\documentclass[12pt]{amsart}
\usepackage{amsmath,amssymb,textcomp,fullpage,hyperref,graphicx,picins}

%\addtolength{\evensidemargin}{-.5in}
%\addtolength{\oddsidemargin}{-.5in}
%\addtolength{\textwidth}{1in}
%\addtolength{\textheight}{1.2in}
%\addtolength{\topmargin}{-.4in}

\newcommand{\bbC}{{\mathbb C}}
\newcommand{\bbN}{{\mathbb N}}
\newcommand{\bbR}{{\mathbb R}}
\newcommand{\bbZ}{{\mathbb Z}}
\newcommand{\ds}{\displaystyle}

\theoremstyle{plain}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}

\theoremstyle{definition}
\newtheorem{example}{Example}[section]
\newtheorem{exercise}{Exercise}[section]
\newtheorem{corollary}[example]{Corollary}
\newtheorem{comment}[example]{Comment}

\newenvironment{myitemize}{
  \begin{list}{$\bullet$}{\setlength{\leftmargin}{16pt}
  \setlength{\labelwidth}{12pt}
  \setlength{\labelsep}{4pt}}
}{
  \end{list}
}

% Following http://tex.stackexchange.com/questions/23521/tabular-vertical-alignment-to-top:
\def\imagetop#1{\vtop{\null\hbox{#1}}}

\begin{document}

\noindent{\scriptsize
  \href{http://www.math.toronto.edu/~drorbn/Copyleft}{\textcopyleft} $\mid$
  \href{http://www.math.toronto.edu/~drorbn}{Dror Bar-Natan}:
  \href{http://www.math.toronto.edu/~drorbn/classes}{Classes}:
  \href{http://www.math.toronto.edu/~drorbn/classes/#1213}{2012-13}:
  \href{http://drorbn.net/index.php?title=12-267}{267 Advanced ODEs}:
  \hfill\today
}
\newline
\vskip 1mm
{\LARGE Qualitative Analysis}

\vskip 1mm

{\scriptsize Based on a 1989 Princeton University handout by George Em Karniadakis.}

{\scriptsize This is a ``first printing'' and it is likely to contain many typos and other mistakes.}

\tableofcontents

\section{Introduction}

\noindent\parbox[t]{0.56\textwidth}{
  \begin{example} \label{exa:intro} Bessel's equation of order 0:
  \vskip 3mm
  \par\includegraphics{J0.pdf}
  \end{example}
}
\parbox[t]{0.44\textwidth}{
  \begin{center}
    \includegraphics[width=1.5in]{BesselStamp.jpg}
    \vskip 5mm
    $\displaystyle x^2y''+xy'+x^2y = 0$
    \vskip 5mm
  \end{center}
  \par
  \begin{myitemize}
  \item Why does it oscillate?
  \item What does the ``period'' approach?
  \item What does the ``amplitude'' approach?
  \end{myitemize}
  A power series, or a numerical approximation, won't help!
}

\newpage
\section{Regular Singular Points}

\parpic[r]{\parbox{0.95in}{
  \includegraphics[width=0.95in]{FrobeniusPhoto.jpg}\newline
  \tiny Ferdinand Georg Frobenius, 1849--1917,
  \href{http://owpdb.mfo.de/detail?photo_id=10587}{Oberwolfach image}
}}
Suppose $0$ is a regular singular point of the equation
\begin{equation} \label{eq:RSP} x^2y'' + xp(x)y' + q(x)y = 0. \end{equation}
(Meaning simply that $p$ and $q$ above have a power series expansion around $0$). Let $p_0=p(0)$ and $q_0=q(0)$, and let $r_1$ and $r_2$ be the roots of the indicial equation $r(r-1)+p_0r+q_0=0$ (if they are real and distinct, assume also that $r_1>r_2$). Then for $x>0$ Equation~\eqref{eq:RSP} has two linearly independent solutions $y_1$ and $y_2$, such that
\[ y_1 = x^{r_1}\left(1+\sum_{n=1}^\infty a_nx^n\right) \]
\picskip{0}\noindent and
\[ y_2 = \begin{cases}
  y_1\log x + x^r\sum_{n=1}^\infty b_nx^n & r_1=r_2=r \\
  cy_1\log x + x^{r_2}\left(1+\sum_{n=1}^\infty b_nx^n\right) & r_1-r_2=N\in\bbN_{>0} \\
  x^{r_2}\left(1+\sum_{n=1}^\infty b_nx^n\right) & \text{otherwise.}
\end{cases} \]

This can be used to deduce qualitative information! The behaviour near $0$ of a power series is dominated by its $0$th term. The cases are:

\[ y\sim\begin{cases}
  ax^{r_1}+bx^{r_2} & r_1-r_2\in\bbR\setminus\bbZ \\
  x^\alpha(a\cos(\beta\log x) + b\sin(\beta\log x)) &
    r_{1,2}=\alpha+i\beta\in\bbC\setminus\bbR \\
  x^r(a+b\log x) & r_1=r_2=r \\
  x^{r_1}(a+bc\log x)+bx^{r_2} & r_1-r_2\in\bbN_{>0}
\end{cases} \]

\begin{example} Bessel's equation of order $\alpha$,
\[ y''+\frac{1}{x}y'+\left(1-\frac{\alpha^2}{x^2}\right) = 0, \]
has indicial equation $r(r-1)+r-\alpha^2=0$ whose solutions are $r_{1,2}=\pm\alpha$. Here are a few possibilities:
\end{example}

\noindent\includegraphics[width=6in]{Bessel.pdf}

\noindent\begin{tabular}{cc}
  \parbox[t]{0.5\textwidth}{
    \begin{example}
      The equation
      \newline\null$\qquad\ds y''-3y'+\left(\frac{13}{2x^2}+\cos x\right)y=0$
      \newline has $r_{1,2}=\frac12\pm\frac52i$.
    \end{example}
  } &
  \imagetop{\includegraphics[width=0.5\textwidth]{OscillatoryRSP.pdf}}
\end{tabular}

\begin{exercise} Determine the behaviour near $x=0$ of solutions of the equation
\[ y''+\left(\frac{1}{2x^2} + \frac{1}{2(1-x^2)}\right)y=0. \]
\end{exercise}

\noindent\parbox{5in}{
  \begin{exercise}Using the change of variable $t=1/x$, study the behaviour of Legendre's equation of order $\alpha$,
  \[ (1-x^2)y''-2xy'+\alpha(\alpha+1)y=0, \]
  for large $x$ and for {\em all} real $\alpha$.
  \end{exercise}
}\hfill
\parbox{1in}{\centering
  \includegraphics[width=0.8in]{Legendre.jpg}
  \newline\tiny Adrien-Marie Legendre
}

\begin{exercise} Find the general solution of Legendre's equation of order $\alpha=0$,
\begin{enumerate}
\item using power series, and,
\item explicitly,
\end{enumerate}
and determine the behaviour of these solutions as $x\to\infty$.
\end{exercise}

\begin{exercise} Show that $x=0$ is a regular singular point of the equation
\[ x^3y''+2(1-\cos x)y'+(\sin x)y = 0 \]
and study the qualitative behaviour of its solutions near that point.
\end{exercise}

\begin{exercise} Show that for any non-zero value of the constant $\beta$, the point $x=\infty$ is a regular singular point of the equation
\[ x^2y''+2xy'+\beta y = 0. \]
Study the behaviour of this equation near $x=\infty$ for $\beta=-\frac34,\frac{3}{16},\frac14, \frac54$. What if $\beta=0$?
\end{exercise}

\begin{exercise} Show that $x=\infty$ is {\em not} a regular singular point for the constant-coefficient equation $y''+ay'+by=0$ for any values of $a$ and $b$ (except $a=b=0$).
\end{exercise}

\section{The Basic Oscillation Theorems}

\noindent\parbox[t]{0.67\textwidth}{
  \begin{theorem} If $q(x)<0$ for every $x$ in some connected subset $I$ of $\bbR$, then any solution of $y''+qy=0$ may have at most one zero on $I$.
  \end{theorem}
  \begin{example} Consider the solutions of $y''-y=0$ with $y(0)=1$ and $y'(0)=c$, for $c\in\{1,0,-0.9,-1,-2\}$.
  \end{example}
  \par\null\hfill\includegraphics[width=0.35\textwidth]{PushoutForce-1.pdf}
}\hfill\imagetop{\includegraphics[width=0.25\textwidth]{PushoutForce-2.pdf}}

\begin{exercise} Solve the equation $y''+\frac{3}{16x^2}y=0$, and decide if its solutions ever  oscillate.
\end{exercise}

\begin{theorem} \label{thm:oscillates} If $q(x)$ is continuous and $q(x)>0$ for all $x\geq A$ and if $\int_A^\infty q(x)dx=\infty$, then any solution to $y''+qy=0$ has infinitely many zeros for $x\geq A$.
\end{theorem}

\begin{proof} Suppose not. Then there is a solution $y$ for which $y(x)>0$ for all $x\geq B$, for some $B\geq A$. If we had $y'(C)\leq 0$ for some $C>B$, then as $y''<0$ and therefore $y'$ is decreasing, we'd have that $y'(x)<0$ for all $x>C$, and therefore there is some $x>C$ with $y(x)=0$. So it must be that $y'(x)>0$ for all $x\geq B$. Now consider $V(x):=-\frac{y'(x)}{y(x)}$. We already know it is negative for all $x\geq B$. Yet
\[ V'=-\frac{y''y-y'^2}{y^2} = \frac{qy^2+y'^2}{y^2} = q+V^2, \]
and hence
\[ V(x)=V(B)+\int_B^xV'(t)dt = V(B)+\int_B^xV^2dt+\int_B^xqdt. \]
But as $\int_B^\infty q(t)dt$ is divergent, the above quantity will become positive for large enough $x$, contradicting the negativity of $V(x)$.
\end{proof}

\begin{example} Solutions of Airy's equation $y''+xy=0$ oscillate for positive $x$ but do not oscillate for negative $x$:
\end{example}

\noindent\hbox{\parbox[t]{0.64\textwidth}{
  \imagetop{\includegraphics{Airy.pdf}}
}\parbox[t]{0.36\textwidth}{
  \null\hfill\includegraphics[width=1.2in]{George_Biddell_Airy_by_Spy.jpg}\hfill\null
  \newline\null\hfill\footnotesize George Biddell Airy,\hfill\null
  \newline\null\hfill 1801--1892\hfill\null
}}

\vskip 3mm

\newpage

In the other direction, we have the following:

\begin{theorem} \label{thm:nooscillations} Let $A>0$ be given. If $q(x)$ is continuous and $q(x)>0$ for all $x\geq A$ and if $\int_A^\infty xq(x)dx<\infty$, and if $y$ is a solution of $y''+qy=0$, then
\begin{enumerate}
\item There is some $B>A$ beyond which $y$ has no zeros.
\item There is a constant $K$ such that
\[ \lim_{x\to\infty}y'(x) = K = \lim_{x\to\infty}\frac{y(x)}{x} \]
\end{enumerate}
\end{theorem}

\begin{comment} I could not prove or find a counterexample to the statement that above, $K$ is always non-zero. If this is true then the first statement above is superfluous as it would immediately follow from the second. I didn't have time to consult with the references, \cite[page 103, problem 28]{CoddingtonLevinson} and~\cite[page 92 Theorem 3]{Coppel}.
\end{comment}

\begin{proof} Find $C>A$ such that $\int_C^\infty xqdx<1$, and assume that $y$ has at least two zeros beyond $C$; let $a$ be the first of those and let $b$ be the second. Let $\alpha=y'(a)$; without loss of generality we may assume that $\alpha>0$. Then $y'(b)<0$ and by convexity we have that on $[a,b]$, $y(x)\leq\alpha(x-a)<\alpha x$. So
\[ \alpha\leq y'(a)-y'(b)=-\int_a^b y''(x)dx=\int_a^b yq dx
  \leq \int_a^b \alpha xqdx \leq \alpha\int_C^\infty xqdx<\alpha,
\]
a contradiction. Therefore $y$ cannot have two further zeros beyond $B$, and (1) is proven.

Now we know that beyond some point $D$, $y$ is non-zero. Without loss of generality it is positive and therefore convex. It therefore lies below any of its tangents, and therefore on $[D,\infty]$ it is bounded by some linear function $\beta x$. Hence for any $a<b$ in $[D,\infty]$,
\[ \left|y'(a)-y'(b)\right| = \left|\int_a^b y''dx\right|
  = \int_a^b yq dx \leq \int_a^b \beta xqdx
  \leq \beta\int_a^\infty xqdx, \]
and the last integral goes to $0$ when $a\to\infty$. Hence $y'(x)$ is a ``Cauchy function'' (the ``function'' analog of a ``Cauchy sequence''), and hence it converges to some limit $K$. The rest follows from L'H\^opital.
\end{proof}

\begin{exercise} Show that solutions of $y''+(\log x)y=0$ oscillate as $x\to\infty$, yet have at most one zero for $0<x<1$.
\end{exercise}

\begin{exercise} Determine the behaviour of solutions of $y''+\frac{x^2-2}{x^2(x^2+1)^2}y=0$ as $x\to\infty$.
\end{exercise}

\begin{exercise} What do the above theorems say about the behaviour of solutions of $y''+\frac{y}{x^2}=0$ near $\infty$? What is their actual behaviour?
\end{exercise}

\begin{exercise} Show that all solutions of $y''+x^\alpha y=0$ are oscillatory for $x>1$ if $\alpha>-1$. For what value of $\alpha$ does Theorem~\ref{thm:nooscillations} apply to determine the large $x$ behaviour of such solutions?
\end{exercise}

\begin{exercise} Let $y$ be the solution of
\[ y''+(x^2-1)^{1/3}y = 0,\qquad y(0)=0, \qquad y'(0)=1. \]
Does $y(x)$ have other zeros for $-\infty<x<\infty$? Does it have infinitely many? What intervals $a<x<b$ cannot contain any other zeros?
\end{exercise}

\begin{exercise} How do solutions of
\[ y''+\frac{1}{(t^2+1)^{3/2}}y = 0 \]
behave as $t\to\infty$? As $t\to-\infty$?
\end{exercise}

\section{Changes of Variables}

\subsection{Changing the Dependent Variable} If $y$ satisfies $y''+p(x)y'+q(x)y=0$ and we set $y=\mu(x) V$, where $\mu$ satisfies $2\mu'+p\mu=0$, then $V$ satisfies $V''+Q(x)V=0$, where $Q=q-\frac14p^2-\frac12p'$. The good news is that $V$ has exactly the same zeros as $y$, so the ``frequency'' of the oscillatory behaviour of $y$ may be studied by studying $V''+Q(x)V=0$. Though note that ``amplitudes'' are modified.

\noindent\begin{tabular}{cc}
  \parbox[t]{0.5\textwidth}{
    \begin{example} \label{exa:BesselChangeDep} For Bessel's equation of order $0$,
      $y''+\frac1xy'+y=0$, which appeared here in Example~\ref{exa:intro},
      setting $V=\sqrt{x}y$ yields the equation $V''+\left(1+\frac{1}{4x^2}\right)V=0$, which oscillates by Theorem~\ref{thm:oscillates}:
    \end{example}
  } &
  \imagetop{\includegraphics[width=0.5\textwidth]{J0V0.pdf}}
\end{tabular}

\subsection{Changing the Independent Variable} If $y$ satisfies $y''+p(x)y'+q(x)y=0$ and we set $z=\nu(x)$, where $\nu$ satisfies $\nu''+p\nu'=0$, then the equation becomes
\[ \frac{d^2y}{dz^2}+Q(z)y=0,
  \qquad\text{where}\qquad
  Q(z)=\frac{q(x(z))}{[\nu'(x(z))]^2}.
\]
The zeros of $y$ get moved by this transformation, so studying the oscillatory behaviour of $y(x)$ as $x\to\infty$ corresponds to studying the oscillatory behaviour of $y(z)$ as $z\to\lim_{x\to\infty}\nu(x)$, and the latter point may or may not be $\infty$. Note though, that the amplitudes of oscillations (if they occur), are unchanged.

\begin{exercise} Bring the Bessel equation of order $0$ to the form $\frac{d^2y}{dz^2}+Q(z)y=0$ by a change of the independent variable and verify once more that its solutions oscillate as $x\to\infty$.
\end{exercise}

\begin{exercise} Try to determine the behaviour of solutions of the equation $y''+y'/x+y/x^3=0$ as $x\to\infty$, first by a change of the dependent variable and then by a change of the independent variable.
\end{exercise}

\begin{example} Under the change of independent variable $z(x)=x^3/3$, the equation $y''-\frac2xy'+y=0$ becomes the equation $\frac{d^2y}{dz^2}+\frac{1}{(3z)^{4/3}}y=0$:

\noindent\includegraphics[width=0.5\textwidth]{Indep1.pdf}\includegraphics[width=0.5\textwidth]{Indep2.pdf}
\end{example}

\begin{exercise} For each of the following equations, decide whether their solutions oscillate for large $x$ (here $n>0$):
\begin{enumerate}
\item $\ds x^2y''+xy'+y=0$.
\item $\ds xy''+(1-x)y'+ny=0$.
\item $\ds y''-2xy'+2ny=0$.
\item $\ds xy''+(2n+1)y'+xy=0$.
\end{enumerate}
\end{exercise}

\begin{exercise}
\begin{enumerate}
\item Study whether solutions of $x^2y''-xy'+5y=0$ oscillate as $x\to\infty$ and as $x\to-\infty$.
\item Do the same for $x^2y''-4xy'+(6-x)y=0$.
\end{enumerate}
\end{exercise}

\begin{exercise} Are there any values of $k$ for which solutions to $(1-x)y''-xy'+ky=0$ oscillate as $x\to\infty$?
\end{exercise}

\begin{exercise} How do solutions of
\[ x(x-1)y''+(3x-\frac12)y'+y=0 \]
behave as $x\to\infty$?
\end{exercise}

\begin{exercise} How do solutions of
\[ y''+\frac{1}{x^2}y'+\frac{1}{4x^4}y=0 \]
behave as $x\to\infty$?
\end{exercise}

\newpage
\section{The Sturm Comparison Theorem}

\parpic[r]{\parbox{1.1in}{\centering
  \imagetop{\includegraphics[width=1.1in]{CharlesSturm.jpg}}
  \footnotesize Charles Sturm, 1803--1855
}}
\begin{theorem} (The Sturm Comparison Theorem) Suppose $y_1$ satisfies $y''_1+q_1y_1=0$ and $y_2$ satisfies $y''_2+q_2y_2=0$ and suppose $q_2>q_1$ in some interval. Then in the open interval between any two zeros of $y_1$ there is a zero of $y_2$ (hence $y_2$ oscillates more rapidly than $y_1$).
\end{theorem}

\begin{proof} Consider $W(x):=y_1(x)y'_2(x)-y_2(x)y_1'(x)$. Then
\[ W' = y_1y_2''-y_2y_1'' = (q_1-q_2)y_1y_2. \]
\picskip{2}
Now argue by contradiction. Suppose $a$ and $b$ are successive zeros of $y_1$, and $a<b$, and that $y_2$ has no zeros on $(a,b)$. On $(a,b)$ the solution $y_1$ is non-zero; without loss of generality, it is positive. This implies that $y_1'(a)>0$ and $y_1'(b)<0$. Also without loss of generality, $y_2> 0$ on $(a,b)$. Then by the above equality and by $q_1<q_2$, it follows that $W$ is decreasing on $(a,b)$. Yet $W(a)=-y_2(a)y_1'(a)\leq 0$ and $W(b)=-y_2(b)y_1'(b)\geq 0$.
\end{proof}

\begin{corollary} \picskip{0}
Assuming $y''+qy=0$, if $q$ is increasing the the distance between successive zeros of $y$ is decreasing, and if $q$ is decreasing then the distance between successive zeros of $y$ is increasing.
\end{corollary}

\begin{proof} Assume for example that $q$ is increasing, and that $a<b$ and $c<d$ are two pairs of successive zeros of $y$, with $c>a$. Then $y_1(x):=y(x+c-a)$ solves $y_1''+q_1y_1=0$, where $q_1(x):=q(x+c-a)$, and quite clearly, $a$ and $d+a-c$ are successive zeros of $y_1$. But $q_1>q$, and for $y$, the next zero after $a$ is $b$, meaning that the next zero of $y_1$ must come before $b$. Namely, $d+a-c<b$, or alternatively, $d-c<b-a$, as required.
\end{proof}

\begin{example} As we have seen in Example~\ref{exa:BesselChangeDep} the Bessel equation of order $0$ is equivalent to the equation $V''+\left(1+\frac{1}{4x^2}\right)V=0$. Hence the distance between successive zeros of the Bessel equation of order $0$ is increasing and by comparison with $v''+v=0$, it converges to $\pi$:
\par\noindent\includegraphics[width=\textwidth]{BesselZeros.pdf}
\end{example}

\newpage

\begin{example} Solutions of Euler's equation $x^2y''+\gamma y=0$ oscillate for $\gamma>\frac14$ but do not oscillate for $\gamma\leq\frac14$:
\end{example}

\noindent\begin{tabular}{cc}
  \parbox[t]{0.32\textwidth}{
    \imagetop{\includegraphics[width=0.32\textwidth]{EulerDensity.jpg}}
    \vskip 5mm
    \null\hfill\includegraphics[width=1.5in]{EulerStamp.jpg}\hfill\null
  } &
  \imagetop{\includegraphics[width=0.64\textwidth]{Euler.jpg}}
\end{tabular}

\begin{corollary} Suppose there exist numbers $\gamma>\frac14$ and $A$ such that for all $x\geq A$ we have $q(x)>\frac{\gamma}{x^2}$. Then every solution of $y''+qy=0$ oscillates infinitely often for $x>A$. However if for all $x\geq A$ we have $q(x)\leq\frac{\gamma}{4x^2}$, then solutions of $y''+qy=0$ have at most one zero for $x\geq A$.
\end{corollary}

\parpic[r]{\includegraphics[width=3.5in]{UltraFine.png}}
\begin{exercise} Construct an equation $y''+qy=0$ whose solutions oscillate, yet so slowly that even the above corollary would not detect these oscillations. [Note that any such equation can be used as a finer comparison criterion than the one in the corollary].
\end{exercise}

\noindent{\bf Hint. } Change the independent variable to slow things down, and then the dependent variable to bring them back to the right form.

\newpage

\begin{exercise} What can you say about the spacing of the zeros of the following equations:
\begin{enumerate}
\item $\ds y''+(x^2-1)^{1/3}y=0$.
\item $y''-(x-x^3)y=0$.
\end{enumerate}
\end{exercise}

\begin{exercise} Let $y$ be a solution of Bessel's equation of order $\alpha$:
\[ y''+\frac{1}{x}y'+\left(1-\frac{\alpha^2}{x^2}\right)y=0. \]
\begin{enumerate}
\item Show that if $\alpha^2<\frac14$ then successive zeros of $y$ are separated by less than $\pi$.
\item Show that if $\alpha^2>\frac14$ then successive zeros of $y$ are separated by more than $\pi$.
\item What if $\alpha^2=\frac14$?
\end{enumerate}
\end{exercise}

\begin{exercise} Show that all solutions of $y''+\left(\frac{1}{4x^2}+e^{-x}\right)y=0$ do not oscillate.
\end{exercise}

\begin{exercise} Study the $x\to\infty$ behaviour of solutions of $y''+\frac3xy'+\left(\frac{1}{x^2}-\frac{1}{2x^4}\right)y=0$.
\end{exercise}

\begin{exercise} For which values of $k$ to all solutions of $(x^2-1)y''+xy'+ky=0$ oscillate as $x\to\infty$?
\end{exercise}

\begin{exercise} Prove that if $q(x)\to L>0$ as $x\to\infty$, then the spacing between successive zeros of solutions of $y''+qy=0$ converges to $\frac{\pi}{\sqrt{L}}$ as $x\to\infty$.
\end{exercise}

\begin{exercise} Prove the ``Sturm Separation Theorem'': If $y_1$ and $y_2$ are two linearly independent solutions of the same equation $y''+p(x)y'+q(x)y=0$, then their zeros alternate. Namely, between any two zeros of $y_1$ there is a zero of $y_2$ and between any two zeros of $y_2$ there is a zero of $y_1$.
\end{exercise}

\section{Amplitudes}

\begin{theorem} \label{thm:amplitudes} Consider a solution $y$ of the equation $y''+py'+qy=0$. If $q>0$ and $q'+2pq>0$ on some interval $[a,b]$ and $y'(a)=0=y'(b)$, then $|y(a)|>|y(b)|$. If instead $q'+2pq<0$ and $y'(a)=0=y'(b)$, then $|y(a)|<|y(b)|$. Similarly for non-strict inequalities.
\end{theorem}

\begin{proof} Consider $F=y^2+\frac{(y')^2}{q}$ and note that $F'=-(q'+2pq)\frac{(y')^2}{q^2}$.
\end{proof}

\begin{example} For Bessel's equation $y''+\frac1xy'+(1-\alpha^2/x^2)y=0$ we have $q'+2pq=2/x>0$, and hence the amplitudes of its oscillations decreases on $x>0$. Yet for $y''+y/x^2=0$ we have $q'+2pq=\frac{-2}{x^3}<0$, and hence the amplitudes of its oscillations increases on $x>0$.
\end{example}

Theorem~\ref{thm:amplitudes} has the following ``opposite'' (really, strengthening):

\begin{proposition} Under the same conditions as in the theorem, let $P$ be some primitive of $p$, meaning $P'=p$. Then
\[ e^{P(a)}\sqrt{q(a)}|y(a)| < e^{P(b)}\sqrt{q(b)}|y(b)| \qquad\text{if}\qquad q'+2pq>0, \]
and
\[ e^{P(a)}\sqrt{q(a)}|y(a)| > e^{P(b)}\sqrt{q(b)}|y(b)| \qquad\text{if}\qquad q'+2pq<0. \]
\end{proposition}

\begin{proof} Use the auxiliary function $G(x)=e^{2P}(qy^2+(y')^2)$.
\end{proof}

\begin{corollary} If $y''+qy=0$ where $q(x)\to L>0$ monotonically as $x\to\infty$, then $y$ oscillates as $x\to\infty$ with amplitudes that approach a finite, non-zero level.
\end{corollary}

\begin{exercise} Describe, as best as you can at this stage, the behaviour as $x\to\infty$ of solutions of the equation $y''+\left(1-\frac{2}{x^2}\right)y=0$.
\end{exercise}

\begin{example} Under the transformation $v=\sqrt{x}y$ Bessel's equation $y''+\frac1xy+\left(1-\frac{\alpha^2}{x^2}\right)y=0$ becomes the equation
\[ v''+\left(1+\frac{1-4\alpha^2}{4x^2}\right)v=0. \]
Thus we see that the oscillations of $v$ increase if $\alpha<\frac12$ and decrease if $\alpha>\frac12$. Further, they approach a constant level --- but this means that the oscillations of $y$ decrease like $\frac{1}{\sqrt{x}}$.
\end{example}

More can and should be said, though perhaps not on this handout.

\section{Irregular Singular Points}

Behaviour of solutions near a finite {\em irregular} singular point $x_0$ can sometimes be studied by the change of variables $t=1/(x-x_0)$. More can and should be said, though perhaps not on this handout.

\begin{thebibliography}{CL}

\bibitem[CL]{CoddingtonLevinson} E.~A.~Coddington and N.~Levinson,
  {\em Theory of Ordinary Differential Equations,}
  McGraw-Hill, New York 1955.

\bibitem[Co]{Coppel} W.~A.~Coppel,
  {\em Stability and Asymptotic Behavior of Differential Equations,}
  Heath, Boston 1965.

\end{thebibliography}

\noindent{\footnotesize
  Dror Bar-Natan, \today;
  \url{http://drorbn.net/index.php?title=12-267}.\newline
  Sources at \url{http://drorbn.net/AcademicPensieve/Classes/12-267/QualitativeAnalysis/}.
}

\newpage
\null
\vfill
\begin{center}\includegraphics[width=6in]{Bessel3D.jpg}\end{center}
\vfill
\[ x^2y''+xy'+(x^2-\alpha^2)y=0 \]
\vfill

\end{document}