\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 $$\label{eq1.1} y'' + (\lambda - q(x)) y = 0$$ 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 $$\label{eq1.2} y(0) \cos (\alpha) + y' (0) \sin (\alpha) = 0 \quad \alpha \in [0, \pi)$$ 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 $$\label{eq1.3} \int^{\infty} (q')^2 |q|^{-5/2} dt < \infty, \quad \int^{\infty} |q''| \ |q|^{-3/2} dt < \infty, \quad \int^{\infty} |q|^{-1/2} dt = \infty,$$ 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 $$\label{eq1.4} v' + v^2 + (\lambda - q) = 0.$$ 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 $$\label{eq1.5} \rho_0' (\lambda) = \frac{1}{\pi} \operatorname{Im} \{v (0, \lambda)\}.$$ 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 $$\label{eq1.6} \rho_0'' (\lambda) = \frac{1}{\pi} \frac{\partial}{\partial \lambda} \{\operatorname{Im} v(0,\lambda) \}.$$ \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 $$\label{eq2.1} v (x, \lambda) = i (\lambda - q (x))^{1/2} + \frac{1}{4} q' (x) (\lambda - q(x))^{-1} + \sum_{n=1}^{\infty} v_n (x, \lambda).$$ 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 $$\label{eq2.2} \begin{gathered} v_1' + \Big(2 i (\lambda - q)^{1/2} + \frac{q'}{2 (\lambda - q)} \Big) v_1 = - Q\\ v_2' + \Big(2 i (\lambda - q)^{1/2} + \frac{q'}{2 (\lambda - q)} \Big) v_2 = - v_1^2\\ v_n' + \Big(2 i (\lambda - q)^{1/2} + \frac{q'}{2 (\lambda - q)} \Big) v_n = - v_{n-1}^2 - 2 v_{n-1} \sum_{m=1}^{n-2} v_m \\ \end{gathered}$$ for $n = 3, 4, \dots$. The required solution to \eqref{eq2.1} is \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} 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 $$\label{eq3.1} \frac{\partial^2 v_1}{\partial \lambda \partial x} + \Big( 2i (\lambda - q)^{1/2} + \frac{q'}{2 (\lambda - q)}\Big) \frac{\partial v_1}{\partial \lambda} = - \frac{\partial Q}{\partial \lambda} - i (\lambda - q)^{-1/2} v_1 + \frac{1}{2} q' (\lambda - q)^{-2} v_1$$ 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 \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} 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 \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} 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 \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} \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 $n = 1, 2, 3, \dots$, $$\label{eq3.5} \big| \frac{\partial v_n}{\partial \lambda} (x, \lambda)\big| \leq \frac{1}{2^{n-1}} (\lambda - q(x))^{1/2} \int_x^{\infty} (\lambda - q (t))^{-1/2} I (t, \lambda) \,dt\,.$$ \end{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. \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} 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}: \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} 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 $$\label{eq4.1} 2(\lambda - q(0)) \int^\infty_0 \frac{q''(t)}{(\lambda - q(t))^2} + \frac{5(q'(t))^2}{4(\lambda - q(t))^3}\, dt < 1.$$ \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. \begin{thebibliography}{00} \bibitem{e1} M. S. P. Eastham; \emph{The asymptotic form of the spectral function in Sturm-Liouville problems with a large potential like $-x^c$ ($c \leq 2$)}. Proc. Royal Soc. Edinburgh 128A (1998), 37-45. \bibitem{g1} D. J. Gilbert, B. J. Harris; \emph{Bounds for the points of spectral concentration of Sturm-Liouville problems}. Mathematika, 47 (2000), 327-337. \bibitem{g2} D. J. Gilbert, B. J. Harris; \emph{Connection formulae for spectral functions associated with singular Sturm-Liouville expressions}. Proc. Royal Soc. Edinburgh, 130A (2000), 25-34. \bibitem{g3} D. J. Gilbert, B. J. Harris, S. M. Riehl; \emph{The spectral function for Sturm-Liouville equations where the potential is of Wigner-von Neumann type or slowly decaying}. Journal of Differential Equations 201, (2004), 139-159. \bibitem{g4} D. J. Gilbert, B. J. Harris, S. M. Riehl; \emph{Higher derivatives of spectral functions associated with one-dimensional Schr\"{o}dinger operators}. Operator Theory: Advances and Applications. 186 (2008), 217-228. \bibitem{h1} B. J. Harris; \emph{The form of the spectral functions associated with Sturm-Liouville equations with large negative potential}. J. Math Analysis and Appl. 222 (1998), 569-584. \bibitem{k1} J. C. Kallenbach; \emph{Ph. D. Dissertation} Northern Illinois University 2007. \bibitem{t1} E. C. Titchmarsh; \emph{Eigenfunction Expansions, part 1}. Clarendon Press: Oxford, 2nd ed. 1962. \end{thebibliography} \end{document}