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数学物理学报, 2019, 39(2): 244-252 doi:

论文

积分-微分算子的迹公式

陈仕荣,

A Trace Formula for Integro-Differential Operators

Chen Shirong,

收稿日期: 2017-07-11  

Received: 2017-07-11  

作者简介 About authors

陈仕荣,E-mail:somxiaorong@163.com , E-mail:somxiaorong@163.com

摘要

该文研究积分-微分算子的迹公式,它在反问题、特征值的数值计算、可积系统理论等有着重要的应用.得到了Dirichlet-Robin边界条件和Dirichlet边界条件下积分-微分算子的迹公式.

关键词: 积分-微分算子 ; 特征值 ; 迹公式 ; 势函数

Abstract

The trace formulae for the integro-differential operator are studied, which have many applications in the inverse problem, the numerical computation of eigenvalues and the theory of integrable system, etc. The trace formula for integro-differential operators with Dirichlet-Robin boundary conditions or Dirichlet boundary conditions are obtained.

Keywords: Integro-differential operator ; Eigenvalue ; Trace formula ; Potential

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本文引用格式

陈仕荣. 积分-微分算子的迹公式. 数学物理学报[J], 2019, 39(2): 244-252 doi:

Chen Shirong. A Trace Formula for Integro-Differential Operators. Acta Mathematica Scientia[J], 2019, 39(2): 244-252 doi:

1 引言

积分-微分算子L:=L(q,M,h,H)定义如下

Ly:=y
(1.1)

满足边界条件

y'\left(0, \lambda\right)-hy\left(0, \lambda\right)=0,
(1.2)

y'\left(\pi, \lambda\right)+Hy\left(\pi, \lambda\right)=0,
(1.3)

其中h, H\in \mathbb{R} \cup\{\infty\}, q(x)M(x, t)是实值函数, q\in L^2\left[0, \pi\right], M\in L^2(D_0), D_0:=\{(x, t):0\leq t\leq x\leq \pi, x, t\in \mathbb{R} \}.h=\infty, H\in \mathbb{R} 时,称问题(1.1)-(1.3)为Dirichlet-Robin边界条件下积分-微分算子;当h=H=\infty时,称问题(1.1)-(1.3)为Dirichlet边界条件下积分-微分算子.

M(x, t)已知的条件下, Kuryshova[6]讨论了积分-微分算子L的逆谱问题及其稳定性; Kuryshova和谢忠村[7]研究了积分-微分算子L的逆结点问题.当M(x, t)=M(x-t)时, Yurko[1]研究了Neumann边界条件y'(0, \lambda)=y'(\pi, \lambda)=0下的积分-微分算子L的逆谱问题,在q(x)已知的条件下,用一组谱重构了M(x);Buterin[2-3]研究了积分-微分算子L的逆谱问题,并建立了M(x)的唯一性定理及逆谱问题的可解性、稳定性理论.王玉华和魏广生[4]用混合谱数据研究了积分-微分算子的逆谱问题.在M\in W_2^{1}(D_0)条件下,王於平[5]证明了积分-微分算子L稠定的结点集能够唯一确定[0, \pi]上的势函数q(x)和积分扰动核M(x),并且给出了这个逆结点问题的解的重构算法.

经典的Sturm-Liouville算子的谱理论也得到深入的研究[8-14].特别地, 1953年, Gelfand和Levitan[14]研究了下面的Sturm-Liouville问题

\begin{eqnarray*} \left\{\begin{array}{ll} -y''+q(x)y=\lambda y, \quad x\in(0, \pi), \\ y'(0)=y'(\pi)=0, \end{array} \right. \end{eqnarray*}

其中q\in C\left[0, \pi\right].他们得到了Neumann边界条件下Sturm-Liouville算子的迹公式

\begin{eqnarray*} \sum\limits_{n=0}^{\infty}\left(\lambda_{n}-n^{2}-\frac{1}{\pi}\int_{0}^{\pi}q(x){\rm d}x\right)= \frac{1}{4}\left(q(0)+q(\pi)\right)-\frac{1}{2\pi}\int_{0}^{\pi}q(x){\rm d}x. \end{eqnarray*}

他们还得到Dirichlet边界条件下Sturm-Liouville算子的迹公式如下

\begin{equation}\label{E1.4} \sum\limits_{n=1}^{\infty}\left(\lambda_{n}-n^{2}-\frac{1}{\pi}\int_{0}^{\pi}q(x){\rm d}x\right)= -\frac{1}{4}\left(q(0)+q(\pi)\right)+\frac{1}{\pi}\int_{0}^{\pi}q(x){\rm d}x. \end{equation}
(1.4)

之后,不少学者对迹公式产生了浓厚的兴趣,得到一些有意义的结果[15-18].当h, H\in \mathbb{R} 时,王於平等[18]得到了积分-微分算子(1.1)-(1.3)的迹公式如下

\begin{eqnarray*} \sum\limits_{n=0}^{\infty}\left(\lambda_n-n^2-\frac{2\omega}{\pi}\right)=\frac{1}{4}\left(q(0)+q(\pi)\right)-\frac{\omega}{\pi} +\frac{1}{2}\int_{0}^{\pi}M(t, t){\rm d}t-\frac{h^2+H^2}{2}, \end{eqnarray*}

其中\omega=h+H+\frac{1}{2}\int_{0}^{\pi}q(x){\rm d}x.

本文研究Dirichlet-Robin边界条件和Dirichlet边界条件下积分-微分算子的迹公式.由于研究积分-微分算子迹公式的需要,在这篇文章中,我们还要假设q\in C\left[0, \pi\right], M\in W_2^{1}(D_0).

2 主要结果及证明

\lambda=\rho^{2}, \tau=\left|{\rm Im}\rho\right|.根据文献[5],当h=\infty时,满足初始条件\varphi(0, \lambda)=0, \varphi'(0, \lambda)=1,方程(1.1)的解\varphi(x, \lambda)表示为

\begin{equation} \varphi\left(x, \lambda\right)=\frac{1}{\rho}\sin (\rho x) +\int_{0}^{x}\varphi\left(t, \lambda\right)\left[\frac{\sin (\rho(x-t))}{\rho}q(t)+\int_{t}^{x}M(\xi, t)\frac{\sin (\rho(x-\xi))}{\rho}{\rm d}\xi\right]{\rm d}t. \end{equation}
(2.1)

M\in W_{2}^{1}(D_0)时,我们得到方程(1.1)的解\varphi(x, \lambda)的渐近式,这个引理在本文主要结果的证明过程中起着重要的作用.

引理2.1  当M\in W_{2}^{1}(D_0)时,则\varphi(x, \lambda)满足下面渐近式

\begin{eqnarray}\label{E2.2} \varphi(x, \lambda)&=&\frac{1}{\rho}\sin(\rho x)-\frac{\cos(\rho x)}{2\rho^2}\int_{0}^{x}q(t){\rm d}t+\frac{\cos(\rho x)}{2\rho^2}\int_{0}^{x}q(t)\cos(2\rho t){\rm d}t\nonumber\\ && +\frac{\sin(\rho x)}{2\rho^2}\int_{0}^{x}q(t)\sin(2\rho t){\rm d}t-\frac{\sin(\rho x)}{2\rho^3}\int_{0}^{x}M(t, t){\rm d}t \\&&-\frac{\sin(\rho x)}{8\rho^3}\left(\int_{0}^{x}q(t){\rm d}t\right)^2 +o\left(\frac{{\rm e}^{\tau x}}{\left|\rho\right|^{3}}\right). \end{eqnarray}
(2.2)

\lambda_{n}(n=0, 1, 2, \cdots)是Dirichlet-Robin边界条件下积分-微分算子的特征值,则对于充分大的n, \lambda_{n}满足下面的渐近式

\begin{equation}\label{E2.3}\rho_{n}=\sqrt{\lambda_{n}}=n+\frac{1}{2}+\frac{\omega_1}{n\pi}+\frac{\kappa_{n, 1}}{n}, \quad H\in\mathbb{R} , \end{equation}
(2.3)

其中\omega_1=H+\frac{1}{2}\int_{0}^{\pi}q(x){\rm d}x, \{\kappa_{n, 1}\}\in l^2.

\lambda_{n}(n=1, 2, \cdots)是Dirichlet边界条件下积分-微分算子的特征值,则对于充分大的n, \lambda_{n}满足下面的渐近式

\begin{equation}\label{E2.4}\rho_{n}=\sqrt{\lambda_{n}}=n+\frac{\int_{0}^{\pi}q(t){\rm d}t}{n\pi}+\frac{\kappa_{n, 2}}{n}, \quad H=\infty, \end{equation}
(2.4)

其中\{\kappa_{n, 2}\}\in l^2.关于(2.3)和(2.4)式的证明,读者可以参考文献[7].

本文的主要结果是两类边界条件下积分-微分算子L的迹公式,即

定理2.1  Dirichlet-Robin边界条件下积分-微分算子L的迹公式为

\begin{equation}\label{E2.5}\sum\limits_{n=0}^{\infty}\left[\lambda_{n}-\left(n+\frac{1}{2}\right)^{2}-\frac{2\omega_1}{\pi}\right]\\=-\frac{1}{4}\left(q(0)-q(\pi)\right)-\frac{\omega_1}{\pi}+\frac{1}{2}\int_{0}^{\pi}M(t, t){\rm d}t-\frac{H^{2}}{2}.\end{equation}
(2.5)

定理2.2  Dirichlet边界条件下积分-微分算子L的迹公式为

\begin{equation}\label{E2.6}\sum\limits_{n=1}^{\infty}\left(\lambda_{n}-n^2-\frac{\int_{0}^{\pi}q(t){\rm d}t}{\pi}\right)=-\frac{1}{4}\left(q(0)+q(\pi)\right)+\frac{1}{\pi}\int_{0}^{\pi}q(t){\rm d}t+\frac{1}{2}\int_{0}^{\pi}M(t, t){\rm d}t.\end{equation}
(2.6)

显然, Dirichlet边界条件下积分-微分算子L的迹公式(2.6)是Dirichlet边界条件下Sturm-Liouville算子的迹公式(1.4)的本质推广.

本文主要结论的证明如下.

引理2.1的证明  由(2.1)式,得

\begin{equation}\label{E2.7}\varphi(x, \lambda)=\frac{1}{\rho}\sin\left(\rho x\right)+O\left(\frac{{\rm e}^{\tau x}}{\rho^2}\right).\end{equation}
(2.7)

由(2.7)式,可得

\begin{eqnarray}\label{E2.8}&&\int_{0}^{x}\varphi(t, \lambda)\frac{\sin(\rho(x-t))}{\rho}q(t){\rm d}t \\&=&\int_{0}^{x}\left(\frac{1}{\rho}\sin\left(\rho t\right)+O\left(\frac{{\rm e}^{\tau t}}{\rho^2}\right)\right)\frac{\sin(\rho(x-t))}{\rho}q(t){\rm d}t \\&=&\frac{1}{\rho^2}\int_{0}^{x}q(t)\cdot\frac{1}{2}\left[\cos\left(2\rho t-\rho x\right)-\cos(\rho x)\right]{\rm d}t+O\left(\frac{{\rm e}^{\tau x}}{\rho^3}\right) \\&=&\frac{1}{2\rho^2}\int_{0}^{x}q(t)\cos\left(2\rho t-\rho x\right){\rm d}t-\frac{1}{2\rho^2}\int_{0}^{x}q(t)\cos(\rho x){\rm d}t+O\left(\frac{{\rm e}^{\tau t}}{\rho^2}\right)+O\left(\frac{{\rm e}^{\tau x}}{\rho^3}\right) \\&=&\frac{1}{2\rho^2}\int_{0}^{x}q(t)\left(\cos(2\rho t)\cos(\rho x)+\sin(2\rho t)\sin(\rho x)\right){\rm d}t-\frac{\cos(\rho x)}{2\rho^2}\int_{0}^{x}q(t){\rm d}t+O\left(\frac{{\rm e}^{\tau x}}{\rho^3}\right) \\&=&\frac{\cos(\rho x)}{2\rho^2}\int_{0}^{x}q(t)\cos(2\rho t){\rm d}t+\frac{\sin(\rho x)}{2\rho^2}\int_{0}^{x}q(t)\sin(2\rho t){\rm d}t \\&&-\frac{\cos(\rho x)}{2\rho^2}\int_{0}^{x}q(t){\rm d}t+O\left(\frac{{\rm e}^{\tau x}}{\rho^3}\right).\end{eqnarray}
(2.8)

因为M\in W_2^{1}(D_0),所以通过分部积分,得

\begin{eqnarray}\label{E2.9}&&\int_{t}^{x}M(\xi, t)\frac{\sin(\rho(x-\xi))}{\rho}{\rm d}\xi \\&=&\frac{1}{\rho^{2}}M(\xi, t)\cos(\rho(x-\xi))\biggr|_{t}^{x}-\frac{1}{\rho^{2}}\int_{t}^{x}M_{\xi}'(\xi, t)\cos(\rho(x-\xi)){\rm d}\xi \\&=&\frac{1}{\rho^{2}}M(x, t)-\frac{1}{\rho^{2}}M(t, t)\cos(\rho(x-t))-\frac{1}{\rho^{2}}\int_{t}^{x}M_{\xi}'(\xi, t)\cos(\rho(x-\xi)){\rm d}\xi.\end{eqnarray}
(2.9)

由(2.9)式,得

\begin{eqnarray}\label{E2.10}&&\int_{0}^{x}\varphi(t, \lambda)\left(\int_{t}^{x}M(\xi, t)\frac{\sin(\rho(x-\xi))}{\rho}{\rm d}\xi\right) {\rm d}t\nonumber\\&=&\int_{0}^{x}\frac{1}{\rho}\sin(\rho t)\left[\frac{1}{\rho^{2}}M(x, t)-\frac{1}{\rho^{2}}M(t, t)\cos(\rho(x-t))\right]{\rm d}t+o\left(\frac{{\rm e}^{\tau x}}{\rho^3}\right)\nonumber\\&=&\frac{1}{\rho^3}\int_{0}^{x}M(x, t)\sin(\rho t){\rm d}t-\frac{1}{\rho^3}\int_{0}^{x}M(t, t)\cdot\frac{1}{2}\left[\sin(\rho x)+\sin(2\rho t-\rho x)\right]{\rm d}t+o\left(\frac{{\rm e}^{\tau x}}{\rho^3}\right)\nonumber\\&=&\frac{1}{\rho^3}\int_{0}^{x}M(x, t)\sin(\rho t){\rm d}t-\frac{1}{2\rho^{3}}\int_{0}^{x}M(t, t)\sin(\rho x){\rm d}t\nonumber\\&&-\frac{1}{2\rho^{3}}\int_{0}^{x}M(t, t)\sin(\rho(2t-x)){\rm d}t+o\left(\frac{{\rm e}^{\tau x}}{\rho^3}\right).\end{eqnarray}
(2.10)

因此(2.8)和(2.10)式证明下式成立

\begin{eqnarray}\label{E2.11}\varphi(x, \lambda)&=&\frac{1}{\rho}\sin(\rho x)-\frac{\cos(\rho x)}{2\rho^2}\int_{0}^{x}q(t){\rm d}t+\frac{\cos(\rho x)}{2\rho^2}\int_{0}^{x}q(t)\cos(2\rho t){\rm d}t\nonumber\\&&+\frac{\sin(\rho x)}{2\rho^2}\int_{0}^{x}q(t)\sin(2\rho t){\rm d}t+O\left(\frac{{\rm e}^{\tau x}}{\rho^3}\right).\end{eqnarray}
(2.11)

再次将(2.11)式代入(2.1)式,通过计算,得\varphi(x, \lambda)的表达式为

\begin{eqnarray}\label{E2.12}\varphi(x, \lambda)&=&\frac{1}{\rho}\sin(\rho x)-\frac{\cos(\rho x)}{2\rho^2}\int_{0}^{x}q(t){\rm d}t+\frac{\cos(\rho x)}{2\rho^2}\int_{0}^{x}q(t)\cos(2\rho t){\rm d}t\nonumber\\&&+\frac{\sin(\rho x)}{2\rho^2}\int_{0}^{x}q(t)\sin(2\rho t){\rm d}t-\frac{\sin(\rho x)}{2\rho^3}\int_{0}^{x}M(t, t){\rm d}t\nonumber\\&&-\frac{\sin(\rho x)}{8\rho^3}\left(\int_{0}^{x}q(t){\rm d}t\right)^2+o\left(\frac{{\rm e}^{\tau x}}{\left|\rho\right|^{3}}\right).\end{eqnarray}
(2.12)

证毕.

定理2.1的证明  由引理2.1中的(2.2)式,得

\begin{eqnarray}\label{E2.13}\varphi'(x, \lambda)&=&\cos(\rho x)-\frac{\sin(\rho x)}{2\rho}\int_{0}^{x}q(t)\cos(2\rho t){\rm d}t+\frac{\cos(\rho x)}{2\rho}\int_{0}^{x}q(t)\sin(2\rho t){\rm d}t\nonumber\\&&+\frac{\sin(\rho x)}{2\rho}\int_{0}^{x}q(t){\rm d}t-\frac{\cos(\rho x)}{2\rho^2}\int_{0}^{x}M(t, t){\rm d}t \\&&-\frac{\cos(\rho x)}{8\rho^2}\left(\int_{0}^{x}q(t){\rm d}t\right)^2+o\left(\frac{{\rm e}^{\tau x}}{|\rho|^{2}}\right).\end{eqnarray}
(2.13)

定义Dirichlet-Robin边界条件下积分-微分算子L特征值的示性函数为

w(\lambda)=\varphi'(\pi, \lambda)+H\varphi(\pi, \lambda).

由(2.2)和(2.13)式,得

\begin{eqnarray}\label{E2.14}w(\lambda)&=&\varphi'(\pi, \lambda)+H\varphi(\pi, \lambda)\nonumber\\&=&\cos(\rho\pi)-\frac{\sin(\rho\pi)}{2\rho}\int_{0}^{\pi}q(t)\cos(2\rho t){\rm d}t+\frac{\cos(\rho \pi)}{2\rho}\int_{0}^{\pi}q(t)\sin(2\rho t){\rm d}t\nonumber\\&&+\frac{\sin(\rho\pi)}{2\rho}\int_{0}^{\pi}q(t){\rm d}t-\frac{\cos(\rho\pi)}{2\rho^2}\int_{0}^{\pi}M(t, t){\rm d}t-\frac{\cos(\rho \pi)}{8\rho^2}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2\nonumber\\&&+\frac{H}{\rho}\sin(\rho\pi)-\frac{H\cos(\rho\pi)}{2\rho^2}\int_{0}^{\pi}q(t){\rm d}t+o\left(\frac{{\rm e}^{\tau\pi}}{\left|\rho\right|^{2}}\right).\end{eqnarray}
(2.14)

\lambda_{n}\left(n=0, 1, 2, \cdots\right)w(\lambda)的零点.令w_{0}(\lambda)=\cos\rho\pi,则

\begin{eqnarray}\label{E2.15}\frac{w(\lambda)}{w_{0}(\lambda)}&=&1-\frac{\tan(\rho\pi)}{2\rho}\int_{0}^{\pi}q(t)\cos(2\rho t){\rm d}t+\frac{1}{2\rho}\int_{0}^{\pi}q(t)\sin(2\rho t){\rm d}t\nonumber\\&&+\frac{\tan(\rho\pi)}{2\rho}\int_{0}^{\pi}q(t){\rm d}t-\frac{1}{2\rho^2}\int_{0}^{\pi}M(t, t){\rm d}t-\frac{1}{8\rho^2}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2\nonumber\\&&+\frac{H}{\rho}\tan(\rho\pi)-\frac{H}{2\rho^2}\int_{0}^{\pi}q(t){\rm d}t+o\left(\frac{{\rm e}^{\tau\pi}}{\left|\rho\right|^{2}\cos(\rho\pi)}\right).\end{eqnarray}
(2.15)

\Gamma_{N_{0}}:=\{\rho:\left|\rho\right|=N_{0}+\frac{3}{4}\}, G_{\delta}=\{\rho :\left|\rho-\left(k+\frac{1}{2}\right)\right|\geq\delta, k=0, \pm1, \pm2, \cdots\},其中N_{0}充分大, \delta>0充分小.则

\begin{eqnarray*}\left|\cos\left(\rho\pi\right)\right|\geq C_{\delta}{\rm e}^{\tau\pi}, \quad \lambda\in C_{\delta}, \quad \left|\rho\right|\geq\rho^{*}, \end{eqnarray*}

其中\rho^{*}=\rho^{*}\left(\delta\right)充分大, C_{\delta}连续地依赖于\delta.因此

\begin{equation}\label{E2.16}\left|o\left(\frac{{\rm e}^{\tau\pi}}{\left|\rho\right|^{2}\cos(\rho\pi)}\right)\right|=o\left(\frac{1}{\left|\rho\right|^2}\right), \quad \left|\rho\right|\geq\rho^{*}.\end{equation}
(2.16)

\mu_{j}=j+\frac{1}{2}\left(j=0, \pm1, \pm2, \cdots\right)w_0(\lambda)的零点.则由留数定理,得

\begin{equation}\label{E2.17}\frac{1}{2\pi{\rm i}}\oint_{\Gamma_{N_{0}}}\lambda\left[\frac{w'(\lambda)}{w(\lambda)}-\frac{w_{0}'(\lambda)}{w_{0}(\lambda)}\right]{\rm d}\lambda=\sum\limits_{n=0}^{N_{0}}\left[2\lambda_{n}-2\left(n+\frac{1}{2}\right)^2\right].\end{equation}
(2.17)

另一方面,由分部积分,得

\begin{eqnarray}\label{E2.18}\frac{1}{2\pi{\rm i}}\oint_{\Gamma_{N_{0}}}\lambda\left[\frac{w'(\lambda)}{w(\lambda)}-\frac{w_{0}'(\lambda)}{w_{0}(\lambda)}\right]{\rm d}\lambda&=&\frac{1}{2\pi{\rm i}}\oint_{\Gamma_{N_{0}}}\lambda {\rm d}\ln\left(\frac{w(\lambda)}{w_{0}(\lambda)}\right) \\&=&-\frac{1}{2\pi{\rm i}}\oint_{\Gamma_{N_{0}}}2\rho \ln\left(\frac{w(\lambda)}{w_{0}(\lambda)}\right){\rm d}\rho.\end{eqnarray}
(2.18)

由(2.17)和(2.18)式,得

\begin{equation}\label{E2.19}\sum\limits_{n=0}^{N_{0}}\left[2\lambda_{n}-2\left(n+\frac{1}{2}\right)^2\right]=-\frac{1}{2\pi{\rm i}}\oint_{\Gamma_{N_{0}}}2\rho \ln\left(\frac{w(\lambda)}{w_{0}(\lambda)}\right){\rm d}\rho.\end{equation}
(2.19)

根据泰勒展式

\ln(1+x)=x-\frac{1}{2}x^2+o(x^2),

则由(2.15)式,得

\begin{eqnarray}\label{E2.20}\ln\left(\frac{w(\lambda)}{w_0(\lambda)}\right)&=&-\frac{\tan(\rho\pi)}{2\rho}\int_{0}^{\pi}q(t)\cos(2\rho t){\rm d}t+\frac{1}{2\rho}\int_{0}^{\pi}q(t)\sin(2\rho t){\rm d}t\nonumber\\&&+\frac{\tan(\rho\pi)}{2\rho}\int_{0}^{\pi}q(t){\rm d}t-\frac{1}{2\rho^2}\int_{0}^{\pi}M(t, t){\rm d}t-\frac{1}{8\rho^2}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2\nonumber\\&&+\frac{H}{\rho}\tan(\rho\pi)-\frac{H}{2\rho^2}\int_{0}^{\pi}q(t){\rm d}t-\frac{1}{2}\cdot\frac{\tan^{2}\rho\pi}{4\rho^2}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2\nonumber\\&&-\frac{1}{2}\cdot\frac{H^2}{\rho^2}\cdot\tan^{2}(\rho\pi)-\frac{1}{2}\cdot\frac{H\tan^{2}(\rho\pi)}{\rho^2}\int_{0}^{\pi}q(t){\rm d}t+o\left(\frac{1}{\left|\rho\right|^2}\right).\end{eqnarray}
(2.20)

所以(2.20)式意味着下式成立

\begin{eqnarray}\label{E2.21}2\rho \ln\left(\frac{w(\lambda)}{w_{0}(\lambda)}\right)&=&-\tan(\rho\pi)\int_{0}^{\pi}q(t)\cos(2\rho t){\rm d}t+\int_{0}^{\pi}q(t)\sin(2\rho t){\rm d}t\nonumber\\&&+\tan(\rho\pi)\int_{0}^{\pi}q(t){\rm d}t-\frac{1}{\rho}\int_{0}^{\pi}M(t, t){\rm d}t-\frac{1}{4\rho}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2\nonumber\\&&+2H\tan(\rho\pi)-\frac{H}{\rho}\int_{0}^{\pi}q(t){\rm d}t-\frac{\tan^{2}(\rho\pi)}{4\rho}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2\nonumber\\&&-\frac{H^2}{\rho}\cdot\tan^{2}(\rho\pi)-\frac{H\tan^{2}(\rho\pi)}{\rho}\int_{0}^{\pi}q(t){\rm d}t+o\left(\frac{1}{\left|\rho\right|}\right).\end{eqnarray}
(2.21)

将(2.21)式代入(2.19)式,得

\begin{eqnarray}\label{E2.22}&&-\frac{1}{2\pi{\rm i}}\oint_{\Gamma_{N_{0}}}2\rho \ln\left(\frac{w(\lambda)}{w_{0}(\lambda)}\right){\rm d}\rho \\&=&\frac{1}{2\pi{\rm i}}\oint_{\Gamma_{N_{0}}}\Biggr\{\tan(\rho\pi)\int_{0}^{\pi}q(t)\cos(2\rho t){\rm d}t-\int_{0}^{\pi}q(t)\sin(2\rho t){\rm d}t\nonumber\\&&-\tan(\rho\pi)\int_{0}^{\pi}q(t){\rm d}t+\frac{1}{\rho}\int_{0}^{\pi}M(t, t){\rm d}t+\frac{1}{4\rho}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2\nonumber\\&&-2H\tan(\rho\pi)+\frac{H}{\rho}\int_{0}^{\pi}q(t){\rm d}t+\frac{\tan^{2}(\rho\pi)}{4\rho}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2\nonumber\\&&+\frac{H^2}{\rho}\cdot\tan^{2}(\rho\pi)+\frac{H\tan^{2}(\rho\pi)}{\rho}\int_{0}^{\pi}q(t){\rm d}t+o\left(\frac{1}{|\rho|}\right)\Biggr\}{\rm d}\rho.\end{eqnarray}
(2.22)

经过计算得

\begin{eqnarray}\label{E2.23}&&{\rm Res}\left[\tan(\rho\pi)\int_{0}^{\pi}q(t)\cos(2\rho t){\rm d}t\right]=\sum\limits_{n=-N_{0}}^{N_{0}}-\frac{1}{\pi}\int_{0}^{\pi}q(t)\cos((2n+1)t){\rm d}t, \nonumber\\&&{\rm Res}\left[\tan(\rho\pi)\right]=\sum\limits_{n=-N_{0}}^{N_{0}}-\frac{1}{\pi}, \quad{\rm Res}\left[\int_{0}^{\pi}q(t)\sin(2\rho t){\rm d}t\right]=0, \quad {\rm Res}\left[\frac{1}{\rho}\right]=1, \nonumber\\&&{\rm Res}\left[\frac{\tan^{2}(\rho\pi)}{\rho}\right]=-\sum\limits_{n=-N_{0}}^{N_{0}}\frac{4}{(2n+1)^{2}\pi^2}.\end{eqnarray}
(2.23)

由(2.19), (2.22)和(2.23)式,得

\begin{eqnarray*}&&\sum\limits_{n=0}^{N_{0}}\left[2\lambda_{n}-2\left(n+\frac{1}{2}\right)^2\right]=-\frac{1}{2\pi{\rm i}}\oint_{\Gamma_{N_{0}}}2\rho \ln\left(\frac{w(\lambda)}{w_{0}(\lambda)}\right){\rm d}\rho\\&=&\sum\limits_{n=-N_{0}}^{N_{0}}-\frac{1}{\pi}\int_{0}^{\pi}q(t)\cos((2n+1)t){\rm d}t+\sum\limits_{n=-N_{0}}^{N_{0}}\frac{1}{\pi}\cdot\int_{0}^{\pi}q(t){\rm d}t+\int_{0}^{\pi}M(t, t){\rm d}t\\&&+\frac{1}{4}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2+2H\sum\limits_{n=-N_{0}}^{N_{0}}\frac{1}{\pi}+H\int_{0}^{\pi}q(t){\rm d}t\\&&-\left[\frac{1}{4}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2+H^2+H\int_{0}^{\pi}q(t){\rm d}t\right]\sum\limits_{n=-N_{0}}^{N_{0}}\frac{4}{(2n+1)^{2}\pi^2}\\&=&-\frac{1}{\pi}\int_{0}^{\pi}q(t)\cos t{\rm d}t-\frac{2}{\pi}\sum\limits_{n=0}^{N_{0}}\int_{0}^{\pi}q(t)\cos (2nt)\cos t{\rm d}t\\&&+\frac{2N_{0}+1}{\pi}\int_{0}^{\pi}q(t){\rm d}t+\int_{0}^{\pi}M(t, t){\rm d}t+\frac{1}{4}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2+\frac{2H(2N_{0}+1)}{\pi}\\&&+H\int_{0}^{\pi}q(t){\rm d}t-\left[\frac{1}{4}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2+H^2+H\int_{0}^{\pi}q(t){\rm d}t\right]\sum\limits_{n=0}^{N_{0}}\frac{8}{(2n+1)^{2}\pi^2}.\end{eqnarray*}

从而得到

\begin{eqnarray}\label{E2.24}&&\sum\limits_{n=0}^{N_{0}}\left[\lambda_{n}-\left(n+\frac{1}{2}\right)^2-\frac{2H+\int_{0}^{\pi}q(t){\rm d}t}{\pi}\right]\nonumber\\&=&-\frac{1}{2\pi}\int_{0}^{\pi}q(t)\cos t{\rm d}t-\frac{1}{\pi}\sum\limits_{n=0}^{N_{0}}\int_{0}^{\pi}q(t)\cos (2nt)\cos t{\rm d}t-\frac{H+\frac{1}{2}\int_{0}^{\pi}q(t){\rm d}t}{\pi}\nonumber\\&&+\frac{1}{2}\int_{0}^{\pi}M(t, t){\rm d}t+\frac{1}{8}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2+\frac{H}{2}\int_{0}^{\pi}q(t){\rm d}t\nonumber\\&&-\left[\frac{1}{4}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2+H^2+H\int_{0}^{\pi}q(t){\rm d}t\right]\sum\limits_{n=0}^{N_{0}}\frac{4}{(2n+1)^{2}\pi^2}.\end{eqnarray}
(2.24)

由(2.3)式知,级数

\sum\limits_{n=0}^{\infty}\left[\lambda_{n}-\left(n+\frac{1}{2}\right)^2-\frac{2H+\int_{0}^{\pi}q(t){\rm d}t}{\pi}\right]

绝对收敛,所以在(2.24)式中,令N_{0}\rightarrow\infty,得

\begin{eqnarray}\label{E2.25}&&\sum\limits_{n=0}^{\infty}\left[\lambda_{n}-\left(n+\frac{1}{2}\right)^2-\frac{2H+\int_{0}^{\pi}q(t){\rm d}t}{\pi}\right]\nonumber\\&=&-\frac{1}{2\pi}\int_{0}^{\pi}q(t)\cos t{\rm d}t-\frac{1}{\pi}\sum\limits_{n=0}^{\infty}\int_{0}^{\pi}q(t)\cos (2nt)\cos t{\rm d}t-\frac{H+\frac{1}{2}\int_{0}^{\pi}q(t){\rm d}t}{\pi}\nonumber\\&&+\frac{1}{2}\int_{0}^{\pi}M(t, t){\rm d}t+\frac{1}{8}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2+\frac{H}{2}\int_{0}^{\pi}q(t){\rm d}t\nonumber\\&&-\left[\frac{1}{4}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2+H^2+H\int_{0}^{\pi}q(t){\rm d}t\right]\sum\limits_{n=0}^{\infty}\frac{4}{(2n+1)^{2}\pi^2}\nonumber\\&=&-\frac{1}{2\pi}\int_{0}^{\pi}q(t)\cos t{\rm d}t-\frac{1}{\pi}\sum\limits_{n=0}^{\infty}\int_{0}^{\pi}q(t)\cos (2nt)\cos t{\rm d}t-\frac{H+\frac{1}{2}\int_{0}^{\pi}q(t){\rm d}t}{\pi}\nonumber\\&&+\frac{1}{2}\int_{0}^{\pi}M(t, t){\rm d}t+\frac{1}{8}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2+\frac{H}{2}\int_{0}^{\pi}q(t){\rm d}t\nonumber\\&&-\frac{1}{2}\left[\frac{1}{4}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2+H^2+H\int_{0}^{\pi}q(t){\rm d}t\right]\nonumber\\&=&-\frac{1}{2\pi}\int_{0}^{\pi}q(t)\cos t{\rm d}t-\frac{1}{\pi}\sum\limits_{n=0}^{\infty}\int_{0}^{\pi}q(t)\cos (2nt)\cos t{\rm d}t\\&&-\frac{H+\frac{1}{2}\int_{0}^{\pi}q(t){\rm d}t}{\pi}+\frac{1}{2}\int_{0}^{\pi}M(t, t){\rm d}t-\frac{H^2}{2}.\end{eqnarray}
(2.25)

b_{2n}=\frac{1}{\pi}\int_{0}^{\pi}q(t)\cos t\cos(2nt) {\rm d}t,则b_{0}=\frac{1}{\pi}\int_{0}^{\pi}q(t)\cos t {\rm d}t,并且

\begin{equation}\label{E2.26}b_{2n}=\frac{1}{2}\cdot\frac{1}{\pi}\int_{0}^{2\pi}q\left(\frac{u}{2}\right)\cos\left(\frac{u}{2}\right)\cos (nu) {\rm d}u.\end{equation}
(2.26)

由(2.26)式知, 2b_{2n}就是q\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)的Fourier展开的系数.根据Fourier级数收敛定理,得到

\begin{equation}\label{E2.27}b_{0}+\sum\limits_{n=0}^{\infty}2b_{2n}=\frac{q(0)\cos0+q(\pi)\cos\pi}{2}=\frac{1}{2}\left(q(0)-q(\pi)\right).\end{equation}
(2.27)

由(2.25)-(2.27)式,得

\begin{eqnarray}&&\sum\limits_{n=0}^{\infty}\left[\lambda_{n}-\left(n+\frac{1}{2}\right)^{2}-\frac{2H+\int_{0}^{\pi}q(t){\rm d}t}{\pi}\right]\nonumber\\&=&-\frac{1}{4}\left(q (0)-q(\pi)\right)-\frac{H+\frac{1}{2}\int_{0}^{\pi}q(t){\rm d}t}{\pi}+\frac{1}{2}\int_{0}^{\pi}M(t, t){\rm d}t-\frac{H^{2}}{2}.\nonumber\end{eqnarray}

证毕.

类似可证定理2.2.

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