\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2016 (2016), No. 283, pp. 1--12.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2016 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2016/283\hfil Eigenvalues of a nonlocal problem] {Structure and asymptotic expansion of eigenvalues of an integral-type nonlocal problem} \author[Z. Zhou, F. Liao \hfil EJDE-2016/283\hfilneg] {Zhong-Cheng Zhou, Fang-Fang Liao} \address{Zhong-Cheng Zhou \newline School of Mathematics and Statistics, Southwest University, Chongqing 400715, China} \email{zhouzc@amss.ac.cn} \address{Fang-Fang Liao\newline Department of Mathematics, Shanghai Normal University, Shanghai 200234, China.\newline Department of Mathematics, Southeast University, Nanjing 210096, China} \email{liaofangfang8178@sina.com} \thanks{Submitted June 24, 2016. Published October 24, 2016.} \subjclass[2010]{34C25, 34D20} \keywords{Eigenvalues; asymptotic expansion; nonlocal problem; Riesz basis} \begin{abstract} We study the structure of eigenvalues of second-order differential equations with nonlocal integral boundary conditions. Moreover, we consider the asymptotic expansion of the eigenvalues and the corresponding eigenfunctions, which shows that the eigenfunctions form a Riesz basis for $L^2([0,1],\mathbb{R})$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{definition}[theorem]{Definition} \allowdisplaybreaks \section{Introduction} In recent years, many researchers studied different kinds of nonlocal boundary-value problems of ordinary differential equations, and in particular focused on the existence and multiplicity of nontrivial solutions for nonlinear nonlocal problems, see for example, \cite{g, gn, kim, ma, mo, bpr07} for multi-point boundary-value problems and \cite{cz, fiz, eh, kt, wi} for general nonlocal boundary-value problems. However, the study on the eigenvalue theory of the corresponding nonlocal linear problems appears to just start. Ma and O'Regan \cite{mo} constructed all real eigenvalues of the problem \begin{equation} \label{ma-o} \begin{gathered} -y''(x)=\lambda y(x),\quad x\in(0,1),\\ y(0)=0, \quad y(1)=\sum_{k=1}^{m}\alpha_ky(\eta_k), \end{gathered} \end{equation} where $m\in\mathbb{N}$, $\alpha=(\alpha_1,\cdots,\alpha_m)\in\mathbb{R}_+^m$ satisfying the nondegeneracy condition $\sum_{k=1}^{m}|\alpha_k|<1$ and $\eta=(\eta_1,\cdots,\eta_m)\in\Delta^m :=\{(\eta_1,\cdots,\eta_m)\in\mathbb{R}^m:0<\eta_1<\cdots<\eta_m<1\}$ are taken as rational. We note that the eigenvalues of \eqref{ma-o} can be analyzed using elementary method because all solutions of \eqref{ma-o} can be found explicitly. However, even for \eqref{ma-o}, as far as we know, the first complete eigenvalue theory was proved in \cite{gsz}. In particular, Gao, Sun and Zhang completely characterized the structure of eigenvalues of \eqref{ma-o} for all $\alpha\in\mathbb{R}_+^m$ and $\eta\in\Delta^m$. Moreover, they gave the complete structure of eigenvalues of general multi-point boundary-value problem \begin{equation} \label{gsz-e} \begin{gathered} -y''(x)+q(x)y(x)=\lambda y(x), \quad x\in(0,1),\\ y(0)=0,\quad y(1)=\sum_{k=1}^{m}\alpha_ky(\eta_k), \end{gathered} \end{equation} where $q\in L^1([0,1],\mathbb{R})$ and $\alpha\in\mathbb{R}_+^m,\eta\in\Delta^m$. Note that problems \eqref{ma-o} and \eqref{gsz-e} are non-symmetry problems. It was proved in \cite{gsz} that \eqref{gsz-e} may admit complex eigenvalues and has always a sequence of real eigenvalues tending to infinity. We will extend the above results to the general nonlocal integral boundary-value problem \begin{equation} \label{os} \begin{gathered} -y''(x)+q(x)y(x)=\lambda y(x),\quad x\in(0,1),\\ y(0)=0,\quad y(1)= \int_0^{1}k(x)y(x)dx, \end{gathered} \end{equation} where $q\in L^1([0,1],\mathbb{R})$ and $k\in C^2([0,1],\mathbb{R})$. We will show that the eigenvalues of the problem \eqref{os} have the similar structure to those of \eqref{gsz-e}. In fact, problem \eqref{os} can be considered as a version of \eqref{gsz-e} with continuous boundary condition of \eqref{gsz-e}. The set of all eigenvalues of \eqref{os} is denoted by $\Sigma_{k}^{q}\in \mathcal{C}$, which is called the spectrum of operator $\mathcal{A}$, where the linear operator $\mathcal{A}: D(\mathcal{A})(\subset L^2([0,1],\mathbb{R}))\to L^2([0,1],\mathbb{R})$ is defined by $$ \mathcal{A}(y)=-y''(x)+q(x)y(x) $$ with $$ D(\mathcal{A})=\big\{y\in H^2(0,1): y(0)=0,\; y(1)= \int_0^{1}k(x)y(x)dx\big\}. $$ When $q\equiv0$, Equation \eqref{os} becomes \begin{equation} \label{be} \begin{gathered} -y''(x)=\lambda y(x),\quad x\in(0,1),\\ y(0)=0,\quad y(1)= \int_0^{1}k(x)y(x)dx. \end{gathered} \end{equation} We can define a linear operator $\mathcal{A}_0: D(\mathcal{A}_0)(\subset L^2([0,1],\mathbb{R}))\to L^2([0,1],\mathbb{R})$ by \begin{align}\label{A0} \mathcal{A}_0(y)=-y''(x), \end{align} with $$ D(\mathcal{A}_0)=\big\{y\in H^2([0,1],\mathbb{R}): y(0)=0,\; y(1)= \int_0^{1}k(x)y(x)dx\big\} $$ and a bounded perturbation linear operator \begin{align}\label{B} (\mathcal{B}_0y)(x)=q(x)y(x), \end{align} on $L^1([0,1],\mathbb{R})$. The eigenvalues of \eqref{be} are exact the eigenvalues of the operator $\mathcal{A}_0$, which can be analyzed using elementary method. However, as far as we know, even for this simple case, the spectrum theory is incomplete in the literature. For some special functions $q$ and $k$, we can adopt the backstepping method (which comes from Krstic) to obtain the existence and explicit expression of eigenvalues via transferring it into well-known eigenvalue problem, see \cite{ks} and related references. Such similar method can be used to prove Theorem \ref{main theorem} for some special functions $q$ and $k$. However, for general function pair $q$ and $k$, this method does not work. One motivation of this paper is to develop a method for general functions $q$ and $k$. The results of \eqref{os} can be used to study the existence of nonlinear differential equations with integral boundary condition. Besides, the stabilization controller design of heat equation by backstepping method strongly depends on the complete spectrum analysis for the problem \eqref{os}, which is another important motivation of this paper. Basically, eigenvalues of \eqref{os} are zeros of some entire functions. To study the distributions of eigenvalues, we will consider \eqref{os} as a perturbation of \eqref{be}. To obtain the existence of infinitely many real eigenvalues as in Theorem \ref{submain}, some properties of almost periodic functions \cite{mo,pt} will be used. To pass the results of \eqref{be} to those with general potentials $q$, many techniques will be exploited. Moreover, some basic estimates for fundamental solutions of \eqref{be} play an important role. This article is organized as follows. In Section 2, we will give some detailed analysis on problem \eqref{be}. In Section 3, after developing some basic estimates, we will prove Theorems \ref{main theorem} and \ref{submain}. In Section 4, we will give the asymptotic expansion for the eigenvalues and eigenfunctions of \eqref{be}. In Section 5, we will prove the existence of eigenvalues for \eqref{os} and corresponding eigenfunctions forming Riesz basis for $L^2([0,1],\mathbb{R})$. \section{Structure of eigenvalues of the zero potential} In this section, we first consider the spectrum for \eqref{be}, which has the zero potential. Let us use $\Sigma_{k}^{0}$ to denote the set of all eigenvalues of \eqref{be}. Let $\lambda\in \mathbb{C}$, the complex solutions of \eqref{be} satisfying $y(0)=0$ are $y(x)=cS_{\lambda}(x)$, where $c\in\mathbb{C}$ and \[ S_{\lambda}(x):=\frac{\sin\sqrt{\lambda}x}{\sqrt{\lambda}} =\sum_{k=0}^{+\infty}\frac{(-1)^k}{(2k+1)!}\lambda^{k}x^{2k+1},~x\in[0,1]. \] Notice that $S_{\lambda}(x)$ is an entire function of $\lambda\in \mathbb{C}$. Define $C_\lambda(x):=\cos(\sqrt{\lambda}x)$ and \begin{align} M_0(\lambda):=S_{\lambda}(1)-\int_0^{1}k(x)S_{\lambda}(x)dx =\frac{\sin\sqrt{\lambda}}{\sqrt{\lambda}}-\int_0^{1}k(x) \frac{\sin\sqrt{\lambda}x}{\sqrt{\lambda}}dx. \end{align} Then $\lambda\in \Sigma_{k}^{0}$ if and only if $\lambda$ satisfies $$ M_0(\lambda)=0 . $$ We recall some properties of almost periodic functions which will be used later. We refer the readers to \cite{F} for more information on almost periodic functions. Suppose that $f:\mathbb{R}\to \mathbb{R}$ is a bounded continuous function. We say that $f$ is almost periodic if for any $\varepsilon>0$, there exists $l_{\varepsilon}>0$ such that for any $a\in \mathbb{R}$, there exists $b\in[a, a+l_{\varepsilon}]$ such that $\|f(\cdot+b)-f(\cdot)\|_{L^{\infty}}<\varepsilon$. If $f:\mathbb{R}\to \mathbb{R}$ is an almost periodic function, then for any $A\in \mathbb{R}$, we have \[ \inf_{u\in[A,\infty)}f(u)=\inf_{u\in\mathbb{R}}f(u),\quad \sup_{u\in[A,\infty)}f(u)=\sup_{u\in\mathbb{R}}f(u). \] Moreover, if $f$ is non-zero and $\bar{f}=\lim_{T\to +\infty}\frac{1}{T}\int_0^{T}f(u)du=0$, then $f$ is oscillatory as $u\to +\infty$. In particular, $f(u)$ has a sequence of positive zeros tending to $+\infty$. \begin{lemma} \label{real eigenvaluea} If $k\in C^2([0,1],\mathbb{R})$, then $\Sigma_{k}^{0}\cap \mathbb{R} =\{\overline{\lambda}_n\}$ which satisfies $$ \overline{\lambda}_1 \leq \overline{\lambda}_2 \leq\cdots \overline{\lambda}_n \leq \cdots,\quad \lim_{n\to +\infty} \overline{\lambda}_n=+\infty. $$ \end{lemma} \begin{proof} Let us first consider possible positive eigenvalues $\lambda=\alpha^2$ of \eqref{be}, where $\alpha>0$. By equation \eqref{be}, we have \[ F(\alpha):=\sin \alpha-\int_0^{1}k(x)\sin (\alpha x) dx=0 \] It is easy to check the function $F(\alpha)$ is a non-zero, almost periodic function and has mean value zero. Therefore, $F(\alpha)$ has many positive zeros tending to $+\infty$, and hence $\Sigma_{k}^{0} $ contains a sequence of positive eigenvalues tending to $+\infty$. Next, we consider possible negative eigenvalues $\lambda=-\alpha^2$ of \eqref{be}, where $\alpha>0$. By the first equality of \eqref{identity} and \eqref{be}, we have \begin{align} \bar{F}(\alpha):=\sinh \alpha-\int_0^{1}k(x)\sinh (\alpha x)dx=0. \end{align} One has $$ \lim_{\alpha\to +\infty} \frac{\bar{F}(\alpha)}{\sinh \alpha}=1, $$ notice that $\bar{F}(\alpha)$ is analytic in $\alpha$, thus $\bar{F}(\alpha)=0$ has at most finitely many positive solutions. Hence $\Sigma_{k}^{0} $ contains at most finitely many negative eigenvalues. Because both $F(\alpha)=0$ and $\bar{F}(\alpha)=0$ have only isolated solutions, the above two cases show that the result holds. \end{proof} Next we show that $\Sigma_{k}^{0}$ contains only real eigenvalues if $ \int_0^1k^2(x)dx\leq 1$. \begin{lemma} \label{lem2.2} Assume $k\in C^2([0,1],\mathbb{R})$ satisfying $ \int_0^1k^2(x)dx\leq 1$. Then $\Sigma_{k}^{0}$ contains only real eigenvalues. Moreover, $ \Sigma_{k}^{0}\subset ( \frac{\pi^2}{4}, +\infty)$. \end{lemma} \begin{proof} Suppose that $\lambda=w^2\in \Sigma_{k}^{0} $, where $w=u+iv$, $u,v \in \mathbb{R}$. We would assert that $v=0$ under the assumption. Otherwise, assume that $v\neq 0$. We have \[ \sin w-\int_0^{1}k(x)\sin (wx) dx=0. \] Note that the following elementary equalities hold for any $u,v\in\mathbb{R}$, \begin{equation} \label{identity} \sin(u+iv)=\sin u\cosh v+i \cos u\sinh v,\quad |\sin(u+iv)|^2=\sin^2 u+ \sinh^2 v. \end{equation} Then \begin{gather*} \sin u\cosh v=\int_0^{1}k(x)\sin (ux)\cosh(vx) dx, \\ \cos u\sinh v=\int_0^{1}k(x)\cos (ux)\sinh(vx) dx. \end{gather*} It follows from H\"{o}lder inequality that \begin{align} 1&=\sin^2 u+\cos^2 u\\ &=\Big(\int_0^{1}k(x)\sin(ux)\frac{\cosh (vx)}{\cosh v} dx\Big)^2 +\Big(\int_0^{1}k(x)\cos(ux)\frac{\sinh (vx)}{\sinh v} dx\Big)^2\\ &< \int_0^{1}k^2(x)dx\int_0^{1}\cos^2 (ux) dx +\int_0^{1}k^2(x)dx\int_0^{1}\sin^2 (ux) dx\\ &=\int_0^{1}k^2(x)dx, \end{align} which is a contradiction. Thus $v=0$. On the other hand, \begin{align} M_0(0)=1-\int_0^{1}xk(x)dx \geq 1-\Big(\int_0^{1}k^2(x)dx\Big)^{1/2}>0, \end{align} hence we have $\Sigma_{k}^{0}\in (0,+\infty)$. Finally, for any $u\in (0, \frac{\pi}{2}]$, by the H\"{o}lder inequality, we know function $F(u)$ satisfies \begin{align*} F(u)&=\sin u-\int_0^{1}k(x)\sin(ux) dx\\ &\geq \sin u-\int_0^{1}|k(x)|\sin(ux) dx\\ &> \sin u-\int_0^{1}|k(x)| dx\sin u \geq 0. \end{align*} Therefore, we obtain that $\Sigma_{k}^{0}\in (\frac{\pi^2}{4},+\infty)$. \end{proof} \section{Structure of eigenvalues of non-zero potentials} Given $q\in L^1((0,1),\mathbb{R})$ and complex parameter $\lambda\in \mathbb{C}$, the fundamental solutions of \eqref{os} are denoted by $y_{m}(x, \lambda, q), m=1,2$, which are solutions satisfying the initial values \begin{align} y_1(0, \lambda, q)=y'_2(0, \lambda, q)=1,\quad y'_1(0, \lambda, q)=y_2(0, \lambda, q)=0. \end{align} Notice that $y_{m}(x, \lambda, q)$ are entire functions of $\lambda\in \mathbb{C}$, To study \eqref{os}, we introduce \begin{align} M_q(\lambda):=y_2(1, \lambda, q)-\int_0^{1}k(x) y_2(x, \lambda, q)dx , \quad \lambda\in \mathbb{C}. \end{align} We use $\Sigma_{k}^{q}$ to denote the set of all eigenvalues of \eqref{os}. Then $\lambda\in \Sigma_{k}^{q}$ if and only if $M_q(\lambda)=0$. We will need the following basic estimates, whose proofs are much similar to those of \cite[Lemma 3.1, Lemma 3.2, Lemma 3.3, Lemma 3.4]{gsz}. Here we only state them without their proofs. \begin{lemma} If $\beta\in (0,1)$, one has \begin{equation} \label{basic estimate} \begin{gathered} \lim_{v\in \mathbb{R},|v|\to +\infty} \frac{|\sin(u+iv)|}{\exp |v|}=\frac{1}{2}, \\ \lim_{v\in \mathbb{R},|v|\to +\infty} \frac{|\sin\beta(u+iv)|}{\exp |v|}=0 \end{gathered} \end{equation} uniformly in $u\in\mathbb{R}$. \end{lemma} \begin{lemma}\label{almost function solution} There exists a constant $c(k)>0$ and a sequence $a_{n}$ of increasing positive numbers such that $a_n\to +\infty$ and $(-1)^n F(a_n)>c(k)$, where $F(u):=\sin u-\int_0^{1}k(x)\sin (ux)dx$. \end{lemma} \begin{lemma} Given $q\in L^1((0,1),\mathbb{R})$ and complex parameter $\lambda\in \mathbb{C}$. Then the following inequalities hold for all $x\in[0,1]$, \begin{gather*} |y_1(x, \lambda, q)-C_{\lambda}(x)| \leq \frac{1}{|\sqrt{\lambda}|}\exp (|\operatorname{Im}\sqrt{\lambda} |x+ \|q\|_{L^1[0,x]}).\\ |y_2(x, \lambda, q)-S_{\lambda}(x)| \leq \frac{1}{|\lambda|}\exp (|\operatorname{Im}\sqrt{\lambda} |x+ \|q\|_{L^1[0,x]}).\\ |y'_1(x, \lambda, q)-C'_{\lambda}(x)| \leq \|q\|\exp (|\operatorname{Im}\sqrt{\lambda}|x+ \|q\|_{L^1[0,x]}).\\ |y'_2(x, \lambda, q)-S'_{\lambda}(x)| \leq \frac{\|q\|}{|\sqrt{\lambda}|}\exp (|\operatorname{Im}\sqrt{\lambda} |x+ \|q\|_{L^1[0,x]}). \end{gather*} \end{lemma} \begin{lemma}\label{estimate of eigenvalue} The following estimate holds for $M_q(\lambda)$, \begin{align} |M_q(\lambda)-M_0(\lambda)|\leq \frac{B}{|w|^2}\exp (|\operatorname{Im} w|), \quad w:=\sqrt{\lambda}\in\mathbb{C}. \end{align} where $$ B=\exp (\|q\|_{L^1[0,1]})+\exp (\|q\|_{L^1[0,1]})\int_0^{1}|k(x)|dx. $$ \end{lemma} \begin{lemma}\label{non eigenvalue} One has $M_q(\lambda)\neq 0$ on $\mathbb{R}$. Consequently, there exists $\lambda_0\in \mathbb{R}$ such that $\lambda_0$ does not belong to $\Sigma_{k}^{q}$. \end{lemma} \begin{proof} Otherwise, we have $M_q(\lambda)\equiv 0$, Notice that \begin{equation} \label{M0} M_0(u^2)\equiv \frac{F(u)}{u}, \quad u>0. \end{equation} Let $\lambda=a_n^2$ in Lemma \ref{almost function solution}, we have $$ \Big|\frac{F(a_n)}{a_n}\Big|= |M_0(a_n^2)|\leq \frac{B}{a_n^2}. $$ Hence, $\lim_{n\to +\infty} |F(a_n)|\leq\lim_{n\to +\infty}\frac{B}{ a_n} =0$, which contradicts Lemma \ref{almost function solution}. \end{proof} \begin{theorem}\label{main theorem} If $q\in L^{1}([0,1],\mathbb{R})$ and $k\in C^2([0,1],\mathbb{R})$, then $\Sigma_{k}^{q}$ is composed of a sequence $\lambda_n=\{\lambda_{n}(q)\}\in\mathbb{C}$ which satisfies $$ \operatorname{Re}\lambda_1 \leq \operatorname{Re}\lambda_2 \leq\cdots \operatorname{Re}\lambda_n \leq \cdots,~~~~ \lim_{n\to +\infty} \operatorname{Re}\lambda_n=+\infty.$$ \end{theorem} \begin{proof} By Lemma \ref{non eigenvalue}, there exists $\lambda_0\in\mathbb{R}$ such that $\lambda_0 \not\in \Sigma_{k}^{q}$, which implies that the problem \begin{equation}\label{yqy} \begin{gathered} -y''(x)+q(x)y(x)-\lambda_0 y(x)=0,\quad x\in(0,1),\\ y(0)=0,\quad y(1)= \int_0^{1}k(x)y(x)dx \end{gathered} \end{equation} has only the trivial solution $y=0$. Let $G_0(x,u)$ be the Green function of \eqref{yqy}. Then $\lambda\in\Sigma_{k}^{q}$ if and only if $\lambda\neq\lambda_0$ and \begin{equation} \begin{gathered} -y''(x)+(q(x)-\lambda_0 )y(x)=(\lambda-\lambda_0)y(x),\\ y(0)=0,\quad y(1)= \int_0^{1}k(x)y(x)\,dx, \end{gathered} \end{equation} has a nontrivial solution $y$. In other words, $\lambda\in\Sigma_{k}^{q}$ if and only if the equation \[ y=(\lambda-\lambda_0)L_q y \] has a non-trivial solution $y$, where $$ L_q y(x):=\int_0^1G_0(x,z)(q(z)-\lambda_0 )y(z)dz. $$ Since $L_q$ is a compact linear operator, one sees that this happens when and only when \[ \frac{1}{\lambda-\lambda_0}\in \sigma(L_q)\subset \mathbb{C}, \] where $\sigma(L_q)$ is the spectrum of $L_q$. Hence $\Sigma_{k}^{q}$ consists of a sequence of eigenvalues which can accumulate only at infinity. For $\lambda\in \mathbb{C}$, denote $$ \lambda=w^2, \quad w=\sqrt{\lambda}=u+iv, \quad u,v\in \mathbb{R}. $$ Suppose that $\lambda\in\Sigma_{k}^{q}$ and $\lambda\neq 0$. Then $M_q(\lambda)=0$ and Lemma \ref{estimate of eigenvalue} imply that \begin{align*} \frac{B}{|w|^2}\exp (|v|) &\geq |M_q(\lambda)-M_0(\lambda)|=|M_0(\lambda)|\\ &=\Big|\frac{\sin w-\int_0^{1}k(x)\sin(wx) dx}{w}\Big|\\ &\geq\frac{|\sin (u+iv)|-\int_0^{1}|k(x)\sin (u+iv)x| dx}{|w|}. \end{align*} We conclude that all non-zero eigenvalues $\lambda\in\Sigma_{k}^{q}$ satisfy \begin{align}\label{inequality} |w|\frac{|\sin (u+iv)|-\int_0^{1}|k(x)\sin (u+iv)x| dx}{\exp |v|}\leq B. \end{align} Let us derive some sequences from estimate \eqref{inequality} for $\lambda\in\Sigma_{k}^{q}$. \smallskip \noindent\textbf{Case 1:} Since $|w|\geq |v|$, it follows from the uniform limits in \eqref{basic estimate} that \begin{align} \lim_{|v|=|\operatorname{Im} w|\to +\infty}|w| \frac{|\sin (u+iv)|-\int_0^{1}|k(x)\sin (u+iv)x | dx}{\exp |v|}=+\infty. \end{align} Thus, there exists a constant $h>0$ such that \begin{align} \lambda\in\Sigma_{k}^{q} \Longrightarrow w=\sqrt{\lambda}\in H_{h}:=\{w\in \mathbb{C} :|\operatorname{Im} w|r\}, \] where $r:=-h^2$. \smallskip \noindent\textbf{Case 2:} Let $\bar{r}>-h^2$, next we assert that \[ \Sigma_{k}^{q}\cap\{\lambda\in \mathbb{C} :\operatorname{Re} \lambda \leq\bar{r}\}=\Sigma_{k}^{q} \cap\{\lambda\in \mathbb{C} :-h^20, \forall n\in\mathbb{N}$. Hence $M_{q}(\lambda)=0 $ has at least one positive solution $\bar{\lambda}_n$ in each interval $(a^2_n,a^2_{n+1}), n\in \mathbb{N}$. Combining with Theorem \ref{main theorem}, we have $\Sigma_{k}^{q}\cap \mathbb{R}$ consists of a sequence of real eigenvalues tending to $+\infty$, hence $\Sigma_{k}^{q}\cap \mathbb{R}$ can be listed as in Theorem \ref{submain}. \end{proof} \section{Asymptotic expansion and Riesz basis} Above we have discussed the structure of eigenvalues of \eqref{os}. In this section, we give the quantity asymptotic estimate for eigenvalues and eigenfunctions of \eqref{be} and \eqref{os}. Moreover, we will show the eigenfunctions forms Resis basis of $L^2([0,1],\mathbb{R})$. We first make some preparations for the main theory. \begin{definition} \label{def4.1} \rm A sequence $\{e_n\}_1^{\infty}\subset L^2([0,1],\mathbb{R})$ is called a basis in $L^2([0,1],\mathbb{R})$ if for any $g\in L^2([0,1],\mathbb{R})$ there exists a unique sequence $\{a_n\}_1^{\infty}$ of real numbers such that $g=\sum_{n=1}^\infty a_n e_n$ in $L^2([0,1],\mathbb{R})$. A basis $\{e_n\}_1^{\infty}$ in $L^2([0,1],\mathbb{R})$ is called a Riesz basis when the series $\sum_{n=1}^\infty a_n e_n$, with real coefficients $a_n$, converges in $L^2([0,1],\mathbb{R})$ if and only if $\sum_{n=1}^\infty a_n^2<\infty $. \end{definition} The following Theorem is very useful in checking the Riesz basis for the generalized eigenfunctions of $\mathcal{A}_0$. \begin{theorem}[\cite{guo02, guo01}] \label{generaliezd basis} Let $\mathcal{T}$ be a densely defined discrete operator, that is $(\lambda I-\mathcal{T})^{-1}$ is compact for some $\lambda$ in a Hilbert space $H$ with $\{z_n\}_1^{+\infty}$ being a Riesz basis for $H$. If there are an $N>0$ and a sequence of generalized eigenvector $\{x_n\}_{N+1}^{+\infty}$ of $\mathcal{T}$ such that $$ \sum_{n=N+1}^{\infty}\|x_{n}-z_{n}\|^2<\infty, $$ then \begin{itemize} \item[(i)] There are an $M>N$ and generalized eigenvectors $\{x_{n0}\}_1^{M}$ of $\mathcal{T}$ such that $\{x_{n0}\}_1^{M}\cup \{x_n\}_{N+1}^{+\infty} $ forms a Riesz basis for $H$. \item[(ii)] Let $\{x_{n0}\}_1^{M}\cup \{x_n\}_{N+1}^{+\infty} $ be eigenvalues $\{\sigma_{n}\}_1^{+\infty}$ of $\mathcal{T}$. Then $\sigma(\mathcal{T})=\{\sigma_{n}\}_1^{+\infty}$, in which $\sigma_{n}$ is counted according to its algebraic multiplicity. \item[(iii)] If there is an $C_0>0$ such that $\sigma_{n}\neq\sigma_{m}$ for all $n,m>C_0$, then there is an $N_0>C_0$ such that all $\sigma_{n}, n>N_0$ are algebraically simple. \end{itemize} \end{theorem} \begin{lemma}\label{eigenvalue asymptotic} The eigenvalues of \eqref{be} have the asymptotic expansion $$ \lambda_n=n^2\pi^2 +2(k(0)-k(1))+O(\frac{1}{n}). $$ \end{lemma} \begin{proof} According to Theorems \ref{main theorem} and \ref{submain}, we know that \eqref{be} has a sequence of eigenvalues. In fact, the eigenvalues $\lambda_{n}$ of \eqref{be} satisfy \begin{align}\label{equality} \sin\sqrt{\lambda_{n}}&=\int_0^{1}k(x)\sin\sqrt{\lambda_{n}}x\,dx\\ &= -\frac{1}{\sqrt{\lambda_{n}}}\Big(k(1)\cos\sqrt{\lambda_{n}}-k(0) -\int_0^{1}k'(x)\cos\sqrt{\lambda_{n}}xdx\Big). \end{align} Therefore, \begin{equation} \label{sqrexpres} \sqrt{\lambda_{n}}=n\pi+O(\frac{1}{\sqrt{\lambda_{n}}} ). \end{equation} At the same time, we know that \begin{gather*} \sin\sqrt{\lambda_{n}}=O(\frac{1}{\sqrt{\lambda_{n}}} ), \\ \cos\sqrt{\lambda_{n}}=1-O(\frac{1}{\sqrt{\lambda_{n}}} ). \end{gather*} Taking them into \eqref{equality}, we have \begin{align} O(\frac{1}{\sqrt{\lambda_{n}}}) = \frac{k(0)}{\sqrt{\lambda_{n}}}-\frac{k(1)}{\sqrt{\lambda_{n}}} +O(\frac{1}{\lambda_{n}}). \end{align} Hence, by \eqref{sqrexpres}, we have \begin{align} \sqrt{\lambda_{n}}-n\pi= \frac{k(0)-k(1)}{\sqrt{\lambda_{n}}} +O(\frac{1}{\lambda_{n}}), \end{align} we can obtain \begin{align} \lambda_n=n^2\pi^2 +2(k(0)-k(1))+O(\frac{1}{n}), \end{align} which completes the proof. \end{proof} \begin{lemma}\label{eigenfunction asymptotic} Let $\{\lambda_n\}_1^{\infty}$ be the eigenvalues of operator $\mathcal{A}_0$. Then the corresponding eigenfunctions $\{y_n\}_1^{\infty}$ have the asymptotic expressions $$ y_n(x)=\sin n\pi x+O(\frac{1}{n}). $$ Moreover, the generalized eigenfunctions of $\mathcal{A}_0$ forms a Riesz basis of $L^2([0,1],\mathbb{R})$. \end{lemma} \begin{proof} According to \eqref{be} and Lemma \ref{eigenvalue asymptotic} for $\lambda_n$, its corresponding eigenfunction has the asymptotic form \begin{align} y_n(x)=\sin \sqrt{\lambda_n}x=\sin n\pi x+O(\frac{1}{n}). \end{align} Next, we show that $\sum_{n=1}^{\infty}\int_0^{1}|\sin(\sqrt{\lambda_n}x)-\sin(n\pi x)|^2 dx<+\infty $. In fact, $$ \int_0^{1}|\sin(\sqrt{\lambda_n}x)-\sin(n\pi x)|^2 dx \leq C \cdot O(\frac{1}{n^2}) $$ by the eigenvalue expansion, where $C$ is a constant number large enough. Therefore, $$ \sum_{n=1}^{+\infty}\int_0^{1}|\sin(\sqrt{\lambda_n}x)-\sin(n\pi x)|^2 dx <+\infty . $$ By Theorem \ref{generaliezd basis}, we know that the generalized eigenfunctions of $\mathcal{A}_0$ forms a Riesz basis of $L^2([0,1],\mathbb{R})$, which completes the proof. \end{proof} For obtaining the asymptotic expansion for the eigenvalue of $\Sigma_k^q$, we show the relationship between $\Sigma_k^q$ and $\Sigma_k^0$. Intuitively, for the problem \eqref{be} and \eqref{os}, if $q$ is a constant, we know $\Sigma_k^q$ is a constant translation of $\Sigma_k^0$, In fact, if $q$ is not a constant, we also know the asymptotic expansion for $\Sigma_k^q$ in terms of $\Sigma_k^0$, which is borrowed from the paper \cite{guo01}. \begin{definition}\label{d calss} \rm A linear operator $\mathcal{A}_0$ in a Hilbert space $H$ is called discrete-type(or $[D]$-class for short), if there are Riesz basis $\{\phi_{n}\}_1^{\infty} $ of $H$, complex series $\{\lambda_n\}_1^{\infty} $ and an integer $N>0$ such that \begin{itemize} \item[(i)] $\lim_{n\to +\infty}| \lambda_n|=\infty$, $\lambda_n \neq \lambda_m$ as $n,m>N$. \item[(ii)] $\mathcal{A}_0\phi_{n}= \lambda_n \phi_{n}$, $n>N$. \item[(iii)] $\mathcal{A}_0[\phi_1,\phi_2, \cdots,\phi_N] \subset[\phi_1,\phi_2, \cdots,\phi_N]$ and $\mathcal{A}_0$ has spectrum $\{\lambda_n\}_1^N$ in \\ $[\phi_1,\phi_2, \cdots,\phi_N]$, where $[\phi_1,\phi_2, \cdots,\phi_N]$ is the linear subspace spanned by $\{\phi_{n}\}_1^N$. \end{itemize} \end{definition} Lemma \ref{eigenfunction asymptotic} show that $\mathcal{A}_0$ defined in \eqref{A0} is a $[D]$-class. The following result can be concluded from the proof of a more general result in \cite{li} (see also \cite{kato} and \cite{ramm}). \begin{theorem}[\cite{li}] \label{perbation} Suppose that $\mathcal{A}_0$ is of $[D]$-class satisfying conditions of definition \ref{d calss} in a Hilbert space $H$. Let $d_n:=\min_{n\neq m}|\lambda_n-\lambda_m|$ and assume that $ \sum_{n>N}^{\infty}d_n^{-2}<\infty$. Then for any linear bounded perturbation operator $\mathcal{B}_0$ on $H$, there are constants $C, L >0$, an integer $M>0$, and eigenpairs $\{\mu_n, \psi_{n}\}_{M}^{\infty} $ of $\mathcal{A}_0+\mathcal{B}_0$ such that \begin{itemize} \item[(i)] $|\mu_n-\lambda_n|\leq C, \forall n\geq M$. \item[(ii)] $\|\psi_{n}-\phi_{n}\|\leq L d_n^{-1}$, $n>M$, and hence $\sum_{M}^{\infty}\|\psi_{n}-\phi_{n}\|^2<\infty$. \end{itemize} \end{theorem} We use Theorem \ref{perbation} for $\mathcal{A}_0, \mathcal{B}_0$, where $\mathcal{A}_0$ is defined by \eqref{A0}, and operator $\mathcal{B}_0$ is a perturbation of $\mathcal{A}_0$, such that $\mathcal{A}=\mathcal{A}_0+\mathcal{B}_0$, we can obtain the following result for $\mathcal{A}$. \begin{theorem} \label{thm4.7} Suppose that $k\in C^2([0,1],\mathbb{R})$, $q\in L^1([0,1],\mathbb{R})$, $\{\mu_n, \psi_{n}\}_1^{\infty}$ are eigenpairs of operator $\mathcal{A}$, $\{\lambda_n, y_{n}\}_1^{\infty}$ are eigenpairs of operator $\mathcal{A}_0$. Then the following results hold. \begin{itemize} \item[(i)] $\mathcal{A}=\mathcal{A}_0+\mathcal{B}_0$ is $[D]$-class. \item[(ii)] The eigenvalue of $\mathcal{A}_0+\mathcal{B}_0$ have asymptotic expansion \[ \mu_n=\lambda_n+O(1),\quad n\to +\infty. \] \item[(iii)] The corresponding eigenfunctions $\{\psi_n(x)\}$ of $\mathcal{A}$ have the asymptotic expansion \begin{equation} \label{py} \psi_{n}(x)=y_{n}(x)+\varepsilon_n(x), \quad n\to +\infty, \end{equation} where $\|\varepsilon_n\|_{L^2([0,1],\mathbb{R})}=O(\frac{1}{n})$. Moreover, \begin{equation} \label{approximation of eigenvalue} \sum_{n=M}^{\infty}\|\psi_{n}-y_{n}\|_{L^{2}(0,1)}^2<\infty. \end{equation} where $y_n(\cdot)$ is the eigenfunctions of {\rm\eqref{be}}. Moreover, the generalized eigenfunctions of $\mathcal{A}$ forms a Riesz basis of $L^2([0,1],\mathbb{R})$. \end{itemize} \end{theorem} \begin{proof} Obviously, (ii), \eqref{py} and \eqref{approximation of eigenvalue} can be obtained according to Theorem \ref{perbation}. Next, we prove that the generalized eigenfunctions of $\mathcal{A}$ form a Riesz basis of $L^2([0,1],\mathbb{R})$. Combined \eqref{approximation of eigenvalue} with Theorem \ref{generaliezd basis}, we know that the generalized eigenfunctions of $\mathcal{A}$ forms a Riesz basis of $L^2([0,1],\mathbb{R})$. Meanwhile, in terms of the definition \ref{d calss} and Theorem \ref{main theorem}, we know (i) also holds, which completes the proof. \end{proof} \subsection*{Acknowledgments} We would like to thank Professor Jifeng Chu for his careful reading of the manuscript and valuable suggestions. Zhongcheng Zhou was supported by National Nature Science Foundation under Grant 11301427 and Fundamental Research Funds for the Central Universities under No. XDJK2014B021. Fangfang Liao was supported by QingLan project of Jiangsu Province. \begin{thebibliography}{99} \bibitem{cz} J. Chu, Z. Zhou; \emph{Positive solutions and eigenvalues of nonlocal boundary-value problems}, Electron. J. 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