%%%%		-*- 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:      03/21/2025
%%%%
%%%% Purpose:   Green's Functions for Sturm-Liouville Boundary Value Problems
%%%%

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\title{Green's Functions for Sturm-Liouville Boundary Value Problems}
\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.

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\subject{Talks}
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%\begin{frame}
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%  \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
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%   15 and 30 frames, all told.

%\section{Introduction}

\begin{frame}\frametitle{Objectives}

In this lesson we will:
\begin{itemize}
\item review the Sturm-Liouville boundary value problem,
\item develop a Green's function for the Sturm-Liouville BVP,
\item present some of the properties of the Green's function for the
  Sturm-Liouville BPV, and
\item use the Green's function to solve a Sturm-Liouville BVP.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Sturm-Liouville Boundary Value Problem}

%\label{eq:green:SLBVP:SLBVP}
\begin{align*}
L[u](x)
	= \left[ p(x) u'(x) \right]' + q(x) u(x)
	&= f(x)
\mbox{ for $a < x < b$}	\\
\alpha_{1} u(a) + \beta_{1} u'(a)
	& = 0	\\
\alpha_{2} u(b) + \beta_{2} u'(b)
	& = 0	,
\end{align*}
where $\alpha_{1}^{2} + \beta_{1}^{2} > 0$
and $\alpha_{2}^{2} + \beta_{2}^{2} > 0$.

\end{frame}
%%

\begin{frame}\frametitle{Particular Solution}

Assume $u_{1}(x)$ and $u_{2}(x)$ are linearly independent solutions to
the homogeneous form of the ordinary differential equation.

\vspace*{0.10in}

Define the Wronskian
\[
W[u_{1}, u_{2}](x)
	= u_{1}(x) u_{2}'(x) - u_{1}'(x) u_{2}(x)
\]
A particular solution to the nonhomogeneous ODE
is
\[
u_{p}(x)
	= -u_{1}(x) \int_{\xi_{1}}^{x} \frac{f(y) u_{2}(y)}{p(y)
          W[u_{1}, u_{2}](y)} \, dy +
u_{2}(x) \int_{\xi_{2}}^{x} \frac{f(y) u_{1}(y)}{p(y)
          W[u_{1}, u_{2}](y)} \, dy	,
\]
where $\xi_{1}$ and $\xi_{2}$ are parameters.

\vspace*{0.10in}

\textbf{Remark}: the denominators of the definite integrals above are constant on
interval $(a, b)$
though the value of the constant depends on the
choice of homogeneous solutions $u_{1}(x)$ and $u_{2}(x)$.

\end{frame}
%%

\begin{frame}\frametitle{General Solution}

The general solution to the nonhomogeneous equation can be written as
\[
u(x)
	= c_{1} u_{1}(x) + c_{2} u_{2}(x) + u_{p}(x)	.
\]
This general solution can be written in the compact form
\[
%\label{eq:green:SLBVP:compact1}
u(x)
	= -\int_{z_{1}}^{x} \frac{f(y) u_{1}(x) u_{2}(y)}{p \, W} \, dy +
\int_{z_{2}}^{x} \frac{f(y) u_{1}(y) u_{2}(x)}{p \, W} \, dy
\]
with the appropriate choice of limits of integration $z_{1}$ and $z_{2}$.
These limits of integration are determined by the boundary conditions.
\[
%\label{eq:green:SLBVP:soln2parts}
u(x)
	= \int_{a}^{x} \frac{f(y) u_{1}(y) u_{2}(x)}{p \, W} \, dy +
\int_{x}^{b} \frac{f(y) u_{1}(x) u_{2}(y)}{p \, W} \, dy
\]

\end{frame}
%%

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

\[
%\label{eq:green:SLBVP:green}
G(x; y)
	= \left\{ \begin{array}{rl}
\frac{u_{1}(x) u_{2}(y)}{p \, W}	& \mbox{if $a \le x < y \le b$}	\\
\frac{u_{1}(y) u_{2}(x)}{p \, W}	& \mbox{if $a \le y < x \le b$.}
\end{array} \right.
\]
$G(x; y)$ permits the solution to the boundary
value problem to be expressed concisely
as
\[
%\label{eq:green:SLBVP:gotGreen}
u(x)
	= \int_{a}^{b} G(x; y) f(y) \, dy	.
\]
\textbf{Remark}: the symbol $y$ is a supplemental variable present in
the Green's function which is eliminated from the final solution by
the process of integration.

\end{frame}
%%

\begin{frame}\frametitle{Differential Operator}

Define a general second-order linear differential operator as
\[
%\label{eq:green:linearoperator:general2ndorder}
L[u]
	= \alpha(x) \frac{d^{2} u}{d x^{2}} + \beta(x) \frac{d u}{d x} + \gamma(x) u	.
\]
Let $\mathcal{D}(L)$ denote the domain of the differential operator
$L$, that is, the set of functions $u$ for which $L[u]$ is defined and
which satisfy the imposed boundary conditions.
The \textbf{adjoint operator} to differential
operator $L$ is denoted $L^{\ast}$ and is the operator satisfying the equation:
\[
%\label{eq:green:linearoperator:adjoint}
\langle L[u], v \rangle
	= \langle u, L^{\ast}[v] \rangle	.
\]
The expressions $\langle L[u], v \rangle$ and $\langle u, L^{\ast}[v]
\rangle$ denote an inner product of functions on interval $[a, b]$.

\end{frame}
%%

\begin{frame}\frametitle{Finding the Adjoint Operator}

Suppose $u(x)$ and $v(x)$ are twice differentiable
functions defined on $[a, b]$, and
\begin{align*}
\left\langle \alpha(x) \frac{d^{2} u}{d x^{2}} + \beta(x) \frac{d u}{d x} +
  \gamma(x) u, v \right\rangle
	&= \langle u, L^{\ast}[v] \rangle	\\
\int_{a}^{b} \alpha(x) \frac{d^{2} u}{d x^{2}} v(x) + \beta(x) \frac{d u}{d x} v(x)
  + \gamma(x) u(x) v(x) \, dx
	&= \int_{a}^{b} u(x) L^{\ast}[v](x) \, dx	.
\end{align*}
Using integration by parts yields
\begin{align*}
\langle L[u], v \rangle
	= {} & \left[ u'(x) \alpha(x) v(x) - u(x) [ \alpha(x) v(x)]' +
               u(x) \beta(x) v(x) \right]_{x=a}^{x=b}	\\
	&+ \int_{a}^{b} \left( [\alpha(x) v(x)]'' - [\beta(x) v(x)]' +
   \gamma(x) v(x) \right) u(x) \, dx	.
\end{align*}
Define the differential operator $L^{\ast}$ as
\[
%\label{eq:green:linearoperator:general2ndorderadjoint}
L^{\ast}[v](x)
	= \frac{d^{2}}{dx^{2}} \left[ \alpha(x) \, v(x) \right] -
   \frac{d}{dx} \left[ \beta(x) \, v(x) \right] + \gamma(x) \, v(x)	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Remarks}

\[
%\label{eq:green:linearoperator:general2ndorderadjoint}
L^{\ast}[v](x)
	= \frac{d^{2}}{dx^{2}} \left[ \alpha(x) \, v(x) \right] -
   \frac{d}{dx} \left[ \beta(x) \, v(x) \right] + \gamma(x) \, v(x)	.
\]
\begin{itemize}
\item $L^{\ast}$ is known as the
\textbf{formal adjoint}
to $L$.
\item The formal adjoint combined with the boundary conditions which function
$v(x)$ must satisfy at $x=a$ and $x=b$ constitute the adjoint.
The expression,
\[
%\label{eq:green:linearoperator:surfaceterms}
\left[ u'(x) \alpha(x) v(x) - u(x) [ \alpha(x) v(x)]' + u(x) \beta(x) v(x) \right]_{x=a}^{x=b}
\]
will be known as the \textbf{surface terms}
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Generalization of Green's Formula}
\fontsize{9pt}{10.8}\selectfont

\[
\langle L[u], v \rangle - \langle u, L^{\ast}[v] \rangle
	= \left[u'(x) \alpha(x) v(x) - u(x) [\alpha(x) v(x)]' + u(x)
          \beta(x) v(x) \right]_{x=a}^{x=b}	.
\]
\begin{itemize}
\item If $L = L^{\ast}$ then the differential operator $L$ is described as
\textbf{formally self-adjoint}.
\item If in addition the functions upon which these transformations operate
satisfy the same boundary conditions, then $L$ is
\textbf{self-adjoint}.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Properties}
\fontsize{9pt}{10.8}\selectfont

%\label{eq:green:SLBVP:SLBVP}
\begin{align*}
L[u](x)
	= \left[ p(x) u'(x) \right]' + q(x) u(x)
	&= f(x)
\mbox{ for $a < x < b$}	\\
\alpha_{1} u(a) + \beta_{1} u'(a)
	& = 0	\\
\alpha_{2} u(b) + \beta_{2} u'(b)
	& = 0	,
\end{align*}
where $\alpha_{1}^{2} + \beta_{1}^{2} > 0$
and $\alpha_{2}^{2} + \beta_{2}^{2} > 0$.

\begin{theorem}
%\label{thm:green:SLBVP:GreensProp}
A Green's function
\[
%\label{eq:green:SLBVP:green}
G(x; y)
	= \left\{ \begin{array}{rl}
\frac{u_{1}(x) u_{2}(y)}{p \, W}	& \mbox{if $a \le x < y \le b$}	\\
\frac{u_{1}(y) u_{2}(x)}{p \, W}	& \mbox{if $a \le y < x \le b$.}
\end{array} \right.
\]
for the Sturm-Liouville boundary value problem has the following properties.
\begin{enumerate}[label=(\roman*)]
\item $G(x; y)$ satisfies the homogeneous ordinary differential
  equation $L[G](x) = 0$ for $x \ne y$.
\item $G(x; y)$ satisfies the boundary conditions of the
  Sturm-Liouville BVP.
\item $G(x; y) = G(y; x)$ for all $a \le x, y \le b$.
\item $G(x; y)$ is continuous for $(x, y) \in (a, b) \times (a, b)$.
\item $G_{x}(x; y)$ is continuous for $x \ne y$ and
has a jump discontinuity across the line $x = y$,
\[
G_{x}(y+; y) - G_{x}(y-; y)
	= \frac{1}{p(y)}	.
\]
\end{enumerate}
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Crucial Property of Green's Function}

\begin{itemize}
\item Green's function satisfies the homogeneous
differential equation $L[G](x) = 0$ when $x \ne y$ but $G_{x}(x;
y)$ does not exist when $x = y$.
\item For $(x, y) \in (a, b) \times (a, b)$, the Green's function
satisfies the nonhomogeneous differential equation
\[
%\label{eq:green:SLBVP:GreenDirac}
L[G](x)
	= \delta(y - x)	,
\]
where $\delta(y - x)$ is the Dirac delta
function.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Example}

Use the Green's function approach to find the solution to the boundary
value problem:
\begin{align*}
u''(x) + 16 u(x)
	&= (x+1)^{2}
\mbox{ for $0 < x < 1$}	\\
u(0)
	&= u'(1)
	= 0	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution (1 of 6)}

Find a solution to
\begin{align*}
G_{xx}(x; y) + 16 G(x; y)
	&= 0
\mbox{ for $0 < x < 1$, $x \ne y$}	\\
G(0; y)
	= G_{x}(1; y)
	&= 0	.
\end{align*}
The general solution is
$G(x; y) = c_{1} \sin (4 x) + c_{2} \cos (4 x)$ and thus the Green's function
may be expressed as
\[
G(x; y)
	= \left\{ \begin{array}{rl}
a_{1} \sin (4 x) + a_{2} \cos (4 x)	& \mbox{if $0 \le x < y \le 1$}	\\
b_{1} \sin (4 x) + b_{2} \cos (4 x)	& \mbox{if $0 \le y < x \le 1$,}
\end{array} \right.
\]
where $a_{1}$, $a_{2}$, $b_{1}$, and $b_{2}$ are functions of the
parameter $y$.

\end{frame}
%%

\begin{frame}\frametitle{Solution (2 of 6)}

The coefficients $a_{1}$, $a_{2}$, $b_{1}$, and $b_{2}$ must be chosen
so that
\begin{itemize}
\item $G(x; y)$ satisfies the same boundary conditions as
solution $u(x)$,
\item $G(x; y)$ is continuous for $(x, y) \in (0, 1) \times
(0, 1)$, and
\item the function $G_{x}(x; y)$ has a jump discontinuity of the form
$G_{x}(y+; y) - G_{x}(y-; y) = 1$.
\end{itemize}
The boundary condition $G(0; y) = 0$ forces $a_{2} = 0$.
The continuity of $G(x; y)$ implies
\[
(a_{1} - b_{1}) \sin (4 y) + b_{2} \cos (4 y)
	= 0	.
\]
The jump discontinuity in the partial derivative implies
\[
(-4 a_{1} + 4 b_{1}) \cos (4 y) - 4 b_{2} \sin (4 y)
	= 1	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Solution (3 of 6)}

\begin{align*}
(a_{1} - b_{1}) \sin (4 y) + b_{2} \cos (4 y)
	&= 0	\\
(-4 a_{1} + 4 b_{1}) \cos (4 y) - 4 b_{2} \sin (4 y)
	&= 1
\end{align*}
Solving this system of two equations yields
\begin{align*}
a_{1} - b_{1}
	&= -\frac{1}{4} \cos (4 y)	\\
b_{2}
	&= -\frac{1}{4} \sin (4 y)	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution (4 of 6)}

The boundary condition $G_{x}(1; y) = 0$ implies
\[
0
	= 4 b_{1} \cos (4) - 4 b_{2} \sin (4)
	= 4 b_{1} \cos (4) + \sin (4 y) \sin (4)
\]
using the value found for $b_{2}$ which in turn determines
\[
b_{1}
	= \frac{-\sin (4 y) \sin (4)}{4 \cos (4)}
\mbox{ and }
a_{1}
	= \frac{-\cos (4y-4)}{4 \cos (4)}	.
\]
Consequently the Green's function for this boundary value problem
takes the form,
\[
G(x; y)
	= -\frac{1}{4 \cos (4)} \left\{ \begin{array}{rl}
\sin (4x) \cos (4y-4)
	& \mbox{if $0 \le x < y \le 1$}	\\
\cos (4x-4) \sin (4y)
	& \mbox{if $0 \le y < x \le 1$.}
\end{array} \right.
\]

\end{frame}
%%

\begin{frame}\frametitle{Solution (5 of 6)}

\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{example03}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Solution (6 of 6)}

\begin{align*}
u(x)
	= {} & \int_{0}^{1} G(x; y) f(y) \, dy	\\
	= {} & \int_{0}^{x} \frac{-\sin (4y) \cos (4x-4)}{4 \cos (4)}
   (y+1)^{2} \, dy	\\
	&+ \int_{x}^{1} \frac{-\sin (4x) \cos (4y-4)}{4 \cos
   (4)} (y+1)^{2} \, dy	\\
	= {} & \frac{-\cos (4x-4)}{4 \cos (4)} \int_{0}^{x} (y+1)^{2}
               \sin (4y) \, dy	\\
	&- \frac{\sin (4x)}{4 \cos (4)} \int_{x}^{1} (y+1)^{2} \cos
   (4y-4) \, dy	\\
	= {} & \frac{1}{16} \left( (x+1)^{2} - \frac{1}{8} \right) -
               \frac{7 \cos (4x-4)}{128 \cos (4)} - \frac{\sin (4x)}{16 \cos (4)}	.
\end{align*}

\end{frame}
%%

%\begin{frame}\frametitle{Homework}

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

%\end{frame}
%%

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\end{document}

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%%%%
