%%%%		-*- 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:   Vibrating Sphere
%%%%

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\title{Vibrating Sphere}
\subtitle{\textit{Partial Differential Equations}} % (optional)


\author % (optional, use only with lots of authors)
{J~Robert~Buchanan}
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%   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}
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% the beginning of each subsection:
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\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 explore the mathematical modeling describing the displacement of
  a circular membrane from equilibrium, and
\item use separation of variables and the spherical coordinate system
  to determine a series solution to the initial boundary value problem.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Model of a Vibrating Sphere}

\begin{itemize}
\item Consider a flexible membrane in the shape of a hollow
sphere of radius $\rho_{0}$.
\item If a point on the surface of the sphere is displaced radially from the
equilibrium radius $\rho_{0}$, the tension in the surrounding area
will attempt to restore the point to the equilibrium radius.
\end{itemize}
Let $u(\varphi, \theta, t)$ be the radial displacement of the point
$(\rho_{0}, \varphi, \theta)$ from equilibrium at time $t$.
\begin{align*}
%\label{eq:apps:vibspherepde}
u_{tt}
	&= c^{2} \Delta u - \omega^{2} u
\mbox{ for $0 < \varphi < \pi$, $-\pi < \theta < \pi$, $t > 0$}	\\
u(\varphi, \theta, 0)
	&= f(\varphi, \theta)
\mbox{ for $0 < \varphi < \pi$, $-\pi < \theta < \pi$}	\\
u_{t}(\varphi, \theta, 0)
	&= g(\varphi, \theta)
\mbox{ for $0 < \varphi < \pi$, $-\pi < \theta < \pi$}
\end{align*}

\end{frame}
%%

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

If $u(\varphi, \theta, t) = F(\varphi, \theta) T(t)$ is a product
solution to the partial differential equation, then
\begin{align*}
F(\varphi, \theta) T''(t)
	&= c^{2} T(t) \left( \Delta F(\varphi, \theta) \right) - \omega^{2}
              F(\varphi, \theta) T(t)	\\
\frac{1}{c^{2}} \left( \frac{T''(t)}{T(t)} + \omega^{2} \right)
	&= \frac{\Delta F(\varphi, \theta)}{F(\varphi, \theta)}	.
\end{align*}
The left-hand side of the equation depends only on $t$ and the
right-hand side depends only on $\varphi$ and $\theta$, thus
both sides of the equation are equal to a constant $-\lambda$.

\vspace*{0.10in}

\uncover<2->{%
The right-hand side of the equation implies the differential equation
\[
\Delta F(\varphi, \theta) + \lambda F(\varphi, \theta)
	= 0
\]
The eigenvalues are $\lambda \equiv \lambda_{n} = n (n+1)$ for $n = 0,
1, 2, \ldots$ and the eigenfunctions are the
spherical harmonics $Y_{n}^{m}(\varphi,
\theta)$ for $m = -n, -n+1, \ldots, n$.
}

\end{frame}
%%

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

The $t$-dependent portion of the product
solution satisfies the ordinary differential equation,
\[
T''(t) + (\omega^{2} + c^{2} n (n+1)) T(t)
	= 0
\]
which has the general solution,
\[
T_{n}(t)
	= c_{1} \cos \left( \sqrt{\omega^{2} + c^{2} n (n+1)} t \right)
	+ c_{2} \sin \left( \sqrt{\omega^{2} + c^{2} n (n+1)} t \right)	.
\]
Thus a product solution has the form
\[
u_{n, m}(\varphi,\theta, t)
	= Y_{n}^{m}( \varphi, \theta )
\left( a_{n, m} \cos ( \sqrt{\omega^{2} + c^{2} \lambda_{n}} t )
	+ b_{n, m} \sin ( \sqrt{\omega^{2} + c^{2} \lambda_{n}} t ) \right)
\]
for $n = 0, 1, 2, \ldots$ and $m = -n, -n+1, \ldots, n$.

\end{frame}
%%

\begin{frame}\frametitle{Series Solution (1 of 2)}
\fontsize{9pt}{10.8}\selectfont

The series solution to the initial boundary value problem has the form
\[
u(\varphi, \theta, t)
	= \sum_{n=0}^{\infty} \sum_{m=-n}^{n} Y_{n}^{m}( \varphi, \theta )
\left( a_{n, m} \cos ( \sqrt{\omega^{2} + c^{2} \lambda_{n}} t )
	+ b_{n, m} \sin ( \sqrt{\omega^{2} + c^{2} \lambda_{n}} t ) \right)	.
\]

\uncover<2->{%
If $t = 0$,
\[
u(\varphi, \theta, 0)
	= f(\varphi, \theta)
	= \sum_{n=0}^{\infty} \sum_{m=-n}^{n} a_{n, m} Y_{n}^{m}(\varphi, \theta)	.
\]
Multiply both sides by the complex conjugate of
$Y_{\hat{n}}^{\hat{m}}(\varphi, \theta) \sin \varphi$ and integrate
with respect to $\theta$ over interval $[-\pi, \pi]$ and with respect
to $\varphi$ over $[0, \pi]$.
The only
nonzero term in the infinite series occurs when $m = \hat{m}$ and $n =
\hat{n}$.
In this case,
\begin{align*}
a_{n, m}
	&= \int_{-\pi}^{\pi} \int_{0}^{\pi} f(\varphi, \theta)
\overline{Y_{n}^{m}(\varphi, \theta)}
\sin \varphi \, d\varphi \, d\theta	\\
	&= \sqrt{\frac{2n+1}{4 \pi} \frac{(n-m)!}{(n+m)!}}
\int_{-\pi}^{\pi} \int_{0}^{\pi} f(\varphi, \theta)
e^{-i m \theta} P_{n}^{m}(\cos \varphi)
\sin \varphi \, d\varphi \, d\theta	.
\end{align*}
}

\end{frame}
%%

\begin{frame}\frametitle{Series Solution (2 of 2)}
\fontsize{9pt}{10.8}\selectfont

Differentiate the solution with respect to $t$ and set
$t = 0$.
\[
u_{t}(\varphi, \theta, 0)
	= g(\varphi, \theta)
	= \sum_{n=0}^{\infty} \sum_{m=-n}^{n}
b_{n, m} \sqrt{\omega^{2} + c^{2} n (n+1)} Y_{n}^{m}(\varphi, \theta)	.
\]
Multiply both sides of the equation by the complex conjugate of
$Y_{\hat{n}}^{\hat{m}}(\varphi,
\theta) \sin \varphi$ and integrate with respect to $\theta$ over interval $[-\pi,
\pi]$ and with respect to $\varphi$ over $[0, \pi]$.
\[
b_{n, m}
	= \frac{1}{\sqrt{\omega^{2} + c^{2} n (n+1)}}
\int_{-\pi}^{\pi} \int_{0}^{\pi} g(\varphi, \theta)
\overline{Y_{n}^{m}(\varphi, \theta)}
\sin \varphi \, d\varphi \, d\theta	.
\]
Note that if $\omega = 0$, then $b_{0, 0} = 0$.

\end{frame}
%%

\begin{frame}\frametitle{Example}

Find a formal series solution describing the radial displacement from
equilibrium of a point on a flexible spherical membrane described by
the initial boundary value problem below.
\begin{align*}
%\label{eq:apps:vibspherepde}
u_{tt}
	&= \Delta u
\mbox{ for $0 < \varphi < \pi$, $-\pi < \theta < \pi$, $t > 0$}	\\
u(\varphi, \theta, 0)
	&= \frac{1}{32} \sin^{2} (2 \varphi) \sin (2 \theta)
\mbox{ for $0 < \varphi < \pi$, $-\pi < \theta < \pi$}	\\
u_{t}(\varphi, \theta, 0)
	&= 0
\mbox{ for $0 < \varphi < \pi$, $-\pi < \theta < \pi$}
\end{align*}
Assume the unperturbed radius of the sphere is $\rho_{0} = 1$.

\end{frame}
%%

\begin{frame}\frametitle{Solution (1 of 5)}
\fontsize{9pt}{10.8}\selectfont

Since the initial velocity is zero, the coefficients $b_{n, m} = 0$ for
all $n$ and $m$.
If $m \ge 0$,
\begin{align*}
a_{n, m}
	&= \frac{1}{32} \int_{-\pi}^{\pi} \int_{0}^{\pi}
\sin^{2} (2 \varphi) \sin (2 \theta)
\overline{Y_{n}^{m}(\varphi, \theta)} \sin \varphi \, d\varphi \, d\theta	\\
	&= \frac{1}{32} \int_{0}^{\pi} \sin^{2} (2 \varphi) \sin \varphi
\int_{-\pi}^{\pi} \sin (2 \theta)
\overline{Y_{n}^{m}(\varphi, \theta)} \, d\theta \, d\varphi	\\
	&= \frac{1}{32} \sqrt{\frac{2n+1}{4 \pi} \frac{(n-m)!}{(n+m)!}}
\int_{0}^{\pi} \sin^{2} (2 \varphi) P_{n}^{m}(\cos \varphi) \sin \varphi \, d\varphi
\int_{-\pi}^{\pi} e^{-i m \theta} \sin (2 \theta) \, d\theta	.
\end{align*}
If $m > 0$,
\[
a_{n,-m}
	= \frac{(-1)^{m}}{32} \sqrt{\frac{2n+1}{4 \pi} \frac{(n-m)!}{(n+m)!}}
\int_{0}^{\pi} \sin^{2} (2 \varphi) P_{n}^{m}(\cos \varphi) \sin \varphi \, d\varphi
\int_{-\pi}^{\pi} e^{i m \theta} \sin (2 \theta) \, d\theta	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Solution (2 of 5)}

Suppose $|m| \ne 2$, then
\[
\int_{-\pi}^{\pi} e^{i m \theta} \sin (2 \theta) \, d\theta
	= \left[ \frac{e^{i m \theta}}{m^{2} - 4} \left( 2 \cos (2 \theta) - i m \sin
              (2 \theta) \right) \right]_{\theta=-\pi}^{\theta=\pi}
	= 0	.
\]
This implies $a_{n, m} = 0$ for $m \ne \pm 2$.
Thus the solution to the vibrating sphere problem takes on the form
\begin{align*}
u(\varphi, \theta, t)
	&= \sum_{n=2}^{\infty} \left[ \cos (\sqrt{n (n+1)} t) \left(
a_{n, -2} Y_{n}^{-2} (\varphi, \theta) +
a_{n, 2} Y_{n}^{2} (\varphi, \theta) \right)
\right]	\\
	&= \sum_{n=2}^{\infty} \left[ \cos (\sqrt{n (n+1)} t) \left(
a_{n, -2} Y_{n}^{2} (\varphi, -\theta) +
a_{n, 2} Y_{n}^{2} (\varphi, \theta) \right)
\right]	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution (3 of 5)}

Using integration by parts
\[
%\label{eq:apps:vibspheremeq2}
\int_{-\pi}^{\pi} e^{\pm 2 i \theta} \sin (2 \theta) \, d\theta
	= \pm \pi \, i	.
\]
Let $m = 2$ then
\[
a_{n, 2}
	= -a_{n, -2}
	= -\frac{\pi \, i}{32} \sqrt{\frac{2n+1}{4 \pi} \frac{(n-2)!}{(n+2)!}}
\int_{0}^{\pi} \sin^{2} (2 \varphi) P_{n}^{2}(\cos \varphi) \sin \varphi \, d\varphi	.
\]
Make the substitution $x = \cos \varphi$,
\begin{align*}
a_{n, 2}
	&= -a_{n, -2}
	= -\frac{\pi \, i}{8} \sqrt{\frac{2n+1}{4 \pi} \frac{(n-2)!}{(n+2)!}}
\int_{-1}^{1} x^{2} (1-x^{2}) P_{n}^{2}(x) \, dx	\\
	&= \frac{\pi \, i}{8} \sqrt{\frac{2n+1}{4 \pi} \frac{(n-2)!}{(n+2)!}}
\int_{-1}^{1} x^{2} (1-x^{2})^{2} \frac{d^{2}}{d x^{2}} \left[ P_{n}(x)
              \right] \, dx	\\
%\label{eq:apps:vibsphereanm}
	&= \left\{ \begin{array}{rl}
\frac{-i \sqrt{\pi}}{14 \sqrt{30}}
	& \mbox{if $n = 2$,}	\\
\frac{-i \sqrt{\pi}}{21 \sqrt{10}}
	& \mbox{if $n = 4$,}	\\
0	& \mbox{otherwise.}
\end{array} \right.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution (4 of 5)}

This result further reduces the infinite series solution to the
vibrating sphere problem to an exact solution containing only a finite
number of terms,
\begin{align*}
u(\varphi, \theta, t)
	= {} & \frac{i \sqrt{\pi}}{14 \sqrt{30}} \cos (\sqrt{6} t) \left(
   Y_{2}^{2}(\varphi, -\theta) - Y_{2}^{2}(\varphi, \theta) \right)	\\
	&+ \frac{i \sqrt{\pi}}{21 \sqrt{10}} \cos (\sqrt{20} t)
   \left( Y_{4}^{2}(\varphi, -\theta) - Y_{4}^{2}(\varphi, \theta) \right)	\\
	= {} & \left( \frac{1}{168} \cos (\sqrt{6} t) P_{2}^{2}(\cos \varphi) +
\frac{1}{420} \cos (2 \sqrt{5} t) P_{4}^{2}(\cos \varphi) \right) \sin (2 \theta)	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution (5 of 5)}

\begin{center}
\resizebox{0.75\textwidth}{!}{\includegraphics{vibsphere}}
\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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