A lista de equações também não lista aquelas no ambiente \align

A lista de equações também não lista aquelas no ambiente \align

Nenhum dos exemplos existentes que encontrei para listar equações inclui as equações no ambiente \align. Como posso incluir também os dois tipos de equações na mesma lista de equações? Aqui está um MWE:

    \documentclass[english]{article}
\setcounter{secnumdepth}{2}
%\setcounter{tocdepth}{1}
\usepackage{amsmath}
\usepackage{tocloft}
\usepackage{xstring}
\usepackage[unicode=true, pdfusetitle,
 bookmarks=true,bookmarksnumbered=false,bookmarksopen=false,
 breaklinks=false,pdfborder={0 0 0},backref=false,colorlinks=false]
 {hyperref}

\makeatletter
\numberwithin{equation}{section}

% we use this for our refernces as well
\AtBeginDocument{\renewcommand{\ref}[1]{\mbox{\autoref{#1}}}}

% redefinition of \equation for convenience
\let\oldequation = \equation
\let\endoldequation = \endequation
\AtBeginDocument{\let\oldlabel = \label}% \AtBeginDocument because hyperref redefines \label
\newcommand{\mynewlabel}[1]{%
  \StrBehind{#1}{eq:}[\Str]% remove "eq:" from labels
  \myequations{\Str}\oldlabel{#1}}
  \renewenvironment{equation}{%
  \oldequation
  \let\label\mynewlabel
}{\endoldequation}

% redefinition of \eqnarray for convenience
\let\oldeqnarray = \eqnarray
\let\endoldeqnarray = \endeqnarray
%\AtBeginDocument{\let\oldlabel = \label}% \AtBeginDocument because hyperref redefines \label
\newcommand{\mynewlabelarray}[1]{%
  \StrBehind{#1}{eq:}[\Str]% remove "eq:" from labels
  \myequations{\Str}\oldlabel{#1}}
  \renewenvironment{eqnarray}{%
  \oldeqnarray
  \let\label\mynewlabelarray
}{\endoldeqnarray}

\newcommand{\listequationsname}{\normalsize  List of Equations}
\newlistof{myequations}{equ}{\listequationsname}
\newcommand{\myequations}[1]{%
      \addcontentsline{equ}{myequations}{\protect\numberline{\theequation}#1}}
\setlength{\cftmyequationsnumwidth}{3em}

\makeatother

\begin{document}
%\tableofcontents
\listofmyequations

\section{Section title}
\begin{equation}
  F=q[E+(v\times B)]
  \label{eq:Force}
\end{equation}

\begin{equation}
  \tau=F\times r
  \label{eq:Torque}
\end{equation}
If the electrical force in \ref{eq:Force} is ignored,
and the remaining magnetic force is used in \ref{eq:Torque},
with the assumption that $v$ is perpendicular to $B$, we find that
\begin{equation}
      \tau=qvBrsin\theta
  \label{eq:Magnetic}
\end{equation}

\begin{align}
  \min_{u_{i}(t),y_i, i=1...N}\!\!\!\!\!\! J(u_i(t),y_i)  &:=  \sum_{i=1}^N \int_0^{T} R_i(u_{i}(t),t) dt \label{eq:objective function}\\
+&  \xi \int_{0}^{T}  \left(\theta\frac{M - I(t)}{M}K(t)  - D(t)\right)^2 dt +  \sum_{i=1}^N \gamma_i y_i  \notag\\
  + & \sum_{i=1}^N p_{i} \int_{T_i}^{T} u_i(t-T_i)dt \notag\\
 +&  h\int_{0}^T \left[\theta\frac{M - I(t)}{M}K(t) - D(t)\right]^+ dt, \notag
\end{align}

subject to
\begin{align}
  K(t)& = \sum_{i=1}^N u_i(t-T_i), & \quad t\in [0,T] \label{eq:2} \\
  u_i(t) &\le  % \theta_i S_i(t) y_i =
  \theta_i (M_i - I_i(t)) y_i,  & \ \ i = 1\ldots N \quad t\in [0,T-T_i] \label{eq:ui}\\
  u_i(t) & =  0, & i = 1\ldots N \quad t\in [T-T_i,T] \label{eq:uio}\\
  \dot{I}_i(t)& = f_i(I_i(t)), &  i = 1 \ldots N \quad t\in [0,T]& \label{eq:dotIi}\\
  \dot{I}(t)& = f(I(t)), &\quad t\in [0,T] \label{eq:dotI}\\
  u_i(t) & \ge  0, & i = 1 \ldots N \quad t\in [0,T]& \label{eq:const5}\\
  K(t)& \ge  0, &\quad t\in [0,T]& \\
  y_i &\in  \{0,1\},  & i = 1 \ldots N& \label{eq:const6}
\end{align}
%where

\begin{equation}\label{eq:capacity constraint}
    u^{j}_i(t) \leq Min_{p\in C^{j}_{i}} \sum_{k \in {K^{j}_{ip}}} u^{j+1}_k(t-T_k)
\end{equation}

\end{document} 

Responder1

Não acho que seja uma boa ideia modificar a \labelmacro manualmente, pois ela é usada e redefinida por vários outros pacotes, incluindo os pacotes hyperrefe cleveref. Eu apenas aplicaria \myequationsas diretivas diretamente às equações que deveriam estar listadas na Lista de Equações. (Quando usado com ambientes gathere align, parece que as \myequationsdiretivas foram emitidasdepoisdiretivas de quebra de linha. Não sei por quê.)

Óh, epor favor não use eqnarrayambientes; use alignem vez disso.

insira a descrição da imagem aqui

\documentclass[english]{article}
\setcounter{secnumdepth}{2}
\usepackage{mathtools} % for \coloneqq and \mathclap macros
\usepackage{tocloft}
\usepackage[unicode=true, pdfusetitle, bookmarks=true, 
    bookmarksnumbered=false, bookmarksopen=false, 
    breaklinks=false, backref=false, 
    colorlinks=true,allcolors=black]
   {hyperref}
\usepackage[noabbrev]{cleveref}

\counterwithin{equation}{section}

\newcommand{\listequationsname}{\normalsize List of Equations}
\newlistof{myequations}{equ}{\listequationsname}
\newcommand{\myequations}[1]{%
      \addcontentsline{equ}{myequations}%
      {\protect\numberline{\theequation}#1}}     
\setlength{\cftmyequationsnumwidth}{2.5em}

\begin{document}
%\tableofcontents
\listofmyequations

\allowdisplaybreaks

\section{Section title}
\begin{gather}
   F=q[E+(v\times B)]
   \label{eq:Force}  \\ \myequations{Force}
   \tau=F\times r
   \label{eq:Torque} 
\end{gather} \myequations{Torque}
If the electrical force in \cref{eq:Force} is ignored, and 
if the remaining magnetic force is used in \cref{eq:Torque}, 
with the assumption that $v$ is perpendicular to~$B$, we 
find that
\begin{equation}
   \tau=qvBr\sin\theta
   \label{eq:Magnetic} 
   \myequations{Magnetic}
\end{equation}

\begin{equation} 
   \label{eq:objective function} 
   \myequations{Objective function}
\begin{aligned}[t] 
\smash[b]{\min_{\mathclap{%
     \substack{u_{i}(t),y_i,\\ i=1,\dots,N}}}}
     \,J(u_i(t),y_i)  
&\coloneqq \sum_{i=1}^N \int_0^{T} R_i(u_{i}(t),t)\, dt 
    \\
&\quad+ \xi \int_{0}^{T} \left(\theta\frac{M - I(t)}{M}K(t) - D(t)\right)^{\!2} dt 
      + \sum_{i=1}^N \gamma_i y_i \\
&\quad+ \sum_{i=1}^N p_{i} \int_{T_i}^{T} u_i(t-T_i)\,dt\\
&\quad+  h\int_{0}^T \left[\theta\frac{M - I(t)}{M}K(t) - D(t)\right]^+ dt,
\end{aligned}
\end{equation}
subject to
\begin{align}
  K(t) &= \sum\nolimits_{i=1}^N u_i(t-T_i), 
    && t\in [0,T] 
       \label{eq:2} \\
  u_i(t) &\le  % \theta_i S_i(t) y_i =
  \theta_i (M_i - I_i(t)) y_i,  
    && i = 1,\ldots, N, 
       \quad t\in [0,T-T_i] 
       \label{eq:ui} \\ \myequations{ui}
  u_i(t) &=  0, 
    && i = 1,\ldots, N, 
       \quad t\in [T-T_i,T] 
       \label{eq:uio} \\ \myequations{uio}
  \dot{I}_i(t) &= f_i(I_i(t)), 
    && i = 1,\ldots,N, 
       \quad t\in [0,T] 
       \label{eq:dotIi} \\ \myequations{dotIi}
  \dot{I}(t) &= f(I(t)), 
    && t\in [0,T] 
       \label{eq:dotI}\\ \myequations{dotI}
  u_i(t) &\ge  0, 
    && i = 1,\ldots,N, 
       \quad t\in [0,T]
       \label{eq:const5} \\ \myequations{const5}
  K(t) &\ge  0, 
    && t\in [0,T] \\
  y_i &\in  \{0,1\},  
    && i = 1,\ldots,N 
      \label{eq:const6}
\end{align} \myequations{const6}
where
\begin{equation}
  \label{eq:capacity constraint}
  \myequations{Capacity constraint}
    u^{j}_i(t) \leq \min_{p\in C^{j}_{i}} \sum_{k \in {K^{j}_{ip}}} u^{j+1}_k(t-T_k)
\end{equation}

\end{document} 

Responder2

Como comentou Mico, você não deve redefinir \label. Se quiser evitar o fornecimento de um rótulo, você pode criar um comando simples que faz as duas coisas. Eu usei \refaqui para pegar o número correto nos alinhamentos ams.

\documentclass[english]{article}
\setcounter{secnumdepth}{2}
\usepackage{mathtools} % for \coloneqq and \mathclap macros
\usepackage{tocloft}
\usepackage[unicode=true, pdfusetitle, bookmarks=true, 
    bookmarksnumbered=false, bookmarksopen=false, 
    breaklinks=false, backref=false, 
    colorlinks=true,allcolors=black]
   {hyperref}
\usepackage[noabbrev]{cleveref}

\counterwithin{equation}{section}

\newcommand{\listequationsname}{\normalsize List of Equations}
\newlistof{myequations}{equ}{\listequationsname}
\newcommand{\myequations}[1]{%
      \addcontentsline{equ}{myequations}%
      {\protect\numberline{\ref{eq:#1}}#1}}     
\setlength{\cftmyequationsnumwidth}{2.5em}

\newcommand\myeq[1]{\label{eq:#1}\myequations{#1}}

\begin{document}
%\tableofcontents
\listofmyequations

\allowdisplaybreaks

\section{Section title}
\begin{gather}
   F=q[E+(v\times B)]
   \myeq{Force}  \\
   \tau=F\times r
   \myeq{Torque} 
\end{gather}
If the electrical force in \cref{eq:Force} is ignored, and 
if the remaining magnetic force is used in \cref{eq:Torque}, 
with the assumption that $v$ is perpendicular to~$B$, we 
find that
\begin{equation}
   \tau=qvBr\sin\theta
   \myeq{Magnetic} 
\end{equation}

\begin{equation} 
   \myeq{objective function} 
\begin{aligned}[t] 
\smash[b]{\min_{\mathclap{%
     \substack{u_{i}(t),y_i,\\ i=1,\dots,N}}}}
     \,J(u_i(t),y_i)  
&\coloneqq \sum_{i=1}^N \int_0^{T} R_i(u_{i}(t),t)\, dt 
    \\
&\quad+ \xi \int_{0}^{T} \left(\theta\frac{M - I(t)}{M}K(t) - D(t)\right)^{\!2} dt 
      + \sum_{i=1}^N \gamma_i y_i \\
&\quad+ \sum_{i=1}^N p_{i} \int_{T_i}^{T} u_i(t-T_i)\,dt\\
&\quad+  h\int_{0}^T \left[\theta\frac{M - I(t)}{M}K(t) - D(t)\right]^+ dt,
\end{aligned}
\end{equation}
subject to
\begin{align}
  K(t) &= \sum\nolimits_{i=1}^N u_i(t-T_i), 
    && t\in [0,T] 
       \label{eq:2} \\
  u_i(t) &\le  % \theta_i S_i(t) y_i =
  \theta_i (M_i - I_i(t)) y_i,  
    && i = 1,\ldots, N, 
       \quad t\in [0,T-T_i] 
       \myeq{ui} \\
  u_i(t) &=  0, 
    && i = 1,\ldots, N, 
       \quad t\in [T-T_i,T] 
       \myeq{uio} \\
  \dot{I}_i(t) &= f_i(I_i(t)), 
    && i = 1,\ldots,N, 
       \quad t\in [0,T] 
       \myeq{dotIi} \\
  \dot{I}(t) &= f(I(t)), 
    && t\in [0,T] 
       \myeq{dotI}\\
  u_i(t) &\ge  0, 
    && i = 1,\ldots,N, 
       \quad t\in [0,T]
       \myeq{const5} \\
  K(t) &\ge  0, 
    && t\in [0,T] \\
  y_i &\in  \{0,1\},  
    && i = 1,\ldots,N 
      \myeq{const6}
\end{align}
where
\begin{equation}
  \myeq{capacity constraint}
    u^{j}_i(t) \leq \min_{p\in C^{j}_{i}} \sum_{k \in {K^{j}_{ip}}} u^{j+1}_k(t-T_k)
\end{equation}

\end{document} 

insira a descrição da imagem aqui

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