%%%%		-*- 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:      03/04/2025
%%%%
%%%% Purpose:   Legendre's Equation
%%%%

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\title{Legendre's Equation}
\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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\the\year}

\subject{Talks}
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%\begin{frame}
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%\section{Introduction}

\begin{frame}\frametitle{Objectives}

In this lesson we will learn:
\begin{itemize}
\item to solve Legendre's differential equation,
\item properties of the Legendre polynomials,
\item the relationships to the associated Legendre polynomials,
\item the associated Legendre functions, and
\item the orthogonality relationships.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Legendre's Differential Equation}

When using the method of separation of variables to solve
Laplace's equation in spherical coordinates, the
following equation arises,
\[
%\label{eq:spc:laplacespherical}
\frac{d^{2} F(\varphi)}{d \varphi^{2}} + \cot \varphi \frac{d
  F(\varphi)}{d \varphi} + \alpha (\alpha+1) F(\varphi) = 0
\]
for $0 < \varphi < \pi$,
where $\alpha$ is a constant.

\vspace*{0.10in}

Make the change of variable $x = \cos \varphi$ and define
$y = f(x) = F(\arccos x)$.
The equation above can be written in the form known as
\textbf{Legendre's differential equation},
\[
%\label{eq:spc:legendre}
(1 - x^{2}) y'' - 2 x y' + \alpha (\alpha + 1) y = 0
\mbox{ for $-1 < x < 1$,}
\]
where $\alpha$ is a constant.

\end{frame}
%%

\begin{frame}\frametitle{Solution to Legendre's Differential Equation}

Assuming a power series solution of the form $y(x) =
\sum_{n=0}^{\infty} a_{n} x^{n}$ produces two linearly independent solutions,
\begin{align*}
y_{1}(x)
	= {} & 1 -  \frac{\alpha (\alpha + 1)}{2!} x^{2} +
\frac{\alpha (\alpha - 2) (\alpha + 1) (\alpha + 3)}{4!} x^{4}	\\
	&+ \sum_{m=3}^{\infty} \frac{\alpha \cdots (\alpha - 2m + 2) (\alpha +
1) \cdots (\alpha + 2m - 1)}{(2m)!} x^{2m}	\\
y_{2}(x)
	= {} & x - \frac{(\alpha - 1)(\alpha + 2)}{3!} x^{3} +
\frac{(\alpha - 1)(\alpha - 3)(\alpha + 2)(\alpha + 4)}{5!} x^{5}	\\
	&+ \sum_{m=3}^{\infty} (-1)^{m} \frac{(\alpha - 1) \cdots (\alpha - 2m + 1)
  (\alpha + 2) \cdots (\alpha + 2m)}{(2m+1)!} x^{2m+1}	.
\end{align*}
If $\alpha \in \mathbb{N} \cup \{ 0 \}$ then these functions are
polynomials of degree $\alpha$.

\end{frame}
%%

\begin{frame}\frametitle{Leading Coefficient Convention}

By convention the leading coefficient $a_{n}$ of a polynomial solution
of degree $n$ to Legendre's differential equation
is chosen to be
\[
%\label{eq:spc:legendreleading}
a_{n} = \frac{(2n)!}{2^{n} (n!)^{2}}	.
\]
In this situation a polynomial of degree $n$ solving Legendre's
differential equation and having the prescribed leading coefficient is called a
\textbf{Legendre polynomial}
and is denoted as $P_{n}(x)$.
The general formula for $P_{n}(x)$ is
\[
%\label{eq:spc:legendreP}
P_{n}(x)
	= \frac{1}{2^{n}} \sum_{k=0}^{\lfloor n/2 \rfloor} \frac{(-1)^{k} (2n -
          2k)!}{k! (n-k)! (n - 2k)!} x^{n-2k}	,
\]
where $\lfloor n/2 \rfloor$ denotes the greatest integer less than or equal to
$n/2$.

\end{frame}
%%

\begin{frame}\frametitle{Table of Legendre Polynomials}

\begin{center}
\begin{tabular}{@{}rcl@{}}
$P_{0}(x)$ & = & $\displaystyle 1$	\\
$P_{1}(x)$ & = & $\displaystyle x$	\\
$P_{2}(x)$ & = & $\displaystyle \frac{1}{2} (3 x^{2} - 1)$	\\
$P_{3}(x)$ & = & $\displaystyle \frac{1}{2} (5 x^{3} - 3x)$	\\
$P_{4}(x)$ & = & $\displaystyle \frac{1}{8} (35 x^{4} - 30 x^{2} + 3)$	\\
$P_{5}(x)$ & = & $\displaystyle \frac{1}{8} (63 x^{5} - 70 x^{3} + 15 x)$	\\
$P_{6}(x)$ & = & $\displaystyle \frac{1}{16} (231 x^{6} - 315 x^{4} + 105 x^{2} - 5)$	\\
\end{tabular}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Illustration}

\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{legendreps}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Properties of Legendre Polynomials}

\begin{theorem}[Rodrigues' Form of Legendre Polynomials]
%\label{thm:spc:legendrerodrigues}
For $n = 0, 1, 2, \ldots$,
\[
%\label{eq:spc:legendrerodrigues}
P_{n}(x)
	= \frac{1}{2^{n} n!} \frac{d^{n}}{d x^{n}} \left[ (x^{2} - 1)^{n} \right]	.
\]
\end{theorem}

\begin{lemma}[Bonnet's Recurrence Relation]
%\label{lma:spc:bonnet}
For $n \in \mathbb{N}$, if $P_{n}(x)$ is the $n$th Legendre polynomial
\[
%\label{eq:spc:bonnet}
(n+1) P_{n+1}(x) + n P_{n-1}(x)
	= (2n+1) x P_{n}(x)	.
\]
\end{lemma}

\begin{theorem}
%\label{thm:spc:legendreorthog}
For $n, m \in \{ 0, 1, 2, \ldots \}$, if $P_{m}(x)$ and $P_{n}(x)$
are the $m$th and $n$th Legendre polynomials respectively, then
\[
%\label{eq:spc:legendreorthog}
\int_{-1}^{1} P_{m}(x) P_{n}(x) \, dx
	= \frac{2}{2n+1} \, \delta_{mn}
\]
where $\delta_{mn}$ is the Kronecker delta function.
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Associated Legendre Functions (1 of 3)}

\begin{theorem}
%\label{thm:spc:assoclegendre}
Let $f(x) = P_{n}(x)$ then 
for $m \in \{0, 1, \ldots, n \}$ the function $f^{(m)}(x)$ solves the
ordinary differential equation,
\[
%\label{eq:spc:assoclegendre}
(1-x^{2}) y'' - 2 x (m+1) y' + (n (n+1) - m (m+1)) y = 0	.
\]
\end{theorem}

\uncover<2->{%
Define the function $g(x) = (1-x^{2})^{m/2}
f^{(m)}(x)$.
Function $g(x)$ solves the ordinary differential equation,
\[
%\label{eq:spc:assoclegendreg}
(1-x^{2}) g''(x) - 2 x g'(x) + \left( n(n+1) - \frac{m^{2}}{1-x^{2}}
\right) g(x) = 0
\]
which is called the \textbf{associated Legendre equation}.
The solutions are referred to as the
\textbf{associated Legendre functions},
and are denoted as $P_{n}^{m}(x) = (-1)^{m} (1-x^{2})^{m/2} P_{n}^{(m)}(x)$
for $n = 0, 1, \ldots$ and $m = 0, 1, \ldots, n$
where $P_{n}(x)$ is the $n$th Legendre polynomial.
}

\end{frame}
%%

\begin{frame}\frametitle{Associated Legendre Functions (2 of 3)}

\begin{center}
\resizebox{1.00\textwidth}{!}{%
\begin{tabular}{@{}ccccc@{}}
{}	& $m=0$ & $m=1$ & $m=2$ & $m=3$	\\
$n=0$	& $1$ & {} & {} & {}	\\
$n=1$	& $x$ & $-\sqrt{1-x^{2}}$ & {} & {}	\\
$n=2$	& $\frac{1}{2} (-1 + 3 x^{2})$ & $-3 x \sqrt{1-x^{2}}$ & $-3
x^{2} + 3$ & {}	\\
$n=3$	& $\frac{1}{2} (-3x + 5 x^{3})$ & $-\frac{3}{2} (-1 + 5 x^{2})
\sqrt{1-x^{2}}$ & $-15 x (-1 + x^{2})$ & $-15 (1-x^{2})^{3/2}$	\\
\end{tabular}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Associated Legendre Functions (3 of 3)}

\[
P_{n}^{m}(x)
	= (-1)^{m} (1-x^{2})^{m/2} P_{n}^{(m)}(x)
\]
The integer parameter $m$ is called the order of the associated
Legendre function.
If $0 < m \le n$ the associated Legendre function for $-m < 0$
is defined as a scalar multiple of the associated Legendre function of
order $m > 0$,
\[
%\label{eq:spc:assoclegendrenegm}
P_{n}^{-m}(x)
	= (-1)^{m} \frac{(n-m)!}{(n+m)!} P_{n}^{m}(x)
\mbox{ for $0 < m \le n$.}
\]

\begin{lemma}
%\label{lma:spc:assoclegendreparity}
If $n+m$ is even then $P_{n}^{m}(x)$ is an even function and if
$n+m$ is odd then $P_{n}^{m}(x)$ is an odd function.
\end{lemma}

\end{frame}
%%

\begin{frame}\frametitle{Properties}

\begin{theorem}
%\label{thm:spc:assoclegendreorthog1}
Suppose $m$ and $n$ are nonnegative integers such that $m \ne n$ and
suppose $k$ is an integer such that $|k| \le \min \{ m, n \}$.
Then
\[
%\label{eq:spc:assoclegendreorthog1}
\int_{-1}^{1} P_{m}^{k}(x) P_{n}^{k}(x) \, dx = 0	.
\]
\end{theorem}

\begin{theorem}
%\label{thm:spc:assoclegendreorthognn}
If $|k| \le n$, then
\[
%\label{eq:spc:assoclegendreorthognn}
\int_{-1}^{1} \left( P_{n}^{k}(x) \right)^{2} \, dx
	= \frac{2}{2n+1} \frac{(n+k)!}{(n-k)!}	.
\]
\end{theorem}


\end{frame}
%%

\begin{frame}\frametitle{Orthogonality}

If $f(x)$ is sufficiently smooth on $[-1, 1]$, the associated Legendre
function expansion of $f(x)$ of order $k$ is
\[
f(x)
	= \sum_{n=k}^{\infty} a_{n} P_{n}^{k}(x)	,
\]
where
\[
%\label{eq:spc:assoclegendrecoeff}
a_{n}
	= \frac{2n+1}{2} \, \frac{(n-k)!}{(n+k)!} \int_{-1}^{1} f(x) P_{n}^{k}(x) \, dx
\]
for $n = k, k+1, k+2, \ldots$.
If $k=0$ this reduces to the Legendre function expansion.

\end{frame}
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%\begin{frame}\frametitle{Homework}

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

%\end{frame}
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