\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2009(2009), No. 27, pp. 1--8.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2009 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2009/27\hfil Regularized trace] {Regularized trace of the Sturm-Liouville operator with irregular boundary conditions} \author[A. Makin\hfil EJDE-2009/27\hfilneg] {Alexander Makin} \address{Alexander Makin \newline Moscow State University of Instrument-Making and Informatics \newline Stromynka 20, Moscow, 107996, Russia} \email{alexmakin@yandex.ru} \thanks{Submitted January 8, 2009. Published February 3, 2009.} \subjclass[2000]{34L05, 34B24} \keywords{Sturm-Liouville operator; eigenvalue problem; spectrum} \begin{abstract} We consider the spectral problem for the Sturm-Liouville equation with a complex-valued potential $q(x)$ and with irregular boundary conditions on the interval $(0,\pi)$. We establish a formula for the first regularized trace of the this operator. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \allowdisplaybreaks \section{Introduction and main result} This paper deals with the eigenvalue problem for the Sturm-Liouville equation $$u''-q(x)u+\lambda u=0 \label{e1.1}$$ on the interval $(0,\pi)$ with the boundary conditions $$u'(0)+(-1)^\theta u'(\pi)+bu(\pi)=0,\quad u(0)+(-1)^{\theta+1}u(\pi)=0,\label{e1.2}$$ where $b$ is a complex number, $b\ne0$, $\theta=0,1$. The goal of this article is to calculate the first-order regularized trace for \eqref{e1.1}-\eqref{e1.2}. The theory of regularized traces of ordinary differential operators has a long history. First, the trace formulas for the Sturm-Liouville operator with the Dirichlet boundary conditions and sufficiently smooth potential $q(x)$ were established in \cite{d1,g1}. Afterwards these investigations were continued in many directions, for instance, the trace formulas for the Sturm-Liouville operator with periodic or antiperiodic boundary conditions were obtained in \cite{l1,s2}, and for regular but not strongly regular ones \cite{n1} similar formulas were found in \cite{m2}. A method for calculating trace formulas for general problems involving ordinary differential equations on a finite interval was proposed in \cite{l3}. The bibliography on the subject is very extensive and we refer to the list of the works in \cite{l2,s1}. The trace formulas can be used for approximate calculation of the first eigenvalues of an operator \cite{s1}, and in order to establish necessary and sufficient conditions for a set of complex numbers to be the spectrum of an operator \cite{s2}. We will prove the following statement. \begin{theorem} \label{thm1} Let $q(x)$ be an arbitrary complex-valued function in $W_1^1(0,\pi)$ and denote by $\lambda_n$ $(n=1,2,\dots)$ the eigenvalues of \eqref{e1.1}-\eqref{e1.2}. Then we have the trace formula \begin{aligned} &\sum_{n=1}^\infty\big\{\lambda_n-n^2-\langle q\rangle -\frac{(-1)^{\theta}(q(\pi)-q(0)-\int_0^\pi q'(t)\cos(2nt)dt)}{\pi b}\big\}\\ &-\frac{\langle q\rangle}{2}+\frac{(q(\pi)-q(0))^2}{2b^2}+\frac{q(\pi)+q(0)}{4}=0, \end{aligned}\label{e1.3} where $\langle q\rangle=\pi^{-1}\int_0^\pi q(t)dt$. \end{theorem} This article is organized as follows. Section 2 is devoted to the analysis of the characteristic determinant. In section 3, we obtain a formula of the regularized trace in explicit form. \section{Analysis of the characteristic determinant} Denote by $c(x,\mu), s(x,\mu)$ $(\mu^2=\lambda)$ the fundamental system of solutions to \eqref{e1.1} with the initial conditions $c(0,\mu)=s'(0,\mu)=1$, $c'(0,\mu)= s(0,\mu)=0$. Simple calculations show that the characteristic equation of \eqref{e1.1}-\eqref{e1.2} can be reduced to the form $\Delta(\mu)=0$, where $$\Delta(\mu)=c(\pi,\mu)-s'(\pi,\mu)-(-1)^\theta bs(\pi,\mu).\label{e2.1}$$ Denote by $\varphi(x,\mu), \psi(x,\mu)$ the system of solutions to \eqref{e1.1} with the initial conditions $\varphi(0,\mu)=\psi(0,\mu)=1$, $\varphi_x'(0,\mu)=i\mu$, $\psi_x'(0,\mu)=-i\mu$. It is readily seen that $$\begin{gathered} \varphi(x,\mu)=\psi(x,-\mu),\quad c(x,\mu)=(\varphi(x,\mu)+\psi(x,\mu))/2,\\ s(x,\mu)=(\varphi(x,\mu)-\psi(x,\mu))/(2i\mu),\\ s'(x,\mu)=(\varphi'(x,\mu)-\psi'(x,\mu))/(2i\mu). \end{gathered}\label{e2.2}$$ For convenience, we introduce $I(x)=\int_0^x q(t)dt$. Asymptotic formulas for the functions $\varphi(x,\mu)$ and $\varphi'(x,\mu)$ were established in \cite{m2}: \begin{aligned} \varphi(x,\mu)&=e^{i\mu x}[1+\frac{1}{2i\mu}I(x)-\frac{1}{8\mu^2}I^2(x)]\\ &\quad -\frac{e^{-i\mu x}}{2i\mu}\int_0^xe^{2i\mu t}q(t)dt+A(x,\mu)+R_1(x,\mu), \end{aligned}\label{e2.3} and \begin{aligned} \varphi_x'(x,\mu)&=i\mu e^{i\mu x}[1+\frac{1}{2i\mu}I(x) -(8\mu^2)^{-1}I^2(x)]\\ &\quad +\frac{e^{-i\mu x}}{2}\int_0^xe^{2i\mu t}q(t)dt+B(x,\mu) +R_3(x,\mu)+R_4(x,\mu)+R_5(x,\mu), \end{aligned}\label{e2.4} where \begin{align*} A(x,\mu)&= (4\mu^2)^{-1}(e^{-i\mu x}\int_0^xe^{2i\mu t}q(t)dt\int_0^tq(s)ds\\ &\quad +e^{i\mu x}\int_0^xe^{-2i\mu t}q(t)dt\int_0^te^{2i\mu s}q(s)ds -e^{-i\mu x}\int_0^xq(t)dt\int_0^te^{2i\mu s}q(s)ds), \end{align*} \begin{aligned} R_1(x,\mu)&= -(4\mu^3)^{-1}\int_0^x(e^{i\mu(x-t)}-e^{-i\mu(x-t)})q(t)dt\\ &\quad\times \int_0^t(e^{i\mu(t-s)}-e^{-i\mu(t-s)})q(s)ds \int_0^s\sin\mu(s-y)q(y)\varphi(y,\mu)dy, \end{aligned} \label{e2.5} \begin{align*} B(x,\mu)&= -(4i\mu)^{-1}[e^{i\mu x}\int_0^xe^{-2i\mu t}q(t)dt \int_0^te^{2i\mu s}q(s)ds\\ &\quad -e^{-i\mu x}\int_0^xe^{2i\mu t}q(t)dt\int_0^tq(s)ds +e^{-i\mu x}\int_0^xq(t)dt\int_0^te^{2i\mu s}q(s)ds], \end{align*} \begin{gather} R_2(x,\mu)= -\frac{1}{16\mu^2}e^{i\mu x}\int_0^x(1+e^{2i\mu(t-x)})q(t)I^2(t)dt,\label{e2.6} \\ R_3(x,\mu)= \frac{1}{2}\int_0^x(e^{i\mu(x-t)}+e^{-i\mu(x-t)})q(t)R_2(t,\mu)dt,\label{e2.7} \\ R_4(x,\mu)= \frac{1}{2}\int_0^x(e^{i\mu(x-t)}+e^{-i\mu(x-t)})q(t)A(t,\mu)dt.\label{e2.8} \end{gather} We need more precise asymptotic formulas for the functions $\varphi(x,\mu)$ and $\varphi'(x,\mu)$. Substituting known the expression \cite{m3} for the function $\varphi(y,\mu)= e^{i\mu y}+O(1/\mu) e^{|Im\mu|y}$ into \eqref{e2.5}, we obtain \begin{align*} R_1(x,\mu)&= -(8i\mu^3)^{-1}\int_0^x(e^{i\mu(x-t)}-e^{-i\mu(x-t)})q(t)dt\\ &\quad \times \int_0^t(e^{i\mu(t-s)}-e^{-i\mu(t-s)})q(s)ds\int_0^s(e^{i\mu(s-y)}-e^{-i\mu(s-y)})q(y)\\ &\quad \times [e^{i\mu y}+O(1/\mu) e^{|Im\mu|y}]dy\\ &=-\frac{e^{i\mu x}}{8i\mu^3}\int_0^x(1-e^{-2i\mu(x-t)})q(t)dt \int_0^t(1-e^{-2i\mu(t-s)})q(s)ds\\ &\quad \times \int_0^s(1-e^{-2i\mu(s-y)})q(y)dy+ O(1/\mu^4) e^{|Im\mu|x}\\ &=-\frac{e^{i\mu x}}{8i\mu^3}\int_0^xq(t)dt\int_0^tq(s)ds\int_0^s q(y)\\ &\quad \times [1-e^{-2i\mu(x-t)}-e^{-2i\mu(t-s)}-e^{-2i\mu(s-y)}+e^{-2i\mu(x-s)}\\ &\quad +e^{-2i\mu(x-t+s-y)}+e^{-2i\mu(t-y)}-e^{-2i\mu(x-y)}]dy +O(1/\mu^4) e^{|Im\mu|x}\\ &=- \frac{e^{i\mu x}}{8i\mu^3} \int_0^xq(t)dt\int_0^tq(s)ds\int_0^s q(y)dy -\frac{1}{8i\mu^3}K(x,\mu) +O(1/\mu^4) e^{|Im\mu|x}, \end{align*} where \begin{align*} K(x,\mu)&= e^{i\mu x}\int_0^xq(t)dt\int_0^tq(s)ds\int_0^s q(y)\\ &\quad \times [-e^{-2i\mu(x-t)}-e^{-2i\mu(t-s)}-e^{-2i\mu(s-y)}+e^{-2i\mu(x-s)}\\ &\quad +e^{-2i\mu(x-t+s-y)}+e^{-2i\mu(t-y)}-e^{-2i\mu(x-y)}]dy. \end{align*} Arguing as in \cite{m1}, we see that $$\int_0^xq(t)dt\int_0^tq(s)ds\int_0^s q(y)dy =\frac{1}{6}I^3(x).$$ The obvious inequality $0\le y\le s\le t\le x$, together with the inequality $|x-2z|\le x$ valid for $0\le z\le x$ and the Riemann lemma \cite{m3}, implies $$K(x,\mu)=o(1)e^{|Im\mu|x},\quad R_1(x,\mu)=-\frac{1}{48i\mu^3}I^3(x)+o(1/\mu^3) e^{|Im\mu|x}.\label{e2.9}$$ Continuing this line of reasoning, we reduce \eqref{e2.6} and \eqref{e2.8} to the form $$R_2(x,\mu)=-\frac{1}{48\mu^2}e^{i\mu x}I^3(x)+o(1/\mu^2) e^{|Im\mu|x},\quad R_4(x,\mu)=o(1/\mu^2) e^{|Im\mu|x}.\label{e2.10}$$ It follows from \eqref{e2.7} and \eqref{e2.9} that $$R_3(x,\mu)=O(1/\mu^3) e^{|Im\mu|x}.\label{e2.11}$$ Combining \eqref{e2.3} and \eqref{e2.9}, we obtain \begin{aligned} \varphi(x,\mu)&=e^{i\mu x}[1+\frac{1}{2i\mu}I(x)-\frac{1}{8\mu^2}I^2(x)]\\ &\quad -\frac{e^{-i\mu x}}{2i\mu}\int_0^xe^{2i\mu t}q(t)dt+A(x,\mu)- \frac{e^{i\mu x}}{48i\mu^3}I^3(x)+o(1/\mu^3) e^{|Im\mu|x}, \end{aligned}\label{e2.12} It follows from \eqref{e2.4}, \eqref{e2.10} and \eqref{e2.11} that \begin{aligned} \varphi_x'(x,\mu) &=i\mu e^{i\mu x}[1+\frac{1}{2i\mu}I(x)-(8\mu^2)^{-1}I^2(x)]\\ &\quad +\frac{e^{-i\mu x}}{2}\int_0^xe^{2i\mu t}q(t)dt+B(x,\mu) -\frac{1}{48\mu^2}e^{i\mu x}I^3(x)+o(1/\mu^2) e^{|Im\mu|x}. \end{aligned}\label{e2.13} Substituting into \eqref{e2.1} the expressions for $c(\pi,\mu)$, $s(\pi,\mu)$, $s'(\pi,\mu)$ of \eqref{e2.2}, \eqref{e2.12}, \eqref{e2.13}, respectively, we obtain \begin{align*} &\Delta(\mu)\\ &=\frac{1}{2}\Big\{ e^{i\mu\pi}[1+\frac{1}{2i\mu}I(\pi)-\frac{1}{8\mu^2}I^2(\pi)]\\ &\quad -\frac{e^{-i\mu\pi}}{2i\mu}\int_0^\pi e^{2i\mu t}q(t)dt+A(\pi,\mu)- \frac{e^{i\pi\mu}}{48i\mu^3}I^3(\pi)+o(1/\mu^3) e^{|Im\mu|\pi}\\ &\quad +e^{-i\mu\pi}[1-\frac{1}{2i\mu}I(\pi)-\frac{1}{8\mu^2}I^2(\pi)]\\ &\quad +\frac{e^{i\mu\pi}}{2i\mu}\int_0^\pi e^{-2i\mu t}q(t)dt+A(\pi,-\mu)+ \frac{e^{-i\pi\mu}}{48i\mu^3}I^3(\pi)+o(1/\mu^3) e^{|Im\mu|\pi}\Big\}\\ &\quad -\frac{1}{2i\mu}\Big\{ i\mu e^{i\mu\pi}[1+\frac{1}{2i\mu}I(\pi) -(8\mu^2)^{-1}I^2(\pi)]\\ &\quad +\frac{e^{-i\mu\pi}}{2}\int_0^\pi e^{2i\mu t}q(t)dt+B(\pi,\mu) -\frac{1}{48\mu^2}e^{i\mu\pi}I^3(\pi)+o(1/\mu^2) e^{|Im\mu|\pi}\\ &\quad -[-i\mu e^{-i\mu\pi}[1-\frac{1}{2i\mu}I(\pi) -(8\mu^2)^{-1}I^2(\pi)]\\ &\quad +\frac{e^{i\mu\pi}}{2}\int_0^\pi e^{-2i\mu t}q(t)dt+B(\pi,-\mu) -\frac{1}{48\mu^2}e^{-i\mu\pi}I^3(\pi) +o(1/\mu^2) e^{|Im\mu|\pi}]\Big\}\\ &\quad -(-1)^\theta\frac{b}{2i\mu}\Big\{ e^{i\mu\pi}[1+\frac{1}{2i\mu}I(\pi)-\frac{1}{8\mu^2}I^2(\pi)]\\ &\quad -\frac{e^{-i\mu\pi}}{2i\mu}\int_0^\pi e^{2i\mu t}q(t)dt +o(1/\mu^2) e^{|Im\mu|\pi}\\ &\quad -[e^{-i\mu\pi}[1-\frac{1}{2i\mu}I(\pi)-\frac{1}{8\mu^2}I^2(\pi)] -\frac{e^{i\mu\pi}}{2i\mu}\int_0^\pi e^{-2i\mu t}q(t)dt +o(1/\mu^2) e^{|Im\mu|\pi}]\Big\}. \end{align*} %\label{e2.14} Define $\Delta_0(\mu)=\frac{-(-1)^\theta b}{2i\mu}(e^{i\pi\mu}-e^{-i\pi\mu})$. Combining like terms in the above expression, gives \begin{aligned} \Delta(\mu)&=\Delta_0(\mu)-\frac{1}{2i\mu}[e^{-i\pi\mu}\int_0^\pi e^{2i\mu t}q(t)dt -e^{i\pi\mu}\int_0^\pi e^{-2i\mu t}q(t)dt]\\ &\quad +\frac{1}{4\mu^2}[e^{i\pi\mu}\int_0^\pi e^{-2i\mu t}q(t)dt\int_0^t q(s)\,ds +e^{-i\pi\mu}\int_0^\pi e^{2i\mu t}q(t)dt\int_0^t q(s)\,ds\\ &\quad -e^{i\pi\mu}\int_0^\pi q(t)dt\int_0^t e^{-2i\mu s}q(s)ds -e^{-i\pi\mu}\int_0^\pi q(t)dt\int_0^t e^{2i\mu s}q(s)ds]\\ &\quad -(-1)^\theta\frac{b}{2i\mu}\{\frac{e^{i\pi\mu} +e^{-i\pi\mu}}{2i\mu}I(\pi) -\frac{e^{i\pi\mu}-e^{-i\pi\mu}}{8\mu^2}I^2(\pi)\\ &\quad -\frac{1}{2i\mu}(e^{-i\pi\mu}\int_0^\pi e^{2i\mu t}q(t)dt +e^{i\pi\mu}\int_0^\pi e^{-2i\mu t}q(t)dt)\}+o(1/\mu^3) e^{|Im\mu|\pi}. \end{aligned}\label{e2.15} \section{Calculation of the regularized trace} First, consider the case $\langle q\rangle=0$. Formula \eqref{e2.15} is then noticeably simplified: $$\Delta(\mu)=\Delta_0(\mu)(1+r(\mu)),\label{e3.1}$$ where \begin{aligned} &r(\mu)\\ &=\frac{1}{\Delta_0(\mu)} \Big\{-\frac{1}{2i\mu}[e^{-i\pi\mu}\int_0^\pi e^{2i\mu t}q(t)dt -e^{i\pi\mu}\int_0^\pi e^{-2i\mu t}q(t)dt]\\ &\quad +\frac{1}{4\mu^2}\big[e^{i\pi\mu}\int_0^\pi e^{-2i\mu t}q(t)dt\int_0^t q(s)\,ds +e^{-i\pi\mu}\int_0^\pi e^{2i\mu t}q(t)dt\int_0^t q(s)\,ds\\ &\quad -e^{i\pi\mu}\int_0^\pi q(t)dt\int_0^t e^{-2i\mu s}q(s)ds -e^{-i\pi\mu}\int_0^\pi q(t)dt\int_0^t e^{2i\mu s}q(s)ds\big] \\ &\quad +(-1)^{\theta+1}\frac{b}{4\mu^2}(e^{-i\pi\mu}\int_0^\pi e^{2i\mu t}q(t)dt+e^{i\pi\mu}\int_0^\pi e^{-2i\mu t}q(t)dt) +o(1/\mu^3) e^{|Im\mu|\pi}\Big\}. \end{aligned}\label{e3.2} Integrating by parts the terms on the right-hand side of \eqref{e3.2}, we have \begin{aligned} r(\mu)&= \frac{(-1)^{\theta+1}}{4b\mu\sin\pi\mu}\{(e^{i\pi\mu} +e^{-i\pi\mu})(q(\pi)-q(0))\\ &\quad -[e^{i\pi\mu}\int_0^\pi e^{-2i\mu t}q'(t)dt +e^{-i\pi\mu}\int_0^\pi e^{2i\mu t}q'(t)dt]\}+\frac{q(\pi)+q(0)}{4\mu^2} +o(1/\mu^2). \end{aligned}\label{e3.3} Denote by $\Gamma_N$ the circle of radius $N+1/2$ centered at the origin. It is well known \cite{m3} that the eigenvalues of \eqref{e1.1}-\eqref{e1.2} form a sequence $\lambda_n=\mu_n^2$, where $\mu_n=n+o(1)$, $n=1,2,\dots$. This asymptotic relation for the eigenvalues implies that, for all sufficiently large $N$, the numbers $\mu_n$ with $n\le N$ are inside $\Gamma_N$, and the numbers $\mu_n$ with $n>N$ are outside $\Gamma_N$. It follows that $$2\sum_{n=1}^N\mu_n^2=\frac{1}{2\pi i}\oint_{\Gamma_N}\mu^2\frac{\Delta'(\mu)}{\Delta(\mu)}d\mu;$$ see \cite{m3}. Obviously, if $\mu\in\Gamma_N$, then $|\Delta_0(\mu)|\ge c_1e^{|Im\mu|\pi}/|\mu|$ $(c_1>0)$. This inequality and the Riemann lemma \cite{m3} imply that $\max_{\mu\in\Gamma_N}|r(\mu)|\to0$ as $N\to\infty$. Combining this with \eqref{e3.1} yields \begin{aligned} \frac{1}{2\pi i}\oint_{\Gamma_N}\mu^2\frac{\Delta'(\mu)}{\Delta(\mu)}d\mu &= \frac{1}{2\pi i}\oint_{\Gamma_N}\mu^2 \big(\frac{\Delta_0'(\mu)}{\Delta_0(\mu)}+\frac{r'(\mu)}{1+r((\mu)}\big)d\mu \\ &= 2\sum_{n=1}^N n^2+\frac{1}{2\pi i}\oint_{\Gamma_N}\mu^2d\ln(1+r(\mu))\\ &= 2\sum_{n=1}^N n^2-\frac{1}{2\pi i}\oint_{\Gamma_N}2\mu\ln(1+r(\mu))d\mu. \end{aligned}\label{e3.4} Expanding $\ln(1+r(\mu))$ by the Maclaurin formula and applying \eqref{e3.3} and the Riemann lemma \cite{m3}, we find that $$\ln(1+r(\mu))= r(\mu)-\frac{(q(\pi)-q(0))^2}{2b^2\mu^2}\cot^2\pi\mu +\frac{o(1)}{\mu^2}\label{e3.5}$$ on $\Gamma_N$. Evidently, $$\lim_{|Im\mu|\to\infty}(\cot^2\pi\mu+1)=0.\label{e3.6}$$ It follows from \eqref{e3.5} and \eqref{e3.6} that \begin{align*} &\frac{1}{2\pi i}\oint_{\Gamma_N}2\mu\ln(1+r(\mu))d\mu\\ &=\frac{1}{2\pi i}\oint_{\Gamma_N}\big[ \frac{(-1)^{\theta+1}}{2b\sin\pi\mu}\{(e^{i\pi\mu}+e^{-i\pi\mu})(q(\pi)-q(0)) \\ &\quad -(e^{i\pi\mu}\int_0^\pi e^{-2i\mu t}q'(t)dt +e^{-i\pi\mu}\int_0^\pi e^{2i\mu t}q'(t)dt)\}\\ &\quad -\frac{(q(\pi)-q(0))^2}{b^2\mu}cot^2\pi\mu+\frac{q(\pi)+q(0)}{2\mu}+\frac{o(1)}{\mu}\big] d\mu\\ &=\sum_{n=-N}^N\Big[(-1)^{\theta+1}\big[(e^{i\pi n}+e^{-i\pi n})(q(\pi)-q(0)) -(e^{i\pi n}\int_0^\pi e^{-2int}q'(t)dt \\ &\quad +e^{-int}\int_0^\pi \int e^{2int}q'(t)dt)\big]\Big] \big/(2b\pi\cos\pi n) \\ &\quad +\frac{1}{2\pi i}\oint_{\Gamma_N}[\frac{(q(\pi)-q(0))^2}{b^2\mu} +\frac{q(\pi)+q(0)}{2\mu}]d\mu+o(1)\\ &=\frac{2(-1)^{\theta+1}}{\pi b}\sum_{n=1}^N(q(\pi)-q(0)\\ &\quad -\int_0^\pi q'(t)\cos(2nt)dt) +\frac{(q(\pi)-q(0))^2}{b^2}+\frac{q(\pi)+q(0)}{2}+o(1). \end{align*} Combining this and \eqref{e3.4}, we obtain \begin{aligned} 2\sum_{n=1}^N\mu_n^2 &=2\sum_{n=1}^N n^2 +\frac{2(-1)^{\theta}}{\pi b} \sum_{n=1}^N(q(\pi)-q(0)-\int_0^\pi q'(t)\cos(2nt)dt)\\ &\quad -\frac{(q(\pi)-q(0))^2}{b^2}-\frac{q(\pi)+q(0)}{2}+o(1). \end{aligned}\label{e3.7} Passing to the limit as $N\to\infty$ in \eqref{e3.7}, we have \begin{aligned} &\sum_{n=1}^\infty\big\{\lambda_n-n^2 -\frac{(-1)^{\theta}(q(\pi)-q(0)-\int_0^\pi q'(t)\cos(2nt)dt)}{\pi b}\big\}\\ &\quad +\frac{(q(\pi)-q(0))^2}{2b^2}+\frac{q(\pi)+q(0)}{4}=0. \end{aligned} \label{e3.8} Now consider the case $\langle q\rangle\ne0$. Let $\tilde q(x)=q(x)-\langle q\rangle$. Then $\langle\tilde q\rangle=0$. Suppose that \eqref{e1.1}-\eqref{e1.2} with potential $\tilde q$ has eigenvalues $\tilde\lambda_n$. Then $\tilde \lambda_n=\lambda_n-\langle q\rangle$. According to \eqref{e3.8}, we have \begin{align*} &\sum_{n=1}^\infty\big\{\tilde\lambda_n-n^2 -\frac{(-1)^{\theta}(\tilde q(\pi)-\tilde q(0) -\int_0^\pi \tilde q'(t)\cos(2nt)dt)}{\pi b}\big\}\\ &\quad +\frac{(\tilde q(\pi)-\tilde q(0))^2}{2b^2}+\frac{\tilde{q}(\pi)+\tilde {q}(0)}{4}=0. \end{align*} Substituting the expressions for $\tilde\lambda_n$ and $\tilde q(x)$ into this equality, we obtain formula \eqref{e1.3}. \subsection*{Acknowledgments} This work was supported by grant 07-01-00158 from the Russian Foundation for Basic Research. \begin{thebibliography}{0} \bibitem{d1} L. A. Dikii. On a formula of Gel'fand-Levitan. Uspekhi Math. Nauk, 8, (1953), 119-123 (in Russian). \bibitem{g1} I. M. Gel'fand and B. M. Levitan. 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