\documentclass[12pt]{article}
\usepackage{amsmath,amsthm,amsfonts,amssymb}
\usepackage{euscript,graphicx,subfigure,times,version}
\usepackage{epsfig}
\usepackage{amsmath,latexsym}
\usepackage{graphicx}
\include{amsmath_def}
\include{symbols}
\excludeversion{hide}
\parindent=0in

\newcommand{\dispsum}[2]{\displaystyle\sum^{#1}_{#2}}

\title{MAT 280: Laplacian Eigenfunctions: Theory, Applications, and Computations\\
Lecture 6: Diffusions on and Vibrations of a Membrane in 2D/3D--III.
3D Ball}
\author{Lecturer: Naoki Saito\\
        Scribe: Eunghyun Lee/Allen Xue}
\date{April 28, 2007}

\begin{document}
\maketitle

In this lecture, we consider diffusions on and vibrations of a
membrane in 3D ball. Basic references for this lecture are
\cite[Sec. 10.3]{STRAUSS}, \cite[Sec. V. 8, V. 9.1, VII.
5]{COURANT-HILBERT} and \cite[Sec. 6.3]{FOLLAND-FOURIER}.

\ \\
Consider a ball $\Omega $ of radius $a$ in $\Rf^{3}$, i.e.,
$\Omega=B^3_a(0)=\{(x,y,z)\cond x^2 +y^2+z^2 <a^2\}$,
$\partial\Omega=S^2_a$. The wave equation for $u(\bx,t)$ on this
domain is as follows:
\begin{equation}\label{wave-equation}
\left\{\begin{array}{ll}
u_{tt}=c^2\varDelta u & \text{in $\Omega$},\\
u=0 & \text{on $\partial\Omega$},\\
u(\bx,0)=f(\bx),\ u_t(\bx,0)=g(\bx) & \text{in $\Omega$}
\end{array}\right.
\end{equation}
By using the method of separation of variables, i.e., setting
$u(\bx,t)=T(t)\cdot v(\bx)$, we get Dirichlet-Laplacian eigenvalue
problem (see Lecture 4 for more details):
\begin{equation}\label{DLeig}
\left\{\begin{array}{ll}
-\varDelta v=\lambda v & \text{in $\Omega$,}\\
v=0 & \text{on $\partial\Omega$.}
\end{array}\right.
\end{equation}

Now consider the spherical (or polar) coordinates in $\Rf^3$, which
introduce three parameters $(r,\phi,\theta)$ as shown in
Fig.~\ref{fig-spherical}, satisfying the followings:
\begin{equation}\nonumber
\left\{\begin{array}{lcl}
x & = & r\sin\theta\cos\phi,\\
y & = & r\sin\theta\sin\phi,\\
z & = & r\cos\theta.
\end{array}\right.
\end{equation}
where $0\leq r<a$, $-\pi\leq\phi<\pi$, $0\leq\theta\leq\pi$.

\begin{figure}[!h]
 \begin{center}
 \includegraphics[width=3.5in, trim=0.5in 0 0 0]{sphel.eps} %[trim=l b r t]
 \caption{Spherical Coordinates in $\Rf^3$.}
 \label{fig-spherical}
\end{center}
\end{figure}

%\begin{center}
%\epsfig{file=page10002.eps,height=5cm,width=7cm,clip= }
%\end{center}

\ \\
Therefore, we can write
\begin{equation}
\begin{array}{lcl}
0 & = & \varDelta v +\lambda v\\
& = & v_{rr} +\frac{2}{r}v_r +\frac{1}{r^2}\Big(\frac{1}{\sin^2
\theta }v_{\phi \phi} +  \frac{1}{\sin \theta }
\big(v_{\theta}\sin\theta\big)_{\theta}\Big)
\end{array}
\end{equation}

We do the following separation of variables $v=R(r)\cdot
Y(\theta,\varphi)$ and obtain:

\begin{equation*}
0= \lambda r^2   + \frac{r^2 R_{rr}+2r R_r}{R} +
\frac{\frac{1}{\sin^2\theta}Y_{\phi\phi}+\frac{1}{\sin \theta }
\big(Y_{\theta}\sin\theta\big)_{\theta}}{Y}
\end{equation*}

Notice that the first two terms in the right-hand side depend on $r$
only and so does the last term in the right-hand side on $\theta$
and $\phi$ only. This means both of them must be constants whose sum
is equal to zero, i.e.,
\[\lambda r^2   + \frac{r^2 R_{rr}+2r R_r}{R}=+\gamma,\qquad \frac{\frac{1}{\sin^2\theta}Y_{\phi\phi}+\frac{1}{\sin \theta }
\big(Y_{\theta}\sin\theta\big)_{\theta}}{Y}=-\gamma.\]

where $\gamma$ is a constant.

\ \\
Then we have

\begin{equation}\label{R-eqn}
 R_{rr} +\frac{2}{r}R_r +\big(\lambda -
\frac{\gamma}{r^2}\big)R=0,\quad \text{with $R(0)$ is finite and
$R(a)=0$.}
\end{equation}

and
\begin{equation}\label{Y-eqn}
\frac{1}{\sin^2\theta}Y_{\phi\phi}+\frac{1}{\sin \theta }
\big(Y_{\theta}\sin\theta\big)_{\theta} +\gamma Y=0.
\end{equation}

\ \\
\eqref{R-eqn} is similar to Bessel's equation, except the
coefficient of $R_r$ term $\frac{2}{r}$, instead of $\frac{1}{r}$.

\ \\
So, let us use the change of dependent variable:
\[w(r)\define\sqrt{r}R(r),\ \ \text{which gives us }\
R=\frac{1}{\sqrt{r}}w,\ \ \ R_r=-\frac{1}{2}r^{-3/2}w+r^{-1/2}w_r.\]

Then \eqref{R-eqn}$\Rightarrow$
\begin{equation}\label{from-R-w}
w_{rr} + \frac{1}{r}w_r +\Big(\lambda -
 \frac{\gamma+\frac{1}{4}}{r^2}\Big)w=0,\quad \text{with $w(0)$ is
 finite and $w(a)=0$.}
\end{equation}

This is now Bessel's differential equation with
$n=\sqrt{\gamma+\frac{1}{4}}$. So,
\[w(r)=J_{\sqrt{\gamma+\frac{1}{4}}}(\sqrt{\lambda}r)\ \text{ or its
constant multiple.}\] $\Rightarrow$
\[R(r)=\frac{J_{\sqrt{\gamma+\frac{1}{4}}}(\sqrt{\lambda}r)}{\sqrt{r}}.\]

$R(a)=0$ forces us to have
$J_{\sqrt{\gamma+\frac{1}{4}}}(\sqrt{\lambda}a) =0$. We will analyze
this later after we proceed a bit more on $Y$. (By doing so, we get
the possible values of $\gamma$.)

\ \\
Notice that the boundary conditions for $Y$ are:
$Y(\theta,\phi+2\pi)=Y(\theta,\phi)$, i.e., $Y$ is $2\pi$ periodic
in $\phi$. And also $Y(0,\phi)$, $Y(\pi,\phi)$ are finite. Such $Y$
satisfying \eqref{Y-eqn} and the above boundary conditions are
called \emph{spherical harmonics} (see \cite[Sec. 2H]{FOLLAND-PDE},
\cite[Sec. IV.2]{STEIN-WEISS}, and \cite{STEELEY}). Also, see
\cite{NCAR}, \cite{libftsh}, and \cite{Guide} for computational
aspect of spherical harmonics.

\ \\
To solve \eqref{Y-eqn} with these boundary conditions, we do one
step of separation variables as $Y(\theta,\phi)=p(\theta)q(\phi)$ to
get
\[\frac{q_{\phi\phi}}{q} + \frac{\sin\theta(p_\theta\sin\theta)_{\theta}}{p} + \gamma \sin^2\theta
=0.\]

Notice that the first term in the left-hand side depends on $\phi$
only, so does the last two terms in the left-hand side on $\theta$
only. So, both of them must be constant:
\[\frac{q_{\phi\phi}}{q}=-\alpha,\qquad \
\frac{\sin\theta(p_\theta\sin\theta)_{\theta}}{p}+\gamma\sin^2\theta=\alpha.\]
where $\alpha$ is a constant.

\ \\
Now let us solve them one by one.\ \\

First:
\[q_{\phi\phi} + \alpha q=0,\qquad \text{with $q$ is $2\pi$ periodic
in $\phi$,}\]
\[\Rightarrow q(\phi) =A\cos m\phi + B\sin m\phi,\qquad
\alpha=m^2.\]

Second:
\begin{equation}\label{f-1}
\frac{1}{\sin\theta}\frac{d}{d\theta}\Big(\sin\theta\frac{dp}{d\theta}\Big)
+ \Big(\gamma - \frac{m^2}{\sin^2\theta}\Big)p = 0.
\end{equation}

Here we have boundary conditions that $p(0)$ and $p(\pi)$ are
finite.

\ \\
Let us introduce a new variable $s=\cos\theta$, i.e., $\sin^2 \theta
= 1-s^2$. Then \eqref{f-1} becomes

\begin{equation}\label{f-2}
\frac{d}{ds}\Big[(1-s^2)\frac{dp}{ds}\Big] + \Big(\gamma -
\frac{m^2}{1-s^2}\Big)p=0,\quad -1\leq s\leq 1.
\end{equation}

where $p(\pm 1)$ are finite. This is the so-called \emph{associated
Legendre equation}.

\ \\
Note that $m=0$ case corresponds to the usual Legendre equation
whose solutions are the Legendre polynomial $P_n(s)$, where $\gamma
=\ell(\ell+1)$ and $\ell\in\mathbb{N}$. The equation \eqref{f-2} can
be solved via the power series as in Bessel's equation. We need to
omit the details of computation, which can be found in ~\cite[Sec.
6.3]{FOLLAND-FOURIER} and ~\cite{STRAUSS}.

\ \\
Now let $P(s)$ be a general solution for the Legendre equation with
general $\gamma$, $\big((1-s^2)P^{'}\big)^{'} +\gamma P =0$. Then
the solution to \eqref{f-2} can be written as
$p(s)=(1-s^2)^{\frac{m}{2}}P^{(m)}(s)$. If $\gamma=\ell(\ell+1)$,
with $\ell\geq m$ and $\ell\in\mathbb{N}$, then $p(s)$ can be
written as
\[p(s)=P^m_{\ell}(s)\define
\frac{(-1)^m}{2^{\ell}\ell!}(1-s^2)^{\frac{m}{2}}\frac{d^{\ell+m}}{ds^{\ell+m}}(s^2-1)^{\ell}\]

Here $P^m_{\ell} (s)$ is called \emph{the associated Legendre
function}, which is merely a polynomial in $s$ with multiplication
of a power of $\sqrt{1-s^2}$. Also notice that $\sqrt{\gamma
+\frac{1}{4}} = \sqrt{\ell^2+\ell +\frac{1}{4}} =
\sqrt{\Big(\ell+\frac{1}{2}\Big)^2} = \ell+ \frac{1}{2}$.

\ \\
Finally, putting everything together, we have
\begin{eqnarray*}
v(r,\theta, \phi) & = & R(r)p(\theta)q(\phi) \\
& = & \frac{J_{\ell+\frac{1}{2}}(\sqrt{\lambda}
r)}{\sqrt{r}}P^m_{\ell}(\cos\theta)(A\cos m\phi +B \sin m\phi)
\end{eqnarray*}

By replacing $\cos$, $\sin$ by complex exponentials, we can also
write a basic solution as
\begin{equation*}
\left\{\begin{array}{lcl} v_{\ell m j}(r,\theta, \phi) & = &
\frac{J_{\ell+\frac{1}{2}}(\sqrt{\lambda_{\ell j}}r)}{\sqrt{r}}P^{|m|}_{\ell}(\cos\theta)\e^{im\phi},\\[8pt]
J_{\ell+\frac{1}{2}}(\sqrt{\lambda_{\ell j}}a) & = & 0,\qquad
\lambda_{\ell 1}<\lambda_{\ell 2}<\cdots,\ \text{for each $\ell$}.
\end{array}\right.
\end{equation*}
where $\ell=0,1,\ldots,\infty$, $m=-\ell,\ldots,0,\ldots,\ell$, and
$j=1,2,\ldots,\infty$. Therefore, for each $(\ell,j)$, there exist
$2\ell+1$ eigenfunctions, i.e., $\lambda_{\ell j}$ has $2\ell+1$
multiplicity.

\ \\
Notice that $v_{\ell mj}$ are orthogonal, i.e.,
\begin{eqnarray*}
\langle v_{\ell mj},v_{\ell^{'}m^{'}j^{'}}\rangle &=& \int
^{\pi}_{-\pi}\int ^{\pi}_{0}\int
^{a}_{0}v_{\ell mj}\,v_{\ell^{'} m^{'} j^{'}}\,r^2\sin\theta \dd{r}\dd{\theta}\dd{\phi} \\
 &=& c\,\delta_{\ell\ell^{'}}\delta_{mm^{'}}\delta_{jj^{'}}
\end{eqnarray*}

Finally, we have
\begin{equation*}
\left\{
\begin{array}{lcl}
u(\bx,t) & = & \dispsum{\infty}{\ell=0}
\dispsum{\infty}{j=0}\dispsum{\ell}{m=-\ell}(A_{\ell m
j}\cos\sqrt{\lambda_{\ell j}}ct +B_{\ell m j}\sin\sqrt{\lambda_{\ell
j}}ct)\cdot v_{\ell m j},\\[8pt]
u(\bx,0) & = & f(\bx)=\dispsum{\infty}{\ell=0}
\dispsum{\infty}{j=0}\dispsum{\ell}{m=-\ell}A_{\ell
m j}v_{\ell m j},\\[8pt]
u_t(\bx,0) & = & g(\bx)=\dispsum{\infty}{\ell=0}
\dispsum{\infty}{j=0}\dispsum{\ell}{m=-\ell}c\sqrt{\lambda_{\ell
j}}B_{\ell m j}v_{\ell m j}.
\end{array}\right.
\end{equation*}

$\Rightarrow A_{\ell m j} = \fr{\langle f, v_{\ell m j}\rangle}{\|
v_{\ell m j}\|^2_2},\qquad c\sqrt{\lambda_{\ell j}}B_{\ell m j}=
\fr{\langle g, v_{\ell m j}\rangle}{\|v_{\ell m j}\|^2_2}$.

\ \\
$A_{\ell m j}$, $B_{\ell m j}$ are complex numbers in general and
more simplification happens by $\e^{im\phi}\rightarrow(\cos
m\phi,\sin m\phi)$.

\ \\
\textbf{Example:}  Let $f(\bx)\equiv 0$ and $g(\bx)=g(r)$.

Then $A_{\ell m j}=0, m=0, \ell=0.$ We also know that
$P^0_0(s)=P_0(s)=1$. Therefore

\begin{eqnarray*}
u(\bx,t)& = &
\sum^{\infty}_{j=1}B_j\sin(\sqrt{\lambda_{0j}}ct)v_{00j}\\[8pt]
& = &
\sum^{\infty}_{j=1}B_j\sin(\sqrt{\lambda_{0j}}ct)\cdot\fr{J_{1/2}(\sqrt{\lambda_{0j}}r)}{\sqrt{r}}
\end{eqnarray*}
with $B_j=\fr{1}{\sqrt{\lambda_{0j}}c}\Int{0}{a}r^2
\fr{J_{1/2}(\sqrt{\lambda_{0j}}r)}{\sqrt{r}}g(r)\dd{r}\Bigg/\Int{0}{a}\fr{J^2_{1/2}(\sqrt{\lambda_{0j}}r)}{r}r^2\dd{r}$.

\ \\
Amazingly, in this case $J_{1/2}(x)=\sqrt{\frac{2}{\pi x}}\sin x$
(see Appendix). Thus
\[u(\bx,t)=\sqrt{\fr{2}{\pi}}\sum^{\infty}_{j=1}
B_j\sin(\sqrt{\lambda_{0j}}ct)\cdot\fr{\sin(\sqrt{\lambda_{0j}}r)}{\sqrt{\lambda_{0j}}r},\]
where $\sqrt{\lambda_{0j}}a=j\pi,\quad j=1,2,\cdots,$ by the
Dirichlet boundary condition.

\ \\
Therefore, $u(\bx,t)=\sqrt{\fr{2}{\pi}}\dispsum{\infty}{j=1}
B_j\sin\left(\fr{c\pi jt}{a}\right)\cdot\fr{\sin\left(\fr{j\pi
r}{a}\right)}{\fr{j\pi r}{a}}.$

\ \\
We get the fundamental frequency
$\sqrt{\lambda_{01}}c=\fr{\pi}{a}c$, which is the same as that of
the 1D string of length $a$.

\ \\
\textbf{Appendix}

To derive the formula $J_{1/2}(x)=\sqrt{\fr{2}{\pi x}}\sin x$, we
need the series definition of the Bessel function of first kind of
order $\alpha$ as follows
\[J_{\alpha}(x)=\sum^{\infty}_{m=0}\fr{(-1)^m}{m!\,\Gamma(m+\alpha+1)}\left(\fr{x}{2}\right)^{2m+\alpha}.\]

Let $\alpha=1/2$. Since for the gamma function $\Gamma(z)$, we have
\[\left\{\begin{array}{l}
\Gamma(z+1)=z\Gamma(z)\\
\Gamma(\frac{1}{2})=\sqrt{\pi}
\end{array}\right.
\]
\[\Rightarrow
\Gamma\left(\fr{1}{2}+m+1\right)=\fr{1}{2}\cdot\left(\fr{1}{2}+1\right)\cdot\
\cdots\ \cdot\left(\fr{1}{2}+m\right).\]

We have
\[\begin{array}{lcl}
J_{1/2}(x) & = &
\sqrt{\fr{2}{x}}\dispsum{\infty}{m=0}\fr{(-1)^m}{m!\,\Gamma\left(\fr{1}{2}+m+1\right)}\left(\fr{x}{2}\right)^{2m+1}\\[10pt]
& = & \sqrt{\fr{2}{\pi
x}}\dispsum{\infty}{m=0}\fr{(-1)^m}{(2m+1)!}x^{2m+1},
\end{array}\]
from which we get $J_{1/2}(x)=\sqrt{\fr{2}{\pi x}}\sin x$.

%solve the Bessel's differential equation
%\[u''+\fr{1}{x}u'+\left(1-\fr{s^2}{x^2}\right)u=0.\]

%Let $u=\fr{v}{\sqrt{x}}$, we have
%\[v''+\left(1-\fr{s^2-1/4}{x^2}\right)v=0.\]

%Here $s=\frac{1}{2}$, then $v''+v=0$, therefore
%\[v(x)=A\cos x+B\sin x\]

%and solution will be
%\[u(x)=\fr{A\cos x}{\sqrt{x}}+\fr{B\sin x}{\sqrt{x}}.\]

%where $A$ must be $0$ to satisfy $u(0)<\infty$ and $B$ will be set
%to $\sqrt{\frac{2}{\pi}}$ to normalize the solution. Notice that

%\[\Int{0}{\infty}\fr{\sin^2 x}{x}\dd{x}=\fr{\pi}{2}.\]

%The solution is actually denoted by $J_s(x)$, where $s=\frac{1}{2}$.


\begin{thebibliography}{10}

\bibitem{STRAUSS}
{\sc W.~A. Strauss}, {\em Partial Differential Equations: An
Introduction}, Brooks/Cole Publishing Company, 1992.

\bibitem{COURANT-HILBERT}
{\sc R. Courant, D. Hilbert}, {\em Methods of Mathematical Physics},
Vol.~I, Wiley-Interscience, 1953.

\bibitem{FOLLAND-FOURIER}
{\sc G.~B. Folland}, {\em Fourier Analysis and Its Applications},
Brooks/Cole Publishing Company, 1992.

\bibitem{FOLLAND-PDE}
{\sc G. B. Folland}: {\em Introduction to Partial Differential
Equations}, 2nd Ed., Princeton Univ.~Press, 1995.

\bibitem{STEIN-WEISS}
{\sc E.~M. Stein, G. Weiss}, {\em Introduction to Fourier Analysis
on Euclidean Spaces}, Princeton Univ. Press, 1971.

\bibitem{STEELEY}
{\sc R. T. Steeley}: ``Spherical Harmonics,'' {\em Amer.
Math. Monthly}, vol.~73, no.~4, pp.~115-121, 1966.

\bibitem{NCAR} Spherepack at NCAR:\\
http://www.cisl.ucar.edu/css/software/spherepack

\bibitem{libftsh}
Software for a fast transform for spherical harmonics by Martin Mohlenkamp: \\
http://www.math.ohiou.edu/~mjm/research/libftsh.html

\bibitem{Guide}
A User's Guide to Spherical Harmonics:\\ http://www.math.ohiou.edu/~mjm/research/uguide.pdf
\end{thebibliography}


\end{document}
