\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 133, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/133\hfil Spectral concentration]
{Spectral concentration in Sturm-Liouville equations with large
negative potential}

\author[B. J. Harris,  J. C. Kallenbach \hfil EJDE-2010/133\hfilneg]
{Bernard J. Harris,  Jeffrey C. Kallenbach}  % in alphabetical order

\address{Bernard J. Harris \newline
Department of Mathematical Sciences, Northern Illinois University,
DeKalb, IL 60115, USA}
\email{harris@math.niu.edu}

\address{Jeffrey C. Kallenbach \newline
Department of Mathematical Sciences, Siena Heights University,
Adrian, MI 49221, USA}
\email{jkallenb@siennaheights.edu}

\thanks{Submitted August 27, 2009. Published September 14, 2010.}
\subjclass[2000]{34L05, 34L20}
\keywords{Spectral theory; Schrodinger equation}

\begin{abstract}
 We consider the spectral function, $\rho_{\alpha} (\lambda)$,
 associated with the linear second-order question
 $$
 y'' + (\lambda - q(x)) y = 0 \quad \text{in } [0, \infty)
 $$
 and the initial condition
 $$
 y(0) \cos (\alpha) + y' (0) \sin (\alpha) = 0, \quad
 \alpha \in [0, \pi).
 $$
 in the case where $q (x) \to - \infty$ as $x \to \infty$.
 We obtain a representation of $\rho_0 (\lambda)$ as a convergent
 series for $\lambda > \Lambda_0$ where $\Lambda_0$ is computable,
 and  a  bound for the points of spectral concentration.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction}

We consider the linear differential equation
\begin{equation}\label{eq1.1}
y'' + (\lambda - q(x)) y = 0
\end{equation}
on the interval $[0, \infty)$ where the potential, $q$,
is a real-valued function of $C^3 [0,\infty)$ and
$q(x) \to - \infty$  as $x \to \infty$.
When augmented with the boundary condition
\begin{equation} \label{eq1.2}
y(0) \cos (\alpha) + y' (0) \sin (\alpha) = 0 \quad \alpha
\in [0, \pi)
\end{equation}

Equation \eqref{eq1.1} leads to
a self-adjoint operator on the Hilbert space $L^2 [0, \infty)$ and an
associated spectral function $\rho_{\alpha} (\lambda)$.  The function
$\rho_{\alpha} (\lambda)$, in particular $\rho_0 (\lambda)$,
is our primary concern here.
For a detailed account of its definition we refer to
\cite{e1,g2,t1}.


It is known that if $q$ satisfies
\begin{equation}\label{eq1.3}
\int^{\infty} (q')^2 |q|^{-5/2} dt < \infty, \quad
\int^{\infty} |q''| \ |q|^{-3/2} dt < \infty, \quad
\int^{\infty} |q|^{-1/2} dt = \infty,
\end{equation}
then $\rho_{\alpha}(\lambda)$ is absolutely continuous on
$(- \infty, \infty)$.  This condition is fulfilled, for example,
when $q(x) = -x^c$ where $0 < c \leq 2$.
In this article, we derive an expression, in the form of a uniformly
absolutely convergent series, for $\rho_0' (\lambda)$ in the case
where $\lambda$ is positive and sufficiently large.  Our results hold
under conditions that are somewhat more restrictive than those of
\eqref{eq1.3}.  In particular, if $q (x) = - x^c$, they hold
in the case $0 < c \leq 1$.

Representations of $\rho_0' (\lambda)$ have been obtained before,
notably in \cite{e1} in the case when $q$ is $m$-times differentiable, but
these have been asymptotic results whereas ours hold for all $\lambda$
greater than some $\Lambda_0$ which is, in principle, computable.

A secondary goal of this article is to establish bounds for  the
points of spectral concentration of \eqref{eq1.1}, \eqref{eq1.3}.
For a discussion of spectral concentration in general we refer to
\cite{g4}, but a point of spectral concentration may broadly be defined
as a value of $\lambda$ which is a local maximum of
$\rho_{\alpha}' (\lambda)$ and is thus a point at which
$\rho_{\alpha} (\lambda)$ is increasing relatively rapidly.
Specifically we show the existence of a $\Lambda_1 \geq \Lambda_0$
for which $\rho_0'' (\lambda)$ is of one sign.

In our analysis we suppose that the parameter $\lambda$ is
positive and that $\lambda - q (x) > 0$ for all $x \in [0,
\infty)$ and $\Lambda_0 \leq \lambda$. This can clearly be done if
$q(x)$ is bounded above.  Our choice of $\Lambda_0$ will be
increased as necessary throughout the paper.

Our main result concerning spectral concentration is the
following.

\begin{theorem} \label{thm1}
If $q \in C^3 [0, \infty)$ satisfies
\begin{itemize}
\item[(i)]
$q(x) \to - \infty$ as $x \to \infty$

\item[(ii)]
$q(x) < 0$ for all $x \in [0, \infty)$

\item[(iii)]
$q' (x) < 0$, $q'' (x) \geq 0$, $ q''' (x) \leq 0$ for all
$x \in [0,\infty)$

\item[(iv)]
$q'' / |q|^{\frac{3}{2}-\varepsilon}$ and
$(q')^2 / |q|^{\frac{5}{2}- \varepsilon} \in L^{1} [0, \infty)$
for some $\varepsilon > 0$

\item[(v)]
$\int_0^{\infty} |q(s)|^{-1/2}
\int_s^{\infty} \frac{|q''|}{|q|^{3/2}} + \frac{(q')^2}{|q|^{5/2}} dt
\, ds <\infty$

\item[(vi)]
$\sup_{x \in [0, \infty)} |q' (x) |\lambda - q(x)|^2 \to
0$ as $\lambda \to \infty$.

\item[(vii)]
$\int_0^{\infty} \frac{q''(t)}{(\lambda - q (t))^2}dt$
and $\int_0^{\infty} \frac{(q' (t))^2}{(\lambda - q(t))^3}dt$
are $o(1)$ as $\lambda \to \infty$
\end{itemize}
Then there exists $\Lambda_1$ such that $\rho_0 (\lambda)$ has no
points of
spectral concentration in $[\Lambda_1 , \infty)$.
\end{theorem}

We note that by writing $\lambda = (\lambda - \lambda_0) + \lambda_0$
in \eqref{eq1.1} condition (ii) effectively  requires that $q$ be
bounded above.

Our principal tool, as in \cite{g1,g3,h1} is the connection between
\eqref{eq1.1} and the Riccati equation
\begin{equation} \label{eq1.4}
v' + v^2 + (\lambda - q) = 0.
\end{equation}
Let $v(x, \lambda)$ be the unique complex-valued solution of
\eqref{eq1.4} which exists for all $x \in [0, \infty)$.
For $\alpha  = 0$ and $\xi \in \mathbb{C}^+, \ v (x, \xi)$ is the
logarithmic derivative with
respect to $x$ of the Weyl solution $u(x, \xi)$ of $y'' +
(\xi - q (x)) y = 0$. That is:
$$
v(x, \xi) =  u' (x, \xi) / u(x, \xi).
$$
It follows that $v (0, \xi) = m (\xi, 0)$ where $m(\xi, 0)$
is the Dirichlet Titchmarsh-Weyl $m$-function.  For the class of
potentials considered, the solution $v(x, \xi)$ of \eqref{eq1.4} is
continuously extendable onto the real $\lambda$-axis as
$\xi = \lambda + i \varepsilon \downarrow \lambda$.
It then follows that
\begin{equation} \label{eq1.5}
\rho_0' (\lambda) = \frac{1}{\pi} \operatorname{Im} \{v (0, \lambda)\}.
\end{equation}

Our strategy then is to identify a suitable solution of \eqref{eq1.4}
which is complex-valued for $\lambda$ real and suitably large.
Consequently we have from \eqref{eq1.5} that
\begin{equation}\label{eq1.6}
\rho_0'' (\lambda) = \frac{1}{\pi} \frac{\partial}{\partial
\lambda} \{\operatorname{Im} v(0,\lambda) \}.
\end{equation}

\section{Preliminaries}


To derive our main result it is convenient to show that the
conditions imposed on the potential, $q$, imply the existence of a
function $I(x, \lambda)$ which satisfies the conclusion of the
following lemma.

\begin{lemma} \label{lemma1}
If $q(x)$ satisfies {\rm (i)--(iv), (vi)} of Theorem \ref{thm1},
then there exists a real-valued function $I(x,\lambda)$
such that $I (x, \lambda) > 0$ for
$x \in [0, \infty)$, $\lambda > 0$ and
\begin{itemize}
\item[(i)]
$I (\cdot, \lambda) \in L^{1} [0,\infty)$

\item[(ii)]
$\frac{I(x, \lambda)}{(\lambda - q (x))^{1/2}}$ is a decreasing
function of $x$ for each $\lambda > 0$

\item[(iii)]
$\int_0^{\infty} I (x,\lambda) dx \to 0$ as $ \lambda
\to \infty$.

\item[(iv)]
$(\lambda - q (x))^{1/2} \big| \int_x^{\infty} e^{2i \int_{x}^{t}
(\lambda - q(s)) \,ds}  \big\{ \frac{q''(t)}{4 (\lambda
- q (t)^{3/2}} + \frac{5(q' (t)^2}{16 (\lambda -
q(t)^{5/2}}\big\} dt \big| \leq I (x, \lambda)$.
\end{itemize}
\end{lemma}


\begin{proof}
We set
\[
I (x, \lambda) : = \frac{q'' (x)}{2(\lambda -
q(x))^{3/2}} + \frac{5 (q' (x))^2}{8 (\lambda - q(x))^{5/2}}
\]
and show that this choice of $I(x, \lambda)$ satisfies (i)--(iv) if
$q$ satisfies the conditions of Theorem \ref{thm1}.
Part (i)  follows from Theorem \ref{thm1} (iv).

Differentiation with respect to $x$
of $(\lambda - q (x))^{-1/2} I (x,y)$ and Theorem \ref{thm1}(iii)
shows each of the terms is decreasing (in $x$) for each $\lambda$
which establishes (ii).

To see (iii) we rewrite  the terms of
$\int_0^{\infty} I (x,\lambda) \,dx$ as
$$
 \frac{1}{2} \int_0^{\infty} \frac{q'' (x)}{(\lambda - q (x))^{\varepsilon} (\lambda - q(x))^{3/2 -
\varepsilon}} dx \leq \frac{1}{2} \lambda^{- \varepsilon}
\int_0^{\infty} \frac{q''(x)}{|q(x)|^{3/2 - \varepsilon}
} \,dx.
$$
The other terms in the sum is treated similarly.

To prove (iv) we note that $\frac{q''}{4 (\lambda - q)^2} +
\frac{5(q')^2}{16 (\lambda - q)^3}$ is decreasing so, by the
Second Mean Value Theorem,
\begin{align*}
&(\lambda - q(x))^{1/2} \big| \int_x^{\infty} e^{2 i \int_{x}^{t}
(\lambda - q (s))^{1/2} ds}
\big\{ \frac{q''}{4 (\lambda - q)^{3/2}} + \frac{5
(q')^2}{16 (\lambda - q)^{5/2}} \big\} \,dx \big|\\
&= (\lambda - q(x))^{1/2} \frac{1}{2} \Big|
\int_x^{\infty}\big\{ 2 (\lambda - q (t))^{1/2} cis
\Big(2 \ \int_x^t (\lambda - q(s))^{1/2} ds \Big) \big\}\\
&\quad \times \big\{ \frac{q''}{4 (\lambda - q)^2} +
\frac{5 (q')^2}{16 (\lambda - q)^3} \big\} dt \Big|\\
&= \frac{1}{2} \big\{ \frac{q'' (x)}{4 (\lambda -
q(x))^{3/2}} + \frac{5}{16} \frac{(q'(x))^2}{(\lambda -
q(x))^{5/2}} \big\}\\
&\quad\times \Big| \int_{\xi_{1}}^{\infty} 2 (\lambda -
q)^{1/2} \cos (2 \int_x^t (\lambda - q(s))^{1/2} ds \,dt\\
&\quad + i \int_{\xi_{2}}^{\infty} 2
(\lambda - q)^{1/2} \sin \big( 2 \int_x^t (\lambda - q(s)\big)^{1/2}
 ds \,dt \Big| \\
&\leq 2 \big\{ \frac{q''(x)}{4 (\lambda - q(x))^{3/2}} +
\frac{5}{16} \frac{q' (x))^2}{(\lambda - q (x))^{5/2}}
\big\} = I (x, \lambda).
\end{align*}
The proof  is  complete.
\end{proof}

To obtain the required complex-valued solution of the Riccati equation
\eqref{eq1.4}, we proceed as in \cite{g1,g3}.  Based on the asymptotic
representation established in \cite{h1}, we seek a solution in the form of
\begin{equation} \label{eq2.1}
v (x, \lambda) = i (\lambda - q (x))^{1/2} + \frac{1}{4} q' (x)
(\lambda - q(x))^{-1} + \sum_{n=1}^{\infty} v_n (x, \lambda).
\end{equation}
Substitution of \eqref{eq2.1} into \eqref{eq1.4} gives
\begin{align*}
&\sum_{n=1}^{\infty} \Big( v_n' + 2 \big\{ i (\lambda -
q)^{1/2} + \frac{1}{4} q' (\lambda - 2)^{-1} \big\} v_n
\Big)\\
& = - Q - v_1^2 - \sum_{n=3}^{\infty} \Big( v_{n-1}^2 + 2
v_{n-1} \sum_{m=1}^{n-2} v_m\Big),
\end{align*}
where
\[
Q:= \frac{q''}{4 (\lambda - q)} + \frac{5
(q')^2}{16 (\lambda - q)^2}.
\]
We choose $v_1, v_2 \dots$ so that
\begin{equation}
\label{eq2.2}
\begin{gathered}
v_1' + \Big(2 i (\lambda - q)^{1/2} +
\frac{q'}{2 (\lambda - q)} \Big) v_1 = - Q\\
v_2' + \Big(2 i (\lambda - q)^{1/2} +
\frac{q'}{2 (\lambda - q)} \Big) v_2 = - v_1^2\\
v_n' + \Big(2 i (\lambda - q)^{1/2} +
\frac{q'}{2 (\lambda - q)} \Big) v_n = - v_{n-1}^2 - 2
v_{n-1} \sum_{m=1}^{n-2} v_m \\
\end{gathered}
\end{equation}
for $n = 3, 4, \dots$.
The required solution to  \eqref{eq2.1} is
\begin{equation}\label{eq2.3}
 \begin{gathered}
v_1 (x, \lambda) = (\lambda - q(x))^{1/2} \int_x^{\infty} (\lambda -
q(t))^{-1/2} e^{2 i \int_{x}^{t} (\lambda - q)^{1/2} ds} Q(t, \lambda)
dt\\
v_2 (x, \lambda) = (\lambda - q(x))^{1/2} \int_x^{\infty} (\lambda -
q(t))^{-1/2} e^{2 i \int_{x}^{t} (\lambda - q)^{1/2} ds} v_1 (t, \lambda)^2
 \,dt\\
\begin{aligned}
v_n (x, \lambda)
& = (\lambda - q(x))^{1/2} \int_x^{\infty} (\lambda -
q(t))^{-1/2} e^{2 i \int_{x}^{t} (\lambda - q)^{1/2} ds}\\
&\quad\times \Big( v_{n-1}^2 + 2 \sum_{m=1}^{n-2} v_m v_{n-1} \Big) \,dt
\end{aligned}
\end{gathered}
\end{equation}
for $n = 3, 4, \dots$.

\begin{lemma}\label{lemma2}
If $\Lambda_0$ is so large that for all $\lambda \geq \Lambda_0$,
$$
9 \int_0^{\infty} I (t,\lambda) \,dt \leq 1
$$
then for $n = 1, 2, 3,\dots $,
$$
 |v_n (x,\lambda)| \leq I (x,\lambda)/2^{n-1} \quad
 \text{for all }  x \in [0, \infty).
$$
\end{lemma}

\begin{proof}
We use induction on $n$.  When $n=1$ this is Lemma \ref{lemma1} (iv).
For $n=2$,
\begin{align*}
|v_2 (x, \lambda)| & \leq  (\lambda - q(x))^{1/2}
\int_x^{\infty} (\lambda - q(t))^{-1/2} I (t, \lambda)^2 \,dt\\
& \leq  I (x, \lambda) \int_0^{\infty} I (t,\lambda) \,dt,
\end{align*}
by Lemma \ref{lemma1} (ii).
If $n \geq 3$ then, by the induction hypothesis:
\begin{align*}
|v_n (x,\lambda)|
& \leq   (\lambda - q(x))^{1/2} \int_x^{\infty}
(\lambda - q(t))^{-1/2} \Big[ \frac{I (t,\lambda)^2}{2^{n-2}}
\sum_{m-1}^{\infty} \frac{1}{2^{m-1}} \Big] \,dt\\
 & \leq  (\lambda - q(x))^{1/2} \int_x^{\infty} (\lambda -
q(t))^{-1/2} \frac{I (t,\lambda)}{2^{n-1}} \big[ \frac{1}{2^{n-2}} +
8 \big] I (t, \lambda) \,dt\\
& \leq  \frac{I (x, \lambda)}{2^{n-1}} \cdot 9 \int_0^{\infty} I (t,
\lambda) \,dt
\end{align*}
and the result follows.
\end{proof}

The uniform, absolute convergence of $\sum_{n=1}^{\infty} v_n
(x,\lambda)$ follows from Lemma \ref{lemma2}.  The uniform absolute
convergence of $\sum_{n=1}^{\infty} v_n' (x,\lambda)$ which
justifies the term differentiation used to derive the series solution,
follows from the bound for the $v_n$ obtained in Lemma \ref{lemma2} and
the representation of the $v_n'$ in \eqref{eq2.2}.  Since, for
example,
\begin{align*}
|v_n' (x, \lambda)|
& \leq  \Big( |2 (\lambda - q(x)^{1/2} | +
\big| \frac{q' (x)}{2 (\lambda - q(x))} \big| \Big) |v_n
(x, \lambda)|\\
&\quad + |v_{n-1} (x,\lambda)|^2 + 2 |v_{n-1}| \sum_{m=1}^{n-2}
|v_m|.\\
& \leq  \Big( 2|\lambda - q|^{1/2} + \big| \frac{q'}{2
(\lambda - q)}\big| \Big) I (x,\lambda) \cdot \frac{1}{2^{n-1}}
\\
&\quad +\Big( \frac{I (x,\lambda)}{2^{n-2}}\Big)^2 + 2 \frac{I
(x)^2}{2^{n-2}} \sum_{m=1}^{\infty} \frac{1}{2^{m-1}} \quad
 \text{for }  n= 3, 4, \dots
\end{align*}
It follows readily that $\sum |v_n' (x, \lambda)|$ is uniformly
absolutely convergent for $x \in [0, \infty)$ and
$\lambda >\Lambda_0$.
We have proved the following result.

\begin{theorem} \label{thm2}
Let $q$ satisfy the conditions of
Theorem \ref{thm1}.  If $\Lambda_0$ is so large that for all
$\lambda \geq \Lambda_0 > 0$,
$9 \int_0^{\infty} I (t,\lambda) \,dt \leq 1$ and
$(\lambda - q(x) > 0$ for all $x \in [0, \infty)$ then
$$
\rho_0' (\lambda) = \frac{1}{\pi} (\lambda - q(0))^{1/2} +
\frac{1}{\pi} \sum_{n=1}^{\infty} Im (v_n (0, \lambda)).
$$
for all $\lambda > \Lambda_0$.
\end{theorem}

\section{Spectral Concentration}

We seek the second derivative of  $\rho_0 (\lambda)$.  Our
strategy is to differentiate the equations of \eqref{eq2.2} with
respect to $\lambda$, justify the equality of the mixed second
order partial derivatives and derive expressions for
$\frac{\partial v_n}{\partial \lambda}$ akin to \eqref{eq2.3}
which we then bound as in Lemma \ref{lemma2}.

Differentiating the first equation of \eqref{eq2.2} with respect
to $\lambda$ gives
\begin{equation} \label{eq3.1}
\frac{\partial^2 v_1}{\partial \lambda \partial x}
 + \Big( 2i (\lambda
- q)^{1/2} + \frac{q'}{2 (\lambda - q)}\Big) \frac{\partial
v_1}{\partial \lambda} = - \frac{\partial Q}{\partial \lambda} - i
(\lambda - q)^{-1/2} v_1 + \frac{1}{2} q' (\lambda - q)^{-2}
v_1
\end{equation}

We note from \eqref{eq2.3} that $v_1 (x, \lambda)$ is continuous and so,
by \eqref{eq2.2}, is $\frac{\partial v_1}{\partial x}$.  It remains to
show that $\frac{\partial v_1}{\partial \lambda}$ is continuous. We do
this by differentiating the first equation of \eqref{eq2.3} under the
integral sign to obtain
\begin{equation}\label{eq3.2}
\begin{aligned}
\frac{\partial v_1}{\partial \lambda}
& =  \frac{1}{2} (\lambda - q(x))^{-1/2} \int_x^{\infty}
e^{2i \int_{x}^{t} (\lambda - q(s))^{1/2}
ds} \big\{ \frac{q''}{4(\lambda - q)^{3/2}} + \frac{5
(q')^2}{16 (\lambda - q)^{5/2}} \big\} dt\\
&\quad + (\lambda - q(x))^{1/2} \int_x^{\infty} 2i
\Big(\int_x^{t}
(\lambda - q (s))^{-1/2} \,ds\Big)
e^{2i \int_{x}^{t} (\lambda - q(s))^{1/2} ds}\\
&\quad \times \left\{ \frac{q''}{4(\lambda - q)^{3/2}} + \frac{5
(q')^2}{16 (\lambda - q)^{5/2}} \right\} dt\\
&\quad + (\lambda - q(x))^{1/2} \int_x^{\infty}
e^{2i \int_{x}^{t} (\lambda - q(s))^{1/2} \,ds}\\
&\quad \times \big\{ -
\frac{3q''}{8} (\lambda - q)^{-5/2} - \frac{25}{32}
(q')^2 (\lambda - q)^{-7/2} \big\} dt
\end{aligned}
\end{equation}
providing that the differentiation under the integral sign is justified.
To ensure that it is, we note that under the conditions of Theorem
\ref{thm1}, the integrand in the expression for $v_1 (x,\lambda)$ in
\eqref{eq2.3} is continuously differentiable
with respect to $\lambda$, and that each term of the integrand in
\eqref{eq3.2} is integrable with respect to $t \in \mathbb{R}^{+}$;
 to see this in the case of the second term, note that by a change in the order of
integration
\begin{align*}
&\big| \int_x^{\infty} \Big( \int_x^t (\lambda - q(s))^{-1/2} \,ds\Big)
e^{2i \int_{x}^{t} (\lambda - q(s))^{1/2} ds}
\big\{ \frac{q''}{4(\lambda - q)^{3/2}} + \frac{5 (q')^2}{16 (\lambda
 - q)^{5/2}} \big\} dt \big|
\\
& \leq \int_x^{\infty} (\lambda - q (s))^{-1/2}
\int_s^{\infty} \frac{q''}{4 (\lambda - q)^{3/2}} +
\frac{5 (q')^2}{16 (\lambda - q)^{5/2}} \,dt \,ds.
\end{align*}
It now follows from \eqref{eq3.2} that $\frac{\partial v_1}{\partial
\lambda}$ is continuous in $x$ and $\lambda$, so the equality of the
mixed partial derivatives is established. We
may therefore replace $\frac{\partial^2 v_1}{\partial \lambda \partial
x}$ by $\frac{\partial^2 v_1}{\partial x \partial \lambda}$ in
\eqref{eq3.1}, then integrate with respect to $x$ to obtain a more
suitable representation of $\frac{\partial v_1}{\partial \lambda}$.
This yields
\begin{equation} \label{eq3.3}
\begin{aligned}
\frac{\partial v_1}{\partial \lambda} (x,\lambda)
& =  (\lambda -
q(x))^{1/2} \int_x^{\infty} (\lambda - q)^{-1/2} e^{2i \int_{x}^{t}
(\lambda - q (s))^{1/2} ds}
 \big\{ \big[ - \frac{q''}{4} (\lambda - q)^{-2} \\
&\quad - 5/8  (q')^2  (\lambda - q)^{-3}\big]
+ [ -i (\lambda - q)^{-1/2} v_1] +
[\frac{q'}{2}(\lambda - q)^{-2} v_1] \big\} dt
\\
&=: I_1 + I_2 + I_3
\end{aligned}
\end{equation}
This provides a convenient first step for an iterative scheme to
establish upper bounds on $\big| \frac{\partial v_n}{\partial
\lambda}\big|$ for $x \geq 0$ and $\lambda$ sufficiently large.
To this end we note that
\begin{gather*}
\big| \frac{q''}{4} (\lambda - q)^{-2} + 5/8
(q')^2 (\lambda - q)^{-3} \big| \leq (\lambda - q)^{-1/2} I
(x,\lambda) \ \text{so}
\\
|I_1| \leq \Big( \sup_{x \in [0, \infty)} (\lambda - q (x))^{-1/2}\Big)
(\lambda - q(x))^{1/2} \int_x^{\infty} (\lambda - q(t))^{-1/2} I (t,
\lambda) dt
\end{gather*}
Also,
\begin{align*}
|I_2|& \leq (\lambda - q(x))^{1/2} \int_x^{\infty} (\lambda - q(t))^{-1}
I (t, \lambda) \,dt
\\
&\leq \Big( \sup_{x \in [0, \infty)} (\lambda - q(x))^{-1/2} \Big)
(\lambda - q(x))^{1/2} \int_x^{\infty} (\lambda - q(t))^{-1/2} I
(t,\lambda) \,dt
\end{align*}
and
$$
|I_3| \leq
\frac{1}{2} \Big( \sup_{x \in [0, \infty)}  |q'' (x)|
(\lambda - q(x))^{-2}\Big) (\lambda - q(x))^{1/2} \int_x^{\infty}
(\lambda - q(t))^{-1/2} I (t,\lambda) dt.
$$
It follows that
\begin{equation}\label{eq3.4}
\begin{aligned}
\big| \frac{\partial v_1}{\partial \lambda} (x, \lambda) \big|
&\leq \big\{ 2 \sup_{x \in [0, \infty)} (\lambda - q(x))^{-1/2}
+ \frac{1}{2} \sup_{x \in [0, \infty)} |q' (x) (\lambda - q(x))^{-2}|
\big\}\\
&\quad \times (\lambda - q(x))^{1/2} \int_{x}^{\infty} (\lambda -
q(t))^{-1/2} I (t, \lambda) \,dt.
\end{aligned}
\end{equation}



\begin{lemma} \label{lemma3}
If $\Lambda_1 \geq \Lambda_0 > 0$ is so large that for all
$\lambda \geq \Lambda_1$,
$$
16 \int_0^{\infty} I (t, \lambda) dt + 2 \sup_{x \in
[0,\infty)} (\lambda - q(x))^{-1/2} + \frac{1}{2}  \sup_{x \in [0,
\infty} |q' (x) (\lambda - q(x))^{-2} | \leq 1,
$$
then for $x \in [0, \infty), \lambda > \Lambda,$ and $n = 1, 2, 3,
\dots$,
\begin{equation} \label{eq3.5}
\big| \frac{\partial v_n}{\partial \lambda} (x, \lambda)\big| \leq
\frac{1}{2^{n-1}} (\lambda - q(x))^{1/2} \int_x^{\infty} (\lambda - q
(t))^{-1/2} I (t, \lambda) \,dt\,.
\end{equation}
\end{lemma}


\begin{proof} We use induction on $n$ to prove the hypothesis:
$\frac{\partial v_n}{\partial \lambda} (x,\lambda)$
is continuous in $x$ and $\lambda$,  for $x \in [0, \infty)$,
$\lambda > \Lambda$
and inequality \eqref{eq3.4} holds.

The case $n=1$ follows from \eqref{eq3.4} since the hypothesis of the
lemma implies the asserted bound.  The case $n=2$  will follow from the
general case, the difference being that some of the series terms are
vacuous.

In the general case, suppose
the induction hypothesis holds for $\frac{\partial
v_1}{\partial \lambda}, \dots, \frac{\partial v_{n-1}}{\partial
\lambda}$.  As in the case for $\frac{\partial v_1}{\partial \lambda}$,
we differentiate \eqref{eq2.2} with respect to $\lambda$, show the
equality of the mixed second order derivatives and obtain an integral
representation for $\frac{\partial v_n}{\partial \lambda}$ which we
bound.



The function $\frac{\partial v_n}{\partial \lambda}$ is continuous
from \eqref{eq2.2} and, differentiating \eqref{eq2.2} with respect
to $\lambda$ shows that
$\frac{\partial^2 v_n}{\partial \lambda \partial x}$ is
continuous if $\frac{\partial v_n}{\partial \lambda}$ is. This we now
show by differentiating \eqref{eq2.3} with respect to $\lambda$ under
the integral.
\begin{equation}\label{eq3.6}
\begin{aligned}
\frac{\partial v_n}{\partial \lambda}
& =  \frac{1}{2} (\lambda - q(x))^{-1} v_n
 - \frac{1}{2} (\lambda - q(x))^{1/2} \int_x^{\infty} (\lambda -
q)^{-3/2} e^{ei \int_{x}^{t} (\lambda - q)^{1/2} ds} \\
&\quad\times \Big( v_{n-1}^{2}
+ 2 \sum_{m=1}^{n-2} v_m v_{n-1}\Big) \,dt\\
&\quad + (\lambda - q(x))^{1/2} \int_x^{\infty} (\lambda - q)^{-1/2}
\big\{ i \int_x^t (\lambda - q(s))^{-1/2} ds \big\} e^{2i
\int_{x}^{t} (\lambda - q)^{1/2} ds}\\
&\quad \times \Big( v_{n-1}^{2} + 2
\sum_{m=1}^{n-2} v_m v_{n-1}\Big) dt\\
&\quad +  (\lambda - q(x))^{1/2} \int_{x}^{\infty} (x - q)^{-1/2} e^{2i
\int_{x}^{t} (\lambda - q)^{1/2} ds}\\
&\quad\times \frac{\partial}{\partial \lambda}
\Big( v_{n-1}^2 + 2 \sum_{m=1}^{n-2} v_m v_{n-1}\Big) \,dt.
\end{aligned}
\end{equation}
The continuity of all but the third term is clear.  This consists of
a sum of terms which are
\begin{align*}
& O\Big((\lambda - q(x))^{1/2}  \int^\infty_x (\lambda - q(t))^{-1/2}
\big\{\int^t_x (\lambda - q(s))^{-1/2}\, ds\big\}
I(t, \lambda)^2 \, dt\Big)\\
& = O\Big((\lambda - q(x))^{1/2} \int^\infty_x (\lambda -
q(s))^{-1/2} \int^\infty_s (\lambda - q(t))^{-1/2} I(t, \lambda)^2\,
dt\, ds\Big) \\
& = O\Big((\lambda - q(x))^{1/2} \int^\infty_x (\lambda - q(s))^{-1}
I(s, \lambda) \int^\infty_0 I(t, \lambda)\, dt\, ds\Big).
\end{align*}
By Lemma \ref{lemma1} (i) and (ii), the continuity of the third term
follows.

By the induction hypothesis the fourth term consists of a sum of
terms each of which is
$$
O\Big( (\lambda - q(x))^{1/2} \int^\infty_x I(t, \lambda)
\int^\infty_t (\lambda - q(s))^{-1/2} I(s, \lambda)\, ds\, dt\Big)
$$
and so is bounded by Lemma \ref{lemma1} (i).

The continuity of $\frac{\partial v_n}{\partial \lambda}$, and hence
of $\frac{\partial^2 v_n}{\partial\lambda \partial x}$ now follows and,
by the equality of the second order mixed partial derivatives, we
have from \eqref{eq2.2}:
\begin{equation}\label{eq3.7}
\begin{aligned}
\frac{\partial v_n}{\partial \lambda}
& = (\lambda - q(x))^{1/2}
\int^\infty_x e^{2i\int^t_x (\lambda - q(s))^{1/2}\, ds} (\lambda -
q(t))^{-1/2}\\
&\quad\times \Big\{ i(\lambda - q)^{-1/2}v_n - \frac{q'}{2(\lambda -
q)^2}v_n
+ 2\sum_{m = 1}^{n - 1} \frac{\partial v_{n - 1}}{\partial \lambda}
v_m + 2 \sum^{n - 2}_{m = 1} v_{n - 1} \frac{\partial v_m}{\partial
\lambda} \Big\}\, dt\\
& =: I_1 + \cdots + I_4
\end{aligned}
\end{equation}
 From Lemma \ref{lemma2},
\begin{align*}
|I_1| & \leq \frac{1}{2^{n - 1}} (\lambda - q(x))^{1/2} \int^\infty_x
(\lambda - q(t))^{-1} I(t, \lambda)\, dt\\
& \leq \frac{1}{2^{n - 1}} \big\{\sup_{0 \leq x < \infty} (\lambda -
q(x))^{-1/2}\big\} (\lambda - q(x))^{1/2}
\int^\infty_x (\lambda - q(t))^{-1/2} I(t, \lambda)\, dt.
\end{align*}
\begin{align*}
|I_2| & \leq \frac{1}{2^{n - 1}} (\lambda - q(x))^{1/2} \int^\infty_x
(\lambda - q(t))^{-1/2} \big| \frac{q'(t)}{2(\lambda -
q(t))^2}\big| I(t, \lambda)\, dt\\
& \leq \frac{1}{2^{n - 1}} \sup_{0 \leq x < \infty}
\big|\frac{q'(x)}{2(\lambda - q(x))^2}\big| (\lambda - q(x))^{1/2}
\int^\infty_x (\lambda - q(t))^{-1/2} I(t, \lambda)\, dt.
\end{align*}
\begin{align*}
|I_3| & \leq 2(\lambda - q(x))^{1/2}\int^\infty_x
 \Big(\int^\infty_t
(\lambda - q(s))^{-1/2} \frac{I(s, \lambda)}{2^{n - 2}}\, ds\Big)
I(t, \lambda)\sum_{m = 1}^{n - 1} \frac{1}{2^{m - 1}}\, dt\\
& \leq \frac{1}{2^{n - 1}} (\lambda - q(x))^{1/2} 8 \int^\infty_x
I(t, \lambda) \int^\infty_t (\lambda - q(s))^{-1/2} I(s, \lambda)\,
ds\, dt\\
& = \frac{8}{2^{n - 1}}(\lambda - q(x))^{1/2} \int^\infty_x (\lambda -
q(s))^{-1/2} I(s, \lambda) \int^s_x I(t, \lambda)\, dt\, ds\\
& \leq \frac{1}{2^{n - 1}} (\lambda - q(x))^{1/2} \int^\infty_x
(\lambda - q(s))^{-1/2} I(s, \lambda)\, ds \big\{ 8
\int^\infty_0 I(t, \lambda)\, dt\big\}.
\end{align*}
\begin{align*}
|I_4| & \leq 2(\lambda - q(x))^{1/2} \int^\infty_x \frac{I(t,
\lambda)}{2^{n - 2}} \int^\infty_t (\lambda - q(s))^{-1/2} I(s,
\lambda) \sum_{m = 1}^{n - 2} \frac{1}{2^{m - 1}}\, ds\, dt\\
& \leq \frac{8}{2^{n - 1}} (\lambda - q(x))^{1/2} \int^\infty_x I(t,
\lambda) \int^\infty_t (\lambda - q(s))^{-1/2} I(s, \lambda)\, ds\,
dt\\
& \leq \frac{1}{2^{n - 1}} (\lambda - q(x))^{1/2} \int^\infty_x
(\lambda - q(s))^{-1/2} I(s, \lambda)\, ds \big\{8\int^\infty_0
I(t, \lambda)\, dt\big\}.
\end{align*}
The result now follows since for $\lambda \geq \Lambda_1$,
$$
16 \int^\infty_0 I(t, \lambda)\, dt + \sup_{0 \leq x < \infty}
\big|\frac{q'(x)}{2(\lambda - q(x))^2}\big| + \sup_{0 \leq x <
\infty} |\lambda - q(x)|^{-1/2} \leq 1.
$$
\end{proof}

\section{Proof of Theorem \ref{thm1}}

If $q$ satisfies the conditions of Theorem \ref{thm1} then there
exists a function $I(x, \lambda)$ satisfying the conclusions of Lemma
\ref{lemma1} and hence Lemmas \ref{lemma2} and \ref{lemma3}.  Thus,
for $\lambda > \Lambda_1$ the following representation of
$\rho''(\lambda)$ holds
$$
\rho''(\lambda) = \frac{1}{2\pi}(\lambda - q(0))^{-1/2} +
\frac{1}{\pi}\sum_{n = 1}^\infty \frac{\partial}{\partial \lambda}
Im (v_n (0, \lambda))
$$
and
\begin{align*}
\big|\rho''(\lambda) - \frac{1}{2\pi}(\lambda - q(0))^{-1/2}\big|
& \leq \frac{1}{\pi} \sum_{n = 1}^\infty
\big|\frac{\partial}{\partial\lambda} v_n(0, \lambda)\big| \\
& \leq \frac{2}{\pi}(\lambda - q(0))^{1/2} \int^\infty_0 (\lambda -
q(t))^{-1/2} I(t, \lambda)\, dt.
\end{align*}
Thus $\rho''(\lambda) > 0$ if $\lambda$ is so large that
$$
4(\lambda - q(0)) \int^\infty_0 (\lambda - q(t))^{-1/2} I(t,
\lambda)\, dt < 1.
$$
With the function $I(t, \lambda)$ from Lemma \ref{lemma1} this
is satisfied if, in addition to the requirements of Lemmas
\ref{lemma2} and \ref{lemma3}, $\lambda$ is so large that
\begin{equation} \label{eq4.1}
2(\lambda - q(0)) \int^\infty_0 \frac{q''(t)}{(\lambda - q(t))^2} +
\frac{5(q'(t))^2}{4(\lambda - q(t))^3}\, dt < 1.
\end{equation}

\subsection*{Acknowledgments}

Some material in the paper is part of the dissertation submitted by
the second author in partial fulfillment of the requirements for
the Ph. D. at Northern Illinois University cite{k1}.
Both authors want to express their gratitude
to Dr. D. J. Gilbert at Dublin Institute of Technology for
her careful reading of the dissertation and helpful suggestions.
The authors are also grateful to the anonymous referee for his/her
helpful comments.

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\end{document}