%%%%		-*- 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:      06/30/2022
%%%%
%%%% Purpose:   Derivations of Common PDEs
%%%%

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\title{Derivations of Common PDEs}
\subtitle{MATH 467 \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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%\section{Introduction}

\begin{frame}\frametitle{Objectives}

In this lesson we will derive first- and second-order PDEs arising as
mathematical models of common physical phenomena.
Applications considered will include
\begin{itemize}
\item gas flow through a pipe,
\item flow of heat energy through a rod,
\item vibration of a flexible string.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Application: Gas Flow}

Consider a gas flowing through a tube whose axis is parallel to the
$x$-axis.
\begin{center}
\resizebox{0.80\textwidth}{!}{\includegraphics{gas}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Assumptions}

\begin{itemize}
\item The \textbf{density} (mass per unit volume) of the gas at
  position $x$ at time $t$ will be $\rho(x, t)$.
\item The \textbf{velocity} of the gas at position $x$ at time $t$
  will be $v(x, t) \mathbf{i}$.
\item The \textbf{cross sectional area} of the tube is a constant $A >
  0$.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Mass and Rate of Change}

The mass of gas in the interval $[x_{0}, x_{0} + \Delta x]$ at time
$t$ can be
expressed as
\[
M(t)
	=
\int_{x_{0}}^{x_{0}+\Delta x} \rho(x, t) A \, dx
	=
A \int_{x_{0}}^{x_{0}+\Delta x} \rho(x, t) \, dx	.
\]
The rate of change of mass in interval $[x_{0}, x_{0} + \Delta x]$ is
\begin{align*}
M'(t)
	&= \frac{d}{dt} \left[ A \int_{x_{0}}^{x_{0}+\Delta x} \rho(x,
   t) \, dx \right]	\\
	&= A \int_{x_{0}}^{x_{0}+\Delta x} \frac{\partial
   \rho}{\partial t}(x, t) \, dx	\\
	&\approx A \int_{x_{0}}^{x_{0}+\Delta x} \frac{\rho(x,
   t+\Delta t) - \rho(x, t)}{\Delta t} \, dx	,
\end{align*}
if $\Delta t$ is small.

\end{frame}
%%

\begin{frame}\frametitle{Net Change in Mass}

The mass of gas passing position $x_{0}$ at time $t$ in an interval
$\Delta t$ is
\[
A \, \rho(x_{0},t) v(x_{0},t) \, \Delta t	.
\]

The net change in mass in the interval $[x_{0}, x_{0} + \Delta x]$
during time interval $\Delta t$ is
\begin{align*}
\lefteqn{A \int_{x_{0}}^{x_{0}+\Delta x} \left[ \rho(x, t + \Delta t)
  - \rho(x, t) \right] \, dx}	\\
	&\approx -A \, \rho(x_{0} + \Delta x, t) v(x_{0} + \Delta x, t) \Delta t +
A \, \rho(x_{0}, t) v(x_{0}, t) \Delta t	.
\end{align*}
Divide both side of this approximation by $A \, \Delta t \, \Delta x$.

\end{frame}
%%

\begin{frame}\frametitle{Continuity Equation}

\begin{align*}
\lefteqn{\frac{1}{\Delta x} 
\int_{x_{0}}^{x_{0}+\Delta x} \frac{1}{\Delta t} \left[ \rho(x, t +
  \Delta t) - \rho(x,t) \right] \, dx
}	\\
	&\approx \frac{1}{\Delta x} \left(
\rho(x_{0}, t) v(x_{0}, t) - \rho(x_{0} + \Delta x, t) v(x_{0} +
                    \Delta x, t)
\right)	\\
	&= \frac{-(\rho(x_{0} + \Delta x, t) v(x_{0} + \Delta x, t) -
   \rho(x_{0}, t) v(x_{0}, t))}{\Delta x}
\end{align*}

\uncover<2->{%
Let $\Delta t \to 0$ and $\Delta x \to 0$ to obtain
\begin{align*}
\rho_{t} &= -\left( \rho \, v \right)_{x}	\\
\rho_{t} + \left( \rho \, v \right)_{x} &= 0
\end{align*}
which is called the \textbf{continuity equation}.
}

\end{frame}
%%

\begin{frame}\frametitle{Solving the Continuity Equation}

If $v(x, t)$ is a known function and the initial density of the gas in
the tube $\rho(x,0) = \rho_{0}(x)$ is known then solving the
continuity equation is equivalent to solving the initial value problem
\begin{align*}
\rho_{t} + v(x,t) \rho_{x} + v_{x}(x,t) \rho &= 0	\\
\rho(x,0) &= \rho_{0}(x)	.
\end{align*}

\vspace*{0.05in}

\textbf{Note}: Since no boundary conditions are specified, assume the
tube is infinitely long in both directions.

\end{frame}
%%

\begin{frame}\frametitle{Case: $v(x, t) = v_{0}$ a Constant}

The initial value problem simplifies to
\begin{align*}
\rho_{t} + v_{0} \rho_{x} &= 0	\\
\rho(x,0) &= \rho_{0}(x)	.
\end{align*}

\begin{enumerate}
\item Confirm that $\rho(x, t) = \rho_{0}(x - v_{0} t)$ is a solution.
\item Interpret the solution.
\end{enumerate}

\end{frame}
%%

\begin{frame}\frametitle{Illustration}

Suppose $\rho_{0}(x) = e^{-x^{2}}$ and $v_{0} = 1$ then the graph of
the solution $\rho(x, t) = e^{-(x - t)^{2}}$ resembles the following.
\begin{center}
\resizebox{0.80\textwidth}{!}{\includegraphics{constantv}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Case: $v(x, t) = \alpha \, x$ Where $\alpha > 0$}

The initial value problem simplifies to
\begin{align*}
\rho_{t} + \alpha \, x \, \rho_{x} + \alpha \, \rho &= 0	\\
\rho(x,0) &= \rho_{0}(x)	.
\end{align*}

\begin{enumerate}
\item Confirm that $\rho(x, t) = \rho_{0}(x \, e^{-\alpha \, t})
  e^{-\alpha t}$ is a solution.
\item Interpret the solution.
\end{enumerate}

\end{frame}
%%

\begin{frame}\frametitle{Illustration}

Suppose $\rho_{0}(x) = e^{-x^{2}}$ and $\alpha = 1$ then the graph of
the solution
\[
\rho(x, t) = e^{-(x \, e^{-t})^{2}} e^{-t} = e^{-t - x^{2} e^{-2 t}}
\]
resembles the following.
\begin{center}
\resizebox{0.80\textwidth}{!}{\includegraphics{valphax}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Application: Heat Flow}

Consider a flow of heat energy in a rod of length $L$ whose axis is
parallel to the $x$-axis.
\begin{center}
\resizebox{0.80\textwidth}{!}{\includegraphics{heatrod}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Assumptions}

\begin{itemize}
\item The sides of the rod are insulated completely.
No heat exchange
takes place with the surrounding media through the lateral surfaces of the rod.
%
\item The cross sectional areas and the mass density (mass per unit
  length) of the rod are a function of $x$ only,
denoted as $A(x)$ and $\rho(x)$
respectively.
%
\item Within each cross-section, the temperature is evenly distributed.
This implies that the temperature distribution is a function of
location $x$ and time $t$ only.
%
\item The cross sectional area $A(x)$ and the density
$\rho(x)$ are both smooth functions of $x$ and
the temperature throughout the rod varies smoothly in both $t$ and $x$.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Definitions}

\begin{description}
\item[$u(x, t)$:] temperature of rod at time $t \ge 0$ and position $x
  \in [0,L]$.
\item[$c(x)$:] specific heat capacity of rod, the amount of heat
  energy per unit mass needed to raise the temperature of the rod by
  one degree,
\item[$Q(x, t)$:] heat generated per unit time per unit volume.
\item[$\phi(x, t)$:] heat flux, the energy per unit time per unit area
  passing location $x$ at time $t$.
\end{description}

\end{frame}
%%

\begin{frame}\frametitle{Partition the Rod}

Consider $[a, b] \subset [0,L]$.
\begin{itemize}
\item Total amount of heat energy
contained in $[a, b]$ at time $t$ is
\[
q(t) = \int_{a}^{b} c(x) \rho(x) A(x) u(x, t) \, dx	.
\]
\item Total amount of heat energy generated
in $[a, b]$ at time $t$ is
\[
\gamma(t) = \int_{a}^{b} Q(x, t) A(x) \, dx	.
\]
\item The net heat flux in $[a, b]$ at time $t$ is
\[
\phi(a, t) A(a) - \phi(b, t) A(b) =
-\int_{a}^{b} \frac{\partial}{\partial x} \left[ \phi(x, t) A(x)
\right] \, dx	.
\]
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Illustration}

\begin{center}
\resizebox{0.80\textwidth}{!}{\includegraphics{heatrodab}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Conservation of Heat Energy}

\resizebox{\textwidth}{!}{$
\left(
\begin{array}{c}
\text{Rate of change with} \\
\text{respect to time of the} \\
\text{total heat energy in} \\
\text{the interval } [a, b]
\end{array}
\right) =
\left(
\begin{array}{c}
\text{Rate at which} \\
\text{heat energy} \\
\text{flows in or out} \\
\text{of the two ends}
\end{array}
\right) +
\left(
\begin{array}{c}
\text{Rate at which} \\
\text{heat energy} \\
\text{is generated in} \\
\text{interval } [a, b]
\end{array}
\right)
$}

\vspace*{0.25in}

This can be expressed mathematically as
\begin{align*}
\frac{d}{dt} \left[ q(t) \right]
	= {} & \phi(a,t) A(a) - \phi(b,t) A(b) + \gamma(t)	\\
\frac{d}{dt} \int_{a}^{b} c(x) \rho(x) A(x) u(x, t) \, dx
	= {} & -\int_{a}^{b} \frac{\partial}{\partial x} \left[ \phi(x, t) A(x)
\right] \, dx	\\
	& +
\int_{a}^{b} Q(x, t) A(x) \, dx
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Important Lemmas (1 of 2)}

\begin{lemma}[Leibniz Integral Rule]
If $f$ and $\partial f/\partial t$ are both
continuous in some region of the $xt$-plane, including the region
where $a(t) \le x \le b(t)$ for $t_{1} \le t \le t_{2}$ and if $a(t)$
and $b(t)$ are continuously differentiable for $t_{1} \le t \le t_{2}$,
then
\begin{align*}
\lefteqn{\frac{d}{dt} \left[ \int_{a(t)}^{b(t)} f(x, t) \, dx \right]}	\\
	&=
\int_{a(t)}^{b(t)} \frac{\partial}{\partial t}  \left[ f(x, t) \right]
\, dx
	+
f(b(t), t) \frac{d b}{d t} - f(a(t), t) \frac{d a}{d t}	.	\nonumber
\end{align*}
\end{lemma}

\end{frame}
%%

\begin{frame}\frametitle{Important Lemmas (2 of 2)}

\begin{lemma}
If $f(x)$ is continuous on an interval $[0,L]$, and
\[
\int_{a}^{b} f(x) \, dx = 0
\]
on any interval $[a, b] \subset [0,L]$, then $f(x) = 0$ for all $x \in
[0,L]$.
\end{lemma}

\end{frame}
%%

\begin{frame}\frametitle{Simplification}

Assuming we can bring the derivative with respect to $t$ inside the
integral the previous equation may be re-written as
\resizebox{\textwidth}{!}{$
\int_{a}^{b} \left( c(x) \rho(x) A(x) \frac{\partial}{\partial t} 
  u(x, t) + \frac{\partial}{\partial x} \left( \phi(x, t) A(x)
\right ) - Q(x, t) A(x) \right) \ dx
	=
0.
$}
\uncover<2->{%
Since the interval $[a, b] \subset [0,L]$ is arbitrary then the
integrand must be 0.
\[
c(x) \rho(x) A(x) \frac{\partial}{\partial t} \left[ u(x, t) \right]
	=
-\frac{\partial}{\partial x} \left[ \phi(x, t) A(x) \right] + Q(x, t)
A(x)
\]
for $t \ge 0$ and $x \in [0,L]$.
}

\end{frame}
%%

\begin{frame}\frametitle{Fourier's Law}

\[
c(x) \rho(x) A(x) \frac{\partial}{\partial t} \left[ u(x, t) \right]
	=
-\frac{\partial}{\partial x} \left[ \phi(x, t) A(x) \right] + Q(x, t)
A(x)
\]
This equation contains two unknowns $u(x, t)$ and $\phi(x, t)$.

\vspace*{0.05in}

\textbf{Fourier's Law of Heat Conduction}: heat energy flows from
warmer regions to cooler regions at a rate proportional to the
difference in the temperatures between the regions.
\[
\phi(x, t) = -K_{0}(x) \frac{\partial u}{\partial x}(x, t)
\]
Function $K_{0}(x) \ge 0$ is called the \textbf{thermal conductivity}
and measures the ability of the rod to conduct heat energy.

\end{frame}
%%

\begin{frame}\frametitle{Simplification}

Using Fourier's Law the conservation of heat energy equation becomes
\[
c(x) \rho(x) A(x) \frac{\partial}{\partial t} \left[ u(x, t) \right]
	=
-\frac{\partial}{\partial x} \left[ -K_{0}(x) \frac{\partial
    u}{\partial x}(x, t)  A(x) \right] + Q(x, t) A(x)	.
\]
\begin{itemize}
\item<2-> If $c$, $\rho$, $A$, and $K_0$ are constant then
\[
\frac{\partial u}{\partial t} =
k \frac{\partial^{2} u}{\partial x^{2}} + \frac{1}{c\rho} Q(x, t)
\]
where $k = K_{0}/(c \, \rho)$ is called the \textbf{thermal
  diffusivity}.
\item<3-> In addition if $Q(x, t) \equiv 0$ then
\[
\frac{\partial u}{\partial t} =
k \frac{\partial^{2} u}{\partial x^{2}}
\]
which referred to as the \textbf{one-dimensional heat equation}.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{IBVPs for the Heat Equation}

\begin{itemize}
\item An \textbf{initial condition} specifies the distribution of
  temperature in the rod at $t=0$.
\[
u(x,0) = f(x) \quad\mbox{for $0 \le x \le L$}
\]
\item There are three different classes of boundary conditions which
  can be specified.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Dirichlet Boundary Conditions}

If the temperatures at the two ends of the rod are specified for $t >
0$ as
\begin{align*}
u(0,t) &= g_{1}(t)	\\
u(L,t) &= g_{2}(t)	.
\end{align*}
These are known as \textbf{Dirichlet boundary conditions} or
\textbf{boundary conditions of the first kind}.

\vspace*{0.05in}

The simplest such BCs are homogeneous:
\begin{align*}
u(0,t) &= 0	\\
u(L,t) &= 0	.
\end{align*}

\vspace*{0.05in}

\textbf{Question}: what physical interpretation can be given to these BCs?

\end{frame}
%%

\begin{frame}\frametitle{Neumann Boundary Conditions}

If the heat fluxes at the two ends of the rod are specified for $t >
0$ as
\begin{align*}
-K_{0} u_{x}(0,t) &= g_{1}(t)	\\
K_{0} u_{x}(L,t) &= g_{2}(t)	.
\end{align*}
These are known as \textbf{Neumann boundary conditions} or
\textbf{boundary conditions of the second kind}.

\vspace*{0.05in}

The simplest such BCs are homogeneous:
\begin{align*}
u_{x}(0,t) &= 0	\\
u_{x}(L,t) &= 0	.
\end{align*}

\vspace*{0.05in}

\textbf{Question}: what physical interpretation can be given to these BCs?

\end{frame}
%%

\begin{frame}\frametitle{Robin Boundary Conditions}

If the rate of heat loss (or gain) between the two ends of the rod and
the surrounding media is proportional to the differences in the
temperatures at the two ends and the surrounding media for $t >
0$ then
\begin{align*}
K_{0} u_{x}(0,t) &= \alpha (u(0,t) - g_{1}(t))	\\
K_{0} u_{x}(L,t) &= -\beta (u(L,t) - g_{2}(t))	.
\end{align*}
These are known as \textbf{Robin boundary conditions} or
\textbf{boundary conditions of the third kind}.

\vspace*{0.05in}

\begin{itemize}
\item The constants $\alpha > 0$ and $\beta > 0$ are called the
\textbf{coefficients of convective heat transfer}.
\item Each boundary condition is equivalent to \textbf{Newton's Law of
    Cooling}.
\item The Dirichlet and Neumann BCs are limiting cases of boundary
  conditions of the third kind.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Example}

Suppose a one-dimensional rod of length $L$ is subject to a constant
heat flux $\phi_{0}$ at $x = 0$ and constant temperature $u_{0}$ at $x
= L$.
The initial temperature distribution of the rod is $f(x)$.
\begin{enumerate}
\item Set up the IBVP modeling this situation.
\item Solve for the steady-state temperature distribution in the rod.
\end{enumerate}

\end{frame}
%%

\begin{frame}<handout:0>\frametitle{Solution (1 of 2)}

IBVP:
\begin{align*}
\frac{\partial u}{\partial t}
	&=
k \frac{\partial^{2} u}{\partial x^{2}}
\mbox{ for $0<x<L$, $t>0$}	\\
u(x, 0) &= f(x) \mbox{ for $0<x<L$}	\\
-K_{0} \frac{\partial u}{\partial x}(0,t)
	&=
\phi_{0} \mbox{ for $t>0$}	\\
u(L,t) &= u_{0} \mbox{ for $t>0$}
\end{align*}
where $k$ is the thermal diffusivity constant for the rod.

\end{frame}
%%

\begin{frame}<handout:0>\frametitle{Solution (2 of 2)}

The steady-state solution is independent of $t$.
\begin{align*}
u_{xx}	&= 0
\mbox{ for $0<x<L$}	\\
-K_{0} u_{x}(0)	&= \phi_{0}	\\
u(L) &= u_{0}
\end{align*}
\begin{itemize}
\item<2-> Integrating twice yields $u(x) = A \, x + B$.
\item<3-> Boundary condition at $x=0$ results in $A =
  -\phi_{0}/K_{0}$.
\item<4-> Boundary condition at $x=L$ produces $B = L \phi_{0}/K_{0} +
  u_{0}$
  and thus
\[
u(x) = \frac{\phi_{0}}{K_{0}} (L - x) + u_{0}	,
\]
is the steady-state solution.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Application: Vibrating String}

\begin{itemize}
\item Consider a flexible string in the $xu$-plane whose horizontal
  coordinates lie in the interval $[0,L]$.
\item Let $u(x, t)$ denote the displacement of the string from
  equilibrium at time $t$ and position $x$.
\item Displacement of the string is confined to the $xu$-plane.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Assumptions}

\begin{itemize}
\item The displacement $u(x, t)$ of the string at all points is small.
%
\item The string is perfectly flexible and offers no
  resistance to bending.
%
\item The density (mass per unit length) of the string is $\rho(x)$.
%
\item The string is under tension (a force) whose magnitude is $T(x, t)$.
The tension acts in the direction tangent to the string at point $(x,u(x, t))$.
%
\item The angle between the tangent line to the string at each point and the
positive $x$-axis will be $\theta(x, t)$.
These angles will be assumed small.
%
\item The vertical component (parallel to the direction
  of displacement) of all the external forces per unit
  length acting on the string is $f(x, t)$.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Illustration}

Consider a small segment of the string with endpoints at $(x,u(x, t))$
and $(x+\Delta x,u(x+\Delta x, t))$.
\begin{center}
\resizebox{0.80\textwidth}{!}{\includegraphics{string}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Newton's Second Law of Motion}

\begin{itemize}
\item Mass of string element: $\displaystyle \rho(x) \sqrt{(\Delta
    x)^{2} + (\Delta u)^{2}}$.
\item Vertical component of tension at $(x,u(x, t))$: $\displaystyle
T(x, t) \sin \theta(x, t)$.
\item Vertical component of tension at $(x+\Delta x,u(x+\Delta x, t))$: $\displaystyle
T(x+\Delta x, t) \sin \theta(x+\Delta x, t)$.
\end{itemize}

\uncover<2->{%
According to Newton's Second Law of Motion,
\begin{align*}
m \, a = {} & \sum F	\\
\rho(x) \sqrt{(\Delta x)^{2} + (\Delta u)^{2}} \frac{\partial^{2}
  u}{\partial t^{2}}
	= {} & T(x+\Delta x, t) \sin \theta(x+\Delta x, t)	\\
	& - T(x, t) \sin \theta(x, t) + f(x, t) \Delta x	.
\end{align*}
Divide both sides by $\Delta x$.
}

\end{frame}
%%

\begin{frame}\frametitle{Simplification}

\begin{align*}
\lefteqn{\rho(x) \sqrt{1 + \left( \frac{\Delta u}{\Delta x} \right)^{2}}
\frac{\partial^{2} u}{\partial t^{2}}}	\\
	&=
\frac{T(x+\Delta x, t) \sin \theta(x+\Delta x, t) -
T(x, t) \sin \theta(x, t)}{\Delta x} + f(x, t)
\end{align*}
Let $\Delta x \to 0$.
\begin{align*}
\rho(x) \sqrt{1 + \left( u_{x} \right)^{2}} \, u_{tt}
	&=
\frac{\partial}{\partial x} \left[ T(x, t) \sin \theta(x, t) \right]
+ f(x, t)	\\
	&=
T_{x} \sin \theta(x, t) + T(x, t) \theta_{x} \cos \theta(x, t) + f(x, t)
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Trigonometric Relationships}

\begin{columns}
\begin{column}{0.5\textwidth}
The following relationships will simplify the last equation further.
\begin{align*}
\tan \theta(x, t) &= u_{x}	\\
\sin \theta(x, t) &= \frac{u_{x}}{\sqrt{1 + \left( u_{x} \right)^{2}}}	\\
\cos \theta(x, t) &= \frac{1}{\sqrt{1 + \left( u_{x} \right)^{2}}}
\end{align*}
\end{column}
%
\uncover<2->{%
\begin{column}{0.5\textwidth}
Assuming $\theta(x, t)$ is small implies $u_{x}$ is small and
\begin{align*}
\sqrt{1 + \left( u_{x} \right)^{2}} &\approx 1	\\
\sin \theta(x, t) &\approx u_{x}	\\
\cos \theta(x, t) &\approx 1	.
\end{align*}
\end{column}}
\end{columns}

\end{frame}
%%

\begin{frame}\frametitle{Simplification}

Using these trigonometric relationships simplifies the previous result
to the form
\[
\rho(x) u_{tt} = T_{x} u_{x} + T(x, t) \theta_{x} + f(x, t)	.
\]

\vspace*{0.05in}

\uncover<2->{%
\textbf{Remarks}:
\begin{itemize}
\item this equation involves three unknowns: $u(x, t)$,
$T(x, t)$, and $\theta(x, t)$.
\item<3-> using the chain rule
\[
u_{xx} = \frac{\partial}{\partial \theta} \left[ u_{x} \right]
\theta_{x} = \frac{\partial}{\partial \theta} \left[ \tan \theta
\right]  \theta_{x} = (\sec^{2} \theta) \theta_{x} \approx \theta_{x}
\]
\item<4-> substituting into the equation above eliminates
  $\theta_{x}$.
\[
\rho(x) u_{tt} = T_{x} u_{x} + T(x, t) u_{xx} + f(x, t)	.
\]
\end{itemize}
}

\end{frame}
%%

\begin{frame}\frametitle{Eliminating the Tension}

The horizontal components of the tension at $(x,u(x, t))$ and
$(x+\Delta x,u(x+\Delta x, t))$ cancel each other since there is no
acceleration parallel to the $x$-axis.
\begin{align*}
T(x+\Delta x,t) \cos \theta(x+\Delta x, t) - T(x, t) \cos \theta(x, t)
	&= 0	\\
\lim_{\Delta x \to 0} \frac{T(x+\Delta x,t) \cos \theta(x+\Delta x, t) - T(x, t) \cos
  \theta(x, t)}{\Delta x}
	&= 0	\\
\frac{\partial}{\partial x} \left[ T(x, t) \cos \theta(x, t) \right]
	&= 0	\\
\uncover<2->{%
T_{x} \cos \theta(x, t) - T(x, t) (\sin \theta(x, t)) \theta_{x} &= 0	\\
\uncover<3->{%
T(x, t) u_{x} \, u_{xx} &\approx T_{x}
}
}
\end{align*}
\uncover<4->{%
Thus the previous equation for the string simplifies to
\[
\rho(x) u_{tt} = T(x, t) \left( u_{x} \right)^{2} u_{xx} + T(x, t)
u_{xx} + f(x, t)	.
\]
}

\end{frame}
%%

\begin{frame}\frametitle{Simplification}

Using the assumption that $u_{x}$ is small, then
\[
\rho(x) u_{tt} = T(x, t) u_{xx} + f(x, t)	.
\]
If the tension and density of the string are constants then
\begin{align*}
\rho_{0} u_{tt} &= T_{0} u_{xx} + f(x, t)	\\
u_{tt} &= \frac{T_{0}}{\rho_{0}} u_{xx} + \frac{1}{\rho_{0}} f(x, t)
  \\
u_{tt} &= c^{2} u_{xx} + \frac{1}{\rho_{0}} f(x, t)	.
\end{align*}
If there are no external forces then
\[
u_{tt} = c^{2} u_{xx}	,
\]
which is called the \textbf{one-dimensional wave equation}.

\uncover<2->{%
\textbf{Question}: what are the units of constant $c$?
}

\end{frame}
%%

\begin{frame}\frametitle{IBVPs for the Wave Equation}

\begin{itemize}
\item An \textbf{initial condition} specifies the
  \textbf{displacement} and \textbf{velocity} of the string
at $t=0$.
\begin{align*}
u(x,0) &= f(x) \quad\mbox{for $0 \le x \le L$}	\\
u_{t}(x,0) &= g(x)
\end{align*}
\item There are three different classes of boundary conditions which
  can be specified.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Dirichlet Boundary Conditions}

If the displacements of the two ends of the string are specified for $t >
0$ as
\begin{align*}
u(0,t) &= f_{1}(t)	\\
u(L,t) &= f_{2}(t)	.
\end{align*}
These are known as \textbf{Dirichlet boundary conditions} or
\textbf{boundary conditions of the first kind}.

\vspace*{0.05in}

The simplest such BCs are homogeneous:
\begin{align*}
u(0,t) &= 0	\\
u(L,t) &= 0	.
\end{align*}

\vspace*{0.05in}

\textbf{Question}: what physical interpretation can be given to these BCs?

\end{frame}
%%

\begin{frame}\frametitle{Neumann Boundary Conditions}

If the slopes of the string at the two ends are specified for $t >
0$ as
\begin{align*}
T_{0} u_{x}(0,t) &= g_{1}(t)	\\
T_{0} u_{x}(L,t) &= g_{2}(t)	.
\end{align*}
These are known as \textbf{Neumann boundary conditions} or
\textbf{boundary conditions of the second kind}.

\vspace*{0.05in}

The simplest such BCs are homogeneous:
\begin{align*}
u_{x}(0,t) &= 0	\\
u_{x}(L,t) &= 0	.
\end{align*}

\vspace*{0.05in}

\textbf{Question}: what physical interpretation can be given to these BCs?

\end{frame}
%%

\begin{frame}\frametitle{Robin Boundary Conditions}

\textbf{Robin boundary conditions} or
\textbf{boundary conditions of the third kind} take the form of
\begin{align*}
T_{0} u_{x}(0,t) &= k_{1} (u(0,t) - d_{1}(t))	\\
T_{0} u_{x}(L,t) &= -k_{2} (u(L,t) - d_{2}(t))	.
\end{align*}

\vspace*{0.05in}

If $d_{1}(t) \equiv 0$ and $d_{2}(t) \equiv 0$ the following
homogeneous boundary conditions arise.
\begin{align*}
T_{0} u_{x}(0,t) &= k_{1} u(0,t)	\\
T_{0} u_{x}(L,t) &= -k_{2} u(L,t)	.
\end{align*}
The Dirichlet ($T_{0} = 0$) and Neumann ($k_{1} = k_{2} = 0$) boundary
conditions are special cases of these BCs.

\end{frame}
%%

\begin{frame}\frametitle{Homework}

\begin{itemize}
\item Read Sections $1.3$ and $1.4$
\item Exercises: 7--11
\end{itemize}

\end{frame}
%%

%%%
\end{document}

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