Content-Type: multipart/mixed; boundary="-------------0312120231577" This is a multi-part message in MIME format. ---------------0312120231577 Content-Type: text/plain; name="03-539.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="03-539.keywords" Trace formulas, non-self-adjoint Hill operators, Dirichlet and Neumann eigenvalues ---------------0312120231577 Content-Type: application/x-tex; name="ssfr.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="ssfr.tex" %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \documentclass[reqno,12pt]{amsart} \usepackage{graphicx,amsmath,amssymb,amsthm} %\usepackage{amsmath,amsthm,amscd,amssymb,latexsym,upref} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %\usepackage{showkeys} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %\headheight=6.15pt %\textheight=9.25in %\textwidth=6.5in %\oddsidemargin=0in %\evensidemargin=0in %\topmargin=-.375in %\input epsf %\epsfxsize=2in %%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %\headheight=8.6pt %\textheight=9.25in %\textwidth=6.65in %\oddsidemargin=0in %\evensidemargin=0in %\topmargin=0.0in %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \numberwithin{equation}{section} \newcommand{\C}{\mathbb C} \newcommand{\R}{\mathbb R} \newcommand{\Z}{\mathbb Z} \newcommand{\N}{\mathbb N} %Derivatives: % \renewcommand{\d}{\prime} \newcommand{\dd}{{\prime \prime}} \newcommand{\ddd}{{\prime \prime \prime}} \newcommand{\dddd}{{\prime \prime \prime \prime}} \renewcommand{\Re}{{\rm Re}\,} \renewcommand{\Im}{{\rm Im}\,} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{untheorem}{Theorem} \newtheorem{unlemma}{Lemma} \newtheorem{conjecture}{Conjecture} \newtheorem{claim}{Claim} \newtheorem{problem}{Problem} \newtheorem{definition}{Definition} \newtheorem*{remark}{Remark} \newtheorem*{remarks}{Remarks} \renewcommand{\theuntheorem}{\hspace*{-4pt}} \renewcommand{\theunlemma}{\hspace*{-4pt}} \renewcommand{\theconjecture}{\hspace*{-4pt}} \renewcommand{\theclaim}{\hspace*{-4pt}} \renewcommand{\theproblem}{\hspace*{-4pt}} \renewcommand{\thedefinition}{\hspace*{-4pt}} \renewcommand{\thefootnote}{} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{document} \title[] {Trace Formulas for Non-Self-Adjoint Periodic Schr\"odinger Operators \\ and some Applications} \author[] {Kwang C.\ Shin} \address{Department of Mathematics, University of Missouri, Columbia, MO 65211} \date{December 10, 2003} \keywords{Trace formulas, non-self-adjoint Hill operators, Dirichlet and Neumann eigenvalues} \footnote{\it 2000 Mathematics Subject Classification. Primary 34A55, 34L15, 34L20.} \begin{abstract} Recently, a trace formula for non-self-adjoint periodic Schr\"odinger operators in $L^2(\R)$ associated with Dirichlet eigenvalues was proved in \cite{FG}. Here we prove a corresponding trace formula associated with Neumann eigenvalues. In addition we investigate Dirichlet and Neumann eigenvalues of such operators. In particular, using the Dirichlet and Neumann trace formulas we provide detailed information on location of the Dirichlet and Neumann eigenvalues for the model operator with the potential $Ke^{2ix}$, where $K\in\C$. \end{abstract} \maketitle \baselineskip = 18pt % \section{Introduction} Consider the differential expression $L$ \begin{equation}\nonumber L=-\frac{d^2}{dx^2}+V(x), \quad x\in\R, \end{equation} where $V$ is a continuous {\it complex-valued} periodic function on $\R$ of period $\pi$. The {\it Hill operator} $H$ in $L^2(\R)$ generated by the differential expression $L$ is defined by $$(Hf)(x)=L(f(x)),\; x\in\R, \; f\in dom(H)=H^{2,2}(\R), $$ where $H^{2,2}(\R)$ denotes the usual Sobolev space. Then $H$ is a densely defined closed operator in $L^2(\R)$ (see, e.g., \cite[Ch. 5]{MSPE}). In addition, we define families of differential operators $H_{x_0}^{D}$ and $H_{x_0}^{N}$ in $L^2([x_0,x_0+\pi])$, $x_0\in\R$, as follows: \begin{align*} (H_{x_0}^D f)(x)&=L(f(x)),\; x\in [x_0,x_0+\pi], \; f\in dom (H_{x_0}^D), \\ (H_{x_0}^N f)(x)&=L(f(x)),\; x\in [x_0,x_0+\pi], \; f\in dom (H_{x_0}^N), \end{align*} where \begin{align*} dom (H_{x_0}^D) &= \big\{f\in L^2([x_0,x_0+\pi]) \, \big|\, f,\,f^\d\in AC([x_0,x_0+\pi]); \\ & \hspace*{4.2cm}f(x_0+)=0=f(x_0+\pi-)\big\}, \\ dom (H_{x_0}^N) &= \big\{f\in L^2([x_0,x_0+\pi]) \, \big|\, f,\,f^\d\in AC([x_0,x_0+\pi]); \\ & \hspace*{4.05cm} f^\d(x_0+)=0=f^\d(x_0+\pi-)\big\}. \end{align*} The $L^2(\R)$ spectrum of the Hill operator $H$ is purely continuous and it is the union of countably many analytic arcs in the complex plane \cite{FSRB}. On the other hand, the spectra of $H_{x_0}^{D}$ and $H_{x_0}^{N}$ are purely discrete. We will denote the {\it Dirichlet eigenvalues} (i.e., the eigenvalues of $H_{x_0}^{D}$) by $\mu_j(x_0)$, $j\in\N$, and the {\it Neumann eigenvalues} (i.e., the eigenvalues of $H_{x_0}^{N}$) by $\nu_k(x_0)$, $k\in\N_0=\N\cup\{0\}$, where we number these eigenvalues in the order of non-decreasing magnitudes. When the potential $V$ is real-valued and periodic, the operators $H$, $H_{x_0}^D$ and $H_{x_0}^N$ are all self-adjoint, and hence the spectra of these operators are subsets of the real line. In this case, the spectrum of $H$ is a countable union of closed intervals $[E_{2m},\, E_{2m+1}]$, $m\in\N_0$, on the real line, where $E_m$ are such that $L\phi=E_m\phi$ has a nontrivial (i.e., nonzero) periodic solution of period $2\pi$. Thus, $E_m$ are eigenvalues of the self-adjoint operator associated with $L$ under periodic boundary conditions at $0$ and $2\pi$, and they are all real. Moreover, \begin{equation} \nu_0(x_0)\leq E_0, \quad E_{2m-1}\leq\mu_m(x_0),\,\nu_m(x_0)\leq E_{2m}, \; m\in\N, \; x_0\in\R \label{int} \end{equation} (see, e.g., \cite[Theorem 3.1.1]{MSPE}, \cite{MW}). Moreover, the following trace formulas hold. \begin{lemma}\label{real_trace} Suppose that $V\in C^1(\R)$ is a periodic function on $\R$ and let $x\in\R$. Then, \begin{align} V(x)&=E_0+\sum_{m=1}^{\infty}\big(E_{2m-1}+E_{2m}-2\mu_m(x)\big), \label{Dt}\\ V(x)&=2\nu_0(x)-E_0+\sum_{m=1}^{\infty}\big(2\nu_m(x)-E_{2m-1} -E_{2m}\big).\label{Nt} \end{align} \end{lemma} Under the hypothesis that $V\in C^3(\R)$ is a real-valued periodic function on $\R$, Trubowitz \cite{TRUB} proved the Dirichlet trace formula (\ref{Dt}). The Neumann trace formula \eqref{Nt} for real-valued potentials is due to McKean and Trubowitz \cite{MT}. In 2001, Gesztesy \cite{FG} extended (\ref{Dt}) to complex-valued periodic $C^1(\R)$ potentials. In Section \ref{trace_formula}, we will prove the trace formula (\ref{Nt}) for complex-valued periodic $C^1(\R)$ potentials, by closely following the methods in \cite{FG}. We refer to \cite[p.\ 121]{GH03} (cf.\ also \cite{GS}) for further references and a detailed history of trace formulas. In the self-adjoint case, Borg \cite{BORG} showed that $V=E_0$ is the only real-valued periodic potential, for which the Hill operator $H$ has the spectrum $[E_0,\infty)$. One can see that if $V\in C^1(\R)$ is real-valued and periodic, the Dirichlet trace formula (\ref{Dt}) yields $V(x)=E_0$ since then $E_{2m-1}=E_{2m}$ and $E_{2m-1}\leq \mu_m(x)\leq E_{2m}$ for $m\in\N$. Extensions of this simple observation to reflectionless potentials can be found in \cite{CGHL}. Unlike the self-adjoint case, there are infinitely many complex-valued potentials that generate a half-line spectrum $[0,\infty)$ as shown by Gasymov \cite{Gasy1} (also, see \cite{Pastur}, \cite{SHIN}). In these cases, even though all $E_m$ are real and $E_{2m-1}=E_{2m}$ for all $m\in\N$, the Dirichlet and Neumann eigenvalues $\mu_m(x)$, $\nu_m(x)$ are not trapped between $E_{2m-1}$ and $E_{2m}$, due to non-self-adjointness of the corresponding operators $H_{x}^D$ and $H_{x}^N$. From the inverse spectral point of view, this explains the highly non-uniqueness property of complex-valued periodic potentials (see, e.g., \cite[p.\ 113]{GH03}). As a concrete example, the potential $V(x)=e^{2ix}$ generates the spectrum $[0,\,\infty)$ (cf.\ \cite{Gasy1}) and one infers $E_0=0$, $E_{2m-1}=E_{2m}=m^2$, $m\in\N$. Marchenko \cite{March} provides the {\it asymptotic} location of $\mu_j(x)$ and $\nu_k(x)$ (see \eqref{E_asymp}, \eqref{mu} and \eqref{asymp_eq} below). However, up to date, the precise location of $\mu_j(x)$ and $\nu_k(x)$ for small $j,\,k$, is not known for this example. The principal application of the trace formulas proved in this paper will be a determination of the location of these eigenvalues. This complements the asymptotic results of Marchenko \cite{March}. In Section \ref{back_ground}, we will briefly review elements of Floquet theory and show that algebraic and geometric multiplicities of the eigenvalues $E_m$ are different for some cases in Theorem \ref{main_estim}. Our principal new result, the Neumann trace formula for periodic potentials will be proved in Section \ref{trace_formula}. In Section \ref{pt_sym}, we will prove the following result: \begin{theorem} If $\overline{V(-x)}=V(x)$, $x\in\R$, then for all $x_0\in\R$, $j\in\N$, and $k\in\N_0$, $\mu_j(\pi-x_0)=\overline{\mu_j(x_0)}$ and $\nu_k(\pi-x_0)=\overline{\nu_k(x_0)}$. \end{theorem} In Section \ref{deift_work}, we will recall some unpublished work by Deift \cite{DEIFT} and review some basic facts on Bessel functions that we use in our final Section \ref{appl}. In Section \ref{appl}, we will use the trace formulas (\ref{Dt}) and (\ref{Nt}) to investigate the location of $\mu_j(x_0)$ and $\nu_k(x_0)$ for the concrete example $V(x)=Ke^{2ix}$, $K\in\C$. More precisely, we will prove the following theorems. \begin{theorem} Suppose that $V(x)=K e^{2ix}$ and $K=|K|e^{2i\varphi_0}$ for some $\varphi_0\in\R$. Then for all $j\in\N$ and $k\in\N_0$, $\mu_j(K,x_0)=\mu_j(|K|, x_0+\varphi_0)$ and $\nu_k(K,x_0)=\nu_k(|K|, x_0+\varphi_0)$. \end{theorem} \begin{theorem}\label{main_estim} Suppose that $V(x)=K e^{2ix}$, $K>0$. Then \begin{itemize} \item[(i)] $(j-1)^2\leq\mu_j(0)\leq j^2$ for all $j\in\N$. \item[(ii)] If $01$ then $M_1=K-2\sum_{m=1}^{\lfloor \sqrt{K}\rfloor}(m^2-\mu_m(0))>0$ and \\ $j^2-\frac{M_1}{2}<\mu_j(0)1$ then $M_2=K-2\nu_0(0)-2\sum_{m=1}^{\lfloor \sqrt{K}\rfloor}(\nu_m(0)-m^2)>0$ and $k^2<\nu_k(0)< k^2+\frac{M_2}{2}$ for all $k\geq \sqrt{K}$. In particular, $\nu_k(0)\not=E_m$ for $k\geq \sqrt{K}$ and $m\in\N_0$. \item[(vii)] If $K>0$ then $(2j-1)^2<\mu_{2j-1}(\pi/2)\leq \mu_{2j}(\pi/2)<(2j)^2$ for all $j\in\N$, and hence $\mu_j(\pi/2)\not=E_m$ for all $j\in\N$ and $m\in\N_0$. Moreover, if $00$ then $(2k)^2<\nu_{2k}(\pi/2)\leq \nu_{2k+1}(\pi/2)<(2k+1)^2$ for all $k\in\N_0$, and hence $\nu_k(\pi/2)\not=E_m$ for all $k\in\N_0$. Moreover, if $00$.} \end{remark} Next, we will provide more precise location of the Dirichlet eigenvalues $\mu_j(0)$. \begin{theorem}\label{mu_real} Suppose that $V(x)=K e^{2ix}$ and $K>0$. Then \begin{itemize} \item[(i)] $(j-1)^2\leq\mu_j(0)\leq j^2$ for all $j\in\N$. \item[(ii)] If $01$ then $M_1=K-2\sum_{m=1}^{\lfloor \sqrt{K}\rfloor}(m^2-\mu_m(0))>0$ and \\ $j^2-\frac{M_1}{2}<\mu_j(0)0$ and $s((2n-1)^2,0,\pi)<0$, $n\in\N$. The continuous function $s(\lambda,0,\pi)$ is real for $\lambda\geq 0$ (in fact, it is real for all $\lambda\in\R$ because if $\lambda<0$ then the two Bessel functions in (\ref{J_eq1}) at $\sqrt{K}\in\R$ are complex conjugates of each other since $\Re(\sqrt{\lambda})=0$). By Lemma \ref{lemma_bessel} (4), $$J_{-n}(u)=(-1)^nJ_{n}(u), \quad n\in\Z$$ and hence \begin{equation} s(n^2,0,\pi)=(-1)^n\left(J_n(\sqrt{K})\right)^2. \end{equation} By Lemma \ref{lemma_bessel} (2) and (3), for each $\sqrt{\lambda}\geq 0$, the zeros of $u\mapsto J_{\sqrt{\lambda}}(u)$ are all real, and positive zeros of these functions are all greater than $\sqrt{\lambda}$. Thus, if $01$ for all $j\geq n+1$, and hence the sum in (\ref{trace_for}) would be divergent. Thus, there is exactly one $\mu_j(0)$ in every interval $((n-1)^2,n^2)$, $n\in\N$. Since the sum is $K$ and since $2j^2-2\mu_j(0)>0$ for all $j\in\N$, we have $$j^2-\frac{K}{2}0$. Then \begin{itemize} \item[(i)] $k^2\leq\nu_k(0)\leq (k+1)^2$ for all $k\in\N_0$. \item[(ii)] If $01$ then $M_2=K-2\nu_0(0)-2\sum_{m=1}^{\lfloor \sqrt{K}\rfloor}(\nu_m(0)-m^2)>0$ and $k^2<\nu_k(0)< k^2+\frac{M_2}{2}$ for all $k\geq \sqrt{K}$. In particular, $\nu_k(0)\not=E_m$ for $k\geq \sqrt{K}$, $m\in\N_0$. \end{itemize} \end{theorem} \begin{proof}[Proof of (ii)] The arguments are very similar to that in the proof of Theorem \ref{mu_real} (ii). However, in the trace formula (\ref{Neumann_trace1}), $\nu_0(x_0)$ is paired with $E_0$, unlike in the Dirichlet case (\ref{Dirichlet_trace1}). Hence, a more careful analysis is needed in the present Neumann case. We recall that $\nu_k(0)$, $k\in\N_0$, are the zeros of \begin{equation} c^\d(\lambda,0,\pi)=\pi K J_{\sqrt{\lambda}}^\d(\sqrt{K})J_{-\sqrt{\lambda}}^\d(\sqrt{K}), \end{equation} a real-valued continuous function on the real line. Moreover, by Corollary \ref{cor_real}, $\nu_k(0)\in\R$ for all $k\in\N_0$. If $00$, that is, there is no negative Neumann eigenvalue since $c^\d(0,0,\pi)\not=0$. The trace formula (\ref{Neumann_trace1}) at $x_0=0$ reads \begin{align} K=Ke^{2i x_0}\Big|_{x_0=0}&= 2\nu_0(0)-E_0+ \sum_{k=1}^{\infty}\left(2\nu_k(0)-E_{2k-1}-E_{2k}\right)\nonumber\\ &= 2\nu_0(0)+\sum_{k=1}^{\infty}\left(2\nu_k(0)-2k^2\right). \label{trace_for1} \end{align} Next, suppose that $J_{\sqrt{\lambda}}^\d(\sqrt{K})=0$ at $\lambda=\nu_0(0)<0$. Then since $J_{\sqrt{\lambda}}^\d(\sqrt{K})$ is an entire function of $\sqrt{\lambda}$, we can write $$J_{\sqrt{\lambda}}^\d(\sqrt{K})=(\sqrt{\lambda}-i\sqrt{|\nu_0(0)|})f(\sqrt{\lambda})$$ for some entire function $f$. Since $$J_{-\sqrt{\nu_0(0)}}^\d(\sqrt{K})=\overline{J_{\sqrt{\nu_0(0)}}^\d(\sqrt{K})} =0,$$ we see that $f(-i\sqrt{|\nu_0(0)|})=0$, and hence $$J_{\sqrt{\lambda}}^\d(\sqrt{K})=(\sqrt{\lambda}-i\sqrt{|\nu_0(0)|})(\sqrt{\lambda}+i\sqrt{|\nu_0(0)|})f_1(\sqrt{\lambda})$$ for some entire function $f_1$. Thus, \begin{align} c^\d(\lambda,0,\pi)&= \pi K J_{\sqrt{\lambda}}^\d(\sqrt{K})J_{-\sqrt{\lambda}}^\d(\sqrt{K})\nonumber\\ &= \pi K(\lambda-\nu_0(0))^2f_1(\sqrt{\lambda})f_1(-\sqrt{\lambda}),\nonumber \end{align} implying $\nu_1(0)<0$. But then (\ref{trace_for1}) would diverge, since there exists at least one $\nu_k(0)$ in the open interval $((n-1)^2,n^2)$ for every $n\in\N$. This is a contradiction, and hence there is no negative Neumann eigenvalue. The previous argument also shows that if $c^\d(n_0^2,0,\pi)=0$ for some $n_0\in \N$, then $c^\d(\lambda,0,\pi)$ has at least a double zero at $\lambda=n_0^2$, while $c^\d(\lambda,0,\pi)$ could have a simple zero at $\lambda=0$. Moreover, still assuming $00$ for all $k\in\N_0$, we infer that \begin{align} &0<\nu_0(0)<\frac{K}{2}, \nonumber \\ &k^2<\nu_k(0)1$. We recall that $\nu_k(x_0)$ is numbered according to $k\in\N_0$. In proving Theorem \ref{mu_real} (i) and (iii), we used Lemma \ref{lemma_bessel} (6), while here we need Lemma \ref{lemma_bessel} (7). The point of Lemma \ref{lemma_bessel} (7) in the proof is that if $c^\d(n^2,0,\pi)=0$ then $c^\d(k^2,0,\pi)\not=0$ for $k=(n-1)$ and $k=(n+1)$. The rest of the proof is similar to that of Theorem \ref{mu_real} (i) and (iii). Hence, we omit further details. \end{proof} Next, we investigate $\mu_j(\pi/2)$, $j\in\N$. \begin{theorem}\label{pi_2} Suppose that $V(x)=K e^{2ix}$ and $K>0$. Then, $$(2j-1)^2<\mu_{2j-1}(\pi/2)\leq \mu_{2j}(\pi/2)<(2j)^2, \quad j\in\N,$$ and hence $\mu_j(\pi/2)\not=E_m$, $m\in\N_0$. Moreover, if $00 \text{ and } f(-(2n-1/2))<0 \text{ for all $n\in\N.$} \end{equation} By the intermediate value theorem this then implies the existence of at least one $\mu_j(\pi/2)$ in every interval of the form $$(-(2n-1)^2, -(2n-1/2)^2),\quad (-(2n-1/2)^2, -(2n)^2), \quad n\in\N.$$ Thus, the convergence of the trace formula \begin{equation} -K=Ke^{2ix_0}\Big|_{x_0=\pi/2}=E_0 +\sum_{j=1}^{\infty}\big(E_{2j-1}+E_{2j}-2\mu_j(\pi/2)\big) \end{equation} implies that there is exactly one eigenvalue $\mu_j(\pi/2)$ in each interval above, and these are all the $\mu_j(\pi/2)$. Since $J_{-n}(i\sqrt{K})=(-1)^n J_n(i\sqrt{K})$, $n\in\Z$, and since by (\ref{f_est}), $f(n)>0$, $n\in\N_0$, we infer from Lemma \ref{lemma_bessel} (4) that \begin{align} f(-n)&=(i\sqrt{K}/2)^nJ_{-n}(i\sqrt{K})\nonumber \\ &=(-1)^n(i\sqrt{K}/2)^{2n}f(n)>0,\quad n\in\N_0. \label{fn_est} \end{align} Next, we show that $f(-2n+1/2)<0$, $n\in\N$. For every $0\leq m\leq 2n-2$, repeated use of the formula $\Gamma(\lambda)=\frac{1}{\lambda}\Gamma(\lambda+1)$, implies \begin{equation}\label{gamma_eq} \Gamma(m-2n+3/2) =\frac{(-2)^{2n-m-1}\Gamma(1/2)}{\prod_{j=1}^{2n-m-1}(4n-2m-(2j+1))}. \end{equation} Thus, from (\ref{f_est}), \begin{align} f(-2n+1/2) &=\sum_{m=0}^{\infty}\frac{\left(\frac{\sqrt{K}}{2} \right)^{2m}}{m!\Gamma(m-2n+1/2+1)}\nonumber\\ &=\sum_{m=0}^{2n-2}(-1)^{m+1} \frac{\prod_{j=1}^{2n-m-1}(4n-2m-(2j+1))}{2^{2n-1}m! \Gamma(1/2)}\left(\frac{K}{2}\right)^{m}\nonumber\\ &\quad+\sum_{m=2n-1}^{\infty}\frac{\left(\frac{\sqrt{K}}{2} \right)^{2m}}{m!\Gamma(m-2n+1/2+1)}. \label{sec_last} \end{align} Moreover, \begin{equation} \sum_{m=2n-1}^{\infty}\frac{\left(\frac{\sqrt{K}}{2}\right)^{2m}}{m!\Gamma(m-2n+1/2+1)} <\frac{\left(\frac{K}{4}\right)^{2n-1}e^{K/4}}{(2n-1)!\sqrt{\pi}}, \quad n\in\N, \end{equation} and hence, if $00$, while (\ref{f_half}) may not hold for some $K>1$. Next, for each $K>1$ and $\frac{1}{K}\leq \varepsilon\leq 1$ we introduce the family of potentials $V(x)=\varepsilon K e^{2ix}$. Again, we use the notation $\mu_j(\varepsilon K,0)$ to indicate the $\varepsilon$-dependence of these eigenvalues. Then, for each $j\in\N$, the function $\varepsilon\mapsto \mu_j(\varepsilon K,0)$ is continuous since $J_{\nu}(u)$ is an entire function of $\nu$ for each fixed $u\not=0$, and an analytic function of $u$ on the positive real axis. When $\varepsilon=\frac{1}{K}$, that is, when $V(x)=e^{2ix}$, we proved above that $$(2j-1)^2<\mu_{2j-1}(1,\pi/2)\leq \mu_{2j}(1,\pi/2)<(2j)^2, \quad j\in\N.$$ So as $\varepsilon$ increases to $1$, the real numbers $\mu_j(\varepsilon K,\pi/2)$ cannot become $(2j-1)^2$ or $(2j)^2$ due to (\ref{fn_est}) which holds for all $K>0$. This completes the proof. \end{proof} Next, we investigate $\nu_k(\pi/2)$ for $k\in\N_0$. \begin{theorem} Suppose that $V(x)=K e^{2ix}$ and $K>0$. Then $$(2k)^2<\nu_{2k}(\pi/2)\leq \nu_{2k+1}(\pi/2)<(2k+1)^2, \quad k\in\N_0,$$ and hence $\nu_k(\pi/2)\not=E_m$, $k\in\N_0$. Moreover, if $00$, \begin{equation} g(n)>0,\quad n\in\Z. \end{equation} Here the identity $$J_{\nu}^\d(u)=\frac{\nu}{u}J_{\nu}(u)-J_{\nu+1}(u)$$ (see \cite[pp.\ 45]{WATSON}) turns out to be useful. Moreover, one can show that if $00$. Then for every $x\in (0,\pi)$, $\mu_j(x),\,\nu_k(x)\not=E_m$. \end{theorem} \begin{proof} {} From (\ref{trans_eq}) one knows that $\mu_j(x)$ are zeros of \begin{equation}\label{J_eq3} s(\lambda,x,x+\pi)=\pi J_{\sqrt{\lambda}}(e^{ix}\sqrt{K})J_{-\sqrt{\lambda}}(e^{ix}\sqrt{K}), \end{equation} and that $\nu_k(x)$ are zeros of \begin{equation}\label{J_eq4} c^\d(\lambda,x,x+\pi)=\pi\sqrt{K}e^{ix} J_{\sqrt{\lambda}}^\d(e^{ix}\sqrt{K}) J_{-\sqrt{\lambda}}^\d(e^{ix}\sqrt{K}). \end{equation} Since all zeros of $J_n(u)$ and $J_n^\d(u)$ are real for $n\geq 0$, we see that if $0