## The Essential Spectrum of a Class of Anti-Triangular Operator Matrices

College of Sciences, Inner Mongolia University of Technology, Hohhot 010051

 基金资助: 国家自然科学基金.  12261065内蒙古自然科学基金.  2021LHMS01004内蒙古自然科学基金.  2022MS01005自治区直属高校基本科研业务费项目.  JY20220151

 Fund supported: the Natural Science Foundation of China.  12261065the National Natural Science Foundation of Inner Mongolia.  2021LHMS01004the National Natural Science Foundation of Inner Mongolia.  2022MS01005the Basic Science Research Fund in the Universities Directly under the Inner Mongolia Autonomus Region.  JY20220151 Abstract

In this paper, the essential spectrum of a class of unbounded unself-adjoint anti-triangular operator matrices is studied. Firstly, we describe the essential spectrum of operator matrices by using the quadratic operator pencil and the properties of its operator entries, and estimate the essential spectrum of the whole operator matrix. On this basis, the accumulation point of the non-real spectrum of the operator matrix is analyzed.

Keywords： Anti-triangular operator matrices ; Essential spectrum ; Accumulation points of spectrum

Hua Rui, Qi Yaru. The Essential Spectrum of a Class of Anti-Triangular Operator Matrices. Acta Mathematica Scientia[J], 2022, 42(6): 1611-1618 doi:

## 1 引言

$\begin{equation} {\cal A}=\Bigg[\begin{array}{cc} 0\; &B \\ -B^{*}\; &-D\\ \end{array}\Bigg]: {\cal D}(B^{*})\times {\cal D}(B)\subset H_{1}\times H_{2}\rightarrow H_{1}\times H_{2}, \end{equation}$

$\begin{equation} \left\{ \begin{array}{l} { } \frac{\partial^{2}u(x, t)}{\partial t^{2}}-\frac{\partial}{\partial x}\big(a(x)\frac{\partial} {\partial x}u(x, t)\big)-\frac{\partial^{2}}{\partial t\partial x}\big(b(x)\frac{\partial}{\partial x}u(x, t)\big)\\ { } +\int_{0}^{1}k(x, y)\frac{\partial^{2}}{\partial t\partial y}u(y, t){\rm d}y=0, (x\in(0, 1), t>0), \\ u(0, t)=u(1, t)=0, (t\geq 0), \\ { } u(x, 0)=u_{0}(x), \frac{\partial}{\partial t}u(x, 0)=u_{1}(x), (0\leq x\leq 1). \end{array} \right. \end{equation}$

$a(x), b(x)\in C^{1}(0, 1),$对于$x\in[0, 1] $$a(x)\geq\lambda_{1}>0, b(x)\geq v_{1}>0,$$ \frac{\partial}{\partial y}k(x, y)\in L^{2}([0, 1]\times[0, 1])$, $\lambda_1, v_1$的定义见下文. 对任意的$p(x)\in L^{2}(0, 1)$满足$\int^{1}_{0}\int^{1}_{0}\frac{\partial}{\partial y}k(x, y)p(x)p(y){\rm d}x{\rm d}y \leq 0.$因此弦振动方程(1.2)可理解为Hilbert空间$L^{2}(0, 1)$中二阶微分方程

$\begin{equation} \nonumber \left\{ \begin{array}{l} \ddot{z}(t)+T_{2}\dot{z}(t)+T_{3}\dot{z}(t)+T_{1}z(t)=0, \\ z(0)=z_{0}, \dot{z}(0)=z_{1} \end{array} \right. \end{equation}$

$\begin{equation} T_{1}=-\frac{\partial}{\partial x}\big(a(x)\frac{\partial}{\partial x}\cdot\big), {\cal D}(T_{1})=\{u\in H^{2}(0, 1), u(0)=u(1)=0\}, \end{equation}$

$\begin{equation} T_{2}=\int^{1}_{0}k(x, y)\frac{\partial}{\partial y}\cdot {\rm d}y, {\cal D}(T_{2})=\{u\in H^{1}(0, 1), u(0)=u(1)=0\}, \end{equation}$

$\begin{equation} T_{3}=-\frac{\partial}{\partial x}\big(b(x)\frac{\partial}{\partial x}\cdot\big), {\cal D}(T_{3})={\cal D}(T_{1})\subset{\cal D}(T_{2}), \end{equation}$

$\begin{equation} \nonumber \left\{ \begin{array}{l} { } \frac{\rm d}{{\rm d}t}\left(\begin{array}{cc} x_{1} \\ x_{2} \\ \end{array}\right)= \Bigg[\begin{array}{cc} 0\; &B \\ -B^{*}\; &-D\\ \end{array}\Bigg]\left(\begin{array}{cc} x_{1} \\ x_{2} \\ \end{array}\right)={\cal B}\left(\begin{array}{cc} x_{1} \\ x_{2} \\ \end{array}\right), \\ x_{1}(0)=Bz_{0}, x_{2}(0)=z_{1}. \end{array} \right. \end{equation}$

$\begin{equation} \nonumber \rho(T):=\{\lambda\in {\Bbb C}| T-\lambda I\ \mbox{ 是单射}, \ {\cal R}(T-\lambda I)=H\}, \end{equation}$

$\liminf \limits _{ n\to \infty }[x_{n}, x_{n}]>0 ,$则称$\lambda $$T$$ \pi_{+}$型谱点, 并将满足上述条件的$\lambda$构成的集合记为$\sigma_{\pi_{+}}(T)$.

## 2 主要结果及其证明

由$0\in\rho(B)$可知$B^{-1}, (B^{*})^{-1}$均为有界算子, 且

${\cal A}$具有有界逆. 进一步, 当$D$为相对于$B$有界的自伴算子且相对界小于1时, ${\cal A}$在Krein空间$(H_{1}\times H_{2}, [\cdot, \cdot])$上是自伴算子. 根据文献[19, 推论6.3] 可知$\sigma({\cal A})$关于实轴对称.

$s\in {\Bbb C}\backslash\sigma_{ess}({\cal A})$, 则${\rm nul}({\cal A}-sI)={\rm nul} L(s) $${\rm def}({\cal A}-sI)={\rm def}L(s) . 其中二次算子族 L(s)=(B^{*})^{-1}(D+s)B^{-1}s+I. 当 s=0 时, 由命题2.1知算子 {\cal A}$$ L(0)$具有有界逆, 故以上结论显然成立. 当$s\neq0$时,

$\begin{equation} {\cal A}^{-1}-\frac{1}{s}I=\Bigg[\begin{array}{cc} I\; &s(B^{*})^{-1} \\ 0\; &I\\ \end{array}\Bigg] \left[\begin{array}{cc} { } M(\frac{1}{s})\; &0 \\ 0\; &{ } -\frac{1}{s}I\\ \end{array}\right] \Bigg[\begin{array}{cc} I\; &0 \\ -sB^{-1}\; &I\\ \end{array}\Bigg], \end{equation}$

$s\in {\Bbb C}\backslash\sigma_{ess}({\cal A})$, 由定义1.3得${\rm nul}({\cal A}-sI)={\rm nul} L(s) $${\rm def}({\cal A}-sI)={\rm def} L(s) . 证毕. 定理2.2 设 {\cal A} 为由(1.1) 式定义的反三角算子矩阵, 若 B^{-1} 为紧算子, 则 进一步, 当 D 为自伴算子时 \sigma_{ess}({\cal A})\subset{\Bbb R} . 根据命题2.1可知 {\cal A}^{-1}\ 存在, 且 B^{-1} 为紧算子, 根据引理1.1知 \begin{equation} \sigma_{ess}({\cal A})=\Big\{\lambda\in{\Bbb C}\backslash\{0\}\Big| \frac{1}{\lambda}\in \sigma_{ess}\Big(-(B^{*})^{-1}DB^{-1}\Big)\Big\}. \end{equation} 进一步, 由 B 为稠定闭算子且 D 为自伴算子, 则 \Big((B^{*})^{-1}DB^{-1}\Big)^{*}=(B^{*})^{-1}DB^{-1} , 再结合(2.2) 式得 \sigma_{ess}({\cal A})\subset {\Bbb R} .证毕. 推论2.1 设 {\cal A} 为由(1.1) 式定义的反三角算子矩阵, 其中 B^{-1}, D 为紧算子, 则 \sigma_{ess}({\cal A})=\emptyset . 由 B^{-1}, D 均为紧算子可知 (B^{*})^{-1}DB^{-1} 为紧算子. 再由文献[21, 定理III.6.26] 及上式(2.2) 得 \sigma_{ess}({\cal A})=\emptyset . 证毕. 下面为了估计 {\cal A} 的本质谱的范围, 先给出如下的引理. 引理2.1 设 {\cal A} 为由(1.1) 式定义的算子. 若 B, DB^{-1} 均为自伴算子, 令 \lambda=\mu+i\sigma,$$ \lambda\neq 0$, 假定存在一个序列$\{(x_{n}, y_{n})^{\top}\}_{n\in {\Bbb N}}\in D({\cal A})$使得

$\begin{equation} \|x_{n}\|^{2}+\|y_{n}\|^{2}=1, \end{equation}$

$\begin{equation} \lim \limits _{ n\to \infty } \|({\cal A}-\lambda I )\left(\begin{array}{cc} x_{n} \\ y_{n} \\ \end{array}\right)\|=0 \end{equation}$

1. $\|y_{n}-B^{-1}\lambda x_{n}\|\rightarrow 0, (n\rightarrow \infty)$;

2. $\liminf \limits _{ n\to \infty }\|x_{n}\|> 0;$

3. 若$\sigma\neq 0,$

4. 若$\sigma=0,$

由(2.4) 式有

$\begin{equation} \lambda x_{n}-By_{n}\rightarrow 0, (n\rightarrow \infty), \end{equation}$

$\begin{equation} Bx_{n}+(\lambda+D)y_{n}\rightarrow0, (n\rightarrow \infty). \end{equation}$

$\begin{equation} \langle DB^{-1}x_{n}, x_{n}\rangle+\frac{2\mu}{\mu^{2}+\sigma^{2}}\langle Bx_{n}, x_{n}\rangle\rightarrow0 (n\rightarrow \infty). \end{equation}$

$\{x_{n}\}\in G_{\lambda}^{\perp}$, 那么存在$\delta>0$使得

$\begin{equation} \nonumber \langle DB^{-1}x_{n}, x_{n}\rangle\geq\Big(-\frac{2\mu}{\mu^{2}+\sigma^{2}}+\delta\Big)\|B\|\|x_{n}\|^{2} =-\frac{2\mu}{\mu^{2}+\sigma^{2}}\|B\|\|x_{n}\|^{2}+\delta\|B\|\|x_{n}\|^{2}. \end{equation}$

$\rho({\cal A})\cap N\neq\emptyset$, 根据文献中的[IV.5.6] 可知${\cal A}-\lambda I$是指标为0的Fredholm算子. 再根据引理1.2可知, 存在有限维的孤立点集$K\subset N$使得$K\subset\sigma_{p, norm}({\cal A})$$N\backslash K\subset\rho({\cal A})$. 故此结论成立.

设$\lambda\in(-\alpha_{1}, 0), G_{\lambda}$满足(2.13) 式, 并且任意序列$\{(x_{n}, y_{n})^{\top}\}_{n\in {\Bbb N}}\subset ( G_{\lambda}\times G_{\lambda})^{\perp}$满足(2.3), (2.4)式. 由引理2.1的(2.5), (2.11) 式知

$\begin{eqnarray} \liminf \limits _{ n\to \infty }\Bigg[\left(\begin{array}{cc} x_{n} \\ y_{n} \\ \end{array}\right), \left(\begin{array}{cc} x_{n} \\ y_{n} \\ \end{array}\right)\Bigg]&=&\liminf \limits _{ n\to \infty }\Big(\langle x_{n}, x_{n}\rangle-\langle y_{n}, y_{n}\rangle\Big){}\\ &\geq&\liminf \limits _{ n\to \infty }\Big(\langle x_{n}, x_{n}\rangle-\lambda^{2}\|B^{-1}\|^{2}\|x_{n}\|^{2}\Big){}\\ &\geq&\liminf \limits _{ n\to \infty }\Big(-\frac{\lambda \langle DB^{-1}x_{n}, x_{n}\rangle+\lambda^{2}\langle B^{-1}x_{n}, x_{n}\rangle}{\|B\|}-\lambda^{2}\|B^{-1}\|^{2}\|x_{n}\|^{2}\Big){}\\ &=&-\lambda\liminf \limits _{ n\to \infty }\Big(\frac{\langle DB^{-1}x_{n}, x_{n}\rangle+2\lambda\|B^{-1}\|\|x_{n}\|^{2}}{\|B\|}\Big). \end{eqnarray}$

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