\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 $$\label{eq2.1} v'_0 = q(x) - v_0^2$$ which is defined on $[0, \infty)$ and satisfies $$\label{eq2.2} \lim_{x \to \infty} x v_0(x) = 0.$$ We further assume that $$\label{eq2.3} (1+t) |v_0(t)| \in L^1 [0, \infty).$$ Under these conditions it will be shown that there exists a sequence of functions $\{ v_n (x, \lambda) \}$ defined recursively as follows: $$\label{eq2.4} v_1 (x, \lambda) := 2i\lambda^{1/2} \int_{x}^{\infty} e^{2i\lambda^{1/2} (t-x) - 2 \int_x^t v_0(s) \, ds} v_0(t)\, dt$$ and $$\label{eq2.5} v_n(x, \lambda) := \int_{x}^{\infty} e^{2i\lambda^{1/2} (t-x) - 2\int_x^t v_0(s)\, ds} \Big( v_{n-1}^2 + 2v_{n-1} \sum_{m=1}^{n-2} v_m \Big) \,dt .$$ \begin{theorem} \label{thm1} Under conditions \eqref{eq2.1}--\eqref{eq2.3} there exists $\Lambda > 0$ so that for $\lambda \in (0, \Lambda)$ $$\label{eq2.6} \rho'_0(\lambda) = \frac{1}{\pi} \big\{ \lambda^{1/2} + \operatorname{Im} \sum_{n=1}^{\infty} v_n(0,\lambda) \big\}.$$ In particular $$\label{eq2.7} \lim_{\lambda \to 0^+} \rho'_0(\lambda) = 0 .$$ \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 $$\label{eq3.1} v' = -\lambda + q - v^2$$ which satisfies $$\label{eq3.2} \lim_{x \to \infty} v(x, \lambda) = i \lambda^{1/2} .$$ Then, from \cite[(4.4)]{ref5}, $$\label{eq3.3} \rho'_0(\lambda) = \frac{1}{\pi} \operatorname{Im} \{ v(\lambda) \} .$$ We try for a solution of \eqref{eq3.1} a series of the form $$\label{eq3.4} v(x, \lambda) = i \lambda^{1/2} + v_0(x) + \sum_{n=1}^{\infty} v_n (x, \lambda).$$ 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 $$\label{eq3.5} v'_1 + (2i \lambda^{1/2} + v_0) v_1 = -2i \lambda^{1/2} v_0$$ and for $n = 2, 3, \ldots$, $$\label{eq3.6} v'_n + 2(i \lambda^{1/2} + v_0) v_n = -v_{n-1}^2 - 2v_{n-1} \sum_{m=1}^{n-2} v_m .$$ 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 $$\label{eq3.7} K := \sup_{0 \leq x \leq t < \infty} \big| e^{-2 \int_x^t v_0(s) \,ds} \big|$$ and suppose there exists $a(x)$ which is a decreasing member of $L^1[0,\infty)$ such that $$\label{eq3.8} |v_1 (x, \lambda)| \leq \lambda^{1/2} a(x)$$ 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 $0 \leq \lambda \leq \Lambda_n \leq \Lambda_{n - 1}$ $$\label{eq4.1} |v_n(x, \lambda)| \leq C_n \lambda^{n/2} a(x).$$ \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 $\lambda \to 0^+$. \begin{thebibliography}{99} \bibitem{ref1} M. S. P. Eastham; \emph{The asymptotic nature of the spectral functions in Sturm-Liouville problems with continuous spectra.} J. Math. Anal. Appl. 213 (1997), 573-582. \bibitem{ref2} M. S. P. Eastham; \emph{On the location of spectral concentration for Sturm-Liouville problems with rapidly decaying potential}. Mathematika 45 (1998), 25-36. \bibitem{ref3} D. J. Gilbert, B. J. Harris; \emph{Bounds for the spectral concentration of Sturm-Liouville problems}. 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