%%%%		-*- 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:      04/02/2025
%%%%
%%%% Purpose:   The Korteweg-de Vries Equation
%%%%

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\title{The Korteweg-de Vries Equation}
\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]
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{
  Department of Mathematics
}
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% - Keep it simple, no one is interested in your street address.

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

\subject{Talks}
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\begin{document}

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

%\begin{frame}
%  \frametitle{Outline}
%  \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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% of between 15min and 45min and the theme suggest that you stick to
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% - Exactly two or three sections (other than the summary).
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% - 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 present the Korteweg-de Vries (KdV) equation,
\item explore the equivalence of several forms of the KdV equation, and
\item find a soliton solution to the KdV equation.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Korteweg-de Vries Equation}

The Korteweg-de Vries equation was initially studied as a mathematical
model of the motion of waves in shallow water.

\vspace*{0.10in}

The Korteweg-de Vries equation has many equivalent mathematical formulations.
\begin{align*}
%\label{eq:nlpde:KdVcanonical}
u_{t} + 6 u \, u_{x} + u_{xxx}
	&= 0	\\
%\label{eq:nlpde:KdV1k}
u_{t} + u \, u_{x} + k \, u_{xxx}
	&= 0	\\
%\label{eq:nlpde:KdVm61}
u_{t} - 6 u \, u_{x} + u_{xxx}
	&= 0	\\
%\label{eq:nlpde:KdV11}
u_{t} + u \, u_{x} + u_{xxx}
	&= 0	\\
%\label{eq:nlpde:KdValphabeta}
u_{t} + \alpha u \, u_{x} + \beta u_{xxx}
	&= 0	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Equivalence}

Suppose $x = A \, X$, $t = B \, T$, and $u = C \, V$ then
\[
u_{t} + 6 u \, u_{x} + u_{xxx}
	= \frac{C}{B} V_{T} + \frac{6 C^{2}}{A} V \, V_{X} + \frac{C}{A^{3}} V_{XXX}
	= 0	,
\]
which is equivalent to
\[
V_{T} + \frac{6 B \, C}{A} V \, V_{X} + \frac{B}{A^{3}} V_{XXX}
	= 0	.
\]
By choosing the constants $A$, $B$, and $C$ appropriately, any form of
the Korteweg-de Vries equation
can be obtained.

\end{frame}
%%

\begin{frame}\frametitle{Traveling Wave Solution}

Set $\xi = x - c \, t$ and substitute $u(x, t) = U(\xi)$ into the
Korteweg-de Vries equation.
\[
\frac{d^{3} U}{d \xi^{3}} + 6 U \frac{dU}{d\xi} - c \frac{dU}{d\xi}
	= 0	.
\]
Integrate both sides with respect to $\xi$.
\[
\frac{d^{2} U}{d \xi^{2}} + 3 U^{2} - c \, U
	= A	,
\]
where $A$ is an arbitrary constant of integration.

\vspace*{0.10in}

\uncover<2->{%
Assume that
$U \to 0$, $U' \to 0$, and $U'' \to 0$ as
$\xi \to \infty$ and as $\xi \to -\infty$.
\[
\frac{d^{2} U}{d \xi^{2}} + 3 U^{2} - c \, U
	= 0	.
\]
Multiply the ordinary differential equation by $U' = dU/d\xi$ and
integrate both sides of the resulting equation.
\[
\frac{1}{2} \left( \frac{dU}{d\xi} \right)^{2} + U^{3} - \frac{1}{2} c
U^{2}
	= B	,
\]
where $B$ is again an arbitrary constant of integration.
}

\end{frame}
%%

\begin{frame}\frametitle{Traveling Wave Solution}

The assumptions that $U \to 0$ and $U' \to 0$
as $\xi \to \infty$ and $\xi \to -\infty$ imply that $B = 0$, and
therefore, after solving for $dU/d\xi$,
\[
%\label{eq:nlpde:kdvtravODE1}
\frac{dU}{d\xi}
	= \pm U \sqrt{c - 2 U}	,
\]
where the $\pm$ sign should be chosen according to the sign of
$dU/d\xi$.

\vspace*{0.10in}

In the following, the $\pm$ sign will be dropped since it
makes no difference in the integration steps to follow.

\end{frame}
%%

\begin{frame}\frametitle{Traveling Wave Solution}

Separate the variables and integrate.
\[
%\label{eq:nlpde:kdvtravODE2}
\int \frac{1}{U \sqrt{c - 2 U}} \, dU
	= \xi + A	,
\]
where $A$ is again a constant.

\vspace*{0.10in}

To evaluate the integral on the left side, make a change of
variable which, while not obvious, is motivated by
identities from hyperbolic trigonometry,
\[
U
	= \frac{c}{2} \sech^{2} \zeta
\implies
dU
	= -c \frac{\sinh \zeta}{\cosh^{3} \zeta} \, d\zeta	.
\]
The substitution results in
\begin{align*}
\int \frac{1}{U \sqrt{c - 2 U}} \, dU
	&= \int \frac{2}{c \sech^{2} \zeta \, \sqrt{c - c \sech^{2} \zeta}}
\frac{-c \sinh \zeta}{\cosh^{3} \zeta} \, d\zeta	\\
	&= -\frac{2}{\sqrt{c}} \int 1 \, d\zeta
	= -\frac{2}{\sqrt{c}} \zeta
	= -\frac{2}{\sqrt{c}} \sech^{-1} \sqrt{\frac{2U}{c}}	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Traveling Wave Solution}

Thus the equation,
\[
-\frac{2}{\sqrt{c}} \sech^{-1} \sqrt{\frac{2U}{c}}
	= \xi + A
\]
is solved for $U = U(\xi)$ to give
\[
U(\xi)
	= \frac{c}{2} \sech^{2} \left( \frac{\sqrt{c}}{2} \xi + A \right)	.
\]
Therefore, the Korteweg-de Vries equation has the following traveling wave solution:
\[
u(x, t)
	=\frac{c}{2} \sech^{2} \left( \frac{\sqrt{c}}{2} (x - c \, t) +
  A \right)	.
\]
The presence of the term $\sqrt{c}$ and the desire to find a
real-valued solution, suggest that $c > 0$.

\end{frame}
%%

\begin{frame}\frametitle{Traveling Wave Solution}

If the constant $c$ is replaced by $4 c^{2}$ for convenience, then
\[
%\label{eq:nlpde:kdvsoliton}
u(x, t)
	= 2 c^{2} \sech^{2} (c (x - 4 c^{2} t) + A)	.
\]
\begin{itemize}
\item This form of solution is a
soliton solution
of the Korteweg-de Vries equation.
\item This soliton solution is positive for all $x$ and $t$.
\item For each fixed $t$ it has a single maximum, and $u(x, t) \to 0$ as
$|x| \to \infty$.
\item All such solutions travel to the right with a speed
depending on the amplitude of the solution.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Illustration}

\begin{center}
\resizebox{0.85\textwidth}{!}{\includegraphics{KdVarray}}
\end{center}
The blue soliton has $c = 3/2$ and $A = 1$ while the yellow soliton curve has
$c = 1$ and $A = 0$.
It is apparent the blue soliton is traveling faster to the right
than the yellow soliton.

\end{frame}
%%

\begin{frame}\frametitle{Other Solutions}

If the assumption that the
traveling wave solution and its derivatives vanish at infinity is
relaxed, then the function $U(\xi)$ satisfies:
\[
%\label{eq:nlpde:kdvtravODE3}
\frac{1}{2} \left( \frac{dU}{d\xi} \right)^{2}
	= -U^{3} + \frac{c}{2} U^{2} + A \, U + B
	\equiv F(U)
\]
where $A$ and $B$ are
arbitrary constants.

\vspace*{0.10in}

Solutions are real if and only if $F(U)$ is positive.
Since the function $F(U)$ is a cubic polynomial (with a negative
leading coefficient), it may have one or three real zeros.

\end{frame}
%%

\begin{frame}\frametitle{Properties of Soliton Solutions}

Soliton solutions maintain their form
even after two solitons propagating at different speeds collide.
The waves seem to simply pass through each other and each keeps its wave profile
with only a slight phase shift in wave position (the peak of the wave
profile).

\end{frame}
%%

\begin{frame}\frametitle{Illustration}

\begin{center}
\resizebox{0.90\textwidth}{!}{\includegraphics{phaseshift}}
\end{center}
When solitons interact they experience a phase shift.
The solid curve depicts the profile of the solution to an initial
value problem in which interaction takes place while the dashed curve
shows the sum of the solitons had there been no interaction.

\end{frame}
%%

%\begin{frame}\frametitle{Homework}

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

%\end{frame}
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