%%%%		-*- 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:   Vibration of a Circular Drumhead
%%%%

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\title{Vibration of a Circular Drumhead}
\subtitle{\textit{Partial Differential Equations}} % (optional)


\author % (optional, use only with lots of authors)
{J~Robert~Buchanan}
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\institute[Millersville University of Pennsylvania]
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{
  Department of Mathematics
}
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% - Keep it simple, no one is interested in your street address.

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

\subject{Talks}
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\begin{document}

\begin{frame}
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%\begin{frame}
%  \frametitle{Outline}
%  \tableofcontents
  % You might wish to add the option [pausesections]
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% 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
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%   15 and 30 frames, all told.

%\section{Introduction}

\begin{frame}\frametitle{Objectives}

In this lesson we will:
\begin{itemize}
\item model the vibration of a circular drumhead,
\item use separation of variables to solve the initial boundary value
  problem, and
\item use a generalized Fourier series to express the formal solution
  describing the displacement of the drumhead.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Assumptions and Model}

A flexible circular membrane of radius $r_{0} > 0$ is centered
at the origin and has its circular edge fixed so that it cannot be
displaced from the $xy$-plane.
Suppose the initial displacement of the membrane for $x^{2} + y^{2} <
r_{0}^{2}$ is described by the function $f(x, y)$ and the initial velocity of
the membrane is $0$.
The displacement $u$ of the
membrane from the $xy$-plane can be
described by the initial boundary value problem:
\begin{align*}
u_{tt}
	&= c^{2} (u_{xx} + u_{yy})
\mbox{ for $x^{2} + y^{2} < r_{0}^{2}$ and $t > 0$}	\\
u(x, y, t)
	&= 0
\mbox{ for $x^{2} + y^{2} = r_{0}^{2}$ and $t > 0$}	\\
u(x, y, 0)
	&= f(x, y)
\mbox{ for $x^{2} + y^{2} < r_{0}^{2}$}	\\
u_{t}(x, y, 0)
	&= 0
\mbox{ for $x^{2} + y^{2} < r_{0}^{2}$.}
\end{align*}
The constant $c > 0$ is the wave speed of vibrations in the membrane
and is directly proportional to the square root of the tension present
in the membrane and inversely proportional to the square root of the
density (mass per unit area) of the membrane.

\end{frame}
%%

\begin{frame}\frametitle{Polar Coordinates}

The circular geometry of the membrane suggests that the solution
may be more easily found using the
polar coordinate system.
Re-writing the problem in polar coordinates
produces
\begin{align*}
%\label{eq:apps:circdrum}
u_{tt}
	&= c^{2} \left( u_{rr} + \frac{1}{r} u_{r} + \frac{1}{r^{2}} u_{\theta\theta} \right)
\mbox{ for $r < r_{0}$, $\theta \in \mathbb{R}$, $t > 0$}	\\
%\label{eq:apps:circdrumbc}
u(r_{0}, \theta, t)
	&= 0
\mbox{ for $\theta \in \mathbb{R}$ and $t > 0$}	\\
%\label{eq:apps:circdrumdisp}
u(r, \theta, 0)
	&= f(r,\theta)
\mbox{ for $r < r_{0}$ and $\theta \in \mathbb{R}$}	\\
u_{t}(r, \theta, 0)
	&= 0
\mbox{ for $r < r_{0}$ and $\theta \in \mathbb{R}$.}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Separation of Variables (1 of 2)}

Assume a product solution of the
form $u(r, \theta, t) = R(r) \Theta(\theta) T(t)$.
\begin{align*}
R(r) \Theta(\theta) T''(t)
	&= c^{2} \left( R''(r) \Theta(\theta) T(t) +
\frac{R'(r) \Theta(\theta) T(t)}{r} +
\frac{R(r) \Theta''(\theta) T(t)}{r^{2}} \right)	\\
%\label{eq:apps:circdrumrtheta}
\frac{T''(t)}{c^{2} T(t)}
	&= \frac{R''(r)}{R(r)} + \frac{1}{r} \frac{R'(r)}{R(r)} + \frac{1}{r^{2}}
              \frac{\Theta''(\theta)}{\Theta(\theta)}	.
\end{align*}
Since the left-hand side of the equation depends on $t$ alone while
the right-hand side depends on $(r, \theta)$, then both sides must be
constant.

\vspace*{0.10in}

Suppose the constant $-\lambda^{2} \le 0$.
\begin{align*}
\frac{R''(r)}{R(r)} + \frac{1}{r} \frac{R'(r)}{R(r)} + \frac{1}{r^{2}}
              \frac{\Theta''(\theta)}{\Theta(\theta)}
	&= -\lambda^{2}	\\
r^{2} \frac{R''(r)}{R(r)} + r \frac{R'(r)}{R(r)} + \lambda^{2} r^{2}
	&= -\frac{\Theta''(\theta)}{\Theta(\theta)}	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Separation of Variables (2 of 2)}

\[
r^{2} \frac{R''(r)}{R(r)} + r \frac{R'(r)}{R(r)} + \lambda^{2} r^{2}
	= -\frac{\Theta''(\theta)}{\Theta(\theta)}
\]
The right-hand side of the equation is a function of $\theta$
only and the solution to the initial boundary value problem should be
$2 \pi$-periodic in $\theta$, thus both sides of the equation are equal
to the constant $m^{2}$ where $m = 0, 1, 2, \ldots$.

\vspace*{0.10in}

\[
\Theta(\theta)
	\equiv \Theta_{m}(\theta)
	= c_{1} \cos (m \theta) + c_{2} \sin (m \theta)	,
\]
where $c_{1}$ and $c_{2}$ are constant coefficients.

\vspace*{0.10in}

\uncover<2->{%
The ordinary differential equation implied by the left-hand side of
the separated variables equation can be written as
\[
r^{2} R''(r) + r R'(r) + (\lambda^{2} r^{2} - m^{2}) R(r)
	= 0	.
\]
This is the parametric form
of Bessel's equation of order $m$.
Thus the general solution is $R(r) = c_{3} J_{m}(\lambda r) + c_{4}
Y_{m}(\lambda r)$.
}

\end{frame}
%%

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

\begin{itemize}
\item The solution should be bounded as $r \to 0^{+}$, thus
$R_{m}(r) = J_{m}(\lambda r)$.
\item The boundary condition $u(r_{0}, \theta, t) = 0$ implies
$J_{m}(\lambda r_{0}) = 0$ for each $m$, and hence
$\lambda \equiv \lambda_{m, n}/r_{0}$
where $\lambda_{m, n}$ is the $n$th zero of the Bessel function of the
first kind of order $m$.
\end{itemize}

\uncover<2->{%
The ordinary differential equation implied by the time
dependent portion of the PDE is
\[
T''(t) + \left( \frac{c \lambda_{m, n}}{r_{0}} \right)^{2} T(t)
	= 0	.
\]
As a result, the time dependent portion of the solution can be written
as
\[
T(t)
	= c_{5} \cos \left( \frac{c \lambda_{m, n} t}{r_{0}} \right) +
        c_{6} \sin \left( \frac{c \lambda_{m, n} t}{r_{0}} \right)        .
\]
}

\end{frame}
%%

\begin{frame}\frametitle{Product Solution}
\fontsize{9pt}{10.8}\selectfont

\begin{align*}
u_{m, n}(r, \theta, t)
	= {} & J_{m}\left( \frac{\lambda_{m, n} r}{r_{0}} \right)
\left( c_{1} \cos (m \theta) + c_{2} \sin (m \theta) \right)	\\
	& {} \left( c_{5} \cos \left( \frac{c \lambda_{m, n} t}{r_{0}} \right) +
c_{6} \sin \left( \frac{c \lambda_{m, n} t}{r_{0}} \right) \right)
\end{align*}
for $m = 0, 1, 2, \ldots$ and $n \in \mathbb{N}$.

\vspace*{0.10in}

\uncover<2->{%
By the Principle of Superposition,
\[
%\label{eq:apps:circdrumsoln}
u(r, \theta, t)
	= \sum_{m=0}^{\infty} \sum_{n=1}^{\infty}
\left[ J_{m}\left( \frac{\lambda_{m, n} r}{r_{0}} \right)
\left( A_{m, n} \cos (m \theta) + B_{m, n} \sin (m \theta) \right)
\cos \left( \frac{c \lambda_{m, n} t}{r_{0}} \right)
\right]
\]
where $A_{m, n}$ and $B_{m, n}$ are coefficients which will be
determined from the initial displacement of the membrane.
}

\end{frame}
%%

\begin{frame}\frametitle{Determining the Coefficients (1 of 2)}

The coefficients can be found by setting $t = 0$.
\[
%\label{eq:apps:circdruminitdisp}
f(r, \theta)
	= \sum_{m=0}^{\infty} \sum_{n=1}^{\infty}
J_{m}\left( \frac{\lambda_{m, n} r}{r_{0}} \right)
\left[ A_{m, n} \cos (m \theta) + B_{m, n}\sin (m \theta) \right]
\]
Multiply both sides by
$\pi^{-1} \cos (k \theta)$ and integrate over $[-\pi, \pi]$.
The only nonzero term in the summation over index $m$ will be the term
for which $m = k$, thus
\[
%\label{eq:apps:circdrumcoeffsaknseries}
\frac{1}{\pi} \int_{-\pi}^{\pi} f(r, \theta) \cos (k \theta) \, d\theta
	= \sum_{n=1}^{\infty} A_{k, n} J_{k}\left( \frac{\lambda_{k, n} r}{r_{0}} \right)	.
\]
Similarly if both sides are
multiplied by
$\pi^{-1} \sin (k \theta)$ and integrated over $[-\pi, \pi]$,
\[
%\label{eq:apps:circdrumcoeffsbknseries}
\frac{1}{\pi} \int_{-\pi}^{\pi} f(r, \theta) \sin (k \theta) \, d\theta
	= \sum_{n=1}^{\infty} B_{k, n} J_{k}\left( \frac{\lambda_{k, n} r}{r_{0}} \right)	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Determining the Coefficients (2 of 2)}

Coefficients can now be found individually using Bessel series
coefficient formulas.
Make the substitution $z = r/r_{0}$.
In the special case where $k=0$,
\[
%\label{eq:apps:circdrumcoeffsa0nseries}
\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(r_{0} z, \theta) \, d\theta
	= \sum_{n=1}^{\infty} A_{0, n} J_{0}( \lambda_{0, n} z)	.
\]
Using the Bessel series coefficient formula,
\[
%\label{eq:apps:circdrumcoeffsa0n}
A_{0, n}
	= \frac{1}{\pi (J_{1}(\lambda_{0, n}))^{2}} \int_{0}^{1} z \,
J_{0} (\lambda_{0, n} z) \left( \int_{-\pi}^{\pi} f(r_{0} z, \theta) \,
d\theta \right) \, dz	.
\]
For $k \in \mathbb{N}$,
\begin{align*}
%\label{eq:apps:circdrumcoeffsakn}
A_{k, n}
	&= \frac{2 \int_{0}^{1} z \, J_{k} (\lambda_{k, n} z)
\left( \int_{-\pi}^{\pi} f(r_{0} z, \theta) \cos (k \theta) \,
d\theta \right) \, dz}{\pi (J_{k+1}(\lambda_{k, n}))^{2}}	\\
%\label{eq:apps:circdrumcoeffsbkn}
B_{k, n}
	&= \frac{2 \int_{0}^{1} z \, J_{k} (\lambda_{k, n} z)
\left( \int_{-\pi}^{\pi} f(r_{0} z, \theta) \sin (k \theta) \,
d\theta \right) \, dz}{\pi (J_{k+1}(\lambda_{k, n}))^{2}}	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Fundamental Frequency}

The motion of the vibrating
membrane can be thought of as the superposition of a number of
modes of vibration.
The $(m, n)$th mode of vibration is given by the expression,
\[
J_{m}\left( \frac{\lambda_{m, n} r}{r_{0}} \right)
\left( A_{m, n} \cos (m \theta) + B_{m, n} \sin (m \theta) \right)
\cos \left( \frac{c \lambda_{m, n} t}{r_{0}} \right)	.
\]
The frequency associated with this mode is
\[
f_{m, n} = \frac{c \lambda_{m, n}}{2 \pi r_{0}}
\]
and therefore the fundamental frequency
corresponding to $m=0$ and $n=1$ is $f_{0, 1} = c \lambda_{0,1}/(2 \pi
r_{0})$.

\end{frame}
%%

\begin{frame}\frametitle{Example}

Find the displacement from equilibrium for the circular drumhead
described in the following initial boundary value problem.
\begin{align*}
%\label{eq:apps:circdrum}
u_{tt}
	&= \left( u_{rr} + \frac{1}{r} u_{r} + \frac{1}{r^{2}} u_{\theta\theta} \right)
\mbox{ for $r < 1$, $\theta \in \mathbb{R}$, $t > 0$}	\\
%\label{eq:apps:circdrumbc}
u(1, \theta, t)
	&= 0
\mbox{ for $\theta \in \mathbb{R}$ and $t > 0$}	\\
%\label{eq:apps:circdrumdisp}
u(r, \theta, 0)
	&= r (1 - r^{2}) \sin \theta
\mbox{ for $r < 1$ and $\theta \in \mathbb{R}$}	\\
u_{t}(r, \theta, 0)
	&= 0
\mbox{ for $r < 1$ and $\theta \in \mathbb{R}$.}
\end{align*}

\end{frame}
%%

\begin{frame}<handout:0>\frametitle{Solution (1 of 3)}

Since $f(r, \theta)$ is an odd function in $\theta$,
the coefficients $A_{k, n} = 0$
for $k = 0, 1, 2, \ldots$ and $n \in \mathbb{N}$.
The orthogonality of the trigonometric functions on $[-\pi, \pi]$
implies $B_{k, n} = 0$ for $k = 2,
3, \ldots$ and $n \in \mathbb{N}$.
Thus the infinite series solution simplifies considerably to the form,
\[
u(r, \theta, t)
	= \sin (\theta) \sum_{n=1}^{\infty}
B_{1, n} J_{1}(\lambda_{1, n} r) \cos (\lambda_{1, n} t)	.
\]

\end{frame}
%%

\begin{frame}<handout:0>\frametitle{Solution (2 of 3)}

The coefficients $B_{1, n}$ for $n \in \mathbb{N}$ can be found by integration
\begin{align*}
B_{1, n}
	&= \frac{2}{\pi (J_{2}(\lambda_{1, n}))^{2}}
\int_{0}^{1} z \, J_{1} (\lambda_{1, n} z)
\left( \int_{-\pi}^{\pi} z (1-z^{2}) \sin^{2} (\theta) \,
d\theta \right) \, dz	\\
	&= \frac{2}{(J_{2}(\lambda_{1, n}))^{2}}
\int_{0}^{1} (z^{2} - z^{4}) \, J_{1} (\lambda_{1, n} z) \, dz
	= \frac{16}{\lambda_{1, n}^{3} J_{2}(\lambda_{1, n})}	.
\end{align*}

\uncover<2->{%
The formal solution can then be expressed as
\[
u(r, \theta, t)
	= 16 \sin (\theta) \sum_{n=1}^{\infty}
        \frac{J_{1}(\lambda_{1, n} r)}{\lambda_{1, n}^{3}
          J_{2}(\lambda_{1, n})} \cos (\lambda_{1, n} t)	.
\]
}

\end{frame}
%%

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

\begin{center}
\resizebox{0.75\textwidth}{!}{\includegraphics{circdrum}}
\end{center}

\end{frame}
%%

%\begin{frame}\frametitle{Homework}

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

%\end{frame}
%%

%%%
\end{document}

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