692 lines
35 KiB
TeX
692 lines
35 KiB
TeX
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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% Sets up the standard "article" document class
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\documentclass[12pt]{article}
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% Loads packages
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% Note that I typically load all packages at once (as below).
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% Sometimes you must load them in a certain order...
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\usepackage{amsmath,
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amsfonts,
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amssymb,
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amsthm,
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enumerate,
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enumitem,
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geometry,
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mleftright,
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nicefrac,
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mathtools,
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xparse,
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ifthen,
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tikz,
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pgfplots,
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floatrow,
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multicol,
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caption,
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xurl}
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% Set up PGF plots (used for graphing)
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\pgfplotsset{compat=newest}
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% Fixes spacing caused by using "\left" or "\right"
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\mleftright
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% Sets the margins of the document.
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% One can do this using plain TeX, but why?
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\geometry{margin=1in}
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% Lines 43-46 set up my use of the cleveref package.
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% This provides a "smarter" cross referencing style
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\usepackage[colorlinks=true]{hyperref}
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\usepackage[sort,capitalise]{cleveref}
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\newcommand{\creflastconjunction}{, and\nobreakspace}
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\crefname{enumi}{item}{items}
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% My personal command macros that I use.
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% Note that "\displaySpace" is use to create the correct spacing for displayed math puctuation (and can be easily updated throughout the entire document by changing line 50).
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\newcommand{\R}{\mathbb{R}}
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\newcommand{\C}{\mathbb{C}}
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\newcommand{\Z}{\mathbb{Z}}
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\newcommand{\N}{\mathbb{N}}
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\newcommand{\Q}{\mathbb{Q}}
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\newcommand{\dpp}{\text{.}}
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\newcommand{\dc}{\text{,}}
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\newcommand{\M}{\mathcal{M}}
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\newcommand{\dd}{{\rm d}}
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\newcommand{\fX}{\mathfrak{X}}
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\newcommand{\fg}{\mathfrak{g}}
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\newcommand{\G}{\mathcal{G}}
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\newcommand{\dexp}{\operatorname{dexp}}
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\newcommand{\ad}{\operatorname{ad}}
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\newcommand{\lr}{\ensuremath{\mkern-1.5mu}}
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% Creates my paired math delimiters for easy use.
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\DeclarePairedDelimiter{\pr}{(}{)}
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\DeclarePairedDelimiter{\br}{[}{]}
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\DeclarePairedDelimiter{\cu}{\{}{\}}
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\DeclarePairedDelimiter{\abs}{\lvert}{\rvert}
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\DeclarePairedDelimiter{\norm}{\lVert}{\rVert}
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\DeclarePairedDelimiter{\vt}{\langle}{\rangle}
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% Creates desired header and footer for document.
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% Note that it uses the adaptively created date macro.
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\setlength{\headheight}{15pt}
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\usepackage{fancyhdr}
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\pagestyle{fancy}
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\fancyhf{}
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\fancyhead[R]{}
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\cfoot{}
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\lfoot{\small{\textsc{Dissemination prohibited. \today}}}
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\rfoot{\thepage}
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% Updates equation and figure numbering.
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\counterwithin{equation}{section}
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\counterwithin{figure}{section}
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% Defines theorem-like environments
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\theoremstyle{definition}
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\newtheorem{definition}{Definition}[section]
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\newtheorem{theorem}[definition]{Theorem}
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\newtheorem{lemma}[definition]{Lemma}
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\newtheorem{problem}[definition]{Problem}
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\newtheorem{remark}[definition]{Remark}
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\newtheorem{assumption}[definition]{Assumption}
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\newtheorem{example}[definition]{Example}
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\renewcommand{\thetheorem}{\theenumi\arabic{theorem}}
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% Helps create better arrows in figures
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\usetikzlibrary{arrows.meta}
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%%%%%%%%%%%%%%%%%%%%%%%%
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%%% Automatic proof headers and endings, local labels
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%%%%%%%%%%%%%%%%%%%%%%%%
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\ExplSyntaxOn
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\NewDocumentEnvironment {athm} {m m} {%
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%\phantomsection
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%\addcontentsline{toc}{subsection}{\texorpdfstring{\cref{#2}}{\ref*{#2}}}
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\begin{#1}\label{#2}\global\def\loc{#2}%
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}{%
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\end{#1}%
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}
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\NewDocumentEnvironment{aproof} {} {%
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\begin{proof}[Proof~of~\cref{\loc}]%
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}{%
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\finishproofthus
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\end{proof}%
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}
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\ExplSyntaxOff
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\newcommand{\eqqref}[1]{\cref{eq:\loc.#1}}
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\newcommand{\eqlabel}[1]{\label{eq:\loc.#1}}
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\newcommand{\finishproofthus}{The proof of \cref{\loc} is thus complete.}
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\NewDocumentEnvironment{asol} {} {%
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\begin{proof}[Solution~to~\cref{\loc}]%
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}{%
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\noindent\finishsolthus
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\end{proof}%
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}
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\newcommand{\finishsolthus}{The solution to \cref{\loc} is thus complete.}
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%%%%%%%%%%%%%%%%%%%%%%%%
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%%% Creating blank space with line breaks
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%%%%%%%%%%%%%%%%%%%%%%%%
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% Note that the code in lines 137-147 is simply a method to
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% create blank space that allows for proof environments to
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% break across pages. This is something that TeX does not
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% like by default, so this is my effort to bypass this.
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\newcommand{\linebreaks}[1]{{\mbox{}\par
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\newcount\foo
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\foo=#1
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\loop
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\phantom{a}
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\advance \foo -1
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\ifnum \foo>0
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\repeat
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\mbox{}\par}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%% %%
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%% Beginning of actual text. %%
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%% %%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\title{A survey of numerical integration techniques on manifolds}
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\author{Joshua Lee Padgett}
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\date{\today}
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\begin{document}
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\maketitle
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\begin{abstract}
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The following set of notes provide a very rough overview and introduction to the ideas and concepts needed to approximate differential equations evolving on homogeneous manifolds.
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This document mostly serves as a means to provide some basic ideas and then allow everyone else to inform me of which concepts need additional details in order to improve understanding.
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\end{abstract}
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\tableofcontents
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\section{Introduction}
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\section{Some basic background material}
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The following are some basic definitions needed to discuss numerical integration on homogeneous manifolds (manifolds acted upon transitively by a Lie group).
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Note that the following should not be viewed as a complete list of all needed (or even relevant) topics and concepts.
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Any set of notes created in this way are usually incomplete, so we should either view them as a starting point from which we commence our journey to total understanding or we should discuss what further details are needed.
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\begin{athm}{definition}{def:manifold}
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Let $m,d \in \N = \{1,2,3,\ldots\}$ with $m \le d$ and let $\M \subseteq \R^d$ with $\M \neq \emptyset$.
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Then $\M$ is an $m$-dimensional smooth manifold if and only if for every $p \in \M$ there exist $\Omega \subseteq \R^d$, $U \subseteq \M$ with $p \in U$, and a smooth function $\varphi \colon \Omega \to \R^d$ such that $\varphi$ is a homeomorphism and $\varphi' \circ \varphi^{-1}$ is injective.
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See, e.g., \url{https://en.wikipedia.org/wiki/Manifold} for more details.
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\end{athm}
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\begin{athm}{definition}{def:tangent}
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Let $\M$ be a manifold, let $p \in \M$, and let $\rho \colon [0,1] \to \M$ be a differentiable function with $\rho(0) = p$ (cf.\ \cref{def:manifold}).
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Then the vector
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\begin{equation}
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\pr[\big]{ \tfrac{\dd }{\dd t} \rho } \lr (t) \Bigr\rvert_{t=0}
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\end{equation}
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is a tangent vector at the point $p$.
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The set of all tangent vectors at $p$ is the tangent space at $p$ and is denoted by $T\M \rvert_p$.
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The collection of all tangent spaces at all points $q \in \M$ is called the tangent bundle of $\M$ and is denoted by $T\M = \cup_{q\in\M} T\M\rvert_{q}$.
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See, e.g., \url{https://en.wikipedia.org/wiki/Tangent_space} for more details.
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\end{athm}
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\begin{athm}{definition}{def:vector_field}
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Let $\M$ be a manifold and let $F \colon \M \to T\M$ be a differentiable function which satisfies for all $p \in \M$ that $F(p) \in T\M\rvert_p$ (cf.\ \cref{def:manifold,def:tangent}).
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Then $F$ is a vector field on $\M$.
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The collection of all vector fields on $\M$ is denoted by $\fX(\M)$.
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See, e.g., \url{https://en.wikipedia.org/wiki/Vector_field} for more details.
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\end{athm}
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\begin{athm}{definition}{def:lie_algebra}
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Let $\fg$ be a vector space and let $[\cdot,\cdot] \colon \fg \times \fg \to \fg$ be a bilinear function which satisfies for all $a,b,c\in\fg$ that $[a,b] = -[b,a]$ and
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\begin{equation}
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[a,[b,c]] + [b,[c,a]] + [c,[a,b]] = 0 \dpp
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\end{equation}
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Then we say that $\fg$ is a Lie algebra and we call the function $[\cdot,\cdot] \colon \fg \times \fg \to \fg$ the Lie bracket on $\fg$.
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See, e.g., \url{https://en.wikipedia.org/wiki/Lie_algebra} for more details.
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\end{athm}
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\begin{athm}{problem}{prob:lie_alg}
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Let $\fg = \mathfrak{so}(3) = \{ a \in \R^{3\times 3} \colon a^* = -a \}$ and let $[ \cdot , \cdot ] \colon \fg \times \fg \to \fg$ satisfy for all $a,b \in \fg$ that $[a,b] = ab - ba$.
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Show that $\fg$ is a Lie algebra (cf.\ \cref{def:lie_algebra}).
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\end{athm}
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\begin{asol}
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\linebreaks{30}
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\end{asol}
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\begin{athm}{problem}{prob:lie_alg2}
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Let $\M$ be a manifold, let $C^\infty(\M)$ be the set of smooth function on $\M$, let $\fg = \fX(\M)$, and let $[ \cdot , \cdot ] \colon \fg \times \fg \to \fg$ satisfy for all $X,Y \in \fX(\M)$, $f \in C^\infty(\M)$ that $[X,Y](f) = (X\circ Y)(f) - (Y\circ X)(f)$ (cf.\ \cref{def:manifold,def:vector_field}).
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Show that $\fg$ is a Lie algebra (cf.\ \cref{def:lie_algebra}).
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\end{athm}
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\begin{asol}
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\linebreaks{30}
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\end{asol}
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\begin{athm}{definition}{def:adjoint}
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Let $\fg$ be a Lie algebra (cf.\ \cref{def:lie_algebra}).
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Then we define $\ad^n \colon \fg \times \fg \to \fg$, $n\in\N_0 = \{0\} \cup \N$, to be the functions which satisfy for all $n \in \N$, $u,v \in \fg$ that $\ad^0(u,v) = v$ and
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\begin{equation}
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\ad^n(u,v) =
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[u,\ad^{n-1}(u,v)]
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\dpp
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\end{equation}
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See, e.g., \url{https://en.wikipedia.org/wiki/Adjoint_representation} for more details.
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\end{athm}
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%\begin{athm}{definition}{def:lie_alg_homo}
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%Let $\fg,\mathfrak{h}$ be Lie algebras and let $\varphi \colon \fg \to \mathfrak{h}$ be a linear function which satisfies for all $u,v\in\fg$ that
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%\begin{equation}
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%\varphi\pr[\big]{ [u,v]_\fg } = \br[\big]{ \varphi(u) , \varphi(v) }_{\mathfrak{h}}
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%\end{equation}
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%(cf.\ \cref{def:lie_algebra}).
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%Then we say that $\varphi$ is a Lie algebra homomorphism.
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%\end{athm}
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\begin{athm}{definition}{def:lie_group}
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Let $\G$ be a manifold (cf.\ \cref{def:manifold}).
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We say that $\G$ is a Lie group if and only if there exists a function $\cdot \colon \G \times \G \to \G$ and $e \in \G$ such that
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\begin{enumerate}[label=(\roman*)]
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\item it holds for all $p,q,r \in \G$ that $p \cdot (q \cdot r) = (p \cdot r) \cdot q$,
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\item it holds for all $p \in \G$ that $p \cdot e = e \cdot p = p$,
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\item it holds for all $p \in \G$ that there exists $p^{-1} \in \G$ such that $p^{-1} \cdot p = e$, and
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\item it holds that $\cdot \colon \G \times \G \to \G$ and $\G \ni p \mapsto p^{-1} \in \G$ are smooth.
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\end{enumerate}
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See, e.g., \url{https://en.wikipedia.org/wiki/Lie_group} for more details.
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\end{athm}
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\begin{athm}{problem}{prob:lie_group}
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Let $\G = \{ a \in \R^{3\times 3} \colon a^{-1} = a^* \}$ and let $\cdot \colon \G \times \G \to \G$ satisfy for all $a,b\in\G$ that $a\cdot b = ab$.
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Show that $\G$ is a Lie group (cf.\ \cref{def:lie_group}).
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\emph{Note:} We have not defined matrix manifolds, but we can easily update this definition to allow for such things (e.g., via representations).
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\end{athm}
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\begin{asol}
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\linebreaks{30}
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\end{asol}
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\begin{athm}{definition}{def:alg_of_group}
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Let $\G$ be a Lie group and let $e \in \G$ satisfy for all $p \in \G$ that $p \cdot e = e \cdot p = p$ (cf.\ \cref{def:lie_group}).
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Then we define the Lie algebra of a Lie group to be $\fg = T\G\rvert_{e}$ (cf.\ \cref{def:lie_algebra,def:tangent}).
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The Lie bracket on $\fg$ is the function $[ \cdot , \cdot ] \colon \fg \times \fg \to \fg$ which satisfies for all $u,v \in \fg$, $g,h \colon [0,1] \to \G$ with $g$ and $h$ being differentiable, $g(0) = h(0) = e$, $g'(0) = u$, and $h'(0) = v$ that
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\begin{equation}
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[u,v] = \frac{\partial^2}{\partial t \partial s} g(t) \cdot h(s) \cdot g^{-1}(t) \bigg\rvert_{t=s=0} \dpp
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\end{equation}
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See, e.g., \url{https://en.wikipedia.org/wiki/Lie_group%E2%80%93Lie_algebra_correspondence} for more details.
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\end{athm}
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\begin{athm}{problem}{prob:lie_group2}
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Let $\G = \operatorname{SO}(3) = \{ a \in \R^{3\times 3} \colon a^{-1} = a^*,\ \det(a)=1 \}$ and let $\cdot \colon \G \times \G \to \G$ satisfy for all $a,b\in\G$ that $a\cdot b = ab$.
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Show that $\fg = \mathfrak{so}(3)$ is the Lie algebra of $\G$ (cf.\ \cref{def:alg_of_group}).
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\end{athm}
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\begin{asol}
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\linebreaks{30}
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\end{asol}
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\begin{athm}{definition}{def:action}
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Let $\M$ be a manifold, let $\G$ be a Lie group, let $e \in \G$ satisfy for all $p \in \G$ that $p \cdot e = e \cdot p = p$, and let $\Lambda \colon \G \times \M \to \M$ be a smooth function which satisfies for all $p,q\in\G$, $y\in\M$ that
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\begin{equation}
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\Lambda(e,y) = y
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\qquad\text{and}\qquad
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\Lambda\pr[\big]{p,\Lambda(q,y)} = \Lambda(p\cdot q,y)
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\end{equation}
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(cf.\ \cref{def:manifold,def:lie_group}).
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Then we say that $\Lambda$ is an (left) action of $\G$ on $\M$.
|
||
|
|
See, e.g., \url{https://en.wikipedia.org/wiki/Lie_group_action} for more details.
|
||
|
|
\end{athm}
|
||
|
|
|
||
|
|
\begin{athm}{definition}{def:right}
|
||
|
|
Let $\G$ be a Lie group, let $e \in \G$ satisfy for all $p \in \G$ that $p \cdot e = e \cdot p = p$, and let $\fg$ be the Lie algebra of $\G$ (cf.\ \cref{def:lie_group,def:alg_of_group}).
|
||
|
|
Then we define the function $R_v \colon \G \to \G$, $v \in \G$, to be the function which satisfies for all $u,v\in\G$ that $R_v(u) = u\cdot v$.
|
||
|
|
Moreover, we have that for all $v \in \G$ it holds that $R_v' = T\rvert_e R_v \colon \fg \to T\G\rvert_v$ (cf.\ \cref{def:tangent}).
|
||
|
|
See, e.g., \url{https://math.stackexchange.com/questions/1740179/differential-of-the-multiplication-and-inverse-maps-on-a-lie-group} for more details.
|
||
|
|
\end{athm}
|
||
|
|
|
||
|
|
\begin{athm}{definition}{def:exp}
|
||
|
|
Let $\G$ be a Lie group, let $e \in \G$ satisfy for all $p \in \G$ that $p \cdot e = e \cdot p = p$, let $\fg$ be the Lie algebra of $\G$, and let $\sigma_a \colon [0,1] \to \G$, $a \in \fg$, be a differentiable function satisfying for all $a\in\fg$, $t \in [0,1]$ that $\sigma_a(0) = e$ and
|
||
|
|
\begin{equation}
|
||
|
|
\sigma_a'(t) = R_{\sigma_a(t)}'(a)
|
||
|
|
\end{equation}
|
||
|
|
(cf.\ \cref{def:lie_algebra,def:lie_group,def:right,def:alg_of_group}).
|
||
|
|
Then we define the exponential map $\exp \colon \fg \to \G$ to be the function which satisfies for all $a \in \fg$ that $\exp(a) = \sigma_a(1)$.
|
||
|
|
See, e.g., \url{https://en.wikipedia.org/wiki/Exponential_map_(Lie_theory)#:~:text=In%20the%20theory%20of%20Lie,tool%20for%20studying%20Lie%20groups.}~for more details.
|
||
|
|
\end{athm}
|
||
|
|
|
||
|
|
\begin{athm}{problem}{prob:exp}
|
||
|
|
Let $d \in \N$, let $\G = \operatorname{GL}(d;\R) = \{ a \in \R^{d\times d} \colon \det(a) \neq 0 \}$, and let $\cdot \colon \G \times \G \to \G$ satisfy for all $a,b\in\G$ that $a\cdot b = ab$.
|
||
|
|
\begin{enumerate}[label=\alph*.]
|
||
|
|
\item Show that $\mathfrak{gl}_d = \R^{d\times d}$ is the Lie algebra of $\G$ (cf.\ \cref{def:alg_of_group}).
|
||
|
|
\item Show that for all $a \in \G$ it holds that
|
||
|
|
%\begin{equation}
|
||
|
|
$\exp(a) = \sum_{k=0}^\infty \tfrac{1}{k!} a^n$
|
||
|
|
%\end{equation}
|
||
|
|
(cf.\ \cref{def:exp}).
|
||
|
|
\item Determine an action of $\G$ on the manifold $\R^d$ (cf.\ \cref{def:action}). Is this action unique?
|
||
|
|
\end{enumerate}
|
||
|
|
\end{athm}
|
||
|
|
|
||
|
|
\begin{asol}
|
||
|
|
\linebreaks{30}
|
||
|
|
\end{asol}
|
||
|
|
|
||
|
|
\begin{athm}{lemma}{lem:1}
|
||
|
|
Let $\M$ be a manifold, let $\G$ be a Lie group, let $e \in \G$ satisfy for all $p \in \G$ that $p \cdot e = e \cdot p = p$, let $\fg$ be the Lie algebra of $\G$, let $\Lambda \colon \G \times \M \to \fg$ be an action, let $T \in (0,\infty)$, $y_0 \in \M$, $a \in \fg$, let $\rho \colon [0,1] \to \G$ be a differentiable function satisfying that $\rho(0) = e$ and $\rho'(0) = a$, and let $y \colon [0,T] \to \M$ be a differentiable function satisfying for all $t\in [0,T]$ that $y(0) = y_0$ and
|
||
|
|
\begin{equation}
|
||
|
|
\pr[\big]{ \tfrac{\dd}{\dd t} y } \lr (t) = \tfrac{\dd}{\dd s} \Lambda\pr[\big]{ \rho(s) , y(t) } \Bigr\rvert_{s=0}
|
||
|
|
\end{equation}
|
||
|
|
(cf.\ \cref{def:manifold,def:lie_algebra,def:lie_group,def:action,%
|
||
|
|
def:alg_of_group}).
|
||
|
|
Then it holds for all $t\in[0,T]$ that
|
||
|
|
\begin{equation}
|
||
|
|
\pr[\big]{ \tfrac{\dd}{\dd t} y } \lr (t) = \Lambda\pr[\big]{ \exp(ta) , y_0 }
|
||
|
|
\end{equation}
|
||
|
|
(cf.\ \cref{def:exp}).
|
||
|
|
\end{athm}
|
||
|
|
|
||
|
|
\begin{aproof}
|
||
|
|
|
||
|
|
\end{aproof}
|
||
|
|
|
||
|
|
\section{Approximating differential equations on manifolds}
|
||
|
|
|
||
|
|
We now begin our exploration of how we may approximate solutions to differential equations evolving on (homogeneous) manifolds.
|
||
|
|
To that end, let $\M$ be a manifold, let $T \in (0,\infty)$, $y_0 \in \M$, let $F \colon [0,T] \times \M \to T\M$, and let $y \colon [0,T] \to \M$ be a differentiable function which satisfies for all $t \in [0,T]$ that $y(0) = y_0$ and
|
||
|
|
\begin{equation}\label{eq:3_1}
|
||
|
|
\pr[\big]{ \tfrac{\dd}{\dd t} y } \lr (t) = F(t,y(t))
|
||
|
|
\end{equation}
|
||
|
|
(cf.\ \cref{def:manifold,def:tangent}).
|
||
|
|
Our goal is to approximate the solution to \cref{eq:3_1} in a manner which preserves the fact that for all $t \in [0,T]$ it holds that $y(t) \in \M$ (you should convince yourself that this is indeed true).
|
||
|
|
While this may seem trivial, it is easy to see that many \emph{classical} numerical methods fail to preserve this property.
|
||
|
|
For an example of such a failure, please see \cref{prob:num1} below.
|
||
|
|
|
||
|
|
\begin{athm}{problem}{prob:num1}
|
||
|
|
Let $T \in (0,\infty)$, $a \in \R^3$, let $y_0 = (y_0^1,y_0^2,y_0^3) \in \R^3$ satisfy that $(y_0^1)^2 + (y_0^2)^2 + (y_0^3)^2 = 1$, let $(\cdot,\cdot) \colon \R^3 \times \R^3 \to \R^3$ satisfy for all $u = (u^1,u^2,u^3) \in \R^3$, $v = (v^1,v^2,v^3)\in\R^3$ that $(u,v) = ( u^2v^3-u^3v^2 , u^3v^1 - u^1v^3 , u^1v^2 - u^2v^1 )$, and let $y \colon [0,T] \to \R^3$ be a differentiable function which satisfies for all $t \in [0,T]$ that $y(0) = y_0$ and
|
||
|
|
\begin{equation}
|
||
|
|
\pr[\big]{ \tfrac{\dd}{\dd t} y } \lr (t) = \pr[\big]{ a, y(t) } \dpp
|
||
|
|
\end{equation}
|
||
|
|
\begin{enumerate}[label=\alph*.]
|
||
|
|
\item Show that for all $t \in [0,T]$ it holds that $y(t) \in \operatorname{SO}(3)$ (cf.\ \cref{prob:lie_group2}).
|
||
|
|
\item Let $w \colon [0,T] \to \R^3$ satisfy for all $t \in [0,T]$ that $w(0) = y_0$ and
|
||
|
|
\begin{equation}\label{eq:3_3}
|
||
|
|
w(t) = w_0 + t \pr[\big]{ a, w_0 } \dpp
|
||
|
|
\end{equation}
|
||
|
|
Show that for all $t \in (0,T]$ it holds that $w(t) \not\in \operatorname{SO}(3)$.
|
||
|
|
\emph{Note:} The function in \cref{eq:3_3} is actually known as Euler's method and is the basic tangent line approximation to a vector field.
|
||
|
|
\end{enumerate}
|
||
|
|
\end{athm}
|
||
|
|
|
||
|
|
\begin{asol}
|
||
|
|
\linebreaks{30}
|
||
|
|
\end{asol}
|
||
|
|
|
||
|
|
In order to circumvent the issues observed in \cref{prob:num1} above, we assume that there exist a Lie group $\G$ with Lie algebra $\fg$ and $e \in \G$ satisfying for all $p \in \G$ that $p \cdot e = e \cdot p = p$, an action $\Lambda \colon \G \times \M \to \fg$, a function $a \colon [0,T] \times \M \to \fg$, and a differentiable function $\rho_{t,x} \colon [0,1] \to \G$, $t\in [0,T]$, $x \in \fg$, which satisfies for all $s \in [0,1]$, $t\in[0,T]$, $x\in\fg$ that $\rho_{t,x}(0) = e$, $\rho_{t,x}'(0) = a(t,x)$, and
|
||
|
|
\begin{equation}\label{eq:2_2}
|
||
|
|
\pr[\big]{ \tfrac{\dd}{\dd t} y } \lr (t) = \tfrac{\dd}{\dd s} \Lambda\pr[\big]{ \rho_{t,a(t,y(t))}(s) , y(t) } \Bigr\rvert_{s=0}
|
||
|
|
\end{equation}
|
||
|
|
(cf.\ \cref{def:manifold,def:lie_algebra,def:lie_group,def:action,%
|
||
|
|
def:alg_of_group,eq:3_1}).
|
||
|
|
|
||
|
|
\begin{athm}{remark}{rem:1}
|
||
|
|
While the above assumptions may seem restrictive, it is important to note that Sophus Lie's third fundamental theorem guarantees that \cref{eq:2_2} will always hold \emph{locally} (cf., e.g., \url{https://en.wikipedia.org/wiki/Lie%27s_third_theorem}).
|
||
|
|
Thus, we may construct such actions locally throughout the interval of interest and then use the semigroup property of the exponential map to ``piece these solutions together.''
|
||
|
|
However, in practice, one rarely runs into such pathological issues (at least, when trying to utilize structure-preserving methods).
|
||
|
|
\end{athm}
|
||
|
|
|
||
|
|
\begin{athm}{definition}{def:dexp}
|
||
|
|
Let $\G$ be a Lie group and let $\fg$ be the Lie algebra of $\G$ (cf.\ \cref{def:lie_group,def:lie_algebra}).
|
||
|
|
Then we define $\dexp \colon \fg \times \fg \to \fg$ to be the function which satisfies for all $u,v \in \fg$ that
|
||
|
|
\begin{equation}
|
||
|
|
\pr[\big]{ \tfrac{\dd}{\dd t} \exp } \lr (u)(v) = R_{\exp(u)}' \circ \dexp_u(v)
|
||
|
|
\end{equation}
|
||
|
|
(cf.\ \cref{def:exp,def:right}).
|
||
|
|
See, e.g., \url{https://en.wikipedia.org/wiki/Derivative_of_the_exponential_map} for more details.
|
||
|
|
\end{athm}
|
||
|
|
|
||
|
|
\begin{athm}{lemma}{lem:2}
|
||
|
|
Let $\G$ be a Lie group and let $\fg$ be the Lie algebra of $\G$ (cf.\ \cref{def:lie_group,def:lie_algebra}).
|
||
|
|
Then it holds for all $u,v \in \fg$ that
|
||
|
|
\begin{equation}
|
||
|
|
\dexp_u(v) = \pr[\big]{ \ad^1(u,v) }^{-1} \br[\Big]{ \exp\pr[\big]{ \ad^1(u,v) } - v }
|
||
|
|
= \sum_{k=0}^\infty \frac{1}{(k+1)!} \ad^k(u,v)
|
||
|
|
\end{equation}
|
||
|
|
(cf.\ \cref{def:dexp,def:exp,def:adjoint}).
|
||
|
|
\end{athm}
|
||
|
|
|
||
|
|
\begin{aproof}
|
||
|
|
|
||
|
|
\end{aproof}
|
||
|
|
|
||
|
|
\begin{athm}{lemma}{lem:2a}
|
||
|
|
Let $\G$ be a Lie group, let $\fg$ be the Lie algebra of $\G$, and let $B_0,B_1,B_2,\ldots \in \R$ satisfy for all $n \in \N_0$ that
|
||
|
|
$B_n = \sum_{i=0}^n \sum_{j=0}^i (-1)^j \binom{i}{j} \nicefrac{j^n}{(i+1)}$
|
||
|
|
(cf.\ \cref{def:lie_group,def:lie_algebra}).
|
||
|
|
Then it holds for all $u,v \in \fg$ that
|
||
|
|
\begin{equation}
|
||
|
|
\begin{split}
|
||
|
|
\dexp_u^{-1}(v)
|
||
|
|
& = \sum_{k=0}^\infty \frac{B_k}{k!} \ad^k(u,v) \\
|
||
|
|
& = v - \tfrac{1}{2} \ad^1(u,v) + \tfrac{1}{12} \ad^2(u,v) - \tfrac{1}{720} \ad^4(u,v) + \tfrac{1}{30240} \ad^6(u,v) - \ldots
|
||
|
|
\end{split}
|
||
|
|
\end{equation}
|
||
|
|
(cf.\ \cref{def:dexp,def:adjoint}).
|
||
|
|
\end{athm}
|
||
|
|
|
||
|
|
\begin{aproof}
|
||
|
|
|
||
|
|
\end{aproof}
|
||
|
|
|
||
|
|
\begin{athm}{remark}{rem:b}
|
||
|
|
The sequence of rational numbers
|
||
|
|
$B_0,B_1,B_2,\ldots \in \R$ which satisfy for all $n \in \N_0$ that
|
||
|
|
\begin{equation}\label{eq:3_8}
|
||
|
|
B_n = \sum_{i=0}^n \sum_{j=0}^i (-1)^j \binom{i}{j} \frac{j^n}{(i+1)}
|
||
|
|
\end{equation}
|
||
|
|
are known as \emph{Bernoulli's numbers} (cf., e.g., \url{https://en.wikipedia.org/wiki/Bernoulli_number}).
|
||
|
|
\cref{eq:3_8} is just one of many representations of these numbers.
|
||
|
|
\end{athm}
|
||
|
|
|
||
|
|
\begin{athm}{lemma}{lem:3}
|
||
|
|
Let $\M$ be a manifold, let $T \in (0,\infty)$, $y_0 \in \M$, let $\G$ be a Lie group, let $e \in \G$ satisfy for all $p \in \G$ that $p \cdot e = e \cdot p = p$, let $\fg$ be the Lie algebra of $\G$, let $\Lambda \colon \G \times \M \to \fg$ be an action, let $a \colon [0,T] \times \M \to \fg$, let $\rho_{t,x} \colon [0,1] \to \G$, $t\in [0,T]$, $x \in \fg$, be a differentiable function which satisfies for all $s \in [0,1]$, $t\in[0,T]$, $x \in \fg$ that $\rho_{t}(0) = e$ and $\rho_{t}'(0) = a(t,x)$, let $y \colon [0,T] \to \M$ be a differentiable function which satisfies for all $t\in[0,T]$ that $y(0) = y_0$ and
|
||
|
|
\begin{equation}
|
||
|
|
\pr[\big]{ \tfrac{\dd}{\dd t} y } \lr (t) = \tfrac{\dd}{\dd s} \Lambda\pr[\big]{ \rho_{t,a(t,x)}(s) , y(t) } \Bigr\rvert_{s=0} \dc
|
||
|
|
\end{equation}
|
||
|
|
and let $\Theta \colon [0,T] \to \fg$ satisfy for all $t \in [0,T]$ that $\Theta(0) = O$ and
|
||
|
|
\begin{equation}
|
||
|
|
\pr[\big]{ \tfrac{\dd}{\dd t} \Theta } \lr (t) = \dexp_{\Theta(t)}^{-1} \pr[\Big]{ a\pr[\big]{ t,y(t) } }
|
||
|
|
= \dexp_{\Theta(t)}^{-1} \pr[\bigg]{ a\pr[\Big]{ t, \Lambda\pr[\big]{ \exp\pr[\big]{ \Theta(t) } , y_0 } } }
|
||
|
|
\end{equation}
|
||
|
|
(cf.\ \cref{def:manifold,def:lie_algebra,def:lie_group,def:action,%
|
||
|
|
def:alg_of_group,def:tangent,def:dexp,def:exp}).
|
||
|
|
Then there exists $T_* \in (0,T]$ such that for all $t \in [0,T_*]$ it holds that
|
||
|
|
\begin{equation}
|
||
|
|
y(t) = \Lambda\pr[\Big]{ \exp\pr[\big]{ \Theta(t) } , y_0 } \dpp
|
||
|
|
\end{equation}
|
||
|
|
\end{athm}
|
||
|
|
|
||
|
|
\begin{aproof}
|
||
|
|
|
||
|
|
\end{aproof}
|
||
|
|
|
||
|
|
Note how the assumptions employed in \cref{lem:4} below differ from, e.g., \cref{lem:3} above.
|
||
|
|
Due to the particular structure we have encountered in our actual project, \cref{lem:4} below utilizes a slightly simplified setting in order to reduce the cumbersome of the ensuing notation.
|
||
|
|
|
||
|
|
\begin{athm}{lemma}{lem:4}
|
||
|
|
Let $\M$ be a manifold, let $T \in (0,\infty)$, $y_0 \in \M$, let $\G$ be a Lie group, let $e \in \G$ satisfy for all $p \in \G$ that $p \cdot e = e \cdot p = p$, let $\fg$ be the Lie algebra of $\G$, let $\Lambda \colon \G \times \M \to \fg$ be an action, let $a \colon [0,T] \to \fg$, let $\rho_{t} \colon [0,1] \to \G$, $t\in [0,T]$, be a differentiable function which satisfies for all $s \in [0,1]$, $t\in[0,T]$ that $\rho_{t}(0) = e$ and $\rho_{t}'(0) = a(t)$, let $y \colon [0,T] \to \M$ be a differentiable function which satisfies for all $t\in[0,T]$ that $y(0) = y_0$ and
|
||
|
|
\begin{equation}
|
||
|
|
\pr[\big]{ \tfrac{\dd}{\dd t} y } \lr (t) = \tfrac{\dd}{\dd s} \Lambda\pr[\big]{ \rho_{t,a(t)}(s) , y(t) } \Bigr\rvert_{s=0} \dc
|
||
|
|
\end{equation}
|
||
|
|
let $B_0,B_1,B_2,\ldots \in \R$ satisfy for all $n \in \N_0$ that
|
||
|
|
$B_n = \sum_{i=0}^n \sum_{j=0}^i (-1)^j \binom{i}{j} \nicefrac{j^n}{(i+1)}$,
|
||
|
|
let $\Theta \colon [0,T] \to \fg$ satisfy for all $t \in [0,T]$ that $\Theta(0) = O$ and
|
||
|
|
\begin{equation}
|
||
|
|
\pr[\big]{ \tfrac{\dd}{\dd t} \Theta } \lr (t) = \dexp_{\Theta(t)}^{-1} \pr[\big]{ a\pr[]{ t } }
|
||
|
|
= \dexp_{\Theta(t)}^{-1} \pr[\big]{ a\pr[]{ t } } \dc
|
||
|
|
\end{equation}
|
||
|
|
and let $\Theta^{[k]} \colon [0,T] \to \fg$, $k \in \N_0$, satisfy for all $k \in \N$, $t \in [0,T]$ that
|
||
|
|
$\Theta^{[0]}(t) = O$ and
|
||
|
|
\begin{equation}
|
||
|
|
\Theta^{[k]}(t) = \int_0^t \dexp_{\Theta^{[k-1]}(\xi)}^{-1} \pr[\big]{ a(\xi) } \, \dd \xi
|
||
|
|
\end{equation}
|
||
|
|
(cf.\ \cref{def:manifold,def:lie_algebra,def:lie_group,def:action,%
|
||
|
|
def:alg_of_group,def:tangent,def:dexp,def:exp}).
|
||
|
|
Then
|
||
|
|
\begin{enumerate}[label=(\roman*)]
|
||
|
|
\item it holds for all $k \in \N$, $t \in [0,T]$ that
|
||
|
|
\begin{equation}
|
||
|
|
\Theta^{[k]}(t)
|
||
|
|
= \sum_{i=0}^\infty \frac{B_i}{i!} \int_0^t \ad^i\pr[\big]{ \Theta^{[k-1]}(\xi), a(\xi) } \, \dd \xi
|
||
|
|
\end{equation}
|
||
|
|
and
|
||
|
|
\item there exists $T_* \in (0,T]$ such that for all $t \in [0,T_*]$ it holds that
|
||
|
|
\begin{equation}
|
||
|
|
\Theta(t) = \lim_{k\to\infty} \Theta^{[k]}(t)
|
||
|
|
\end{equation}
|
||
|
|
\end{enumerate}
|
||
|
|
(cf.\ \cref{def:adjoint}).
|
||
|
|
\end{athm}
|
||
|
|
|
||
|
|
\begin{aproof}
|
||
|
|
|
||
|
|
\end{aproof}
|
||
|
|
|
||
|
|
\begin{athm}{remark}{rem:3}
|
||
|
|
Armed with \cref{lem:4}, we are now in a position to construct structure-preserving numerical approximations to \cref{eq:3_1}.
|
||
|
|
These approximations will proceed via two steps.
|
||
|
|
First, we will choose some $m \in \N_0$ and utilize $\Theta^{[m]}$ in our approximation method: this requires proving a so-called \emph{convergence result} for the Picard iterates.
|
||
|
|
Next, we will construct an appropriate approximation to the exponential map: this will be accomplished via certain \emph{rational function} approximation techniques.
|
||
|
|
These two issues will be discussed further as these notes continue.
|
||
|
|
\end{athm}
|
||
|
|
|
||
|
|
We now (informally) continue the line of thinking developed in \cref{lem:4} above.
|
||
|
|
First, note that direct calculations yield that for all $t \in [0,T]$ it holds that
|
||
|
|
\begin{equation}\label{expand1}
|
||
|
|
\Theta^{[1]}(t) = \int_0^t a(\xi_1) \, \dd \xi_1 \dc
|
||
|
|
\end{equation}
|
||
|
|
\begin{equation}\label{expand2}
|
||
|
|
\begin{split}
|
||
|
|
\Theta^{[2]}(t)
|
||
|
|
& = \int_0^t a(\xi_1) \, \dd \xi_1
|
||
|
|
- \tfrac{1}{2} \int_0^t \left[ \int_0^{\xi_1} a(\xi_2) \, \dd \xi_2 , a(\xi_1) \right] \, \dd \xi_1 \\
|
||
|
|
& \qquad + \tfrac{1}{12} \int_0^t \left[ \int_0^{\xi_1} a(\xi_2) \, \dd \xi_2 , \left[ \int_0^{\xi_1} a(\xi_2) \, \dd \xi_2 , a(\xi_1) \right] \right] \, \dd \xi_1 + \ldots \dc
|
||
|
|
\end{split}
|
||
|
|
\end{equation}
|
||
|
|
and
|
||
|
|
\begin{equation}\label{expand3}
|
||
|
|
\begin{split}
|
||
|
|
\Theta^{[3]}(t)
|
||
|
|
& = \int_0^t a(\xi_1) \, \dd \xi_1
|
||
|
|
- \tfrac{1}{2} \int_0^t \left[ \int_0^{\xi_1} a(\xi_2) \, \dd \xi_2 , a(\xi_1) \right] \, \dd \xi_1 \\
|
||
|
|
& \qquad + \tfrac{1}{12} \int_0^t \left[ \int_0^{\xi_1} a(\xi_2) \, \dd \xi_2 , \left[ \int_0^{\xi_1} a(\xi_2) \, \dd \xi_2 , a(\xi_1) \right] \right] \, \dd \xi_1 \\
|
||
|
|
& \qquad + \tfrac{1}{4} \int_0^t \left[ \int_0^{\xi_1} \left[ \int_0^{\xi_2} a(\xi_3) \, \dd\xi_3 , a(\xi_2) \right] \, \dd\xi_2, a(\xi_1) \right] \, \dd\xi_1 \\
|
||
|
|
& \qquad - \tfrac{1}{24} \int_0^t \left[ \int_0^{\xi_1} \left[ \int_0^{\xi_2} a(\xi_3) \, \dd\xi_3 , \left[ \int_0^{\xi_2} a(\xi_3) \, \dd\xi_3 , a(\xi_2) \right] \right] \, \dd\xi_2 , a(\xi_1) \right] \, \dd \xi_1 \\
|
||
|
|
& \qquad - \tfrac{1}{24} \int_0^t \left[ \int_0^{\xi_1} \left[ \int_0^{\xi_2} a(\xi_3) \, \dd\xi_3 , a(\xi_2) \right] \, \dd \xi_2 , \left[ \int_0^{\xi_1} a(\xi_2) \, \dd\xi_2 , a(\xi_1) \right]\right] \, \dd\xi_1 \\
|
||
|
|
& \qquad - \tfrac{1}{24} \int_0^t \left[ \int_0^{\xi_1} a(\xi_2) \, \dd \xi_2 , \left[ \int_0^{\xi_1} \left[ \int_0^{\xi_2} a(\xi_3) \, \dd\xi_3 , a(\xi_2) \right] \, \dd \xi_2 , a(\xi_1) \right] \right ] \, \dd \xi_1 + \ldots \dpp
|
||
|
|
\end{split}
|
||
|
|
\end{equation}
|
||
|
|
Observe that we could have continued these calculations, but things become increasingly tedious (however, things can be simplified considerably via the use of graph theory).
|
||
|
|
%%%Next, note that \cref{expand1,expand2,expand3} seem to indicate that one could construct $H_k \colon [0,T] \to \fg$, $k\in\N_0$, which satisfy for all $k \in \N_0$, $t\in[0,T]$ that $H_k$ is a linear combination of terms including exactly $k+1$ integrals (or, equivalently, $k$ Lie brackets) and
|
||
|
|
%%%\begin{equation}
|
||
|
|
%%%\Theta(t) = \sum_{k=0}^\infty H_k(t) \dpp
|
||
|
|
%%%\end{equation}
|
||
|
|
Next, note that even when we keep all (infinitely many) terms in \cref{expand2} or \cref{expand3}, we are only able to approximate the function $\Theta$ with limited accuracy (this is a direct consequence of the Picard-Lindel{\"o}f theorem (cf., e.g., \url{https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem}).
|
||
|
|
That is, let $\norm{\cdot} \colon \fg \to [0,\infty)$ be an appropriate norm.
|
||
|
|
Then, there exists $C_k \in \R$, $k\in\N_0$, and $f \colon \N_0 \to [0,\infty)$ such that for all $k\in\N_0$, $t \in [0,T_*]$ it holds that
|
||
|
|
\begin{equation}
|
||
|
|
\norm[\big]{ \Theta(t) - \Theta^{[k]}(t) } \le C_k t^{f(k)} \dpp
|
||
|
|
\end{equation}
|
||
|
|
This indicates that keeping \emph{all} terms in the low-level expansions is not ideal.
|
||
|
|
This motivates the result in \cref{lem:5} below.
|
||
|
|
|
||
|
|
\begin{athm}{lemma}{lem:5}
|
||
|
|
Let $\M$ be a manifold, let $T \in (0,\infty)$, $y_0 \in \M$, let $\G$ be a Lie group, let $e \in \G$ satisfy for all $p \in \G$ that $p \cdot e = e \cdot p = p$, let $\fg$ be the Lie algebra of $\G$, let $\Lambda \colon \G \times \M \to \fg$ be an action, let $a \colon [0,T] \to \fg$, let $\rho_{t} \colon [0,1] \to \G$, $t\in [0,T]$, be a differentiable function which satisfies for all $s \in [0,1]$, $t\in[0,T]$ that $\rho_{t}(0) = e$ and $\rho_{t}'(0) = a(t)$, let $y \colon [0,T] \to \M$ be a differentiable function which satisfies for all $t\in[0,T]$ that $y(0) = y_0$ and
|
||
|
|
\begin{equation}\label{lem:5_ode}
|
||
|
|
\pr[\big]{ \tfrac{\dd}{\dd t} y } \lr (t) = \tfrac{\dd}{\dd s} \Lambda\pr[\big]{ \rho_{t,a(t)}(s) , y(t) } \Bigr\rvert_{s=0} \dc
|
||
|
|
\end{equation}
|
||
|
|
let $B_0,B_1,B_2,\ldots \in \R$ satisfy for all $n \in \N_0$ that
|
||
|
|
$B_n = \sum_{i=0}^n \sum_{j=0}^i (-1)^j \binom{i}{j} \nicefrac{j^n}{(i+1)}$,
|
||
|
|
let $\Theta \colon [0,T] \to \fg$ satisfy for all $t \in [0,T]$ that $\Theta(0) = O$ and
|
||
|
|
\begin{equation}
|
||
|
|
\pr[\big]{ \tfrac{\dd}{\dd t} \Theta } \lr (t) = \dexp_{\Theta(t)}^{-1} \pr[\big]{ a\pr[]{ t } }
|
||
|
|
= \dexp_{\Theta(t)}^{-1} \pr[\big]{ a\pr[]{ t } } \dc
|
||
|
|
\end{equation}
|
||
|
|
and let $\Omega^{[k]} \colon [0,T] \to \fg$, $k \in \N_0$, satisfy for all $k \in \N$, $t \in [0,T]$ that
|
||
|
|
$\Omega^{[0]}(t) = O$ and
|
||
|
|
\begin{equation}\label{eq:3_23}
|
||
|
|
\Omega^{[k]}(t) = \sum_{i=0}^{k-1} \frac{B_i}{i!} \int_0^t \ad^i\pr[\big]{ \Omega^{[k-1]}(\xi), a(\xi) } \, \dd \xi
|
||
|
|
\end{equation}
|
||
|
|
(cf.\ \cref{def:manifold,def:lie_algebra,def:lie_group,def:action,%
|
||
|
|
def:alg_of_group,def:tangent,def:dexp,def:exp,def:adjoint}).
|
||
|
|
Then
|
||
|
|
\begin{enumerate}[label=(\roman*)]
|
||
|
|
\item\label{i1} there exists $T_* \in (0,T]$ such that for all $t \in [0,T_*]$ it holds that
|
||
|
|
\begin{equation}
|
||
|
|
\Theta(t) = \lim_{k\to\infty} \Omega^{[k]}(t)
|
||
|
|
\end{equation}
|
||
|
|
and
|
||
|
|
\item\label{i2} it holds for all $k \in \N_0$, $t \in [0,T_*]$, and $\norm{\cdot} \colon \fg \to [0,\infty)$ which satisfy for all $u,v \in \fg$, $c \in \R$ that $\norm{u+v} \le \norm{u} + \norm{v}$, $\norm{cu} = \abs{c}\norm{u}$, and $\norm{u} = 0$ if and only if $u=O$ that there exists $C \in \R$ such that
|
||
|
|
\begin{equation}\label{eq:3_25}
|
||
|
|
\norm[\big]{ \Theta(t) - \Omega^{[k]}(t) } \le C t^{k+1} \dpp
|
||
|
|
\end{equation}
|
||
|
|
\end{enumerate}
|
||
|
|
\end{athm}
|
||
|
|
|
||
|
|
\begin{aproof}
|
||
|
|
|
||
|
|
\end{aproof}
|
||
|
|
|
||
|
|
It is now important to emphasize what is implied by \cref{lem:5} above.
|
||
|
|
\begin{enumerate}[label=\arabic*.]
|
||
|
|
\item \Cref{i1} in \cref{lem:5} above demonstrates that we can truncate the Picard iterates and still obtain the desired convergence to the true solution. \emph{Note:} We have purposely avoided the issue of discussing\slash defining what a limit means in a Lie algebra, but hopefully this is clear.
|
||
|
|
\item \Cref{i2} in \cref{lem:5} above shows that we may choose our Picard iterate based on the amount of accuracy we desire.
|
||
|
|
This can be seen in the right-hand side of \cref{eq:3_25}.
|
||
|
|
\item It should be clear that \cref{i2} in \cref{lem:5} holds true if we can prove the result for one such $\norm{\cdot} \colon \fg \to [0,\infty)$ as we are in a finite-dimensional setting.
|
||
|
|
\item Something which is hidden in the details (and often misunderstood) is the constant ``$C$'' in \cref{i2} in \cref{lem:5}.
|
||
|
|
This constant is a function of the underlying smoothness of the function $a$, as it will be the result of trying to bound the nested commutators within the Picard iterates.
|
||
|
|
For our project we will not need to concern ourselves with this constant too much; we only need to understand its role and how it can be controlled.
|
||
|
|
\end{enumerate}
|
||
|
|
|
||
|
|
\begin{athm}{problem}{prob:1}
|
||
|
|
Verify \cref{i2} in \cref{lem:5} for the case $k=1$. Feel free to impose as much regularity (i.e., increase the assumptions on the objects introduced in \cref{lem:5}) if you feel this will help.
|
||
|
|
\end{athm}
|
||
|
|
|
||
|
|
\begin{asol}
|
||
|
|
\linebreaks{30}
|
||
|
|
\end{asol}
|
||
|
|
|
||
|
|
\begin{athm}{problem}{prob:2}
|
||
|
|
Verify \cref{i2} in \cref{lem:5} in the case where $k=2$, $T = 1$, $\M = \G = \operatorname{SO}(3)$ and $\Lambda \colon \G \times \G \to \fg$ satisfies for all $a,b \in \G$ that $\Lambda(a,b) = ab$.
|
||
|
|
If you want to simplify things further, you may let $\alpha \in \fg$ and choose $a \colon [0,T] \to \fg$ to satisfy for all $t \in [0,T]$ that $a(t) = t\alpha$.
|
||
|
|
\end{athm}
|
||
|
|
|
||
|
|
\begin{asol}
|
||
|
|
\linebreaks{30}
|
||
|
|
\end{asol}
|
||
|
|
|
||
|
|
\subsection{A first attempt at a numerical approximation}
|
||
|
|
|
||
|
|
We will now (somewhat informally) motivate the construction of a numerical approximation to, e.g., \cref{lem:5_ode}.
|
||
|
|
To that end, we will commence by fixing $k=1$.
|
||
|
|
This and \cref{eq:3_23} yield that for all $t \in [0,T_*]$ it holds that
|
||
|
|
\begin{equation}\label{eq:3_26}
|
||
|
|
\Omega^{[1]}(t) = B_0 \int_0^t \ad^0\pr[\big]{ \Omega^{[0]}(\xi) , a(\xi) } \, \dd \xi
|
||
|
|
= \int_0^t a(\xi) \, \dd \xi \dpp
|
||
|
|
\end{equation}
|
||
|
|
We now apply an approximation to the integral in \cref{eq:3_26} (since, in general, we will not be able to compute the integral exactly) to obtain that for all $t\in[0,T_*]$ it holds that
|
||
|
|
\begin{equation}\label{eq:3_27}
|
||
|
|
\Omega^{[1]}(t) \approx t a(0) \dpp
|
||
|
|
\end{equation}
|
||
|
|
|
||
|
|
\begin{athm}{problem}{prob:3}
|
||
|
|
Let $T,L \in (0,\infty)$, let $\fg$ be a Lie algebra, let $\norm{\cdot} \colon \fg \to [0,\infty)$ satisfy for all $u,v \in \fg$, $c \in \R$ that $\norm{u+v} \le \norm{u} + \norm{v}$, $\norm{cu} = \abs{c}\norm{u}$, and $\norm{u} = 0$ if and only if $u=O$, and let $a \colon [0,T] \to \fg$ satisfy for all $s,t \in [0,T]$ that $\norm{a(t) -a(s)} \le L\abs{t-s}$ (cf.\ \cref{def:lie_algebra}).
|
||
|
|
Show that for all $t\in[0,T]$ it holds that
|
||
|
|
\begin{equation}
|
||
|
|
\norm*{ \int_0^t a(\xi) \, \dd \xi - t a(0) } \le \frac{Lt^2}{2} \dpp
|
||
|
|
\end{equation}
|
||
|
|
\end{athm}
|
||
|
|
|
||
|
|
\begin{asol}
|
||
|
|
\linebreaks{30}
|
||
|
|
\end{asol}
|
||
|
|
|
||
|
|
Combining \cref{lem:3,lem:5,eq:3_27} demonstrate that for all $t\in[0,T_*]$ it holds that
|
||
|
|
\begin{equation}
|
||
|
|
y(t) = \Lambda\pr[\Big]{ \exp\pr[\big]{ \Theta(t) } , y_0 } \approx \Lambda\pr[\Big]{ \exp\pr[\big]{ \Omega^{[1]}(t) } , y_0 } = \Lambda\pr[\Big]{ \exp\pr[\big]{ t a(0) } , y_0 } \dpp
|
||
|
|
\end{equation}
|
||
|
|
Provided that the group action, $\Lambda$, and $\exp(ta(0))$ can be computed with relative ease, we have arrived at a reasonable numerical approximation.
|
||
|
|
Moreover, the approximations employed thus far have been linear approximations performed on objects in the Lie algebra---thus, our approximation will still lie in the original manifold for all $t\in[0,T_*]$.
|
||
|
|
|
||
|
|
\begin{athm}{problem}{prob:4}
|
||
|
|
Develop a similar approximation using $\Omega^{[2]}$ (cf.\ \cref{lem:5}).
|
||
|
|
Note that in this case, you will likely need to impose additional regularity assumptions (be sure to only impose precisely enough to derive your result).
|
||
|
|
\end{athm}
|
||
|
|
|
||
|
|
\begin{asol}
|
||
|
|
\linebreaks{30}
|
||
|
|
\end{asol}
|
||
|
|
|
||
|
|
\subsection{Approximating the exponential map}
|
||
|
|
|
||
|
|
|
||
|
|
|
||
|
|
\end{document}
|