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\begin{document}
\title[Effective Pr\"ufer Angles]{Effective Pr\"ufer Angles and
Relative Oscillation Criteria}
\author[H. Kr\"uger]{Helge Kr\"uger}
\address{Department of Mathematics\\ Rice University\\ Houston\\ TX 77005\\ USA}
\email{\href{mailto:helge.krueger@rice.edu}{helge.krueger@rice.edu}}
\urladdr{\href{http://www.mat.univie.ac.at/~helge/}{http://www.mat.univie.ac.at/\~{}helge/}}
\author[G. Teschl]{Gerald Teschl}
\address{Faculty of Mathematics\\
Nordbergstrasse 15\\ 1090 Wien\\ Austria\\ and International Erwin Schr\"odinger
Institute for Mathematical Physics, Boltzmanngasse 9\\ 1090 Wien\\ Austria}
\email{\href{mailto:Gerald.Teschl@univie.ac.at}{Gerald.Teschl@univie.ac.at}}
\urladdr{\href{http://www.mat.univie.ac.at/~gerald/}{http://www.mat.univie.ac.at/\~{}gerald/}}
%\thanks{{\it To appear in ...}}
\thanks{{\it Research supported by the Austrian Science Fund (FWF) under Grant No.\ Y330}}
\keywords{Sturm--Liouville operators, oscillation theory}
\subjclass[2000]{Primary 34C10, 34B24; Secondary 34L20, 34L05}
\begin{abstract}
We present a streamlined approach to relative oscillation criteria based
on effective Pr\"ufer angles adapted to the use at the edges of the essential
spectrum.
Based on this we provided a new scale of oscillation criteria for general
Sturm--Liouville operators which answer the question whether a perturbation
inserts a finite or an infinite number of eigenvalues into an essential
spectral gap. As a special case we recover the Gesztesy--\"Unal criterion
(which works below the spectrum and contains classical criteria by Kneser, Hartman, Hille, and
Weber) and the well-known results by Rofe-Beketov including the extensions by Schmidt.
\end{abstract}
\maketitle
\section{Introduction}
In this article we want to use relative oscillation theory
and apply it to obtain criteria for when an edge of the essential spectrum
is an accumulation point of eigenvalues for general Sturm--Liouville operators
\be
\tau = \left(-\frac{d}{dx} p \frac{d}{dx} + q\right),
\quad \text{on}\quad (a,b).
\ee
Without loss of generality we will assume that $a\in\R$ is a regular endpoint
and that $b$ is limit point. Furthermore, we always assume the usual
local integrability assumptions on the coefficients (see Section~\ref{sec:res}).
We will assume that $H_0$ is a given background operator associated with
$\tau_0 = (-\frac{d}{dx} p_0 \frac{d}{dx} + q_0)$ (think e.g.\ of a periodic
operator) and that $E$ is a boundary point of the essential spectrum of $H_0$
(which is not an accumulation point of eigenvalues). Then we want to know
when a {\em small} perturbation $\tau_1 = (-\frac{d}{dx} p_1 \frac{d}{dx} + q_1)$
gives rise to an infinite number of eigenvalues accumulating at $E$.
By relative oscillation theory, this question reduces to the question of when
a given operator $\tau_1-E$ is relatively oscillatory with respect to $\tau_0-E$
(cf.\ Section~\ref{sec:relosc}).
In the simplest case $\tau_0 = - \frac{d^2}{dx^2}$, $E=0$, Kneser \cite{kn} showed that
the borderline case is given by ($p_1=p_0=1$)
\be
q_1(x) = \frac{\mu}{x^2},
\ee
where the critical constant is given by $\mu_c=-\frac{1}{4}$. That is,
for $\mu<\mu_c$ the perturbation is oscillatory and for $\mu>\mu_c$ it
is nonoscillatory. In fact, later on Hartman \cite{hrt1}, Hille \cite{hl1}, and Weber \cite{wbr}
gave a whole scale of criteria addressing the case $\mu=\mu_c$. Recently this result was further
generalized by Gesztesy and \"Unal \cite{gu}, who showed that for a general Sturm--Liouville
operators (with $p_1=p_0$) the borderline case is given by
\be \label{guebc}
q_1(x) = q_0(x) + \frac{\mu}{p_0(x) u_0(x)^2 v_0(x)^2},
\ee
where the critical constant is again $\mu_c=-\frac{1}{4}$. Here
$u_0$ is a minimal positive solution of $\tau_0 u =0$ and $v_0$ is a second
linearly independent solution with Wronskian $W(u_0,v_0)=1$.
Since for $p_0=1$, $q_0=0$ we have $u_0=1$ and $v_0=x$, this result contains
Kneser's result as a special case. Moreover, they also provided a scale
of criteria for the case $\mu=\mu_c$.
While Kneser's result is classical, the analogous question for a periodic background $q_0$
(and $p_0=1$) was answered much later by Rofe-Beketov in a series of papers
\cite{rb1}--\cite{rb5} in which he eventually showed that the borderline case is again given by
\be\label{rbbc}
q_1(x) = q_0(x) + \frac{\mu}{x^2},
\ee
where the critical constant $\mu_c$ can be expressed in terms of the Floquet
discriminant. His result was recently extended by Schmidt \cite{kms} to the case
$p_0=p_1 \ne 1$ and Schmidt also provided the second term in the
case $\mu=\mu_c$.
Of course these results raise the natural question, if there is a generalization of the
Gesztesy--\"Unal result which holds inside any essential spectral gap (and not just the lowest).
Clearly (\ref{guebc}) makes no sense, since above the lowest edge of the essential spectrum,
all solutions of $\tau_0 u = E u$ have an infinite number of zeros.
However, in the periodic background case, as in the constant background case, there
is one solution $u_0$ which is bounded and a second solution $v_0$ which grows like $x$.
Hence, at least formally, the Gesztesy--\"Unal result explains why the borderline
case is given by (\ref{rbbc}). However, their proof has positivity of $H_0-E$ as
the main ingredient and thus cannot be generalized to the case above the
infimum of the spectrum.
Based on the original ideas of Rofe-Beketov and the extensions by Schmidt,
the purpose of this paper is to show that if $u_0$, $v_0$ are two linearly
independent solutions of $\tau_0 u = E u$ with Wronskian $W(u_0,v_0)=1$ such that
$u_0(x)$ grows like $\alpha(x)>0$ and $v_0(x) - \beta(x) u_0(x)$ also grows like
$\alpha(x)$ for some monotone $\beta(x) \lessgtr 0$, then the borderline case is given by $(p_0=p_1)$
\be
q_1(x) = q_0(x) + \frac{\mu\, \beta'(x)}{\alpha(x)^2 \beta(x)^2},
\ee
where the critical constant $\mu_c$ can be expressed in terms of averages
of $u_0(x)/\alpha(x)$. By virtue of d'Alembert's formula, this reduces to
(\ref{guebc}) for $E$ below the spectrum, where we can set $\alpha=u_0$ and
$\beta=\frac{v_0}{u_0}=\int p_0^{-1} u_0^{-2}$.
Our approach also allows us to provide the
full scale of criteria in the case $\mu=\mu_c$ and we will handle
the case $p_0 \ne p_1$. In particular, we will recover all previously mentioned
results as special cases and at the same time extend them to the case $p_0\ne p_1$.
\section{Main results}
\label{sec:res}
In this section we will summarize our main results. The proofs will be given in
Section~\ref{sec:epa}.
We will consider Sturm--Liouville operators on $L^{2}((a,b), r\,dx)$
with $-\infty \le a**0.
\end{equation}
We will use $\tau$ to describe the formal differentiation expression and
$H$ for the operator given by $\tau$ with separated boundary conditions at
$a$ and/or $b$.
If $a$ (resp.\ $b$) is finite and $q,p^{-1},r$ are in addition integrable
near $a$ (resp.\ $b$), we will say $a$ (resp.\ $b$) is a \textit{regular}
endpoint.
Without loss of generality we will assume that $a\in\R$ is a regular endpoint
and that $b$ is limit point.
We begin with the case where $E$ is the infimum of the spectrum of $H_0$.
Suppose that $\tau_0-E$
has a positive solution and let $u_0$ be the corresponding minimal positive solution
of $(\tau_0-E) u_0= 0$ near $b$, that is,
$$
\int^b \frac{dt}{p_0(t)u_0(t)^2} = \infty.
$$
Recall that by d'Alembert's formula there is a second linearly independent solution
\be\label{eq:dAl}
v_0(x) = u_0(x) \int_a^x \frac{dt}{p_0(t) u_0(t)^2}
\ee
satisfying $W(u_0,v_0)=1$.
\begin{theorem}\label{thm:gu}
Suppose $\tau_0-E$ has a positive solution and let $u_0$ be a minimal positive solution.
Define $v_0$ by d'Alembert's formula (\ref{eq:dAl}) and suppose
$$
\lim_{x\to b} p_0 v_0\, p_0 u_0' \Delta p = \lim_{x\to b} p_0 \Delta p = 0.
$$
Then $\tau_1-E$ is oscillatory if
\be
\limsup_{x\to b} p_0 v_0^2 ( u_0^2 \Delta q + (p_0 u_0')^2 \Delta p) < -\frac{1}{4}
\ee
and nonoscillatory if
\be
\liminf_{x\to b} p_0 v_0^2 ( u_0^2 \Delta q + (p_0 u_0')^2 \Delta p) > -\frac{1}{4}.
\ee
\end{theorem}
\begin{remark}
(i). If $u_0$ is a positive solution which is not minimal near $b$,
that is $\int^b p_0(t)^{-1} u_0(t)^{-2} dt<\infty$, then
$$
v_0(x) = u_0(x) \int_x^b \frac{dt}{p_0(t) u_0(t)^2}
$$
is a minimal positive solution.
(ii). Clearly, the requirement that $\tau_0-E$ has a positive solution can be weakened to
$\tau_0-E$ being nonoscillatory. In fact, after increasing $a$ beyond the last zero
of some solution, we can reduce the nonoscillatory case to the positive one.
\end{remark}
\noindent
The special case $\Delta p=0$ is the Gesztesy--\"Unal oscillation criterion \cite{gu}. In fact,
it is only the first one in a whole scale of oscillation criteria. To get the remaining ones,
we start by demonstrating that Kneser's classical result together with all its generalizations
follows as a special case.
To see this, we recall the iterated logarithm $\log_n(x)$ which is defined recursively via
$$
\log_0(x) = x, \qquad \log_n(x) = \log(\log_{n-1}(x)).
$$
Here we use the convention $\log(x)=\log|x|$ for negative values of $x$. Then
$\log_n(x)$ will be continuous for $x>\E_{n-1}$ and positive for $x>\E_n$, where
$\E_{-1}=-\infty$ and $\E_n=\E^{\E_{n-1}}$. Abbreviate further
$$
L_n(x) = \frac{1}{\log_{n+1}'(x)} = \prod_{j=0}^n \log_j(x),\qquad
Q_n(x) = -\frac{1}{4} \sum_{j=0}^{n-1} \frac{1}{L_j(x)^2}.
$$
\begin{corollary} \label{cor:khwh}
Fix some $n\in\N_0$ and $(a,b)=(\E_n,\infty)$. Let
$$
p_0(x) = 1, \qquad
q_0(x) = Q_n(x).
$$
and suppose
\be\label{eq:khwh}
p_1(x) = 1 +o\Big(\frac{x}{L_n(x)}\Big).
\ee
Then $\tau_1$ is oscillatory if
\be
\limsup_{x\to\infty} L_n(x)^2 \left( \Delta q(x) + \frac{\delta_n}{4 x^2} \Delta p(x) \right) < -\frac{1}{4}
\ee
and nonoscillatory if
\be
\liminf_{x\to\infty} L_n(x)^2 \left( \Delta q(x) + \frac{\delta_n}{4 x^2} \Delta p(x) \right) > -\frac{1}{4},
\ee
where $\delta_n = 0$ for $n=0$ and $\delta_n =1$ for $n \ge 1$.
\end{corollary}
\begin{proof}
Observe
$$
u_0(x)= \sqrt{L_{n-1}(x)},\qquad
v_0(x) = u_0(x) \log_n(x) = \sqrt{\log_n(x) L_n(x)}
$$
(where we set $L_{-1}(x)=1$) and check $q_0= \frac{u_0''}{u_0}$. Then
$$
p_0 v_0^2 ( u_0^2 \Delta q + (p_0 u_0')^2 \Delta p) = L_n(x)^2
\left( \Delta q(x) + \frac{1}{4} \Big(\sum_{j=0}^{n-1} \frac{1}{L_j(x)}\Big)^2
\Delta p(x) \right)
$$
where $\sum_{j=0}^{n-1} \frac{1}{L_j(x)} = 0$ for $n=0$ and
$\sum_{j=0}^{n-1} \frac{1}{L_j(x)} = x^{-1}+ o(x^{-1})$ for $n \ge 1$.
\end{proof}
\noindent
The special case $n=0$ and $\Delta p=0$ is Kneser's classical result \cite{kn}.
The extension to $n\in\N_0$ and $\Delta p=0$ is due to Weber \cite{wbr}, p.53--62, and
was later rediscovered by Hartman \cite{hrt1} and Hille \cite{hl1}.
In fact, there is an analogous scale of oscillation criteria which contains
Theorem~\ref{thm:gu} as the first one $n=0$:
\begin{theorem}\label{thm:gun}
Fix $n\in\N_0$.
Suppose $\tau_0-E$ has a positive solution and let $u_0$ be a minimal positive solution.
Define $v_0$ by d'Alembert's formula (\ref{eq:dAl}) and suppose
$$
p_0 v_0\, p_0 u_0' \Delta p = o\big(\frac{(v_0/u_0)^2}{L_n(v_0/u_0)^2}\big), \qquad
p_0 \Delta p = o\big(\frac{(v_0/u_0)^2}{L_n(v_0/u_0)^2}\big).
$$
Then $\tau_1-E$ is oscillatory if
\be
\limsup_{x\to b} L_n(\frac{v_0}{u_0})^2 \left( p_0 u_0^2 ( u_0^2 \Delta q + (p_0 u_0')^2 \Delta p)
-Q_n(\frac{v_0}{u_0}) \right) < -\frac{1}{4}
\ee
and nonoscillatory if
\be
\liminf_{x\to b} L_n(\frac{v_0}{u_0})^2 \left( p_0 u_0^2 ( u_0^2 \Delta q + (p_0 u_0')^2 \Delta p)
-Q_n(\frac{v_0}{u_0}) \right) > -\frac{1}{4}.
\ee
\end{theorem}
\noindent
The special case $\Delta p=0$ is again due to \cite{gu}. The special case $\tau_0=-\frac{d^2}{dx^2}$
gives again Corollary~\ref{cor:khwh}, however, under the (for $n>0$) somewhat stronger condition
$\lim_{x\to\infty} x^{-2} L_n(x)^2 \Delta p(x) =0$.
Moreover, there is even a version which takes averaged (rather than pointwise) deviations from
the borderline case:
\begin{theorem}\label{thm:gua}
Suppose $\tau_0-E$ has a positive solution and let $u_0$ be a minimal positive solution.
Define $v_0$ by d'Alembert's formula (\ref{eq:dAl}) and suppose
$$
p_0 v_0^2 (u_0^2 \Delta q + (p_0 u_0')^2 \Delta p) =O(1), \qquad
\lim_{x\to b} p_0 v_0\, p_0 u_0' \Delta p = \lim_{x\to b} p_0 \Delta p = 0,
$$
and $\rho= (p_0 u_0 v_0)^{-1}$ satisfies $\rho=o(1)$ and (\ref{condrho}).
Then $\tau_1-E$ is oscillatory if
\be
\limsup_{x\to b} \frac{1}{\ell} \int_x^{x+\ell} p_0(t) v_0^2(t)
\left( u_0(t)^2 \Delta q(t) + (p_0(t) u_0'(t))^2 \Delta p(t)\right) dt < -\frac{1}{4}
\ee
and nonoscillatory if
\be
\liminf_{x\to b} \frac{1}{\ell} \int_x^{x+\ell} p_0(t) v_0^2(t)
\left( u_0(t)^2 \Delta q(t) + (p_0(t) u_0'(t))^2 \Delta p(t)\right) dt > -\frac{1}{4}.
\ee
\end{theorem}
\noindent
Again we have
\begin{corollary}
Fix some $n\in\N_0$ and $(a,b)=(\E_n,\infty)$. Let
$$
p_0(x) = 1, \qquad
q_0(x) = Q_n(x)
$$
and suppose
$$
q_1(x) = Q_n(x) + O\Big(\frac{1}{L_n(x)^2}\Big), \qquad
p_1(x) = 1 + \begin{cases} o(1), & n=0,\\
O\Big(\frac{x^2}{L_n(x)^2}\Big), & n\ge 1.
\end{cases}
$$
Then $\tau_1$ is oscillatory if
\be
\inf_{\ell>0} \limsup_{x\to\infty} \frac{1}{\ell} \int_x^{x+\ell} L_n(t)^2 \left( \Delta q(t) +
\frac{\delta_n}{4 t^2} \Delta p(t) \right) dt < -\frac{1}{4}
\ee
and nonoscillatory if
\be
\sup_{\ell>0} \liminf_{x\to\infty} \frac{1}{\ell} \int_x^{x+\ell} L_n(t)^2 \left( \Delta q(t) +
\frac{\delta_n}{4 t^2} \Delta p(t) \right) dt > -\frac{1}{4},
\ee
where $\delta_n = 0$ for $n=0$ and $\delta_n =1$ for $n \ge 1$.
\end{corollary}
\noindent
To the best of our knowledge this result is new even in the special case $n=0$,
in which we have that $\tau_1$ with $q_1= O(x^{-2})$ and $p_1 = 1 + o(1)$
is oscillatory if
\be
\inf_{\ell>0} \limsup_{x\to\infty} \frac{1}{\ell} \int_x^{x+\ell} t^2 q_1(t) dt < -\frac{1}{4}
\ee
and nonoscillatory if
\be
\sup_{\ell>0} \liminf_{x\to\infty} \frac{1}{\ell} \int_x^{x+\ell} t^2 q_1(t) dt > -\frac{1}{4}.
\ee
There is also a scale of criteria given in Theorem~\ref{thm:guan} which contains
Theorem~\ref{thm:gua} as the special case $n=0$.
Our next aim is to extend these result to the case where we are not necessarily at the infimum of the
spectrum of $H_0$. We will again assume that there is a {\em minimal} solution $u_0$ (i.e., one
solution with minimal growth) such that all other solutions are of the form
$v_0= \ti{v}_0 + \beta u_0$, where $\ti{v}_0$ grows like $u_0$ and $\beta$ is some positive or negative
function, which measures how much faster $v_0$ grows on average with respect to $u_0$.
For example, in the case of periodic operators we will have that $u_0$ (and hence $\ti{v}_0$)
is bounded and $\beta(x)= \pm x$ (the sign depending on wether we are at a lower or upper edge of
the spectrum). Moreover, since expressions like $\liminf p_0 u_0^2 v_0^2 \Delta q$
will just be zero if $u_0$ (and $v_0$) have zeros, we will average over some interval.
But first of all we will state our growth condition more precisely:
\begin{definition}\label{def:regedge}
A boundary point $E$ of the essential spectrum of $H_0$ will be called admissible if there
is a minimal solution $u_0$ of $(\tau_0-E) u_0= 0$ and a second linearly independent
solution $v_0$ with $W(u_0,v_0)=1$ such that
$$
\begin{pmatrix} u_0 \\ p_0 u_0' \end{pmatrix} = O(\alpha), \quad
\begin{pmatrix} v_0 \\ p_0 v_0' \end{pmatrix} - \beta\begin{pmatrix} u_0 \\ p_0 u_0' \end{pmatrix} =
o(\alpha \beta)
$$
for some weight functions $\alpha>0$, $\beta\lessgtr 0$, where $\beta$ is absolutely continuous such
that $\rho=\frac{\beta'}{\beta}>0$ satisfies $\rho(x)=o(1)$ and (\ref{condrho}).
\end{definition}
\noindent
Clearly, two solutions as in Definition~\ref{def:regedge} can always be found if one chooses
$\alpha$ to grow faster than any solution. However, such a choice will only produce nonoscillatory
perturbations! Hence, in order to get finite critical coupling constants below, the {\em right} choice
for $\alpha$ and $\beta$ will be crucial. Also note that sign of $\beta$ will be chosen depending
on whether $E$ is a lower or upper boundary of the essential spectrum (i.e., if the essential spectral
gap starts below or above $E$).
Note that a second linearly independent solution $v_0$ with $W(u_0,v_0)=1$ can be
obtained by Rofe-Beketov's formula
\begin{align*}
v_0(x) =&
u_0(x) \int^x \frac{(q_0(t)+p_0(t)^{-1} - E\, r(t))(u_0(t)^2 - (p_0(t) u_0'(t))^2)}{(u_0(t)^2 + (p_0(t) u_0'(t))^2)^2} dt\\
& - \frac{p_0(x) u_0'(x)}{u_0(x)^2 + (p_0(x) u_0'(x))^2}
\end{align*}
(the case $p_0\ne 1$ is due to \cite{kms}). In fact, this formula can be used to show that
these assumptions are satisfied for certain almost periodic potentials (see \cite[Sect.~6.4]{rbk}).
Now, we come to our main result.
\begin{theorem}\label{thm:main}
Suppose $E$ is an admissible boundary point of the essential spectrum of $\tau_0$,
with $u_0$, $v_0$ and $\alpha$, $\beta$ as in Definition~\ref{def:regedge}.
Furthermore, suppose that we have
\be
\Delta q, \Delta p = O\big(\frac{\beta'}{\alpha^2 \beta^2}\big).
\ee
Then $\tau_1-E$ is relatively oscillatory with respect to $\tau_0-E$ if
\be
\inf_{\ell>0} \limsup_{x\to b} \frac{1}{\ell} \int_x^{x+\ell} \frac{\beta(t)^2}{\beta'(t)}
\left( u_0(t)^2 \Delta q(t) + (p_0(t) u_0'(t))^2 \Delta p(t)\right) dt < -\frac{1}{4}
\ee
and relatively nonoscillatory with respect to $\tau_0-E$ if
\be
\sup_{\ell>0} \liminf_{x\to b} \frac{1}{\ell} \int_x^{x+\ell} \frac{\beta(t)^2}{\beta'(t)}
\left( u_0(t)^2 \Delta q(t) + (p_0(t) u_0'(t))^2 \Delta p(t)\right) dt > -\frac{1}{4}.
\ee
\end{theorem}
\noindent
Note that the growth conditions from Definition~\ref{def:regedge} on the derivatives $p_0 u_0'$
and $p_0 v_0'$ are not needed if $\Delta p=0$. Similarly, the growth conditions on $u_0$ an
$v_0$ are not needed if $\Delta q=0$.
In the case where $\Delta q$ and $\Delta p$ have precise asymptotics we have:
\begin{corollary}
Suppose
\be
\Delta q = \mu \frac{\beta'}{\alpha^2 \beta^2} (1 + o(1)), \qquad
\Delta p = \nu \frac{\beta'}{\alpha^2 \beta^2} (1 + o(1)).
\ee
Then $\tau_1-E$ is relatively oscillatory with respect to $\tau_0-E$ if
\be\label{eq:reloscs}
\inf_{\ell>0} \limsup_{x\to b} \frac{1}{\ell} \int_x^{x+\ell}
\left( \mu \frac{u_0(t)^2}{\alpha(t)^2} + \nu \frac{(p_0(t) u_0'(t))^2}{\alpha(t)^2}\right) dt < -\frac{1}{4}
\ee
and relatively nonoscillatory with respect to $\tau_0-E$ if
\be\label{eq:relosci}
\sup_{\ell>0} \liminf_{x\to b} \frac{1}{\ell} \int_x^{x+\ell}
\left( \mu \frac{u_0(t)^2}{\alpha(t)^2} + \nu \frac{(p_0(t) u_0'(t))^2}{\alpha(t)^2}\right) dt > -\frac{1}{4}.
\ee
\end{corollary}
\noindent
As pointed out in \cite{kt}, the precise asymptotic requirement can be removed by a
simple Sturm-type comparison argument (see Lemma~\ref{lem:monowron} below).
In the special case where $p_0$, $q_0$, and $r$ are periodic
functions, one has $\alpha(x)=1$,
$\beta(x)=\pm x$ (with the plus sign if $E$ is a lower band edge and the minus
sign if $E$ is an upper band edge) and can take $\ell$ to be the period.
Then
$$
C_q = \frac{1}{\ell} \int_x^{x+\ell} u_0(t)^2 dt, \qquad
C_p = \frac{1}{\ell} \int_x^{x+\ell} (p_0(t) u_0'(t))^2 dt,
$$
are constants and (\ref{eq:reloscs}) respectively
(\ref{eq:relosci}) just read
$$
\mu C_q + \nu C_p \lessgtr -\frac{1}{4}.
$$
In the special case $p_0=p_1=1$ we recover Rofe-Beketov's well-known result \cite{rb3}--\cite{rb5}
since one can show (see Section~\ref{sec:floq})
$$
C_q = \frac{2 |D'(E)|}{\ell^2}
$$
for $r(x)=1$, where $D$ is the Floquet discriminant.
In the special case $\Delta p=0$ we recover the recent extension by Schmidt \cite{kms}.
If $p_0$, $q_0$ are almost periodic and there exists an almost periodic
solution at the band edge $E$, then $E$ is an admissible band edge
($\alpha(x)=1$, $\beta(x)=\pm x$) after Lemma~6.5
in \cite{rbk}. By taking $\ell\to\infty$ in our formulas we recover the
oscillation criteria by Rofe-Beketov (\cite[Thm.~6.12]{rbk}).
In \cite{rbk}, it is furthermore shown that if the spectrum of the operator
$H_0$ has a band-structure, obeying some growth condition, then there
exist almost periodic solutions at the band edge and a formula for
the critical coupling constant in terms of the band edges is provided.
Clearly, as before we can get a whole scale of criteria:
\begin{theorem}\label{thm:mainn}
Fix $n\in\N_0$.
Suppose $E$ is an admissible boundary point of the essential spectrum of $\tau_0$,
with $u_0$, $v_0$ and $\alpha$, $\beta$ as in Definition~\ref{def:regedge}.
Furthermore, suppose that we have
\be
\Delta q, \Delta p = O\big(\frac{\beta'}{\alpha^2\beta^2}\big).
\ee
Abbreviate
\be
Q = \frac{1}{\beta'} \left( u_0^2 \Delta q + (p_0 u_0')^2 \Delta p\right).
\ee
Then $\tau_1-E$ is relatively oscillatory with respect to $\tau_0-E$ at $b$ if
\be
\inf_{\ell>0} \limsup_{x\to b} \frac{L_n(\beta(x))^2}{\beta(x)^2} \left(\frac{1}{\ell} \int_{x}^{x+\ell}
\beta(t)^2 Q(t) dt - \beta(x)^2 Q_n(\beta(x)) \right) < -\frac{1}{4}
\ee
and relatively nonoscillatory with respect to $\tau_0-E$ at $b$ if
\be
\sup_{\ell>0} \liminf_{x\to b} \frac{L_n(\beta(x))^2}{\beta(x)^2} \left(\frac{1}{\ell} \int_{x}^{x+\ell}
\beta(t)^2 Q(t) dt - \beta(x)^2 Q_n(\beta(x)) \right) > -\frac{1}{4}.
\ee
\end{theorem}
As a consequence we get:
\begin{corollary}
Let $\tau_0$ be periodic with $r(x)=1$ and let $n\in\N_0$. Define
$$
\mu_c= -\frac{2 |D|'(E)}{\ell^2},
$$
and suppose
\be
q_1 = q_0 + \frac{1}{\mu_c} \Big(Q_n + \frac{\mu}{L_n^2} \Big)+
o\Big(\frac{1}{L_n^2}\Big), \qquad
p_1 = 1 + o\Big(\frac{1}{L_n^2}\Big).
\ee
Then $\tau_1-E$ is relatively oscillatory with respect to $\tau_0-E$ if
\be
\mu < -\frac{1}{4}
\ee
and relatively nonoscillatory with respect to $\tau_0-E$ if
\be
\mu > -\frac{1}{4}.
\ee
\end{corollary}
\noindent
Again the special case $n=1$ and $\Delta p=0$ is due to \cite{kms}.
The assumption $k(x)=1$ can be dropped, but then $\mu_c$ can no longer
be expressed in terms of the derivative of the Floquet discriminant.
\section{Relative Oscillation Theory in a Nutshell}
\label{sec:relosc}
In this section, we give a brief overview of relative oscillation theory, as developed by us
in \cite{kt} respectively \cite{kt2}. In particular we refer to these papers for further results
and proofs.
Our main object will be the (modified) Wronskian
\be
W_x(u_0,u_1)= u_0(x)\, p_1(x)u_1'(x) - p_0(x)u_0'(x)\, u_1(x)
\ee
of two functions $u_0$, $u_1$ and its zeros. Here we think of $u_0$ and $u_1$ as two
solutions of two different Sturm--Liouville equations
\be \label{stlij}
\tau_j = \frac{1}{r} \Big(- \frac{d}{dx} p_j \frac{d}{dx} + q_j \Big), \qquad j=0,1.
\ee
Under these assumptions $W_x(u_0,u_1)$ is absolutely continuous and satisfies
\be\label{dwr}
W'(u_0, u_1) = (q_1 - q_0) u_0 u_1 +
\left(\frac{1}{p_0} - \frac{1}{p_1}\right) p_0 u_0' p_1 u_1'.
\ee
Next we recall the definition of Pr\"ufer variables $\rho_u$, $\theta_u$
of an absolutely continuous function $u$:
\be\label{eq:defprue}
u(x)=\rho_u(x)\sin(\theta_u(x)), \qquad
p(x) u'(x)=\rho_u(x) \cos(\theta_u(x)).
\ee
If $(u(x), p(x) u'(x))$ is never $(0,0)$ and $u, pu'$ are absolutely continuous, then $\rho_u$ is
positive and $\theta_u$ is uniquely determined once a value of
$\theta_u(x_0)$ is chosen by requiring continuity of $\theta_u$.
Notice that
\be \label{wpruefer}
W_x(u,v)= -\rho_u(x)\rho_v(x)\sin(\Delta_{v,u}(x)), \qquad
\Delta_{v,u}(x) = \theta_v(x)-\theta_u(x).
\ee
Hence the Wronskian vanishes if and only if the two Pr\"ufer angles differ by
a multiple of $\pi$.
We take two solutions $u_j$, $j=1,2$, of $\tau_j u_j =\lam_j u_j$ and
associated Pr\"ufer variables $\rho_j$, $\theta_j$.
%Since we can
%replace $q \to q - \lam r$ it is no restriction to assume $\lam_0=\lam_1=0$.
%We remark, that in (\ref{eq:defprue}) one has to take $p_j$ as $p$ for $u_j$,
%$j=0,1$.
We will call the total difference
\be\label{eq:wprueferwron}
\#_{(c,d)}(u_0,u_1) = \ceil{\Delta_{1,0}(d) / \pi} - \floor{\Delta_{1,0}(c) / \pi} -1
\ee
the number of weighted sign flips in $(c,d)$, where we have written $\Delta_{1,0}(x)=
\Delta_{u_1,u_0}$ for brevity.
One can interpret $\#_{(c,d)}(u_0,u_1)$ as the weighted sign flips of the Wronskian
$W_x(u_0,u_1)$, where a sign flip is counted as $+1$ if $q_0-q_1$ and $p_0-p_1$ are positive
in a neighborhood of the sign flip, it is counted as $-1$ if $q_0-q_1$ and $p_0-p_1$ are negative
in a neighborhood of the sign flip.
After these preparations we are now ready for
\begin{definition}\label{def:wsf}
For $\tau_0$, $\tau_1$ possibly singular Sturm--Liouville operators as in (\ref{stlij})
on $(a,b)$, we define
\be
\underline{\#}(u_0,u_1) = \liminf_{d \uparrow b,\,c \downarrow a} \#_{(c,d)}(u_0,u_1)
\quad\mbox{and}\quad
\overline{\#}(u_0,u_1) = \limsup_{d \uparrow b,\,c \downarrow a} \#_{(c,d)}(u_0,u_1),
\ee
where $\tau_j u_j = \lam_j u_j$, $j=0,1$.
We say that $\#(u_0,u_1)$ exists, if
$\overline{\#}(u_0,u_1)=\underline{\#}(u_0,u_1)$, and write
\be
\#(u_0,u_1) = \overline{\#}(u_0,u_1)=\underline{\#}(u_0,u_1).
\ee
in this case.
\end{definition}
\noindent
One can show that $\#(u_0,u_1)$ exists
if $p_0-p_1$ and $q_0-\lam_0r- q_1+\lam_1r$ have the same definite sign
near the endpoints $a$ and $b$.
We recall that in classical oscillation theory
$\tau$ is called oscillatory if a solution of $\tau u = 0$ has infinitely many
zeros.
\begin{definition}\label{def:relosc}
We call $\tau_1$ relatively nonoscillatory with respect to
$\tau_0$, if the quantities $\underline{\#}(u_0, u_1)$ and
$\overline{\#}(u_0, u_1)$ are finite for all solutions
$\tau_j u_j = 0$, $j = 0,1$.
We call $\tau_1$ relatively oscillatory with respect to
$\tau_0$, if one of the quantities $\underline{\#}(u_0, u_1)$ or
$\overline{\#}(u_0, u_1)$ is infinite for some solutions
$\tau_j u_j = 0$, $j = 0,1$.
\end{definition}
\noindent
It turns out that this definition is in fact independent of the solutions
chosen. Moreover, since a Sturm-type comparison theorem holds
for relative oscillation theory, we have
\begin{lemma}\label{lem:monowron}
If $\tau_1$ is relatively oscillatory with respect to $\tau_0$ for $p_1 \geq p_0$, $q_1 \leq q_0$
then the same is true for any $\tau_2$ with $p_2 \geq p_1$, $q_2 \leq q_1$.
Similarly, if $\tau_1$ is relatively nonoscillatory with respect to $\tau_0$ for $p_1 \geq p_0$,
$q_1 \leq q_0$ then the same is true for any $\tau_2$ with $p_1 \leq p_2 \leq p_0$,
$q_1 \leq q_2 \leq q_0$.
\end{lemma}
The connection between this definition and the spectrum is given by:
\begin{theorem} \label{thm:rosc}
Let $H_j$ be self-adjoint operators associated with $\tau_j$, $j=0,1$. Then
\begin{enumerate}
\item
$\tau_0-\lam_0$ is relatively nonoscillatory with respect to $\tau_0-\lam_1$
if and only if $\dim\Ran P_{(\lam_0,\lam_1)}(H_0)<\infty$.
\item
Suppose $\dim\Ran P_{(\lam_0,\lam_1)}(H_0)<\infty$ and
$\tau_1-\lam$ is relatively nonoscillatory with respect to $\tau_0-\lam$ for one
$\lam \in [\lam_0,\lam_1]$. Then it is relatively nonoscillatory for
all $\lam\in[\lam_0,\lam_1]$ if and only if $\dim\Ran P_{(\lam_0,\lam_1)}(H_1)<\infty$.
\end{enumerate}
\end{theorem}
\noindent
We will now give a first nonoscillation theorem.
%For a practical application of this theorem one needs of course criteria
%when $\tau_1-\lam$ is relatively nonoscillatory with respect to $\tau_0-\lam$
%for $\lam$ inside an essential spectral gap.
\begin{lemma}\label{lem:nonoscingap2}
Let $H_0$ be bounded from below. Suppose $a$ is regular ($b$ singular) and
\begin{enumerate}
\item
$\lim_{x\to b} r(x)^{-1} (q_0(x) - q_1(x)) = 0$, $\frac{q_0}{r}$ is bounded near $b$, and
\item
$\lim_{x\to b} p_1(x) p_0(x)^{-1} = 1$.
\end{enumerate}
Then $\sig_{ess}(H_0)=\sig_{ess}(H_1)$ and
$\tau_1 - \lam$ is relatively nonoscillatory with respect
to $\tau_0 - \lam$ at $b$ for every $\lam \in \R \backslash\sigma_{ess}(H_0)$.
The analogous result holds for $a$ singular and $b$ regular.
\end{lemma}
We furthermore note that we have.
\begin{lemma}\label{lem:estproj}
Let $H_0$, $H_1$ be self-adjoint operators associated with $\tau_0$, $\tau_1$,
respectively, and separated boundary conditions. Suppose
that $(\lam_0, \lam_1)\subseteq\R\backslash(\sig_{ess}(H_0)\cup\sig_{ess}(H_1))$, then
\begin{align}
\nonumber
\dim\Ran P_{(\lam_0, \lam_1)} (H_1) &-
\dim\Ran P_{(\lam_0, \lam_1)} (H_0)\\
& \leq \underline{\#}(\psi_{1,\mp} (\lam_1), \psi_{0,\pm} (\lam_1)) -
\overline{\#}(\psi_{1,\mp} (\lam_0), \psi_{0,\pm} (\lam_0)) + 2,
\end{align}
respectively,
\begin{align}
\nonumber
\dim\Ran P_{(\lam_0, \lam_1)} (H_1) &-
\dim\Ran P_{(\lam_0, \lam_1)} (H_0)\\
& \geq \overline{\#}(\psi_{1,\mp} (\lam_1), \psi_{0,\pm} (\lam_1)) -
\underline{\#}(\psi_{1,\mp} (\lam_0), \psi_{0,\pm} (\lam_0)) - 2.
\end{align}
\end{lemma}
Turning the above inequalities into equalities, was one of the topics of
\cite{kt} and \cite{kt2}.
\section{Effective Pr\"ufer angles and relative oscillation criteria}
\label{sec:epa}
As in the previous section, we will consider two Sturm--Liouville
operators $\tau_j$, $j=0,1$, and corresponding self-adjoint operators $H_j$, $j=0,1$.
Now we want to answer the question, when a boundary point $E$ of the essential
spectrum of $H_0$ is an accumulation point of eigenvalues of $H_1$.
By Theorem~\ref{thm:rosc} we need to investigate if $\tau_1-E$ is relatively oscillatory
with respect to $\tau_2-E$ or not, that is, if the difference of Pr\"ufer angels
$\Delta_{1,0}=\theta_1-\theta_0$ is bounded or not.
Hence the first step is to derive an ordinary differential equation for $\Delta_{1,0}$. While
this can easily be done, the result turns out to be not very effective for our purpose.
However, since the number of weighted sign flips $\#_{(c,d)}(u_0,u_1)$ is all we
are eventually interested in, any {\em other} Pr\"ufer angle which gives the same
result will be as good:
\begin{definition}
We will call a continuous function $\psi$ a Pr\"ufer angle for the Wronskian $W(u_1,u_0)$, if
$\#_{(c,d)}(u_1,u_0) = \ceil{\psi(d) / \pi} - \floor{\psi(c) / \pi} -1$ for any $c,d\in(a,b)$.
\end{definition}
Hence we will try to find a more effective Pr\"ufer angle $\psi$ than $\Delta_{1,0}$ for the Wronskian
of two solutions. The right choice was found by Rofe-Beketov \cite{rb2}--\cite{rb5} (see also the recent
monograph \cite{rbk}):
Let $u_0, v_0$ be two linearly independent solutions of $(\tau_0-\lam) u = 0$ with $W(u_0,v_0)=1$
and let $u_1$ be a solution of $(\tau_1-\lam) u = 0$. Define $\psi$ via
\be\label{def:psi}
W(u_0,u_1)= -R \sin(\psi), \qquad W(v_0,u_1)= -R \cos(\psi).
\ee
Since $W(u_0,u_1)$ and $W(v_0,u_1)$ cannot vanish simultaneously, $\psi$ is a well-defined
absolutely continuous function, once one value at some point $x_0$ is fixed.
\begin{lemma}
The function $\psi$ defined in (\ref{def:psi}) is a Pr\"ufer angle for the Wronskian $W(u_0,u_1)$.
\end{lemma}
\begin{proof}
Since $W(u_0,u_1)= -R \sin(\psi) = -\rho_{u_0}\rho_{u_1}\sin(\Delta_{1,0})$ it suffices
to show that $\psi = \Delta_{1,0} \mod 2\pi$ at each zero of the Wronskian. Since we can assume
$\theta_{v_0}-\theta_{u_0}\in(0,\pi)$ (by $W(u_0,v_0)=1$), this follows by
comparing signs of $R\cos(\psi)= \rho_{v_0}\rho_{u_1}\sin(\theta_{u_1}-\theta_{v_0})$.
\end{proof}
\begin{lemma}\label{lem:lemusualprue}
Let $u_0, v_0$ be two linearly independent solutions of $(\tau_0-\lam) u = 0$ with $W(u_0,v_0)=1$
and let $u_1$ be a solution of $(\tau_1-\lam) u = 0$.
Then the Pr\"ufer angle $\psi$ for the Wronskian $W(u_0,u_1)$ defined in (\ref{def:psi})
obeys the differential equation
\be\label{eq:diff1wronski}
\psi' = -\Delta q \big(u_0 \cos (\psi) - v_0 \sin(\psi)\big)^2
- \Delta p \big(p_0 u_0' \cos(\psi) - p_0 v_0' \sin(\psi)\big)^2,
\ee
where
$$
\Delta p = \frac{1}{p_0} - \frac{1}{p_1}, \qquad
\Delta q = q_1 -q_0.
$$
\end{lemma}
\begin{proof}
Observe $R \psi' = -W(u_0,u_1)' \cos(\psi) + W(v_0,u_1)' \sin(\psi)$
and use (\ref{dwr}), (\ref{def:psi}) to evaluate the right hand side.
\end{proof}
\begin{remark}
Special cases of the {\em phase equation} (\ref{eq:diff1wronski}) have been used
in the physics literature before (\cite{ba}, \cite{ca}). Moreover, $\psi$ was originally
not interpreted as Pr\"ufer angle for Wronskians, but defined via
\be
\begin{pmatrix} u_1 \\ p_1 u_1' \end{pmatrix} =
\begin{pmatrix} v_0 & u_0 \\ p_0 v_0' & p_0 u_0' \end{pmatrix}
\begin{pmatrix} -R \sin(\psi) \\ R \cos(\psi)\end{pmatrix}.
\ee
Augmenting the definition
$$
\begin{pmatrix} u_0 & u_1 \\ p_0 u_0' & p_1 u_1' \end{pmatrix} =
\begin{pmatrix} v_0 & u_0 \\ p_0 v_0' & p_0 u_0' \end{pmatrix}
\begin{pmatrix} 0 & -R \sin(\psi) \\ 1 & R \cos(\psi)\end{pmatrix},
$$
and taking determinants shows $W(u_0,u_1)= -R \sin(\psi)$. Similarly we obtain
$W(v_0,u_1)= -R \cos(\psi)$ and hence this definition is equivalent to (\ref{def:psi}).
\end{remark}
\noindent
In the case $p_0=p_1$ equation (\ref{eq:diff1wronski}) can be interpreted as the Pr\"ufer
equation of an associated Sturm--Liouville equation with coefficients given rather implicitly
by means of a Liouville-type transformation of the independent variable. Hence a standard
oscillation criterion of Hille and Wintner \cite[Thm~2.12]{sw} can be used.
This is the original strategy by Rofe-Beketov (see \cite[Sect.~6.3]{rbk}).
In fact, using the transformation $\eta=\tan(\psi)$ it is straightforward to check that $\psi$ satisfies
(\ref{eq:diff1wronski}) if $\eta$ satisfies the Riccati equation
\be
\eta' = -\Delta q \big(u_0 - v_0 \eta\big)^2
- \Delta p \big(p_0 u_0' - p_0 v_0' \eta\big)^2.
\ee
Hence we obtain
\begin{lemma}
Suppose $\Delta p =0$ and $\Delta q<0$. Then $\tau_1$ is relatively (non)oscillatory with
respect to $\tau_0$ if and only if the Sturm--Liouville equation associated with
$$
p^{-1} = -\Delta q\, v_0^2 \exp(2 \int \Delta q\, u_0 v_0), \qquad
q =-\Delta q\, u_0^2\, \exp(-2 \int\Delta q\, u_0 v_0)
$$
is (non)oscillatory.
\end{lemma}
\begin{proof}
Making another transformation $\phi= \exp(-2 \int \Delta q\, u_0 v_0) \eta$
we can eliminate the linear term to obtain the Riccati equation
$$
\phi' = q + \frac{1}{p} \phi^2
$$
for the logarithmic derivative $\phi=\frac{p u'}{u}$ of solutions of the above Sturm--Liouville
equation.
\end{proof}
\noindent
Clearly, an analogous result holds for the case where $\Delta q=0$ and $\Delta p <0$.
Since most oscillation criteria are for the case $p=1$, a Liouville-type transformation
is required before they can be applied. Nevertheless, in order to handle the general case
$\Delta q\ne 0$ and $\Delta p \ne 0$ we will use a more direct approach.
However, even though equation (\ref{eq:diff1wronski}) is strikingly simple, it is still not well suited for
a direct analysis, since in general $u_0$ and $v_0$ will have different growth behaviour (e.g.,
for $\tau_0=-\frac{d^2}{dx^2}$ we have $u_0(x)=1$ and $v_0(x)=x$ at the boundary of the spectrum).
In order to fix this problem Schmidt \cite{kms} proposed to use yet another Pr\"ufer angle $\varphi$
given by the Kepler transformation
\be
\cot(\psi) = \alpha \cot(\varphi) + \beta,
\ee
where $\alpha\lessgtr 0$ and $\beta$ are arbitrary absolutely continuous functions. It is straightforward to
check that there is a unique choice for $\varphi$ such that it is again absolutely continuous and
satisfies $\floor{\frac{\psi}{\pi}} = \floor{\frac{\varphi}{\pi}}$:
\be\label{def:varphi}
\varphi = \begin{cases}
\sgn(\alpha) n \pi, & \psi = n\pi,\\
\sgn(\alpha) n\pi + \arccot(\alpha^{-1}(\cot(\psi) - \beta)), & \psi\in(n\pi,(n+1)\pi),
\end{cases} \qquad n\in\Z,
\ee
where the branch of $\arccot$ is chosen to have values in $(0,\pi)$.
The differential equation for $\varphi$ reads as follows:
\begin{lemma}
Let $u_0, v_0$ be two linearly independent solutions of $(\tau_0-\lam) u = 0$ with $W(u_0,u_1)=1$
and let $u_1$ be a solution of $(\tau_1-\lam) u = 0$.
Moreover, let $\alpha\lessgtr 0$ and $\beta$ be arbitrary absolutely continuous functions.
Then $\sgn(\alpha) \varphi$, with $\varphi$ defined in (\ref{def:varphi}), is a
Pr\"ufer angle $\varphi$ for the Wronskian $W(u_0,u_1)$ and
obeys the differential equation
\begin{align} \label{deqvp}
\nn\varphi' =& \frac{\alpha'}{\alpha} \sin(\varphi) \cos(\varphi) + \frac{\beta'}{\alpha} \sin^2(\varphi) \\
& - \frac{\Delta q}{\alpha}\big(\alpha\, u_0 \cos(\varphi) - (v_0 - \beta\, u_0) \sin(\varphi) \big)^2\\ \nn
& - \frac{\Delta p}{\alpha}\big(\alpha\, p_0 u_0' \cos(\varphi) - (p_0 v_0' - \beta\, p_0 u_0')
\sin(\varphi) \big)^2.
\end{align}
\end{lemma}
\begin{proof}
Rewrite (\ref{eq:diff1wronski}) as
$$
\frac{\psi'}{\sin(\psi)^2} =
-\Delta q \big(u_0 \cot(\psi) - v_0)\big)^2
-\Delta p \big(p_0 u_0' \cot(\psi) - p_0 v_0')\big)^2.
$$
On the other hand one computes
$$
\frac{\psi'}{\sin(\psi)^2} = -(\cot(\psi))' = -\left(\alpha \cot(\varphi) + \beta\right)'
= \alpha \frac{\varphi'}{\sin(\varphi)^2} - \alpha' \cot(\varphi) - \beta'
$$
and solving for $\varphi'$ gives (\ref{deqvp}).
\end{proof}
\noindent
We will mainly be interested in the special case $\alpha=\beta$, where
\begin{align}\label{eq:prueang2}
\varphi' =& \frac{\beta'}{\beta} \left(\sin^2(\varphi) + \sin(\varphi) \cos(\varphi) \right)\\ \nn
& - \beta\, \Delta q\big(u_0 \cos(\varphi) - \frac{1}{\beta} (v_0 - \beta\, u_0) \sin(\varphi) \big)^2\\ \nn
& - \beta\, \Delta p\big(p_0 u_0' \cos(\varphi) - \frac{1}{\beta} (p_0 v_0' - \beta\, p_0 u_0')
\sin(\varphi) \big)^2.
\end{align}
Note that if $\beta<0$ then not $\varphi$, but $-\varphi$ is a Pr\"ufer angle. However, this choice
will avoid case distinctions later on.
Now we turn to applications of this result. As a warm up we will treat the
case where $E$ is the infimum of the spectrum of $H_0$ and prove Theorem~\ref{thm:gu}.
\begin{proof}[Proof of Theorem~\ref{thm:gu}]
Since $\tau_0-E$ is nonoscillatory, $\tau_1-E$ is relatively (non)oscillatory
with respect to $\tau_0-E$ if and only if $\tau_1-E$ is (non)oscillatory.
Set $\beta= \frac{v_0}{u_0} = \int p_0^{-1} u_0^{-2} dt$ and $\rho=\frac{\beta'}{\beta}=
\frac{1}{p_0 u_0 v_0}$. Now observe that (\ref{eq:prueang2}) reads
\begin{align*}
\varphi' =& \rho \big( \sin^2(\varphi) + \sin(\varphi) \cos(\varphi)
- p_0 v_0^2 u_0^2 \Delta q \cos^2(\varphi)\\ & -
p_0 v_0^2 \Delta p(p_0 u_0' \cos(\varphi) - \frac{1}{v_0} \sin(\varphi))^2 \big)\\
=& \rho \left( \sin^2(\varphi) + \sin(\varphi) \cos(\varphi)
- p_0 v_0^2 (u_0^2 \Delta q + (p_0 u_0')^2 \Delta p) \cos^2(\varphi) \right)
+o(\rho)
\end{align*}
and use Corollary~\ref{cor:boundsol}. The latter is applicable since $\rho>0$ and
$\int^b \rho(x) dx = \int^b \frac{\beta'(x) dx}{\beta(x)} =
\lim_{x\to b} \log(\beta(x)) =\infty$.
\end{proof}
\noindent
Now note that Corollary~\ref{cor:khwh} in turn gives us an criterion when the differential equation
for our Pr\"ufer angle has bounded solutions:
\begin{lemma}\label{lem:odebnd}
Fix some $n\in\N_0$, let $Q$ be a locally integrable on $(a,b)$ and suppose $\beta\lessgtr 0$
is absolutely continuous with $\rho=\frac{\beta'}{\beta}>0$ locally bounded and
$\lim_{x\to b} |\beta(x)|=\infty$. Then all solutions of the differential equation
\be
\varphi' = \rho \left(\sin^2(\varphi) + \sin(\varphi) \cos(\varphi) -
\beta^2 Q \cos^2(\varphi) \right) + o\big(\frac{\rho \beta^2}{L_n(\beta)^2}\big)
\ee
tend to $\infty$ if
$$
\limsup_{x\to b} L_n(\beta(x))^2 \left(Q(x) - Q_n(\beta(x))\right) < -\frac{1}{4}
$$
and are bounded above if
$$
\liminf_{x\to b} L_n(\beta(x))^2 \left(Q(x) - Q_n(\beta(x))\right) > -\frac{1}{4}.
$$
In the last case all solutions are bounded under the additional assumption
$Q =Q_n(\beta) + O(L_n(\beta)^{-2})$.
\end{lemma}
\begin{proof}
The case $n=0$ is Lemma~\ref{lem:boundsol} and hence we can assume $n\ge 1$.
By a change of coordinates $y=\beta(x)$ we can reduce everything to the case $\beta(x)=x$
(and $b=\infty$).
Now we start by showing that
\begin{align*}
\varphi' =& \frac{1}{x} \left(\Big(1-\frac{A x^2}{L_n(x)} \Big) \sin^2(\varphi)
+ \sin(\varphi) \cos(\varphi) -
x^2 \Big(Q_n + \frac{B}{4 L_n(x)^2}\Big) \cos^2(\varphi) \right)\\ & + o\Big(\frac{x}{L_n(x)^2}\Big)
\end{align*}
has only bounded solutions if $A+B>-1$ and only unbounded solutions (tending
to $\infty$) if $A+B<-1$. Since the error term $o(x L_n(x)^{-2})$ can be bounded
by $\eps x L_n(x)^{-2}(\sin^2(\varphi)+\cos^2(\varphi))$ it suffices to show this for one equation
in this class by an easy sub/super-solution argument: If $A+B<-1$, then any
solution of one equation with slightly smaller $A$ and $B$ is a sub-solution and hence forces the
solution to go to $\infty$. Similarly, If $A+B>-1$, then any
solution of one equation with slightly smaller $A$ and $B$ is a sub-solution and any
solution of one equation with slightly larger $A$ and $B$ is a super-solution, which together
bound the solutions.
To see the claim for one equation in this class note that unboundedness (boundedness) of solutions
is equivalent to $\tau_1 = -d^2/dx^2 + Q$ being relatively (non)oscillatory with respect to
$\tau_0 = -d^2/dx^2$. Hence it suffices to choose $\alpha=x(1+ A x^2 L_n^{-2})$,
$\beta= x$ and $Q = Q_n + (A+B) / (4L_n^2)$ in (\ref{deqvp}) and
invoke Corollary~\ref{cor:khwh}.
Finally, the claim from the lemma follows from this result together with another sub/super-solution
argument.
\end{proof}
\noindent
The special cases $n=0,1$ are essentially due to Schmidt (\cite[Prop.~3 and~4]{kms}).
With this result, we can now prove Theorem~\ref{thm:gun}:
\begin{proof}[Proof of Theorem~\ref{thm:gun}]
Set $\beta= \frac{v_0}{u_0} = \int p_0^{-1} u_0^{-2} dt$ and
$Q= p_0 u_0^2 (u_0^2 \Delta q + (p_0 u_0')^2 \Delta p)$.
As in the proof of Theorem~\ref{thm:gu}, (\ref{eq:prueang2}) reads
$$
\varphi' = \rho \left( \sin^2(\varphi) + \sin(\varphi) \cos(\varphi)
- \beta^2 Q \cos^2(\varphi) \right) +o\big(\frac{\rho\beta^2}{L_n(\beta)^2}\big)
$$
and invoking Lemma~\ref{lem:odebnd} finishes the proof (note that $\psi$ and hence also $\varphi$
is always bounded from below, since $\tau_0$ is nonoscillatory).
\end{proof}
\noindent
Moreover, by averaging we obtain
\begin{theorem}\label{thm:guan}
Suppose $\tau_0-E$ has a positive solution and let $u_0$ be a minimal positive solution.
Define $v_0$ by d'Alembert's formula (\ref{eq:dAl}) and abbreviate
\be
Q(x) = p_0(x) u_0^2(x) \left( u_0(x)^2 \Delta q(x) + (p_0(x) u_0'(x))^2 \Delta p(x)\right), \qquad
\beta(x)= \frac{v_0(x)}{u_0(x)}.
\ee
Suppose
$$
\beta^2 Q =O(1), \qquad
p_0 v_0\, p_0 u_0' \Delta p = o\Big(\frac{\beta^2}{L_n(\beta)}\Big), \quad
p_0 \Delta p = o\Big(\frac{\beta^2}{L_n(\beta)}\Big),
$$
and $\rho= (p_0 u_0 v_0)^{-1}$ satisfies $\rho=o(1)$ and (\ref{condrho}).
Then $\tau_1-E$ is oscillatory if
\be
\inf_{\ell>0} \limsup_{x\to b} \frac{L_n(\beta(x))^2}{\beta(x)^2} \left(\frac{1}{\ell} \int_{x}^{x+\ell}
\beta(t)^2 Q(t) dt - \beta(x)^2 Q_n(\beta(x)) \right) < -\frac{1}{4}
\ee
and nonoscillatory if
\be
\sup_{\ell>0} \liminf_{x\to b} \frac{L_n(\beta(x))^2}{\beta(x)^2} \left(\frac{1}{\ell} \int_{x}^{x+\ell}
\beta(t)^2 Q(t) dt - \beta(x)^2 Q_n(\beta(x)) \right) > -\frac{1}{4}.
\ee
\end{theorem}
\begin{proof}
Derive the differential equation for $\varphi$ as in the proof of Theorem~\ref{thm:gu} and
then take averages using Corollary~\ref{cor:aver}. Observe that the error
term is preserved by monotonicity of $\frac{\beta^2}{L_n(\beta)^2}$ and (\ref{condrho}).
\end{proof}
\noindent
Now we turn to the case above the infimum of the essential spectrum.
\begin{proof}[Proof of Theorem~\ref{thm:mainn}]
Observe that (\ref{eq:prueang2}) reads
$$
\varphi' = \frac{\beta'}{\beta} \left( \sin^2(\varphi) + \sin(\varphi) \cos(\varphi)
- \beta^2 Q \cos^2(\varphi) \right) +o\Big(\frac{\rho \beta^2}{L_n(\beta)^2}\Big).
$$
Average over a length $\ell$ using Corollary~\ref{cor:aver} and observe that the error
term is preserved by monotonicity of $\frac{\beta^2}{L_n(\beta)^2}$ and (\ref{condrho}).
Now apply Lemma~\ref{lem:odebnd}.
\end{proof}
\begin{corollary}
Suppose
\be
\rho= o(\frac{\beta^2}{L_n(\beta)^2}), \quad\text{and}\quad
\frac{1}{\ell} \int_x^{x+\ell} \frac{u_0(t)^2}{\alpha(t)^2} dt = C + o\Big(\frac{\beta^2}{L_n(\beta)^2}\Big)
\ee
for some $\ell>0$. Furthermore, assume
\be
\Delta q = \frac{\beta'}{\alpha^2 C} \Big(Q_n(\beta) + \frac{\mu}{L_n(\beta)^2} \Big)+
o\Big(\frac{\beta'}{\alpha^2 L_n(\beta)^2)}\Big), \qquad
\Delta p = o\Big(\frac{\beta'}{\alpha^2 L_n(\beta)^2}\Big).
\ee
Then $\tau_1-E$ is relatively oscillatory with respect to $\tau_0-E$ if
\be
\mu < -\frac{1}{4}
\ee
and relatively nonoscillatory with respect to $\tau_0-E$ if
\be
\mu > -\frac{1}{4}.
\ee
\end{corollary}
\begin{proof}
It is sufficient to show that
$$
\frac{1}{\ell} \int_x^{x+\ell} \left( \frac{\beta(t)^2}{L_j(\beta(t))^2}
- \frac{\beta(x)^2}{L_j(\beta(x))^2} \right) \frac{u_0(t)^2}{\alpha(t)^2} dt =
o\Big(\frac{\beta^2(x)}{L_n(\beta(x))^2}\Big)
$$
for $j = 0, \dots, n$. Since $u_0 \alpha^{-1}$ is bounded, this follows since
by the mean value theorem and monotonicity of $\beta$ we have
$$
\frac{1}{\ell} \int_x^{x+\ell} \left|\frac{\beta(t)^2}{L_j(\beta(t))^2} - \frac{\beta(x)^2}{L_j(\beta(x))^2} \right| dt
\le \frac{\beta(x)^2}{L_j(\beta(x))^2} \sum_{k=1}^{j} \frac{\beta(x)}{L_k(\beta(x))} \sup_{t\in[0,\ell]}\rho(x+t),
$$
finishing the proof.
\end{proof}
\noindent
Note that the assumptions hold for periodic operators by choosing $\ell$ to be the period.
Furthermore, inspection of the proof shows that if $|\beta|\to\infty$, then
$\rho=o(\beta^2 L_n(\beta)^{-2})$ can be replaced by $\rho=O(\beta^2 L_n(\beta)^{-2})$.
\section{Appendix: Averaging ordinary differential equations}
\label{sec:ode}
In Section~\ref{sec:epa} we have reduced everything to the question if certain
ordinary differential equation have bounded solutions or not. In this section
we collect the required results for these ordinary differential equations
All proofs are elementary and we give them for the sake of completeness.
\begin{lemma}\label{lem:boundsol}
Suppose $\rho(x)>0$ (or $\rho(x)<0$) is not integrable near $b$. Then the equation
\be
\varphi'(x) = \rho(x) \bigg(A \sin^2\varphi(x) + \cos\varphi(x)\sin\varphi(x)
+ B \cos^2\varphi(x)\bigg) + o(\rho(x))
\ee
has only unbounded solution if $4 A B > 1$ and only bounded solution if $4 A B < 1$.
In the unbounded case we have
\be
\varphi(x) = \left(\frac{\sgn(A)}{2} \sqrt{4 A B-1} + o(1) \right) \int^x \rho(t) dt.
\ee
\end{lemma}
\begin{proof}
By a straightforward computation we have
$$
A \sin^2(\varphi) + \sin(\varphi) \cos(\varphi) + B \cos^2(\varphi) = \frac{A + B}{2}
+ \frac{\sqrt{1 + (A - B)^2}}{2} \cos(2(\varphi-\varphi_0)).
$$
for some constant $\varphi_0= \varphi_0(A,B)$. Hence $\psi(x)=\varphi(x)-\varphi_0$ satisfies
\be\label{eq:psiode}
\psi'(x) = \rho(x) \bigg(\frac{A + B}{2} + \frac{\sqrt{1 + (A - B)^2}}{2} \cos(2\psi(x)) \bigg)
+ o(\rho(x))
\ee
If $4 A B < 1$, we have $|A + B| < \sqrt{1 + (A - B)^2}$ from which it follows that the
right hand side of our differential equation is strictly negative for $\varphi(x) \pmod \pi$ close to $\pi/2$ and strictly positive if $\varphi(x) \pmod \pi$ close to $0$. Hence any solution remains in such
a strip.
If $4 A B > 1$, we have $|A + B| > \sqrt{1 + (A - B)^2}$ and thus the right hand side is always
positive, $\psi'(x) \ge C \rho(x)$, if $A,B>0$ and always negative, $\psi'(x) \le -C \rho(x)$, if $A,B<0$.
Since $\rho$ is not integrable by assumption, $\psi$ is unbounded.
In order to derive the asymptotics, rewrite (\ref{eq:psiode}) as
$$
\psi'(x) = \rho(x)\left(\frac{C + D}{2} \cos^2(\psi(x)) + \frac{C - D}{2} \sin^2(\psi(x))\right) + o(\rho(x)),\\
$$
where $C = A + B$ and $D = \sqrt{1 + (A-B)^2}$.
Now, introduce
$$
\ti{\psi}(x) = \arctan\left(\sqrt{\frac{C -D}{C + D}} \tan(\psi(x))\right)
$$
and observe $|\psi - \ti{\psi}|<\pi$. Moreover,
$$
\ti{\psi}'(x) = \frac{\rho(x)}{2} \sgn(C + D) \sqrt{C^2 - D^2} + o(\rho(x)).
$$
Hence the claim follows since by assumption $4 A B > 1$, which implies
$\sgn(C + D) = \sgn(A)$.
\end{proof}
\noindent
We will also need the case where $A=1$ and $B$ depends on $x$ but not necessarily converge to a
limit as $x\to b$. However, by a simple sub/super-solution argument we obtain
from our lemma
\begin{corollary}\label{cor:boundsol}
Suppose $\rho(x)>0$ is not integrable near $b$. Then all solutions of the equation
\be
\varphi' = \rho \bigg(\sin^2(\varphi) + \sin(\varphi)\cos(\varphi) - B \cos^2(\varphi)\bigg)
+ o(\rho)
\ee
tend to $\infty$ if $B(x)\le B_0$ for some $B_0$ with
$B_0 < -\frac{1}{4}$ and are bounded below if $B(x)\ge B_0$ for some
$B_0$ with $B_0 > -\frac{1}{4}$.
\end{corollary}
\noindent
In addition, we also need to look at averages:
Let $\ell > 0$, and denote by
\be
\ol{g}(x) = \frac{1}{\ell} \int_x^{x+\ell} g(t) dt.
\ee
the average of $g$ over an interval of length $\ell$.
\begin{lemma}\label{lem:aver}
Let $\varphi$ obey the equation
\be
\varphi'(x) = \rho(x) f(x) + o(\rho(x)),
\ee
where $f(x)$ is bounded. If
\be\label{condrho}
\frac{1}{\ell} \int_0^\ell \left|\rho(x+t) -\rho(x) \right| dt = o(\rho(x))
\ee
then
\be
\ol{\varphi}'(x) = \rho(x) \ol{f}(x) + o(\rho(x))
\ee
Moreover, suppose $\rho(x)=o(1)$. If $f(x)= A(x) g(\varphi(x))$, where $A(x)$ is bounded
and $g(x)$ is bounded and Lipschitz continuous, then
\be
\ol{f}(x)= \ol{A}(x) g(\ol{\varphi}) + o(1).
\ee
\end{lemma}
\begin{proof}
To show the first statement observe
\begin{align*}
\ol{\varphi}'(x) &= \frac{\varphi(x+\ell)-\varphi(x)}{\ell} = \frac{1}{\ell} \int_x^{x+\ell}
\rho(t) f(t) dt + o(\rho(x))\\
&= \rho(x) \ol{f}(x) + \frac{1}{\ell} \int_x^{x+\ell}
(\rho(t) -\rho(x)) f(t) dt + o(\rho(x)).
\end{align*}
Now the first claim follows from (\ref{condrho}) since $f$ is bounded. Note that
(\ref{condrho}) implies that the $o(\rho)$ property is preserved under averaging.
To see the second, we use
\begin{align*}
\ol{f}(x) &= \frac{1}{\ell} \int_x^{x+\ell} A(t) g(\varphi(t)) dt\\
&= \ol{A}(x) g(\ol{\varphi}(x)) + \frac{1}{\ell} \int_x^{x+\ell} A(t) (g(\varphi(t)) - g(\ol{\varphi}(x))) dt.
\end{align*}
Since $g$ is Lipschitz we can use the mean value theorem together with
$$
|\varphi(x+t)) - \ol{\varphi}(x)| \le C \sup_{0 \le s \le \ell} \rho(x+s)
$$
to finish the proof.
\end{proof}
\noindent
Condition (\ref{condrho}) is a strong version of saying that $\ol{\rho}(x)=\rho(x)(1+o(1))$ (it is
equivalent to the latter if $\rho$ is monotone). It will be typically fulfilled if $\rho$
decreases (or increases) polynomially (but not exponentially). For example, the condition holds if
$\sup_{t\in[0,1]} \frac{\rho'(x+t)}{\rho(x)} \to 0$.
Furthermore, note that if $A(x)$ has a limit, $A(x)=A_0+o(1)$, then $\ol{A}(x)$ can be replaced by
the limit $A_0$ and we have the next result
\begin{corollary}\label{cor:aver}
Let $\varphi$ obey the equation
\be
\varphi' = \rho \bigg(A \sin^2(\varphi) + \sin(\varphi) \cos(\varphi)
+ B \cos^2(\varphi)\bigg) + o(\rho)
\ee
with $A, B$ bounded functions and assume that $\rho=o(1)$ satisfies (\ref{condrho}).
Then the averaged function $\ol{\varphi}$ obeys the equation
\be
\ol{\varphi}' = \rho \bigg(\ol{A} \sin^2(\ol{\varphi}) + \sin(\ol{\varphi}) \cos(\ol{\varphi})
+ \ol{B} \cos^2(\ol{\varphi})\bigg) + o(\rho).
\ee
\end{corollary}
\noindent
Note that in this case $\varphi$ is bounded (above/below) if and only if $\ol{\varphi}$ is bounded
(above/below).
\section{Appendix: Periodic operators}
\label{sec:floq}
We will now suppose that $r(x)$, $p_0(x)$, $q_0(x)$ are $\ell$-periodic functions. The purpose of
this section is to recall some basic facts from Floquet theory in order to compute the critical
coupling constant for periodic operators in terms of the derivative of the Floquet discriminant.
A classical reference with further details is \cite{ea}.
Denote by $c(z,x)$, $s(z,x)$ a fundamental system of solutions of $\tau u = z u$
corresponding to the initial conditions $c(z,0)=p_0(0) s'(z,0)=1$,
$s(z,0)= p_0(0) c'(z,0)=0$. We then call
\be
M(z) =
\begin{pmatrix} c(z,\ell) & s(z,\ell) \\
p_0(\ell) c'(z,\ell) & p_0(\ell) s'(z,\ell) \end{pmatrix}
\ee
the monodromy matrix. Constancy of the Wronskian, $W(c(z),s(z)) = 1$, implies
$\det M(z)=1$ and defining the Floquet discriminant by
$$
D(z) = \tr(M(z)) = c(z,\ell) + p_0(\ell) s'(z,\ell),
$$
the eigenvalues are called Floquet multipliers,
\be
\rho_\pm(z) =\frac{D(z) \pm \sqrt{D(z)^2-4}}{2}, \qquad \rho_+(z)\rho_-(z)=1,
\ee
(where the branch of the square root is chosen such that $|\rho_+(z)|<1$).
In particular, there are two solutions
\be
u_\pm(z,x) = c(z,x) + m_\pm(z) s(z,x),
\ee
the Floquet solutions, satisfying
\be
\begin{pmatrix} u_\pm(z,\ell) \\ p(\ell) u_\pm'(z,\ell) \end{pmatrix} =
\rho_\pm(z) \begin{pmatrix} u_\pm(z,0) \\ p(0) u_\pm'(z,0) \end{pmatrix} =
\rho_\pm(z) \begin{pmatrix} 1 \\ m_\pm(z) \end{pmatrix} .
\ee
Here
\be
m_\pm(z) = \frac{\rho_\pm(z) - c(z,\ell)}{s(z,\ell)}
\ee
are the Weyl $m$-functions. The functions are $u_\pm(z,x)$ are exponentially decaying
as $x\to\pm\infty$ if $|\rho_+|<1$, that is, $|D(\lam)| \leq 2$. Hence the spectrum of $H_0$ is
purely absolutely continuous and given by
\be
\sig(H_0) = \{ \lam\in\R\,|\, |D(\lam)| \leq 2 \} = \bigcup_{n=0}^\infty [E_{2n},E_{2n+1}].
\ee
\begin{lemma}
We have
$$
\int_0^\ell u_+(z,t) u_-(z,t) r(t) dt = -\frac{2 \dot{D}(z)}{s(z,\ell)},
$$
where the dot denotes a derivative with respect to $z$.
\end{lemma}
\begin{proof}
Let $u(z,x)$, $v(z,x)$ be two solutions of $\tau u = z u$, which are smooth, then
integrating (\ref{dwr}) with $u_0=u(z)$ and $u_1=v(z_1)$, dividing by $z_1-z$ and
taking $z_1\to z$ gives
$$
W_\ell(\dot{v}(z),u(z)) - W_0(\dot{v}(z),u(z)) = \int_0^\ell u(z,t) v(z,t) r(t) dt.
$$
Now choose $u(z)=u_-(z)$ and $v(z)=u_+(z)$ and evaluate the Wronskians.
\end{proof}
\noindent
The Wronskian of $u_+$ and $u_-$ is given by
$$
W(u_-(z),u_+(z))= m_+(z)-m_-(z)= \frac{2 \sqrt{D(z)^2-4}}{s(z,\ell)}
$$
and hence they are linearly independent away from the band
edges $E_n$. At an band edge $E_n$ we have
$u_-(E_n,x)=u_+(E_n,x) \equiv u(E_n,x)$ and a second linearly independent
solution is given by
$$
s(E_n,x), \qquad W(s(E_n), u(E_n)) = 1.
$$
Here we assume without loss of generality that $s(E_n,\ell) \ne 0$ (which can always
be achieved by shifting the base point $x_0=0$ if necessary).
It is easy to check that $s(E_n,x+\ell) = \sig_n s(E_n,x) + s(E_n,\ell) u(E_n,x)$, where
$\sig_n=\rho_\pm(E_n) = \sgn(D(E_n))$. In particular, $s(E_n,x)$ is of the form
$$
s(E_n,x)= \ti{s}(E_n,x) + \frac{\sig_n s(E_n,\ell)}{\ell} x\, u(E_n,x), \qquad
\ti{s}(E_n,x+\ell)= \sig_n \ti{s}(E_n,x)
$$
and thus $u(E_n,x)$, $s(E_n,x)$ satisfy the requirements of Definition~\ref{def:regedge} with
$\alpha(x)=1$ and $\beta(x) = \sgn(D(E_n)) s(E_n,\ell) \ell^{-1} x$. In particular,
$\beta(x)>0$ for an upper band edge $E_{2m}$ and $\beta(x)<0$ for a lower band
edge $E_{2m+1}$. Setting
$$
u_0(x) = \sqrt{\frac{|s(E_n,\ell)|}{\ell}} u(E_n,x), \qquad
v_0(x) = \sqrt{\frac{\ell}{|s(E_n,\ell)|}} s(E_n,x)
$$
we have $\beta(x)=\sgn(D(E_n) s(E_n,\ell)) x$ and $\ell^{-1}\int_0^\ell u_0(t)^2 r(t) dt =
2|\dot{D}(E_n)| \ell^{-2}$.
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**