%%%%		-*- 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:      02/14/2025
%%%%
%%%% Purpose:   Laplace's Equation on a Disk
%%%%

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\title{Laplace's Equation on a Disk}
\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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\subject{Talks}
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%\begin{frame}
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  % 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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% - Exactly two or three sections (other than the summary).
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%   15 and 30 frames, all told.

%\section{Introduction}

\begin{frame}\frametitle{Laplace's Equation on a Disk}

Consider the Dirichlet boundary value problem:
\begin{align*}
\Delta u
	&= 0
\mbox{ for $x^{2} + y^{2} < a^{2}$}	\\
u(x, y)
	&= \phi(x, y)
\mbox{ for $x^{2} + y^{2} = a^{2}$.}
\end{align*}

\textbf{Remark}: since the boundary of $\Omega$ is not a rectangle, we
cannot use separation of variables directly.
Instead we must convert to polar coordinates before using separation
of variables.

\vspace*{0.10in}

\uncover<2->{%
In polar coordinates $(r, \theta)$ the Laplacian operator can be
expressed as
\[
\Delta u
	= u_{rr} + \frac{1}{r} u_{r} + \frac{1}{r^{2}} u_{\theta\theta}	,
\]
for $r > 0$.
}

\end{frame}
%%

\begin{frame}\frametitle{Dirichlet BVP in Polar Coordinates}

\begin{align*}
u_{rr} + \frac{1}{r} u_{r} + \frac{1}{r^{2}} u_{\theta\theta}
	&= 0
\mbox{ for $0 < r < a$ and $-\infty < \theta < \infty$}	\\
u(a, \theta)
	&= f(\theta) = \phi(a \cos \theta, a \sin \theta)
\end{align*}

\vspace*{0.10in}

\textbf{Remarks}:
\begin{itemize}
\item The boundary conditions are periodic in $\theta$ with period $2
  \pi$.
\item The solution $u(r, theta)$ should be $2\pi$-periodic in
  $\theta$.
\item The Laplacian is not defined for $r = 0$, but we wish for the
  solution to remain finite as $r \to 0^{+}$.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Separation of Variables in Polar Coordinates}

Assume $u(r, \theta) = R(r) T(\theta)$, then
\begin{align*}
R''(r) T(\theta) + \frac{1}{r} R'(r) T(\theta) + \frac{1}{r^{2}} R(r)
  T''(\theta)
	&= 0	\\
r^{2} \left( \frac{R''(r) T(\theta)}{R(r) T(\theta)} + \frac{1}{r} \frac{R'(r)
  T(\theta)}{R(r) T(\theta)} + \frac{1}{r^{2}} \frac{R(r)
  T''(\theta)}{R(r) T(\theta)} \right)
	&= 0 \cdot r^{2}	\\
\frac{r^{2} R''(r)}{R(r)} + \frac{r \, R'(r)}{R(r)}
	=
-\frac{T''(\theta)}{T(\theta)}
	&= c
\end{align*}
where $c$ is a constant.

\end{frame}
%%

\begin{frame}\frametitle{Implied Ordinary Differential Equations}

\[
T''(\theta) + c \, T(\theta)
	= 0
\]
\begin{description}
\item[case $c=0$:] the only nontrivial $2 \pi$-periodic solution is
\[
T_{0}(\theta)
	= A_{0}
\]
\item[case $c = \lambda^{2} > 0$:] the only nontrivial $2
  \pi$-periodic solution is
\[
T_{n}(\theta)
	= A_{n} \cos (n \theta) + B_{n} \sin (n \theta)
\]
and $c = \lambda_{n}^{2} = n^{2}$ for $n \in \mathbb{N}$.
\end{description}

\end{frame}
%%

\begin{frame}\frametitle{Euler's Equation}

The implied ordinary differential equation for $R(r)$ is
\[
r^{2} R''(r) + r \, R'(r) - c \, R(r)
	= 0	,
\]
which is known as \textbf{Euler's equation}.
\begin{description}
\item[case $c=0$:]
\begin{align*}
0
	&= r^{2} R''(r) + r \, R'(r)	\\
R_{0}(r)
	&= C_{0} \ln r + D_{0}
\end{align*}
\item[case $c = n^{2} > 0$:]
\begin{align*}
0
	&= r^{2} R''(r) + r \, R'(r) - n^{2} \, R(r)	\\
R_{n}(r)
	&= C_{n} r^{-n} + D_{n} r^{n}
\end{align*}
\end{description}
In order for the product solution to be bounded as $r \to 0^{+}$ we
must choose $C_{0} = C_{1} = \cdots = 0$.

\end{frame}
%%

\begin{frame}\frametitle{Product Solution}

Define the function
\[
u_{n}(r, \theta)
	= R_{n}(r) T_{n}(\theta)
	= r^{n} \left[ A_{n} \cos (n \theta) + B_{n} \sin (n \theta) \right]
\]
for $n = 0, 1, \ldots$.

\vspace*{0.10in}

By the Principle of Superposition the function
\[
u(r, \theta)
	= A_{0} + \sum_{n=1}^{\infty} r^{n} \left[ A_{n} \cos (n
  \theta) + B_{n} \sin (n \theta) \right]
\]
satisfies the Laplacian on the disk for any choice of constants
$A_{0}$, $A_{n}$, and $B_{n}$.

\vspace*{0.10in}

Fourier series techniques can be used to satisfy the boundary condition.

\end{frame}
%%

\begin{frame}\frametitle{Boundary Condition}

Since $u(a, \theta) = f(\theta)$ then
\begin{align*}
f(\theta)
	&= A_{0} + \sum_{n=1}^{\infty} a^{n} \left[ A_{n} \cos (n
  \theta) + B_{n} \sin (n \theta) \right]	\\
\uncover<2->{%
A_{0}
	&= \frac{1}{2 \pi} \int_{-\pi}^{\pi} f(\theta) \, d\theta
	= \frac{\alpha_{0}}{2}
  \\
a^{n} A_{n}
	&= \frac{1}{\pi} \int_{-\pi}^{\pi} f(\theta) \cos (n
                  \theta) \, d\theta
	= \alpha_{n}	\\
a^{n} B_{n}
	&= \frac{1}{\pi} \int_{-\pi}^{\pi} f(\theta) \sin (n
                  \theta) \, d\theta
	= \beta_{n}
}
\end{align*}
\uncover<3->{%
and
\[
u(r, \theta)
	= \frac{\alpha_{0}}{2} + \sum_{n=1}^{\infty} \left(
  \frac{r}{a} \right)^{n} \left[ \alpha_{n} \cos (n \theta) +
  \beta_{n} \sin (n \theta) \right]	.
\]
}

\end{frame}
%%

\begin{frame}\frametitle{Example}

Solve Laplace's equation on the unit disk with the following Dirichlet
boundary condition.
\begin{align*}
\Delta u
	&= 0
\mbox{ for $x^{2} + y^{2} < 1$}	\\
u(1, \theta)
	&= \pi - \theta
\mbox{ for $-\pi < \theta < \pi$}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution}

The solution takes the form
\[
u(r, \theta)
	= \frac{\alpha_{0}}{2} + \sum_{n=1}^{\infty} r^{n} \left[
  \alpha_{n} \cos (n \theta) + \beta_{n} \sin (n \theta) \right]
\]
where
\begin{align*}
\alpha_{0}
	&= \frac{1}{\pi} \int_{-\pi}^{\pi} (\pi - \theta) \, d\theta
	= 2 \pi	\\
\alpha_{n}
	&= \frac{1}{\pi} \int_{-\pi}^{\pi} (\pi - \theta) \cos (n \theta) \,
              d\theta
	= 0	\\
\beta_{n}
	&= \frac{1}{\pi} \int_{-\pi}^{\pi} (\pi - \theta) \sin (n \theta) \,
              d\theta
	= \frac{2 (-1)^{n}}{n}	.
\end{align*}
\uncover<2->{%
\[
u(r, \theta)
	= \pi + \sum_{n=1}^{\infty} \frac{2 (-1)^{n} r^{n}}{n} \sin (n \theta)
\]
}

\end{frame}
%%

\begin{frame}\frametitle{Illustration}

\[
u(r, \theta)
	= \pi + \sum_{n=1}^{\infty} \frac{2 (-1)^{n} r^{n}}{n} \sin (n \theta)
\]
\begin{center}
\resizebox{0.70\textwidth}{!}{\includegraphics{example02}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Summing the Series}

It is possible to sum the Fourier series by recalling the geometric
series and using some complex number arithmetic.

\begin{itemize}
\item Geometric Series: $\displaystyle \sum_{n=0}^{\infty} a \, r^{n} =
\frac{a}{1-r}$ if $|r| < 1$.
\item Euler's Identity: $e^{i \, \theta} = \cos \theta + i \, \sin \theta$.
\item If $z = a + i \, b$ then $|z| = \sqrt{z \, \overline{z}} =
  \sqrt{a^{2} + b^{2}}$.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Complex Arithmetic (1 of 3)}

Suppose $z = a + i \, b$ where $a, b \in \mathbb{R}$ and $i = \sqrt{-1}$.
The geometric series in $z$ is
\[
\sum_{n=0}^{\infty} (-z)^{n}
	= \frac{1}{1+z}
\mbox{ (if $|z|<1$).}
\]
Integrate both sides of this equation with respect to $z$.
\begin{align*}
\sum_{n=0}^{\infty} \frac{(-1)^{n} z^{n+1}}{n+1}
	&= \ln(1+z)
\mbox{ (and re-index)}	\\
-\sum_{n=1}^{\infty} \frac{(-z)^{n}}{n}
	&= \ln(1+z)
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Complex Arithmetic (2 of 3)}

In polar coordinate form $z = r \, e^{i \, \theta}$ where $r =
\sqrt{a^{2} + b^{2}} = |z|$ and $\theta = \tan^{-1} (b/a)$.
\begin{align*}
-\sum_{n=1}^{\infty} \frac{(-z)^{n}}{n}
	&= \ln(1+z)	\\
\uncover<2->{%
-\sum_{n=1}^{\infty} \frac{1}{n} (-r e^{i \theta})^{n}
	&= \ln(1+r e^{i \theta})
\mbox{ (if $|r|<1$)}	\\
-\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} r^{n} e^{i n \theta}
	&= \ln(1+r e^{i \theta})
\mbox{ (find imaginary part)}	\\
-\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} r^{n} \Im{e^{i n \theta}}
	&= \Im{\ln(1+r e^{i \theta})}	\\
\uncover<3->{%
-\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} r^{n} \sin (n \theta)
	&= \Im{\ln(1 + r \cos \theta + i r \sin \theta)}
}
}
\end{align*}
\uncover<4->{%
\textbf{Remark}:
we must determine the natural logarithm of a complex number so that we
may find its imaginary part.
}

\end{frame}
%%

\begin{frame}\frametitle{Natural Logarithm of a Complex Number}

Suppose $z = a + i b \in \mathbb{C}$ and $w = \alpha + i \beta \in
\mathbb{C}$ such that
\[
e^{w} = z
\]
then $w = \ln z$.
\uncover<2->{%
\begin{align*}
e^{\alpha + i \beta}
	&= a + i b	\\
e^{\alpha} e^{i \beta}
	&= a + i b	\\
e^{\alpha} ( \cos \beta + i \sin \beta )
	&= a + i b
\end{align*}
Thus $a = e^{\alpha} \cos \beta$ and $b = e^{\alpha} \sin \beta$ and
\begin{align*}
a^{2} + b^{2}
	&= e^{2 \alpha}
\implies \alpha = \ln \sqrt{a^{2} + b^{2}}	\\
\frac{b}{a}
	&= \tan \beta
\implies \beta = \tan^{-1} \frac{b}{a}	.
\end{align*}
Thus
\[
\ln z
	= \ln \sqrt{a^{2} + b^{2}} + i \tan^{-1} \frac{b}{a}	.
\]
}

\end{frame}
%%

\begin{frame}\frametitle{Complex Arithmetic (3 of 3)}

Returning to a previous equation:
\begin{align*}
-\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} r^{n} \sin (n \theta)
	&= \Im{\ln(1 + r \cos \theta + i r \sin \theta)}	\\
-\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} r^{n} \sin (n \theta)
	&= \tan^{-1} \left( \frac{r \sin \theta}{1 + r \cos \theta} \right)
  \\
2 \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} r^{n} \sin (n \theta)
	&= -2 \tan^{-1} \left( \frac{r \sin \theta}{1 + r \cos \theta} \right)
  \\
u(r, \theta)
	= \pi + 2 \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} r^{n} \sin (n \theta)
	&= \pi - 2 \tan^{-1} \left( \frac{r \sin \theta}{1 + r \cos \theta} \right)	.
\end{align*}
Thus we have summed the infinite series.

\end{frame}
%%

\begin{frame}\frametitle{Laplace's Equation on a Sector of an Annulus}

Find the solution to the following boundary value problem.
\begin{align*}
\Delta u
	&= 0
\mbox{ for $1 < x^{2} + y^{2} < 4$
with $x > 0$ and $y > 0$}	\\
u(x,0)
	&= 0	\\
u(0,y)
	&= 0	\\
u(x, y)
	&= 2 x y
\mbox{ for $x^{2} + y^{2} = 1$}	\\
u(x, y)
	&= \left( \frac{\pi}{2} - \tan^{-1}
             \frac{y}{x} \right) \tan^{-1} \frac{y}{x}
\mbox{ for $x^{2} + y^{2} = 4$}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Polar Coordinates}

\begin{align*}
u_{rr} + \frac{1}{r} u_{r} + \frac{1}{r^{2}} u_{\theta\theta}
	&= 0
\mbox{ for $1 < r < 2$ with $0 < \theta < \pi/2$}	\\
u(r, 0)
	&= 0	\\
u(r, \pi/2)
	&= 0	\\
u(1, \theta)
	&= \sin (2 \theta)	\\
u(2, \theta)
	&= \theta \left( \frac{\pi}{2} - \theta \right)
\end{align*}
We will again use separation of variables.

\end{frame}
%%

\begin{frame}\frametitle{Separation of Variables}

Assuming $u(r, \theta) = R(r) T(\theta)$ then
\begin{align*}
R''(r) T(\theta) + \frac{1}{r} R'(r) T(\theta) + \frac{1}{r^{2}} R(r)
  T''(\theta)
	&= 0	\\
\frac{r^{2} R''(r)}{R(r)} + \frac{r R'(r)}{R(r)}
	&= -\frac{T''(\theta)}{T(\theta)} = c	.
\end{align*}
This implies the following boundary value problem for the angular
factor of the solution.
\begin{align*}
T''(\theta) + c T(\theta)
	&= 0	\\
T(0)
	&= 0	\\
T(\pi/2)
	&= 0
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Eigenfunctions and Eigenvalues}

The only nontrivial solutions to
\begin{align*}
T''(\theta) + c T(\theta)
	&= 0	\\
T(0)
	&= 0	\\
T(\pi/2)
	&= 0
\end{align*}
are
\[
T_{n}(\theta)
	= \sin (2 n \theta)
\]
with $c = \lambda_{n}^{2} = 4 n^{2}$ for $n \in \mathbb{N}$.

\vspace*{0.10in}

\uncover<2->{%
Using the eigenvalues in the radial factor of the product solution
yields
\begin{align*}
r^{2} R''(r) + r \, R'(r) - 4 n^{2} R(r)
	&= 0	\\
R_{n}(r)
	&= A_{n} r^{-2n} + B_{n} r^{2n}	.
\end{align*}
}

\end{frame}
%%

\begin{frame}\frametitle{Product Solution}

Define the product solution
\[
u_{n}(r, \theta)
	= R_{n}(r) T_{n}(\theta)
	= \left( A_{n} r^{-2n} + B_{n} r^{2n} \right) \sin (2 n \theta)	.
\]
By the Principle of Superposition a linear combination of product
solutions will also solve Laplace's equation and satisfy the boundary
conditions at $\theta = 0$ and $\theta = \pi/2$.
\[
u(r, \theta)
	= \sum_{n=1}^{\infty} \left( A_{n} r^{-2n} + B_{n} r^{2n} \right) \sin (2 n \theta)
\]
\uncover<2->{%
Now we must choose the coefficients $A_{n}$ and $B_{n}$ so that the
boundary conditions at $r = 1$ and $r = 2$ are satisfied.
Fourier series will be employed.
}

\end{frame}
%%

\begin{frame}\frametitle{Boundary Condition at $r = 1$}

\begin{align*}
u(1, \theta)
	&= \sum_{n=1}^{\infty} \left( A_{n} + B_{n} \right)
                  \sin (2 n \theta)	\\
\sin (2 \theta)
	&= \sum_{n=1}^{\infty} \left( A_{n} + B_{n} \right)
                  \sin (2 n \theta)
\end{align*}
The last equation implies the system of equations
\begin{align*}
A_{1} + B_{1}
	&= 1	\\
A_{n} + B_{n}
	&= 0
\mbox{ for $n \ge 2$.}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Boundary Condition at $r = 2$}
\fontsize{9pt}{10.8}\selectfont

\begin{align*}
u(2, \theta)
	&= \sum_{n=1}^{\infty} \left( A_{n} 2^{-2n}+ B_{n} 2^{2n} \right) \sin (2 n \theta)	\\
\theta \left( \frac{\pi}{2} - \theta \right)
	&= \sum_{n=1}^{\infty} \left( A_{n} 2^{-2n} + B_{n} 2^{2n} \right) \sin (2 n \theta)
\end{align*}
Multiply both sides of the last equation by $\sin (2 m \theta)$ and
integrate over $[0, \pi/2]$.
\begin{align*}
\int_{0}^{\pi/2} \theta \left( \frac{\pi}{2} - \theta \right) \sin (2
  m \theta) \, d\theta
	&= \sum_{n=1}^{\infty} \left( A_{n} 2^{-2n} + B_{n} 2^{2n} \right)
              \int_{0}^{\pi/2} \sin (2 n \theta) \sin (2 m \theta) \,
              d\theta	\\
\frac{1 - (-1)^{m}}{4 m^{3}}
	&= \frac{\pi/2}{2} \left( A_{m} 2^{-2m} + B_{m} 2^{2m} \right)	\\
\frac{1 - (-1)^{m}}{\pi m^{3}}
	&= A_{m} 2^{-2m} + B_{m} 2^{2m}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Systems of Equations (1 of 3)}

\begin{align*}
A_{1} + B_{1}
	&= 1	\\
\frac{1}{4} A_{1} + 4 B_{1}
	&= \frac{2}{\pi}
\end{align*}
which implies
$\displaystyle A_{1} = \frac{16 \pi - 8}{15 \pi}$
and
$\displaystyle B_{1} = \frac{8 - \pi}{15 \pi}$.

\end{frame}
%%

\begin{frame}\frametitle{Systems of Equations (2 of 3)}

For $n = (2k+1)$ (\textit{i.e.}, $n$ odd and greater than 1)
\begin{align*}
A_{2k+1} + B_{2k+1}
	&= 0	\\
2^{-4k-2} A_{2k+1} + 2^{4k+2} B_{2k+1}
	&= \frac{2}{\pi (2k+1)^{3}}
\end{align*}
which implies
\begin{align*}
A_{2k+1}
	&= \frac{2^{4k+3}}{\pi (2k+1)^{3} (2^{4(2k+1)} - 1)}	\\
%
B_{2k+1}
	&= -\frac{2^{4k+3}}{\pi (2k+1)^{3} (2^{4(2k+1)} - 1)}	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Systems of Equations (3 of 3)}

For $n = 2k$ ($n$ even)
\begin{align*}
A_{2k} + B_{2k}
	&= 0	\\
2^{-4k} A_{2k} + 2^{4k} B_{2k}
	&= 0
\end{align*}
which implies $A_{2k} = B_{2k} = 0$.

\uncover<2->{%
\begin{align*}
u(r, \theta)
	= {} & \left( \frac{(16 \pi - 8) r^{-2}}{15 \pi}  + \frac{(8 - \pi)
                  r^{2}}{15 \pi} \right) \sin (2 \theta)	\\
	&+ \sum_{n=1}^{\infty}
\frac{2^{4k+3} (r^{-4k-2} - r^{4k + 2})}{\pi (2k+1)^{3} (2^{4(2k+1)} - 1)} \sin
            ((4n+2) \theta)
\end{align*}
}

\end{frame}
%%

\begin{frame}\frametitle{Illustration}

\begin{center}
\resizebox{0.80\textwidth}{!}{\includegraphics{example03}}
\end{center}

\end{frame}
%%

%\begin{frame}\frametitle{Homework}

%\begin{itemize}
%\item Read Section $6.3$--$6.4$
%\item Exercises: 10--15
%\end{itemize}

%\end{frame}
%%

%%%
\end{document}

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