticity} &&
    \eqmathbox{\frac{\widebar{D}\boldsymbol{\xi}'}{Dt}
            = \mathbf{0}} & \label{eq:linear_EOM_vector_vorticity}
\end{flalign}
\end{subequations}
\begin{subequations}
\begin{flalign}
& \myeqlabel{Mass} &&
    \eqmathbox{\frac{\widebar{D}\rho'}{Dt} + \bar{\rho}\,\nabla\cdot\boldsymbol{u}'
            = \mathbf{0}} & \label{eq:linear_EOM_vector_mass}\\
& \textbf{Momen\rlap{tum}} &&
    \eqmathbox{\frac{\widebar{D}\mathbf{u}'}{Dt} + \frac{1}{\bar{\rho}} \nabla p'
            = \mathbf{0}} & \label{eq:linear_EOM_vector_mom}\\
& \myeqlabel{Energy} &&
    \eqmathbox{c_p\frac{\widebar{D}}{Dt}(\bar{\rho}\,T' + \rho'\widebar{T}) + \nabla \cdot (\boldsymbol{u'}\bar{p} + \bar{\boldsymbol{u}}p')
            = \dot{q}'\bar{\rho} + \bar{\dot{q}}\rho' }& \label{eq:linear_EOM_vector_energy}\\
& \myeqlabel{Entropy} &&
    \eqmathbox{\frac{\widebar{D}s'}{Dt}
            = \frac{\dot{q}'}{\bar{\rho}\,\bar{T}}} & \label{eq:linear_EOM_vector_entropy}\\
& \myeqlabel{Vorticity} &&
    \eqmathbox{\frac{\widebar{D}\boldsymbol{\xi}'}{Dt}
            = \mathbf{0}} & \label{eq:linear_EOM_vector_vorticity}
\end{flalign}
\end{subequations}

\end{document}

Best Answer

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[Tex/LaTex] Number only some equations in an equation array

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