%%%%		-*- Mode: LaTeX -*-
%%%% Author:    J. Robert Buchanan
%%%% Address:   Department of Mathematics
%%%%		Millersville University
%%%%            P.O. Box 1002
%%%%            Millersville, PA 17551-0302
%%%% Phone:     717-871-7305
%%%% FAX:	717-871-7948
%%%%
%%%% Email:     Robert.Buchanan@millersville.edu
%%%%
%%%% Date:      03/05/2025
%%%%
%%%% Purpose:   Modeling a Swinging Chain
%%%%

\documentclass[10pt]{beamer}

\mode<presentation>
{
  %\usetheme{Warsaw}
  \usetheme{default}
  %\usetheme{CambridgeUS}
  % or ...

  %\setbeamercovered{transparent}
  % or whatever (possibly just delete it)
}


\usepackage[english]{babel}
% or whatever

\usepackage[latin1]{inputenc}
% or whatever

\usepackage{times}
\usepackage[T1]{fontenc}
% Or whatever. Note that the encoding and the font should match. If T1
% does not look nice, try deleting the line with the fontenc.


\title{Modeling a Swinging Chain}
\subtitle{\textit{Partial Differential Equations}} % (optional)


\author % (optional, use only with lots of authors)
{J~Robert~Buchanan}
% - Use the \inst{?} command only if the authors have different
%   affiliation.

\institute[Millersville University of Pennsylvania]
% (optional, but mostly needed)
{
  Department of Mathematics
}
% - Use the \inst command only if there are several affiliations.
% - Keep it simple, no one is interested in your street address.

\date{ % (optional)
Fall
%Spring
%Summer
\the\year}

\subject{Talks}
% This is only inserted into the PDF information catalog. Can be left
% out. 



% If you have a file called "university-logo-filename.xxx", where xxx
% is a graphic format that can be processed by latex or pdflatex,
% resp., then you can add a logo as follows:

% \pgfdeclareimage[height=0.5cm]{university-logo}{university-logo-filename}
% \logo{\pgfuseimage{university-logo}}



% Delete this, if you do not want the table of contents to pop up at
% the beginning of each subsection:
\AtBeginSubsection[]
{
  \begin{frame}<beamer>
    \frametitle{Outline}
    \tableofcontents[currentsection,currentsubsection]
  \end{frame}
}


% If you wish to uncover everything in a step-wise fashion, uncomment
% the following command: 

%\beamerdefaultoverlayspecification{<+->}


\begin{document}

\begin{frame}
  \titlepage
\end{frame}

%\begin{frame}
%  \frametitle{Outline}
%  \tableofcontents
  % You might wish to add the option [pausesections]
%\end{frame}


% Since this a solution template for a generic talk, very little can
% be said about how it should be structured. However, the talk length
% of between 15min and 45min and the theme suggest that you stick to
% the following rules:  

% - Exactly two or three sections (other than the summary).
% - At *most* three subsections per section.
% - Talk about 30s to 2min per frame. So there should be between about
%   15 and 30 frames, all told.

%\section{Introduction}

\begin{frame}\frametitle{Objectives}

In this lesson we will:
\begin{itemize}
\item model the motion of a swinging chain, and
\item solve the partial differential equation modeling a swinging chain.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Developing a Mathematical Model}

\begin{itemize}
\item Consider a chain or string with one end fixed at the origin of a
rectangular coordinate system, with the positive $x$-axis oriented in
the downward direction.
\item The length of the chain is $L$ and the linear density of the chain is
$\rho(x)$.
\item The chain is subject to gravitational acceleration $g$ (again in the
downward, positive $x$-direction).
\item If the chain is unperturbed by other forces, it will remain still,
occupying the interval $[0, L]$ along the $x$-axis.
\item If the chain is moved by a small amount from this equilibrium
position, let $u(x, t)$ denote the displacement of the chain from the
$x$-axis at position $x$ and time $t$.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Tension in the Chain}

The tension at position $x$ is
\[
%\label{eq:apps:chaintension}
T(x)
	= g \int_{x}^{L} \rho(s) \, ds	.
\]
Proceeding as in the development of the model for a vibrating string,
\[
%\label{eq:apps:chaingeneral}
\rho(x) u_{tt}
	= -g \rho(x) u_{x} + T(x) u_{xx} + f(x, t)
\]
where $f(x, t)$ represents any external forces on the chain.

\vspace*{0.10in}

\uncover<2->{%
If $f(x, t) = 0$ the initial, boundary value problem for the swinging
chain is
\begin{align*}
\rho(x) u_{tt}
	&= ( T(x) u_{x} )_{x}	\\
u(0, t)
	&= 0
\mbox{ for $t > 0$}	\\
u(x, 0)
	&= f(x)
\mbox{ and }
u_{t}(x, 0)
	= g(x)
\mbox{ for $0 < x < L$.}
\end{align*}
}

\end{frame}
%%

\begin{frame}\frametitle{Constant Linear Density}

Suppose $T(x) = g \rho (L-x)$ where $\rho$ is the constant linear
density.

\vspace*{0.10in}

This simplifies the partial differential equation to
\[
%\label{eq:apps:chainsimple}
u_{tt}
	= -g u_{x} + g (L-x) u_{xx}	.
\]

\vspace*{0.10in}

Assuming a product solution of the form $u(x, t) = X(x) T(t)$,
\[
%\label{eq:apps:chainsep}
\frac{T''(t)}{g T(t)}
	= -\frac{X'(x)}{X(x)} + (L-x) \frac{X''(x)}{X(x)}
	= \gamma	,
\]
where $\gamma$ is a constant.
Thus the variables can be separated.

\vspace*{0.10in}

The boundary value problem for $X(x)$ can be expressed as
%\label{eq:apps:chainbvp}
\begin{align*}
(L-x) X''(x) - X'(x) - \gamma \, X(x)
	& = 0
\mbox{ for $0 < x < L$}	\\
X(0)
	& = 0	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Change of Variable}

\textbf{Remark}: any physically
meaningful solution to the initial boundary value problem of the
swinging chain must be bounded.
This implies that $\gamma < 0$.

\vspace*{0.10in}

Let $L - x = \alpha \xi^{\beta}$ where $\alpha$ and $\beta$ are
constants.
Making the change of variable results in
\[
%\label{eq:apps:chainchngvar}
\xi^{2-\beta} \frac{d^{2} X}{d \xi^{2}} + \xi^{1-\beta} \frac{d X}{d
  \xi} - \alpha \beta^{2} \gamma X
	= 0	.
\]
If $\beta = 2$, $\alpha = -1/(4 \gamma)$, and both sides of
the ODE are multiplied by $\xi^{2}$ then
Bessel's equation of order zero results.
\[
\xi^{2} \frac{d^{2} X}{d \xi^{2}} + \xi \frac{d X}{d \xi} + \xi^{2} X
	= 0	.
\]
Consequently the general solution can be expressed as $X(\xi) = c_{1}
J_{0}(\xi) + c_{2} Y_{0}(\xi)$.

\end{frame}
%%

\begin{frame}\frametitle{Bounding the Solution}

\begin{itemize}
\item The solution must be bounded as $x \to L^{-}$ which is equivalent to
$\xi \to 0^{+}$ and implies $c_{2} = 0$.
\item In terms of the original independent variable
$X(x) = c_{1} J_{0}(2 \sqrt{-\gamma (L-x)})$.
\item Choose $\gamma = -\lambda^{2}$ with $\lambda > 0$ and
impose the boundary condition at $u(0, t) = 0$ for all $t > 0$
implying $J_{0}(2 \lambda \sqrt{L}) = 0$ which in turn implies that $2
\lambda \sqrt{L}$ is a zero of the Bessel function of the first kind
of order zero.
\item Solve the time-dependent portion of separated equation to yield
  $T(t) = c_{3} \cos \left( \sqrt{g} \lambda t \right) + c_{4} \sin
  \left( \sqrt{g} \lambda t \right)$.
\item Hence the product solutions indexed by $\lambda > 0$ take the form
\[
u_{\lambda}(x, t)
	= \left[ A_{\lambda} \cos \left( \sqrt{g} \lambda t \right) +
          B_{\lambda} \sin \left( \sqrt{g} \lambda t \right) \right]
        J_{0}(2 \lambda \sqrt{L-x})	,
\]
where $A_{\lambda}$ and $B_{\lambda}$ are constants.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Principle of Superposition}
\fontsize{9pt}{10.8}\selectfont

The formal
solution to the swinging chain can be written as the generalized Fourier series:
\[
%\label{eq:apps:chainsoln}
u(x, t)
	= \sum_{n=1}^{\infty} \left[ A_{n}
\cos \left( \frac{\lambda_{0,n}}{2} \sqrt{\frac{g}{L}} t \right) + B_{n} \sin
\left( \frac{\lambda_{0,n}}{2} \sqrt{\frac{g}{L}} t \right) \right]
J_{0} \left( \lambda_{0,n} \sqrt{1-\frac{x}{L}} \right)	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Example}

Find the formal solution to the initial boundary value problem of the
perturbed chain released from rest.
\begin{align*}
u_{tt}
	&= -g u_{x} + g (L-x) u_{xx}	\\
u(0, t)
	&= 0
\mbox{ for $t > 0$}	\\
u(x, 0)
	&= x
\mbox{ and }
u_{t}(x, 0)
	= 0
\mbox{ for $0 < x < L$.}
\end{align*}

\end{frame}
%%

\begin{frame}<handout:0>\frametitle{Solution (1 of 4)}
\fontsize{9pt}{10.8}\selectfont

\[
%\label{eq:apps:chainsoln}
u(x, t)
	= \sum_{n=1}^{\infty} \left[ A_{n}
\cos \left( \frac{\lambda_{0,n}}{2} \sqrt{\frac{g}{L}} t \right) + B_{n} \sin
\left( \frac{\lambda_{0,n}}{2} \sqrt{\frac{g}{L}} t \right) \right]
J_{0} \left( \lambda_{0,n} \sqrt{1-\frac{x}{L}} \right)	.
\]
Since the initial velocity of the chain is zero then $B_{n} = 0$ for
$n \in \mathbb{N}$.
Setting $t = 0$,
\[
u(x, 0)
	= x
	= \sum_{n=1}^{\infty} A_{n} J_{0} \left( \lambda_{0, n}
          \sqrt{1-\frac{x}{L}} \right)	.
\]
Recall the change of variable.
\[
\xi = \sqrt{1-\frac{x}{L}}
\iff
x = L (1 - \xi^{2})
\]
The initial displacement of the chain can be expressed as
\[
L (1 - \xi^{2})
	= \sum_{n=1}^{\infty} A_{n} J_{0} \left( \lambda_{0, n} \xi \right)	.
\]

\end{frame}
%%

\begin{frame}<handout:0>\frametitle{Solution (2 of 4)}

Using the orthogonality of the Bessel functions,
\begin{align*}
A_{n}
	&= \frac{2 \int_{0}^{1} L (1 - \xi^{2}) J_{0}(\lambda_{0, n} \xi)
  \xi \, d\xi}{(J_{1}(\lambda_{0, n}))^{2}}	\\
	&= \frac{2 L}{(J_{1}(\lambda_{0, n}))^{2}} \left(
\frac{1}{\lambda_{0, n}^{2}} \int_{0}^{\lambda_{0, n}} u J_{0}(u) \, du
              - \frac{1}{\lambda_{0, n}^{4}} \int_{0}^{\lambda_{0, n}}
   u^{2} (u  J_{0}(u)) \, du \right)	\\
	&= \frac{2 L}{(J_{1}(\lambda_{0, n}))^{2}} \left(
\frac{2}{\lambda_{0, n}^{4}} \int_{0}^{\lambda_{0, n}} u^{2} J_{1}(u) \, du
\right)	\\
	&= \frac{4 L}{\lambda_{0, n}^{4} (J_{1}(\lambda_{0, n}))^{2}}
\left[ u^{2} J_{2}(u) \right]_{u=0}^{u=\lambda_{0, n}}	\\
	&= \frac{4 L J_{2}(\lambda_{0, n})}{\lambda_{0, n}^{2} (J_{1}(\lambda_{0, n}))^{2}}	\\
	&= \frac{8 L}{\lambda_{0, n}^{3} J_{1}(\lambda_{0, n})}	.
\end{align*}

\end{frame}
%%

\begin{frame}<handout:0>\frametitle{Solution (3 of 4)}

We can numerically approximate as many of the zeros of the Bessel
function of the first kind of order zero as necessary.

\vspace*{0.10in}

The formal solution takes the form,
\[
u(x, t)
	= 8 L \sum_{n=1}^{\infty} \frac{1}{\lambda_{0, n}^{3} J_{1}(\lambda_{0, n})}
J_{0}\left( \lambda_{0, n} \sqrt{1 - \frac{x}{L}} \right)	.
\]

\end{frame}
%%

\begin{frame}<handout:0>\frametitle{Solution (4 of 4)}

\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{chain}}
\end{center}

\end{frame}
%%

%\begin{frame}\frametitle{Homework}

%\begin{itemize}
%\item Read Section $x.y$
%\item Exercises: 
%\end{itemize}

%\end{frame}
%%

%%%
\end{document}

%%%%
%%%% EOF
%%%%
