commit cc05feffed5b5e65cebe04df59e1c6d774b2d4ae Author: alistair Date: Wed Oct 6 01:59:17 2021 +1000 initial diff --git a/.gitignore b/.gitignore new file mode 100644 index 0000000..4811c59 --- /dev/null +++ b/.gitignore @@ -0,0 +1,23 @@ +*.aux +*.glo +*.idx +*.log +*.toc +*.ist +*.acn +*.acr +*.alg +*.bbl +*.blg +*.dvi +*.glg +*.gls +*.ilg +*.ind +*.lof +*.lot +*.maf +*.mtc +*.mtc1 +*.out +*.synctex.gz diff --git a/img/dsprlin.pdf b/img/dsprlin.pdf new file mode 100644 index 0000000..8915c2c Binary files /dev/null and b/img/dsprlin.pdf differ diff --git a/img/dsprlin.pdf_tex b/img/dsprlin.pdf_tex new file mode 100644 index 0000000..f545b23 --- /dev/null +++ b/img/dsprlin.pdf_tex @@ -0,0 +1,64 @@ +%% Creator: Inkscape 1.1 (c68e22c387, 2021-05-23), www.inkscape.org +%% PDF/EPS/PS + LaTeX output extension by Johan Engelen, 2010 +%% Accompanies image file 'dsprlin.pdf' (pdf, eps, ps) +%% +%% To include the image in your LaTeX document, write +%% \input{.pdf_tex} +%% instead of +%% \includegraphics{.pdf} +%% To scale the image, write +%% \def\svgwidth{} +%% \input{.pdf_tex} +%% instead of +%% \includegraphics[width=]{.pdf} +%% +%% Images with a different path to the parent latex file can +%% be accessed with the `import' package (which may need to be +%% installed) using +%% \usepackage{import} +%% in the preamble, and then including the image with +%% \import{}{.pdf_tex} +%% Alternatively, one can specify +%% \graphicspath{{/}} +%% +%% For more information, please see info/svg-inkscape on CTAN: +%% http://tug.ctan.org/tex-archive/info/svg-inkscape +%% +\begingroup% + \makeatletter% + \providecommand\color[2][]{% + \errmessage{(Inkscape) Color is used for the text in Inkscape, but the package 'color.sty' is not loaded}% + \renewcommand\color[2][]{}% + }% + \providecommand\transparent[1]{% + \errmessage{(Inkscape) Transparency is used (non-zero) for the text in Inkscape, but the package 'transparent.sty' is not loaded}% + \renewcommand\transparent[1]{}% + }% + 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PDF/EPS/PS + LaTeX output extension by Johan Engelen, 2010 +%% Accompanies image file 'exspr.pdf' (pdf, eps, ps) +%% +%% To include the image in your LaTeX document, write +%% \input{.pdf_tex} +%% instead of +%% \includegraphics{.pdf} +%% To scale the image, write +%% \def\svgwidth{} +%% \input{.pdf_tex} +%% instead of +%% \includegraphics[width=]{.pdf} +%% +%% Images with a different path to the parent latex file can +%% be accessed with the `import' package (which may need to be +%% installed) using +%% \usepackage{import} +%% in the preamble, and then including the image with +%% \import{}{.pdf_tex} +%% Alternatively, one can specify +%% \graphicspath{{/}} +%% +%% For more information, please see info/svg-inkscape on CTAN: +%% http://tug.ctan.org/tex-archive/info/svg-inkscape +%% +\begingroup% + \makeatletter% + \providecommand\color[2][]{% + \errmessage{(Inkscape) Color is used for the text in Inkscape, but the package 'color.sty' is not loaded}% + \renewcommand\color[2][]{}% + }% 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\includegraphics{.pdf} +%% To scale the image, write +%% \def\svgwidth{} +%% \input{.pdf_tex} +%% instead of +%% \includegraphics[width=]{.pdf} +%% +%% Images with a different path to the parent latex file can +%% be accessed with the `import' package (which may need to be +%% installed) using +%% \usepackage{import} +%% in the preamble, and then including the image with +%% \import{}{.pdf_tex} +%% Alternatively, one can specify +%% \graphicspath{{/}} +%% +%% For more information, please see info/svg-inkscape on CTAN: +%% http://tug.ctan.org/tex-archive/info/svg-inkscape +%% +\begingroup% + \makeatletter% + \providecommand\color[2][]{% + \errmessage{(Inkscape) Color is used for the text in Inkscape, but the package 'color.sty' is not loaded}% + \renewcommand\color[2][]{}% + }% + \providecommand\transparent[1]{% + \errmessage{(Inkscape) Transparency is used (non-zero) for the text in Inkscape, but the package 'transparent.sty' is not loaded}% + \renewcommand\transparent[1]{}% + }% + 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parent latex file can +%% be accessed with the `import' package (which may need to be +%% installed) using +%% \usepackage{import} +%% in the preamble, and then including the image with +%% \import{}{.pdf_tex} +%% Alternatively, one can specify +%% \graphicspath{{/}} +%% +%% For more information, please see info/svg-inkscape on CTAN: +%% http://tug.ctan.org/tex-archive/info/svg-inkscape +%% +\begingroup% + \makeatletter% + \providecommand\color[2][]{% + \errmessage{(Inkscape) Color is used for the text in Inkscape, but the package 'color.sty' is not loaded}% + \renewcommand\color[2][]{}% + }% + \providecommand\transparent[1]{% + \errmessage{(Inkscape) Transparency is used (non-zero) for the text in Inkscape, but the package 'transparent.sty' is not loaded}% + \renewcommand\transparent[1]{}% + }% + \providecommand\rotatebox[2]{#2}% + \newcommand*\fsize{\dimexpr\f@size pt\relax}% + \newcommand*\lineheight[1]{\fontsize{\fsize}{#1\fsize}\selectfont}% + \ifx\svgwidth\undefined% + \setlength{\unitlength}{320.75952194bp}% + \ifx\svgscale\undefined% + \relax% + \else% + \setlength{\unitlength}{\unitlength * \real{\svgscale}}% + \fi% + \else% + \setlength{\unitlength}{\svgwidth}% + \fi% + \global\let\svgwidth\undefined% + \global\let\svgscale\undefined% + \makeatother% + \begin{picture}(1,0.47394926)% + \lineheight{1}% + \setlength\tabcolsep{0pt}% + \put(0,0){\includegraphics[width=\unitlength,page=1]{linearelasticspring.pdf}}% + \put(0.3100374,0.23765981){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\lineheight{41.66666794}\smash{\begin{tabular}[t]{l}$\ell_0$\end{tabular}}}}% + \put(0.50493296,0.23765981){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\lineheight{41.66666794}\smash{\begin{tabular}[t]{l}$\ell_0$\end{tabular}}}}% + \put(0.69395617,0.23765981){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\lineheight{41.66666794}\smash{\begin{tabular}[t]{l}$\ell_0$\end{tabular}}}}% + 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$\ell_0$ + $\ell$ + $\ell$ + $\ell$ + + $t=0$ + $t>0$ + + diff --git a/img/pss.png b/img/pss.png new file mode 100644 index 0000000..4ed526e Binary files /dev/null and b/img/pss.png differ diff --git a/img/ref.pdf b/img/ref.pdf new file mode 100644 index 0000000..7be3f18 Binary files /dev/null and b/img/ref.pdf differ diff --git a/img/ref.pdf_tex b/img/ref.pdf_tex new file mode 100644 index 0000000..7232f7a --- /dev/null +++ b/img/ref.pdf_tex @@ -0,0 +1,68 @@ +%% Creator: Inkscape 1.1 (c68e22c387, 2021-05-23), www.inkscape.org +%% PDF/EPS/PS + LaTeX output extension by Johan Engelen, 2010 +%% Accompanies image file 'ref.pdf' (pdf, eps, ps) +%% +%% To include the image in your LaTeX document, write +%% \input{.pdf_tex} +%% instead of +%% \includegraphics{.pdf} +%% To scale the image, write +%% \def\svgwidth{} +%% \input{.pdf_tex} +%% instead of +%% \includegraphics[width=]{.pdf} +%% +%% Images with a different path to the parent latex file can +%% be accessed with the `import' package (which may need to be +%% installed) using +%% \usepackage{import} +%% in the preamble, and then including the image with +%% \import{}{.pdf_tex} +%% Alternatively, one can specify +%% \graphicspath{{/}} +%% +%% For more information, please see info/svg-inkscape on CTAN: +%% http://tug.ctan.org/tex-archive/info/svg-inkscape +%% +\begingroup% + \makeatletter% + \providecommand\color[2][]{% + \errmessage{(Inkscape) Color is used for the text in Inkscape, but the package 'color.sty' is not loaded}% + \renewcommand\color[2][]{}% + }% + \providecommand\transparent[1]{% + \errmessage{(Inkscape) Transparency is used (non-zero) for the text in Inkscape, but the package 'transparent.sty' is not loaded}% + \renewcommand\transparent[1]{}% + }% + \providecommand\rotatebox[2]{#2}% + \newcommand*\fsize{\dimexpr\f@size pt\relax}% + \newcommand*\lineheight[1]{\fontsize{\fsize}{#1\fsize}\selectfont}% + \ifx\svgwidth\undefined% + \setlength{\unitlength}{458.7824015bp}% + \ifx\svgscale\undefined% + \relax% + \else% + \setlength{\unitlength}{\unitlength * \real{\svgscale}}% + \fi% + \else% + \setlength{\unitlength}{\svgwidth}% + \fi% + \global\let\svgwidth\undefined% + \global\let\svgscale\undefined% + \makeatother% + \begin{picture}(1,0.81060728)% + \lineheight{1}% + \setlength\tabcolsep{0pt}% + \put(0,0){\includegraphics[width=\unitlength,page=1]{ref.pdf}}% + \put(0.00408424,-0.08695084){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\lineheight{41.66666794}\smash{\begin{tabular}[t]{l}t = 0\end{tabular}}}}% + \put(-0.00325675,0.78079842){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\lineheight{41.66661453}\smash{\begin{tabular}[t]{l}$t = 0$\end{tabular}}}}% + \put(-0.00325675,0.35380866){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\lineheight{41.66661453}\smash{\begin{tabular}[t]{l}$t > 0$\end{tabular}}}}% + \put(0.07430359,0.70738274){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\lineheight{41.66661453}\smash{\begin{tabular}[t]{l}$A = 0$\end{tabular}}}}% + \put(0.42223477,0.70738274){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\lineheight{41.66661453}\smash{\begin{tabular}[t]{l}$A$\end{tabular}}}}% + \put(0.7721386,0.70738274){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\lineheight{41.66661453}\smash{\begin{tabular}[t]{l}$A=l_0$\end{tabular}}}}% + \put(0,0){\includegraphics[width=\unitlength,page=2]{ref.pdf}}% + \put(0.52303844,0.34355145){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\lineheight{41.66661453}\smash{\begin{tabular}[t]{l}$U(A,t) = X(A,t) - A$\end{tabular}}}}% + \put(0.53324261,0.00576636){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\lineheight{41.66661453}\smash{\begin{tabular}[t]{l}$X(A,t)$\end{tabular}}}}% + \put(0,0){\includegraphics[width=\unitlength,page=3]{ref.pdf}}% + \end{picture}% +\endgroup% diff --git a/img/ref.svg b/img/ref.svg new file mode 100644 index 0000000..9bce604 --- /dev/null +++ b/img/ref.svg @@ -0,0 +1,253 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + t = 0 + $t = 0$ + $t > 0$ + $A = 0$ + $A$ + $A=l_0$ + + + + $U(A,t) = X(A,t) - A$ + $X(A,t)$ + + + + + diff --git a/notes.tex b/notes.tex new file mode 100644 index 0000000..3203555 --- /dev/null +++ b/notes.tex @@ -0,0 +1,2726 @@ + +\documentclass[12pt,ragged2e,oneside,a4paper]{book} + +% For embeddeing the source file +\usepackage{attachfile} +\usepackage{caption} + +\renewcommand{\captionfont}{\small} + + +\usepackage{tocloft} + +\usepackage{enumitem} +\usepackage{marginnote} +\usepackage{graphicx} +% for different enumerate styles +\usepackage{multicol} +\usepackage[utf8]{inputenc} +\usepackage[T1]{fontenc} +\usepackage{amsmath} +\usepackage{amsthm} +\usepackage{bookmark} +\usepackage{pgfplots} +\usepackage{import} +\usepackage{xcolor} +\usepackage{soul} +\newcommand{\mathcolorbox}[2]{\colorbox{#1}{$\displaystyle #2$}} + +\usepgfplotslibrary{external} +\tikzexternalize +\pgfplotsset{width=8cm,compat=1.9} + +\usepackage[pdf]{graphviz} +% for unnumbered theorems + +\usepackage{mathtools} +% for := + +\usepackage{amssymb} +% for mathbb + +\usepackage[english]{babel} + +\usepackage[top=2cm, bottom=3cm, outer=7cm, inner=1cm, heightrounded, +marginparwidth=5cm, marginparsep=0.3cm]{geometry} + +\title{MATH3102 Lecture Notes} +\author{Alistair Michael} + +\graphicspath{{images/}} + +\newcommand{\sforall}{\enspace\forall} + +\renewcommand{\labelenumi}{\roman{enumi})} +\newcommand{\dsum}{\displaystyle\sum} +\newcommand{\ds}{\displaystyle} +\newcommand{\lsum}{\underline{\int_a^b}} +\newcommand{\usum}{\overline{\int_a^b}} +\newcommand{\rieman}{\int_a^b} +\newcommand{\Arg}{\text{Arg}\,} +\newcommand{\Log}{\text{Log}\,} +\newcommand{\Int}{\text{Int}\,} +\newcommand{\Ext}{\text{Ext}\,} +\newcommand{\p}{\partial} +\newcommand{\pd}{\partial} + +\newtheoremstyle{theoremstyle} % name + {\topsep} % Space above + {\topsep} % Space below + {} % Body font + {} % Indent amount + {\itshape} % Theorem head font + {} % Punctuation after theorem head + {.5em} % Space after theorem head + {} % Theorem head spec (can be left empty, meaning ‘normal’) + +\theoremstyle{theoremstyle} +\newtheorem{thm}{Theorem}[subsection] +\newtheorem{lemma}[thm]{Lemma} +\newtheorem{corr}{Corrolary}[thm] +\newtheorem{claim}{Claim}[thm] + +\newtheorem{definition}{Definition}[subsection] +\newtheorem{prop}{Proposition}[subsection] +\newtheorem{ex}{Example}[subsection] + +\newtheoremstyle{ittheoremstyle} % name + {\topsep} % Spaceabove + {\topsep} % Space below + {\itshape} % Body font + {} % Indent amount + {\scshape} % Theorem head font + {.} % Punctuation after theorem head + {.5em} % Space after theorem head + {} % Theorem head spec (can be left empty, meaning ‘normal’) + +\theoremstyle{ittheoremstyle} + + +\newcommand{\incfig}[1]{% + \def\svgwidth{\columnwidth} + \import{./img/}{#1.pdf_tex} +} + +\newcommand{\incfigw}[2]{% + \begin{center} + \def\svgwidth{#2\columnwidth} + \import{./img/}{#1.pdf_tex} + \end{center} +} + + +\newcommand{\N}{\mathbb N} +\newcommand{\R}{\mathbb R} +\newcommand{\Z}{\mathbb Z} +\newcommand{\C}{\mathbb C} +\newcommand{\Q}{\mathbb Q} +\newcommand{\contra}{\Rightarrow\Leftarrow} + +\setlength{\parskip}{0.5em} + +\setlength{\parindent}{0em} + +\newcommand{\nombreindice}{Contents By Week} +\newlistof{lecture}{lecn}{\nombreindice} + +\newcounter{weekno} +\newcounter{lecno} + +% definition of the commands used for the Spanish ToC; +% \captce for chapters, \sectce for sections and +% \ssectce for subsections +\newcommand\captce[1]{% + \addcontentsline{lecn}{chapter}{\protect\makebox[1.3em][l]{\thechapter}#1}} + \newcommand\week[1]{\stepcounter{weekno}\setcounter{lecno}{0} + \addcontentsline{lecn}{section}{\protect\makebox[4.8em][l]{Week \theweekno}#1}} + \newcommand\lecture[1]{\stepcounter{lecno} +\addcontentsline{lecn}{subsection}{\protect\makebox[6em][l]{Lecture \thelecno}#1} +\marginnote{(Lecture \theweekno.\thelecno)} +} + +\renewcommand{\cfttoctitlefont}{\normalfont\bfseries\Large} +\renewcommand{\cftlecntitlefont}{\normalfont\bfseries\Large} + +\begin{document} + + + +\maketitle + +This document has its \LaTeX{} source attached. \attachfile[description=Document +text source,icon=Paperclip]{notes.tex} + +\tableofcontents +\listoflecture + +\pagebreak + + +\week{} +\chapter{Introduction} +\lecture{} + +Applied mathematics is about creating and analyzing mathematical models of +phenomena and processes in the real world. + +The textbook is M. Holmes \emph{Introduction to Foundations of Applied +Mathematics, Springer.} + + +\includegraphics[height=0.5\linewidth]{ab.pdf} + + +\textbf{Modelling:} Creating models from words to equations, identifying key +variables and processes. Models are as simple as possible. + +\emph{Every model is wrong, ... some are useful.} + +Limited insight and predictive power in highly complex systems or systems that +are very sensitive to the initial conditions. + +\section{Topics} + +\begin{itemize} + \item Dimensional analysis + \item Perturbation methods + \item Traffic flow models + \item Continuum mechanics in 1D + \item Elastic and viscoelastic materials + \item Continuum mechanics in 3D + \item Fluid flows +\end{itemize} + + +\chapter{Dimensional Analysis} + +\textbf{Dimensional Estimate:} Qualitative prediction of the order of magnitude +of key quantities without using a model. + +\textbf{Non-dimensionalisation:} Reducing the numbner of effective indpendent +parameters. + +So to create a dimensional estimate for the period of a pendulum $P$, we +understand it may depend on time, distance ($L$), velocity, mass and gravity. + +$P = f(L, v_0, M, g)$ + +How can we combine these paramaters to get the correct dimension as a result? + +$$\frac L g = \frac {m} {m / s^2}$$ + +So as an estimate we can say $P = const \times \sqrt{ \frac L g }$. + +\lecture{} + +\begin{ex} +Ex. Energy of the Bomb + +We want to determine the energy released by a bomb using dimensional analysis. + +We know the radius of the shockwave increases with time, at a rate which depends on the +air density and the energy released. + +\textbf{Radius $R$:} $d$ + +\textbf{Energy:} $m \frac {d^2}{t^2}$ + +\textbf{Air Density ($\rho$):} $\frac {m}{d^3}$ + +So $\frac \rho E = \frac {t^2} {d^5} \implies t^2 \frac E \rho = d^5 $ + +So $R(t) = C \cdot \frac E \rho ^{\frac 15} \cdot t^{\frac 25}$ +\end{ex} + +\subsubsection{Non-dimensionalization} + +\begin{enumerate} + \item[1.] Introduce new units to each independent quantity to reduce the number + of parameters. + \item[2.] Identify relative magnitude of processes and identify dominant + terms and small perturbations. + \item[3.] Choose the non-dimensional form of the equation so that all terms + are either order 1 or small (perturbation). +\end{enumerate} + +\begin{ex} Pendulum + + $\phi$ is angle to overtical, $l$ is length of the pendulum. + + + + \begin{align*} + ml \frac {d^2 \phi}{dt^2} &= -mg \sin(\phi) \\ + \frac {d^2 \phi}{dt^2} &= -\frac gl \sin(\phi) \\ + \end{align*} + + $\frac gl$ can be eliminited by choosing a new time unit. + + Call $t' = \frac t {t_0}$ the nondimensional time unit. + + $g$ = distance / time$^2$. $l$ = distance, $\phi = $ 1. $\sqrt{{\frac lg}} + = $ time $\implies $ choose $t_0 = \sqrt{\frac lg} $ + + \marginnote{$\phi$ is a ratio of two distances it doesnt have a dimension.} + $$\frac {d^2 \phi}{d{t'}^2} = -\sin (\phi)$$ + + Which has no parameters. +\end{ex} + +\begin{ex} Damped Pendulum + + \begin{align*} + ml \frac {d^2\phi}{dt^2} &= - mg\sin(\phi) - kl\frac {d\phi}{dt} \\ + \frac {d^2\phi}{dt^2} &= - \frac gl\sin(\phi) - \frac km\frac {d\phi}{dt} \\ + \end{align*} + + So there are two parameters $\frac km$ and $\frac gl$. + + So choose a new time unit $t = t' t_0$ to make one of the coefficients + disappear. And make $t_0 = \sqrt{\frac lg} $ + + \begin{align*} + \frac {d^2\phi}{dt^2} &= - \frac gl\sin(\phi) {t_0}^2 - \frac km\frac + {d\phi}{dt} t_0 \\ + \frac {d^2\phi}{dt^2} &= - \sin(\phi)- \frac km\frac + {d\phi}{dt} \sqrt{\frac lg} \\ + \frac {d^2\phi}{dt^2} &= - \sin(\phi)- \alpha\frac + {d\phi}{dt} + \end{align*} + + Natural time unit proportional to the period. + + The solution depends on a single parameter $\alpha = (\frac km)\sqrt{\frac l g}$ +\end{ex} + +\begin{ex} + Rabbits and Foxes. + + \begin{align*} + \frac {dR}{dt} &= rR \left(1- \frac RK\right) - pF \frac R{R + S} \\ + \frac {dF}{dt} &= \gamma p F \frac {R}{R + S} - \delta F \\ + \end{align*} + + We want to reduce the parameters: it is a six-dimensional parameter space + for a model of two species. + + \marginnote{So the name of the game is choosing units to absorb all the coefficients.} + Choose natural units for the variables $R$, $F$ and $t$: which look like the + most concrete ones. + + \begin{align*} + R &= R_0 R' \\ + F &= F_0 F' \\ + t &= t_0 t' \\ + \end{align*} + + Exercise. +\end{ex} + +\lecture{} + +\subsubsection{Non-Dimensional Variables and Parameters} + +Where we have a variable $x$ and introduce $x = x_0 x'$ such that $x_0$ +minimises the number of parameters in the model, we call $x'$ a non-dimensional +variable. + +\begin{ex} Cancer Growth (PDE) + + $$\frac {\partial C}{\partial t} = rC \left(1 - \frac CK\right) + D \Delta C$$ + + \begin{itemize} + \item First term is growth from cell division. + \item Second term is diffusion due to random cell movements. + \item $\Delta$ is the laplacian. $\sum \frac {\partial^2}{\partial {x_i}^2}$ + \end{itemize} + + + What are the dimensions of $r$ and $D$? + + \begin{itemize} + \item $r = \frac 1 {time}$ + \item $D = \frac {dist^2}{time}$ + \end{itemize} + + We want to get something with the dimension $\frac {dist}{time}$. + + So $v \sim \sqrt{{rD}}$. + + So use $t_0 = \frac 1r$ $x_0 = \sqrt{{\frac Dr}}$ and $C_0 = K$. + + \begin{align*} + C &= C'K \\ + t &= \frac {t'}r \\ + x &= x'\sqrt{\frac Dr} \\ + \\ + \frac {\partial C'}{\partial t'} &= C' (1 - C') + \Delta' C' + \end{align*} + +\end{ex} + +\chapter{Perturbation Methods} + + Transform the original model into non-dimensional form and identify a small + non-dimensional parameter $\varepsilon < 1$. + + Assume that in the special case $\varepsilon = 0$, that the model simplifies + into a problem that is fully solvable analytically. + + This method gives good approximations for small epsilons. + + \begin{ex} + $m \frac {d^2 y}{dt^2} = -mg $, $y(0) = 0$, $\frac {dy}{dt}|_{t = 0} = v_0$. + + Introduce a nondimensional variables for $t$ and height $y$. + + $y = y_0 t'$ and $t = t_0 t'$ so choose $t_0 = v_0 / g$ and $y_0 = + {v_0}^2 / g$ which leads to the non-dimensional problem + + $$\frac {d^2 y'}{dt'^2} = -1,\quad\quad y'(0)=0,\quad\quad \frac + {dy'}{dt'}|_{t' = 0} = 1$$ + + Its still fully solvable: $y'(t') = - \frac 12 {t'}^2 + t'$. If we want + to generalise it, for example saying gravitational acceleration is no + longer constant. Where $R$ is the radius of the Earth then + + $$\frac {d^2y}{dt^2} = - g \frac {R^2}{(R + y)^2}$$ + + Using the same non-dimensionalisation as before we get: + + $$\frac {d^2 y'}{dt'^2} = -\left(1 + \frac{{v_0}^2}{Rg}y'\right)^{-2},\quad y'(0)=0,\quad \frac + {dy'}{dt'}|_{t' = 0} = 1$$ + + So the small non-dimensional parameter is $\varepsilon = \frac {{v_0}^2}{Rg}$ + + So can find approximate solutions when $\varepsilon \ll 1$ is close to $0$. + + + \end{ex} + + + + \section{Perturbation Methods} + + \week{} + \lecture{} + + \begin{ex} + Solve $x^2 + 2\varepsilon x - 1 = 0$ with perturbation methods. + + The equation has a small non-dimensional parameter $\varepsilon$. + + It can be solved exactly: $x_{1,2} = -\varepsilon \pm + \sqrt{{\varepsilon^2} + 1}$ but we will use it as an example. + See it is easily solvable when $\varepsilon = 0$. + + When $\varepsilon \ll 1$ we can use a taylor series expansion (which is + only valid around $0$ when $\varepsilon \ne 0$). + + { + $$x_{1,2}(\varepsilon) \approx \pm 1 - \varepsilon \pm \frac 12 \varepsilon^2 \mp \frac 18 \varepsilon^4 + \cdots$$ + \captionof{figure}{Taylor Series Expansion}} + + Look for an approximation of the form: $x(\varepsilon) = c_0 + x_1 + \varepsilon + x_2 \varepsilon^2$ + + Assume $\varepsilon \ll 1$ and $x_0, x_1, x_2,\dots$ are unknown + coefficients of magnitude $\sim 1$. We want to be able to order the + terms by the magnitude according only to the power of $\varepsilon$. + + In $(x_0 + x_1\varepsilon + \dots)^2 + 2\varepsilon(x_0 + x_1\varepsilon + + \dots) - 1 = 0$ it is clear that the $x_0$ terms dominate, and we can + take out the dominant terms and group them by the powers of epsilon. + + The largest terms where $\varepsilon ^ 0$ + + $${x_0}^2 - 1 = 0 \quad \quad x_0 = \pm 1$$ + + The next largest terms, terms proportional to $\varepsilon^1$: + + $$2x_0 x_1 \varepsilon + 2\varepsilon x_0 = 0 \quad \quad x_1 = -1$$ + + Terms proportional to $\varepsilon^2$: + + $${x_1}^2 \varepsilon^2 + 2 x_0 x_2 \varepsilon^2 + 2x_1 \varepsilon^2 = + 0 \quad \quad x_2 = \pm \frac 12$$ + + We can neglect the rest assuming $\varepsilon^3$ is insignificantly + small. + + So we have the approximate solution: + + \begin{align*} + x(\varepsilon) &= x_0 + x_1 \varepsilon + x_2 \varepsilon^2 + \cdots \\ + x(\varepsilon) &= \pm 1 - \varepsilon \pm \frac 12 \varepsilon^2 + \cdots + \end{align*} + \marginnote{Which matches the taylor series.} + + + All we need to be able to solve these kinds of problems is to be able to + solve for the coefficients along the way and use the approximation of + $\varepsilon$ close to $0$: we do not need the explicit form that we + have in this example. + + + \end{ex} + + \textbf{Perturbation method: Solve $F(y; \varepsilon) = 0$ for $\varepsilon \ll 1$ assuming + that $F(y; 0) = 0$ has an exact solution $y_0$ to find an approximate + analytical solution.} + + \begin{definition}{Perturbation Method}{} + + Look for a solution in the form of an asymptotic expansion + + $$y(\varepsilon) = y_0 + y_1 f_1(\varepsilon) + y_2f_2(\varepsilon) + + \cdots $$ + + Where $1 \gg f_1(\varepsilon) \gg f_2(\varepsilon) \gg \cdots$ when + $\varepsilon \ll 1$ (for example a power series like in Example 2.0.1). + + \end{definition} + + \begin{ex} + Find an asymptotic solution for $x^3 + \varepsilon x - 1 = 0$ when + $\varepsilon \ll 1$ + + \marginnote{These problems are artificially simple but it is fine for the + coefficients to contain parameters.} + + Special case: $\varepsilon = 0,\; x^3 - 1 = 0$ has the exact solution $x = + 1$. + + $(x_0 + x_1 \varepsilon + x_2\varepsilon^2 \cdots) ^3 + \varepsilon(x_0 + + x_1\varepsilon + x_2 \varepsilon^2 \cdots) - 1 = 0$ + + \begin{align*} + e^0 &: x_0^3 - 1 = 0 &\implies x_0 = 1 \\ + e^1 &: 3x_0^2x_1 + x_0 = 0, \; 3x_1 + 1 = 0 &\implies x_1 = \frac 13 \\ + e^2 &: 3x_0^2x_2 + 3x_1^2x_0 + x_1 = 0,\; 3x_2 + \frac 13 - \frac 13 = + 0 &\implies x_2 = 0\\ + e^3 &: {...} &\implies x_3 = \frac 1 {81} + \end{align*} + + + { + $$x = 1 - \frac 13 \varepsilon + \frac 1 {81}\varepsilon^3 + \cdots$$ + \captionof{figure}{The Asymptotic Series Solution} + } + + \end{ex} + + \begin{ex} + Point mass in a non-constant gravitational field: + + \begin{equation} + \frac {d^2y'}{dt'^2} = - \left(1 + \frac {{v_0}^2}{Rg}y'\right)^{-2}, y'(0) + = 0, \frac {dy'}{dt'}(t = 0) = 1 + \end{equation} + + Small non-dimensional parameter $\varepsilon = \frac {{v_0}^2}{Rg}$. It + is not solvable for nonzero $\varepsilon$ but can be approximated when + $\varepsilon \ll 1$. + + \marginnote{You could drop the primes after the change of variables to + make the notation less confusing.} + + $$\varepsilon = 0 \implies \frac {d^2y'}{dt'^2} = -1 $$ + + $y'(t') = - \frac 1 2 t'^2 + t'$ + + $$\frac {d^2y'}{dt'^2}\left(1 + \frac {{v_0}^2}{Rg}y'\right)^{2} = -1 $$ + + \marginnote{Rearranging (1)} + + \begin{align*} + \varepsilon^0:& \frac {d^2y_0}{dt'^2} = -1, y_0(0) = 0, \frac + {dy_0}{dt'}(t'=0) = 1 && y_0(t') = - \frac 12 t'^2 + t' \\ + \varepsilon^1:& \frac {d^2y_1}{dt'^2} + 2y_0 \frac {d^2y_0}{dt'^2} = + 0 &&\implies \frac{d^2 y_1}{2t'^2} = -t'^2 + 2t' \\ + &\text{And solve with initial condition} + && \implies y_1(t') = - \frac 13 t'^3 + \frac 1 {12}t'^4 + \end{align*} + + $$y'(t') = - \frac 12 t'^2 + t' - \varepsilon \frac {t'^3(4 - t')}{12} + + \cdots$$ + + \end{ex} + + \subsection{Singular Perturbation} + + \lecture{} + + There are some cases where $\varepsilon = 0$ and small $\varepsilon \ne 0$ + are actually qualitatively different (according to the model). + + A typical case is when setting $\varepsilon = 0$ results in a change of + order of the equation (i.e. when the epsilon is on the highest order term). + These are called \emph{singular perturbation} problems, and the standard + method does not work. + + \begin{ex} $\varepsilon ^2 + x - 1 = 0, \; \varepsilon << 1$ + + If we use the regular perturbation method: + + \[ + x(\varepsilon) = x_0 + x_1 \varepsilon + x_2 \varepsilon^2 + \dots + \quad\quad\qquad (*) + \] + + \marginnote{Substitute $x(\varepsilon)$ into the equation.} + $\varepsilon(x_0 + x_1 \varepsilon + \dots) ^2 + (x_0 + x_1\varepsilon + + \dots) - 1 = 0$ + + + Largest $\varepsilon^0$: $x_0 - 1 = 0 \implies x_0 = 1$. + + \marginnote{Using solution $x_0 = 1$} + $\varepsilon^1: $ $x_0^2 + x_1 = 0 \implies x_1 = -1$ + + So $x = 1 - \varepsilon + 2\varepsilon^2 + \dots$ + + What is the problem with this approach? This is a quadratic equation but + we only got \textbf{one solution.} This is exactly solvable, so we know + the solution should be $x_{1,2} = \frac 1 {2\varepsilon}(-1 \pm + \sqrt{{1-4\varepsilon}})$. + + \begin{center} + \begin{tikzpicture}{testp} + \begin{axis}[xlabel = $\varepsilon$, ylabel=$x$, xtick={0,0.1}] + \addplot[domain=0:0.1, samples=90,color=blue,]{ 1/(2 * x) * (-1 + sqrt(1 -4 * x))}; + \addplot[domain=0:0.1, samples=90,color=blue]{ 1/(2 *x) * (-1 - sqrt(1 - 4* x))}; + \end{axis} + \end{tikzpicture} + \end{center} + + Graphing this solution we can see we have large negative numbers in $x$ as + $\varepsilon$ approaches zero so the assumption about the form of the + solution $(*)$ is not correct: all out coefficients are not order $1$. + + When we set $\varepsilon = 0$ it reduces the order of the equation to + $1$. + \end{ex} + + \subsubsection{Solving Singular Perturbation Problems} + + Use a transformation of variables $x = e^\alpha z$, where + $\alpha$ is determined based on the condition that the highest order + term participates in the dominant part of the equation when $\varepsilon + \to 0$. Then solve it as a regular perturbation problem. + + \begin{ex} Solving $\varepsilon x^2 + x - 1 = 0$. + + \begin{align*} + x = \varepsilon^\alpha z &\implies \varepsilon^{2\alpha + 1}z^2 + + \varepsilon^\alpha z - 1 = 0 + \end{align*} + + We need to match terms of the same order, and there always needs to + be at least two of them so they can cancel out to zero. So we need + to decide which powers of epsilon match up. + + Either: + + \begin{itemize} + \item $2 \alpha + 1 = 0$ + \item $\alpha = 0$, or + \item $2 \alpha + 1 = \alpha$ + \end{itemize} + + + Case 1: + \begin{align*} + 2 \alpha + 1 &= 0 \implies \alpha = \frac 12 \\ + \varepsilon^0 z^2 + \varepsilon ^{-\frac 12} - 1 &= 0\\ + \varepsilon^0 z^2 - 1 &= 0 + \end{align*} + + When $\varepsilon \to 0$ the first and last terms are of the same + magnitude but the middle is larger: even though we chose the first + and last to be the dominant terms. This doesnt work so we have to + drop this case. + + Case 2: $\alpha = 0$ + + \begin{align*} + \varepsilon z^2 + \varepsilon^0z - 1 &= 0 \\ + \varepsilon z^2 + z - 1 &= 0 \\ + \end{align*} + + This is not even a transformation and it is stilla singular + perturbation problem. + + Case 3: $2\alpha + 1 = \alpha \implies \alpha = -1$ + + \begin{align*} + \varepsilon^{-1} z^2 + \varepsilon^{-1} - 1 &= 0 \\ + \implies z^2 + z - \varepsilon &= 0 \\ + \end{align*} + + So now the third term $-1$ is small compared to the very large + $\varepsilon^{-1}$ when $\varepsilon \to 0$. If we set $\varepsilon + = 0$ we still have two roots. + \begin{itemize} + \item The highset order term is part of the dominant balance. + \item We don't loose one of the solutions when $\varepsilon = + 0$. + \end{itemize} + + + Substitute $z = z_0 + z_1 \varepsilon + \dots$ to get + \begin{align*} + (z_0 + z_1\varepsilon + z_2\varepsilon^2+\dots)^2 + (z_0 + z_1 + \varepsilon + z_2 \varepsilon^2) - \varepsilon &= 0 \\ + z_0^2 + z_0 = 0 \implies z_0 &= 0, -1 \\ + 2z_0 z_1 + z_1 - 1 = 0 \implies z_1 &= 1,-1 \\ + \dots + \end{align*} + + Two solutions: $z = \varepsilon - \varepsilon^2 + \dots$ and $z = -1 + - \varepsilon + \varepsilon^2 + \dots$. + + Now do the backwards transformation $x = e^{-1}z$: + \begin{align*} + x &= 1 - \varepsilon + \cdots \\ + x &= -\varepsilon^{-1} - 1 + \varepsilon + \cdots + \end{align*} + \end{ex} + + \hrule + \lecture{} + + Generally considering an $f(x;\varepsilon)$, $x = ?$ when $\varepsilon + \ll 1$ epsilon allowing us to approximate a solution. + \begin{itemize} + \item Substitute a series approximation: $x = x_0 + x_1 \varepsilon + + x_2 \varepsilon^2 + \cdots$ + \item Use larger numbers of terms and eventually truncate when there + is enough accuracy. + \end{itemize} + + Boundary conditions: + + Eg $\varepsilon y'' 2y' + 2y = 0$ want to find $y$ in $[0,1]$ with + boundary conditions for example $y(0) = 0$ and $y(1) = 1$. + + Note $\varepsilon$ multiplies the term with the highset derivative. If + $\varepsilon = 0$ then it becomes a first order ode and cannot satisfy + both boundary conditions, so it is a singular perturbation problem. + + In this case we can directly solve it for reference: + + \begin{center} + \begin{tikzpicture}{testp} + \begin{axis}[width=0.9\linewidth,height=2in,xlabel = $x$, ylabel=$y$, xtick={0,0.5}] + \addplot[domain=0:0.5, samples=200,color=blue]{-2.71964267074 * + (exp(-18.994 * x) - exp(-1.0553 * x ) }; + \addlegendentry{$\varepsilon= 0.1$} + \addplot[domain=0:0.5, samples=200,color=red]{-2.73205 * + (exp(-198.995 * x) - exp(-1.00505 * x ) }; + \addlegendentry{$\varepsilon= 0.1$} + \addplot[domain=0:0.5, samples=200,color=green]{-2.71964267074 * + (exp(-1999 * x) - exp(-1.0005 * x ) }; + \addlegendentry{$\varepsilon= 0.1$} + \end{axis} + \end{tikzpicture} + \end{center} + + You can see there is a sharp transition point in the solution + called `boundary layers' + + \subsubsection{Boundary Layers, Matched Asymptotic Expansions} + + For a singular perterbation in ODE boundary value problems, for example + $\varepsilon y '' + 2y' + 2y = 0 $ where $y(0) = 0, y(1) = 1$, obviously + this is singular since $\varepsilon = 0$ makes the highest order term + disappear. Standard perterbation method is only good far from the sharp + transition point shown above---the `boundary layers'. + + Matched asymptotic expansions makes an approximation that is valid in + the entire domain. + + \begin{itemize} + \item Use rescaling $x = \varepsilon^{\alpha} z$. We want to create + a transformation that absorbs the singularity so we can solve + the transformed equation using the normal perterbation method + and then transform it back. + + + So if we were to set $\varepsilon = 0$ the order should not + change. + \item Chain rule transforms derivatives: + + $Y'(z) = \frac{dY}{dx} \frac {dx}{dz} = y'(x) \varepsilon^a$ + + $Y''(z) = y''(x) \varepsilon^{2a} $ + \item After change of variables to the rescaled equation do a + regular perturbation so we still get a nonzero second derivative + term. + + $\varepsilon y''(x) 2y'(x) + 2y(x) = 0 \to$ + + Rescaled equation: $\varepsilon^{1-2\alpha}Y''(z) + \varepsilon^{-\alpha}2Y'(z) + + 2Y(z) = 0$ + + \week{}\lecture{} + \item Determine $\alpha$ + + Choose an alpha so the term with the highest derivative is part + of the dominant balance. + \begin{enumerate} + \item $1 - 2 \alpha = -\alpha \implies \alpha = 1$ + \item $1 - 2\alpha = 0 \implies \alpha = \frac 12$ + \item $-\alpha = 0 \implies \alpha = 0$ (no rescaling) + \end{enumerate} + + \begin{enumerate} + \item $1 - 2 \alpha = -\alpha \implies \alpha = 1$ + \begin{align*} + \varepsilon ^{-1} Y''(z) + \varepsilon^{-1}2Y'(z) + + 2Y(z) &= 0 \implies \\ + Y''(z) + 2Y'(zz) + \varepsilon2Y(z) &= 0 + \end{align*} + The two largest terms (with + $\varepsilon^{-1}$) can cancel out and the other is a + small correction + + So this one is okay. + + \item $1 - 2\alpha = 0 \implies \alpha = \frac 12$ + + $Y''(z ) + \varepsilon^{-\frac 12} 2Y'(z) + 2Y(z)$ + + The largest term (the middle term) does not have another + term of the same order which can balance it out to reach + a possible solution, so this cannot produce a solution. + \end{enumerate} + + \item So we choose $\alpha = 1$, and can now sovle with + perterbation. + + \emph{Boundary condition:} We only expect this to give a good + solution near $x=$, not for the whole domain, so we can only use the boundary + condition at $x = 0$. + + $\alpha = 1 \implies Y''(z) + 2Y'(z) + \varepsilon2Y(z) = + 0,\quad Y(0) = 0$ + + \begin{align*} + Y(z) &= Y_0(z) + \varepsilon Y_1(z) + \cdots \\ + 0 &= (Y_0(z) + \varepsilon Y_1(z) + \dots)'' + 2(Y_0(z) + + \varepsilon Y_1(z) + \dots)' + \varepsilon(Y_0(z) + + \varepsilon Y_1(z) + \dots) \\ + \end{align*} + \begin{itemize} + \item [$e^0$] We get $Y_0'' + 2Y_0'(z) = 0$, which is + solvable: + + $Y_0(z) = e^{\lambda z} \implies \lambda^2 + 2\lambda = + 0 \implies \lambda_1 = 0, \lambda_2 = -2$ so + + $y_0(z) = c(1 - e^{-2z})$, with an unknown constant $c$. + + Transforming back: $x = \varepsilon z \implies$ + + \marginnote{The boundary layer solution, which is only + valid around $x = 0$.} + $$y_0(x) = c(1 - e^{\frac{-2x} \varepsilon})$$ + \end{itemize} + + \item Matching the two solutions: + + Boundary layer (sometimes called inner solution), valid near $x = 0$. $y_{BL}(x) = c(1 - e^{\frac{-2x} + \varepsilon})$. + + \marginnote{$y_{out}$ obtained before using normal perterbation} + Outer solution: $y_{out} = e^{1 - x}$. Valid away from $x = 0$. + + We want to combine them to crate an approximation that is valid + for the whole domain. + + Find the \textbf{common limit}: + + $y_{BL} (x, \varepsilon \to 0) = c$ + + $Y_{out}(x \to 0) = e$. + + This implies $e^1 = c$, $y_{BL} = e(1 - e^{\frac{-2x}{\varepsilon}})$ + + \item Combine to get composite solution: + \begin{align*} + y(x) &= y(x)_{OUT} + y(x)_{BL} - \text{common limit} \\ + &= y(x) e^{1-x} + e(1 - e^{-2x/\varepsilon}) - e \\ + &= e^{1-x} - e^{1 - \frac {2x}e} + \end{align*} + + + \end{itemize} + + By solving directly we know the exact solution for $\varepsilon = 0.1$ + is + + $-2.874 \left(e^{-18.944 x}-e^{-1.056 x}\right)$ + + \begin{center} + \begin{tikzpicture}{testp} + \begin{axis}[width=0.9\linewidth,height=2in,xlabel = $x$, ylabel=$y$, xtick={0,0.5}] + \addplot[domain=0:0.5, samples=200,color=red]{exp(1-x) }; + \addlegendentry{$y_{out}(x)$} + \addplot[domain=0:0.5, samples=200,color=blue]{ e * (1 - + exp((-2 * x) / + 0.1))}; + \addlegendentry{$y_{BL}(x)$} + \addplot[domain=0:0.5, samples=200,color=green]{exp(1-x) - exp(1 - + 2 * x / 0.1)}; + \addlegendentry{Combined} + \addplot[domain=0:0.5, samples=200,color=orange]{-2.71964267074 * + (exp(-18.994 * x) - exp(-1.0553 * x ) }; + \addlegendentry{Exact} + \end{axis} + \end{tikzpicture} + \end{center} + + + \lecture{} + + Mathematica. + + \lecture{} + + \subsubsection{Methods of Multiple Scales} + + \begin{ex} + Nonlinear pendulum without damping: + + $ml\theta'' = -mg \sin(\theta)$ + + In nondimensiona form: + $\theta '' = \sin(\theta),\quad\theta(t) = ?$ + + To do perterbation we add an assumption, $\theta(0) = \varepsilon$, + $\theta'(0) = 0$, and a small initial angle $\epsilon \ll 1$ + + Trying regular perterbation: $\theta(t) = \theta_0 (t) + \varepsilon + \theta_1(t) + \cdots. $ + Use taylor expansion of $\sin(\theta)$ around $0$ for small + oscillations. $\sin(\theta) \approx \theta - \frac {\theta^3} 6 + + \cdots $ So when oscillation is small you can use the taylor series to + represent $\sin(\theta)$. + + Sub the expansion in the initial conditions: $\theta_0(0) + + \varepsilon \theta_1(0) + ... = \varepsilon. $ so $\theta_0(0) = 0$, + $\theta_1(0) = 1$. And + + $\theta_0'(0) + \varepsilon\theta_1'(0) + ... = 0 \implies + \theta'_0(0) = 0, \theta_1'(0) = 0$. + + \begin{align*} + \varepsilon^0 :&& \theta_0'' = -\theta_0' + \frac 16 + (\theta_0^3) + \cdots = -\sin(\theta_0) \\ + &&\theta_0(0) = 0\quad \theta_0'(0) = 0 \\ + &&\theta_0(t) = 0 \\ + \varepsilon^1:&& \theta_1'' = -\theta_1 \\ + && \theta_1(0) = 1, \theta_1'(0) = 0 \\ + \theta_1(t) = \cos(t) \implies \theta(t) = + \varepsilon \cos(t) + \end{align*} + + The approximation is not uniform in time. + + Solving for the next power of $\varepsilon$... + + $$\theta(t) = \varepsilon\cos(t) - \frac 1 {16} \varepsilon^3 t \sin(t) + + \cdots$$ + + For very large $t$ the second term may be larger than the first + which is a problem since we expect the order to be based only on the + power of $\varepsilon$. + + \begin{center} + \begin{tikzpicture}{testp} + \begin{axis}[width=0.9\linewidth,height=2in,xlabel = $t$, + xtick={180,190,200}] + \addplot[domain=0:50, samples=200,color=blue]{ exp(-0.1/(2*x)) * + sin(1/2 * sqrt(4 - 0.1^2) * deg(x)) }; + \addlegendentry{Exact} + \addplot[domain=0:50, samples=200,color=red]{ (1 - 0.1*x/2) * + sin(deg(x)) }; + \addlegendentry{Approx} + \end{axis} + \end{tikzpicture} + \end{center} + + There are two timescales in the problem the oscillation period and the + slow drift in the phase of the oscillation which is not described by + the first term so we can add parameters to the time to split it into + multiple timescales. + + \end{ex} + + \week{} + \lecture{} + + The ordering of the terms, their relative magnitude, may change in a + way which violates our assumptions. This is what the method of multiple + scales addresses. + + There are two characteristic timescales: oscillation period, + the time of decay. + + \subsubsection{Method of multiple scales} + + \begin{itemize} + \marginnote{(the equation is already non-dimensional)} + \item Introduce two new nondimensional time variables to rescale. + + $t_1 = t$ and $t_2 = \varepsilon^\alpha t$ so $y(t) \implies Y(t_1,t_2)$ + + \item Use the chain rule: + + $$\frac d{dt} = \frac {\partial}{\partial t_1} + + \varepsilon^\alpha \frac \partial{\partial t_2}$$ + + $$\frac{d^2}{dt^2} = \frac{\partial^2}{\partial {t_1}^2} + + 2\varepsilon^\alpha \frac {\partial^2}{\partial t_1\partial t_2} + + \varepsilon^{2\alpha} \frac {\partial^2}{\partial {t_2}^2}$$ + + \item Initial conditions: + + $Y(0,0) = 0, \quad \frac{\partial Y}{\partial t_1}(0,0) + + \varepsilon^\alpha \frac {\partial Y}{\partial t_2}(0,0) = 1$ + \item Substitute + + $$Y(t_1,t_2) = Y_0(t_1,t_2) + \varepsilon Y_1(t_1,t_2) + + O(\varepsilon^2)$$ + \item + + Largest:$\varepsilon^0: \frac{\partial^2 Y_0}{\partial t_1^2} + Y_0 = + 0$ + + $Y_0(0,0) = 0, \quad \frac{\partial Y_0} {\partial t_1}(0,0) = 1$ + + We know there is a general solution: + + $$Y_0(t_1, t_2) = A(t_2) \sin(t_1) + B(t_2)\cos(t_2)$$ + + Where $B(0) = 0$ and $A(0) = 1$ is the initial condition, by + susbstituting the initial conditions in the general solution. + + We will choose $A(t)$ and $B(t)$ in such a way as to avoid + resonancnes. + + + In order for the friction terms to be balanced by the mixed + derivatives choose $\alpha = 1$, (slow time: $t_2 = \varepsilon^1 + t$) + + + Terms $\varepsilon^1$: + + $\frac {\partial^2 Y_1} {\partial t_1^2} + Y_1 = - \frac {\partial + y_0} {\partial t_1} - 2 \frac {\partial ^2 Y_0} {\partial t_1 + \partial t_2}$ + + Substitute + + $\frac {\partial^2 Y_1} {\partial t_1^2} + Y_1 = - A(t_2)\cos(t_1) + + B(t_2) \sin(t_1) - 2A'\cos(t_1) + 2B'(t_2) \sin(t_1)$ + + Group together forcing terms. + + $\frac {\partial^2 Y_1} {\partial t_1^2} + Y_1 = [- A(t_2) + 2A'(t_2)]\cos(t_1) + + [B(t_2) \sin(t_1) + 2B'(t_2)] \sin(t_1)$ + + To avoid resonant forcing which leads to secular terms we need $A + 2A' + = 0$ and $B + 2B' = 0$. So we need to solve these differential + equations: + + $$A(t_2) = A(0) e^{\frac{-t_2}2} \quad \quad B(t_2) = B(0) + e^{\frac{-t_2}2}$$ + + \item Transform back and substitute + + $y(t) = e^{\frac {-\varepsilon t}2 } \sin(t)$ + + \end{itemize} + + +\chapter{Traffic Flow Models} + \lecture{} + + \textbf{Agent based models:} rules for individual cars and how they + interact can be stochastic. can be used for simulations, not suitable + for mathematical analysis. + + \textbf{Conntinuum model:} car density function and velocity + distribution etc. Treated as a continuous variable. + + \begin{definition}{}{} Density function (1D) + + $\rho(x,t) = \frac {\text{The number of cars in } [x - \Delta x, x + \Delta x]} { + \text{distance } 2\Delta x}$ + + The $\Delta x$ needs to be large compared to the size of the cars but + small compared to the region. + \end{definition} + + \begin{definition}{}{} Flux $J(x,t)$ + + The number of cars that passing a position $x$ per unit time + $\frac {[t-\Delta t, t + \Delta t]} {2\Delta t}$. + + For constant velocity of cars with distantce $d$ $J = \frac v {l + d} = + v \rho$. + \end{definition} + + \begin{definition}{}{} Balance Law + + Conservation law for cars along the region $[x - \Delta x, x + + \Delta x]$ of the road. + + Initial cars + cars going in - cars going out. + + $$J(x - \Delta x, t) \Delta t - J(x + \Delta x , t) \Delta t = + \frac{J(x - \Delta x, t) - J( x + \Delta x, t)}{2\Delta x}$$ + + Taking the limit $\Delta t \to 0$ and $\Delta x \to 0$ + + $$\frac {\partial \rho}{\partial t} = - \frac {\partial J}{\partial + x}$$ + + The flux can also be written as $J = \rho v$ so we have the equation + + $$\frac {\partial \rho}{\partial t} + \frac {\partial (\rho + v)}{\partial x} = 0$$ + \end{definition} + + In order to obtain a closed system we need a relationship between the + velocity and density of traffic. + + \begin{definition}{}{} + Constitutive law + + $v(\rho)$ is known as the constitutive law + + It is natural to expect $v(\rho)$ is monotonously decreasing: as the + density of cars increases the traffic slows down. And $v(0) = v_M$ when + there are no other cars around traffic moves at the maximum allowable + velocity. + + This can be based on observations from real traffic. + \end{definition} + + The simplest constitutive law is greenshield's law assumes a linear + function: + + $$v(\rho) = v_M\left(1 - \frac \rho{\rho_M}\right)$$ + + substituting into the balance law this gives + + $$\frac {\partial \rho}{\partial t} + c(\rho) \frac {\partial (\rho + )}{\partial x} = 0 \quad\quad \text{where} $$ + + + $$c(\rho) = v_M\left(1 - \frac {2\rho}{\rho_M}\right)$$ + + If you assume an almost uniform density and they propogate like waves + you can get wave velocity, which is the velocity of propogation of + perterbations in the density of traffic: + + $$c(\rho) = v(\rho) + \rho v'(\rho)'$$ + + $$\frac {\partial \rho} { \partial t} + c(\rho) \frac {\partial \rho}{\partial x} = + 0$$ + + \subsubsection{Solving} + + Consider the simplest case, which is consntant velocity: $v(\rho) = a$. + But we don't assume density is uniform: $v$ is independent of the density, + which is not realistic for traffic it + is just a simpler maths problem; advection equation for fluid flow. + + The solution is $\rho(x,t) = f(x - at)$, can check this with + substitution. + + $$\frac {\partial \rho}{\partial t} + a \frac {\partial \rho }{\partial x} + = 0$$ + + Initial: $\rho (x,0) = f(x) $ + + Boundary: $\rho (x = 0, t) = g(t)$. + + Don't need boundary conditions on both sides of the domain, because it + only goes in one direction. + + Solution : $$\rho(x,t) = \begin{cases}f(x - at) & x > at \\ g(t - \frac + )xa & x < at\end{cases}$$ + + Nonconstant velocity: + + $$\frac {\partial \rho}{\partial t} + c (\rho) \frac {\partial + \rho}{\partial x} = 0, \quad \rho(x,t=0) = f(x)$$ + + The density changes according to the velocity. + +\week{} + +\lecture{} + +\begin{ex} +\end{ex} + +\begin{itemize} + \item uniform positive velocity + \item density linearly linked to velocity according to greenfields law +\end{itemize} + + + +Perterbations propagate along the roads according to wave equation: +areformation of hte balance equation: + +$$\frac {\partial \rho}{\partial t} + c (\rho) \frac {\partial \rho}{\partial x} += 0$$ + +Where $c(\rho) = J' (\rho) = v + \rho v'(\rho)$. + +\textbf{Special case with constatnt velocity} gives the travelling wave +solution: $\rho(x,t) = f(x - ct)$ the density distribution is transported along hte x axis with + constatnt velocity without any distortion. The shape is exactly the + same. This is when the cars dont care about the density. + + \textbf{Method of characteristic trajectories}: $x(t) = x_0 + ct$. +find th trajectory ending at $(x,t) \implies x_0 = x(t) - ct$; + The density is conserved along the + lines $x(t) = x_0 + c(\rho(x_0))t$. (A transformation into an arbitrary + space such that there are lies where $\rho$ is constant to make it easy to + solve.) + + +The same method works for non-constant velocity, we know the waves propogate at +constant speed (because of our assumption) but we don't know the slope. + +For any $(x,t)$ find an $x_0$ such that $x = x_0 + c(f(x_0))t$ then $\rho(x,t) += \rho(x_0, 0) = f(x_0)$. + +\emph{characteristics:} points $x_0$ that move to the point $(x,t)$. + +\begin{itemize} + \item There may be regions with no characteristics + \item there may be regions with multipl eintersecing characteristics +\end{itemize} + +\section{Rankine-Hugoniot condition:} + +\begin{itemize} + \item If there are discontinities in the initial conditions of density + ($\rho$), you can + end up with diverging or overlapping characteristic lines and there can + be no $\rho_l < \rho_r$ or no $\rho_l > \rho_r$ unique solutions for a particular region. +\end{itemize} + + \begin{center} + \begin{tikzpicture}{testp} + \begin{axis}[xmin=0, xmax=10, ymin=0, ymax=10, ticks=none, xlabel=x, ylabel=t] + \addplot[domain=0:7.5,samples=90,color=red]{ 2.5 * (x - 5)}; + \addplot[domain=0:20,samples=90,color=black,very thick,dotted]{ 2 * x}; + \addplot[domain=0:20,samples=90,color=black,very thick,dotted]{ 2 * (x - 1)}; + \addplot[domain=0:20,samples=90,color=black,very thick,dotted]{ 2 * (x - 2)}; + \addplot[domain=0:20,samples=90,color=black,very thick,dotted]{ 2 * (x - 3)}; + \addplot[domain=0:20,samples=90,color=black,very thick,dotted]{ 2 * (x - 4)}; + \addplot[domain=0:20,samples=90,color=black,very thick,dotted]{ 2 * (x - 5)}; + \addplot[domain=0:20,samples=90,color=black,very thick,dotted]{ 3 * (x - 5)}; + \addplot[domain=0:20,samples=90,color=black,very thick,dotted]{ 3 * (x - 6)}; + \addplot[domain=0:20,samples=90,color=black,very thick,dotted]{ 3 * (x - 7)}; + \addplot[domain=0:20,samples=90,color=black,very thick,dotted]{ 3 * (x - 8)}; + \addplot[domain=0:20,samples=90,color=black,very thick,dotted]{ 3 * (x - 9)}; + \end{axis} + \end{tikzpicture} + \end{center} + + +You can integrate the balance equation to get information about the +discontinuity. Integrate $x$ in $[s(t) - \varepsilon, s(t) + \varepsilon]$, +givin + +$$\int_{s - \varepsilon}^{s + \varepsilon} \frac {\partial \rho}{\partial t} dx ++ J(\rho_r) - J(\rho_l) = 0$$ + +Can differentiate this again using \textbf{Leibniz rule}: the moving integrating bounds +changes the value of the integral. + +If they are constant: + +$$\frac d {dt} \int_\alpha ^\beta f(x,t) dx = \int_\alpha^\beta \frac {\partial +f} {\partial t} dx$$ + +Otherwise + +$$\cdots$$ + +Specifically for the non-constant functions $\alpha(t) = s - \varepsilon$ and $\beta(t) = s + +\varepsilon$ + +$$\frac d {dt} \int_{x - \varepsilon}^{s + \varepsilon} \rho dx - s'(t) \rho_r + +s'(t) + \rho_l + J(\rho_r) - J(\rho_l) = 0$$ + + +$$\frac d {dt} \int_{x - \varepsilon}^{s + \varepsilon} \rho dx = \frac d {dt} +(\epsilon \rho_l + \epsilon \rho _r) = 0$$ + +\marginnote{cool} +$$s'(t) = \frac {J(\rho_r) - J(\rho l)} {\rho_r - \rho_L} \quad\quad\quad(*)$$ + +From this we can also show that the discontinuity moves with the velocity + +$$s'(t) = \frac 1 {\rho_r - \rho _l } \int_{\rho_l} ^{\rho_r} c(\rho) d\rho$$ + + +\begin{itemize} + \item This solution is not limited to greenshield's law it gives + caharacteristic lines for intersections of the characteristic lines and + they are related to the function $c(\rho)$ and the way these boundaries + move through traffic (because of hte moving $s$ in the integral before). +\end{itemize} + +Under \textbf{Greenshield's law} it becomes the average speed of the wave velocity: + +$$s'(t) = \frac 12 (c_R + c_L)$$ + + +We want to solve these kinds of initial conditions cause the models often +produce steps (singularities in the solution) or 'shocks' between fast and slow +moving regions (because the faster cars `run into' the slow cars and teh +slowdown region gets smaller until it is a jump). + +note that $c(\rho)$ is not strictly positive: perturbation waves can move +backwards through the traffic, eg if a car suddenly stops on a highway. + + + \lecture{} + + \textbf{Initial Conditions $\rho_l > \rho_r$} + + \begin{center} + \begin{tikzpicture}{testp} + \begin{axis}[xmin=0, xmax=10, ymin=0, ymax=10, ticks=none, xlabel=x, ylabel=t] + \addplot[domain=0:20,samples=90,color=black,very thick,dotted]{ 3 * x}; + \addplot[domain=0:20,samples=90,color=black,very thick,dotted]{ 3 * (x - 1)}; + \addplot[domain=0:20,samples=90,color=black,very thick,dotted]{ 3 * (x - 2)}; + \addplot[domain=0:20,samples=90,color=black,very thick,dotted]{ 3 * (x - 3)}; + \addplot[domain=0:20,samples=90,color=black,very thick,dotted]{ 3 * (x - 4)}; + \addplot[domain=0:20,samples=90,color=black,very thick,dotted]{ 3 * (x - 5)}; + \addplot[domain=0:20,samples=90,color=black,very thick,dotted]{ 2 * (x - 5)}; + \addplot[domain=0:20,samples=90,color=black,very thick,dotted]{ 2 * (x - 6)}; + \addplot[domain=0:20,samples=90,color=black,very thick,dotted]{ 2 * (x - 7)}; + \addplot[domain=0:20,samples=90,color=black,very thick,dotted]{ 2 * (x - 8)}; + \addplot[domain=0:20,samples=90,color=black,very thick,dotted]{ 2 * (x - 9)}; + \end{axis} + \end{tikzpicture} + \end{center} + + The initial jump is dissolved into a transition zone that gets wider + over time: ``expansion fan''. In this case the Rankine-Huginot + condition cant be used, since the jump discontinuity doesnt persist for + greater $t$. + + \begin{ex} Left side maximum density and zero velocity, right side + maximum velocity and zero density. + + \begin{itemize} + \item Smooth out the initial step function with a linear interpolation + + $$f(x) = \begin{cases} \rho_M & x < -\varepsilon \\ + \rho_M \frac{\varepsilon - x}{2\varepsilon} & -\varepsilon < x < + \varepsilon \\ + 0 & \varepsilon < x + \end{cases}$$ + + In the transition zone you have $-\varepsilon < x < \varepsilon$ + and greenshields law to calculate wave velocity $c(f(x)) = V_M \frac x + \varepsilon$. + + solve stuff + \end{itemize} + + \end{ex} + + \week{} + \lecture{} + \chapter{Continuum Mechanics} + + Describing the movement and deformation dymaics of materials represented + as a continuous medium (rather than representations as connected + particles). For example elastic solids, flowing gases and liquids. + + + \incfig{ref} + + + \subsection{Coordinates} + + Consider two different coordinate systems: +\begin{itemize} \item [] \textbf{Material coordinates:} label a pooint within the + continous material and follow its movement along hte + trajectories $X(A,t)$ where $A$ is the intial location: $X(A,0) + = A$. (If you painted a dot on a piece of jelly). + \item [] \textbf{Spatial coordinates:} Fixed coordinate system $x$ + of an external observer. Does not follow the movement of the + material. + \end{itemize} + + \textbf{Material Coordinates:} + + \begin{itemize} + \item Trajectory $X(A,t)$ where $X(A,0) = A$ + + \item Displacement: $U(A,t) = X(A,t) - A$ + + \item Velocity: $V(A,t) = \frac {\partial U(A,t)} {\partial t}$ + \end{itemize} + + \textbf{Spatial Coordinates:} + + \begin{itemize} + \item Displacement: $u(x,t)$ of the material $x$ at time $t$. + To relate it to $U(A,t) $ find $A$ at $t = 0$ by solving $X(a,t) = x$ for + $A = a(x,t)$ (a in the spatial coordinates). + + \item $u (x,t) = U(a(x,t) , t)$ + \item $v(x,t) = \frac {\partial U(a(x,t),t)} {\partial t} = V(a(x,t),t)$ which is not equal to $\frac {\partial + u}{\partial t}$ (there is an additional dependence on time, we + evaluate by traversing the trajectory function $A$). + \end{itemize} + + Example: + + Consider a material described by the trajectories $X(A,t) = A \frac{1 + + 2t}{1 + t}$. Initial domain: $0 < A < l_0$. + + The displacement from $A$ of a material point at time $t$ is + + $$U(A,t) = X (A,t) - A = A \frac t {1 + t}$$ + + \marginnote{Can check this is different from $\frac{\partial u}{\partial + t}$. However it is the same as $\frac {\partial X}{\partial t}$ since $A$ + is a constant.} + + The velocity: + $$V(A,t) = \frac {\partial U(A,t)}{\partial t} = A \frac 1{(1 + t)^2}$$ + + The domain in material coordinates is $0 < A < l_0$, in spatial + coordinates $X(0,t) = 0 < x < X(l_0,t) = l_0 \frac {1 + 2t}{1 + t}$ + + If final coordinate is $x$ at $t$ find the initial coordinate $A$ + + $$x = X(A,t) \implies A = x \frac {1 + t}{1 + 2t}$$ + + The displacement of a material point at $x$ at $t$ relative to its + initial position: + + + $$u(x,t) = U(x\frac{1 + 2t}{1 + t}, t) = x \frac t {1 + 2t}$$ + + $$v(x,t) = V(x \frac {1 + 2t}{1 + t}, t) = x \frac 1 {(1 + t)(1 + 2t)}$$ + + \subsection{Material Derivative} + + Consider a scalar quantity that describes a propertiy transported by the + material. In spatial coordinates the distribution is described by + $f(x,t) $ and in material coordinates is $F(A,t)$. They must satisfy the + relationship $f(X(A,t),t) = F(A,t)$. + + \emph{What is the rate of change of this property in a material + coordinate system (along the trajectories of movement of the material + points).} + + \begin{align*} + \frac {\partial F}{\partial t} &= \frac {\partial f}{\partial t} + + \frac {\partial f} {\partial x} \frac {\partial X(A,t)} {\partial t} \\ + &= + \frac {\partial f}{\partial t} + V(A,t) \frac {\partial f}{\partial x} + \\ &= + \frac {\partial f}{\partial t} + v(x,t) \frac {\partial f} {\partial x} + \end{align*} + + Material Derivative: $$\frac D {Dt} \equiv \frac \partial{\partial t} + + v \frac \partial {\p x} \equiv \frac {\partial F}{\partial t}$$ + + \lecture{} + + \subsection{Continuity Equation} + + The law of conservation of mass + + $$\frac {\partial \rho}{\partial t} + \frac {\partial(v \rho)}{\partial + x} = 0$$ + + This is not sufficient to solve, we need another equation to describe + the velocity field $v(x,t)$. We use a continuum mechanics version of + newtons law: conservation of momentum. + + \subsection{Momentum Equation} + + The motion of a point particle under forces: $ma = \sum F_i$. + The rate of change of momentum: + + $$\frac d {dt} mv = F$$ + + Typoes of forces in continuum mechanics + + \begin{itemize} + \item external surface forces + \begin{itemize} + \item Act along the boundary of the domain + \end{itemize} + \item external body forces + \begin{itemize} + \item Act on every material point inside the domain, for + example gravity. + \end{itemize} + \item internal forces + \begin{itemize} + \item Stress ($\tau(x,t) = $ force / area) generated by different parts of + the material acting on eachother. For example restoring + force in elastic material. + \end{itemize} + \end{itemize} + + + Consider a segment of material between $A_L$ and $A_R$ with trajectories + $X(A_L,t) = \alpha (t) $ and $X(A_R,t) = \beta(t)$. + + The rate of change of the momentum of this segment is the sum of the + forces acting on it, external body force and internal surface force. + \marginnote{$S$ is the surface force. } + + $$\frac d {dt} \int_{\alpha(t)}^{\beta (t)} S \rho v \; dx = + \int_{\alpha(t)}^{\beta(t)} S\rho f \; dx + S\tau(\beta,t) - S + \tau(\alpha, t)$$ + + $$\frac d {dt}\int_{\alpha(t)}^{\beta(t)} \rho v \; dx = + \int_{\alpha(t)}^{\beta(t)} \rho f \; dx + \tau(\beta,t) - \tau(\alpha, + t)$$ + + $$\beta'(t)\rho(\beta9t))v(\beta(t)) - + \alpha'(t)\rho(\alpha(t))v(\alpha(t)) = \int_{\alpha(t)}^{\beta(t)} \rho + f \; dx +\int_{\alpha(t)}^{\beta(t)} \frac {\partial \tau}{\partial x}\; + dx$$ + + $\cdots$ + + $$\frac {\partial(\rho v)}{\partial t} + \frac {\partial (\rho + v^2)}{\partial x} = \rho f + \frac {\partial \tau}{\partial x}$$ + + $$\rho v_t + v \rho_t + \rho v_x + v \frac {\partial \rho v}{\partial + x} = \rho f +_ \frac {\partial \tau} {\partial x}$$ + + $\cdots$ + + Then you get the momentum equation as: + + $$\rho \frac {Dv}{Dt} = \rho f + \frac {\partial \tau}{\partial x}$$ + + \begin{center} + Material Derivative $=$ body force $+$ internal stress + \end{center} + + You can think of this as a generalisation of Newton's law with internal + forces in addition to the external forces. + + \subsubsection{In Material Coordinates} + + \begin{itemize} + \item Density $R(A,t) = R_0(A) \left(1 + \frac {\partial U}{\p + A}\right)^{-1} \quad (*)$ + \item Velocity $V(A,t)$ + \item Stress $T(A,t)$ + \item Body force $F(A,t)$ + \end{itemize} + + All considered along the trajectories $X(A,t)$. + + Using the transformation + + $$\frac {\partial \tau}{\partial x} = \frac{\partial T}{\partial + A}\left(\frac {\p U}{\p A} + 1\right)^{-1}$$ + + We need to apply this to the stress, and momentum equation. + + \begin{align*} + R(A,t) \frac {\p V}{\p t} &= RF + \frac {\p T}{\p A}\left(\frac {\p + U}{\p A} + 1\right)^{-1} \\ + R \left(\frac {\p + U}{\p A} + 1\right)\frac {\p V}{\p t} &= R\left(\frac {\p + U}{\p A} + 1\right)F + \frac {\p T}{\p A} + \\ + \marginnote{By the continuum equation in material coordinates $(*)$} + \\ + R_0 \frac {\p V}{\p t} &= R_0 F + \frac {\p T}{\p A} + \end{align*} + + \lecture{} + + \subsection{Constsitutive Laws} + + \subsubsection{Constitutive Law for Elastic Materials} + + Here, stress ($T$) is generated as a result of the displacement of the + material elemeents $U(A,t)$. However $T \ne T(U)$ since then a constant + displacement does not generate zero stress. Stress is generated by the + relative elongation of material elements: \emph{strain}. + + \subsubsection*{Strain in 1D} + + Consider a small section $A, A + \Delta$ at $t = 0$. At time $t$, + + \begin{align*} + A &\to A + U(A, t) \\ + A + \Delta &\to A + \Delta + U(A + \Delta,t) + \end{align*} + + After substituting the relative elongation (local strain) becomes + + $$\frac {l(t) - l_0}{l_0} = \frac {U(A + \Delta, t) - U(A,t)}{\Delta}$$ + + Then since $\Delta \ll 1$ we can use ethe taylor series approximation + for $U(A + \Delta, t)$, $U(A + \Delta, t) \approx U(A) + \frac {\p + U}{\p A} \Delta + \cdots$ + + $$\frac {l(t) - l_0}{l_0} = \frac {U(A + \Delta, t) - U(A,t)}{\Delta} = + \frac {\p U}{\p A}$$ + + Hence in elastic materials the stress is a function of strain + + $$T = f(\frac {\p U}{\p A})$$ + + $f(0) = 0$, but it is otherwise determined by the properties of the + material. + + Assuming a \textbf{linear elastic material}: $T = E\frac {\p U}{\p A}$, where $E$ is a + constant. + \marginnote{We can often choose a region around which a linear + approximation is a good solution} + + Then we get the momentum equation in this form; + + $$R_0 \frac {\p^2 U}{\p t^2} = R_0 F + E \frac {\p^2 U}{\p A^2}$$ + + \emph{Note:} we still don't know about the external forces $F$. + + \begin{ex} + Consider an elastic bungie cord with length $l_0$: how long is it + when stretched by its own weight hanging free in vertical position. + + Initial position $0 < A < l_0$ at $t = 0$. + + Boundary conditions: $U(0,t) = 0$ at the fixed end. + + $T(l_0, t) = 0$ at the free end, $\implies$ $E \frac {\p U}{\p A}(A = l_0, t) = 0$ + + $U(0) = 0$, $U'(l_0) = 0$. + + $$U(A) = - \frac {R_0 g}{2E} + C_1 A$$ + + $$U'(A) = - \frac {R_0 g}E A + C_1$$ + + $$C_1 = \frac {R_0 g}E l_0$$ + + $$U(A) = \frac {R_0 g}{2 E} A(2 l_0 - A)$$ + + final length is $l = l_0 + U(l_0) = l_0 + \frac {R_0 g} {2E} {l_0}^2$ + + We can also calculate the stress distribution along the cord: + + $$T(A) = E U'(A) = R_0 g(l_0 - A)$$ + + The density distribution (non-constant because it is now stretyched.) + + $$R(A) = \frac {R_0} {1 + \frac {R_0 g}E (l_0 - A)}$$ + + Note we are working in material coordinates, if we want them as a + function of spatial coordinates you need to invert $x = A + U(A)$. + + \end{ex} + + + \week{} + \lecture{} + +\section{Review of Continuum Mechanics in 1D} + + \incfig{ref} + + + \subsubsection*{Material coordinate} + + \begin{itemize} + \item $t$ time + \item $A$ cross section + \item $X(A,t)$ a spatial position + \item + $U(A,t) = X(A,t) - X(A,0)$ a displacement at a time from the + \item + $V = \p_t U(A,t) $ velocity + + \item $R(A,t) = \rho(X(A,t), t)$ density + \item $T(A,t) = \tau(X(A,t), t)$ stress + \item $F(A,t) = f(X(A,t), t)$ body force + \end{itemize} + + \marginnote{bad handwriting idk if these are right} + \subsubsection*{Spatial coordinate} + \begin{itemize} + \item $t$ time + \item $a(x,t) = X^{-1}$ + \item $x$ spatial position + \item $u(X,t) = U(a(x,t) ,t)$ displacement + \item $v = D_t u = \frac {\partial_t u}{1 - u_X}$ + + \item $\rho(x,t)$ density ($\frac {mass}{volume}$) + \item $\tau(x,t)$ stress ($\frac {force}{area}$) + \item $f(x,t)$ body force ($\frac {force}{mass}$) + \end{itemize} + + \subsubsection*{Governing equations} + + + $$\p_t f + \p _x(v \rho) = v \quad{\text{continuity equation}}$$ + + $$\rho(\p_t v + v v_x) = \rho f + \p _X \tau = D_t v + \quad{\text{momentum eqn}}$$ + + Have the unknowns $R, V,T$ and only two equations so need a + constitutive equation for $\tau$ or $T$. This can be found using a creep + experiment: stretch a material an amount and see how much force is + requried, or apply an amount of force and see how much it streteches. + + $$\frac{\ell} {\ell_0} = \lambda \quad\quad\text{Extension ratio}$$ + + \begin{itemize} + \item Typically extension grows with stress. + \end{itemize} + + At steady state $\p_{t t} U = 0 \implies \p _A T = 0$ stress $T$ is + constant along the material (from the momentum equation) so we cannot have $T = T(U)$ stress is a + function of displacement. + + + $$ +\begin{array}{|l|l|l|} +\hline \text { Name } & \text { Ratio } & \text { Definition } \\ +\hline \text { Lagrangian Strain } & \left(\ell-\ell_{0}\right) / \ell_{0} & \epsilon=U_{A} \\ +\hline \text { Eulerian Strain } & \left(\ell-\ell_{0}\right) / \ell & \epsilon_{e}=u_{x} \\ +\hline \text { Green Strain } & \left(\ell^{2}-\ell_{0}^{2}\right) /\left(2 \ell_{0}^{2}\right) & \epsilon_{g}=U_{A}+\frac{1}{2} U_{A}^{2} \\ +\hline \text { Almansi Strain } & \left(\ell^{2}-\ell_{0}^{2}\right) /\left(2 \ell^{2}\right) & \epsilon_{a}=u_{x}-\frac{1}{2} u_{x}^{2} \\ +\hline \text { Midpoint Strain } & 2\left(\ell-\ell_{0}\right) /\left(\ell+\ell_{0}\right) & \epsilon_{m}=U_{A} /\left(1+\frac{1}{2} U_{A}\right) \\ +\hline \text { Hencky Strain } & \ln \left(\ell / \ell_{0}\right) & \epsilon_{h}=\ln \left(1+U_{A}\right) \\ +\hline +\end{array} +$$ + + +\lecture{} +Eg For lagrangian strain: $\epsilon \sim \frac {\ell - \ell_0}{\ell_0}$ + +\begin{align*} + \frac{l - l_0}{l_0} &= \frac {X(A + \Delta A, t) - X(A - \Delta t, t) - + 2\Delta A - A \times A}{2 \Delta A} \\ + &= \frac {X(A + \Delta A, t) - (A + \Delta A) - (X(A - + \Delta t, t) - (A - \Delta A))}{2\Delta A} \\ + & \quad\quad \text{Taylor expansion} \\ + &= \frac {U(A,t) + \Delta A U_A(A,t) + O(\Delta A^2) - (U(A,t) + - \Delta A(U_A(A,t) + O(\Delta A)^2)) } {2\Delta A}\\ + &= U_A(A,t) + O(\Delta A) \\ + &\implies \lim_{\Delta A \to 0} \frac {\ell - + \ell_0}{\ell_0} = U_A(A,t) =: \epsilon \\ +\end{align*} + +$\implies$ that for elastic materials + + $$T = T(\epsilon) = T(U_A) $$ + +Now the momentum equation is given by +\marginnote{(non-linear) wave equation} + +$$R_0 U_{tt} = R_0 F + T'(U_A) U_{AA}$$ + +$$\lambda = \frac {\ell }{\ell_0} = \frac {\ell - \ell_0 + \ell_0} {\ell_0} \sim +\epsilon + 1 $$ + +$\eimplies$ the reference state has $\lambda = 1 \Leftrightarrow \epsilon = 0$ + +For a small enough $\epsilon$ any material will exhibit a linear stress--strain +relation. + +$$T(\epsilon) = E\epsilon - E\p_A U$$ + +Wher $E$ is young's modulus also known as the elastic modulus. + +Physical dimensions: + +\begin{itemize} + \item[] $[T] = \frac {force}{area}$ + \item[] $[\epsilon] = [\frac {\ell - \ell_0}{\ell_0}] = 1$ + \item[] $[E] = \frac{force}{area}$ +\end{itemize} + +\begin{center} +\begin{tabular}{l|l|l} +\hline Material & Young's modulus $(\mathrm{GPa})$ & Density $\left(\mathrm{kg} / \mathrm{m}^{3}\right)$ \\ +\hline Diamond & 1000 & 3500 \\ +\hline Stainless steel & 200 & 8030 \\ +\hline Glass & 65 & 2600 \\ +\hline White oak & 12 & 770 \\ +\hline Beeswax & $0.2$ & 960 \\ +\hline Rubber & $0.007$ & 1200 \\ +\hline Silica aerogel & $0.001$ & 100 \\ +\hline +\end{tabular} +\end{center} + +The momentum equation: $R_0 U_{tt} = R_0 F + EU_{AA}$ multiplied by $\frac +1{R_0}$ gives $U{tt} = F + \frac E {R_0} U_{AA}$, which is the linear constant +coefficient wave equation. So we can understand that the characteristic frequencies +are proportional to wave speed $c = \sqrt{\frac E{R_0}}$. For example a high +elastic modulus in the body of a brass instrument has a high wave speed so +expresses those frequencies more. + +Another way of imagining this is if the linear elastic material is imagined as a +chain of beads connected by Hookean springs (at reference length $l_0$), then $F = K(\ell - \ell_0)$ where +$K$ is the spring constant. + +\incfig{linearelasticspring} + + +\begin{align*} + T &=E \frac {\ell - \ell_0} {\ell_0} \\ + &\text{multiply by $\sigma$, cross sectional area} \\ + \sigma T &= \frac { \sigma E}{ \ell_0 } (\ell - \ell_0) \\ + &= K +\end{align*} + +Spring constant is $\frac {\sigma E}{\ell_0}$ + +So we can use \textbf{the creep test} to find young's modulus. + +We need to solve the system of equations + +$$\begin{cases} 0 = EU_{AA} \\ 0 = U(0) \\ U(\ell_0) = \ell - \ell_0 \end{cases}$$ + +\begin{align*} + U &= \alpha A = B \\ + U(0) &= 0 = \beta \\ + U(\ell_0) &= \ell - \ell_0 = \alpha\ell_0 \\ + \implies U(A) = \frac {\ell - \ell_0} A +\end{align*} + +Material density + +\begin{align*} + R &= \frac {R_0}{1 + U_A} \\ + \implies R &= \frac{R_0} {1 + \frac {\ell - \ell_0} {\ell_0}} = \frac {R_0} {\frac + \ell {\ell_0}} = \frac {\ell_0}{\ell} R_0 +\end{align*} + +\lecture{} + +\subsubsection{Bungee rope at steady state} + +\begin{itemize} + \item [] +$\ell$ is the length at rest while hanging: extended undner its own weight. +\item[] +$\ell_0$ is the length of the reference configuration: bungee rope under no +force, at rest +\end{itemize} + +$R_0 U_{t t} = R_0 F + EU_{ AA }$ + +$R_0 U_{t t} = 0$ + +$R_0 F$ is body force + +The body force is obviously gravity so $F = g = 9.81 m/{s^2}$ + +So the momentum equation is $U = R_0 g + EU_{A A}$ + +We dont nkow the value of $U$ at $A = \ell_0$ but we know the boundary +condition: $T(\ell_0) = 0 \implies E U_A(\ell_0) = 0$, and $U(0) = 0$ + +Solve + +$$\begin{cases}0 = R_0 g + EU_{A A}\quad\quad(*)\\ U(0) = 0 \\ U_A(A = \ell_0) = 0 \end{cases}$$ + +THe easiest way is to integrate $(*)$ on $(A, \ell_0)$ to get $0 = R_0 g(\ell_0 +- A) + E(U_A(\ell_0) - U_A(A))$ + +$\implies U_A = \frac {R_0}E g(\ell_0 - A) \quad \quad (* * )$ + +Ie, the tension on the bungee cord decreases linearly along the bungee cord +since $U \propto T$. (This makes sence since as you go down the cord there is +less weight beneath each cross section.) + + \begin{center} + \begin{tikzpicture}{testp} + \begin{axis}[ytick={0}, xtick={0.1},xlabel=$A$, + xticklabels={$\ell_0$},yticklabels={$\frac {R_0}E + g\ell_0$},] + \addplot[domain=0:0.1, samples=90,color=blue,]{ -x / 4}; + \end{axis} + \end{tikzpicture} + \end{center} + + Integrate $(* * ) $ on $(0, A)$ to get + + $U = \frac {R_0} E g (\ell_0 A - \frac {A^2}{2})$ + + $U(\ell_0) = \frac{R_0 g}{E} \frac {{\ell_0}^2} 2$ + + \begin{center} + \begin{tikzpicture}{hy} + \begin{axis}[ytick={1}, yticklabels={$\frac{R_0 g}{E} \frac + {{\ell_0}^2} 2$}, xtick={0,2},xlabel=$A$, + xticklabels={0, $2\ell_0$}, + ] + \addplot[domain=0:2, samples=90,color=blue,]{ -(x -1)^2 + 1}; + \end{axis} + \end{tikzpicture} + \end{center} + + So clearly $\ell$ grows like ${\ell_0}^2$ because of the cord's own weight. + + + +\section{Hyperelastic Materials} + +\begin{itemize} + \item Nonlinear stress-strain relations. + \item Energy balance +\end{itemize} + + +For example for rubber the stress strain raltion looks something like + + + +\begin{figure}[htpb] + \centering + \begin{tikzpicture}{testppop} + \begin{axis}[ytick=none,xtick=none,xlabel=$\epsilon$, ylabel=$T$,] + \addplot[domain=-1:2, samples=90,color=blue,]{ (x -0)^3 + 0}; + \end{axis} +\end{tikzpicture} + \caption{Stress-Strain for rubber} +\end{figure} + +\begin{align*} + T(\epsilon) &= \psi'(\epsilon) \text{ where } \\ + \psi(\epsilon) &:= \text{ Strain energy density} \\ + \text{For example} &T(\epsilon) = K\epsilon^3 \\ + T(\epsilon) &= K\epsilon^3 \implies \psi(\epsilon) = K \frac {\epsilon^4}4 \\ + T(\epsilon) &= E\arctan(\epsilon) \\ + T(\epsilon) &= E \epsilon + K \epsilon^3 \\ + &\implies \psi(\epsilon) = E \frac {\epsilon^2} 2 + K \frac + {\epsilon^4} 4 +\end{align*} + +Implies the momenntum equation + +\begin{align*} + R_0 U{z z} &= R_0 F + \p_A \psi'(U_A) \\ + &= R_0 F + \psi'' (U_A) U_{A A} +\end{align*} + +Typically need that $\psi'' > 0$ +so that the problem is well posed: $T(\epsilon) = \psi'(\epsilon) $ is monotone +in $\epsilon$. + +\subsubsection*{Energy balance} + +For a $A_\ell $ and $A_r$. (note $\sigma$ is the cross sectional area). + +Multiply momentum equatino by $\sigma U_t$. + +\begin{align*} + \sigma R_0 U_t U_{tt} &= \sigma R_0 F V + \sigma \p_A \psi'(U_A) U_z \\ +\text{integrate over} A_\ell \text{ to } A_r \\ + \sigma R_0 \int_{A_\ell}{A_r} U_t U_{T t} \: dA &= \sigma R_0 + \int_{A_\ell}^{A_R} FV \: dA + \sigma \int_{A_\ell} ^{A_r} \p_A \psi'(U_A) + U_z \: dA \\ + &\text{Now integrate by parts} + \\ + \sigma R_0 \int_{A_\ell}^{A_r} \frac d {dt} \frac {U_t^2} 2 \: da &= \sigma + R_0 \int_{A_\ell} ^{A_r} FV \:dA + r[\psi'(U_A) U_t]_{A_\ell}^{A_R} - \sigma + \int_{A_l}^{A_r} \psi'(U_A) U_{At} \: da \\ + ... \\ + \implies +\end{align*} +$$\frac d{dt} \int_{A_\ell}^{A_r} \left[\sigma R_0 \frac {V^2}{2} + \sigma +\psi(U_A)\right] \: dA = \sigma \left(R_0 \int_{A_\ell}^{A_r} FV \: da + +TV|_{A_r} - TV |_{A_\ell}\right) $$ + +Where we can see + +\begin{itemize} + \item +$\sigma R_0 \frac {V^2}{2}$ : kinetic energy + \item + $ \sigma \psi(U_A)$ : strain energy + \item +$R_0 \int_{A_\ell}^{A_r} FV \: da $ : Power delivered by body forces +\item +$TV|_{A_r} - TV |_{A_\ell} $ : Power delivered by force acting on the tips + +\end{itemize} + +\section{Time-dependent elastic models} + +For a linearly elastic material: + +$$\begin{cases} R(A,t) = \frac {R_0} {1 + U_A} \quad(i)\\ + R_0 U_{t t} = R_0 F + EU_{AA}\quad(ii)\end{cases}$$ + +These are not coupled. You should solve $(ii)$ then $(i)$ + +From math2100: The wave equation on $\R$ + +$$(*) \begin{cases} +U_{t t} - c^2 U_{A A} = 0\\ +U(t = 0, A) = f(A)\\ +\p_t U(t = 0, A) = \rho(A) +\end{cases}$$ + +You can get the solution using the d'Alembert solution to the wave equation. +This superimposes two waves, one that moves right and one that moves left at +the same speed. + +$$U(t,a) = \frac 12 (f(A - ct) + f(A + ct)) + \frac 1 {2c} \int_{A - ct}^{A + +ct} g(\hat A) \: d\hat A$$ + +Remark: This formula also gives a solution to $(*)$ on an interval with derichlet zero +boundary conditions ($U(t, A = 0) = 0 = U(t, A = \ell)$: when there is no + +displacement at the endpoints of $(0,\ell)$). + +To get this initial condition you can extend $f,g$ as odd $2\ell$ periodic +functions. + +\begin{figure}[htpb] + \centering +\incfig{exten} + \caption{Extended function that is periodic on $2\ell$} +\end{figure} + +\begin{ex} + + With a slinky: + + $A = 0$ and $A = 10 = \ell_0$ + + Intially move the loop of $A = 4$ to the left by $\Delta A = 2$, then + release it. + + \incfigw{exspr}{0.5} + + How do you write $U(t = 0, A) = U_0(A)$? + +\end{ex} + +\week{} +\lecture{} + +We know +\begin{align*} + U_0(A = 0) &= 0 \\ + U_0(A = 10) &= 0 \\ + U_0(A = 4) &= -2 \\ +\end{align*} + +At $t = 0$ the slinky satisfies $EU_{A A} = 0 \implies U_A$ is a constant. + +\begin{align*} + U_A &= \frac {-2 - 0}{4 - 0} = -\frac 1 2 \implies U = -\frac 12 A \\ + U_A &= \frac {0 - (-2)}{10 - 4} = \frac 13 \implies U = \frac 13 (A - 10) \\ + U_0 &=\begin{cases} -\frac 12 A & 0 \le A \le 4 \\ \frac 13 (A - 10) & 4 < A + \le 10\end{cases} \\ +\end{align*} +The time-dependent problem is given by + +\begin{align*} + U_{t t} &= \frac E{R_0} U_{A A} \\ + U(0) &= 0 = U(10) \\ + U(t = 0, A) &=\begin{cases} -\frac 12 A & 0 \le A \le 4 \\ \frac 13 (A - 10) + & 4 < A \le 10 \end{cases} \\ + \p_t U(t = 0, A) &= 0 +\end{align*} + +$\implies$ the solution is given by D'Alembert's formula where $\rho=0$ and $f$ + is the $2\ell$ periodic odd extension of $U_0$. + + + \begin{ex} + Finite-length bar under external forcing (resonance) + + $U_{t t} = \frac E {R_0} U_{A A} + F(A,t) $ on the interval $(0,\ell)$. + + $U_A(t = 0, A = 0) = 0 = U_A(t = 0, A = \ell) $ + + $F(A,t) = a(t) \cos(\kappa_n A) $ where $a(t) = \sin(\omega t)$ + + $\omega$ is angular frequency of the forcing. + + $\kappa_n = \frac {n \pi}\ell$ where $n$ is the wave number of the specific + mode of the forcing term. + + \begin{center} + \begin{tikzpicture}{hy} + \begin{axis}[xtick={3.141},xticklabels={$\ell$}, ytick=none,] + \addplot[domain=0:3.141 , samples=90, color=green]{ cos(deg(x))}; + \addlegendentry{$n = 1$} + \addplot[domain=0:3.141 , samples=90, color=black]{ cos(2 * deg(x))}; + \addlegendentry{$n = 2$} + \addplot[domain=0:3.141 , samples=90, color=blue]{ cos(3 * deg(x))}; + \addlegendentry{$n = 3$} + \end{axis} + \end{tikzpicture} + \end{center} + + + We can solve with Laplace transform to turn the pde into an ODE, transform + the entire system of equations to get rid of the time derivatives + + +$$ +F(s)=L(f(t))=\int_{0}^{\infty} e^{-s t} f(t) d t +$$ + +and the opposite + +$$ +L(t f(t))=-\frac{d F(s)}{d s} +$$ + +\begin{align*} + (*) \begin{cases} + s^2 \hat U - c^2 \hat U_{A A} &= \hat F (A,s) = \hat a(s) \cos(\kappa_n A) \\ + &\text{Boundary conditions:} \\ + \hat U_A(s, A=0) &= 0 = \hat U_A(s, A = \ell) +\end{cases} +\end{align*} + +To solve $(*)$ first consider the homogeneous equation: $\hat U_{hom} - \frac +{c^2}{s^2} \hat U_{A A} = 0$ + +$$\hat U_{hom} = \alpha \exp(\frac cs A) + \beta \exp(-\frac cs A)$$ + +Find a particular solution : + +$$\hat U _{part} = \gamma \sin(\kappa_n A) + \delta \cos(\kappa_n A)$$ + + +\begin{align*} + \implies \p_{A A} U_{part} &= -\kappa_n^2 (\gamma \sin(\kappa_n A) + \delta +\cos(\kappa_n A)) \\ + s^2 \hat U_{part} - c^2 \p_{A A} \hat U_{part} &= (s^2 + c^2 + \kappa_n^2)(\sin(\kappa_n a) + \delta \cos(\kappa_n A)) \\ + &= \hat a(s) \cos(\kappa_n A) \\ + \implies \gamma = 0 \text{ and} \delta = \frac {\hat a(s)}{ s^2 + c^2 + \kappa_n^2} +\end{align*} + +$\implies $ the general solution of $(*)$ is + +$$ +\hat U = \frac {\hat a(s)}{ s^2 + c^2 + \kappa_n^2} \cos(\kappa_n A) + \alpha \exp(\frac cs A) + \beta \exp(-\frac cs A) +$$ + +Boundary conditions: + +\begin{align*} + \hat U_A &= -\kappa_n \frac {\hat a}{s^2 + \kappa_n^2 c^2} \sin(\kappa_n A ) + + \frac cs (\alpha \exp(\frac cs A) - \beta \exp(-\frac cs A)) \\ + U = \hat U_A(A = 0) &= \frac cs (\alpha - \beta) \implies \alpha = \beta \\ + U = \hat U_A(A = \ell) &= \frac cs (\alpha \exp(\frac cs \ell) - \beta + \exp(-\frac cs \ell))\alpha \\ + & \text{since }\ell \ne 0, \; \alpha \exp(\frac cs \ell) - \beta + \exp(-\frac cs \ell) + \implies \alpha = 0 \\ + &\implies \hat U = \frac {\hat a} {s^2 \kappa_n^2 + c^2} \cos (\kappa_n A) +\end{align*} + + \lecture{} + +Now we compute the inverse laplace transform using the convolution theorem. + +\begin{align*} + \hat U &= \frac {\hat a} {s^2 \kappa_n^2 c^2} \cos (\kappa_n A) \\ + &\text{We have to use the convolution theorem} \\ + L^{-1}(\hat U) &= \cos(\kappa_n A) L^{-1}(\frac 1 {s^2 + \kappa_n^2 + c^2}) \times u(t) + \quad\quad \text{(note $\cos(\kappa_n A)$ does not depend on $s$)} \\ + L^{-1}(\frac 1 {s^2 + \kappa_n^2 + c^2}) \times u(t) &= \frac 1 {k_n c} \int _0^t \sin(\omega(t - \hat t)) + \sin(\kappa_n c \hat t) \: :d\hat t \\ + &= \frac 1 {\kappa_n c} \begin{cases} \frac 1 {\omega^2 + \kappa_n^2 c^2}(\omega \sin(\kappa_n ct) - \kappa_n c + \sin(\omega t)) & \omega \ne \kappa_n c\\ + -\frac 12 + \mathcolorbox{yellow}{t} + \cos(\kappa_n ct) + \frac 1 {2\kappa_n c} + \sin(\kappa_n ct) & \omega = \kappa_n c + \end{cases} \\ +\end{align*} + +$\implies \omega = \kappa_n c$ is the resonance cases; the amplitude of $U$ +grows linearly in time (due to the highlighted \mathcolorbox{yellow}{$t$}). The +reason is $\kappa_n c$ are the eigenmodes of the structure. + + +\marginnote{Remember $R_0$ is density, $E$ is elastic modulus.} + +Since the eigenfrequenceies $\kappa_n c$ are proportional to $c = \sqrt{\frac E +{R_0}}$ one can probe the material propertiy $\sqrt{\frac E {R_0}}$ through an +acoustic signal (frequancies of resonances in the material will be $\kappa_n +\sqrt{\frac E {R_0}}$). + + \end{ex} + + \subsection{Viscoelastic Materials} + + \begin{itemize} + \item Introducing damping to the stress-strain relation. + \end{itemize} + + \begin{ex} + + For example, considering the slinky again. It is unrealistic for it to + oscillate forever, instead some sort of damping, eg air-external friction, + internal friction (resistance to deformation) will slow it down. + + If you have a spring hanging freely with a weight and it is in equilibrium, you + have the spring force $F_s = -K_s u = mg$, and a damping force $F_d-c\dot u = 0$. + When it moves the damping force creates a resistance. Resitive force is proportional + to the velocity: $F_d = -c \dot u$. + + Newtond's Law: $m \ddot u = F_d + F_s$ gives the damped oscillation equation + + $$m \ddot u + c\dot u + K u = 0$$ + + These systems provide a framework to model viscoelasticity. + \end{ex} + + \subsubsection{Kelvin-Voigt element} + + You have a (horizontally) a spring and dashpot in parallel. + + $$F = - (F_s + F_d)$$ + + So displacements are equal: $u_s = u_d = u$. + + Spring + constant $K$. + + $$F_s = -K u$$ + $$F_d = -c \dot u$$ + + To create a constitutive law for stress, identify $F$ by stress $T$ and $u$ b + strain $\epsilon$. Also replace $\dot F$ by $\partial_t T$ (stress rate), and + $\dot u$ by $\p_t \epsilon$ (strain rate). + + $$T = E(\epsilon + \tau_1 \p_t \epsilon ) $$ + + $E$ elastic modulus + + $\tau_1$ dissipation time of the strain. + + \subsubsection{Maxwell element} + + Spring and dashpot in series. + +Displacements sum to $u$: $u = u_s + u_d$ + +Forces are equal: $F = - F_s = -F_d$ + +Constitutive law: + +$F = -F_s= ku_s \implies u_s = \frac FK \implies \dot u_s = +\frac {\dot F }K $ + +$F = -F_d = c \dot u_d \implies \dot u_d = \frac FC$ + +So +$\dot u = \dot u _s + \dot u_d = \frac {\dot F}k + \frac F c$ + +$(* * ) \quad\quad c \dot u = \frac ck \dot F + F$ + +Which gives the constituive law + +$$T + \tau_0 \p_t T = E\tau_1 \p_t \epsilon$$ + + +\lecture{} + +\subsubsection{Standard Linear Model} + +\incfigw{dsprlin}{0.7} + +To model the spring and despot in series use the resulf for the Maxwell element +$(* * ) $. + +$$c_1 u' = \frac {c_1}{k_1} F_1' + F_1 = \frac {c_1}{k_1} (F' - k_0 u') + F - +k_0 u$$ + +$$\implies k_0 u + u(c_1 + \frac {k_0 }{k_1} c_1) = F + \frac {c_1}{k_1} F'$$ + +Rewrite this as + +$$F + a_1 F' = a_2 u + a_3 u'$$ + +Where $a_1 = \frac {c_1} {k_1}$, $a_2 = k_0$ and $a_3 = c_1 + \frac +{k_0}{k_1}c_1$ so that $a_3 = c_1 + \frac {k_0}{k_1} c_1 > \frac {k_0}{k_1} c_1 += a_1 a_2$ + +This gives the constitutive law + +$$T + \tau_0 T_t = E(\epsilon + \tau_1 \p_t \epsilon)$$ + +\begin{align*} + \tau_0 &= a_1 = \frac {c_1}{k_1} \\ + E &= a_2 = k_0 \\ + E T_1 &= a_3 +\end{align*} + +And as we showed + +$ET_1 = a_3 > a_1 a_2 = ET_0$ + +$E = a_2 = k_0$ + +Which $\implies \tau_1 > \tau_0$, where $\tau_0$ and $\tau_1$ are dissipation +times, $E$ the elastic modulus. + + +$$\begin{cases} + R = \frac {R_0}{1 + U_A} \\ + R_0 U_{t t} = R_0 F + T_A \\ + T + \tau_0 T_t = E(U_A + \tau_1 U_{At}) +\end{cases}$$ + +\subsubsection{Summary} + +\emph{Standard linear model} +$$T + \tau_0 \p_t T = E(U_A + \tau_1 \p_t \epsilon)$$ + +\emph{Kelvin-Voigt model} +$$T = E(\epsilon + \tau_1 \p_t \epsilon ) $$ + +Note the above two models involve the strain (in the right hand side), so they will eventually go back to +the reference configuration. These are viscoelastic solids. + +\emph{Maxwell Model} +$$T + \tau_0 \p_t T = E\tau_1 \p_t \epsilon$$ + +This is a visco-elastic fluid since it only includes the strain rate, it will +not return to the reference configuration. It is elastic on fast time scales, +but on a long timescale you have a creeping motion that deforms the fluid. + +\subsubsection{Relaxation Function} + +This unifies the above three models, by adding constant $C$ to the standard linear model. + +$$T + \tau_0 \p_t T = E(C\epsilon \tau_1 \p_t \epsilon)$$ + +\begin{itemize} + \item[] $\tau_0 =0, C = 1$ Kelvin-Voigt model + \item[] $C = 0$ Maxwell model + \item[] $C = 1$ Standard linear model +\end{itemize} + +Solving +$$\begin{cases} T + \tau_0 \p_t T = h \\ +T(t=0) = 0 +\end{cases}$$ + +where $h = E(C\epsilon + \tau_1 \p_t \epsilon)$, (and assuming $\epsilon(0) = 0$) + + +$$T = \int_0^t E(C + (\frac {\tau_1}{\tau_0} - c)e^{\frac {-(t - +\hat t)}{\tau_0}})\p_t \epsilon{\hat t}\: d\hat t$$ + +Where the relaxation function + +$$G(t - \hat t) = E(C + (\frac {\tau_1}{\tau_0} - c)e^{\frac {-(t - +\hat t)}{\tau_0}})$$ + +\begin{itemize} + \item Standard linear model + $$G(t) = E(1 + (\frac {\tau_1}{\tau_0} - 1) e^{\frac {-t}{\tau_0}})$$ + \item Maxwell element + + $$G(t) = E \frac {\tau_1}{\tau_0} e^{\frac {-t}{\tau_0}}$$ + + \item Kelvin-Voigt + + Note cannot simpoly sat $\tau_0$ to zero but it is attainable using a + different method. + + $$T = E(\epsilon + \tau_1 \p_t \epsilon) = \int_0^1 E(\p_t \epsilon(\hat + t) + \tau_1 \rho_0 (t - \hat t) \p _t \epsilon(\hat t) ) \: d \hat t + \implies$$ + + $$G(t) = E(1 + \tau_1 \delta_0(t))$$ +\end{itemize} + +A general formulation for constitutive relation for stress ($T$) is + +$$T = \int_0^t G(t - \hat t) \epsilon_t (\hat t) \: d\hat t = G * \p_t \epsilon $$ + +Here $G \ge 0$ can be choses quite arbitrarily (want it to be integrable, decay +to zero). + +This is the ``visco-elastic law of relaxation type''. + +\week{} +\lecture{} + +Instead you may also write the stress as an integral against the strain and +integrate by parts in $T_2$ using $\epsilon (0) = 0$: + +$$T_2 = \int_0^t E \frac {T_1}{T_0} E^{-\frac {t - \hat t}{\tau_0}} \p_t +\epsilon(\hat t) \: dt = + E\frac {T_1}{T_0} \epsilon(t) - \int_0^t E\frac {T_1}{T_2} e^{-t - \hat t} +{\tau_0} \epsilon(\hat t) \: d \hat t$$ + +For example for the standard linear model ($C = 1$) this implies + +$$T = T_1 + T_2 = \underbrace{E \frac {T_1}{T_0} +\epsilon(t)}_{\begin{minipage}[adjusting]{5em}instantaneous elastic stress \end{minipage}} +- \underbrace{\frac E {\tau_0} \int_0^t \left( \frac {\tau_1}{\tau_0} - 1\right ) e^{\frac +{t - \hat t}{\tau_0}} \epsilon(\hat t) \: d \hat +t}_{\begin{minipage}{5em}damping\end{minipage}} +$$ + +{The computations clearly get complicated so often you can't solve +viscoelastic models explicitly very easiliy} + +\begin{ex} + Maxwell-viscoelastic bar occupying $0 \le A < \infty$ with body force $F = + 0$. + + Momentum equation: + + $$(*) \qquad\quad R_{0} U_{t t} - T_A = \p_A (G \star \epsilon_t) = \int_0^t G(t - \hat t) + U_{AA_t}$$ + + Where $G(t) = E \frac {\tau_1}{\tau_0} e^{\frac {-t} {\tau_0}}$. + + We have the ``far field condition'' on the right, the displacement vanishes at + infinity: + + $$\lim_{A \to \infty} U(A,t) = 0$$ + + $$U(t=0, A) = 0 = \p_t U(t = 0, A)$$ + + How to enforce $U_A(t , A=0) = F(t) $ experimentally? + + Apply a force to the left endpoint: $T(A = 0, t) = \int_0^t G(t - \hat t) F_t(\hat t) \: d\hat t $ + + Use laplace transform on $(*)$, $\hat U = L(U)$ + + $$\begin{cases} + R_0 s^2 \hat U = L(G) s \hat U_{A A} $ where $L(G) = \fraC {\hat E}{s + + \frac 1 {\tau_0}} \\ + \hat U_A (A = 0, s) = \hat F(s) \\ + \lim_{A \to \infty} \hat U(A, s) = 0 + \end{cases}$$ + + + \begin{align*} + R_0 s \hat U = \frac {\hat E}{s + \frac 1 {\tau_0} } \hat U_{A A} + &\implies \frac {R_0 s}{\hat E}(s + \frac 1 {\tau_0}) \hat U = \hat U + _{A A} \\ + &\implies \hat U = \alpha (s) e ^{\omega (s) A} + \beta(s) e^{-\omega(s) A} + \qquad {\text{ $($ODE for }\hat U)}\\ + \text{where } \omega(s) &=\sqrt{\frac {R_0} s {\hat E} (s + \frac 1 + {\tau_0})} \\ + \end{align*} + + \marginnote{laplace transform only works when the real part of $s$ is large + enough compared to the imaginary part.} + + Note that $Re(\omega) > 0$ if $Re(s) $ is large enough. + Therefore $\alpha(s) = $ to guarantee that $\lim_{A \to \infty} \hat U(A) = + 0$. + + Find the other constant: + + \begin{align*} + \implies & \hat U(A,s) = \beta(s) e^{-\omega(s) A} \\ + \implies & \hat U_A(A,s) = -\omega(s) \beta(s) e^{-\omega(s) A} \\ + {\text{2nd boundary cond. }} \hat F(s) &= \hat U_A(A = 0, s) \\ + &=-\omega9s) \beta(s) \\ + \implies \beta(s) &= \frac {-\hat F(s)}{\omega(s)} \\ + \implies \hat U(A, s) &= -\frac 1 {-\omega (s)} \hat F(s) e^{-\omega(s) A} \\ + \end{align*} + + Now using the convolution theorem to inverse laplace transform: + + \begin{align*} + U(A,t) &= F(t) \star \underbrace{L^{-1}\left(-\frac 1 {-\omega (s)} e^{-\omega(s) A} + \right)}_{(* *)} \\ + (* *) &= \sqrt{\frac {\hat E}{R_0}} + e^{\frac {-t}{\tau_0}}I_0 \left(\frac {-1}{2\tau_0} \sqrt{t^2 - \frac {R_0}{\hat E} + A^2}\right) \times H\left(t - \sqrt{\frac {R_0}{\hat E}} A\right)\\ + I_0 & {\text{ is the Bessel function of the 1st kind }} \\ + H & \text{ is the heaviside function } \\ + U &= \int_0^t Q(A,\hat t) H(\hat t - \sqrt{\frac {R_0}{\hat E}} A) F (t - \hat t) + \hat H \\ + &=\int_{\sqrt{\frac {R_0}{\hat E}}A}^t Q(A, \hat t) F(t - \hat t) \: d\hat t \times H\left(t - \sqrt + {\frac{R_0}{\hat E}}A\right) \\ + \implies & U \text{ is zero wherever } t < \sqrt{\frac {R_0}A} , \\ + \implies & A > t\sqrt {\frac {\haat E} {R_0}} = t \sqrt{\frac {E \tau_1}{\tau_0} + \frac 1 {R_0}} \\ + \text{For } F(t) &= \sin(t) \\ + \end{align*} + + The speed of propagation in the linearly elastic material $c = \sqrt{\frac E {R_0}}$ is smaller + than the propagation in the viscoelastic model $\sqrt{\frac E{R_0} \frac {\tau_1}{\tau_0} }$ + $(\tau_1 > \tau_0)$. And for $\tau_0 \to 0$ (no memory, purely viscous) the speed of propagation + become infinite just like in the diffusion equation. +\end{ex} + +\lecture{} + +\subsubsection{Dynamic (complex) modulus} + +\begin{itemize} + \item Dynamic mechanical analysis + \item ie ``The {Vibration test}'' +\end{itemize} + +A bar of infinite length connected by a rod to a wheel that forces the bar periodically. We are +aiming to determine the properties of the bar, so do not assume anything about its elastic +properties. + + +$U(A = 0, t) = a \sim(\omega t)$ apply a sinusoidal displacement at the tip. + +$\lim_{A \to \infty} U(A,t) = 0$. + +$T + \tau_0 \p_t T = E(\epsilon + \tau_1 n\p _t \epsilon) = E(U_a + \tau_1 U_{At}) \qquad (*)$ + +$R_0 U_{t t} = T_a \qquad(* *)$ + +it is easier to write everything in complex coordinates: + +$U(A = 0, t) = e^{i \omega t}$, $(\sin(\omega t))$ is the complex part + +$U(A,t) = \bar U(A) e^{i \omega t}$ + +$T(A,t) = \bar T (A) e^{i \omega t}$ + +Substitute in $(*)$ + +\begin{align*} + \bar T(A) e^{i \omega t} (1 + i\omega \tau_0)) &= E \bar U_A (A) e^{i \omega t} (1 + \tau_1 i + \omega) \\ + \bar T(A) &= E \frac {1 + i \omega \tau_1}{1 + i \omega \tau_0} \bar U_A(A) = E^* \bar U_A(A) +\end{align*} + +Which is the complex stress-strain relation under oscillatory conditions (because it looks very +similar to the linear law for it but in terms of complex numbers). + +\begin{align*} + E^* &= E \frac {1 + i\omega \tau_1}{1 + i \omega \tau_0} = E \frac {(1 + i \omega \tau_1)(1 - i + \omega \tau_0)}{(1 + i \omega \tau_0)(1 - i \omega \tau_0)} \\ + &= E \frac {1 + \omega^2 \tau_0 \tau_1}{1 + \omega^2 {\tau_0} ^2} + i E \frac {\omega (\tau_1 + - \tau_0)}{1 + \omega^2 {\tau_0}^2} \\ + &= E' + i E'' \\ + E': & \text{ storage modulus (`the elastic part') } \\ + E'': & \text{ loss modulus (`the viscous part') } \\ + & \text{from the momentum equation } (* * ) : R_0 U_{t t} = T_A \\ + R_0 \bar U(A) (-\omega^2) e^{i \omega t} &= \bar T_A(A) e^{i \omega t} + = E* \bar U_{A A} (A) e^{i \omega t} \\ + \frac {-\omega^2 R_0}{E^*} \bar U(A) &= \bar U_{A A} (A) \\ + \text{ make } \kappa &= \sqrt{\frac {\omega^2 R_0} {E^*}} \\ + &\text{The general solution is: } + \bar U(A) = \alpha e^{i \kappa A} + \beta e^{i \kappa A} \\ + \kappa^2 &= \frac {\omega^2 R_0} E \left(\frac {1 + \omega^2 \tau_0 \tau_1}{1 + \omega^2 + \tau_1^2} + i \frac {\omega(\tau_0 - \tau_1)} {1 + \omega^2 \tau_1^2}\right) \\ + Im(\kappa) < 0 &\implies Re (i\kappa) > 0\\ + & \implies \alpha = 0 \text{ to satisfy } \lim_{A \to \infty} \bar U(A) = 0 \\ + U(A = 0, t) &= \alpha e^{i \omega t} \\ + & \implies \bar U(0) e^{i \omega t} = \beta e^{-i \kappa 0} e^{i \omega t} + = \beta e^{i \omega t} \\ + & \implies a = \beta \\ + &\implies \bar U(A) = a e^{i \kappa A} +\end{align*} + +Now want the stress at the endpoint $A = 0$ + +\begin{align*} + T(A = 0, t) &= \bar T(A = 0) e^{i \omega t} = E^* \bar U_A(A = 0) e^{i \omega t} \\ + &= E^*(-i) \kappa a e^{i \omega t}\\ +\end{align*} + +\begin{itemize} + \item You increase the freqency forcing and for every frequency you measure the phase + difference. +\end{itemize} + + +\lecture{} + +What is the displacement $U$ and stress $T$ at the left end point. + +$U(A = 0, t) = \bar U(A = 0) e^{i \omega t} a e^{i \kappa 0}e^{i \omega t} = a e^{i \omega _t}$ + +$T(A = 0, t) = \bar T(A = 0) e^{i \omega t} = E^* (-i) \kappa a e^{i \omega t}$ + +$T$ In polar coordinates: +\begin{align*} + -i &= e^{-\frac \pi 2}, \\ + \kappa a e^{i \omega t} &= \kappa a e^{i \omega t} \quad\text{and, } \\ + E^* &= re^{i \delta} = E^* \sqrt{R_0 \frac {\omega^2}{E^*}} = \sqrt{R_0 \omega^2 +E^*} \\ + r^2 e^{2 i \delta} &= R_0 \omega^2 E^* = R_0 \omega^2(E' + iE'') \\ + & \text{take the argument } 2\delta = \arctan \frac {E''}{E'} \\ + \implies \delta &= \frac 12 \arctan {\frac {\omega(\tau_1 - \tau_0)}{1 + + \omega^2 \tau_0 \tau_1}} \\ + \implies T(A = 0, t) &= a r \exp(i (\omega t - \frac \pi 2 + \delta)) +\end{align*} + +Phase difference: $\delta - \frac \pi 2$ + +\subsubsection*{Interpretation} + +\begin{enumerate} + \item +If you have an elastic material, then $\tau_1 = \tau_0 = 0$, so $\delta = 0$ and +the phase difference is $-\frac \pi 2$ independently of the frequency $\omega$ +\item +if you have $\tau_0 = 0$ (the Kelvin-Voigt model of a viscoelastic solid) $\delta = \frac 12 +\arctan(\omega \tau_1)$, which is a monotone function in $\omega$ +\item + In general $\delta$ has a maximum of $\omega = \frac 1 {\sqrt{\tau_0 + \tau_1}}$ + \begin{align*} + g &= \frac {\omega (\tau_1 - \tau_0)}{1 + \omega^2 \tau_0 \tau_1} \\ + g' &= \frac {\sqrt{\tau_1 - \tau_0}}{1 + \omega_2 \tau_0 \tau_1} - \frac + {\omega(\tau_1 - \tau_0)}{(1 + \omega^2 \tau_0 \tau_1)^2 } 2 \omega + \tau_0 \tau_1 = 0\\ + 0 &= 1 + \omega^2 \tau_0 \tau_1 - 2\omega^2 \tau_0 \tau_1 \\ + \omega &= \frac 1 {\sqrt {\tau_0 \tau_1}} + \end{align*} +\end{enumerate} + + +\pagebreak +\chapter{Continuum Mechanics in 3D} + +\begin{description} + \item[Material Coordinates] (lagrangian) + + \item $A \in \R^3$ the material coordinate/position at $t = 0$ + \item $X(A,t) \in \R^3$, $X(A,0) = A$ + \item $U(A,t) = X(A,t) - A \in \R^3$ displacement + \item $V(A,t) = \p_t U(A,t) $ velocity + \item[Spatial/Eulerian coordinates] $x\in \R^3$ + \item $u(x,t) = U(a(x,t) , t) \Leftrightarrow u(X(A,t) , t) = U(A,t)$ + \item $v(x,t) = V(a(x,t) , t) \Leftrightarrow v(X(A,t) , t) = V(A,t)$ +\end{description} + +Usually we use material coordinates for solids and spatial coordinates for +fluids. + +\subsection{Deformation Gradient} + +Consider $A_0 \in \R^3$ and $A = A_0 + \Delta A \in \R^3$ + +\marginnote{Use the taylor expansion only keeping the first derivative term} +$x = X(A_0 + \Delta A, t) = \underbrace{X(A_0, t)}_{X_0} + \underbrace{\nabla_A +X(A_0,t)}_{F(A_0, t)} \Delta A + o(\Delta A)$ + +\begin{definition}{}{} Deformation gradient + $$\nabla _A X(A_0,t) = F(A_0, t) $$ + +$x - x_0 \sim F(A_0, t) (A - A_0)$ + +eg at $t = 0$ $X(A,t = 0) = A)$, and $F = \nabla_A X = \nabla_A A = \mathbb I$ + +\end{definition} + +\subsubsection{Impenetrability} + +The assumption is you can always go from material coordinates to spatial +coordinates and back: so the $A$ must always have an inverse, so $\det F \ne 0$. + +$$J = \det F = \det \underbrace{\nabla_A X}_{\text{Jacobi matrix}} \ne 0$$ + +This is essentially the impenetrability assumption from 1D continuum mechanics in +three dimensions. + +At $t = 0$, $X(A, t= 0) = A \implies \F = \mathbb I \implies J = 1$ + +Since $J(t = 0) = 1$, $J \ne 0$ always implies $J(t) > 0$ + + +\pagebreak +\appendix + +\chapter{Laplace Tranform Review}\label{laplace} + +\marginnote{Taken from MATH2100 Workbook} + +$$ +F(s)=L(f(t))=\int_{0}^{\infty} e^{-s t} f(t) d t +$$ +Powers +$$ +L\left(t^{n}\right)=\frac{n !}{s^{n+1}} \quad L\left(t^{a}\right)=\frac{\Gamma(a+1)}{s^{a+1}} +$$ +Sine and Cosine +$$ +L(\cos (\alpha t))=\frac{s}{s^{2}+\alpha^{2}} \quad \text { and } \quad L(\sin (\alpha t))=\frac{\alpha}{s^{2}+\alpha^{2}} +$$ +Dirac Delta function +$$ +L(\delta(t-k))=e^{-k s} +$$ +Linearity +$$ +L(a g(t)+b h(t))=a G(s)+b H(s) +$$ +First Shifting Theorem +$$ +L\left(e^{a t} f(t)\right)=F(s-a) +$$ +Second Shifting Theorem +$$ +L(f(t-k) u(t-k))=e^{-k s} F(s) +$$ +Convolution Theorem +$$ +L\left(\int_{0}^{t} f(\tau) g(t-\tau) d \tau\right)=F(s) G(s) +$$ +Differentials +$$ +\begin{gathered} +L\left(f^{(n)}(t)\right)=s^{n} F(s)-s^{n-1} f(0)-s^{n-2} \dot{f}(0)-\ldots-s f^{n-2}(0)-f^{n-1}(0) \\ +L(t f(t))=-\frac{d F(s)}{d s} +\end{gathered} +$$ +Periodic Functions $f(t+p)=f(t)$ +$$ +L(f(t))=\frac{1}{1-e^{-s p}} \int_{0}^{p} e^{-s t} f(t) d t +$$ + + + + + +\end{document}