More stuff

This commit is contained in:
Shakil Rafi 2024-02-15 18:55:42 -06:00
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\documentclass{article}
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\title{Numerical Analysis of DiffEq HW 1}
\author{Shakil Rafi}
\begin{document}
\maketitle
\textbf{1.1} We first need to establish a lemma:
\newtheorem{lem}{Lemma}
\begin{lem}
$hf(t_{{n}+\frac{1}{2}},\frac{1}{2}(y(t_n)+y(t_{n+1})) = hf(t_{n+\frac{1}{2}},y(t_{n+\frac{1}{2}})) +\eta$ where $\eta$ is $\mathcal{O}(h^3)$.
\end{lem}
\begin{align*}
||\eta|| = h||f(t_{n+\frac{1}{2}},\frac{1}{2}(y(t_n)+y(t_{n+1})) - f(t_{n+\frac{1}{2}},y(t_{n+\frac{1}{2}})|| \\
\leq h\lambda||\frac{1}{2}(y(t_n)+y(t_{n+1})) = y(t_{n+\frac{1}{2}}) \\
\leq h\frac{\lambda}{2}|y(t_n)+y(t_n+1) - 2y(t_{n+\frac{1}{2}})||
\end{align*}
Then Taylor expansion around $y(t_{n})$ gives us:
\begin{align*}
&||\eta|| \leq \frac{h\lambda}{2}||y+(y+hy'-2y(y+\frac{1}{2}hy')+\mathcal{O}(h^2)||\\
&= \frac{h\lambda}{2}||y+y+hy^2-2y^2-hyy'+\mathcal{O}(h^2)|| \\
&=\mathcal{O}(h^3)
\end{align*}
\textbf{Proving the convergence of the implicit midpoint rule}:
The implicit midpoint rule is:
\begin{align}
y_{n+1}=y_n+hf(t_n+\frac{h}{2},\frac{1}{2}(y_n+y_{n+1})).
\end{align}
Substituting the exact value then gives us:
\begin{align}
y(t_{n+1}) = y(t_n)+hf(t_n+\frac{h}{2}, \frac{1}{2}(y(t_n)+y(t_{n+1}))+\mathcal{O}(h^2)
\end{align}
Following closely the proof of $(1.9)$ and $(1.4)$ we subtract $(2)$ from $(1)$, and applying the lemma to get:
\begin{align*}
e_{n+1} &= e_n +h(f(t_{n+\frac{1}{2}},\frac{1}{2}(y_n+y_{n+1})) -f(t_{n+\frac{1}{2}},\frac{1}{2}(y(t_n)+y(t_{n+\frac{1}{2}})) +\mathcal{O}(h^3) \\
||e_{n+1}|| &\leq ||e_{n}|| +\frac{\lambda h}{2}(||e_{n+1}||+||e_n||)+\mathcal{O}(h^3)\\
\end{align*}
We then reproduce the steps in Iserles:
\begin{align*}
||e_{n+1}||-\frac{\lambda h}{2}||e_{n-1}|| &\leq ||e_n|| + \frac{\lambda h}{2}||e_n|| + \mathcal{O}(h^3) \\
||e_{n+1}|| &\leq \bigg(\frac{1+\frac{\lambda h}{2}}{1-\frac{\lambda h}{2}}\bigg)||e_n|| + \bigg(\frac{c}{1+\frac{\lambda}{2}}\bigg) + \mathcal{O}(h^3) \\
\end{align*}
And similar to Iserles we claim that:
\begin{align*}
||e_n|| \leq \frac{c}{\lambda}\bigg[ \bigg(\frac{1+\frac{\lambda h}{2}}{1-\frac{\lambda h}{2}}\bigg)^n -1 \bigg]h^2
\end{align*}
We will use induction to prove this step. Clearly it holds for $n=0$. Assume that the above inequality holds upto and including $n\in \mathbb{N}$, then for $n+1$ we have:
\begin{align*}
||e_{n+1}|| &\leq \bigg(\frac{1+\frac{\lambda h}{2}}{1-\frac{\lambda h}{2}}\bigg)\frac{c}{\lambda}\bigg[ \bigg(\frac{1+\frac{\lambda h}{2}}{1-\frac{\lambda h}{2}}\bigg)^n -1 \bigg]h^2+ \bigg(\frac{c}{1+\frac{\lambda}{2}}\bigg) + \mathcal{O}(h^3) \\
&= \frac{c}{\lambda} \bigg[\bigg(\frac{1+\frac{\lambda h}{2}}{1-\frac{\lambda h}{2}}\bigg)^{n+1}\bigg]h^2 - \frac{c}{\lambda}\bigg(\frac{1+\frac{\lambda h}{2}}{1-\frac{\lambda h}{2}}\bigg)h^2 +\mathcal{O}(h^3) \\
&= ||e_n|| \leq \frac{c}{\lambda}\bigg[ \bigg(\frac{1+\frac{\lambda h}{2}}{1-\frac{\lambda h}{2}}\bigg)^{n+1} -1 \bigg]h^2
\end{align*}
\textbf{To show that the $\theta$ method is convergent}, we define the theta method as:
\begin{align*}
y_{n+1} &= y_{n}+h[\theta f(t_n,y_n)+(1-\theta)f(t_{n+1},y_{n+1})]
\end{align*}
Whereas substituting exact solutions gives us:
\begin{align*}
y(t_{n+1}) y(t_n) +h[\theta f(t_n,y(t_n))+(1-\theta)f(t_{n+1},y(t_{n+1}))]+\mathcal{O}(h^3)
\end{align*}
Once again subtracting $(3)$ from $(4)$ gives us:
\begin{align*}
e_{n+1} &= e_n +h[\theta f(t_n,y_n)-\theta f(t_n,y(t_n))+(1-\theta)f(t_{n+1},y(t_{n+1}))+(1-\theta)f(t_{n+1},y(t_{n+1}))] +\mathcal{O}(h^3)\\
||e_{n+1}|| &\leq ||e_n||+h[\theta \lambda ||e_n||+(1-\theta)\lambda ||e_{n+1}||]+\mathcal{O}(h^3) \\
||e_{n+1}|| &\leq \bigg(\frac{1+h\theta \lambda}{1-h(1-\theta)\lambda} \bigg)||e_n||+\frac{c}{1-h(1-\theta)\lambda} +ch^3 \quad \text{for some }c
\end{align*}
Similar to the trapezoid method we will argue:
\begin{align*}
||e_n|| \leq \frac{c}{\lambda} \bigg[ \bigg(\frac{1+h\theta \lambda}{1-h(1-\theta)\lambda} \bigg) ^n -1 \bigg] h^2
\end{align*}
We argue via induction. Clearly we have that it is true for $n=0$ as at that point the exact and approximate solutions are the same. Assume now that it is true upto and including $n$. We need to prove for $n+1$:
\begin{align*}
||e_{n+1}|| &\leq \bigg(\frac{1+h\theta \lambda}{1-h(1-\theta)\lambda} \bigg)||e_n||+\frac{c}{1-h(1-\theta)\lambda} +ch^3\\
||e_{n+1}|| &\leq \bigg(\frac{1+h\theta \lambda}{1-h(1-\theta)\lambda} \bigg)\bigg(\frac{c}{\lambda} \bigg[ \bigg(\frac{1+h\theta \lambda}{1-h(1-\theta)\lambda} \bigg) ^n -1 \bigg] h^2\bigg)+\frac{c}{1-h(1-\theta)\lambda} +ch^3
\end{align*}
Now observe that $\theta$ varies between $1$ and $0$. Thus $\bigg(\frac{1+h\theta \lambda}{1-h(1-\theta)\lambda} \bigg)$ varies between $1+h\lambda$ and $\frac{1}{1-h\lambda}$, both of which are bigger than one. As such we can continue and say:
\begin{align*}
||e_n|| \leq \frac{c}{\lambda} \bigg[ \bigg(\frac{1+h\theta \lambda}{1-h(1-\theta)\lambda} \bigg) ^{n+1} -1 \bigg] h^2
\end{align*}
\textbf{1.2a} Let $y'=Ay$, and let $e_n=y_n-y(nh)$
We want to prove using induction:
\begin{align*}
||e_n||_2 \leq ||y_0||_2 \max_{\lambda \in \sigma(A)}|(1-h\lambda)^n-e^{nh\lambda}|
\end{align*}
But before that we make an observation:
\begin{align*}
e_n = y_n-y(nh)
\end{align*}
And since we are using Euler method, we can say:
\begin{align*}
y(t_{n+1}) &= y(t_n) + hy'(t) + \mathcal{O}(h^2) \\
y_{n+1} - y(t_{n-1}) &= y_n - y(nh) +h[(f(t_n,y(t_n))-f(t_n,y_n))]
\end{align*}
Substitution $A$ gives us:
\begin{align*}
e_{n+1} &= e_n+h[Ay_n - Ay(nh)]+\mathcal{O}(h^2) \\
||e_{n+1}|| &\leq ||e_n||_2 +h\lambda ||e_n||_2+\mathcal{O}(h^2)\\
||e_{n+1}|| &\leq ||e_n||_2(1+h\lambda) +\mathcal{O}(h^2)
\end{align*}
For the induction part we observe that the statement clearly holds true for $n=0$ since then we get:
\begin{align*}
||e_0||_2 &\leq ||y_0||_2 \max_{\lambda \in \sigma(A)}|(1+h\lambda)^0 - e^0| \\
0 &\leq 0
\end{align*}
Using it as our base case, assume the inequality holds for upto and including $n$ the for $||e_{n+1}||$ we have:
\begin{align*}
||e_{n+1}||_2 &\leq ||e_n||_2(1+h\lambda)+\mathcal{O}(h^2) \\
||e_{n+1}||_2 &\leq ||y_0||_2 \max_{\lambda \in \sigma(A)}|(1+h\lambda)^n - e^{nh\lambda}|(1+h\lambda)+\mathcal{O}(h^2x)\\
&\leq ||y_0||_2\max_{\lambda \in \sigma(A)}|(1+h\lambda)^{n+1}-[e^{nh\lambda}+h\lambda e^{nh\lambda}]| \\
&\leq ||y_0||_2\max_{\lambda \in \sigma(A)}|(1+h\lambda)^{n+1}-e^{(n+1)h\lambda}|
\end{align*}
\textbf{1.2b} From the hint, we first seek to prove $1+x \leq e^x$. Let $f(x) = e^x-x-1$, then $f'(x) = e^x-1$ and $f''(x) = e^x$. There is a global minimum of $0$ and this function is concave up, and so $f(x) > 0$ over all $x$ and hence $e^x \geq 1+x$.
\newline
Following the hint again we seek to prove that $1+x+\frac{x^2}{2} \geq e^x$. Observe the series expansion of $e^x$ is $e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}\mathcal{O}(x^3)$. Thus:
\begin{align*}
1+x+\frac{x^2}{2} +\frac{x^3}{3!}+\mathcal{O}(h^4) - 1-x-\frac{x^2}{2} \\
\cancel{1}+\cancel{x}+\cancel{\frac{x^2}{2}} +\frac{x^3}{3!}+\mathcal{O}(h^4) - \cancel{1}-\cancel{x}-\cancel{\frac{x^2}{2}} \\
\frac{x^3}{3!} + \mathcal{O}(h^4) < 0 \text{ as } x\in[-1,0]
\end{align*}
We use a similar logic for the last part of the hint. Observe that: $(a-b)^n = \Sigma^n_{i=0}{n \choose i}a^{n-i}b^i = a^n -na^{n-1}b + ...$. So:
\begin{align*}
\Sigma^n_{i=1}{n \choose i}a^{n-i}b^i &- a^n+na^{n-1}b \\
\Sigma^n_{i=2}{n \choose i}a^{n-i}b^i &\geq 0 \text{ as a is close to being 1 and b is small}
\end{align*}
For the actual proof, let $a=e^x$ and $b=\frac{1}{2}x^2$. We then get:
\begin{align*}
e^{nx} - \frac{1}{2}nx^2e^{(n-1)x} &\leq (e^x-\frac{x^2}{2})^n \leq (1+x)^n \leq e^{nx}
\end{align*}
\textbf{1.4}
\end{document}

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\documentclass[12pt]{article}
\usepackage{amsmath,
mleftright,
amssymb,
amsthm,
nicefrac,
etoolbox,
xparse,
geometry,
enumitem,
mathtools,
bbm
}
\mleftright
\usepackage[colorlinks=true]{hyperref}
\geometry{margin=1in}
\usepackage[sort,capitalize]{cleveref}
\newcommand{\creflastconjunction}{, and\nobreakspace}
\crefformat{equation}{(#2#1#3)}
\crefname{enumi}{item}{items}
\crefname{equation}{}{}
\theoremstyle{plain}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\newtheorem{setting}[theorem]{Setting}
\newtheorem{conjecture}[theorem]{Conjecture}
\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}
\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\numberwithin{equation}{section}
%%%
\input{commands.tex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
%
%
%
\title{MLP starting ideas}
%
%
%
\author{
%%%Joshua Lee Padgett$^{1,2}$
%%%\bigskip
%%%\\
%%%\small{$^1$ Department of Mathematical Sciences, University of Arkansas,}
%%%\vspace{-0.1cm}\\
%%%\small{Arkansas, USA, e-mail: \texttt{padgett@uark.edu}}
%%%\smallskip
%%%\\
%%%\small{$^2$ Center for Astrophysics, Space Physics, and Engineering Research,}
%%%\vspace{-0.1cm}\\
%%%\small{Baylor University, Texas, USA, e-mail: \texttt{padgett@uark.edu}}
}
%
%
%
\date{\today}
%
%
%
\maketitle
\begin{abstract}
%
Abstract goes here\dots
%
\end{abstract}
%
%
%
\tableofcontents
%
%
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Introduction}
\label{sec:intro}
%
%
%
Add an appropriate introduction\dots
%
%
%
%
%
\section{Multilevel Picard approximations for the heat equation}
\label{sec:mlp}
%
%
%
\begin{athm}{theorem}{th:1}
%
%
Let $T,\kappa, \delta \in (0,\infty)$,
$\Theta = \bigcup_{n\in\N}\! \Z^n$,
%
let $\smallU_d \in C^{1,2}([0,T]\times \R^d,\R)$, $d\in\N$, satisfy for all $d\in\N$, $t \in [0,T]$, $x=(x_1,\allowbreak x_2,\allowbreak \dots, \allowbreak x_d)\in\R^d$ that
%
\begin{equation}\label{eq:1}
\abs{ \smallU_d(t,x)} \le \kappa d^\kappa \pr[\big]{ 1 + \textstyle\sum_{k=1}^d \abs{ x_k } }^\kappa
\qquad
\text{and}
\qquad
\pr[]{\tfrac{\partial}{\partial t}\smallU_d}(t,x) = \pr[]{\Delta_x \smallU_d}(t,x) \dc
\end{equation}
%
let $(\Omega, \cF ,\P)$ be a probability space,
%
let $W^{d,\theta} \colon [0,T] \times \Omega\to \R^d$, $d\in\N$, $\theta\in\Theta$, be independent standard Brownian motions,
%
let $\mlp{\littleM}{d}{\theta} \colon [0,T] \times \R^d \times \Omega \to \R$, $d,\littleM \in \Z$, $\theta \in \Theta$, satisfy for all $d,\littleM \in \N$, $\theta \in \Theta$, $t \in [0,T]$, $x \in \R^d$ that
%
\begin{align}
& \mlp{\littleM}{d}{\theta}(t,x)
=
\frac{1}{\littleM} \biggl[ \SmallSum{k=1}{\littleM} \smallU_d \pr[\big]{ 0,x + \sqrt{2}\,W_{t}^{d,(\theta,0,-k)} } \biggr], \nonumber
\end{align}
%
and for every $d,n,\littleM \in \N$ let $\cost{n}{\littleM}{d} \in \N$ be the number of function evaluations of $\smallU_d(0,\cdot)$ and the number of realizations of scalar random variables which are used to compute one realization of $\mlp{\littleM}{d}{0}(T,0) \colon \Omega \to \R$.
%
%
Then there exist $c\in\R$ and $\fR \colon \N \times (0,1] \to \N$ such that for all $d \in \N$, $\varepsilon \in (0,1]$ it holds that
%
\begin{equation}
\textstyle
\pr[\Big]{\E\br[\big]{\abs{\smallU_d(T,0) - \mlp{\fR(d,\varepsilon)}{d}{0}(T,0)}^2}}^{\!\!\nicefrac{1}{2}} \le \varepsilon
\qquad
\text{and}
\qquad
\cost{\fR(d,\varepsilon)}{\fR(d,\varepsilon)}{d} \le c d^c \varepsilon^{-(2+\delta)}
\dpp
\end{equation}
%
%
\end{athm}
%
%
%
\section{Stochastic solutions to parabolic partial differential equations}
\label{sec:sfp}
%
%
%
\begin{athm}{lemma}{lem:feynman-kac_1}
%
%
Let $T \in (0,\infty)$,
%
let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space,
%
let $\smallU_d \in C^{1,2}([0,T]\times\R^d,\R)$, $d\in\N$, satisfy for all $d\in\N$, $t\in[0,T]$, $x\in\R^d$ that
%
\begin{equation}\label{eq:pde_1}
\pr[]{\tfrac{\partial}{\partial t}\smallU_d}(t,x)
+ \pr[]{\Delta_x \smallU_d}(t,x)
= 0 \dc
\end{equation}
%
let $\fwpr^d \colon [0,T] \times \Omega \to \R^d$, $d\in\N$, be standard Brownian motions,
%
and
let $\cX^{d,t,x} \colon [t,T] \times \Omega \to \R^d$, $d\in\N$, $t\in[0,T]$, $x\in\R^d$, be a stochastic process with continuous sample paths satisfying that for all $d\in\N$, $t\in[0,T]$, $s \in [t,T]$, $x\in\R^d$ we have $\mathbb{P}$-a.s.\ that
%
\begin{equation}\label{eq:stochastic_process_1}
\cX_{s}^{d,t,x}
= x + \int_t^s \sqrt{2} \dx \fwpr_r^d
= x + \sqrt{2} \, \fwpr_{t-s}^d
\dpp
\end{equation}
%
Then for all $d\in\N$, $t\in[0,T]$, $x\in\R^d$ it holds that
%
\begin{equation}\label{eq:feynman-kac_sol_1}
\smallU_d(t,x)
=
\E\br[\bpig]{ \smallU_d\pr[\big]{ T , \cX_{T}^{d,t,x} } } \dpp
\end{equation}
%
%
\end{athm}
%
%
%
\begin{aproof}
%
%
\end{aproof}
%
%
%
\newcommand{\Hess}{\operatorname{Hess}}
\newcommand{\Trace}{\operatorname{Trace}}
\begin{athm}{lemma}{lem:feynman-kac_2}
%
%
Let $T \in (0,\infty)$,
%
let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space,
%
let $\sigma_d \colon \R^d \to \R^{d\times d}$, $d\in\N$, be infinitely often differentiable functions,
%
let $\smallU_d \in C^{1,2}([0,T]\times\R^d,\R)$, $d\in\N$, satisfy for all $d\in\N$, $t\in[0,T]$, $x\in\R^d$ that
%
\begin{equation}\label{eq:pde_2}
\pr[]{\tfrac{\partial}{\partial t}\smallU_d}(t,x)
+ \Trace \pr[\pig]{ \sigma(x) \br[]{ \sigma(x) }^* \pr[]{\Hess_x \smallU_d}(t,x) }
= 0 \dc
\end{equation}
%
let $\fwpr^d \colon [0,T] \times \Omega \to \R^d$, $d\in\N$, be standard Brownian motions,
%
and
let $\cX^{d,t,x} \colon [t,T] \times \Omega \to \R^d$, $d\in\N$, $t\in[0,T]$, $x\in\R^d$, be a stochastic process with continuous sample paths satisfying that for all $d\in\N$, $t\in[0,T]$, $s\in[t,T]$, $x\in\R^d$ we have $\P$-a.s.\ that
%
\begin{equation}\label{eq:stochastic_process_2}
\cX_{s}^{d,t,x}
= x + \int_s^t \sqrt{2} \, \sigma\pr[]{ \cX_r^{d,t,x} } \dx \fwpr_r^d
\dpp
\end{equation}
%
Then for all $d\in\N$, $t\in[0,T]$, $x\in\R^d$ it holds that
%
\begin{equation}\label{eq:feynman-kac_sol_2}
\smallU_d(t,x)
=
\E\bpigl[ \smallU_d\pr[\big]{ T , \cX_{T}^{d,t,x} } \bpigr] \dpp
\end{equation}
%
%
\end{athm}
%
%
%
\begin{aproof}
%
%
\end{aproof}
%
%
%
\begin{athm}{lemma}{lem:feynman-kac_3}
%
%
Let $T \in (0,\infty)$,
%
%%%let $\vt{\cdot,\cdot} \colon \bigcup_{d\in\N} \pr[]{ \R^d \times \R^d} \to \bigcup_{d\in\N}\!\R^d$ satisfy for all $d\in\N$, $v = (v_1,\dots,v_d), w = (w_1,\dots,w_d) \in \R^d$ that $\vt{v,w} = \sum_{k=1}^d v_k w_k$,
%
let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space,
%
let $\mu_d \in \R^d \to \R^d$, $d\in\N$, be infinitely often differentiable functions,
%
let $\smallU_d \in C^{1,2}([0,T]\times\R^d,\R)$, $d\in\N$, satisfy for all $d\in\N$, $t\in[0,T]$, $x\in\R^d$ that
%
\begin{equation}\label{eq:pde_3}
\pr[]{\tfrac{\partial}{\partial t}\smallU_d}(t,x)
+ \pr[]{\Delta_x \smallU_d}(t,x)
+ \br[]{ \mu_d(x) }^* \pr[]{ \nabla_x \smallU_d }(t,x)
= 0 \dc
\end{equation}
%
let $\fwpr^d \colon [0,T] \times \Omega \to \R^d$, $d\in\N$, be standard Brownian motions,
%
and
let $\cX^{d,t,x} \colon [t,T] \times \Omega \to \R^d$, $d\in\N$, $t\in[0,T]$, $x\in\R^d$, be a stochastic process with continuous sample paths satisfying that for all $d\in\N$, $t\in[0,T]$, $s\in[t,T]$, $x\in\R^d$ we have $\P$-a.s.\ that
%
\begin{equation}\label{eq:stochastic_process_3}
\cX_{s}^{d,t,x}
= x + \int_s^t \mu_d\pr[]{ \cX_r^{d,t,x} } \dx r + \int_s^t \sqrt{2} \dx \fwpr_r^d
\dpp
\end{equation}
%
Then for all $d\in\N$, $t\in[0,T]$, $x\in\R^d$ it holds that
%
\begin{equation}\label{eq:feynman-kac_sol_3}
\smallU_d(t,x)
=
\E\bpigl[ \smallU_d\pr[\big]{ T , \cX_{T}^{d,t,x} } \bpigr] \dpp
\end{equation}
%
%
\end{athm}
%
%
%
\begin{aproof}
%
%
\end{aproof}
%
%
%
\begin{athm}{lemma}{lem:feynman-kac_4}
%
%
Let $T \in (0,\infty)$,
%
%
let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space,
%
let $\alpha_d \in \R^d \to \R$, $d\in\N$, be infinitely often differentiable functions,
%
let $\smallU_d \in C^{1,2}([0,T]\times\R^d,\R)$, $d\in\N$, satisfy for all $d\in\N$, $t\in[0,T]$, $x\in\R^d$ that
%
\begin{equation}\label{eq:pde_4}
\pr[]{\tfrac{\partial}{\partial t}\smallU_d}(t,x)
+ \pr[]{\Delta_x \smallU_d}(t,x)
+ \alpha_d(x) \smallU_d(t,x)
= 0 \dc
\end{equation}
%
let $\fwpr^d \colon [0,T] \times \Omega \to \R^d$, $d\in\N$, be standard Brownian motions,
%
and
let $\cX^{d,t,x} \colon [t,T] \times \Omega \to \R^d$, $d\in\N$, $t\in[0,T]$, $x\in\R^d$, be a stochastic process with continuous sample paths satisfying that for all $d\in\N$, $t\in[0,T]$, $s\in[t,T]$, $x\in\R^d$ we have $\P$-a.s.\ that
%
\begin{equation}\label{eq:stochastic_process_4}
\cX_{s}^{d,t,x}
= x + \int_s^t \sqrt{2} \dx \fwpr_r^d
\dpp
\end{equation}
%
Then for all $d\in\N$, $t\in[0,T]$, $x\in\R^d$ it holds that
%
\begin{equation}\label{eq:feynman-kac_sol_4}
\smallU_d(t,x)
=
\E\br[\bpig]{ \exp\pr[\big]{ \textstyle\int_t^T \alpha_d( \cX_r^{d,t,x} ) \dx r } \smallU_d\pr[\big]{ T , \cX_{T}^{d,t,x} } } \dpp
\end{equation}
%
%
\end{athm}
%
%
%
\begin{aproof}
%
%
\end{aproof}
%
%
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%\bibliographystyle{acm}
%%%\bibliography{bibfile}
\end{document}

3
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@ -0,0 +1,3 @@
\contentsline {section}{\numberline {1}Introduction}{1}{section.1}%
\contentsline {section}{\numberline {2}Multilevel Picard approximations for the heat equation}{1}{section.2}%
\contentsline {section}{\numberline {3}Stochastic solutions to parabolic partial differential equations}{2}{section.3}%

21
MLP Ideas/bibfile.bib Normal file
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@ -0,0 +1,21 @@
% Encoding: UTF-8
@article{grohs2018proof,
title={A proof that artificial neural networks overcome the curse of dimensionality in the numerical approximation of {B}lack-{S}choles partial differential equations},
author={Grohs, Philipp and Hornung, Fabian and Jentzen, Arnulf and Von Wurstemberger, Philippe},
journal={arXiv:1809.02362},
note={\emph{To appear in Mem.\ Amer.\ Math.\ Soc.}},
year={2018},
pages={124 pages},
}
@article{rio2009moment,
title={Moment inequalities for sums of dependent random variables under projective conditions},
author={Rio, Emmanuel},
journal={J.\ Theor.\ Probab.},
volume={22},
number={1},
pages={146--163},
year={2009},
publisher={Springer}
}

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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% %%
%% Preamble of document %%
%% %%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Sets up the standard "article" document class
\documentclass[12pt]{article}
% Loads packages
% Note that I typically load all packages at once (as below).
% Sometimes you must load them in a certain order...
\usepackage{amsmath,
amsfonts,
amssymb,
amsthm,
enumerate,
enumitem,
geometry,
mleftright,
nicefrac,
mathtools,
xparse,
ifthen,
tikz,
pgfplots,
floatrow,
multicol,
caption}
% Set up PGF plots (used for graphing)
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% Fixes spacing caused by using "\left" or "\right"
\mleftright
% Sets the margins of the document.
% One can do this using plain TeX, but why?
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% Lines 43-46 set up my use of the cleveref package.
% This provides a "smarter" cross referencing style
\usepackage[colorlinks=true]{hyperref}
\usepackage[nosort,capitalise]{cleveref}
\newcommand{\creflastconjunction}{, and\nobreakspace}
\crefname{enumi}{item}{items}
% My personal command macros that I use.
% 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).
\newcommand{\R}{\mathbb{R}}
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\newcommand{\Z}{\mathbb{Z}}
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\newcommand{\pq}{\overrightarrow{PQ}}
\newcommand{\dotp}{\boldsymbol{\cdot}}
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% Creates my paired math delimiters for easy use.
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\DeclarePairedDelimiter{\abs}{\lvert}{\rvert}
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% Creates desired header and footer for document.
% Note that it uses the adaptively created date macro.
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% Updates equation and figure numbering.
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\newtheorem{remark}[definition]{Remark}
\renewcommand{\thetheorem}{\theenumi\arabic{theorem}}
% Helps create better arrows in figures
\usetikzlibrary{arrows.meta}
%%%%%%%%%%%%%%%%%%%%%%%%
%%% Automatic proof headers and endings, local labels
%%%%%%%%%%%%%%%%%%%%%%%%
\ExplSyntaxOn
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%%% Creating blank space with line breaks
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% Note that the code in lines 137-147 is simply a method to
% create blank space that allows for proof environments to
% break across pages. This is something that TeX does not
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\ifnum \foo>0
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\noindent}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% %%
%% Beginning of actual text. %%
%% %%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\title{A survey of direct methods for sparse linear systems}
\author{Joshua Lee Padgett}
\date{\today}
\begin{document}
\maketitle
\begin{abstract}
The following problems are meant to be additional practice which may help students understand some of the more difficult material in \emph{A survey of direct methods for sparse linear systems} by Davis et al.
\end{abstract}
\tableofcontents
\section{Solving sparse triangular systems}
We begin by introducing a few definitions from graph theory.
These definitions and concepts are in no way complete (in fact, there are much better ways of formulating these ideas if we have more background); however, they will suffice for our current needs.
Students who are interested in obtaining a deeper understanding of the material in Davis et al.\ should consult a textbook on introductory graph theory.
\begin{athm}{definition}{def:graph}
Let $n \in \N = \{1,2,3,\ldots\}$. A \emph{graph} $G = (V,E)$ is a set of \emph{nodes} $V = \{v_1,v_2,\allowbreak \ldots,\allowbreak v_n\}$ and a set of \emph{edges} $E = \{(v_i,v_j) \colon v_i,v_j \in V \}$ connecting those nodes.
\end{athm}
\begin{athm}{definition}{def:directed}
Let $n \in \N = \{1,2,3,\ldots\}$, let $V = \{v_1,v_2,\ldots,v_n\}$ be a set of vertices, let $E = \{(v_i,v_j) \colon v_i,v_j \in V \}$ be a set of edges for $V$, and let $G = (V,E)$ be a graph (cf.\ \cref{def:graph}).
We say that $G$ is an \emph{undirected} graph if for all $i,j \in \{1,2,\ldots,n\}$ it holds that $(v_i,v_j) = (v_j,v_i)$.
We say that $G$ is a directed graph, otherwise.
\end{athm}
\begin{athm}{definition}{def:path}
Let $n \in \N = \{1,2,3,\ldots\}$, let $V = \{v_1,v_2,\ldots,v_n\}$ be a set of vertices, let $E = \{(v_i,v_j) \colon v_i,v_j \in V \}$ be a set of edges for $V$, and let $G = (V,E)$ be a graph (cf.\ \cref{def:graph}).
For every $i,j \in \{1,2,\ldots,n\}$ with $i \neq j$ a \emph{path} $v_i \rightsquigarrow v_j$ is a sequence of nodes $(v_i,v_{i+1},\ldots,v_j)$ which satisfies for all $k \in \{i,i+1,\ldots,j-1\}$ that $(v_k,v_{k+1}) \in E$.
Moreover, the \emph{length} of the path is $i-j+1$ (i.e., the number of vertices in the sequence).
\end{athm}
Note that \cref{def:path} is not completely rigorous because our definition is enforcing that ``$v_i$ comes before $v_j$ in the path.'' This is not a requirement --- we have simply presented \cref{def:path} in this way for convenience. We can circumvent this issue by introducing sub-indices to allow for arbitrary ordering. However, this should not cause any confusion for our purposes (and interested students should attempt this generalization themselves).
Moreover, observe
that \cref{def:path} essentially states that a path is a sequence of vertices in a graph where one can ``trace'' the sequence by following consecutive edges.
\begin{athm}{definition}{def:cycle}
Let $n \in \N = \{1,2,3,\ldots\}$, let $V = \{v_1,v_2,\ldots,v_n\}$ be a set of vertices, let $E = \{(v_i,v_j) \colon v_i,v_j \in V \}$ be a set of edges for $V$, and let $G = (V,E)$ be a graph (cf.\ \cref{def:graph}).
For every $i \in \{1,2,\ldots,n\}$ a \emph{cycle} is a path $v_i \rightsquigarrow v_i$ where the first and last vertex are the same (cf.\ \cref{def:path}).
If $G$ has no cycles, then $G$ is said to be \emph{acyclic}.
\end{athm}
We first provide a worked example of a situation related to Section 3 of Davis et al.
Note that the situation in \cref{prob:0} is much simpler than that depicted in Figure 3.2 of Davis et al.\ (to provide more clarity).
\begin{athm}{problem}{prob:0}
Let $A \in \R^{5 \times 5}$ satisfy
\begin{equation}\label{eq:01}
A =
\begin{pmatrix}
1 & & & & & \\
& 1 & & & & \\
& -2 & 1 & & \\
& 4 & 7 & 1 & \\
& & 3 & 9 & 1 \\
\end{pmatrix}
\end{equation}
(where empty spaces in $A$ are assumed to represent entries of zero)
and let $b \in \R^{5 \times 1}$ satisfy
\begin{equation}\label{eq:02}
b = \begin{pmatrix}
0 & 4 & 0 & 1 & 0
\end{pmatrix}^T \, .
\end{equation}
\begin{enumerate}[label=\alph*.]
\item\label{0_a} Does $A$ seem to be \emph{sparse}? Does $b$ seem to be \emph{sparse}?
\item\label{0_b} What is the compressed-column form of $A$ (cf.\ page 8 of Davis et al.)?
\item\label{0_c} Let $V = \{1,2,3,4,5\}$, let $E = \{(j,i) \colon A_{ij} \neq 0,\ i,j \in V \}$ (cf.\ \cref{eq:1}), and let $G_A$ be the directed acyclic graph which satisfies $G_A = (V,E)$ (cf.\ \cref{def:graph,def:directed,def:cycle}).
Draw the graph $G_A$ --- be sure to label all nodes appropriately and demonstrate the correct directionality of the edges.
\item\label{0_d} Let $\mathcal{B} \subseteq \{1,2,3,4,5\}$ be the set which satisfies
\begin{equation}\label{eq:03}
\mathcal{B} = \cu[\big]{ i \in \{1,2,3,4,5\} \colon b_i \neq 0 }
\end{equation}
(cf.\ \cref{eq:2}), let $\operatorname{Reach}_A \colon \{1,2,3,4,5\} \to \{1,2,3,4,5\}$ be the function which satisfies for all $i \in \{1,2,3,4,5\}$ that
\begin{equation}\label{eq:04}
\operatorname{Reach}_A(i) = \cu[\big]{ j \in \{1,2,3,4,5\} \colon \exists \, i \rightsquigarrow j,\ i,j \in G_A }
\end{equation}
(cf.\ \cref{def:path,1_c}),
and let $\mathcal{X} \subseteq \{1,2,3,4,5\}$ satisfy
\begin{equation}\label{eq:05}
\mathcal{X} = \cup_{i \in \mathcal{B}} \operatorname{Reach}_A(i) \, .
\end{equation}
Compute $\mathcal{X}$ using a ``depth-first'' ordering to determine the topology of the set (cf.\ page 11 of Davis et al.).
\end{enumerate}
\end{athm}
\begin{aproof}
We solve each part of \cref{prob:0} separately.
\begin{enumerate}[label=\alph*.]
\item Without ``muddying the water'' with unnecessary details, it should be clear that the objects in \cref{eq:01,eq:02} are indeed sparse.
The solution to \cref{0_a}\ is thus complete.
\item The compressed-column form of $A$ is given by:
\begin{verbatim}
int p[] = { 0, 1, 4, 7, 10 } ;
int i[] = { 0, 1, 2, 3, 2, 3, 4, 3, 4, 4 } ;
double A[] = { 1, 1, -2, 4, 1, 7, 3, 1, 9, 1 } ;
\end{verbatim}
(again, see page 6 of Davis et al.).
\begin{quote}
\textbf{Aside:} Note that the above is written with a \texttt{C}-style formatting --- can you update the style to reflect the syntax of a different programming language?
\end{quote}
The solution to \cref{0_b}\ is thus complete.
\item A possible depiction of the graph $G_A$ is given in \cref{fig:1} below.
\begin{figure}[H]
\centering
\begin{tikzpicture}[scale=2,vertex/.style={draw,circle}, arc/.style={draw,thick,->}]
\node[vertex] (2) at (0,0) {2};
\node[vertex] (3) at (1,-0.5) {3};
\node[vertex] (4) at (-1,-0.5) {4};
\node[vertex] (5) at (1,-1.4) {5};
\draw[-{Latex[width=3mm]}] (2) -- (3) ;
\draw[-{Latex[width=3mm]}] (2) -- (4) ;
\draw[-{Latex[width=3mm]}] (3) -- (4) ;
\draw[-{Latex[width=3mm]}] (3) -- (5) ;
\draw[-{Latex[width=3mm]}] (4) -- (5) ;
\end{tikzpicture}
\captionsetup{format=hang,labelfont=bf,width=0.75\linewidth}
\caption{A sample depiction of the directed graph $G_A$.}\label{fig:1}
\end{figure}
Note that we say that \cref{fig:1} is a \emph{possible} depiction of $G_A$ because the configuration of the vertices in a drawing of $G_A$ is arbitrary (i.e., not specified by \cref{def:graph}).
The solution to \cref{0_c}\ is thus complete.
\item First, note that $\mathcal{B}$ is the set of indices associated to nonzero values in $b$ (cf.\ \cref{eq:01,eq:03}).
Thus, it holds that
\begin{equation}
\mathcal{B} = \{2,4\}
\end{equation}
(where here we use the actual position, rather than the, e.g., associated position in a \texttt{C}-style array).
Next, observe that \cref{eq:04} is the set of all $j \in \{1,2,3,4,5\}$ \emph{reachable} from $i \in \{1,2,3,4,5\}$ with $i \neq j$.
Moreover, \cref{eq:05} indicates that
\begin{equation}\label{eq:05a}
\mathcal{X} = \operatorname{Reach}_A(2) \cup \operatorname{Reach}_A(4) \, ,
\end{equation}
where the topology of the set is determined via a depth-first search.
To clarify this issue, we have marked the elements of $\mathcal{B}$ in \cref{fig:2} below.
\begin{figure}[H]
\centering
\begin{tikzpicture}[scale=2,vertex/.style={draw,circle}, arc/.style={draw,thick,->}]
\node[vertex,red] (2) at (0,0) {2};
\node[vertex] (3) at (1,-0.5) {3};
\node[vertex,red] (4) at (-1,-0.5) {4};
\node[vertex] (5) at (1,-1.4) {5};
\draw[-{Latex[width=3mm]}] (2) -- (3) ;
\draw[-{Latex[width=3mm]}] (2) -- (4) ;
\draw[-{Latex[width=3mm]}] (3) -- (4) ;
\draw[-{Latex[width=3mm]}] (3) -- (5) ;
\draw[-{Latex[width=3mm]}] (4) -- (5) ;
\end{tikzpicture}
\captionsetup{format=hang,labelfont=bf,width=0.75\linewidth}
\caption{A sample depiction of the directed graph $G_A$.
The elements of the set $\mathcal{B}$ are indicated by red nodes.}\label{fig:2}
\end{figure}
In addition, note that \cref{fig:2,eq:04} ensure that
\begin{equation}\label{eq:06}
\operatorname{Reach}_A(2) = \{2,3,5\} \, .
\end{equation}
Moreover, observe that \cref{fig:2,eq:04} assure that
\begin{equation}\label{eq:07}
\operatorname{Reach}_A(4) = \{4,5\} \, .
\end{equation}
Combining \cref{eq:05a,eq:06,eq:07} with the depth-first search topology (i.e., that we order the ``lists'' by traversing the set $\mathcal{B}$ in reverse and we omit repeated indices) yields
\begin{equation}
\mathcal{X} = \{4,5,2,3\} \, .
\end{equation}
\begin{quote}
\textbf{Aside:} It is important to note that there is some confusion from Figure 3.2 of Davis et al.
In Davis et al.\ the graph is more complicated, hence, there are parts of the graph ``$G_L$'' which have no affect on $\mathcal{X}$ --- which is perfectly fine!
This is important to notice because it helps one understand that there is a difference between the graph of the matrix and the set $\mathcal{X}$.
\end{quote}
The solution to \cref{0_d}\ is thus complete.
\end{enumerate}
\end{aproof}
\addcontentsline{toc}{subsection}{\texorpdfstring{\cref{rem1}}{Remark 1.6}}
\begin{remark}\label{rem1}
It is worth noting that in \cref{0_c}\ of \cref{prob:0} we could have potentially included edges of the form, e.g., $(1,1)$ since there are nonzero entries along the main diagonal of $A$. In some branches of graph theory this is done (and is represented by a ``looping'' edge). However, in numerical linear algebra, this edge is almost always omitted as it does not aid in the symbolic study of the underlying algorithms (think about why this is the case).
\end{remark}
Now, attempt to complete \cref{prob:1} yourself.
The problem is fairly similar to \cref{prob:0} with an additional programming issue presented in \cref{1_e}.
\begin{athm}{problem}{prob:1}
Let $A \in \R^{8 \times 8}$ satisfy
\begin{equation}\label{eq:1}
A =
\begin{pmatrix}
1 & & & & & & & & \\
& 1 & & & & & & & \\
& 3 & 1 & & & & & \\
& 1 & 2 & 1 & & & & \\
& & 2 & 3 & 1 & & & \\
& & 1 & 2 & 3 & 1 & & \\
& & & & 4 & & 1 & \\
& & & 1 & & & & 1
\end{pmatrix}
\end{equation}
(where empty spaces in $A$ are assumed to represent entries of zero)
and let $b \in \R^{8 \times 1}$ satisfy
\begin{equation}\label{eq:2}
b = \begin{pmatrix}
0 & 0 & 1 & 3 & 4 & 0 & 0 & 9
\end{pmatrix}^T \, .
\end{equation}
\begin{enumerate}[label=\alph*.]
\item Does $A$ seem to be \emph{sparse}? Does $b$ seem to be \emph{sparse}?
\item What is the compressed-column form of $A$ (cf.\ page 8 of Davis et al.)?
\item\label{1_c} Let $V = \{1,2,\ldots,8\}$, let $E = \{(j,i) \colon A_{ij} \neq 0 \}$ (cf.\ \cref{eq:1}), and let $G_A$ be the directed acyclic graph which satisfies $G_A = (V,E)$ (cf.\ \cref{def:graph,def:directed,def:cycle}).
Draw the graph $G_A$ --- be sure to label all nodes appropriately and demonstrate the correct directionality of the edges.
\item\label{1_d} Let $\mathcal{B} \subseteq \{1,2,\ldots,8\}$ be the set which satisfies
\begin{equation}
\mathcal{B} = \cu[\big]{ i \in \{1,2,\ldots,8\} \colon b_i \neq 0 }
\end{equation}
(cf.\ \cref{eq:2}), let $\operatorname{Reach}_A \colon \{1,2,\ldots,8\} \to \{1,2,\ldots,8\}$ be the function which satisfies for all $i \in \{1,2,\ldots,8\}$ that
\begin{equation}
\operatorname{Reach}_A(i) = \cu[\big]{ j \in \{1,2,\ldots,8\} \colon \exists \, i \rightsquigarrow j,\ i,j \in G_A }
\end{equation}
(cf.\ \cref{def:path,1_c}),
and let $\mathcal{X} \subseteq \{1,2,\ldots,8\}$ satisfy
\begin{equation}
\mathcal{X} = \cup_{i \in \mathcal{B}} \operatorname{Reach}_A(i) \, .
\end{equation}
Compute $\mathcal{X}$ using a ``depth-first'' ordering to determine the topology of the set (cf.\ page 11 of Davis et al.).
\item\label{1_e} Write a computer program which finds the $x \in \R^{8 \times 1}$ (Do you understand that this $x$ is unique? Can you prove this?) which satisfies
\begin{equation}
Ax = b
\end{equation}
(cf.\ \cref{eq:1,eq:2}) using your work from \cref{1_d}\ and Algorithm 3.2 from Davis et al.
\end{enumerate}
\end{athm}
\begin{aproof}
\linebreaks{20}
\end{aproof}
\section{The Cholesky factorization method}
Let $n \in \N$, $b \in \C^{n\times 1}$, $A \in \C^{n\times n}$ satisfy that $A$ is \emph{sparse}.
The problem of finding $x \in \C^{n\times 1}$ such that
\begin{equation}\label{2_1}
Ax = b
\end{equation}
using direct methods starts by factoring the matrix $A$.
This factorization typically is carried out by going through a symbolic phase (this phase typically depends only on the nonzero pattern of $A$) and the numerical phase (the phase which actually produces the desired factorization).
The solve phase uses standard triangular system solvers (i.e., is not the difficult part of solving \cref{2_1}).
It is worth noting that symbolic phase is much faster than the numerical phase. Moreover, a well spent symbolic phase increases the efficiency of the numerical phase.
Loosely speaking, one should think of the symbolic phase as the effort spend trying to understand the nonzero pattern of the solution vector by studying the nonzero pattern of the original matrix $A$ and the chosen factorization method.
For simplicity of exposition, Davis et al.\ chooses to motivate the idea of symbolic analysis via the well-known Cholesky decomposition method.
\begin{athm}{problem}{prob:2.1}
Let $A \in \R^{3\times 3}$ satisfy
\begin{equation}
A =
\begin{pmatrix}
25 & 15 & -5 \\
15 & 18 & 0 \\
-5 & 0 & 11
\end{pmatrix} \, .
\end{equation}
\begin{enumerate}[label=\alph*.]
\item Does the Cholesky decomposition of $A$ exist? (One possible method to determine this is to show that $A$ is \emph{Hermitian} and \emph{positive-definite}.)
\item Compute that Cholesky decomposition of $A$. That is, find $L \in \R^{3\times 3}$ which satisfies for all $i , j \in \{1,2,3\}$ with $i < j$ that $L_{ij} = 0$ and
\begin{equation}
L L^T = A \, .
\end{equation}
\end{enumerate}
\end{athm}
\begin{aproof}
\linebreaks{20}
\end{aproof}
\begin{athm}{remark}{rem2}
Something mentioned in Davis et al., but not yet explored, is the fact that we can use a Cholesky factorization even if no such factorization exists for a given matrix $A$. This is done by permuting the matrix into a preferable form (cf.\ page 394 of Davis et al.).
For now, we will not worry with such issues (in fact, Davis et al.\ postpones such issues until Section 8 of the article).
\end{athm}
We now attempt to better understand the concept of \emph{elimination trees}.
The algorithm for constructing elimination trees is provided in Section 4.1 of Davis et al.
\begin{athm}{problem}{prob:2.2}
Let $A \in \R^{5 \times 5}$ satisfy
\begin{equation}\label{eq:2_2_1}
A =
\begin{pmatrix}
1 & & & & \\
& 1 & -2 & 4 & \\
& -2 & 5 & -1 & 3 \\
& 4 & -1 & 66 & 30 \\
& & 3 & 30 & 91
\end{pmatrix}
\end{equation}
(where empty spaces in $A$ are assumed to represent entries of zero).
\begin{enumerate}[label=\alph*.]
\item What is the compressed-column form of $A$ (cf.\ page 8 of Davis et al.)?
\item Let $V = \{1,2,3,4,5\}$, let $E = \{(j,i) \colon A_{ij} \neq 0 \}$ (cf.\ \cref{eq:2_2_1}), and let $G_A$ be the directed acyclic graph which satisfies $G_A = (V,E)$ (cf.\ \cref{def:graph,def:directed,def:cycle}).
Draw the graph $G_A$ --- be sure to label all nodes appropriately and demonstrate the correct directionality of the edges.
\item Does the Cholesky decomposition of $A$ exist? (One possible method to determine this is to show that $A$ is \emph{Hermitian} and \emph{positive-definite}.)
\item Compute that Cholesky decomposition of $A$. That is, find $L \in \R^{5\times 5}$ which satisfies for all $i , j \in \{1,2,3,4,5\}$ with $i < j$ that $L_{ij} = 0$ and
\begin{equation}
L L^T = A \, .
\end{equation}
Be sure to mark the \emph{fill-in entries} of $L$ (or, at least, designate them in some way).
\item Construct the elimination tree of $A$. Do you understand how this aids in the computation of the Cholesky factorization?
\end{enumerate}
\end{athm}
\begin{aproof}
\linebreaks{20}
\end{aproof}
\begin{athm}{problem}{prob:2.3}
Let $A \in \R^{8 \times 8}$ satisfy
\begin{equation}\label{eq:2_3_1}
A =
\begin{pmatrix}
1 & & & & & & & \\
& 1 & 3 & 1 & & & & \\
& 3 & 10 & 5 & 2 & 1 & & \\
& 1 & 5 & 6 & 7 & 4 & & 1 \\
& & 2 & 7 & 14 & 11 & 4 & 3 \\
& & 1 & 4 & 11 & 15 & 12 & 2 \\
& & & & 4 & 12 & 17 & \\
& & & 1 & 3 & 2 & & 2
\end{pmatrix}
\end{equation}
(where empty spaces in $A$ are assumed to represent entries of zero).
\begin{enumerate}[label=\alph*.]
\item What is the compressed-column form of $A$ (cf.\ page 8 of Davis et al.)?
\item Let $V = \{1,2,\ldots,8\}$, let $E = \{(j,i) \colon A_{ij} \neq 0 \}$ (cf.\ \cref{eq:2_3_1}), and let $G_A$ be the directed acyclic graph which satisfies $G_A = (V,E)$ (cf.\ \cref{def:graph,def:directed,def:cycle}).
Draw the graph $G_A$ --- be sure to label all nodes appropriately and demonstrate the correct directionality of the edges.
\item Does the Cholesky decomposition of $A$ exist? (One possible method to determine this is to show that $A$ is \emph{Hermitian} and \emph{positive-definite}.)
\item Compute that Cholesky decomposition of $A$. That is, find $L \in \R^{8\times 8}$ which satisfies for all $i , j \in \{1,2,\ldots,8\}$ with $i < j$ that $L_{ij} = 0$ and
\begin{equation}
L L^T = A \, .
\end{equation}
Be sure to mark the \emph{fill-in entries} of $L$ (or, at least, designate them in some way).
\item Construct the elimination tree of $A$. Do you understand how this aids in the computation of the Cholesky factorization?
\end{enumerate}
\end{athm}
\begin{aproof}
\linebreaks{20}
\end{aproof}
\end{document}

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\contentsline {section}{\numberline {1}Solving sparse triangular systems}{1}{section.1}%
\contentsline {subsection}{\cref {def:graph}}{1}{section*.2}%
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\contentsline {subsection}{\cref {rem1}}{4}{equation.1.10}%
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\contentsline {section}{\numberline {2}The Cholesky factorization method}{6}{section.2}%
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\begin{thebibliography}{1}
\bibitem{berndt2001introduction}
{\sc Berndt, R.}
\newblock {\em An introduction to symplectic geometry}.
\newblock No.~26. American Mathematical Soc., 2001.
\end{thebibliography}

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\BOOKMARK [1][-]{section.1}{Introduction}{}% 1
\BOOKMARK [1][-]{section.2}{Some basic background material}{}% 2
\BOOKMARK [1][-]{section.3}{Approximating differential equations on manifolds}{}% 3
\BOOKMARK [2][-]{subsection.3.1}{A first attempt at a numerical approximation}{section.3}% 4
\BOOKMARK [2][-]{subsection.3.2}{Approximating the exponential map}{section.3}% 5

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%% %%
%% Preamble of document %%
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% Sets up the standard "article" document class
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\title{A survey of numerical integration techniques on manifolds}
\author{Joshua Lee Padgett}
\date{\today}
\begin{document}
\maketitle
\begin{abstract}
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.
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.
\end{abstract}
\tableofcontents
\section{Introduction}
\section{Some basic background material}
The following are some basic definitions needed to discuss numerical integration on homogeneous manifolds (manifolds acted upon transitively by a Lie group).
Note that the following should not be viewed as a complete list of all needed (or even relevant) topics and concepts.
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.
\begin{athm}{definition}{def:manifold}
Let $m,d \in \N = \{1,2,3,\ldots\}$ with $m \le d$ and let $\M \subseteq \R^d$ with $\M \neq \emptyset$.
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.
See, e.g., \url{https://en.wikipedia.org/wiki/Manifold} for more details.
\end{athm}
\begin{athm}{definition}{def:tangent}
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}).
Then the vector
\begin{equation}
\pr[\big]{ \tfrac{\dd }{\dd t} \rho } \lr (t) \Bigr\rvert_{t=0}
\end{equation}
is a tangent vector at the point $p$.
The set of all tangent vectors at $p$ is the tangent space at $p$ and is denoted by $T\M \rvert_p$.
The collection of all tangent spaces at all points $q \in \M$ is called the tangent bundle of $\M$ and is denoted by $T\M = \cup_{q\in\M} T\M\rvert_{q}$.
See, e.g., \url{https://en.wikipedia.org/wiki/Tangent_space} for more details.
\end{athm}
\begin{athm}{definition}{def:vector_field}
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}).
Then $F$ is a vector field on $\M$.
The collection of all vector fields on $\M$ is denoted by $\fX(\M)$.
See, e.g., \url{https://en.wikipedia.org/wiki/Vector_field} for more details.
\end{athm}
\begin{athm}{definition}{def:lie_algebra}
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
\begin{equation}
[a,[b,c]] + [b,[c,a]] + [c,[a,b]] = 0 \dpp
\end{equation}
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 $\fg$.
See, e.g., \url{https://en.wikipedia.org/wiki/Lie_algebra} for more details.
\end{athm}
\begin{athm}{problem}{prob:lie_alg}
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 $[a,b] = ab - ba$.
Show that $\fg$ is a Lie algebra (cf.\ \cref{def:lie_algebra}).
\end{athm}
\begin{asol}
\linebreaks{30}
\end{asol}
\begin{athm}{problem}{prob:lie_alg2}
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}).
Show that $\fg$ is a Lie algebra (cf.\ \cref{def:lie_algebra}).
\end{athm}
\begin{asol}
\linebreaks{30}
\end{asol}
\begin{athm}{definition}{def:adjoint}
Let $\fg$ be a Lie algebra (cf.\ \cref{def:lie_algebra}).
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
\begin{equation}
\ad^n(u,v) =
[u,\ad^{n-1}(u,v)]
\dpp
\end{equation}
See, e.g., \url{https://en.wikipedia.org/wiki/Adjoint_representation} for more details.
\end{athm}
%\begin{athm}{definition}{def:lie_alg_homo}
%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
%\begin{equation}
%\varphi\pr[\big]{ [u,v]_\fg } = \br[\big]{ \varphi(u) , \varphi(v) }_{\mathfrak{h}}
%\end{equation}
%(cf.\ \cref{def:lie_algebra}).
%Then we say that $\varphi$ is a Lie algebra homomorphism.
%\end{athm}
\begin{athm}{definition}{def:lie_group}
Let $\G$ be a manifold (cf.\ \cref{def:manifold}).
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
\begin{enumerate}[label=(\roman*)]
\item it holds for all $p,q,r \in \G$ that $p \cdot (q \cdot r) = (p \cdot r) \cdot q$,
\item it holds for all $p \in \G$ that $p \cdot e = e \cdot p = p$,
\item it holds for all $p \in \G$ that there exists $p^{-1} \in \G$ such that $p^{-1} \cdot p = e$, and
\item it holds that $\cdot \colon \G \times \G \to \G$ and $\G \ni p \mapsto p^{-1} \in \G$ are smooth.
\end{enumerate}
See, e.g., \url{https://en.wikipedia.org/wiki/Lie_group} for more details.
\end{athm}
\begin{athm}{problem}{prob:lie_group}
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 $a\cdot b = ab$.
Show that $\G$ is a Lie group (cf.\ \cref{def:lie_group}).
\emph{Note:} We have not defined matrix manifolds, but we can easily update this definition to allow for such things (e.g., via representations).
\end{athm}
\begin{asol}
\linebreaks{30}
\end{asol}
\begin{athm}{definition}{def:alg_of_group}
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}).
Then we define the Lie algebra of a Lie group to be $\fg = T\G\rvert_{e}$ (cf.\ \cref{def:lie_algebra,def:tangent}).
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
\begin{equation}
[u,v] = \frac{\partial^2}{\partial t \partial s} g(t) \cdot h(s) \cdot g^{-1}(t) \bigg\rvert_{t=s=0} \dpp
\end{equation}
See, e.g., \url{https://en.wikipedia.org/wiki/Lie_group%E2%80%93Lie_algebra_correspondence} for more details.
\end{athm}
\begin{athm}{problem}{prob:lie_group2}
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 $a\cdot b = ab$.
Show that $\fg = \mathfrak{so}(3)$ is the Lie algebra of $\G$ (cf.\ \cref{def:alg_of_group}).
\end{athm}
\begin{asol}
\linebreaks{30}
\end{asol}
\begin{athm}{definition}{def:action}
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
\begin{equation}
\Lambda(e,y) = y
\qquad\text{and}\qquad
\Lambda\pr[\big]{p,\Lambda(q,y)} = \Lambda(p\cdot q,y)
\end{equation}
(cf.\ \cref{def:manifold,def:lie_group}).
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}

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\contentsline {section}{\numberline {1}Introduction}{1}{section.1}%
\contentsline {section}{\numberline {2}Some basic background material}{1}{section.2}%
\contentsline {section}{\numberline {3}Approximating differential equations on manifolds}{8}{section.3}%
\contentsline {subsection}{\numberline {3.1}A first attempt at a numerical approximation}{16}{subsection.3.1}%
\contentsline {subsection}{\numberline {3.2}Approximating the exponential map}{18}{subsection.3.2}%

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## The following will produce a very simple implementation of Euler's
## method for solving the following ODE:
##
## y'(t) = y(t), t > 0,
## y(0) = 0
##
## For now, the code will explicit use the forcing term f(y) = y, but will
## later be generalized.
##
import math
import numpy as np # shortcut name for numpy
from matplotlib import pyplot as plt # shortcut name for pyplot
from matplotlib.animation import FuncAnimation
N = 100 # Number of time steps
t0 = 0 # Initial time
tf = 1 # Final time
h = (tf - t0)/(N-1) # Numerical step size
t = np.linspace(t0,tf,N) # Temporal grid
y = np.zeros([N]) # Empty vector for storing solution values
z = np.zeros([N]) # Empty vector for storing error values
w = np.zeros([N]) # Empty vector for true solution
y[0] = 1 # Initial value of the problem
# The following is the actual implementation of Euler's method
for i in range(1,N):
y[i] = y[i-1] + h*y[i-1]
# The following constructs the true solution
for i in range(0,N):
w[i] = math.exp(t[i])
# The following computes the error between the approximation
# and the true solution, y = exp(t)
for i in range(0,N):
z[i] = abs( y[i] - w[i] )
# The following creates strings to generate the titles in the plots
s1 = "Euler Approximation \n for h = "
s = "%1.2f" % h
title1 = s1 + s
s2 = "Absolute Error \n for h = "
title2 = s2 + s
# The following generates the desired plots
fig, axs = plt.subplots(ncols=2,nrows=2)
gs = axs[1,0].get_gridspec()
# remove the underlying axes
for ax in axs[-1,0:]:
ax.remove()
axbig = fig.add_subplot(gs[-1,0:])
axs[0,0].plot(t,y,'--',label="Approx.")
axs[0,0].plot(t,w,label="True")
axs[0,0].set_title(title1)
axs[0,0].set_xlabel('t')
axs[0,0].set_ylabel('y')
axs[0,0].legend()
axs[0,1].plot(t,z)
axs[0,1].set_title(title2)
axs[0,1].set_xlabel('t')
axs[0,1].set_ylabel('y')
# The following is a (fairly complicated) bit of code that allows
# you to make a video of the pointwise error at each of the points
# in an ever refining domain.
xdata, ydata = [], []
ln, = plt.plot([], [], 'ro')
def out(ii):
NN = 5*ii
tt = np.linspace(t0,tf,NN)
hh = tt[1]-tt[0]
ww = np.zeros([NN])
yy = np.zeros([NN])
zz = np.zeros([NN])
yy[0] = 1
for i in range(1,NN):
yy[i] = yy[i-1] + hh*yy[i-1]
for i in range(0,NN):
ww[i] = math.exp(tt[i])
for i in range(0,NN):
zz[i] = abs( yy[i] - ww[i] )
return hh,tt,zz
def init():
axbig.set_xlim(t0,tf)
axbig.set_ylim(0,0.15)
axbig.set_xlabel('t')
axbig.set_ylabel('Error')
return ln,
def update(ii):
hh,xdata,ydata = out(ii)
ln.set_data(xdata, ydata)
ss = "Error for h = "
ss1 = "%1.4f" % hh
title3 = ss + ss1
axbig.set_title(title3)
return ln,
ani = FuncAnimation(fig,update,frames=range(1,20),interval=150,init_func=init)
fig.tight_layout()
plt.show()

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\relax
\providecommand\hyper@newdestlabel[2]{}
\providecommand\HyperFirstAtBeginDocument{\AtBeginDocument}
\HyperFirstAtBeginDocument{\ifx\hyper@anchor\@undefined
\global\let\oldcontentsline\contentsline
\gdef\contentsline#1#2#3#4{\oldcontentsline{#1}{#2}{#3}}
\global\let\oldnewlabel\newlabel
\gdef\newlabel#1#2{\newlabelxx{#1}#2}
\gdef\newlabelxx#1#2#3#4#5#6{\oldnewlabel{#1}{{#2}{#3}}}
\AtEndDocument{\ifx\hyper@anchor\@undefined
\let\contentsline\oldcontentsline
\let\newlabel\oldnewlabel
\fi}
\fi}
\global\let\hyper@last\relax
\gdef\HyperFirstAtBeginDocument#1{#1}
\providecommand\HyField@AuxAddToFields[1]{}
\providecommand\HyField@AuxAddToCoFields[2]{}
\gdef \@abspage@last{2}

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\title{Reformulation without f}
\author{Shakil Rafi}
\begin{document}
\maketitle
\begin{lemma}
Let $T \in (0,\infty)$, let $\lp \Omega, \mathcal{F}, \mathbb{P}\rp $ be a probability space, let $\alpha_d \in \R^d \rightarrow \R$, $d \in \N$, be infinitely often differentiable functions, let $u_d \in C^{1,2} \lp \lb 0,T \rb \times, \R^d, \R \rp$, $d \in \N$, satisfy for all $d \in \N$, $t \in \lb 0,T\rb$, $x \in \R^d$ that:
\begin{align}
\lp \frac{\partial}{\partial t} u_d \rp \lp t,x \rp + \lp \Delta_x u_d \rp \lp t,x \rp + \alpha_x \lp x \rp u_d \lp t,x \rp = 0
\end{align}
Let $\mathcal{W}^d: \lb 0,T \rb \times \Omega \rightarrow \R^d$, $d \in \N$ be standard Brownian motions, and let $\mathcal{X}^{d,t,x}: \lb t,T\rb \times \Omega \rightarrow \R^d$, $d \in \N$, $t\in \lb 0,T \rb$, $s \in \lb t,T\rb$, $x\in \R^d$ we have $\mathbb{P}$-a.s. that:
\begin{align}
\mathcal{X}^{d,t,x}_s = x+ \int^t_s \sqrt{2} d\mathcal{W}^d_r
\end{align}
Then for all $d \in \N$, $t\in \lb 0,T\rb$, $x\in \R^d$ it holds that:
\begin{align}
u_d\lp t,x \rp + \mathbb{E} \lb \exp \lp \int^T_t \alpha_x \lp \mathcal{X}^{d,t,x}_r \rp dr \rp u_d \lp T, \mathcal{X}^{d,t,x}_T \rp \rb
\end{align}
\end{lemma}
\begin{proof}
Let $T \in [0,\infty)$, and let $\lp \Omega, \mathcal{F}, \mathbb{P} \rp$ be a probability space. For all $d \in \N$, let $V \in C^{1,1} \lp \R^d \times [0,T],\R \rp $ be $V(x,t) = \alpha_d(x)$, let $\sigma_d : \R^d \rightarrow \R^{d\times d}$ be given by $\sigma_d(x) = \diag_d \lp \sqrt{2} \rp$, let $\mu_d: \R^d \rightarrow \R^d$ be given by $\mu_d(x) = \mymathbb{0}_d$, and finally let $f(t,x) = 0$. By Feynman-Kac and substituting the above, the following expression:
\begin{align}\label{0.0.4}
\lp \frac{\partial}{\partial t} u_d \rp \lp t,x \rp + \frac{1}{2}\Trace\lp \sigma(t,x) \lb \sigma(t,x) \rb^* \lp \Hess_x(u_d \rp \lp t,x \rp \rp + \la \mu(t,x), \lp \nabla_x u_d \rp \lp t,x\rp \ra +V(t,x)u_d(t,x) \nonumber\\ +f(t,x) = 0
\end{align}
is rendered:
\begin{align}\label{0.0.5}
\lp \frac{\partial}{\partial t} u_d \rp \lp t,x \rp + \lp \Delta_x u_d \rp \lp t,x\rp + \alpha_d(x)u_d(x) = 0
\end{align}
Note then that Feyman-Kac sates that the solution to (\ref{0.0.4}) can be written as:
\begin{align}
u(t,x) = \E \lb \int^T_t e^{\int^r_t V (\mathcal{X}_t,\tau) d\tau}f(\mathcal{X}_r, r )dr + e^{-\int^T_t V(\mathcal{X}_\tau,\tau)d\tau } u(\mathcal{X}_T,T) \rb
\end{align}
Where $\mathcal{X}$ is an $\lp \Omega, \mathcal{F}, \mathbb{P} \rp$-adapted stochastic process given by:
\begin{align}
\mathcal{X}_t = x + \int^t_s \mu_d \lp \mathcal{X} \rp dr+\int^t_s\sqrt{2}d\mathcal{W}_r^d
\end{align}
Note then that the substitutions then yield that the solution to (\ref{0.0.5}) is given by:
\begin{align}
u(t,x) = \E \lb \exp \lp \int^T_t \alpha_d \lp \mathcal{X} \rp dr \rp u_d \lp T,\mathcal{X}^{d,t,x}_T \rp \rb
\end{align}
\end{proof}
\end{document}

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\documentclass[11pt]{report}
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\title{Reformulation without f}
\author{Shakil Rafi}
\begin{document}
\maketitle
\begin{theorem}
Consider the following diffeq:
\begin{align}
\lp \frac{\partial }{\partial t} u \rp \lp t,x \rp + \la \mu \lp t,x \rp, \nabla_x \lp t,x \rp \ra + \frac{1}{2} \lp \Trace \lp \Hess_x u_d \rp \lp x \rp \rp -V\lp t,x \rp u \lp t,x\rp +f\lp t,x \rp = 0
\end{align}
\end{theorem}
\end{document}

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\title{Reformulation without f}
\author{Shakil Rafi}
\begin{document}
\maketitle
\begin{lemma}
Let $\Theta = \bigcup_{n\in\N}$, $d,m,\mathfrak{d} \in \N$, $T \in (0,\infty)$, $\act \in C \lp \R,\R\rp$, $\mathfrak{I}, \mathbf{F}, \mathbf{G} \in \neu$ satisfy $\lay\lp \mathfrak{I} \rp = \lp 1,\mathfrak{d},1\rp$, $\real_{\act} \lp \mathfrak{I} \rp = \mathbb{I}_1$, $\real_{\act} \lp \mathbf{F} \rp \in C\lp \R, \R \rp$, and $\real_{\act} \lp \mathbf{G} \rp \in C \lp \R^d, \R \rp$, for every $\theta \in \Theta$ let $\mathcal{U}^\theta: \lb 0,T \rb \rightarrow \lb 0,T \rb$ and $\mathcal{W}^\theta : \lb 0,T \rb \rightarrow \R^d$, be functions, for every $\theta \in \Theta$, $n\in \N_0$ let $U^\theta_n: [0,T] \times \R^d \rightarrow \R$ satisfy for all $t\in [0,T]$, $x \in \R^d$, that:
\begin{align*}
&U^\theta_n(t,x) = \frac{\mymathbb{1}_\N (n)}{M^n} \lb \sum^{m^n}_{k=1} \lp \real_{\act} \lp \mathbf{G}\rp \rp \lp x+\mathcal{W}^{\lp \theta,0,-k \rp}_{T-t} \rp \rb \\
&+\sum^{n-1}_{i=1} \frac{T-t}{M^{n-i}} \lb \sum^{M^{n-i}}_{k=1} \lp \lp \real_{\act} \lp \mathbf{F}\rp \circ U_i^{(\theta,i,k)} \rp -\mymathbb{1}_\N \lp i \rp \lp \real_{\act} \lp \mathbf{F}\rp \circ U^{\lp \theta,-i,k \rp} \rp \rp \lp \mathcal{U}_t^{\lp \theta,i,k \rp},x+\mathcal{W}^{\lp\theta,i,k\rp}_{\mathcal{U}_t^{\lp \theta,i,k\rp}-t} \rp \rb
\end{align*}
and let $\mathbf{U}^\theta_{n,t} \in \neu$, $t \in [0,T]$, $n\in \Z$, $\theta \in \Theta$, satisfy for all $\theta \in \Theta$, $n\in \N$, $t\in [0,T]$ that $\mathbf{U}^\theta_{0,t} = \lp \lp 0 \:0\:\cdots \:0 \rp,0 \rp \in \R^{1 \times d} \times \R$ and:
\begin{align*}
\mathbf{U}^\theta_{n,t} &= \lb \oplus^{M^n}_{k=1} \lp \frac{1}{M^n} \circledast \lp \mathbf{G}\bullet \aff_{\mathbb{I}_d, \mathcal{W}^{\lp \theta,0,-k \rp }_{T-t}} \rp \rp \rb \\ &\boxplus_{\mathfrak{I}} \lb \boxplus_{i=0,\mathfrak{I}} \lb \lp \frac{T-t}{M^{n-i}}\rp \circledast \lp \boxplus^{M^{n-i}}_{k=1,\mathfrak{I}}\lp \lp \mathbf{F}\bullet \mathbf{U}^{\lp \theta,i,k \rp}_{i,\mathcal{U}_t^{\lp \theta,i,k \rp}}\rp \bullet \aff_{\mathbb{I}_d,\mathcal{W}^{\lp \theta,i,k \rp}_{\mathcal{U}^{\lp \theta,i,k \rp}-t}} \rp \rp \rb \rb \\
&\boxplus_{\mathfrak{I}}\lb \boxplus^{n-1}_{i=0,\mathfrak{I}}\lb \lp \frac{(t-T)\mymathbb{1}_\N \lp i \rp }{M^{n-i}} \rp \circledast \lp \boxplus^{M^{n-i}}_{k=1,\mathfrak{I}}\lp \lp \mathbf{F}\bullet \mathbf{U}^{\lp \theta,-i,k \rp }_{\max\{i-1,0\},\mathcal{U}_t^{\lp \theta,i,k \rp}} \rp \bullet \aff_{\mathbb{I}_d,\mathcal{W}^{\lp \theta,i,k \rp }_{\mathcal{U}_t^{\lp \theta,i,k \rp}-t}}\rp \rp \rb \rb
\end{align*}
\end{lemma}
\begin{theorem}\label{thm1}
Let $p,q,r,L,C,\alpha_0,\alpha_1,\beta_0,\beta_1, T \in [0,\infty)$, $\mathfrak{q} \in [2,\infty)$, $\act \in C(\R,\R)$, $\mathfrak{I} \in \neu$. $\lp \mathbf{F}_{d,\ve} \rp _{\lp d,\ve \rp \in \N_0 \times (0,1]} \subsetneq \neu$. For every $d \in \N_0$ let $f_d \in C \lp \R^{\max\{d,1\}},\R \rp$, for every $d \in \N$ let $\nu_d: \mathcal{B} \lp \R^d \rp \rightarrow [0,1]$ be a probability measure, and assume for all $d \in \N_0$, $v,w \in \R$, $x \in \R^{\max\{d,1\}}$, $\ve \in (0,1]$ that $\lp \int_{\R^d}\left\|y\right\|^{pq\mathfrak{q}} \nu_d \lp dy\rp \rp ^{\frac{1}{pq\mathfrak{q}}}\les Cd^r$, $\hid\lp \mathfrak{I} \rp=1$, $\real_{\act} \lp \mathfrak{I} \rp = \id_\R $, $\real_{\act} \lp \mathbf{F}_{d,\ve} \rp \in C \lp \R^{\max\{d,1\}},\R \rp $, $\max \{ \left|f_0(v)-f_0(w)\right|, \left|\lp \real_{\act} \lp \mathbf{F}_{0,\ve} \rp \rp \lp x \rp - \lp \real_{\act} \lp \mathbf{F}_{0,\ve} \rp \rp \lp x \rp \right|\} \les L \left|v-w\right|$, $\ve^{\alpha_{\min\{d,1\}}}\dep \lp \mathbf{F}\rp_{d,\ve}+\ve^{\beta_{\min\{d,1\}}}\left\| \lay \lp \mathbf{F}_{d,\ve} \rp \right\|_{\max} \les C \lp \max \{x,1\}^p \rp$, and:
\begin{align}
\ve \left| \lp \real_{\act} \lp \mathbf{F}_{d,\ve}\rp \rp \lp x \rp \right| + \left|f_d\lp x\rp - \lp \real_{\act}\lp \mathbf{F}_{d,\ve} \rp \rp \lp x \rp \right| \les \ve C \lp \max \{x,1\}\rp ^p \lp 1+ \left\|x \right\| \rp ^{pq}
\end{align}
It is then the case that for every $d \in \N$, there exists a $u_d \in C \lp \lb 0,T \rb \in \R^d,\R \rp$ with the following properties:
\begin{enumerate}[label = (\roman*)]
\item $u_d$ is polynomially growing.
\item $u_d$ is a viscosity solution.
\item $u_d$ is a solution to:
\begin{align}
\lp \frac{\partial}{\partial t} u_d \rp \lp t,x \rp +\frac{1}{2}\Trace\lp \sigma_d \lp x \rp \lb \sigma_d \lp x \rp \rb ^*\lp \Hess_x u_d \rp \lp t,x\rp \rp + \la u_d \lp x \rp, \lp \nabla_xu_d \rp \lp t,x \rp \ra + \alpha_d(x)u_d(t,x)= 0 \nonumber
\end{align}
with $u_d \lp T,x \rp =g_d \lp x \rp$ for $\lp t,x \rp \in \lp 0,T \rp \times \R^d$, and
\item there exist $\lp \mathbf{U}_{d,t,\ve} \rp _{ \lp d,t,\ve \rp \in \N \times \lb 0,T \rb \times \lp 0,1\rb}$ and $\eta \in \lp \eta_\delta \rp _{\delta \in \lp 0,\infty \rp}: \R \rightarrow \R$ such that for all $d \in \N$, $t \in \lb 0,T \rb$, $\ve \in (0,1]$, $\delta \in \lp 0,\infty \rp $ it holds that $\real_{\act} \lp \mathbf{U}_{d, t,\ve} \rp \in C \lp \R^d, \R \rp$, $\param \lp \mathbf{U}_{d,t,\ve} \rp \les \eta_\delta d^{p\lp 7+4q+ \lp 2+q \rp \delta \rp } \ve ^{-\lp 4+2\delta + \max\{\alpha_0,\alpha_1\}+2\max\{\beta_0,\beta_1\} \rp}$ and:
\begin{align}
\lp \int_{\R^d} \left|u_d\lp t,x \rp -\lp \real_{\act} \lp \mathbf{U}_{d,t,\ve} \rp \rp \lp x \rp \right|^\mathfrak{q} \nu_d \lp dx\rp \rp ^{\frac{1}{\mathfrak{q}}} \les \ve
\end{align}
\end{enumerate}
\end{theorem}
\begin{proof}
The proof of Theorem \ref{thm1} is thus complete
\end{proof}
\end{document}

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