michal-h21
commented
1 week ago There are two issues:
eq:section4-sh4isequationand notrdfequation- this version of
\RdfDrawGraphhas issues with labels that contain-. In fact, Graphviz has a problem, but we need to remove them. So here is the fixed version:
\documentclass[12pt,landscape]{article}
\usepackage{amssymb,amsfonts,amsmath,amsthm}
\usepackage{accents}
\newcommand{\setT}{{\mathord{\mathbb T}}}
\makeatletter
\newcommand{\sbullet}{%
\hbox{\fontfamily{lmr}\fontsize{.4\dimexpr(\f@size
pt)}{0}\selectfont\textbullet}}
\DeclareRobustCommand{\mathbullet}{\accentset{\sbullet}}
\makeatother
\newcommand{\eqdef}{\overset{\mbox{\tiny{def}}}{=}}
\usepackage{thmtools}
\newtheorem{thm}{Theorem}
\newtheorem{defn}{Definition}
\newtheorem{lem}{Lemma}
\newtheorem{cor}{Corollary}
\newtheorem{prop}{Proposition}
\newtheorem{rem}{Remark}
\theoremstyle{definition}
\usepackage{xr-hyper}
\usepackage[colorlinks]{hyperref}
\externaldocument[A-]{nordstroem-jee}
\usepackage{cleveref}
% \usepackage{cleveref-nameref}
% \usepackage{cleveref-fix}
\Crefname{defn}{Definition}{Definitions}
\crefname{thm}{Theorem}{Theorems}
\Crefname{lem}{Lemma}{Lemmas}
\Crefname{prop}{Proposition}{Propositions}
\crefname{prop}{Proposition}{Propositions}
\crefname{cor}{Corollary}{Corollaries}
\Crefname{cor}{Corollary}{Corollaries}
\crefname{rem}{Remark}{Remarks}
\Crefname{rem}{Remark}{Remarks}
\usepackage{tikz}
\usepackage{shellesc}
\usetikzlibrary{graphs,graphdrawing,calc}
\usegdlibrary{layered}
\usepackage[T1]{fontenc}
\usepackage{rdfref}
\robustify\int
\NewDocumentEnvironment{rdfequation}{o +b}{
\begin{equation}
#2
\IfBlankF{#1}{
\rdflabel{#1}
\AssignProperty{#1}{eq:value}{$#2$}
}
\end{equation}
}{}
\newwrite\graphwrite
\makeatletter
\ExplSyntaxOn
\newcommand\graphout[1]{\protected@write\graphwrite{\let\owrite\write\def\write{\immediate\owrite}}{#1}}
% #1 name of the graph (default \jobname-graph) -- it needs to be different than name of any existing file!
% #2 RDF property to be printed
% #3 list of allowed rdf:type for objects
% #4 list of allowed rdf:type for subjects
% don't forget to call latex with --shell-escape
\NewDocumentCommand\RdfDrawGraph{O{\jobname-graph} O{rdfs:label} m m}{%
\immediate\openout\graphwrite=#1.dot\relax%
\graphout{digraph hello\@charlb} % save graph header
\RdfLoopReferences{#3}{#4}{%
\edef\objectlabel{\currentobject}%
\edef\subjectlabel{\currentsubject}%
\regex_replace_all:nnN{[:\-]}{_}\objectlabel%
\regex_replace_all:nnN{[:\-]}{_}\subjectlabel%
\graphout{\objectlabel [shape="rectangle",texlbl="\GetProperty{\currentobject}{#2}"]}
\graphout{\subjectlabel [shape="rectangle",texlbl="\GetProperty{\currentsubject}{#2}"]}
\graphout{\objectlabel~->~\subjectlabel} %
}
\graphout{\@charrb}% save footer
\immediate\closeout\graphwrite%
\ShellEscape{dot2tex~--figonly~--autosize~-t~raw~-o~#1.tmp~#1.dot }%
\InputIfFileExists{#1.tmp}{}%
}
\ExplSyntaxOff
\makeatother
\begin{document}
% we need to declare some properties for the eq: prefix
\AddRdfType{eq}{
\AddPropertyEx{rdf:type}{eq:equation}
\AddPropertyEx{doc:pageNo}{\thepage}
% this will be used in the graph label, change to your liking
\AddPropertyEx{rdfs:label}{Equation \theequation}
}
\makeatletter
\AddRdfType{sec}{
\AddPropertyEx{rdf:type}{sec:sectioning}
\AddPropertyEx{doc:pageNo}{\thepage}
% this will be used in the graph label, change to your liking
\AddPropertyEx{rdfs:label}{Section \@currentlabel}
}
\makeatother
\renewenvironment{proof}[1][\proofname]{{\bfseries Proof of #1.}}{\qed}.
\section{Proof of an important proposition}
\rdflabel{sec:first}
\begin{rdfequation}[eq:wave:1]
\begin{aligned}
% \label{eq:wave:1}
&\partial_t^2 u+2\varkappa \partial_t u -\Delta u=\mathord{e^{-\varkappa t}} a(t,x)(1+u)^3\\
% \label{eq:wave:2}
& (u(0,x),\partial_t u(0,x))=(f(x),g(x)).
\end{aligned}
\end{rdfequation}
\begin{thm}[Local existence]
\rdflabel{thr:section3-fourier:1}
Let $m>\frac{5}{2}$,
$a(t,\cdot)\in L^\infty([0,\infty); H^m(\setT^3)$,
$ f\in H^{m+1}(\setT^3)$ and $ g\in H^{m}(\setT^3)$, then there exists a positive
$T$ and a unique solution $u$ to the Cauchy problem
\cref{eq:wave:1}--\cref{eq:wave:2} such that
\begin{equation*}
u\in L^\infty([0,T]; H^{m+1}(\setT^3)\cap C^{0,1}([0,T]; H^{m}(\setT^3),
\end{equation*}
where $C^{0,1}$ is a Lipschitz continuous function.
\end{thm}
\begin{prop}[A Nonlinear estimate]
\rdflabel{prop:1}
Let $m>\frac{3}{2}$ and $a \in H^{m}$, then there is a universal constant
$C(A)$, depending just on the constants of multiplications and embedding, such
that
\begin{equation}
\label{eq:estimate:3}
\left\|a(1+u)^3 \right\|_{H^{m}}, \left\|a(1+u)^3 \right\|_{L^\infty}\leq C(A) \|a\|_{H^{m}}
\end{equation}
for all $u\in H^m$ with $\|u\|_{H^m}\leq A$.
\end{prop}
\begin{rdfequation}[eq:energy:1]
% \rdflabel{eq:energy:1}
E(t)\eqdef \langle V(t),V(t)\rangle_m=\|\partial_t u_h+\varkappa u_h\|_{H^m}^2+\|\partial_x u_h\|_{ H^m}^2.
\end{rdfequation}
\begin{rdfequation}[eq:Gronwall:2]
\rdflabel{eq:Gronwall:2}
\begin{split}
\sqrt{E(t)} & \leq e^{-\varkappa(t-t_0)}\sqrt{E(t_0)} +
\varkappa^2\int_{t_0}^te^{-\varkappa(t-s)} \left\Vert u_h(s) \right\Vert_{H^{m}}ds
\\ & + \int_{t_0}^te^{-\varkappa (t-s)}e^{-\varkappa s} \left\Vert
a(s,\cdot) \left( 1+u(s) \right)^3 \right\Vert_{H^m}ds
\end{split}.
\end{rdfequation}
\begin{prop}[Estimate for the gradient]
\rdflabel{prop:9}
Let
$ \partial_x u=(\partial_1 u,\partial_2u,\partial_3 u)^{\intercal}$ and
$u\in \mathbullet H^{m+1}$ and $m\geq0$ be an integer.
Then the following holds
\begin{equation*}
\rdflabel{eq:norm:36}
\left\|{u}\right\|_{\mathbullet H^{m+1}}=\|\partial_x u\|_{\mathbullet H^{m}}.
\end{equation*}
\end{prop}
\begin{subequations}
\begin{align}
\rdflabel{eq:section4-sh12}
& \widehat{u}_0^{\prime\prime} + 2\varkappa \widehat{u}_0^{\prime} =
e^{-\varkappa t}\widehat F_0 \\
\rdflabel{eq:section4-sh32}
& \partial_t^2 u_h +2\varkappa \partial_t u_h -\Delta u_h =
\mathord{e^{-\varkappa t}} \left( a(t,x)(1+u)^3 -\widehat{F}_{0} \right),
\end{align}
\end{subequations}
\begin{equation}
\rdflabel{eq:section5-sh:33}
\left\{\begin{array}{ll}
u(x,0)=f(x), & \partial_t u(x,0)=g(x),\\
f\in \mathbullet H^{m+1}, & g\in \mathbullet H^m.
\end{array}\right.
\end{equation}
\begin{prop}[A priori estimates]
\rdflabel{prop:2}
Let $ 0<\varkappa<1$, $1<\beta<\frac{1}{\varkappa}$ and set $\alpha= \|V(0)\|_{H^m}$.
Assume the solution $u=\widehat u_0+u_h$ to
\cref{eq:section4-sh12}--\cref{eq:section4-sh32} with initial data
\cref{eq:section5-sh:33} exists for $t\in [0,T]$.
If $\mathord{\left\|{a(t,\cdot)}\right\|_{ H^m}}$ is sufficiently small,
then there exists a $T^{+}$, $0<T^{+}\leq T$, such that
\begin{equation}
\rdflabel{eq:section4-sh:2}
\sup_{[0,T^{+}]}\mathord{\left\|{u(t)}\right\|_{ H^m}}\leq \alpha\beta
\end{equation}
and
\begin{rdfequation}[eq:section4-sh:1]
% \rdflabel{eq:section4-sh:1}
E(T^{+})\leq E(0).
\end{rdfequation}
\end{prop}
\section{Proof}
\rdflabel{sec:proof}
\begin{proof}[~\autoref{prop:2}]
We start with the proof of inequality \autoref{eq:section4-sh:2}.
Recall that although we have written $\alpha= \|V(0)\|_{H^m}$, the initial
data are in the homogeneous Sobolev space and therefore by
\Cref{prop:9} and \cref{eq:energy:1}, we obtain
\begin{equation}
\|u(0)\|_{H^{m+1}}=\|f\|_{H^{m+1}}=\|\partial_x f\|_{H^{m}}\leq
\|V(0)\|_{H^m}=\alpha.
\end{equation}
Hence, since $ \beta > 0$, it follows from the existence
\Cref{thr:section3-fourier:1} and the continuity property of the
corresponding solutions, that there exists $0<T^+\leq T$ such that
\begin{rdfequation}
\rdflabel{eq:section4-sh4}
\sup_{[0,T^+]}\| u(t)\|_{ H^m}\leq \alpha\beta.
\end{rdfequation}
We now turn to inequality \cref{eq:section4-sh:1}.
For $t\in [0,T^+]$ we observe, using inequality \rdfref{eq:Gronwall:2}
that
\begin{equation}
\label{eq:Gronwall:1}
\begin{split}
\sqrt{E(t)} & \leq e^{-\varkappa t}\sqrt{E(0)} + \varkappa^2\int_{0}^te^{-\varkappa (t-s)} \alpha\beta ds + \int_{0}^te^{-\varkappa (t-s)}e^{-\varkappa s} \left\Vert a(s,\cdot) \left( 1+u(s) \right)^3 \right\Vert_{H^m} ds\\%
& \leq e^{-\varkappa t}\sqrt{E(0)} +\varkappa \left(1 -e^{-\varkappa t}\right)\alpha\beta+ te^{-\varkappa t}\sup_{[0,t]}\left\Vert a(s,\cdot) \left( 1+u(s) \right)^3 \right\Vert_{H^m}.
\end{split}
\end{equation}
A simple algebraic manipulation shows us that, $E(t)\leq E(0)$, if
\begin{equation}
\varkappa \left(e^{\varkappa t}-1\right)\alpha\beta+ t\sup_{[0,t]}\left\Vert a(s,\cdot) \left(1+u(s) \right)^3 \right\Vert_{H^m}\leq \left(e^{\varkappa t}-1\right)\sqrt{E(0)},
\end{equation}
or equivalently
\begin{rdfequation}[eq:section4-sh:7]
% \label{eq:section4-sh:7}
t\sup_{[0,t]}\left\Vert a(s,\cdot) \left(1+u(s) \right)^3 \right\Vert_{H^m}\leq \left(e^{\varkappa t}-1\right)\alpha\left(\beta-\varkappa\right).
\end{rdfequation}
Since for $s\in [0,T^+]$, we can conclude that
$\|u(s)\|_{H^m}\leq \alpha\beta\leq 2\alpha \beta$, we can apply
\Cref{prop:1} with $A=2\alpha \beta$, that results in
\begin{rdfequation}[eq:energy:3]
% \rdflabel{eq:energy:3}
\left\Vert a(s,\cdot) \left(1+u(s) \right)^3 \right\Vert_{H^m}\leq C(2\alpha\beta)\|a(s,\cdot)\|_{H^m}.
\end{rdfequation}
We now set
\begin{rdfequation}[eq:energy:4]
% \rdflabel{eq:energy:4}
\epsilon_0=\frac{\varkappa\alpha(\beta-\varkappa)}{C(2\alpha\beta)}.
\end{rdfequation}
Since $\beta-\varkappa>0$, $\epsilon_0>0$, therefore we can demand the smallness
condition
\begin{rdfequation}[eq:energy:5]
% \rdflabel{eq:energy:5}
\sup_{[0,\infty)}\|a(t,\cdot)\|_{H^m}\leq \epsilon_0.
\end{rdfequation}
We now let $t=T^+$ in inequality \rdfref{eq:section4-sh:7}, then by
inequality \rdfref{eq:energy:3}, condition \rdfref{eq:energy:5}, with
\rdfref{eq:energy:4} in section~\rdfpageref{sec:first}., we conclude that
\begin{rdfequation}[eq:section4-sh3]
% \label{eq:section4-sh3}
\begin{split}
T^+\sup_{[0,T^+]}\left\Vert a(s,\cdot) \left(1+u(s) \right)^3 \right\Vert_{H^m} &
\leq T^+ C(2\alpha\beta)\sup_{[0,T^+]}\|a(s,\cdot)\|_{H^m}\leq T^+
C(2\alpha\beta)\epsilon_0 \\ & = \varkappa T^+\alpha(\beta-\varkappa)\leq (e^{\varkappa
T^+}-1)\alpha(\beta-\varkappa),
\end{split}
\end{rdfequation}
holds and consequently \autoref{eq:section4-sh3} implies inequality
\autoref{eq:section4-sh:7}.
This proves \cref{eq:section4-sh:1} and completes the proof of the
proposition.
In the last step, we used the elementary inequality $x\leq e^{x}-1$.
\end{proof}
%\RdfLoopReferences{}{eq:equation,sec:sectioning}{%
\RdfLoopReferences{}{eq:equation,sec:sectioning}{%
\GetProperty{\currentobject}{rdfs:label} -> \GetProperty{\currentsubject}{rdfs:label}, %
}
\RdfDrawGraph[\jobname-graph][eq:value]{}{eq:equation}
\end{document}
ouboub
Hi
Sorry, but it seems that I have not really understood the syntax for generating a dependency graphs for equations. I took a real example of a proof, changed
equationstordfequationthe same forlabelsandreferences. Here is the exampleHowever,
lualatex --shell-escape --output-dir="build/" a-priori.texorlualatex --shell-escape a-priori.texcompiles without a problem, but no graph is generated.What do I miss?