%%%%		-*- 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:      03/28/2025
%%%%
%%%% Purpose:   Solving Elliptic PDEs Using Green's Functions
%%%%

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\title{Solving Elliptic PDEs Using Green's Functions}
\subtitle{\textit{Partial Differential Equations}} % (optional)


\author % (optional, use only with lots of authors)
{J~Robert~Buchanan}
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{
  Department of Mathematics
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% Since this a solution template for a generic talk, very little can
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%\section{Introduction}

\begin{frame}\frametitle{Objectives}

In this lesson we will:
\begin{itemize}
\item solve Poisson's equation in two dimensions,
\item solve Poisson's equation in three dimensions,
\item introduce the method of images, and
\item solve Helmholtz's equation in two dimensions.
\end{itemize}

\end{frame}
%%

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

Consider Poisson's equation on a domain $D \subset \mathbb{R}^{2}$
with boundary $\partial D$.
%\label{eq:green:poisson:BVP}
\begin{align*}
L[u]
	=\Delta u
	&= \phi
\mbox{ for $(x, y) \in D$}	\\
u
	&= f
\mbox{ for $(x, y) \in \partial D$}
\end{align*}
If $G$ is the Green's function, then
\[
\iint_{D} G \phi \, dA
	= \oint_{\partial D} (u_{\mathbf{N}} G - u G_{\mathbf{N}}) \,
        ds + \iint_{D} L^{\ast}[G] u \, dA	.
\]
Since $L^{\ast}[G] = \delta(\xi-x, \eta-y)$ then this equation
simplifies further to
\[
\iint_{D} G \phi \, dA
	= \oint_{\partial D} (u_{\mathbf{N}} G - u G_{\mathbf{N}}) \,
        ds + u(x, y)	.
\]
The boundary conditions specified above are of the Dirichlet type so
$u = f$ on $\partial D$.
However, $u_{\mathbf{N}}$ is not specified on $\partial D$, so
assume $G = 0$ on $\partial D$.
\[
%\label{eq:green:poisson:BVPsoln}
u(x, y)
	= \iint_{D} G \phi \, dA + \oint_{\partial D} f G_{\mathbf{N}}
        \, ds	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Principal Solution}

\[
G(x, y; \xi, \eta)
	= U(x, y; \xi, \eta) + g(x, y; \xi, \eta)
\]
The principal solution must satisfy the partial differential equation,
\[
\Delta U
	= U_{\xi\xi} + U_{\eta\eta}
	= \delta(\xi-x, \eta-y)
\]
on $D$.
This equation has a physical interpretation in the setting of the
gravitational potential.
Imagine a unit point mass is fixed at $(x, y)$, then $U$ is
the gravitational potential at $(\xi, \eta)$ induced by the point mass.

\end{frame}
%%

\begin{frame}\frametitle{Polar Coordinates}

Convert this partial differential equation to polar
coordinates with the origin at $(x, y)$.
\begin{align*}
\xi - x
	&= r \cos \theta	\\
\eta - y
	&= r \sin \theta
\end{align*}
Since the gravitational potential induced at $(\xi, \eta)$ will depend
only on the distance from $(x, y)$ to $(\xi, \eta)$ which is coordinate $r$,
the Laplacian of $U$ can be written as
\[
U_{rr} + \frac{1}{r} U_{r}
	= \frac{1}{r} (r U_{r})_{r}
	= \delta(r)	.
\]
When $r > 0$ the right hand side is $0$,
so a general solution is of the form $U(r) = c_{1} \ln r + c_{2}$.
\end{frame}
%%

\begin{frame}\frametitle{Principal Solution at $r = 0$}

Are there values of the constants
$c_{1}$ and $c_{2}$ which permit the equation $\Delta U = \delta(r)$
to be satisfied?

\vspace*{0.10in}

Define $U_{\epsilon}(r) = c_{1} \ln (r + \epsilon) +
c_{2}$ for $\epsilon > 0$.
\[
%\label{eq:green:poisson:principalgreen}
\Delta U_{\epsilon}
	= \frac{1}{r} \left[ r (U_{\epsilon})_{r} \right]_{r}
	= \frac{c_{1} \epsilon}{r (r+\epsilon)^{2}}
\]
Integrate this Laplacian over $\mathbb{R}^{2}$ and set the result
equal to the integral of the Dirac delta function over
$\mathbb{R}^{2}$.
\[
\iint_{\mathbb{R}^{2}} \Delta U_{\epsilon} \, dA
	= \iint_{\mathbb{R}^{2}} \delta(\xi-x, \eta-y) \, dA
\]
which implies $2 \pi c_{1} = 1$ or
$c_{1} = 1/(2 \pi)$.
Constant $c_{2}$ is arbitrary and for convenience is chosen to be zero.

\end{frame}
%%

\begin{frame}\frametitle{Limit of the Principal Solution}

If $r > 0$,
\[
\Delta U(r)
	= \lim_{\epsilon \to 0^{+}} \Delta U_{\epsilon}(r)
	= \lim_{\epsilon \to 0^{+}} \frac{\epsilon}{2 \pi r (r+\epsilon)^{2}}
	= 0	.
\]
When $r = 0$ the expression for $\Delta U_{\epsilon}$ is indeterminate
of the form $0/0$.
Applying l'H\^{o}pital's rule produces
\[
\lim_{\epsilon \to 0^{+}} \frac{\epsilon}{2 \pi r (r+\epsilon)^{2}}
	= \lim_{\epsilon \to 0^{+}} \frac{1}{4 \pi r (r+\epsilon)}
	= \frac{1}{4 \pi r^{2}}	
\]
which ``blows up'' like the Dirac delta function at $r = 0$.
Hence the principal solution $U(r) = \frac{1}{2 \pi} \ln r$.

\vspace*{0.10in}

Hence we have,
\[
G(x, y; \xi, \eta)
	= \frac{1}{2 \pi} \ln r + g(x, y; \xi, \eta)	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Completing the Green's Function}

The function $g$ must be chosen to satisfy the partial differential
equation $L^{\ast}[g] = \Delta g = 0$ and $G = 0$ on the boundary of
domain $D$.
The solution is tied to the geometry of region $D$.

\vspace*{0.10in}

We will explore using the \textbf{method of images} to find the
Green's function $G$ on a disk of radius $R$.
\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{diskD}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Method of Images}

Suppose a unit point mass is placed at location $Q$ (called the field
point) and another ``negative'' unit mass is placed
at $Q'$ so that along the boundary of disk $D$ the gravitational
potential induced by the two masses vanishes.
Such a potential will satisfy the requirements of a Green's function
on the disk.
\[
%\label{eq:green:poisson:preimagegreen}
G
	= \frac{1}{4 \pi} \ln \left( \frac{R^{2}}{\rho^{2}} \,
          \frac{r^{2}}{(r')^{2}} \right)
\]
Using the change of variables $\xi = \hat{\rho} \cos
\hat{\theta}$ and $\eta = \hat{\rho} \sin \hat{\theta}$ then
\[
%\label{eq:green:poisson:imagegreen}
G
	= \frac{1}{4 \pi} \ln \frac{R^{2} (\hat{\rho}^{2} + \rho^{2} -
          2 \hat{\rho} \rho \cos(\hat{\theta} - \theta))}{R^{4} +
          \hat{\rho}^{2} \rho^{2} - 2 R^{2} \hat{\rho} \rho \cos
          (\hat{\theta} - \theta)}	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Illustration}

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

\end{frame}
%%

\begin{frame}\frametitle{Solution to Poisson's Equation on the Disk}

Note that
$\phi \equiv \phi(\hat{\rho}, \hat{\theta})$, $f \equiv
f(\hat{\theta})$, $dA = \hat{\rho} \, d\hat{\rho} \, d\hat{\theta}$,
and that $ds = R \, d\hat{\theta}$.

\vspace*{0.10in}

The directional derivative $G_{\mathbf{N}}$ where $\mathbf{N}$ is the
unit outward normal vector to $\partial D$ is the partial derivative
of $G$ with respect to $\hat{\rho}$ evaluated at $\hat{\rho} = R$.
\[
%\label{eq:green:poisson:BVPpolarsoln}
u(\rho, \theta)
	= \int_{0}^{2 \pi} \int_{0}^{R} G(\hat{\rho}, \hat{\theta})
        \phi(\hat{\rho}, \hat{\theta}) \hat{\rho} \, d\hat{\rho} \, d\hat{\theta} +
\int_{0}^{2 \pi} f(\hat{\theta}) \left[ \frac{\partial G}{\partial
    \hat{\rho}} \right]_{\hat{\rho} = R} R \, d\hat{\theta}	.
\]
The integral is
often difficult to evaluate exactly, but can often be approximated
numerically to any desired precision.

\end{frame}
%%

\begin{frame}\frametitle{Example}

Consider the Poisson boundary value problem,
\begin{align*}
\Delta u
	&= 1
\mbox{ for $0 \le r < 1$ and $0 \le \theta < 2 \pi$}	\\
u(1, \theta)
	&= \sin \theta
\mbox{ for $0 \le \theta < 2 \pi$.}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution (1 of 2)}

\begin{align*}
u(\rho, \theta)
	= {} & \frac{1}{4 \pi} \int_{0}^{2 \pi} \int_{0}^{1}
               \hat{\rho} \ln \left( \frac{\hat{\rho}^{2} + \rho^{2} -
               2 \hat{\rho} \rho \cos (\hat{\theta} - \theta)}{1 +
               \hat{\rho}^{2} \rho^{2} - 2 \hat{\rho} \rho \cos
               (\hat{\theta} - \theta)} \right) \, d\hat{\rho} \,
               d\hat{\theta}	\\
	&+ \frac{1}{2 \pi} \int_{0}^{2 \pi}  \frac{(1 - \rho^{2}) \sin
   \hat{\theta}}{1 + \rho^{2} - 2 \rho \cos (\hat{\theta} - \theta)} \, d\hat{\theta}	.
\end{align*}

\end{frame}
%%

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

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

\end{frame}
%%

\begin{frame}\frametitle{Poisson's Equation in Half-Space}

Consider Poisson's equation defined on a solid region $D$ in
three-dimensional space
%\label{eq:green:poisson:3D}
\begin{align*}
\Delta u
	&= \phi
\mbox{ for $(x, y, z) \in D$}	\\
u
	&= f
\mbox{ for $(x, y, z) \in \partial D$.}
\end{align*}
Using the Divergence Theorem,
\[
%\label{eq:green:poisson:3DBVPsoln}
u(x, y, z)
	= \iiint_{D} G \phi \, dV + \iint_{\partial D} f
        G_{\mathbf{N}} \, dS	.
\]
The Green's function $G$ in many geometries can be found using the method
of images.

\end{frame}
%%

\begin{frame}\frametitle{Poisson's Equation in Half-Space}

\begin{itemize}
\item Let $D = \{ (x, y, z) \, | \, z \ge 0 \}$.
\item Let the point $\mathbf{x} = (x, y, z)$ be fixed with $z > 0$ and let
point $\bm{\xi} = (\xi, \eta, \zeta)$ be any other point in $\mathbb{R}^{3}$.
\item Denote the distance from $\mathbf{x}$ to $\bm{\xi}$
by $\rho = ((x-\xi)^{2} + (y-\eta)^{2} + (z-\zeta)^{2})^{1/2}$.
\end{itemize}
The principal solution $U$ for this geometry must satisfy the partial
differential equation $\Delta U = \delta(\rho)$.

\vspace*{0.10in}

Consider the Laplacian in spherical coordinates.
$U$ will depend only on distance from point $\mathbf{x}$ and
satisfies
\[
U_{\rho\rho} + \frac{2}{\rho} U_{\rho}
	= \delta(\rho)	.
\]
When $\rho > 0$,
\[
U(\rho)
	= \frac{c_{1}}{\rho} + c_{2}	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Poisson's Equation in Half-Space}

Let $\epsilon > 0$ and define $U_{\epsilon}(\rho) =
c_{1}/(\rho+\epsilon) + c_{2}$ and set
\[
\iiint_{\mathbb{R}^{3}} \Delta U_{\epsilon} \, dV
	= \iiint_{\mathbb{R}^{3}} \delta(\rho) \, dV	.
\]
That is
\[
\iiint_{\mathbb{R}^{3}} \frac{-2 c_{1} \epsilon}{\rho
  (\rho+\epsilon)^{3}} \, dV
	= 1	,
\]
or in spherical coordinates
\[
c_{1} \int_{0}^{2 \pi} \int_{0}^{\pi} \int_{0}^{\infty} \frac{\epsilon
  \rho}{(\rho + \epsilon)^{3}} \sin \varphi \, d\rho \, d\varphi \, d\theta
	= -\frac{1}{2}
\iff        
c_{1}
	= -\frac{1}{4 \pi}	.
\]
For convenience the arbitrary constant $c_{2}$ will be chosen to be zero.
\[
U(\rho)
	= \lim_{\epsilon \to 0^{+}} U_{\epsilon}(\rho)
	= -\frac{1}{4 \pi \rho} 
\]

\end{frame}
%%

\begin{frame}\frametitle{Method of Images}

Include an additional unit point mass
so that $G = 0$ on $\partial D$.
The image point denoted $\mathbf{x}'$ will be the reflection of
$\mathbf{x}$ across the plane where $z = 0$, therefore $\mathbf{x}' = (x, y, -z)$.
Define $\rho' = ((x-\xi)^{2} + (y-\eta)^{2} + (z+\zeta)^{2})^{1/2}$,
and then let
\[
%\label{eq:green:poisson:halfspacegreen}
G(\mathbf{x}; \bm{\xi})
	= \frac{1}{4 \pi \rho'} - \frac{1}{4 \pi \rho}	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Green's Function}

$G = 0$ on $\partial D$ since for
every point $\mathbf{x}_{0} = (x_{0}, y_{0}, 0) \in \partial D$
\begin{align*}
G
	&= \frac{1}{4 \pi ((x_{0}-x)^{2} + (y_{0}-y)^{2} +
z^{2})^{1/2}} - \frac{1}{4 \pi  ((x_{0}-x)^{2} + (y_{0}-y)^{2} + z^{2})^{1/2}}	\\
	&= 0	.
\end{align*}
The unit outward normal of the Green's function along $\partial D$ is
$G_{\mathbf{N}} = -[ \frac{\partial G}{\partial \zeta} ]_{\zeta=0}$.
\begin{align*}
-\frac{\partial G}{\partial \zeta}
	&= \frac{1}{4 \pi (\rho')^{2}} \, \frac{\partial
   \rho'}{\partial \zeta} - \frac{1}{4 \pi \rho^{2}} \, \frac{\partial
   \rho}{\partial \zeta}
	= \frac{z+\zeta}{4 \pi (\rho')^{3}} - \frac{z-\zeta}{4 \pi \rho^{3}}	\\
\left[ -\frac{\partial G}{\partial \zeta} \right]_{\zeta=0}
	&= \frac{z}{2 \pi \rho^{3}}	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Result}

The solution to Poisson's equation
where $D$ is the half-space with $z \ge 0$
can be expressed as the integral,
\[
%\label{eq:green:poisson:3DBVPsolnalt}
u(x, y, z)
	= \frac{1}{4 \pi} \iiint_{\mathbb{R}^{3}} \left(
          \frac{1}{\rho'} - \frac{1}{\rho} \right) \phi \, dV +
\frac{z}{2 \pi} \iint_{\mathbb{R}^{2}} \left( \frac{1}{\rho^{3}} \right) f \, dS	.
\]
If $\phi = 0$ then the solution to Laplace's equation on the
half-space is
\[
%\label{eq:green:laplace:3DBVPsoln}
u(x, y, z)
	= \frac{z}{2 \pi} \iint_{\mathbb{R}^{2}} \left(
          \frac{1}{\rho^{3}} \right) f \, dS	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Solving Helmholtz's Equation}

Consider the Helmholtz equation in a domain $D \subset
\mathbb{R}^{2}$ with boundary $\partial D$,
\begin{align*}
L[u]
	= \Delta u + k^{2} u
	&= \phi
\mbox{ for $(x, y) \in D$}	\\
u
	&= f
\mbox{ for $(x, y) \in \partial D$.}
\end{align*}
We must find the principal solution $U(x, y;
\xi, \eta)$ to the differential equation
\[
\Delta U + k^{2} U
	= \delta(\xi - x, \eta - y)	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Polar Coordinates}

If a unit point mass is located at $(x, y)$ then the function $U$
depends again only on $r = ((\xi-x)^{2} + (\eta-y)^{2})^{1/2}$ and
hence the PDE can be re-written as
\begin{align*}
\frac{1}{r} ( r U_{r} )_{r} + k^{2} U
	&= \delta(r)	\\
r ( r U_{r} )_{r} + k^{2} r^{2} U
	&= r^{2} \delta(r)	\\
r^{2} U_{rr} + r U_{r} + k^{2} r^{2} U
	&= r^{2} \delta(r)	.
\end{align*}
This is the
parametric form of Bessel's equation of order
zero with linearly independent solutions to the homogeneous form, 
$J_{0}(k r)$ and $Y_{0}(k r)$.

\end{frame}
%%

\begin{frame}\frametitle{Dirac Delta Behavior}

Function $J_{0}(k r)$ is finite at $r
= 0$; however, $Y_{0}(k r)$ does grow unbounded as $r
\to 0^{+}$.

\vspace*{0.10in}

Let $D_{\epsilon}$ be a disk of radius $\epsilon > 0$ centered at $(x, y)$.
Integrate both sides of the equation $\delta(r) = \Delta U + k^{2} U$
over the disk.
\begin{align*}
\iint_{D_{\epsilon}} \delta(r) \, dA
	&= c_{1} \iint_{D_{\epsilon}} (\Delta Y_{0}(k r) + k^{2}
   Y_{0}(k r)) \, dA	\\
1
	&= 2 \pi c_{1} \left( \int_{0}^{\epsilon} r (\Delta Y_{0}(k r)) \, dr +
   k^{2} \int_{0}^{\epsilon} r Y_{0}(k r) \, dr \right)
\end{align*}
For small $r > 0$,
\[
Y_{0}(k r)
	\approx \frac{2}{\pi} \ln r	.
\]
The second integral vanishes as $\epsilon \to 0^{+}$ since
\[
\int_{0}^{\epsilon} r Y_{0}(k r) \, dr
	\approx \frac{2}{\pi} \int_{0}^{\epsilon} r \ln r \, dr
	= \frac{1}{\pi} \left( \epsilon^{2} \ln \epsilon -
   \frac{1}{2} \epsilon^{2} \right)
	\to 0
\]
as $\epsilon \to 0^{+}$.
We can show $U(r) = \frac{1}{4} Y_{0}(k r)$.

\end{frame}
%%

\begin{frame}\frametitle{Green's Function for Helmholtz Operator}

The Green's function for the Helmholtz operator can be written as the
sum of the principal solution $U$ and a function $g(x, y; \xi, \eta)$
such that
\[
L^{\ast}[g]
	= g_{\xi\xi} + g_{\eta\eta} + k^{2} g
	= 0	.
\]
Since $G = 0$ on $\partial R$ then $g = -\frac{1}{4} Y_{0}(k r)$ on
$\partial R$.

\vspace*{0.10in}

Let $D$ be a disk centered at the origin with radius $R$.
Let $\xi = \hat{\rho} \cos \hat{\theta}$
and $\eta = \hat{\rho} \sin \hat{\theta}$, then the boundary value problem
can be written as
%\label{eq:green:helmholtz:polarg}
\begin{align*}
g_{\hat{\rho}\hat{\rho}} + \frac{1}{\hat{\rho}} g_{\hat{\rho}} +
  \frac{1}{\hat{\rho}^{2}} g_{\hat{\theta}\hat{\theta}} + k^{2} g
	&= 0	\\
g(R, \hat{\theta})
	&= -\frac{1}{4} Y_{0}\left( k \sqrt{R^{2} + \rho^{2} - 2 R \rho \cos
          (\hat{\theta} - \theta)} \right)	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Product Solution}

Assume a product solution of the form $g(\hat{\rho}, \hat{\theta}) =
\Rho(\hat{\rho}) \Theta(\hat{\theta})$ then
\[
\frac{\hat{\rho}^{2} \Rho''(\hat{\rho}) + \hat{\rho} \Rho'(\hat{\rho}) + (k
  \hat{\rho})^{2} \Rho(\hat{\rho})}{\Rho(\hat{\rho})}
	= \frac{-\Theta''(\hat{\theta})}{\Theta(\hat{\theta})}
	= n^{2}	,
\]
where $n$ is a nonnegative integer since the $\hat{\theta}$-dependent
factor of the solution must be $2 \pi$-periodic.
\[
\hat{\rho}^{2} \Rho''(\hat{\rho}) + \hat{\rho} \Rho'(\hat{\rho}) + [ (k \hat{\rho})^{2} -
n^{2} ] \Rho(\hat{\rho})
	= 0
\]
This is the Bessel equation of order $n$ with a solution defined on the
disk of radius $R$ of the form
\[
\Rho_{n}(\hat{\rho})
	= J_{n}(k \hat{\rho})
\mbox{ for $n = 0, 1, \ldots$.}
\]

\end{frame}
%%

\begin{frame}\frametitle{Product Solution}

\[
\Theta_{n}(\hat{\theta})
	= a_{n} \cos (n \hat{\theta}) + b_{n} \sin (n \hat{\theta})	.
\]
Consequently the term $g$ can be expressed as
\[
%\label{eq:green:helmholtz:polargsoln}
g(\rho, \theta; \hat{\rho}, \hat{\theta})
	= \frac{a_{0}}{2} J_{0}(k \hat{\rho}) + \sum_{n=1}^{\infty} \left[ J_{n}(k
          \hat{\rho}) (a_{n} \cos (n \hat{\theta}) + b_{n} \sin (n
          \hat{\theta})) \right]	.
\]
The coefficients $a_{0}$, $a_{n}$, and $b_{n}$ can be found using the
Euler-Fourier formulas.

\end{frame}
%%

\begin{frame}\frametitle{Example}

Suppose a circular membrane of radius $R = 1$ is centered at the origin.
The membrane is subject to a loading force and boundary conditions as
described by the following boundary value problem.
\begin{align*}
\Delta u + u
	&= \rho (1-\rho) \sin \theta
\mbox{ for $0 \le \rho < 1$ and $0 \le \theta < 2 \pi$}	\\
u(1, \theta)
	&= 0
\mbox{ for $0 \le \theta < 2 \pi$.}
\end{align*}
Find the displacement $u(\rho, \theta)$ of the membrane.

\end{frame}
%%

\begin{frame}\frametitle{Solution (1 of 2)}

The principal solution is
\[
U(\rho, \theta; \hat{\rho}, \hat{\theta})
	= \frac{1}{4} Y_{0}(\sqrt{\hat{\rho}^{2} + \rho^{2} - 2
          \hat{\rho} \rho \cos(\hat{\theta} - \theta)})	.
\]
The function
\[
%\label{eq:green:helmholtz:polargsoln}
g(\rho, \theta; \hat{\rho}, \hat{\theta})
	= \frac{a_{0}}{2} J_{0}(\hat{\rho}) + \sum_{n=1}^{\infty}
        \left[ J_{n}(\hat{\rho}) (a_{n} \cos (n \hat{\theta}) + b_{n}
          \sin (n \hat{\theta})) \right]	.
\]
The Green's function is
\[
G(\rho, \theta; \hat{\rho}, \hat{\theta})
	= U(\rho, \theta; \hat{\rho}, \hat{\theta}) + g(\rho, \theta;
        \hat{\rho}, \hat{\theta})	.
\]
The solution to the boundary value problem is
\[
u(\rho, \theta)
	= \int_{0}^{2 \pi} \int_{0}^{1} \left( U(\rho, \theta;
          \hat{\rho}, \hat{\theta}) + g(\rho, \theta;
          \hat{\rho}, \hat{\theta}) \right) (1-\hat{\rho})
        \sin \hat{\theta} \hat{\rho} \, d\hat{\rho} \,d\hat{\theta}	.
\]

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