%%%%		-*- 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:      03/18/2025
%%%%
%%%% Purpose:   Duhamel's Principle
%%%%

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\title{Duhamel's Principle}
\subtitle{\textit{Partial Differential Equations}} % (optional)


\author % (optional, use only with lots of authors)
{J~Robert~Buchanan}
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%   affiliation.

\institute[Millersville University of Pennsylvania]
% (optional, but mostly needed)
{
  Department of Mathematics
}
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% - Keep it simple, no one is interested in your street address.

\date{ % (optional)
Fall
%Spring
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\the\year}

\subject{Talks}
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% the beginning of each subsection:
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\begin{document}

\begin{frame}
  \titlepage
\end{frame}

%\begin{frame}
%  \frametitle{Outline}
%  \tableofcontents
  % You might wish to add the option [pausesections]
%\end{frame}


% 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
% the following rules:  

% - Exactly two or three sections (other than the summary).
% - At *most* three subsections per section.
% - Talk about 30s to 2min per frame. So there should be between about
%   15 and 30 frames, all told.

%\section{Introduction}

\begin{frame}\frametitle{Objectives}

In this lesson we will:
\begin{itemize}
\item introduce Duhamel's principle, and
\item show how Duhamel's principle can be used to solve the
  nonhomogeneous heat equation with homogeneous boundary and initial
  conditions.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Initial Boundary Value Problem}

%\label{eq:NH:sourceon}
\begin{align*}
u_{t}
	&= \kappa \, u_{xx} + g(x, t)
\mbox{ for $0 < x < L$ and $t > 0$}	\\
u(0, t)
	&= 0
\mbox{ and }
u(L, t) = 0
\mbox{ for $t > 0$}	\\
u(x, 0)
	&= 0
\mbox{ for $0 < x < L$.}
\end{align*}
\textbf{Remark}: the function
$g(x, t)$ represents the rate of temperature change in the medium
caused by an internal heat source.

\end{frame}
%%

\begin{frame}\frametitle{Motivation of Duhamel's Principle}

\begin{itemize}
\item Suppose the interval $[0, T]$ is partitioned into subintervals
$\{ [s_{i-1}, s_{i}] \}_{i=1}^{n}$ where
\[
0 = s_{0} \le s_{1} \le \cdots \le s_{i-1} \le s_{i} \le \cdots \le s_{n} = T	.
\]
\item For $i = 1, \ldots, n$ define $\Delta s_{i} = s_{i} - s_{i-1}$.
\item Define the functions $\mathbbm{1}_{i}(t)$ as
\[
\mathbbm{1}_{i}(t)
	= \left\{ \begin{array}{ll}
1	& \mbox{if $s_{i-1} \le t < s_{i}$,}	\\
0	& \mbox{otherwise.}
\end{array} \right.
\]
\end{itemize}
Now consider the initial boundary value problem
%\label{eq:NH:sourcesupport}
\begin{align*}
u_{t}
	&= \kappa \, u_{xx} + \mathbbm{1}_{i}(t) g(x, t)
\mbox{ for $0 < x < L$ and $t > 0$}	\\
u(0, t)
	&= 0
\mbox{ and }
u(L, t) = 0
\mbox{ for $t > 0$}	\\
u(x, 0)
	&= 0
\mbox{ for $0 < x < L$.}
\end{align*}
\textbf{Remark}: the nonhomogeneity in the PDE is only present during
the interval $s_{i-1} \le t \le s_{i}$.

\end{frame}
%%

\begin{frame}\frametitle{Motivation of Duhamel's Principle}

%\label{eq:NH:sourcesupport}
\begin{align*}
u_{t}
	&= \kappa \, u_{xx} + \mathbbm{1}_{i}(t) g(x, t)
\mbox{ for $0 < x < L$ and $t > 0$}	\\
u(0, t)
	&= 0
\mbox{ and }
u(L, t) = 0
\mbox{ for $t > 0$}	\\
u(x, 0)
	&= 0
\mbox{ for $0 < x < L$.}
\end{align*}
\begin{itemize}
\item For $i = 1, \ldots, n$ let $u_{i}(x, t)$ be the solution to the
corresponding initial boundary value problem.
\item For $t$ in the interval $[0, T]$, $\sum_{i=1}^{n}
\mathbbm{1}_{i}(t) g(x, t) = g(x, t)$ and thus
by the Principle of Superposition,
$\sum_{i=1}^{n} u_{i}(x, t) = u(x, t)$ where $u(x, t)$ solves the original
initial boundary value problem for $0 \le t \le T$.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Motivation of Duhamel's Principle}

%\label{eq:NH:sourcesupport}
\begin{align*}
u_{t}
	&= \kappa \, u_{xx} + \mathbbm{1}_{i}(t) g(x, t)
\mbox{ for $0 < x < L$ and $t > 0$}	\\
u(0, t)
	&= 0
\mbox{ and }
u(L, t) = 0
\mbox{ for $t > 0$}	\\
u(x, 0)
	&= 0
\mbox{ for $0 < x < L$.}
\end{align*}
\begin{itemize}
\item Consider the three intervals, 
$[0, s_{i-1})$, $[s_{i-1}, s_{i})$ and $[s_{i}, T]$.
\item Since the boundary and initial conditions are zero
$u_{i}(x, t) = 0$ for $0 \le t < s_{i-1}$ and
$0 \le x \le L$.
\item At $t = s_{i}$ the temperature distribution in the medium is
approximately $g(x, s_{i-1}) \Delta s_{i}$.
\item For $t \ge s_{i}$ the heat source is turned off and a new initial
boundary value problem can be considered.
\item For the new problem, there is a nonzero,
initial temperature distribution approximated as $g(x, s_{i-1}) \Delta
s_{i}$.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{New Initial Boundary Value Problem}

%\label{eq:NH:aftersource}
\begin{align*}
u_{t}
	&= \kappa \, u_{xx}
\mbox{ for $0 < x < L$ and $t > s_{i}$}	\\
u(0, t)
	&= 0
\mbox{ and }
u(L, t) = 0
\mbox{ for $t > s_{i}$}	\\
u(x, s_{i})
	&= g(x, s_{i-1}) \Delta s_{i}
\mbox{ for $0 < x < L$.}
\end{align*}
For $t \ge s_{i}$ the contribution to the temperature distribution
in the rod from the switching on of the heat source during the time
interval $[s_{i-1}, s_{i})$ is approximately $v(x, t) \Delta s_{i}$
where $v(x, t)$ is the solution to the following initial value
problem:
%\label{eq:NH:aftersourcescaled}
\begin{align*}
v_{t}
	&= \kappa \, v_{xx}
\mbox{ for $0 < x < L$ and $t > s_{i}$}	\\
v(0, t)
	&= 0
\mbox{ and }
v(L, t) = 0
\mbox{ for $t > s_{i}$}	\\
v(x, s_{i})
	&= g(x, s_{i-1})
\mbox{ for $0 < x < L$.}
\end{align*}
The dependence of $v$ on the time at which the heat
source is turned on, is expressed as $v(x, t; s)$.

\end{frame}
%%

\begin{frame}\frametitle{New Initial Boundary Value Problem}

%\label{eq:NH:parameters}
\begin{align*}
v_{t}	&= \kappa \, v_{xx}
\mbox{ for $0 < x < L$ and $t > s_{i}$}	\\
v(0, t; s_{i-1})
	&= v(L, t; s_{i-1}) = 0
\mbox{ for $t > s_{i}$}	\\
v(x, s_{i}; s_{i-1})
	&= g(x, s_{i-1})
\mbox{ for $0 < x < L$.}
\end{align*}
\begin{itemize}
\item $u_{i}(x, t) = v(x, t; s_{i-1}) \Delta s_{i}$ where $v(x, t; s_{i-1})$
solves the IBVP above.
\item At $t =T$ the solution $u$ to the original IBVP is
\[
u(x, T) \approx \sum_{i=1}^{n} v(x, T; s_{i-1}) \Delta s_{i}	.
\]
\item As the norm of the partition approaches zero,
\[
u(x, T)
	= \lim_{n \to \infty} \sum_{i=1}^{n} v(x, T; s_{i-1}) \Delta s_{i}
	= \int_{0}^{T} v(x, T; s) \, ds	.
\]
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Summary}

Since $T$ is arbitrary, the variable can be replaced by $t$ and
$v(x, t; s)$ is a solution to the following initial boundary value problem:
%\label{eq:NH:parameters0}
\begin{align*}
v_{t}
	&= \kappa \, v_{xx}
\mbox{ for $0 < x < L$ and $t > s$}	\\
v(0, t; s)
	&= v(L, t; s) = 0
\mbox{ for $t > s$}	\\
v(x, s; s)
	&= g(x, s)
\mbox{ for $0 < x < L$.}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Change of Variable}

Define $\hat{v}(x, t-s; s) = v(x, t; s)$, then
$v_{t} = \hat{v}_{t}$ and $v_{xx} = \hat{v}_{xx}$ and hence
$\hat{v}(x, t; s)$ solves the initial boundary value problem:
%\label{eq:NH:vhatIBVP}
\begin{align*}
\hat{v}_{t}
	&= \kappa \, \hat{v}_{xx}
\mbox{ for $0 < x < L$ and $t > 0$}	\\
\hat{v}(0, t; s)
	&= \hat{v}(L, t; s) = 0
\mbox{ for $t > 0$}	\\
\hat{v}(x, 0; s)
	&= g(x, s)
\mbox{ for $0 < x < L$.}
\end{align*}
The solution $u(x, t)$ becomes
\[
%\label{eq:NH:defnuint}
u(x, t)
	= \int_{0}^{t} v(x, t; s) \, ds
	= \int_{0}^{t} \hat{v}(x, t-s; s) \, ds	.
\]

\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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