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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 17, pp. 1--5.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/17\hfil The form of the spectral function]
{The form of the spectral function associated with Sturm-Liouville
problems for small values of the spectral parameter}

\author[B. J. Harris \hfil EJDE-2013/17\hfilneg]
{B. J. Harris} 

\address{Bernie J. Harris \newline
Department of Mathematical Sciences,
Northern Illinois University,
DeKalb, IL 60115-2888, USA}
\email{harris@math.niu.edu}

\thanks{Submitted July 9, 2012. Published January 21, 2013.}
\subjclass[2000]{34E05}
\keywords{Sturm Liouville equation; spectral function; 
small eigenparameter}

\begin{abstract}
 We study the linear second-order differential equation
 $$
 -y'' + q(x) y = \lambda y
 $$
 where, amongst other conditions,  $q \in L^1[0,\infty)$. 
 We obtain a convergent series expansion for the spectral 
 function which is valid for small values of $\lambda$.
 We also derive an asymptotic representation.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks


\section{Introduction}\label{sec:1}

We consider the linear, second-order differential equation
\begin{gather}\label{eq1.1}
-y'' + q(x) y = \lambda y \text{ for } x \in [0, \infty), \\
\label{eq1.2} 
y(0) = 0
\end{gather}
in the case where $q$ is a real-valued member of $L^1[0, \infty)$.  It
is well known, see for example \cite{ref5} that under these
circumstances the spectral function $\rho_0(\lambda)$ associated with
\eqref{eq1.1}, \eqref{eq1.2} is such that $\rho'_0(\lambda)$ exists
and is continuous on $(0, \infty)$.  In recent years many papers have
investigated the form of $\rho_0(\lambda)$ for large values of
$\lambda$.  In particular we mention the asymptotic results in
\cite{ref1,ref2} and the explicit representations derived
in \cite{ref3,ref4,ref6} which are valid for all
$\lambda \geq \Lambda_0$ where $\Lambda_0$ is computable.  In
\cite{ref4} the condition $q \in L^1[0, \infty)$ was relaxed to the
requirement that $q$ be of Wigner-von Neumann type or be slowly
decreasing.  The situation for small values of $\lambda$ is somewhat
more complicated as the form of the derived series will show.  In particular the
conditions on $q$ and the form of the series representation are in
terms of the solution of a particular Riccati equation.  A necessary
condition for the existence of such a solution on $(0, \infty)$ is
the finiteness of $\int_0^{\infty} (1 + t)^2 q(t)\, dt$.  It follows
that the results require $q$ to be small at infinity.  A consequence
of our main result is a representation of $\lim_{\lambda \to 0^+}
\rho'_0(\lambda)$. We also, in \S \ref{sec:4}, show that the
convergent series may be truncated and an asymptotic representation
obtained.

\section{Results} \label{sec:2}

We assume the existence of a solution, $v_0(x)$, of the Riccati equation
\begin{equation}\label{eq2.1}
v'_0 = q(x) - v_0^2
\end{equation}
which is defined on $[0, \infty)$ and satisfies
\begin{equation}\label{eq2.2}
\lim_{x \to \infty} x v_0(x) = 0.
\end{equation}
We further assume that
\begin{equation}\label{eq2.3}
(1+t) |v_0(t)| \in L^1 [0, \infty).
\end{equation}
Under these conditions it will be shown that there exists a sequence of
functions $\{ v_n (x, \lambda) \}$ defined recursively as follows:
\begin{equation}\label{eq2.4}
v_1 (x, \lambda) :=  2i\lambda^{1/2} \int_{x}^{\infty} e^{2i\lambda^{1/2}
(t-x) - 2 \int_x^t v_0(s) \, ds} v_0(t)\, dt
\end{equation}
and
\begin{equation}\label{eq2.5}
v_n(x, \lambda) := \int_{x}^{\infty} e^{2i\lambda^{1/2} (t-x) - 2\int_x^t
v_0(s)\, ds} \Big( v_{n-1}^2 + 2v_{n-1} \sum_{m=1}^{n-2} v_m \Big)
\,dt .
\end{equation}

\begin{theorem} \label{thm1}
Under conditions \eqref{eq2.1}--\eqref{eq2.3} there exists $\Lambda > 0$ 
so that for $\lambda \in (0, \Lambda)$
\begin{equation}\label{eq2.6}
\rho'_0(\lambda) = \frac{1}{\pi} \big\{ \lambda^{1/2} + \operatorname{Im}
\sum_{n=1}^{\infty} v_n(0,\lambda) \big\}.
\end{equation}
In particular
\begin{equation}\label{eq2.7}
\lim_{\lambda \to 0^+} \rho'_0(\lambda) = 0 .
\end{equation}
\end{theorem}

\begin{example}\label{examp2.1} \rm
If $q(x) := -e^{-x} (1 - e^{-x})$ then it is easy to see that 
$v_0(x) = e^{-x}$ satisfies \eqref{eq2.1}, \eqref{eq2.2}, and \eqref{eq2.3} and
$\lim_{\lambda \to 0^+} \rho'_0(\lambda) = 0$.
\end{example}

\begin{remark} \label{note2.1}\rm
If $v_0$ satisfies \eqref{eq2.1} then
$$
(1+t)^2 v'_0(t) = (1+t)^2 q(t) - (1+t)^2 v_0(t)^2
$$
and an integration by parts and \eqref{eq2.2} gives
$$
-v_0(0) - 2 \int_0^{\infty} (1+t) v_0(t) \,dt = \int_0^{\infty} (1+t)^2
q(t) \,dt - \int_0^{\infty} (1+t)^2 v_0(t)^2.
$$
The boundedness of $\int_0^{\infty} (1+t)^2 q(t)\,dt$ now
follows from \ref{eq2.1}--\eqref{eq2.3}.
\end{remark}

\begin{remark} \label{note2.2}\rm
It is shown below that the requirements \eqref{eq2.1}--\eqref{eq2.3}
ensure that $v_0(x)$ is real-valued.
\end{remark}

\section{Proof of Theorem \ref{thm1}}
\label{sec:3}

Following the analysis employed in \cite{ref5}, we seek a solution of the
Riccati equation
\begin{equation}\label{eq3.1}
v' = -\lambda + q - v^2
\end{equation}
which satisfies
\begin{equation}\label{eq3.2}
\lim_{x \to \infty} v(x, \lambda) = i \lambda^{1/2} .
\end{equation}
Then, from \cite[(4.4)]{ref5},
\begin{equation}\label{eq3.3}
\rho'_0(\lambda) = \frac{1}{\pi} \operatorname{Im} \{ v(\lambda) \} .
\end{equation}
We try for a solution of \eqref{eq3.1} a series of the form
\begin{equation}\label{eq3.4}
v(x, \lambda) = i \lambda^{1/2} + v_0(x) + \sum_{n=1}^{\infty} v_n (x,
\lambda).
\end{equation}
If term by term differentiation of the terms of the series of \eqref{eq3.4}
is justified, substitution of \eqref{eq3.4} into \eqref{eq3.1} leads to a
choice of the $\{ v_n \}$ such that
\begin{equation}\label{eq3.5}
v'_1 + (2i \lambda^{1/2} + v_0) v_1 = -2i \lambda^{1/2} v_0
\end{equation}
and for $n = 2, 3, \ldots$,
\begin{equation}\label{eq3.6}
v'_n + 2(i \lambda^{1/2} + v_0) v_n = -v_{n-1}^2 - 2v_{n-1}
\sum_{m=1}^{n-2} v_m .
\end{equation}
It is straightforward to check that the functions defined in \eqref{eq2.4}
and \eqref{eq2.5} satisfy \eqref{eq3.5} and \eqref{eq3.6}. We now bound the
$\{ v_n \}$ and show that the series $\sum v'_n$ is absolutely uniformly 
convergent on compact subsets of $[0,\infty)$ which is sufficient to 
justify the term by term differentiation.

\begin{lemma}\label{lem1}
Let
\begin{equation}\label{eq3.7}
K := \sup_{0 \leq x \leq t < \infty} \big| e^{-2 \int_x^t v_0(s) \,ds}
\big|
\end{equation}
and suppose there exists $a(x)$ which is a decreasing member of 
$L^1[0,\infty)$ such that
\begin{equation}\label{eq3.8}
|v_1 (x, \lambda)| \leq \lambda^{1/2} a(x)
\end{equation}
for $x \in [0, \infty)$ and $\lambda \in [0, \Lambda]$ where $\Lambda$ is so
small that $10K \lambda^{1/2} \int_0^{\infty} a(t) \,dt \leq 1$ for $\lambda
\in [0, \Lambda]$.
Then $|v_n(x, \lambda)| \leq \frac{\lambda^{1/2} a(x)}{2^{n-1}}$ for $x \in
[0, \lambda)$ and $\lambda \in [0, \Lambda]$.
\end{lemma}

\begin{proof}
We use induction on $n$. When $n=1$, the result follows from the hypothesis
\eqref{eq3.8}. Suppose now the result is true for all subscripts up to the
$(n-1)$st. Then from \eqref{eq2.4}, \eqref{eq3.7}, and the induction
hypothesis:
\begin{align*}
|v_n(x, \lambda)| &\leq K \int_x^{\infty} |v_{n-1}|^2 + 2|v_{n-1}|
\sum_{m=1}^{n-2} |v_m| \, dt \\
&\leq K \int_x^{\infty} \frac{\lambda a(t)^2}{2^{2n-4}} + \frac{2\lambda
a(t)^2}{2^{n-2}} \sum_{m=1}^{n-2} \frac{1}{2^{m-1}} \, dt \\
&\leq \frac{\lambda^{1/2}a(x)}{2^{n-1}} \lambda^{1/2} \big\{
\frac{1}{2^{n-3}} + 8 \big\} \int_0^{\infty} a(t) \, dt
\end{align*}
since $a(\cdot)$ is a decreasing function. The result now follows from the
choice of $\Lambda$.

It may now be seen from the Lemma and \eqref{eq3.6} that the series 
$\sum v'_n$ is absolutely uniformly convergent which justifies the
 term by term differentiation.
To complete the proof of the theorem we observe that, since 
$v_0(\cdot) \in L^1 [0, \infty)$, there exists a $K$ which satisfies 
\eqref{eq3.7} and also, from \eqref{eq2.4}, that
$$
|v_1 (x, \lambda)| \leq 2 \lambda^{1/2} K \int_x^{\infty} |v_0(t)| \, dt.
$$
We now choose $a(x) := 2K \int_x^{\infty} |v_0(t)| \,dt$ and note that
$$
\int_0^{\infty} a(x) \, dx = \int_0^{\infty} 2K \int_x^{\infty}
|v_0(t)|\, dt \, dx = 2K \int_0^{\infty} t|v_0(t)|\, dt.
$$
The first part of the theorem now follows.

It remains to show that, under the assumptions \eqref{eq2.1}--\eqref{eq2.3},
$v_0$ is real-valued. Suppose not; if $v_0(t) = u(t) + iw(t)$ then upon
substitution into \eqref{eq2.1} and the separation of real and imaginary
parts we see that
$$
w' = -2uw
$$
whence
$$
w(t) = C e^{-2 \int_0^t u(s)\, ds}
$$
The requirement $\lim_{t \to \infty} v_0(t) = 0$ then requires either
$C=0$ or $\lim_{t \to \infty} \int_0^t u(s) \, ds = \infty$.
But the latter case contradicts \eqref{eq2.3} which requires 
that $(1+t) v_0(t)$ and hence $(1+t)u(t) \in L^1 [0, \infty)$, 
so the only possibility is that $v_0$ is real-valued.
\end{proof}

\section{An asymptotic expansion} \label{sec:4}

The bounds derived in Lemma \ref{lem1} lead to estimates for the
$\{v_n\}$ which show that $\sum_{n = 1}^{\infty} v_n(x, \lambda)$ is
uniformly, absolutely convergent for $x \in [0, \infty)$ and $0 \leq
\lambda < \Lambda$ for some $\Lambda$ which is, in principle at
least, computable.  In terms of $\lambda$ however the bounds are all
of order $\lambda^{1/2}$.  We now show that the terms of the series
are decreasing asymptotically with increasing powers of $\lambda$.

\begin{lemma} \label{lem2}
With $K$ as in \eqref{eq3.7} and with $v_1$ satisfying \eqref{eq3.8}
there exist sequences of constants $\{C_n\}$ and $\{\Lambda_n\}$ so
that for $x \in [0, \infty)$ and $0 \leq \lambda \leq \Lambda_n \leq
\Lambda_{n - 1}$
\begin{equation}
\label{eq4.1}
|v_n(x, \lambda)| \leq C_n \lambda^{n/2} a(x).
\end{equation}
\end{lemma}

\begin{proof}
We proceed by induction.
From \eqref{eq2.5},
\begin{align*}
|v_2(x, \lambda)| & \leq \int_x^{\infty} e^{-2\int_x^t v_0(s)\, ds}
|v_1(t, \lambda)|^2\, dt \\
& \leq \lambda K \int_x^{\infty} a(t)^2\, dt \leq \lambda a(x) K
\int_0^{\infty} a(t)\, dt
\end{align*}
Suppose the result is true up to $n \geq 2$, then from \eqref{eq2.5}:
\begin{align*}
|v_{n + 1} (x, \lambda)| 
& \leq K \int_x^{\infty} |v_n|^2 + 2|v_n|
 \sum_{m = 1}^{n - 1} |v_m|\, dt \\
& \leq K \int_x^{\infty} C^2_n \lambda^n a(t)^2 + 2 C_n \lambda^{n/2} a(t)
 \sum_{m = 1}^{n - 1} C_m \lambda^{m/2} a(t)\, dt \\
& \leq K\lambda^{\frac{n + 1}{2}} a(x) 
\big\{ C^{\frac{n - 1}{2}}_n \lambda - 2 C_n \sum_{m = 1}^{n - 1}
C_m \lambda^{\frac{m - 1}{2}} \big\} \int_0^{\infty} a(t)\, dt
\end{align*}
since the $\Lambda_n$ form a decreasing sequence.  The result now
follows.
\end{proof}

In consequence of Lemma \ref{lem2} we have that for every $N$, 
$$
\rho'_0(\lambda) = \frac{1}{\pi} \big\{\lambda^{1/2} 
+ \operatorname{Im} \sum_{n =1}^N v_n(0, \lambda)\big\} 
+ O\left(\lambda^{\frac{N +1}{2}}\right)
$$
 as $\lambda \to 0^+$ and, in particular,
$$
\rho'_0(\lambda) = \frac{\lambda^{1/2}}{\pi} \big\{1 + 2
\int_0^{\infty} \cos (2\lambda^{1/2} t) e^{-2\int_0^t v_0(s)\, ds}
v_0(t)\, dt\big\} + O(\lambda)
$$
  as  $\lambda \to 0^+$.

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\bibitem{ref6} B. J. Harris;
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\end{thebibliography}

\end{document}