%%%%		-*- 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:      02/18/2025
%%%%
%%%% Purpose:   Boundary Value Problems
%%%%

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\title{Two-Point 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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\subject{Talks}
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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 learn:
\begin{itemize}
\item to classify second order, two-point boundary value problems,
\item to place boundary value problems in self-adjoint form,
\item properties of the eigenvalues and eigenfunctions of boundary
  value problems.
\end{itemize}

\vspace*{0.10in}

This enables us to generalize the method of separation of variables.

\end{frame}
%%

\begin{frame}\frametitle{Background}

When solving the heat equation for $0 \le x \le L$ with homogeneous
boundary conditions we had to solve the induced \textbf{boundary value
  problem}:
\begin{align*}
X''(x) + \lambda \, X(x)
	&= 0	\\
X(0)
	&= 0	\\
X(L)
	&= 0	.
\end{align*}
We saw that nontrivial solutions existed only if
\begin{align*}
\lambda = \lambda_{n}
	&= \frac{n^{2} \pi^{2}}{L^{2}}	\\
X(x) = X_{n}(x)
	&= \sin \frac{n \pi x}{L}
\mbox{ for $n \in \mathbb{N}$.}
\end{align*}
The set $\left\{ \lambda_{n} \right\}_{n=1}^{\infty}$ is called the set of
\textbf{eigenvalues} and
$\left\{ X_{n}(x) \right\}_{n=1}^{\infty}$ is the corresponding set of
\textbf{eigenfunctions}.

\end{frame}
%%

\begin{frame}\frametitle{Generalization}

Consider the following second-order,
linear homogeneous ODE of the form
\[
P(x) y''(x) + Q(x) y'(x) + R(x) y(x)
	= 0
\mbox{ for $a < x < b$.}
\]
Suppose that $P(x)$, $Q(x)$, and $R(x)$ are continuous on interval
$[a, b]$ and that $P(x) \ne 0$ for all $x$ in $[a, b]$.

\vspace*{0.10in}

\uncover<2->{%
\begin{itemize}
\item If $Q(x) = P'(x)$ then the ODE can be written as
\[
\left[ P(x) y'(x) \right]' + R(x) y(x)
	= 0	.
\]
\item If $Q(x) \ne P'(x)$ multiply both sides of the ODE by
\[
r(x)
	= \frac{1}{P(x)} e^{\int_{a}^{x} Q(s)/P(s) \, ds}
	= \frac{p(x)}{P(x)}	.
\]
\end{itemize}
}

\end{frame}
%%

\begin{frame}\frametitle{Self-Adjoint Form}

\begin{align*}
P(x) y''(x) + Q(x) y'(x) + R(x) y(x)
	&= 0	\\
r(x) P(x) y''(x) + r(x) Q(x) y'(x) +
r(x) R(x) y(x)
	&= 0	\\
p(x) y''(x) + p'(x) y'(x) + q(x) y(x)
	&= 0	\\
\left[ p(x) y'(x) \right]' + q(x) y(x)
	&= 0
\end{align*}
where $q(x) = r(x) R(x)$.

\vspace*{0.10in}

An ODE written in this last form is called \textbf{self-adjoint}.

\end{frame}
%%

\begin{frame}\frametitle{Example}

Suppose $0 < a \le x \le b$ and express the following ODE in
self-adjoint form.
\[
x \, y''(x) + (1 - x) y'(x) + y(x)
	= 0
\]

\vspace*{0.10in}

\uncover<2->{%
Note that
\[
r(x)
	= \frac{1}{x} e^{\int_{a}^{x} (1-s)/s \, ds}
	= \frac{1}{a} e^{-x + a}	.
\]
Multiplying both sides of the ODE by $r(x)$ produces,
\begin{align*}
x \, y''(x) + (1 - x) y'(x) + y(x)
	&= 0	\\
\frac{x}{a} e^{-x + a} \, y''(x) + \frac{1}{a} e^{-x + a} (1 - x)
  y'(x) + \frac{1}{a} e^{-x + a} y(x)
	&= 0	\\
\left[ x \, e^{-x + a} y'(x) \right]' + e^{-x + a} y(x)
	&= 0	.
\end{align*}
}

\end{frame}
%%

\begin{frame}\frametitle{Introduction of Operator Notation}

Given that the ODE:
\[
P(x) y''(x) + Q(x) y'(x) + R(x) y(x)
	= 0
\]
can be written in the self-adjoint form:
\[
\left[ p(x) y'(x) \right]' + q(x) y(x)
	= 0
\]
define the \textbf{linear operator},
\[
L[y]
	= \left[ p(x) y' \right]' + q(x) y	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Sturm-Liouville BVP (1 of 2)}

Consider a second-order linear ODE and boundary conditions:
\begin{align*}
\left[ p(x) y' \right]' + (q(x) + \lambda r(x)) y
	&= 0
\mbox{ for $a < x < b$}	\\
\alpha_{1} y(a) + \beta_{1} y'(a)
	&= 0	\\
\alpha_{2} y(b) + \beta_{2} y'(b)
	&= 0	,
\end{align*}
where $\alpha_{1}^{2} + \beta_{1}^{2} \ne 0$ and
$\alpha_{2}^{2} + \beta_{2}^{2} \ne 0$.

\vspace*{0.10in}

\textbf{Remarks}:
\begin{itemize}
\item This is known as a \textbf{Sturm-Liouville BVP} with
  \textbf{separated boundary conditions}.
\item The unknown constant $\lambda$ is the \textbf{eigenvalue
    parameter}.
\item The function $r(x)$ is called the \textbf{weight function}.
\item The solution $y(x)$ is the \textbf{eigenfunction} corresponding
  to $\lambda$.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Sturm-Liouville BVP (2 of 2)}

The boundary value problem can be concisely written as,
\begin{align*}
L[y] + \lambda r(x) y
	&= 0
\mbox{ for $a < x < b$}	\\
\alpha_{1} y(a) + \beta_{1} y'(a)
	&= 0	\\
\alpha_{2} y(b) + \beta_{2} y'(b)
	&= 0	.
\end{align*}

\vspace*{0.10in}

\textbf{Remarks}: if
\begin{itemize}
\item $p(a) = 0$ or $p(b) = 0$, or
\item $p(a) = \pm\infty$, $p(b) = \pm\infty$, $q(a) = \pm\infty$,
  $q(b) = \pm\infty$, $r(a) = \pm\infty$, or $r(b) = \pm\infty$, or
\item $a = \pm\infty$ or $b = \pm\infty$, then
\end{itemize}
the BVP is \textbf{singular}.
Otherwise the BVP is \textbf{regular}.

\end{frame}
%%

\begin{frame}\frametitle{Example}

Find the eigenvalues and eigenfunctions of the regular Sturm-Liouville
BVP:
\begin{align*}
y''(x) + \frac{1}{x} y'(x) + \frac{\lambda}{x^{2}} y(x)
	&= 0
\mbox{ for $1 < x < 2$}	\\
y(1)
	&= 0	\\
y(2)
	&= 0	.
\end{align*}

\vspace*{0.10in}

\textbf{Hint}: make the change of variable $z = \ln x$.

\end{frame}
%%

\begin{frame}\frametitle{Solution (1 of 5)}

\begin{align*}
\frac{dy}{dx}
	&= \frac{dy}{dz} \, \frac{dz}{dx}
	= \frac{1}{x} \frac{dy}{dz}
	= e^{-z} \frac{dy}{dz}	\\
\frac{d^{2} y}{dx^{2}}
	&= \frac{d}{dx} \left[ e^{-z} \frac{dy}{dz} \right]
	= e^{-2 z} \frac{dy}{dz} + e^{-2 z} \frac{d^{2} y}{dz^{2}}
\end{align*}

\vspace*{0.10in}

Substituting into the BVP yields:
\begin{align*}
e^{-2 z} \frac{d^{2} y}{dz^{2}} + e^{-2 z} \frac{dy}{dz} + e^{-z}
  e^{-z} \frac{dy}{dz} + \lambda (e^{-z})^{2} y
	&= 0
\mbox{ for $0 < z < \ln 2$}	\\
y''(z) + 2 y'(z) + \lambda y(z)
	&= 0
\mbox{ for $0 < z < \ln 2$}	.
\end{align*}

\end{frame}
%%

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

If $\lambda = 1$ the general solution can be expressed as
\[
y(z)
	= (C_{1} + C_{2} z) e^{-z}
\implies
y(x)
	= (C_{1} + C_{2} \ln x) \frac{1}{x}	.
\]
Using the boundary conditions:
\begin{align*}
y(1)
	&= 0 = (C_{1} + C_{2} \ln 1) \frac{1}{1}
\implies C_{1} = 0	\\
y(2)
	&= 0 = \frac{C_{2}}{2} \ln 2
\implies C_{2} = 0	.
\end{align*}

\end{frame}
%%

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

If $\lambda < 1$ the general solution can be expressed as
\begin{align*}
y(z)
	&= C_{1} e^{(-1 - (1 - \lambda)^{1/2}) z} + C_{2} e^{(-1 + (1 - \lambda)^{1/2}) z}	\\
y(x)
	&= C_{1} x^{(-1 - (1 - \lambda)^{1/2})} + C_{2} x^{(-1 + (1 - \lambda)^{1/2})}	.
\end{align*}
Using the boundary conditions:
\begin{align*}
y(1)
	&= 0
	= C_{1} + C_{2}
\implies C_{1} = -C_{2}	\\
y(2)
&= 0
	= C_{1} 2^{(-1 - (1 - \lambda)^{1/2})} - C_{1} 2^{(-1 + (1 - \lambda)^{1/2})}
\implies C_{1} = C_{2} = 0	.
\end{align*}

\end{frame}
%%

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

If $\lambda > 1$ the general solution can be expressed as
\begin{align*}
y(z)
	&= C_{1} e^{(-1 - i (\lambda - 1)^{1/2}) z} + C_{2} e^{(-1 + i
           (\lambda - 1)^{1/2}) z}	\\
y(x)
	&= \frac{1}{x} \left( C_{1} \cos ((\lambda - 1)^{1/2} \ln x) +
           C_{2} \sin ((\lambda - 1)^{1/2} \ln x) \right)	.
\end{align*}
Using the boundary conditions:
\begin{align*}
y(1)
	&= 0
	= C_{1}	\\
y(2)
	&= 0
	= \frac{C_{2}}{2} \sin ((\lambda - 1)^{1/2} \ln 2)	.
\end{align*}

\begin{itemize}
\item If $C_{2} = 0$ then there are no nontrivial solutions.
\item If $C_{2} \ne 0$ then $(\lambda - 1)^{1/2} \ln 2 = n \, \pi$ for
  $n \in \mathbb{N}$.
\end{itemize}

\end{frame}
%%

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

Finally, the eigenvalues are
\[
\lambda
	= \lambda_{n}
	= 1 + \frac{n^{2} \pi^{2}}{(\ln 2)^{2}}
\]
with corresponding eigenfunctions
\[
y(x)
	= y_{n}(x)
	= \frac{1}{x} \sin \left( \frac{n \, \pi}{\ln 2} \ln x\right)
\]
for $n \in \mathbb{N}$.

\end{frame}
%%

\begin{frame}\frametitle{Nontrivial Solutions (1 of 3)}

Let $y_{1}(x)$ and $y_{2}(x)$ be any two linearly independent
solutions to the BVP:
\begin{align*}
L[y] = \left[ p(x) y' \right]' + q(x) y
	&= 0
\mbox{ for $a < x < b$}	\\
\alpha_{1} y(a) + \beta_{1} y'(a)
	&= 0	\\
\alpha_{2} y(b) + \beta_{2} y'(b)
	&= 0	,
\end{align*}
where $\alpha_{1}^{2} + \beta_{1}^{2} \ne 0$ and $\alpha_{2}^{2} +
\beta_{2}^{2} \ne 0$.

\vspace*{0.10in}

If $y(x)$ is any nontrivial solution to the BVP then there exist
constants $C_{1}$ and $C_{2}$ such that
\[
y(x) = C_{1} y_{1}(x) + C_{2} y_{2}(x)	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Nontrivial Solutions (2 of 3)}

Since $y(x)$ satisfies the boundary conditions at $x = a$ and $x = b$,
then
\begin{align*}
\left[ \begin{array}{cc}
\alpha_{1} y(a) + \beta_{1} y'(a)	\\
\alpha_{2} y(b) + \beta_{2} y'(b)
\end{array} \right]
	&= \left[ \begin{array}{cc}
0	\\ 0
\end{array} \right]	\\
\left[ \begin{array}{cc}
\alpha_{1} y_{1}(a) + \beta_{1} y_{1}'(a)	&
\alpha_{1} y_{2}(a) + \beta_{1} y_{2}'(a)	\\
\alpha_{2} y_{1}(b) + \beta_{2} y_{1}'(b)	&
\alpha_{2} y_{2}(b) + \beta_{2} y_{2}'(b)
\end{array} \right]
\left[ \begin{array}{c} c_{1} \\ c_{2} \end{array} \right]
	&= \left[ \begin{array}{c} 0 \\ 0 \end{array} \right]	.
\end{align*}
The matrix equation has nontrivial solutions if and only if
\[
\left| \begin{array}{cc}
\alpha_{1} y_{1}(a) + \beta_{1} y_{1}'(a)	&
\alpha_{1} y_{2}(a) + \beta_{1} y_{2}'(a)	\\
\alpha_{2} y_{1}(b) + \beta_{2} y_{1}'(b)	&
\alpha_{2} y_{2}(b) + \beta_{2} y_{2}'(b)
\end{array} \right|
	= 0	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Nontrivial Solutions (3 of 3)}

We can summarize these results as the following theorem.
\begin{theorem}
The BVP:
\begin{align*}
\left[ p(x) y' \right]' + q(x) y
	&= 0
\mbox{ for $a < x < b$}	\\
\alpha_{1} y(a) + \beta_{1} y'(a)
	&= 0	\\
\alpha_{2} y(b) + \beta_{2} y'(b)
	&= 0	,
\end{align*}
where $\alpha_{1}^{2} + \beta_{1}^{2} \ne 0$ and $\alpha_{2}^{2} +
\beta_{2}^{2} \ne 0$ has a nontrivial solution if and only if for any
two linearly independent solutions $y_{1}$ and $y_{2}$ of the BVP,
\[
\left| \begin{array}{cc}
\alpha_{1} y_{1}(a) + \beta_{1} y_{1}'(a)	&
\alpha_{1} y_{2}(a) + \beta_{1} y_{2}'(a)	\\
\alpha_{2} y_{1}(b) + \beta_{2} y_{1}'(b)	&
\alpha_{2} y_{2}(b) + \beta_{2} y_{2}'(b)
\end{array} \right|
	= 0	.
\]
In this case all solutions are $u(x) = C \, y(x)$ where $y$ is any
nontrivial solution of the BVP and $C$ is any constant.
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Example}

Find a nontrivial solution to the BVP problem if a nontrivial solution
exists.
\begin{align*}
x^{2} y''(x) + x \, y(x) - 4 y(x)
	&= 0
\mbox{ for $1 < x < 2$}	\\
2 y(1) + y'(1)
	&= 0	\\
y(2) + y'(2)
	&= 0
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution}

The ODE is Euler's equation for which two linearly independent
solutions are $y_{1}(x) = x^{2}$ and $y_{2}(x) = x^{-2}$.

\vspace*{0.10in}

Suppose $y(x) = C_{1} x^{2} + C_{2} x^{-2}$ then to satisfy the
boundary conditions:
\begin{align*}
0
	&= 2 y(1) + y'(1) = 2 C_{1} + 2 C_{2} + 2 C_{1} - 2 C_{2}
\implies C_{1} = 0	\\
0
	&= y(2) + y'(2) = \frac{C_{2}}{4} - \frac{2 C_{2}}{8}
\end{align*}
which is true for any choice of $C_{2}$.

\vspace*{0.10in}

Every nontrivial solution to the BVP has the form $y(x) = C/x^{2}$
where $C \ne 0$ is a constant.

\end{frame}
%%

\begin{frame}\frametitle{Lagrange's Identity}

\begin{lemma}[Lagrange's Identity]
Let $L$ be the operator defined as
\[
L[y]
	= \left[ p(x) y' \right]' + q(x) y
\]
and let $u$ and $v$ be any twice continuously differentiable functions on
$(a, b)$, then
\[
u \, L[v] - v \, L[u]
	= \frac{d}{dx} \left[ p(x) \left( u(x)
    \frac{dv}{dx} - v(x) \frac{du}{dx} \right) \right]	.
\]
\end{lemma}

\end{frame}
%%

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

\begin{lemma}[Green's Formula]
Let $L$ be the operator defined as
\[
L[y]
	= \left[ p(x) y' \right]' + q(x) y
\]
and let
$u$ and $v$ be any twice continuously differentiable functions on
$(a, b)$, then
\[
\int_{a}^{b} \left( u L[v] - v L[u] \right) \, dx
	= \left[ p(x) \left( u(x) \frac{dv}{dx} - v(x) \frac{du}{dx} \right)
\right]_{x=a}^{x=b}	.
\]
\end{lemma}

\end{frame}
%%

\begin{frame}\frametitle{Picone's Identity}

\begin{lemma}[Picone's Identity]
Suppose that $u(x)$ and $v(x)$ are solutions to the following ordinary
differential equations:
\begin{align*}
\left[ p_{1}(x) u' \right]' + q_{1}(x) u
	&= 0	\\
\left[ p_{2}(x) v' \right]' + q_{2}(x) v
	&= 0	.
\end{align*}
For all $x$ such that $v(x) \ne 0$, Picone's Identity holds,
\[
\left[ \frac{u}{v} ( p_{1} u' v - p_{2} u v' ) \right]'
	= (q_{2} - q_{1}) u^{2} + (p_{1} - p_{2}) (u')^{2} + \frac{p_{2} (u' v -
              u v')^{2}}{v^{2}}	.
\]
\end{lemma}

\end{frame}
%%

\begin{frame}\frametitle{Multiplicity of Eigenvalues}

Consider the ODE:
\[
L[y](x) + \lambda r(x) y(x) = \left[ p(x) y'(x) \right]' + (q(x) +
\lambda r(x)) y(x)
	= 0	.
\]

\begin{definition}
An eigenvalue $\lambda$ is said to be \textbf{simple} if there is only
one linearly independent  eigenfunction $\phi$ corresponding to
$\lambda$.
If there exist $k$ linearly independent
eigenfunctions corresponding to $\lambda$, the eigenvalue is said to
be of \textbf{multiplicity} $k$.
\end{definition}

\end{frame}
%%

\begin{frame}\frametitle{Eigenvalues of Sturm-Liouville BVPs}

\begin{theorem}
All eigenvalues of the regular Sturm-Liouville boundary value
problem with separated boundary conditions:
\begin{align*}
\left[ p(x) y'(x) \right]' + (q(x) + \lambda r(x)) y(x)
	&= 0
\mbox{ for $a < x < b$}	\\
\alpha_{1} y(a) + \beta_{1} y'(a)
	&= 0	\\
\alpha_{2} y(b) + \beta_{2} y'(b)
	&= 0	,
\end{align*}
with $\alpha_{1}^{2} + \beta_{1}^{2} \ne 0$ and
$\alpha_{2}^{2} + \beta_{2}^{2} \ne 0$ are simple.
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Inner Products and Weight Functions}

\begin{definition}
Let $f(x)$ and $g(x)$ be two integrable functions on the interval
$[a, b]$, and let $\rho(x)$ be a positive, continuous function
on $[a, b]$.
The \textbf{weighted inner product of $f$ and $g$ with respect to $\rho$ on
$[a, b]$} denoted as $\langle \cdot, \cdot \rangle_{\rho}$ is
\[
\langle f, g \rangle_{\rho} = \int_{a}^{b} f(x) g(x) \rho(x) \, dx	.
\]
\end{definition}

\end{frame}
%%

\begin{frame}\frametitle{Orthogonality}

\begin{theorem}
Suppose $y_{\lambda}(x)$ and $y_{\mu}(x)$ are
solutions to the BVP:
\begin{align*}
\left[ p(x) y'(x) \right]' + (q(x) + \nu r(x)) y(x)
	&= 0
\mbox{ for $a < x < b$}	\\
\alpha_{1} y(a) + \beta_{1} y'(a)
	&= 0	\\
\alpha_{2} y(b) + \beta_{2} y'(b)
	&= 0	,
\end{align*}
corresponding to eigenvalues $\nu = \lambda$ and
$\nu = \mu$ respectively.
If $\lambda \ne \mu$, then 
$y_{\lambda}(x)$ and $y_{\mu}(x)$ are orthogonal with respect to the weight
function $r(x)$.
\end{theorem}

\vspace*{0.10in}

\textbf{Remark}:
eigenfunctions corresponding to distinct eigenvalues of 
a Sturm-Liouville boundary value problem are orthogonal. 

\end{frame}
%%

\begin{frame}\frametitle{Eigenvalues are Real}

\begin{theorem}
The eigenvalues of the Sturm-Liouville boundary value problem:
\begin{align*}
\left[ p(x) y'(x) \right]' + (q(x) + \nu r(x)) y(x)
	&= 0
\mbox{ for $a < x < b$}	\\
\alpha_{1} y(a) + \beta_{1} y'(a)
	&= 0	\\
\alpha_{2} y(b) + \beta_{2} y'(b)
	&= 0	,
\end{align*}
are all real and the corresponding eigenfunctions are real-valued
functions (except possibly for a complex-valued constant
multiplicative factor).
\end{theorem}

\end{frame}
%%

%\begin{frame}\frametitle{Homework}

%\begin{itemize}
%\item Read Sections $7.1$ and $7.2$
%\item Exercises: 1--7
%\end{itemize}

%\end{frame}
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\end{document}

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