%%%%		-*- 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 General Boundary Value Problems
%%%%

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\title{Green's Functions for General 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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Fall
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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
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%\section{Introduction}

\begin{frame}\frametitle{Objectives}

In this lesson we will:
\begin{itemize}
\item present the general $n$th-order linear differential operator,
\item present the general $n$th-order linear boundary value problem,
\item describe the properties of a Green's function in the setting of
  a general $n$th-order linear boundary value problem,
\item use a Green's function to solve an $n$th-order linear boundary
  value problem,
\item introduce functionals, distributions, and symbolic differentiation.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{$n$th-Order Boundary Value Problem}
\fontsize{9pt}{10.8}\selectfont

\[
%\label{eq:green:linearoperator:generalOP}
L[\cdot]
	= a_{n}(x) \frac{d^{n}}{dx^{n}}[\cdot] + a_{n-1}(x)
        \frac{d^{n-1}}{dx^{n-1}}[\cdot] + \cdots + a_{1}(x)
        \frac{d}{dx}[\cdot] + a_{0}(x) [\cdot]
\]
where the coefficient functions $a_{i}(x)$ for $i = 0, 1, \ldots, n$
are defined and continuous on the interval $[a, b]$ with $a_{n}(x) \ne 0$ for $a \le
x \le b$.
$\mathcal{D}(L)$ will be the functions $u(x)$ with piecewise
continuous $n$th derivatives on $[a, b]$.
An $n$th order linear boundary value problem is
%\label{eq:green:linearoperator:generalBVP}
\begin{align*}
L[u]
	&= -f
\mbox{ for $a < x < b$}	\\
B_{1}[u]
	&= \sum_{j=0}^{n-1} (a_{1, j} u^{(j)} (a) + b_{1, j} u^{(j)}(b))
	= 0	\\
{}	&\vdotswithin{=} {}	\\
B_{n}[u]
	&= \sum_{j=0}^{n-1} (a_{n, j} u^{(j)} (a) + b_{n, j} u^{(j)}(b))
	= 0
\end{align*}
where $a_{i, j}$ and $b_{i, j}$ for $i = 1, \ldots, n$ and $j =
0, \ldots, n-1$ are constants and $B_{i}$ denotes a
linear combination of function $u(x)$ and its derivatives through
order $n-1$.
The boundary conditions are assumed to be linearly independent.
In boundary condition $B_{k}[u]$ at least one of $a_{k, j}$ or $b_{k,
  j}$ is nonzero for all $k = 1, 2, \ldots, n$.

\end{frame}
%%

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

By definition the Green's function $G(x; y)$,
\begin{itemize}
\item satisfies the ordinary differential equation $L[G] = 0$
  for $(x, y) \in (a, b) \times (a, b)$ when $x \ne y$,
$G(x; y)$ and its derivatives with respect to $x$ up to order $n$
exist and are continuous for $a < x < y$ and $y < x < b$.
\item $G(x; y)$ satisfies the boundary conditions of
  the BVP.
\item For $k = 0, 1, \ldots, n-2$ the derivative
  $\frac{\partial^{k}}{\partial x^{k}} [ G(x; y) ]$ is continuous
  across the line where $x = y$, in other words
\[
\lim_{x \to y^{+}} \frac{\partial^{k}}{\partial x^{k}} [ G(x; y) ] -
\lim_{x \to y^{-}} \frac{\partial^{k}}{\partial x^{k}} [ G(x; y) ]
	= 0
\mbox{ for $k = 0, 1, \ldots, n-2$.}
\]
\item $\frac{\partial^{n-1}}{\partial x^{n-1}} [ G(x; y) ]$
has a jump discontinuity crossing the line $x = y$, specifically
\[
\lim_{x \to y^{+}} \frac{\partial^{n-1}}{\partial x^{n-1}} [ G(x; y) ] -
\lim_{x \to y^{-}} \frac{\partial^{n-1}}{\partial x^{n-1}} [ G(x; y) ]
	= \frac{-1}{a_{n}(y)}	.
\]
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Existence and Uniqueness}

\begin{theorem}
%\label{thm:green:linearoperator:generalBVP}
If the homogeneous boundary value problem of
the $n$th-order boundary value problem (that is, the case where
$f(x) = 0$ on $(a, b)$) has only the zero solution, then there exists
a unique Green's function $G(x; y)$ associated with the boundary value
problem.
\end{theorem}

\vspace*{0.10in}

\uncover<2->{%
\begin{theorem}
%\label{thm:green:linearoperator:useGreen}
If the homogeneous boundary value problem of
(again, when $f(x) = 0$ on
$(a, b)$) has only the zero solution, then the solution to the
nonhomogeneous form can
be expressed as the integral
\[
%\label{eq:green:linearoperator:endresult}
u(x)
	= \int_{a}^{b} G(x; y) f(y) \, dy	.
\]
where $G(x; y)$ is the Green's function of the associated homogeneous
boundary value problem.
\end{theorem}
}

\end{frame}
%%

\begin{frame}\frametitle{Example}

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

\end{frame}
%%

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

Green's function
must satisfy $G_{xxx}(x; y) = 0$ on $[0, 1] \times [0, 1]$ where $x \ne y$.
The general form of the Green's function is
\[
G(x; y)
	= \left\{ \begin{array}{rl}
\alpha_{1}(y) + \alpha_{2}(y) x + \alpha_{3}(y) x^{2}
	& \mbox{if $0 \le x < y \le 1$}	\\
\beta_{1}(y) + \beta_{2}(y) x + \beta_{3}(y) x^{2}
	& \mbox{if $0 \le y < x \le 1$.}
\end{array} \right.
\]
The coefficients must be chosen to satisfy
the boundary conditions, the continuity of $G(x; y)$ and $G_{x}(x; y)$
on $[0, 1] \times [0, 1]$, and the jump
discontinuity of $G_{xx}(x; y)$ at $x = y$.

\end{frame}
%%

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

The first partial derivative of $G(x; y)$ with respect to $x$ can be expressed as
\[
G_{x}(x; y)
	= \left\{ \begin{array}{rl}
\alpha_{2}(y) + 2 \alpha_{3}(y) x
	& \mbox{if $0 \le x < y \le 1$}	\\
\beta_{2}(y) + 2 \beta_{3}(y) x
	& \mbox{if $0 \le y < x \le 1$.}
\end{array} \right.
\]
The boundary condition $G_{x}(0; y) = 0$ implies $\alpha_{2}(y) = 0$.
The second partial derivative of $G(x; y)$ with respect to $x$ can be expressed as
\[
G_{xx}(x; y)
	= \left\{ \begin{array}{rl}
2 \alpha_{3}(y)	& \mbox{if $0 \le x < y \le 1$}	\\
2 \beta_{3}(y)	& \mbox{if $0 \le y < x \le 1$.}
\end{array} \right.
\]
The boundary condition $G_{xx}(1; y) = 0$ implies $\beta_{3}(y) = 0$.

\end{frame}
%%

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

$G(x; y)$ now has the simplified form of
\[
G(x; y)
	= \left\{ \begin{array}{rl}
\alpha_{3}(y) x^{2}
	& \mbox{if $0 \le x < y \le 1$}	\\
\beta_{1}(y) + \beta_{2}(y) x
	& \mbox{if $0 \le y < x \le 1$.}
\end{array} \right.
\]
Continuity of $G(x; y)$ on $(0, 1) \times (0, 1)$ implies
\[
\lim_{x \to y^{-}} G(x; y)
	= \lim_{x \to y^{+}} G(x; y)
\iff
\alpha_{3}(y) y^{2}
	= \beta_{1}(y) + \beta_{2}(y) y
\]
and similarly continuity of $G_{x}(x; y)$ on $(0, 1) \times (0, 1)$ implies
\[
\lim_{x \to y^{-}} G_{x}(x; y)
	= \lim_{x \to y^{+}} G_{x}(x; y)
\iff
2 \alpha_{3}(y) y
	= \beta_{2}(y)	.
\]

\end{frame}
%%

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

The jump discontinuity in $G_{xx}(x; y)$ across the line $x = y$ requires
\begin{align*}
G_{xx}(y+; y) - G_{xx}(y-; y)
	= \frac{-1}{1} = -1
	&\iff 0 - 2 \alpha_{3}(y) = -1	\\
	&\iff \alpha_{3}(y) = \frac{1}{2}	.
\end{align*}
This results yields $\beta_{2}(y) = y$ and $\beta_{1}(y) = -\frac{1}{2} y^{2}$.
Finally the Green's function is
\[
G(x; y)
	= \left\{ \begin{array}{rl}
\frac{1}{2} x^{2}
	& \mbox{if $0 \le x < y \le 1$}	\\
x y - \frac{1}{2} y^{2}
	& \mbox{if $0 \le y < x \le 1$}
\end{array} \right.
\]

\end{frame}
%%

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

\begin{align*}
u(x)
	&= \int_{0}^{1} G(x; y) (-e^{y}) \, dy	\\
	&= \int_{0}^{x} \left( \frac{y^{2}}{2} - x \, y \right) (-e^{y}) \, dy
	+ \int_{x}^{1} \frac{-x^{2}}{2} (-e^{y}) \, dy 	\\
	&= -1 - x - \frac{x^{2}}{2} e^{x} + e^{x} + \frac{x^{2}}{2}
   e^{x} - \frac{e x^{2}}{2}	\\
	&= -1 - x - \frac{e}{2} x^{2} + e^{x}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Symbolic Functions, Test Functions, and
    Functionals}

\begin{itemize}
\item Symbolic functions are not required to have numerical values for all real
  number inputs.
\begin{example}
$\delta(x) = 0$ for $x \ne 0$, but $\delta(0)$ has no
defined value.
\end{example}
\item<2-> Symbolic functions will yield values when multiplied by a family of
suitable \textbf{test functions} and integrated
over the real numbers.
\item Test functions consist of
ordinary functions which have continuous derivatives of all orders and
which vanish outside of some closed and bounded interval.
These functions are denoted $C _{0}^{\infty}$ and are called
functions with compact support.
\item<3-> A \textbf{functional} $F$ is a mapping from the
vector space of test functions to the real (or complex) numbers.
\item<4-> A functional $F$ defined on a vector space of test functions is called
a \textbf{linear functional} if whenever
$\phi_{1}$ and $\phi_{2}$ are test functions and $k$ is a real or complex scalar,
then $F(\phi_{i})$ is a real or complex number and the following
equation is satisfied.
\[
%\label{eq:green:inverse:distribution}
F(\phi_{1} + k \, \phi_{2})
	= F(\phi_{1}) + k \, F(\phi_{2})
\]
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Distributions}

If $F(\phi) = \langle f, \phi \rangle$ is continuous, then it is referred to as a
\textbf{distribution}.
Every distribution has a derivative which is itself another distribution.
For example, suppose $f$ is a fixed, differentiable function from an inner product
space, $\phi$ is a test function and
\[
F(\phi)
	= \langle f, \phi \rangle
	= \int_{-\infty}^{\infty} f(x) \phi(x) \, dx	.
\]
Using integration by parts,
\begin{align*}
\int_{-\infty}^{\infty} f'(x) \phi(x) \, dx
	&= \left[ f(x) \phi(x) \right]_{x \to -\infty}^{x \to \infty}
   - \int_{-\infty}^{\infty} f(x) \phi'(x) \, dx	\\
	&= -\int_{-\infty}^{\infty} f(x) \phi'(x) \, dx	.
\end{align*}
Thus the derivative of $F$ (which will be denoted $F'$) can be
expressed as
\[
F'(\phi)
	= -\langle f, \phi' \rangle
	= -\int_{-\infty}^{\infty} f(x) \phi'(x) \, dx	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Symbolic Differentiation}

The Dirac delta function can be defined as the distribution with the
property that $\langle \delta, \phi \rangle = \phi(0)$ for all test
functions $\phi$.
The derivative of $\delta(x)$ is then the distribution for which
\[
%\label{eq:green:linearoperator:Diracprime}
\int_{-\infty}^{\infty} \delta'(x) \phi(x) \, dx
	= -\phi'(0)	,
\]
for all test functions.
This process of determining the derivative of a distribution is
called \textbf{symbolic differentiation}.

\end{frame}
%%

\begin{frame}\frametitle{Inverse of a Linear Transformation}

The inverse of a linear
differentiation transformation $L$ is an integral transformation with a
kernel $G(x; y)$ of the form
\[
L^{-1}[f]
	= \int_{a}^{b} G(x; y) f(y) \, dy	.
\]
If the kernel is a distribution and $u$ and $v$ are test
functions satisfying the equation $L[u] = v$ then
\begin{align*}
u
	= L^{-1} L[u]
	= L^{-1}[v]
	&= \int_{a}^{b} G(x; y) v(y) \, dy	\\
L \, L^{-1}[v]
	&= L \left[ \int_{a}^{b} G(x; y) v(y) \, dy \right]	.
\end{align*}
Since the linear transformation $L$ performs differentiation with
respect to variable $x$, the operator may be pulled inside
the integral formally.
This implies
\[
v(x)
	= \int_{a}^{b} L[G(x; y)] v(y) \, dy	.
\]
If this equation holds for all test functions $v$ then $L[G(x; y)] = \delta(y-x)$.

\end{frame}
%%

\begin{frame}\frametitle{Example}

Find the inverse of the
second-order differentiation transformation $L[\cdot] \equiv
\frac{d^{2}}{d x^{2}}[\cdot]$.

\end{frame}
%%

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

The inverse of $L$ is an integral
operator with a kernel $G(x; y)$ satisfying
\[
L[G(x; y)]
	= \frac{d^{2}}{d x^{2}} \left[ G(x; y) \right]
	= \delta(y-x)	.
\]
We can show
\[
\frac{d}{d x} \left[ G(x; y) \right]
	= \int \delta(y-x) \, dx
	= -H(y-x) + A(y)
\]
where $H$ is the Heaviside function and $A(y)$ is an arbitrary function of $y$.
\[
G(x; y)
	= \int (-H(y-x) + A(y)) \, dx
	= (y-x) H(y-x) + x A(y) + B(y)
\]
where $B(y)$ is another arbitrary function of $y$.
The arbitrary functions $A$ and $B$ can be determined from boundary
conditions.

\end{frame}
%%

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

Suppose we have boundary
conditions $u(a) = u(b) = 0$, then the adjoint will be the operator
$L^{\ast}$ satisfying the equation
\[
\int_{a}^{b} L[u](x) v(x) \, dx
	= \int_{a}^{b} u(x) L^{\ast}[v](x) \, dx	.
\]
Applying the operator $L$ and integrating by parts yield
\begin{align*}
\int_{a}^{b} L[u](x) v(x) \, dx
	&= \int_{a}^{b} u''(x) v(x) \, dx	\\
	&= \left[ u'(x) v(x) \right]_{x=a}^{x=b} - \int_{a}^{b} u'(x)
   v'(x) \, dx	\\
	&= -\left[ u(x) v'(x) \right]_{x=a}^{x=b} + \int_{a}^{b} u(x)
   v''(x) \, dx	\\
	&= \int_{a}^{b} u(x) L[v](x) \, dx	.
\end{align*}
Hence $L$ is self-adjoint.

\end{frame}
%%

%\begin{frame}\frametitle{Homework}

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

%\end{frame}
%%

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%%%% EOF
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