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数学物理学报, 2025, 45(1): 54-73

加权Laplace算子Dirichlet特征值问题的一个万有不等式及其应用

杨贵诚1, 温杨哲2, 毛井,2,3,*

1湖北商贸学院 武汉 430074

2湖北大学数学与统计学学院, 应用数学湖北省重点实验室 武汉 430062

3湖北大学智能感知系统与安全教育部重点实验室 武汉 430062

A Universal Inequality for the Dirichlet Eigenvalue Problem of the Weighted Laplacian and its Application

Yang Guicheng1, Wen Yangzhe2, Mao Jing,2,3,*

1Hubei Business College, Wuhan 430074

2Faculty of Mathematics and Statistics, Key Laboratory of Applied Mathematics of Hubei Province, Hubei University, Wuhan 430062

3Key Laboratory of Intelligent Sensing System and Security (Hubei University), Ministry of Education, Wuhan 430062

通讯作者: * 毛井, E-mail:jiner120@163.com

收稿日期: 2024-06-3   修回日期: 2024-07-26  

基金资助: 国家自然科学基金(11801496)
国家自然科学基金(11926352)
霍英东教育基金会高等院校青年教师基金(161004)
应用数学湖北省重点实验室和智能感知系统与安全教育部重点实验室(湖北大学)

Received: 2024-06-3   Revised: 2024-07-26  

Fund supported: NSFC(11801496)
NSFC(11926352)
Fok Ying-Tung Education Foundation(161004)
Key Laboratory of Intelligent Sensing System and Security(Hubei University),Ministry of Education

摘要

该文研究了欧氏空间中具有光滑边界的有界区域 Ω 上加权 Laplace 算子 Lϕ 的 Dirichlet 特征值问题. 在加权函数 ϕ 满足一定约束条件的前提下, 利用变分法, 并在恰当地构造测试函数的基础上, 可以得到该特征值问题的一个万有不等式.

关键词: 加权 Laplace 算子; Dirichlet 特征值问题; Green 公式

Abstract

In this paper, we study the Dirichlet eigenvalue problem of the weighted Laplace operator Lϕ on a bounded domain Ω with a smooth boundary in n-dimensional Euclidean space. Under the premise that the weighted function ϕ satisfies certain constraints, a universal inequality of the eigenvalue problem can be obtained by using the variational method and constructing the test function appropriately.

Keywords: weighted Laplace operator; Dirichlet eigenvalue problem; Green formula

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本文引用格式

杨贵诚, 温杨哲, 毛井. 加权Laplace算子Dirichlet特征值问题的一个万有不等式及其应用[J]. 数学物理学报, 2025, 45(1): 54-73

Yang Guicheng, Wen Yangzhe, Mao Jing. A Universal Inequality for the Dirichlet Eigenvalue Problem of the Weighted Laplacian and its Application[J]. Acta Mathematica Scientia, 2025, 45(1): 54-73

1 引言

2016 年, 成庆明教授在其发表在第 6 届世界华人数学家大会文集里的文献 [7] 中回顾了有界开区域上 Laplace 算子 Dirichlet 特征值的万有不等式 (这里 "万有" 二字的内在含义是特征值所满足的不等式不依赖于区域的选取) 研究的一些重要进展 (包含他本人和杨洪苍研究员合作的部分研究工作 [9], 并建议道对万有不等式的深入研究将有助于尝试解决大数学家 George Pölya 教授 [15]提出的著名的 Pölya 猜想. 该猜想至今仍旧是公开问题. 故而, 有界开区域上 Laplace 算子 Dirichlet 特征值的万有不等式是值得深入研究的, 其重要性由此可见一斑.

给定一个 n-维完备黎曼流形 Mn, ΩMnMn 上的有界带边区域, 所谓的固定薄膜问题实际上是考虑 Ω 上 Laplace 算子 Δ如下的 Dirichlet 特征值问题

{Δu+λu=0 在 Ω 内, u=0 在 Ω 上. 
(1.1)

该问题本质上是要找到所有可能的实数 λ 使得(1.1)存在非平凡解. 对于系统(1.1)而言, 当 Mn 取成是 2-维欧氏平面 R2 时, 它可以用来描述边界固定的薄膜的振动, 这也是该系统被称为 "固定薄膜问题" 的原因所在. 众所周知 (可参阅文献 [5]):

• Dirichlet 特征值问题 (1.1) 仅有离散谱, 谱中的元素 (即特征值) 可以按照单调不减的方式排成如下的序列

0<λ1(Ω)<λ2(Ω)λ3(Ω),

并且序列中的每个特征值 λi(Ω) 重复其重数次, 这里 λi(Ω) 的重数就是其特征空间的维数 (是有限维的), 而特征空间实际上是由 λi(Ω) 的特征函数所张成的向量空间. 对于特征值 λi(Ω) 而言, 其特征函数就是将 λ=λi(Ω) 代入 (1.1) 式中时所有可能的非平凡解 u. 利用变分原理, 第 (k+1) 个特征值 λk+1(Ω),k=1,2,3,, 可以被刻画为

λk+1(Ω)=inf

这里 u_i, i=1,2, \cdots, k, 是属于特征值 \lambda_i(\Omega) 的特征函数, dv 表示黎曼体积元, W_0^{1, 2}(\Omega) 是具有紧致支撑的 Sobolev 空间. 如果没有特别说明, 在本论文中不同处的相同符号具有相同的意义. 下文中, 为了书写方便, 在不引起混淆的情况下, 我们将简记特征值 \lambda_i(\Omega)\lambda_i, 这个简写的约定也将会使用到本文将讨论的其它类型的特征值问题上. 特别地, 第一个 Dirichlet 特征值的变分刻画为

\lambda_1(\Omega)=\inf\left\{\left.\frac{\int_{\Omega}|\nabla u|^2 {\rm d} v}{\int_{\Omega} u^2 {\rm d} v} \right\rvert\, u \in W_0^{1,2}(\Omega), u \neq 0\right\}.

对于问题 (1.1), 当 M^n 取成是 2-维欧氏平面 \mathbb{R}^{2} 时, 上世纪 50 年代, Payne, Pölya 和 Wein-berger [14] 得到了前三个 Dirichlet 特征值的如下万有不等式

\begin{matrix}\label{eq_1} \frac{{{\lambda _3}{\rm{ + }}{\lambda _2}}}{{{\lambda _1}}} \le 6 \text {. } \end{matrix}
(1.2)

上世纪 60 年代末期, Thompson[17] 注意到文献 [14] 中的技巧也可以发展到高维的情形, 并因此而得到了

\begin{matrix}\label{eq_2} \frac{{{\lambda _2}{\rm{ + }}{\lambda _3}{\rm{ + }} \cdots {\rm{ + }}{\lambda _{n + 1}}}}{{{\lambda _1}}} \leq n + 4. \end{matrix}
(1.3)

在这之后, M S Ashbaugh, R D Benguria, G N Hile, M H Protter, 杨洪苍研究员等国内外同行陆续对万有不等式 (1.2)-(1.3) 进行了不少有意义的拓展[1,2,4,10,19]. 2011 年, 陈大广教授和合作者[6,定理 2.1、 推论 2.2、 推论 3.1]将上述同行的工作做了进一步的拓展, 并通过他们文章中的图表 2 直观地展示了他们结果相对于已有结果的优势.

对于给定的黎曼流形 \left(M^n, g\right) 以及定义在 M^n 上的光滑实值函数 \phi, 可以自然地定义光滑度量测度空间 \left(M^n, g, {\rm e}^{-\phi} {\rm d} v\right), 其实质上是对黎曼流形 \left(M^n, g\right) 赋予一个共形测度 {\rm e}^{-\phi} {\rm d} v, 这里 {\rm d} v 同前面一样表示与度量 g 相关的黎曼体积元. 令 \Omega\left(M^n, g, {\rm e}^{-\phi} {\rm d} v\right) 中的有界带边区域, 在 \Omega 上可以定义如下两个算子

\begin{eqnarray*} \begin{aligned} &{\rm{Ri}}{{\rm{c}}_\phi }:= {\rm{Ric}} + {\rm{Hess}}\phi, \\ &\mathbb L_\phi : =\Delta - \langle\nabla \phi, \nabla \cdot\rangle, \end{aligned} \end{eqnarray*}

这里 Ric, Hess 分别代表 M^n 上的 Ricci 曲率张量和 Hessian 算子, \nabla 表示 M^n 上梯度算子, \left\langle { \cdot, \cdot } \right\rangle 表示由度量 g 所诱导的内积. 事实上, 算子 {\rm{Ri}}{{\rm{c}}_\phi }, \mathbb {L}_\phi 分别被称为 Bakry-Émery Ricci 曲率 (或加权 Ricci 曲率) 和加权 Laplace 算子 (或 Witten-Laplace 算子). 虽然形式上看光滑度量测度空间 \left(M^n, g, {\rm e}^{-\phi}{\rm d} v\right) 相对于黎曼流形 \left(M^n, g\right) 只是添加了一个共形密度, 但是它同经典的黎曼几何还是有不少不一样的表现. 这里我们想通过简单地回顾国际知名华人几何学家韦国芳教授和合作者关于加权曲率有界的假设条件下的若干比较定理的文献 [18]中给出的简洁例子来说明研究光滑度量测度空间里的几何问题的必要性和意义. 具体地, 考虑度量测度空间 \left( \mathbb{R}^n, g, {\rm e}^{ - \phi }{\rm d} v \right), 这里 {\rm d} v 代表同欧氏度量 g 相关的体积元, 如果加权函数为 \phi(x)=\frac{\lambda}{2}|x|^2, \lambda>0, 则有 \rm Hess_\phi =\lambda g, \operatorname{Hess}_{\phi}=\lambda g, \operatorname{Ric}_{\phi}=\lambda g. 故而不同于黎曼流形的 Ricci 曲率有严格正常数下界的情形, 加权 Ricci 曲率有正常数下界的度量测度空间没有必要一定是紧致的. 因此, 研究光滑度量测度空间上的几何问题是必要的、很有意义的.

给定一个光滑度量测度空间 \left(M^{n}, g, \mathrm{e}^{-\phi} \mathrm{d} v\right)以及其上的有界带边区域 \Omega, 可以考虑如下的 Dirichlet 特征值问题

\begin{equation}\label{ld-2} \left\{\begin{aligned} \begin{array}{ll} \mathbb {L}_\phi u + \lambda_\phi u= 0\qquad & \text{在} \ \Omega \ \text{内}, \\ u = 0\qquad\qquad\quad\quad & \text{在} \ \partial\Omega \ \text{上}. \\ \end{array} \end{aligned}\right. \end{equation}
(1.4)

不难验证算子 \mathbb {L}_\phi 关于内积

\left( {f,h} \right): = \int_\Omega f h{{\rm e}^{ - \phi }}{\rm d} v

是自伴的, 这里 f, h \in \widetilde{W}_0^{1,2}(\Omega), \widetilde{W}_0^{1, 2}(\Omega) 是具有紧致支撑的光滑函数空间在范数

\|f\|_{\widetilde W_0^{1,2}(\Omega )}:=\left(\int_{\Omega} f^2 {\rm e}^{-\phi}{\rm d} v+\int_{\Omega}|\nabla f|^2 {\rm e}^{-\phi}{\rm d} v\right)^{1 / 2}

下的完备化. 采用类似于文献[5] 中的一些经典讨论 (如离散谱的存在性、Rayleigh 定理、Max-min 定理等), 不难知晓

&#x02022; 特征值问题 (1.4) 仅有离散谱, 谱中的元素 (即特征值) 可以按照单调不减的方式排成如下的序列

0<\lambda_{1, \phi}(\Omega)<\lambda_{2, \phi}(\Omega) \leq \lambda_{3, \phi}(\Omega) \leq \cdots \rightarrow \infty,

并且序列中的每个特征值 \lambda_{i,\phi}(\Omega) 重复其重数次, 这里 \lambda_{i, \phi}(\Omega) 的重数就是其特征空间的维数 (是有限维的). 利用变分原理, 第 (k+1) 个特征值 \lambda_{k+1}(\Omega), k=1,2,3, \cdots, 可以被如下刻画

\lambda_{k+1, \phi}(\Omega)=\left\{\left.\frac{\int_{\Omega}|\nabla u|^2 {\rm e}^{-\phi} {\rm d} v}{\int_{\Omega} u^2 {\rm e}^{-\phi} {\rm d} v} \right\rvert\, u \in \widetilde{W}_0^{1,2}(\Omega), u \neq 0, \int_{\Omega} u u_i {\rm e}^{-\phi}{\rm d} v=0\right\},

这里 u_i, i=1,2, \cdots, k, 是属于特征值 \lambda_{i, \phi}(\Omega) 的特征函数. 特别地, 第一个 Dirichlet 特征值的变分刻画为

\lambda_{1, \phi}(\Omega)=\left\{\left.\frac{\int_{\Omega}|\nabla u|^2 {\rm e}^{-\phi} {\rm d} v}{\int_{\Omega} u^2 {\rm e}^{-\phi} {\rm d} v} \right\rvert\, u \in \widetilde{W}_0^{1,2}(\Omega), u \neq 0\right\}.

以上这些事实在某些论文里有提及, 如文献 [p2276-2277].

本文集中考虑 (当 M^n 取成是 n-维欧氏空间 \mathbb{R}^n 时) Dirichlet 特征值问题 (1.4), 并试图将文献 [6] 的结论拓展到加权 Laplace 算子的情形.

本文主要是给出了 \mathbb{R}^2\mathbb{R}^n(n \geq 3) 上的加权 Laplace 算子 Dirichlet 特征值问题的一个万有不等式. 设 \Omega 是欧氏空间 \mathbb{R}^n 中的有界带边区域, 其边界 \partial\Omega 是逐段光滑的, \phi 是定义在 \Omega 上的光滑实值函数且 {\rm d}\mu = {{\rm e}^{ - \phi }}{\rm d} v , 此时 {\rm d}v \Omega 上由欧氏空间中标准度量所诱导的体积元. 用 \Delta\nabla 分别表示欧氏空间上的 Laplace 算子和梯度算子, 可定义加权 Laplace 算子 \mathbb {L}_\phi= \Delta - \left\langle {\nabla \phi, \nabla \cdot } \right\rangle .

考虑 \mathbb{R}^n 中有界区域 \Omega 上加权 Laplace 算子的 Dirichlet 特征值问题

\begin{equation}\label{jld} \left\{\begin{aligned} \begin{array}{ll} \mathbb {L}_\phi u+\lambda_\phi u = 0\qquad\quad &\text{在} \ \Omega\ \text{内}, \\ u = 0\qquad\qquad\quad\quad & \text{在} \ \partial\Omega \ \text{上}. \\ \end{array} \end{aligned}\right. \end{equation}
(1.5)

我们知道特征值问题 (1.5) 有离散谱, 且谱中的元素 (即特征值) 可按照单调不减的方式排成如下的序列

0 < {\lambda _{1,\phi}} < {\lambda _{2,\phi}} \le {\lambda _{3,\phi}}\le \cdots \le {\lambda _{k,\phi}} \le \cdots \to \infty,

其中每一个 {\lambda _{i,\phi}} 是实值的并且重复其重数次. 设 {u_i} 是对应 {\lambda _{i,\phi}} 的特征函数.

本文运用变分法, 通过构造恰当的测试函数, 得到了如下几个结论

定理1.1 假设 \Omega\mathbb{R}^2 中有界区域, 考虑 \Omega 上的加权 Laplace 算子的 Dirichlet 特征值问题 (1.5). 若取 \phi=-C{\left|X\right|^2}, 其中 C 是常数, 则我们有

(i) 当 C<0 时, 如果 \frac{{{\lambda_{2,\phi}}}}{{{\lambda_{1,\phi}}}}<2-\frac{{{\lambda_{1,\phi}}}}{{{\lambda _{3,\phi}}}}, 则有

\label{eq-3} \frac{{{\lambda _{3,\phi}}}}{{{\lambda _{1,\phi}}}} < 5 - \frac{16C}{{{\lambda _{1,\phi}}}}.

\frac{{{\lambda _{2,\phi}}}}{{{\lambda _{1,\phi}}}} \ge 2 - \frac{{{\lambda _{1,\phi}}}}{{{\lambda _{3,\phi}}}} 时, 有

\label{eq-4} \frac{{{\lambda _{3,\phi }} + {\lambda _{2,\phi }}}}{{{\lambda _{1,\phi }}}} \le 6 - \frac{{16C}}{{{\lambda _{1,\phi }}}}.

(ii) 当 C>0 时, 如果 \frac{{{\lambda _{2,\phi}}}}{{{\lambda _{1,\phi}}}} < 2 - \frac{{{\lambda _{1,\phi}}}}{{{\lambda _{3,\phi}}}}, 则有

\label{eq-5} \frac{{{\lambda _{3,\phi}}}}{{{\lambda _{1,\phi}}}} < 5.

\frac{{{\lambda _{2,\phi}}}}{{{\lambda _{1,\phi}}}} \ge 2 - \frac{{{\lambda _{1,\phi}}}}{{{\lambda _{3,\phi}}}}, 有

\begin{matrix}\label{eq-6} {\lambda _{3,\phi}} + {\lambda _{2,\phi}} \le 6{\lambda _{1,\phi}}. \end{matrix}
(1)

注1.1 研究某些泛函的极值问题具有重要的意义. n-维欧氏空间中经典的等周问题之一就是研究 \mathbb{R}^n 中体积固定的区域的面积泛函的极值, 即

F\left( \Omega \right) = \min\limits_\Omega \left\{ {\left| {\partial \Omega } \right|;\left| \Omega \right| = \text{常数}} \right\},

其中 \Omega\mathbb{R}^n 中的有界区域. 这个问题的答案便是经典的几何等周不等式:

定理1.2[5]\Omega\subset \mathbb{R}^n 是固定体积的有界带边区域, 设 B\mathbb{R}^n 中体积等于 |\Omega| 的球体, 则我们有 |\partial \Omega|\geq |\partial B|, 其中等号成立当且仅当 \Omega=B.

类似的, 针对 \mathbb{R}^{n} 中固定体积的有界区域 \Omega 上 Laplace 算子的 Dirichlet 特征值问题, 我们可以定义相应的特征值泛函, 并考虑对应的特征值泛函的极值问题, 即

{F_k}\left( \Omega \right) = \min \limits_\Omega\left\{ {{\lambda _k(\Omega)};\left| \Omega \right| = \text{常数}} \right\},

我们把由上述极值问题得到的不等式称为谱等周不等式. 特别地, 对于 k=1 的情形, Faber [3] 和 Krahn [15] 证明了如下的谱等周不等式: 假设 \Omega\mathbb{R}^n 中固定体积的有界区域, 考虑 Laplace 算子的 Dirichlet 特征值问题 (1.1), 对于 \Omega 上的第一个特征值 {\lambda _1}\left( \Omega \right) {\lambda _1}\left( \Omega \right) \ge {\lambda _1}\left( B\right) 成立, 当且仅当 \Omega=B 时等式成立, 其中 B 是一个体积等于 \left| \Omega \right| 的球体, {\lambda _1}(B) 是球体 B 上 Laplace 算子的第一个 Dirichlet 特征值.

类似的, 对于加权 Laplace 算子的 Dirichlet 特征值可以提出类似的极值问题, 设 \Omega\subset\mathbb{R}^n 是加权体积固定的有界带边区域, 我们可以考虑如下特征值泛函的极值问题:

{F_k}\left( \Omega \right) = \min\limits_\Omega \left\{ {{\lambda _{k,\phi}};\left| \Omega \right|_\phi = \text{常数}} \right\},

其中 {\left| \Omega \right|_\phi }{\rm{ = }}\int_\Omega {{{\rm e}^{ - \phi }}} {\rm d}v, 我们称 {\left| \Omega \right|_\phi }\Omega 的加权体积.

考虑上述泛函极值问题也是有意义的, Morgan 及其合作者在 2008 年文献[16]证明了下面的谱等周不等式:

定理1.3[23] 假设 \Omega\mathbb{R}^n 中加权体积固定的有界区域, 即 {\left| \Omega\right|_\phi} 是常数, \phi = - C{\left| X \right|^2}, 其中 C 为非负常数, X\Omega 所对应的位置向量, 考虑特征值问题 (1.5), 我们有

{\lambda _{1,\phi}}\left( \Omega \right) \ge {\lambda _{1,\phi}}\left({B_R}\left( o \right) \right)

等号成立当且仅当 \Omega = {B_R}\left( o \right), 其中 {B_R}(o) 是一个以 o 为圆心、半径为 R、加权体积等于 {\left| \Omega \right|_\phi } 的球. {\lambda _{1,\phi}}\left( {B_R}\left( o \right) \right) 是球 {B_R}\left( o \right) 上加权 Laplace 算子的第一 Dirichlet 特征值.

接下来我们可以利用定理1.3, 对不等式(1.6) 的左边求最小值, 可以得到

\mathop {\rm min}\limits_\Omega ({\lambda _{3,\phi }}\left( \Omega \right) + {\lambda _{2,\phi }}\left( \Omega \right)) \le 6{\lambda _{1,\phi }}\left( \Omega \right),

进一步有

\mathop {\rm min}\limits_\Omega ({\lambda _{3,\phi }}\left( \Omega \right) + {\lambda _{2,\phi }}\left( \Omega \right)) \le 6\mathop {\rm min}\limits_\Omega {\lambda _{1,\phi }}\left( \Omega \right).

于是在满足定理 1.3 的相关假设条件下, 我们得到 \min \limits_\Omega{\lambda _{1,\phi }}\left( \Omega \right){\rm{ = }}{\lambda _{1,\phi }}\left({B_R}\left( o \right)\right), 于是不等式 (1.6) 可以变为

\mathop {\rm min}\limits_\Omega ({\lambda _{3,\phi }}\left( \Omega \right) + {\lambda _{2,\phi }}\left( \Omega \right)) \le 6{\lambda _{1,\phi }}\left( B_R(o) \right).

注1.2 当空间是 ( {{\mathbb{R}^2},g,{{\rm e}^{ - \frac{{{{\left| X \right|}^2}}}{4}}}{\rm d} v} ), 体积元为 {\rm d}v = {\rm d}x{\rm d}y, 此时的欧氏度量为 g ={\rm d}{x^2} +{\rm d}{y^2}. 这时有 {\rm Ric}_\phi={\rm Ric}+{\rm Hess } \phi=\frac{1}{2}g, 故而该空间就可以看成是 Einstein 流形的拓展. 事实上, 给定一个光滑度量测度空间, 若满足 \rm Ri{c_\phi } = \lambda g, 其中 \lambda 是常数, 那么该空间称为 Ricci 孤立子, 可以看成是 Einstein 流形的一种推广.

对于 \mathbb{R}^n(n\geq 3) 的情况, 我们有下面的结论:

定理1.4 假设 \Omega\mathbb{R}^n 中的有界区域, 边界 \partial\Omega 是逐段光滑的, 若取 \phi = -C{\left| X \right|^2}, 其中 C 是常数.

C<0 时, 则有: \frac{{{\lambda_{2,\phi}} + \cdots + {\lambda _{n + 1,\phi}}}}{{{\lambda_{1,\phi}}}} \le \left( n + 4 \right) - \frac{{8nC}}{{{\lambda_{1,\phi}}}};

C>0 时, 则有: \frac{{{\lambda _{2,\phi}} + \cdots + {\lambda _{n + 1,\phi}}}}{{{\lambda _{1,\phi}}}} \le n + 4.

注1.3 类似的, 我们利用定理 1.3 对定理 1.4 (iv) 式的左边求最小值可以得到

\mathop {\rm min}\limits_\Omega ({\lambda _{2,\phi }}\left( \Omega \right) + \cdots + {\lambda _{n + 1,\phi }}\left( \Omega \right)) \le \left( {n + 4} \right)\mathop {\rm min}\limits_\Omega {{\lambda _{1,\phi }}\left( \Omega \right)}.

于是根据定理 1.3, 我们得到 \min \limits_\Omega{\lambda _{1,\phi }}\left( \Omega \right){\rm{ = }}{\lambda _{1,\phi }}\left({B_R}\left( o \right)\right), 于是不等式 (iv) 可以变为:

\mathop {\rm min}\limits_\Omega ({\lambda _{2,\phi }}\left( \Omega \right) + \cdots + {\lambda _{n + 1,\phi }}\left( \Omega \right)) \le \left( {n + 4} \right)\lambda _{1,\phi}(B_R(o)).

注1.4 我们近期注意到曾令忠老师和孙和军老师在文献[20] 中考虑了等距浸入到高维欧氏空间中的完备光滑度量测度空间里的有界区域上加权 Laplace 算子的 Dirichlet 特征值问题, 并且借助于该度量测度空间作为欧氏空间子流形时的平均曲率、加权函数的梯度和 Laplacian, 他们成功地得到了 Dirichlet 特征值所满足的一个特征值不等式, 将 M Levitin 和 L Parnovski 的重要工作[12]拓展到了加权 Laplace 算子的情形. 此外, 利用得到的特征值不等式, 他们还成功得到了当完备光滑度量测度空间取更为特殊的梯度 Ricci 孤立子、自收缩子、柱体等几何空间时加权 Laplace 算子的 Dirichlet 特征值所满足的若干万有不等式. 值得指出的是, 如果文献 [20,定理 1.4]中的完备自收缩子就取成是 n-维欧氏空间 \mathbb{R}^n, 当加权函数进一步取成 \phi=-C|X|^2=\frac{1}{2}|X|^2, 本文定理 1.4 中的万有不等式是可以被文献[27,定理 1.4] 所包含的, 但是本文其它情形下的万有不等式利用文献 [20] 中的技术手段和结论是得不到的, 特别是在定理 1.1 中的假设情形 \frac{\lambda_{2,\phi}}{\lambda_{1,\phi}}<2-\frac{\lambda_{1,\phi}}{\lambda_{3,\phi}} 下, 我们的结果是更优的.

2 定理 1.1 的证明

为了方便后续的计算, 首先给出一个引理:

引理2.1\Omega 是欧氏空间 \mathbb{R}^{n} 上的有界区域, 其边界 \partial\Omega 是分段光滑, 则对于 \Omega 上任意的光滑函数 f,g,\phi\in {C^\infty }\left( M \right), f\left| {_{\partial M} = } \right.g\left| {_{\partial M} = } \right.0, 有

\int_\Omega {\left\langle {\nabla f,\nabla g} \right\rangle } {\rm d}\mu = - \int_\Omega {g{\mathbb {L}_\phi }f} {\rm d}\mu = - \int_\Omega {f{\mathbb {L}_\phi }g} {\rm d}\mu,

其中{\rm d}\mu = {{\rm e}^{ - \phi }}{\rm d} v .

\mathbb {L}_\phi 的定义以及 Green 公式和 Dirichlet 边值条件

\begin{equation*} \begin{aligned} \int_\Omega {f{\mathbb {L}_\phi }g} {\rm d}\mu =&\int_\Omega {f{{\rm e}^{ - \phi }}\Delta g{\rm d}v - \int_\Omega {f{{\rm e}^{ - \phi }}\left\langle {\nabla \phi,\nabla g} \right\rangle } }{\rm d}v \\ =&- \int_\Omega {\left\langle {\nabla \left( {f{{\rm e}^{ - \phi }}} \right),\nabla g} \right\rangle }{\rm d}v - \int_\Omega {f{{\rm e}^{ - \phi }}\left\langle {\nabla \phi,\nabla g} \right\rangle } {\rm d}v \\ =& - \int_\Omega {\left\langle {\nabla f,\nabla g} \right\rangle } {\rm d}\mu. \end{aligned} \end{equation*}

在接下来的计算中, 为了不引起混淆, 我们有时会把 \left\langle ~~~,~~~ \right\rangle 记成 \cdot 或省略不写. 对于平面上的加权 Laplace 算子的 Dirichlet 特征值问题 (1.5) (n=2), 设 u_{k} 是对应于特征值 \lambda_{k,\phi} 的标准特征函数, 即 u_{k} 满足

\begin{eqnarray*} \left\{\begin{aligned} &{\mathbb {L}_\phi }{u_k}+ {\lambda _{k,\phi }}{u_k} = 0\qquad\quad&& \text{在} \ \Omega\ \text{上},\\ &{u_k}= 0\qquad\qquad\quad\quad\quad\quad &&\text{在} \ \partial\Omega\ \text{上}, \\ &\int_\Omega {{u_i}{u_j}{\rm d}\mu = {\delta _{ij}}}. \end{aligned}\right. \end{eqnarray*}

从引理 2.1 及类似于文献 [6] 的计算方法, 我们得到下面的 (2.1)-(2.10) 式, 即: 对于 h \in {C^2}(\Omega ) \cap C(\bar \Omega ), 定义

\begin{eqnarray*} \left\{\begin{aligned} &{{a_{ij}} = \int_\Omega h {u_i}{u_j}{\rm d}\mu },\\ &{{b_{ij}} = \int_\Omega {{u_j}} \nabla h\nabla {u_i}{\rm d}\mu },\\ &{{d_{ij}} = \int_\Omega {{u_i}} {u_j}{\mathbb {L}_\phi }h{\rm d}\mu }. \end{aligned}\right. \end{eqnarray*}

由 Green 公式, 我们有

\begin{matrix}\label{2eq-1} &2b_{ij}=(\lambda_{i,\phi}-\lambda_{j,\phi})a_{ij}-d_{ij} \end{matrix}
(2.1)

\begin{matrix}\label{2eq-2} -2\int_{\Omega} {hu_{i}\nabla h\nabla u_{i}}{\rm d}\mu=\int_\Omega {u_i^2h{\mathbb {L}_\phi }h{\rm d}\mu }+\int_{\Omega}{|\nabla h|^{2}u_{i}^{2}}{\rm d}\mu. \end{matrix}
(2.2)

因为 \{u_{i}\}_{i=1}^\inftyL^{2}(\Omega) 上完备的标准正交基, 我们有

\begin{eqnarray*} \begin{aligned} - 2\int_\Omega {h{u_i}\nabla h\nabla {u_i}{\rm d}\mu } {\rm{ }}& = - 2\sum\limits_{j = 1}^\infty {{a_{ij}}{b_{ij}}} {\rm{ }},\\ \int_\Omega {hu_i^2{\mathbb {L}_\phi }h{\rm d}\mu } + \int_\Omega {{{\left| {\nabla h} \right|}^2}} u_i^2{\rm d}\mu &= \sum\limits_{j = 1}^\infty {{a_{ij}}{d_{ij}}} + \int_\Omega {{{\left| {\nabla h} \right|}^2}} u_i^2{\rm d}\mu. \end{aligned} \end{eqnarray*}

从 (2.1) 和 (2.2) 式, 可以得到

\begin{matrix}\label{2eq-3} {\int_\Omega {\left| {\nabla h} \right|} ^2}u_i^2{\rm d}\mu = \sum\limits_{j = 1}^\infty {\left( {{\lambda _{j,\phi}} - {\lambda _{i,\phi}}} \right)a_{ij}^2}. \end{matrix}
(2.3)

由(2.3) 式知

\sum\limits_{j = 1}^\infty {\left( {{\lambda _{j,\phi }} - {\lambda _{i,\phi }}} \right)a_{ij}^2} \geq-\sum\limits_{j = 1}^k {\left( {{\lambda _{k+1,\phi}} - {\lambda _{j,\phi}}} \right)a_{ij}^2 + \left( {{\lambda _{k + 1,\phi}} - {\lambda _{i,\phi}}} \right)\sum\limits_{j = 1}^\infty {a_{ij}^2} },

\begin{matrix}\label{2eq-4} \begin{aligned} \left( {{\lambda _{k + 1,\phi }} - {\lambda _{i,\phi }}} \right)\sum\limits_{j = 1}^\infty {a_{ij}^2}\leq \int_\Omega {{{\left| {\nabla h} \right|}^2}} u_i^2{\rm d}\mu + \sum\limits_{j = 1}^k {\left( {{\lambda _{k + 1,\phi }} - {\lambda _{j,\phi }}} \right)a_{ij}^2}.\\ \end{aligned} \end{matrix}
(2.4)

x, y\mathbb{R}^2 上的坐标函数, 则 {u_x} = \frac{{\partial {u_1}}}{{\partial x}},{u_y} = \frac{{\partial {u_1}}}{{\partial y}}. 同时, 对于 j = 1, 2,\cdots, 以及实数 \alpha>\frac{1}{2}, 定义

\begin{eqnarray*} \begin{aligned} &{a_j} = \int_\Omega {x{u_1}} {u_j}{\rm d}\mu, {b_j} = \int_\Omega {y{u_1}} {u_j}{\rm d}\mu, {d_j} = \int_\Omega {{u_1^\alpha }} {u_j}{\rm d}\mu, \\ &{A_j} = \int_\Omega {{u_x}} {u_j}{\rm d}\mu, {B_j} = \int_\Omega {{u_y}} {u_j}{\rm d}\mu. a_{ij}=\int_\Omega h u_i u_j {\rm d}\mu. \end{aligned} \end{eqnarray*}

由 (2.1) 式和 \Delta x = \Delta y = 0, 我们有

\begin{matrix}\label{2eq-5} \begin{aligned} 2{A_j} &= \left( {{\lambda _{1,\phi}} - {\lambda _{j,\phi}}} \right){a_j} + \int_\Omega {{u_1}{u_j}{\phi _x}{\rm d}\mu }, \\ 2{B_j} &= \left( {{\lambda _{1,\phi}} - {\lambda _{j,\phi}}} \right){b_j} + \int_\Omega {{u_1}{u_j}{\phi _y}{\rm d}\mu }. \end{aligned} \end{matrix}
(2.5)

我们可以选择合适的坐标系使得 {a_1}={a_2}={b_1}=0, 由 {u_1}\Omega 上不变号, 不失一般性, 我们假设在 \Omega {u_1} 恒为正.

h 分别为 xy 代入 (2.3) 和 (2.4) 式, 我们可以得到

\begin{equation}\label{2eq-6} \left\{\begin{aligned} &\sum\limits_{j = 3}^\infty {\left( {{\lambda _{j,\phi}} - {\lambda _{1,\phi}}} \right)a_j^2 = 1},\\ &\left( {{\lambda _{k+1,\phi}} - {\lambda _{1,\phi}}} \right)\sum\limits_{j = 3}^\infty {a_j^2 \le 1 + \sum\limits_{j = 3}^k {\left( {{\lambda _{k + 1,\phi}} - {\lambda _{j,\phi}}} \right)} } a_j^2; \end{aligned}\right. \end{equation}
(2.6)
\begin{equation}\label{2eq-7} \left\{\begin{aligned} &\sum\limits_{j = 2}^\infty {\left( {{\lambda _{j,\phi}} - {\lambda _{1,\phi}}} \right)b_j^2 = 1},\\ &\left( {{\lambda _{l + 1,\phi}} - {\lambda _{1,\phi}}} \right)\sum\limits_{j = 2}^\infty {b_j^2 \le 1 + \sum\limits_{j = 2}^l {\left( {{\lambda _{l + 1,\phi}} - {\lambda _{j,\phi}}} \right)} } b_j^2. \end{aligned}\right. \end{equation}
(2.7)

接着取 h = {u_1^{\alpha - 1}}, 其中 \alpha > \frac{1}{2} 是实常数. 由 Green 公式, 可以得到

\begin{matrix}\label{2eq-8} & {\int_\Omega {\left| {\nabla {u_1^{(\alpha - 1)}}} \right|} ^2}{u_1^2}{\rm d}\mu = \frac{{{{\left( {\alpha - 1} \right)}^2}}}{{2\alpha - 1}}{\lambda _{1,\phi}}\int_\Omega {{u_1^{2\alpha }}} {\rm d}\mu = \sum\limits_{j = 2}^\infty {\left( {{\lambda _{j,\phi}} - {\lambda _{1,\phi}}} \right)} d_j^2. \end{matrix}
(2.8)

{\beta _j} = \frac{{{d_j}}}{{\sqrt {\frac{{{{\left( {\alpha - 1} \right)}^2}}}{{2\alpha - 1}}{\lambda _{1,\phi}}\int_\Omega {{u_1^{2\alpha }}} {\rm d}\mu } }}, j = 2, \cdots,

然后, 通过 (2.8) 式可以得到

\begin{matrix}\label{2eq-9} \sum\limits_{j = 2}^\infty {\left( {{\lambda _{j,\phi}} - {\lambda _{1,\phi}}} \right)} \beta _j^2 = 1. \end{matrix}
(2.9)

再根据 (2.9) 式可得

\begin{matrix}\label{2eq-10} & \left( {{\lambda _{t + 1,\phi}} - {\lambda _{1,\phi}}} \right)\sum\limits_{j = 2}^\infty {\beta _j^2} \le 1 + \sum\limits_{j = 2}^t {\left( {{\lambda _{t + 1,\phi}} - {\lambda _{j,\phi}}} \right)} \beta _j^2. \end{matrix}
(2.10)

A\left( \alpha \right) = \frac{{\int_\Omega {{u_1^{2\alpha }}{\rm d}\mu } }}{{{{\left( {\int_\Omega {{u_1^{\alpha + 1}}} {\rm d}\mu } \right)}^2}}} = \frac{1}{{1 - \frac{{{{\left( {\alpha - 1} \right)}^2}}}{{2\alpha - 1}}{\lambda _{1,\phi}}\sum\limits_{j = 2}^\infty {\beta _j^2}}},

通过 Green 公式, 有

\begin{eqnarray*} {d_1}&= - \int_\Omega {\left( {\alpha + 1} \right)x{u^\alpha }{u_x}} {\rm d}\mu + \int_\Omega {x{u^{\alpha + 1}}{\phi _x}} {\rm d}\mu\\& = \int_\Omega {xu_1\left( { - \left( {\alpha + 1} \right){u_1^{\alpha - 1}}{u_x} + {u_1^\alpha }{\phi _x}} \right)} {\rm d}\mu. \end{eqnarray*}

同理可得

{d_1}= \int_\Omega {yu_1\left( { - \left( {\alpha + 1} \right){u_1^{\alpha - 1}}{u_y} + {u_1^\alpha }{\phi _y}} \right)} {\rm d}\mu.

由 Schwarz 不等式, 得

\begin{matrix}\label{2eq-11} & d_1^2 \le {\int_\Omega {\left( {xu_1} \right)} ^2}{\rm d}\mu {\int_\Omega {\left( { - \left( {\alpha + 1} \right){u_1^{\alpha - 1}}{u_x} + {u_1^\alpha }{\phi _x}} \right)} ^2}{\rm d}\mu \end{matrix}
(2.11)

\begin{matrix}\label{2eq-12} & d_1^2 \le {\int_\Omega {\left( {yu_1} \right)} ^2}{\rm d}\mu {\int_\Omega {\left( { - \left( {\alpha + 1} \right){u_1^{\alpha - 1}}{u_y} + {u_1^\alpha }{\phi _y}} \right)} ^2}{\rm d}\mu. \end{matrix}
(2.12)

再次运用 Green 公式, 我们发现

\begin{matrix}\label{2eq-13} & \int_\Omega {{u_1^{2\alpha - 2}}\left( {u_x^2 + u_y^2} \right)} {\rm d}\mu = \int_\Omega {\nabla u_1 \cdot \nabla \left( {\frac{{{u_1^{2\alpha - 1}}}}{{2\alpha - 1}}} \right)} {\rm d}\mu = \frac{{{\lambda _{1,\phi}}}}{{2\alpha - 1}}\int_\Omega {{u_1^{2\alpha }}} {\rm d}\mu. \end{matrix}
(2.13)

有了以上的准备, 我们先建立下列引理.

引理2.2\Omega\mathbb{R}^2 上的有界区域, \left| {\nabla \phi } \right| \le M (M 是常数), 则对于加权 Laplace 算子的 Dirichlet 特征值问题 (1.5) (n=2), 有

\begin{eqnarray*}\label{2eq-14} \begin{aligned} &\frac{{{\lambda _{k + 1,\phi}} - {\lambda _{1,\phi}}}}{{1 + \sum\limits_{j = 3}^k {\left( {{\lambda _{k + 1,\phi}} - {\lambda _{j,\phi}}} \right)} a_j^2}} + \frac{{{\lambda _{l + 1,\phi}} - {\lambda _{1,\phi}}}}{{1 + \sum\limits_{j = 2}^l {\left( {{\lambda _{l + 1,\phi}} - {\lambda _{j,\phi}}} \right)} b_j^2}}\\ \le \,& \frac{{{{\left( {\alpha + 1} \right)}^2}{\lambda _{1,\phi }}}}{{2\alpha - 1}}\frac{{\int_\Omega {{u_1^{2\alpha }}} {\rm d}\mu }}{{{{\left( {\int_\Omega {{u_1^{\alpha + 1}}} {\rm d}\mu } \right)}^2}}} + \frac{{\int_\Omega {{u_1^{2\alpha }}{{\left| {\nabla \phi } \right|}^2}} {\rm d}\mu }}{{{{\left( {\int_\Omega {{u_1^{\alpha + 1}}} {\rm d}\mu } \right)}^2}}} + \frac{{\alpha + 1}}{\alpha }\frac{{\int_\Omega {{u_1^{2\alpha - 1}}{\mathbb {L}_\phi }\phi } {\rm d}\mu }}{{{{\left( {\int_\Omega {{u_1^{\alpha + 1}}} {\rm d}\mu } \right)}^2}}}, \end{aligned} \end{eqnarray*}

\begin{equation*}\label{2eq-15} \int_\Omega {{u_1^\alpha }{u_y}} {\rm d}\mu = \sum\limits_{j = 1}^\infty {{d_j}{B_j}} = \frac{1}{{1 + \alpha }}\int_\Omega {{u_1^{\alpha + 1}}{\phi _y}} {\rm d}\mu \end{equation*}

和 (2.5) 式知

\begin{equation}\label{2eq-16} \sum\limits_{j = 2}^\infty {\left( {{\lambda _{j,\phi}} - {\lambda _{1,\phi}}} \right)} {\beta _j}{b_j} = \frac{{\frac{{\alpha - 1}}{{\alpha + 1}}\int_\Omega {{u_1^{\alpha + 1}}{\phi _y}} {\rm d}\mu }}{{\sqrt {\frac{{{{\left( {\alpha - 1} \right)}^2}}}{{2\alpha - 1}}{\lambda _{1,\phi}}\int_\Omega {{u_1^{2\alpha }}} {\rm d}\mu } }}. \end{equation}
(2.14)

利用 (2.7), (2.9), (2.14) 式和 Schwarz 不等式, 有

\begin{equation}\label{2eq-17} \begin{split} \left( {{\lambda _{2,\phi}} - {\lambda _{1,\phi}}} \right)\beta _2^2 & \le 1 - \left( {{\lambda _{2,\phi }} - {\lambda _{1,\phi }}} \right)b_2^2 + \frac{{\left( {2\alpha - 1} \right)\int_\Omega {{u_1^{2\alpha }}} {\rm d}\mu \int_\Omega {{u_1^2}\phi _y^2} {\rm d}\mu }}{{{{\left( {1 + \alpha } \right)}^2}{\lambda _{1,\phi }}\int_\Omega {{u_1^{2\alpha }}} {\rm d}\mu }}\\ &= 1 - \left( {{\lambda _{2,\phi}} - {\lambda _{1,\phi}}} \right){ b_2^2} + \frac{{\left( {2\alpha - 1} \right){M^2}}}{{{{\left( {1 + \alpha } \right)}^2}{\lambda _{1,\phi}}}}. \end{split} \end{equation}
(2.15)

对于 j\ge3, 做正交变换 \widetilde x = ax + by, \widetilde y = cx + {\rm d}y, 其中 ad - bc= 1, 使得

\widetilde {{a_j}} = \int_\Omega{\widetilde xu_1{u_j}} {\rm d}\mu=0, \widetilde {{b_j}} = \int_\Omega {\widetilde yu_1{u_j}} {\rm d}\mu = \sqrt {a_j^2 + b_j^2}.

与 (2.14) 式的做法类似, 有

\begin{equation}\label{2eq-18} \sum\limits_{j = 2}^\infty {\left( {{\lambda _{j,\phi}} - {\lambda _{1,\phi}}} \right)} {\beta _j}\widetilde{{b_j}} = \frac{{\frac{{\alpha - 1}}{{\alpha + 1}}\int_\Omega {{u_1^{\alpha + 1}}\left( {c{\phi _x} + d{\phi _y}} \right)} {\rm d}\mu }}{{\sqrt {\frac{{{{\left( {\alpha - 1} \right)}^2}}}{{2\alpha - 1}}{\lambda _{1,\phi}}\int_\Omega {{u_1^{2\alpha }}} {\rm d}\mu } }}. \end{equation}
(2.16)

于是,

\begin{matrix}\label{2eq-19} \left( {{\lambda _{j,\phi}} - {\lambda _{1,\phi}}} \right)\beta _j^2 \le 1 - \left( {{\lambda _{j,\phi}} - {\lambda _{1,\phi}}} \right){\widetilde {{b_j}}^2} + \frac{{\left( {2\alpha - 1} \right){M^2}}}{{{{\left( {1 + \alpha } \right)}^2}{\lambda _{1,\phi}}}}. \end{matrix}
(2.17)

\widetilde y = cx + {\rm d}y 代入到 (2.3) 式, 有

\begin{eqnarray*}\label{2eq-20} \begin{aligned} {\int_\Omega {\left| {\nabla \left( {cx + {\rm d}y} \right)} \right|} ^2}{u_1^2}{\rm d}\mu = \sum\limits_{j = 1}^\infty {\left( {{\lambda _{j,\phi}} - {\lambda _{1,\phi}}} \right)} {\left( {\int_\Omega {\left( {cx + {\rm d}y} \right)u_1{u_j}} {\rm d}\mu } \right)^2}=1. \end{aligned} \end{eqnarray*}

结合 (2.6), (2.7), (2.11)-(2.13), (2.16) 和 (2.17) 式可得

\begin{equation*}\label{2eq-21} \begin{aligned} &\frac{{{\lambda _{k + 1,\phi}} - {\lambda _{1,\phi}}}}{{1 + \sum\limits_{j = 3}^k {\left( {{\lambda _{k + 1,\phi}} - {\lambda _{j,\phi}}} \right)} a_j^2}} + \frac{{{\lambda _{l + 1,\phi}} - {\lambda _{1,\phi}}}}{{1 + \sum\limits_{j = 2}^l {\left( {{\lambda _{l + 1,\phi}} - {\lambda _{j,\phi}}} \right)} b_j^2}}\\ \le \,& \frac{{{\lambda _{k + 1,\phi}} - {\lambda _{1,\phi}}}}{{\left( {{\lambda _{k + 1,\phi}} - {\lambda _{1,\phi}}} \right)\sum\limits_{j = 3}^\infty {a_j^2} }} + \frac{{{\lambda _{l + 1,\phi}} - {\lambda _{1,\phi}}}}{{\left( {{\lambda _{l + 1,\phi}} - {\lambda _{1,\phi}}} \right)\sum\limits_{j = 2}^\infty {b_j^2} }}\\ \le \,& \frac{{{{\int_\Omega {\left( { - \left( {\alpha + 1} \right){u_1^{\alpha - 1}}{u_x} + {u_1^\alpha }{\phi _x}} \right)} }^2}{\rm d}\mu + {{\int_\Omega {\left( { - \left( {\alpha + 1} \right){u_1^{\alpha - 1}}{u_y} + {u_1^\alpha }{\phi _y}} \right)} }^2}{\rm d}\mu }}{{d_1^2}}\\ \le \,& \frac{{{{\left( {\alpha + 1} \right)}^2}{\lambda _{1,\phi }}}}{{2\alpha - 1}}\frac{{\int_\Omega {{u_1^{2\alpha }}} {\rm d}\mu }}{{{{\left( {\int_\Omega {{u_1^{\alpha + 1}}} {\rm d}\mu } \right)}^2}}} + \frac{{\int_\Omega {{u_1^{2\alpha }}{{\left| {\nabla \phi } \right|}^2}} {\rm d}\mu }}{{{{\left( {\int_\Omega {{u_1^{\alpha + 1}}} {\rm d}\mu } \right)}^2}}} + \frac{{\alpha + 1}}{\alpha }\frac{{\int_\Omega {{u_1^{2\alpha}}{\mathbb {L}_\phi }\phi } {\rm d}\mu }}{{{{\left( {\int_\Omega {{u_1^{\alpha + 1}}} {\rm d}\mu } \right)}^2}}}. \end{aligned} \end{equation*}

引理2.3 对于加权 Laplace 算子的 Dirichlet 特征值问题 (1.5) (n=2), 加权函数 \phi = -C{\left| X \right|^2}, 其中 X\Omega 的位置向量. 则当 C>0 时, 有

\frac{{{\lambda _{k + 1,\phi }} - {\lambda _{1,\phi }}}}{{1 + \sum\limits_{j = 3}^k {\left( {{\lambda _{k + 1,\phi }} - {\lambda _{j,\phi }}} \right)} a_j^2}} + \frac{{{\lambda _{l + 1,\phi }} - {\lambda _{1,\phi }}}}{{1 + \sum\limits_{j = 2}^l {\left( {{\lambda _{l + 1,\phi }} - {\lambda _{j,\phi }}} \right)} b_j^2}}\le 4{\lambda _{1,\phi }} - \frac{{{\lambda _{1,\phi }}\left( {{\lambda _{*,\phi }} - {\lambda _{1,\phi }}} \right)}}{{{\lambda _{*,\phi }}}};

C<0 时, 有

\frac{{{\lambda _{k + 1,\phi }} - {\lambda _{1,\phi }}}}{{1 + \sum\limits_{j = 3}^k {\left( {{\lambda _{k + 1,\phi }} - {\lambda _{j,\phi }}} \right)} a_j^2}} + \frac{{{\lambda _{l + 1,\phi }} - {\lambda _{1,\phi }}}}{{1 + \sum\limits_{j = 2}^l {\left( {{\lambda _{l + 1,\phi }} - {\lambda _{j,\phi }}} \right)} b_j^2}} \le 4{\lambda _{1,\phi }} - \frac{{{\lambda _{1,\phi }}\left( {{\lambda _{*,\phi }} - {\lambda _{1,\phi }}} \right)}}{{{\lambda _{*,\phi }}}}-16C,

其中

{\lambda _ {*,\phi} } - {\lambda _{1,\phi}} = \frac{{{\lambda _{t + 1,\phi}} - {\lambda _{1,\phi}}}}{{1 + \sum\limits_{j = 2}^t {\frac{{{\lambda _{t + 1,\phi}} - {\lambda _{j,\phi}}}}{{{\lambda _{j,\phi}} - {\lambda _{1,\phi}}}}\left( {1 - \left( {{\lambda _{j,\phi}} - {\lambda _{1,\phi}}} \right)(b_j^2+a_j^2) + \frac{{{M^2}}}{{3{\lambda _{1,\phi}}}}} \right)} }}.

\phi = -C{\left| X \right|^2},0 \le \left| X \right| \le {\rm diam}\left( \Omega \right) , 有

\begin{eqnarray*}\label{2eq-22} \nabla \phi = - 2CX,\Delta \phi = - 4C,{\mathbb {L}_\phi }\phi = \Delta \phi - \left\langle {\nabla \phi, \nabla \phi } \right\rangle = - 4C - 4{C^2}{\left| X \right|^2}. \end{eqnarray*}

这里 \left| {\nabla \phi } \right| \le M,M = 2\left| C \right|{\rm diam}\left( \Omega \right), 再根据引理 2.2 有

\begin{equation*} \begin{aligned} &\frac{{{\lambda _{k + 1,\phi }} - {\lambda _{1,\phi }}}}{{1 + \sum\limits_{j = 3}^k {\left( {{\lambda _{k + 1,\phi }} - {\lambda _{j,\phi }}} \right)} a_j^2}} + \frac{{{\lambda _{l + 1,\phi }} - {\lambda _{1,\phi }}}}{{1 + \sum\limits_{j = 2}^l {\left( {{\lambda _{l + 1,\phi }} - {\lambda _{j,\phi }}} \right)} b_j^2}}\\ \le \,& \frac{{{{\left( {\alpha + 1} \right)}^2}{\lambda _{1,\phi }}}}{{2\alpha - 1}}\frac{{\int_\Omega {{u_1^{2\alpha }}} {\rm d}\mu }}{{{{\left( {\int_\Omega {{u_1^{\alpha + 1}}} {\rm d}\mu } \right)}^2}}} + \frac{{\int_\Omega {4{C^2}{{\left| X \right|}^2}{u_1^{2\alpha }}} {\rm d}\mu }}{{{{\left( {\int_\Omega {{u_1^{\alpha + 1}}} {\rm d}\mu } \right)}^2}}} + \frac{{\alpha {\rm{ + }}1}}{\alpha }\frac{{\int_\Omega {{u_1^{2\alpha }}\left( { - 4C - 4{C^2}{{\left| X \right|}^2}} \right)} {\rm d}\mu }}{{{{\left( {\int_\Omega {{u_1^{\alpha + 1}}} {\rm d}\mu } \right)}^2}}}\\ =\, &\frac{{{{\left( {\alpha + 1} \right)}^2}{\lambda _{1,\phi }}}}{{2\alpha - 1}}\frac{{\int_\Omega {{u_1^{2\alpha }}} {\rm d}\mu }}{{{{\left( {\int_\Omega {{u_1^{\alpha + 1}}} {\rm d}\mu } \right)}^2}}} - \frac{{4{C^2}}}{\alpha }\frac{{\int_\Omega {{u_1^{2\alpha }}{{\left| X \right|}^2}} {\rm d}\mu }}{{{{\left( {\int_\Omega {{u_1^{\alpha + 1}}} {\rm d}\mu } \right)}^2}}} - \frac{{4\left( {\alpha {\rm{ + }}1} \right)C}}{\alpha }\frac{{\int_\Omega {{u_1^{2\alpha }}} {\rm d}\mu }}{{{{\left( {\int_\Omega {{u_1^{\alpha + 1}}} {\rm d}\mu } \right)}^2}}}\\ =\,& 4{\lambda _{1,\phi }} - \frac{{{\lambda _{1,\phi }}\left( {{\lambda _{ *, \phi }} - {\lambda _{1,\phi }}} \right)}}{{{\lambda _{ *, \phi }}}} - 4C\frac{{\left( {{\lambda _{ *, \phi }} + {\lambda _{1,\phi }}} \right)\left( {3{\lambda _{ *, \phi }} - {\lambda _{1,\phi }}} \right)}}{{{\lambda _{ *, \phi }}^2}}, \end{aligned} \end{equation*}

其中 \alpha = \frac{{2{\lambda _ {*,\phi} }}}{{{\lambda _ {*,\phi} } + {\lambda _{1,\phi}}}}, 并且在 \alpha > \frac{1}{2} 情况下, 0 < \frac{{\left( {{\lambda _{*,\phi }} + {\lambda _{1,\phi }}} \right)\left( {3{\lambda _{*,\phi }} - {\lambda _{1,\phi }}} \right)}}{{\lambda _{*,\phi }}^2} < 4.

由上式可以推出

C>0 时, - 4C( {1 +\frac{{{\lambda _{1,\phi }}}}{{{\lambda _{*,\phi }}}}} )( {3 - \frac{{{\lambda _{1,\phi }}}}{{{\lambda _{*,\phi }}}}} ) \le 0;

C<0 时, - 4C( {1+ \frac{{{\lambda _{1,\phi }}}}{{{\lambda _{*,\phi }}}}} )( {3 - \frac{{{\lambda _{1,\phi }}}}{{{\lambda _{*,\phi }}}}} ) \le -16C .

将这两种情况代入上式完成引理的证明.

有了上述准备, 我们下面证明定理 1.1.

\bf{ 定理 1.1 的证明} 令引理 2.3 中的 k = l = t = 2, 有

\begin{matrix}\label{2eq-23} {\lambda _{3,\phi }}{\rm{ - }}{\lambda _{1,\phi }} &\le 4{\lambda _{1,\phi }} - 16C - \left( {{\lambda _{3,\phi }}{\rm{ - }}{\lambda _{1,\phi }}} \right) \times \bigg( \frac{1}{{1 + \left( {{\lambda _{3,\phi }}{\rm{ - }}{\lambda _{2,\phi }}} \right)b_2^2}}\\ &~~~ + \frac{{{\lambda _{1,\phi }}}}{\lambda _{3,\phi }} + {\lambda _{1,\phi }}\frac{{{\lambda _{3,\phi }}{\rm{ - }}{\lambda _{2,\phi }}}}{{{\lambda _{2,\phi }}{\rm{ - }}{\lambda _{1,\phi }}}}\left( {1{\rm{ - }}\left( {{\lambda _{2,\phi }}{\rm{ - }}{\lambda _{1,\phi }}} \right)b_2^2 + \frac{{{M^2}}}{{3{\lambda _{1,\phi }}}}} \right) \bigg). \end{matrix}
(2.18)

由几何平均值不等式, 我们有

\begin{matrix}\label{2eq-24} &\frac{1}{{\frac{{{\lambda _{3,\phi }}}}{{{\lambda _{1,\phi }}}} + \frac{{{\lambda _{3,\phi }} - {\lambda _{2,\phi }}}}{{{\lambda _{2,\phi }} - {\lambda _{1,\phi }}}}\left( {1 - \left( {{\lambda _{2,\phi }} - {\lambda _{1,\phi }}} \right)b_2^2 + \frac{1}{3}\frac{{{M^2}}}{{{\lambda _{1,\phi }}}}} \right)}} + \frac{1}{{1 + \left( {{\lambda _{3,\phi }} - {\lambda _{2,\phi }}} \right)b_2^2}}\\ =\,& \frac{{\frac{{{\lambda _{3,\phi }}}}{{{\lambda _{1,\phi }}}} + \frac{{{\lambda _{3,\phi }} - {\lambda _{1,\phi }}}}{{{\lambda _{2,\phi }} - {\lambda _{1,\phi }}}} + \frac{1}{3}\frac{{{\lambda _{3,\phi }} - {\lambda _{2,\phi }}}}{{{\lambda _{2,\phi }} - {\lambda _{1,\phi }}}}\frac{{{M^2}}}{{{\lambda _{1,\phi }}}}}}{{\left( {1 + \left( {{\lambda _{3,\phi }} - {\lambda _{2,\phi }}} \right)b_2^2} \right)\left( {\frac{{{\lambda _{3,\phi }}}}{{{\lambda _{1,\phi }}}} + \frac{{{\lambda _{3,\phi }} - {\lambda _{2,\phi }}}}{{{\lambda _{2,\phi }} - {\lambda _{1,\phi }}}}\left( {1 - \left( {{\lambda _{2,\phi }} - {\lambda _{1,\phi }}} \right)b_2^2 + \frac{1}{3}\frac{{{M^2}}}{{{\lambda _{1,\phi }}}}} \right)} \right)}} \\ \ge \,& \frac{4}{{\frac{{{\lambda _{3,\phi}}}}{{{\lambda _{1,\phi}}}} + \frac{{{\lambda _{3,\phi}} - {\lambda _{1,\phi}}}}{{{\lambda _{2,\phi }} - {\lambda _{1,\phi}}}} + \frac{1}{3}\frac{{{\lambda _{3,\phi}} - {\lambda _{2,\phi }}}}{{{\lambda _{2,\phi }} - {\lambda _{1,\phi}}}}\frac{{{M^2}}}{{{\lambda _{1,\phi} }}}}}. \end{matrix}
(2.19)

因此, 由 (2.18) 和 (2.19)式, 当 C<0 时, 对于 \frac{{{\lambda _{2,\phi}}}}{{{\lambda _{1,\phi}}}} < 2 - \frac{{{\lambda _{1,\phi}}}}{{{\lambda _{3,\phi}}}}, 我们可以推出

\begin{equation*} \begin{aligned} {\lambda _{3,\phi }} - {\lambda _{1,\phi }}{\rm{ }} &< 4{\lambda _{1,\phi }} - \left( {{\lambda _{3,\phi }} - {\lambda _{1,\phi }}} \right)\frac{4}{{\frac{{{\lambda _{3,\phi }}}}{{{\lambda _{1,\phi }}}} + \frac{{{\lambda _{3,\phi }} - {\lambda _{1,\phi }}}}{{{\lambda _{2,\phi }} - {\lambda _{1,\phi }}}} + \frac{1}{3}\frac{{{\lambda _{3,\phi }} - {\lambda _{2,\phi }}}}{{{\lambda _{2,\phi }} - {\lambda _{1,\phi }}}}\frac{{{M^2}}}{{{\lambda _{1,\phi }}}}}} - 16C\\ &< 4{\lambda _{1,\phi }} - \left( {{\lambda _{3,\phi }} - {\lambda _{1,\phi }}} \right)\frac{4}{{\frac{{{\lambda _{3,\phi }} - {\lambda _{1,\phi }}}}{{{\lambda _{2,\phi }} - {\lambda _{1,\phi }}}}\left( {2 + \frac{1}{3}\frac{{{M^2}}}{{{\lambda _{1,\phi }}}}} \right)}} - 16C,\\ \end{aligned} \end{equation*}

整理一下可推出定理 1.1 的第一个式子.

对于 \frac{{{\lambda _{2,\phi }}}}{{{\lambda _{1,\phi}}}} \ge 2 - \frac{{{\lambda _{1,\phi}}}}{{{\lambda _{3,\phi}}}}, 不等式 (2.18) 可以被重写成

\begin{equation*} \begin{split} &{\lambda _{3,\phi }}{\rm{ - }}{\lambda _{1,\phi }}{\rm{ + }}{\lambda _{2,\phi }}{\rm{ - }}{\lambda _{1,\phi }}{\rm{ + }}\frac{{\left( {{\lambda _{3,\phi }}{\rm{ - }}{\lambda _{2,\phi }}} \right)\left( {1{\rm{ - }}\left( {{\lambda _{2,\phi }}{\rm{ - }}{\lambda _{1,\phi }}} \right)b_2^2} \right)}}{{1 + \left( {{\lambda _{3,\phi }}{\rm{ - }}{\lambda _{2,\phi }}} \right)b_2^2}}\\ \le\, & 4{\lambda _{1,\phi }} - \frac{{{\lambda _{3,\phi }}{\rm{ - }}{\lambda _{1,\phi }}}}{{\frac{{{\lambda _{3,\phi }}}}{{{\lambda _{1,\phi }}}}}} -16C + \frac{{\left( {{\lambda _{3,\phi }}{\rm{ - }}{\lambda _{1,\phi }}} \right)\frac{{{\lambda _{3,\phi }}{\rm{ - }}{\lambda _{2,\phi }}}}{{{\lambda _{2,\phi }}{\rm{ - }}{\lambda _{1,\phi }}}}\left( {1{\rm{ - }}\left( {{\lambda _{2,\phi }}{\rm{ - }}{\lambda _{1,\phi }}} \right)b_2^2 + \frac{1}{3}\frac{{{M^2}}}{{{\lambda _{1,\phi }}}}} \right)}}{{\frac{{{\lambda _{3,\phi }}}}{{{\lambda _{1,\phi }}}}\left( {\frac{{{\lambda _{3,\phi }}}}{{{\lambda _{1,\phi }}}} + \frac{{{\lambda _{3,\phi }}{\rm{ - }}{\lambda _{2,\phi }}}}{{{\lambda _{2,\phi }}{\rm{ - }}{\lambda _{1,\phi }}}}\left( {1{\rm{ - }}\left( {{\lambda _{2,\phi }}{\rm{ - }}{\lambda _{1,\phi }}} \right)b_2^2 + \frac{1}{3}\frac{{{M^2}}}{{{\lambda _{1,\phi }}}}} \right)} \right)}}.\\ \end{split} \end{equation*}

由 (2.7) 式, 我们可以得到 \left( {{\lambda _{2,\phi }}{\rm{ - }}{\lambda _{1,\phi }}} \right)b_2^2 \le 1. 因此, 对于 \frac{{{\lambda _{2,\phi }}}}{{{\lambda _{1,\phi }}}} \ge 2 - \frac{{{\lambda _{1,\phi }}}}{{{\lambda _{3,\phi }}}}, 我们有

\begin{equation*}\label{2eq-25} \frac{{\left( {{\lambda _{3,\phi }}{\rm{ - }}{\lambda _{1,\phi }}} \right)\frac{{{\lambda _{3,\phi }}{\rm{ - }}{\lambda _{2,\phi }}}}{{{\lambda _{2,\phi }}{\rm{ - }}{\lambda _{1,\phi }}}}}}{{\frac{{{\lambda _{3,\phi }}}}{{{\lambda _{1,\phi }}}}\left( {\frac{{{\lambda _{3,\phi }}}}{{{\lambda _{1,\phi }}}} + \frac{{{\lambda _{3,\phi }}{\rm{ - }}{\lambda _{2,\phi }}}}{{{\lambda _{2,\phi }}{\rm{ - }}{\lambda _{1,\phi }}}}\left( {1{\rm{ - }}\left( {{\lambda _{2,\phi }}{\rm{ - }}{\lambda _{1,\phi }}} \right)b_2^2 + \frac{1}{3}\frac{{{M^2}}}{{{\lambda _{1,\phi }}}}} \right)}\right)}} \le \frac{{{\lambda _{3,\phi }}{\rm{ - }}{\lambda _{2,\phi }}}}{{1 + \left( {{\lambda _{3,\phi }}{\rm{ - }}{\lambda _{2,\phi }}} \right)b_2^2}} \end{equation*}

\begin{equation*}\label{2eq-26} \frac{{\left( {{\lambda _{3,\phi }}{\rm{ - }}{\lambda _{1,\phi }}} \right)\frac{{{\lambda _{3,\phi }}{\rm{ - }}{\lambda _{2,\phi }}}}{{{\lambda _{2,\phi }}{\rm{ - }}{\lambda _{1,\phi }}}}\frac{1}{3}\frac{{{M^2}}}{{{\lambda _{1,\phi }}}}}}{{\frac{{{\lambda _{3,\phi }}}}{{{\lambda _{1,\phi }}}}\left( {\frac{{{\lambda _{3,\phi }}}}{{{\lambda _{1,\phi }}}} + \frac{{{\lambda _{3,\phi }}{\rm{ - }}{\lambda _{2,\phi }}}}{{{\lambda _{2,\phi }}{\rm{ - }}{\lambda _{1,\phi }}}}\left( {1{\rm{ - }}\left( {{\lambda _{2,\phi }}{\rm{ - }}{\lambda _{1,\phi }}} \right)b_2^2 + \frac{1}{3}\frac{{{M^2}}}{{{\lambda _{1,\phi }}}}} \right)} \right)}}{\rm{ }}\le \frac{{{\lambda _{3,\phi }}{\rm{ - }}{\lambda _{1,\phi }}}}{{\frac{{{\lambda _{3,\phi }}}}{{{\lambda _{1,\phi }}}}}}.\\ \end{equation*}

由上面两个不等式可以推出 {\lambda _{3,\phi }} + {\lambda _{2,\phi }} \le 6{\lambda _{1,\phi }}-16C. 进一步可以推出 C>0 时的结论.

3 定理 1.4 的证明

u_{k} 满足

\begin{equation*}\label{neq-27} \left\{\begin{aligned} & {\mathbb {L}_\phi }u_{k}+\lambda_{k,\phi}u_{k}=0\qquad\quad && \text{在} \ \Omega\ \text{内},\\ &u_k= 0\qquad\qquad\quad\quad\quad\quad && \text{在} \ \partial\Omega \ \text{上},\\ &\int_\Omega {{u_i}{u_j}{\rm d}\mu = {\delta _{ij}}}. \end{aligned}\right. \end{equation*}

{x_1}, \cdots, {x_n}\mathbb{R}^n 的坐标函数, 记 An \times n 的矩阵 \left( {{A_{\alpha \beta }}} \right),

{A_{\alpha \beta }}=\int_\Omega {{x_\alpha }} {u_1}{u_{\beta + 1}}{\rm d}\mu.

由 Gram-Schmidt 正交化得存在一个上三角矩阵 R = \left( {{R_{\alpha \beta }}} \right) 和一个正交矩阵 Q = \left( {{q_{\alpha \beta }}} \right) 使得 R = QA, 因此当 1 \le \beta < \alpha \le n 时,

{R_{\alpha \beta }} = \sum\limits_{\gamma = 1}^n {{q_{\alpha \gamma }}} {A_{\gamma \beta }} = \int_\Omega {\sum\limits_{\gamma = 1}^n {{q_{\alpha \gamma }}} {x_\gamma }{u_1}{u_{\beta + 1}}{\rm d}\mu } = 0.

定义

\begin{equation*}\label{neq-28} \left\{\begin{aligned} &{a_{\alpha j}} = \int_\Omega {{z_\alpha }{u_1}{u_j}{\rm d}\mu }, 1 \le j \le \alpha \le n,\\[2mm] &b_{\alpha j}=\int_{\Omega} u_j \nabla z_\alpha \cdot \nabla u_1 {\rm d}\mu,\\[2mm] &{d_{\alpha j}} = \int_\Omega {{u_1}{u_j}} {\mathbb {L}_\phi }{z_\alpha }{\rm d}\mu. \end{aligned}\right. \end{equation*}

{y_\alpha } = \sum\limits_{\gamma = 1}^n {{q_{\alpha \gamma }}} {x_\gamma }, 当 1 \le \beta < \alpha \le n

\int_\Omega {{y_\alpha }{u_1}{u_{\beta + 1}}{\rm d}\mu } = \int_\Omega {\sum\limits_{\gamma = 1}^n {{q_{\alpha \gamma }}} {x_\gamma }{u_1}{u_{\beta + 1}}{\rm d}\mu } = 0.

1 \le \alpha \le n{z_\alpha }{\rm{ = }}{y_\alpha } - y_\alpha ^{\left( 0 \right)},y_\alpha ^{\left( 0 \right)} = \int_\Omega {{y_\alpha }} u_1^2{\rm d}\mu. 则当 1 \le j \le \alpha \le n

\begin{equation}\label{neq-29} {a_{\alpha j}} = 0. \end{equation}
(3.1)

于是对于具有 a_{\alpha j} 的项, 只需要考虑 j>\alpha 即可, 由 Green 公式可得

- {\lambda _{j,\phi}}{a_{\alpha j}} = \int_\Omega {{\mathbb {L}_\phi }} \left( {{z_\alpha }{u_1}} \right){u_j}{\rm d}\mu = 2{b_{\alpha j}} - {\lambda _{1,\phi}}{a_{\alpha j}} - {d_{\alpha j}},

\begin{equation*}\label{neq-30} 2b_{\alpha j}=\left(\lambda_{1,\phi}-\lambda_{j,\phi}\right) a_{\alpha j}- {d_{\alpha j}}. \end{equation*}

因为 \{u_{i}\}_{i=1}^\inftyL^{2}(\Omega) 上完备的标准正交基, 有

\begin{equation*}\label{neq-31} z_\alpha u_1=\sum\limits_{j=\alpha+1}^{\infty} a_{\alpha j} u_j \text {, }\left\|z_\alpha u_1\right\|^2=\sum\limits_{j=\alpha+1}^{\infty} a_{\alpha j}^2,\\ \end{equation*}

进而有

\begin{equation*}\label{neq-32} \begin{aligned} 2\int_\Omega {{z_\alpha }} {u_1}\nabla {z_\alpha }\cdot\nabla {u_1} {\rm d}\mu &= 2\sum\limits_{j = \alpha + 1}^\infty {{a_{\alpha j}}} {b_{\alpha j}}\\ &= \sum\limits_{j = \alpha + 1}^\infty {\left( {{\lambda _{1,\phi}} - {\lambda_{j,\phi}}} \right)} a_{\alpha j}^2 - \sum\limits_{j = \alpha + 1}^\infty {a_{\alpha j}}{d_{\alpha j}}. \end{aligned} \end{equation*}

另一方面, 由 Green 公式可得

\begin{equation*} \begin{aligned} - 2\int_\Omega {{z_\alpha }} {u_1}\nabla {z_\alpha }\cdot\nabla {u_1}{\rm d}\mu &= - \frac{1}{2}\int_\Omega \nabla z_\alpha ^2\cdot\nabla u_1^2{\rm d}\mu \\ &= \frac{1}{2}\int_\Omega u _1^2{\mathbb {L}_\phi }z_\alpha ^2{\rm d}\mu = 1 + \sum\limits_{j = \alpha + 1}^\infty {{a_{\alpha j}}{d_{\alpha j}}}, \end{aligned} \end{equation*}

进一步得出

\begin{equation}\label{neq-33} \sum\limits_{j=\alpha+1}^{\infty}\left(\lambda_{j,\phi}-\lambda_{1,\phi}\right) a_{\alpha j}^2=1. \end{equation}
(3.2)

对于任意的正整数 k, 有

\sum\limits_{j=\alpha+1}^{\infty}\left(\lambda_{j,\phi}-\lambda_{1,\phi}\right) a_{\alpha j}^2\geq\sum\limits_{j=\alpha+1}^k\left(\lambda_{j,\phi}-\lambda_{k+1,\phi}\right) a_{\alpha j}^2+\left(\lambda_{k+1,\phi}-\lambda _{1,\phi}\right) \sum\limits_{j=\alpha+1}^{\infty} a_{\alpha j}^2.

因此, 可以推出

\begin{equation}\label{neq-34} \left(\lambda_{k+1,\phi}-\lambda_{1,\phi}\right)\left\|z_\alpha u_1\right\|^2 \leq 1+\sum\limits_{j=\alpha+1}^k\left(\lambda_{k+1,\phi}-\lambda_{j,\phi}\right) a_{\alpha j}^2. \end{equation}
(3.3)

d_j=\int_{\Omega} u_1^t u_j {\rm d}\mu, 可知 u_1^t=\sum\limits_{j=1}^{\infty} d_j u_j, \left\|u_1^t\right\|^2=\int_{\Omega} u_1^{2 t}{\rm d}\mu=\sum\limits_{j=1}^{\infty} d_j^2. 对于任意的函数 f \in C^2(\Omega) \cap C(\bar{\Omega}), 由 Green 公式可得

\begin{matrix}\label{neq-36} - 2\int_\Omega f {u_1}\nabla f\cdot\nabla {u_1}{\rm d}\mu = \int_\Omega u _1^2f{\mathbb {L}_\phi }f{\rm d}\mu + \int_\Omega | \nabla f{|^2}u_1^2{\rm d}\mu, \end{matrix}
(3.4)
\begin{equation}\label{neq-37} \int_\Omega {{{\left| {\nabla u_1^{t - 1}} \right|}^2}} u_1^2{\rm d}\mu = \frac{{{{(t - 1)}^2}}}{{2t - 1}}{\lambda _{1,\phi}}\int_\Omega u _1^{2t}{\rm d}\mu = \sum\limits_{j = 2}^\infty {\left( {{\lambda _{j,\phi}} - {\lambda _{1,\phi}}} \right)} d_j^2, \end{equation}
(3.5)

其中第一个等式是在实常数 t>\frac{1}{2} 时通过 Green 公式得出, 第二个等式是令 f=u_1^{t-1} 代入到 (3.4) 式得到. 令

\begin{equation}\label{neq-38} \beta_j=\frac{d_j}{\sqrt{\frac{(t-1)^2}{2 t-1} \lambda_{1,\phi} \int_{\Omega} u_1^{2 t}{\rm d}\mu}}, \end{equation}
(3.6)

则有

\begin{equation}\label{neq-39} \sum\limits_{j=2}^{\infty}\left(\lambda_{j,\phi}-\lambda_{1,\phi}\right) \beta_j^2=1. \end{equation}
(3.7)

对于任意的正整数 l, 有

\sum\limits_{j=2}^{\infty}\left(\lambda_{j,\phi}-\lambda_{1,\phi}\right) \beta_j^2 \geq \sum\limits_{j=2}^l\left(\lambda_{j,\phi}-\lambda_{l+1,\phi}\right) \beta_j^2+\left(\lambda_{l+1,\phi}-\lambda_{1,\phi}\right) \sum\limits_{j=2}^{\infty} \beta_j^2.

因此, 可以推出

\begin{equation}\label{neq-40} \left(\lambda_{l+1,\phi}-\lambda_{1,\phi}\right) \sum\limits_{j=2}^{\infty} \beta_j^2 \leq 1+\sum\limits_{j=2}^l\left(\lambda_{l+1,\phi}-\lambda_{j,\phi}\right) \beta_j^2. \end{equation}
(3.8)

B(t)=\frac{\int_{\Omega} u_1^{2 t}{\rm d}\mu}{\left(\int_{\Omega} u_1^{t+1}{\rm d}\mu \right)^2},

从 (3.4)-(3.8) 式得

\begin{equation}\label{neq-41} \frac{(t-1)^2}{2 t-1} \lambda_{1,\phi} \sum\limits_{j=1}^{\infty} \beta_j^2=1. \end{equation}
(3.9)

因为

\left(\int_{\Omega} u_1^{t+1} {\rm d}\mu \right)^2=d_1^2=\frac{(t-1)^2}{2 t-1} \lambda_{1,\phi} \beta_1^2 \int_{\Omega} u_1^{2 t} {\rm d}\mu,

根据 B(t) 的定义, 有

\begin{equation}\label{neq-42} B(t) = \frac{{\int_\Omega u _1^{2t}{\rm d}\mu }}{{{{\left( {\int_\Omega u _1^{t + 1}{\rm d}\mu } \right)}^2}}} = \frac{1}{{\frac{{{{(t - 1)}^2}}}{{2t - 1}}{\lambda _{1,\phi }}\beta _1^2}} = \frac{1}{{1 - \frac{{{{(t - 1)}^2}}}{{2t - 1}}{\lambda _{1,\phi }}\sum\limits_{j = 2}^\infty {\beta _j^2} }}. \end{equation}
(3.10)

由 Green 公式得

\begin{equation*} \begin{aligned} {d_1} &= \int_\Omega u _1^{t + 1}{\rm d}\mu = \frac{1}{2}\int_\Omega u _1^{t + 1}\Delta z_\alpha ^2{\rm d}\mu\\ &= - (t + 1)\int_\Omega {{z_\alpha }} {u_1}\left( {u_1^{t - 1}\nabla {z_\alpha }\cdot\nabla {u_1} - \frac{1}{{t + 1}}u_1^t\nabla {z_\alpha }\cdot\nabla \phi } \right){\rm d}\mu. \end{aligned} \end{equation*}

由 Schwarz 不等式得

\begin{equation*}\label{neq-43} d_1^2 \le {(t + 1)^2}\int_\Omega {{{\left( {{z_\alpha }{u_1}} \right)}^2}} {\rm d}\mu \int_\Omega {{{\left( {u_1^{t - 1}\nabla {z_\alpha }\cdot\nabla {u_1} - \frac{1}{{t + 1}}u_1^t\nabla {z_\alpha }\cdot\nabla \phi } \right)}^2}} {\rm d}\mu. \end{equation*}

因此, 有

\begin{equation}\label{neq-44} \sum\limits_{\alpha = 1}^n {\frac{1}{{{{\left\| {{z_\alpha }{u_1}} \right\|}^2}}}}\le \frac{{{{(t + 1)}^2}}}{{2t - 1}}{\lambda _{1,\phi}}B(t) + \frac{{\int_\Omega u _1^{2t}{{\left| {\nabla \phi } \right|}^2}{\rm d}\mu }}{{{{\left( {\int_\Omega u _1^{t + 1}{\rm d}\mu } \right)}^2}}} + \frac{{\frac{{t + 1}}{t}\int_\Omega u _1^{2t}{\mathbb {L}_\phi }\phi {\rm d}\mu }}{{{{\left( {\int_\Omega u _1^{t + 1}{\rm d}\mu } \right)}^2}}}. \end{equation}

引理3.1\Omega\mathbb{R}^n 上的有界区域, 若 \phi = {{{-C{\left| X \right|}^2}}}, \left| {\nabla \phi } \right| \le M (M 是常数), C 是常数. 对于加权 Laplace 算子的 Dirichlet 特征值问题 (1.5), 有

\begin{equation*}\label{neq-45} \begin{split} & \sum\limits_{\alpha = 1}^n {\frac{{{\lambda _{k + 1,\phi}} - {\lambda _{1,\phi }}}}{{1 + \sum\limits_{j = \alpha + 1}^k {\left( {{\lambda _{k + 1,\phi }} - {\lambda _{j,\phi }}} \right)} a_{\alpha j}^2}}} \\ \le \,& \frac{{{{\left( {t + 1} \right)}^2}{\lambda _{1,\phi }}}}{{2t - 1}}\frac{{\int_\Omega {{u_1^{2t}}} {\rm d}\mu }}{{{{\left( {\int_\Omega {{u_1^{t + 1}}} {\rm d}\mu } \right)}^2}}} + \frac{{\int_\Omega {{u_1^{2t}}{{\left| {\nabla \phi } \right|}^2}} {\rm d}\mu }}{{{{\left( {\int_\Omega {{u_1^{t + 1}}} {\rm d}\mu } \right)}^2}}} + \frac{{t + 1}}{t}\frac{{\int_\Omega {{u_1^{2t }}{\mathbb {L}_\phi }\phi } {\rm d}\mu }}{{{{\left( {\int_\Omega {{u_1^{t + 1}}} {\rm d}\mu } \right)}^2}}}, \end{split} \end{equation*}

对于 j \geq 2, 做正交变换 {\widetilde z_\alpha } = \sum\limits_{\beta = 1}^n {{h_{\alpha \beta }}} {z_\beta }(1 \le \alpha \le n), \operatorname{det}\left(h_{\alpha \beta}\right)=1 使得

\begin{equation*} \widetilde{a}_{\alpha j}=\int_{\Omega} \widetilde{z}_\alpha u_1 u_j {\rm d}\mu =\left\{\begin{array}{ll} \sqrt {\sum\limits_{\beta = 1}^{j - 1} {a_{\beta j}^2} }, & \alpha=1, \\ 0, & \alpha \geq 1. \end{array}\right. \end{equation*}

\widetilde{z}_\alpha 的定义和 (3.1) 式得

\begin{equation*} \begin{aligned} 0 & =-\frac{2}{t+1} \int_{\Omega} u_1^{t+1} \Delta \widetilde{z}_\alpha {\rm d}\mu = \frac{2}{{t + 1}}\int_\Omega {\nabla \left( {u_1^{t + 1}{{\rm e}^{ - \phi }}} \right)} \cdot\nabla {\widetilde z_\alpha }{\rm d} v\\ &= 2\sum\limits_{p = 1}^\infty {{d_p}\int_\Omega {{u_p}\nabla {u_1}} \cdot\nabla {{\widetilde z}_\alpha }{\rm d}\mu } - \frac{2}{{t + 1}}\int_\Omega {u_1^{t + 1}\nabla \phi } \cdot\nabla {\widetilde z_\alpha }{\rm d}\mu \\ &= \sum\limits_{p = 1}^\infty {\sum\limits_{\beta = 1}^n {{d_p}} } {h_{\alpha \beta }}\left( {\left( {{\lambda _{1,\phi}} - {\lambda _{p,\phi}}} \right){a_{\beta p}} - {d_{\beta p}}} \right)- \frac{2}{{t + 1}}\int_\Omega {u_1^{t + 1}\nabla \phi } \cdot\nabla {\widetilde z_\alpha }{\rm d}\mu \\ & =\sum\limits_{p=1}^{\infty}\left(\lambda_{1,\phi}-\lambda_{p,\phi}\right) d_p \widetilde{a}_{\alpha p} - \sum\limits_{p = 1}^\infty {\sum\limits_{\beta = 1}^n {{d_p}} } {h_{\alpha \beta }}{d_{\beta p}}- \frac{2}{{t + 1}}\int_\Omega {u_1^{t + 1}\nabla \phi } \cdot\nabla {\widetilde z_\alpha }{\rm d}\mu. \end{aligned} \end{equation*}

因此, 由 (3.6) 式得

\begin{equation}\label{neq-46} \sum\limits_{p = 2}^\infty {\left( {{\lambda _{p,\phi}} - {\lambda _{1,\phi}}} \right)} {\beta _p}{\widetilde a_{\alpha p}} = \frac{{\frac{{t - 1}}{{t + 1}}\int_\Omega {u_1^{t + 1}\nabla \phi } \cdot\nabla {{\widetilde z}_\alpha }{\rm d}\mu }}{{\sqrt {\frac{{{{\left( {t - 1} \right)}^2}}}{{2t - 1}}{\lambda _{1,\phi}}\int_\Omega {u_1^{2t}} {\rm d}\mu } }}. \end{equation}
(3.12)

这时

\begin{matrix}\label{neq-47} 1 & =-2 \int_{\Omega} \widetilde{z}_\alpha u_1 \nabla \widetilde{z}_\alpha \cdot \nabla u_1 {\rm d}\mu- \int_\Omega {u_1^2{{\widetilde z}_\alpha }} {\mathbb {L}_\phi }{\widetilde z_\alpha }{\rm d}\mu \\ & =-2 \sum\limits_{p=1}^{\infty} \widetilde{a}_{\alpha p} h_{\alpha \beta} b_{\beta p} -\sum\limits_{p = 1}^\infty {{{\widetilde a}_{\alpha p}}} {\widetilde d_{\alpha p}} =\sum\limits_{p=2}^{\infty}\left(\lambda_{p,\phi}-\lambda_{1,\phi}\right) \widetilde{a}_{\alpha p}^2. \end{matrix}
(3.13)

因此, 由 (3.12) 式和 Schwarz 不等式可得

\begin{equation*} \begin{aligned} &{\bigg( {\left( {{\lambda _{j,\phi }} - {\lambda _{1,\phi }}} \right){\beta _j}{{\widetilde a}_{\alpha j}} - \frac{{\frac{{t - 1}}{{t + 1}}\int_\Omega {u_1^{t + 1}\nabla \phi } \cdot\nabla {{\widetilde z}_\alpha }{\rm d}\mu }}{{\sqrt {\frac{{{{\left( {t - 1} \right)}^2}}}{{2t - 1}}{\lambda _{1,\phi }}\int_\Omega {u_1^{2t}} {\rm d}\mu } }}} \bigg)^2} \\ \le & \bigg( {\sum\limits_{p = 2,p \ne j}^\infty {\left( {{\lambda _{p,\phi }} - {\lambda _{1,\phi }}} \right)} \beta _p^2} \bigg)\bigg( {\sum\limits_{p = 2,p \ne j}^\infty {\left( {{\lambda _{p,\phi }} - {\lambda _{1,\phi }}} \right)} \widetilde a_{\alpha p}^2} \bigg). \end{aligned} \end{equation*}

则根据 (3.9) 和 (3.13) 式可得出

\begin{equation}\label{neq-48} \left(\lambda_{j,\phi}-\lambda_{1,\phi}\right) \beta_j^2 \leq 1-\left(\lambda_{j,\phi}-\lambda_{1,\phi}\right) \widetilde{a}_{\alpha j}^2+ \frac{{\left( {2t - 1} \right){M^2}}}{{{{\left( {t + 1} \right)}^2}{\lambda _{1,\phi}}}}. \end{equation}
(3.14)

从 (3.3), (3.8), (3.10), (3.11) 和 (3.14) 式, 有

\begin{equation*} \begin{aligned} &\sum\limits_{\alpha = 1}^n {\frac{{{\lambda _{k + 1,\phi }} - {\lambda _{1,\phi }}}}{{1 + \sum\limits_{j = \alpha + 1}^k {\left( {{\lambda _{k + 1,\phi }} - {\lambda _{j,\phi }}} \right)} a_{\alpha j}^2}}} \\ \leq & \sum\limits_{\alpha=1}^n \frac{1}{\left\|z_\alpha u_1\right\|^2} \le \frac{{{{\left( {t + 1} \right)}^2}{\lambda _{1,\phi }}}}{{2t - 1}}\frac{{\int_\Omega {{u_1^{2t}}} {\rm d}\mu }}{{{{\left( {\int_\Omega {{u_1^{t + 1}}} {\rm d}\mu } \right)}^2}}} + \frac{{\int_\Omega {{u_1^{2t}}{{\left| {\nabla \phi } \right|}^2}} {\rm d}\mu }}{{{{\left( {\int_\Omega {{u_1^{t + 1}}} {\rm d}\mu } \right)}^2}}} + \frac{{t + 1}}{t}\frac{{\int_\Omega {{u_1^{2t}}{\mathbb {L}_\phi }\phi } {\rm d}\mu }}{{{{\left( {\int_\Omega {{u_1^{t + 1}}} {\rm d}\mu } \right)}^2}}}. \end{aligned} \end{equation*}

为了证明定理 1.4, 类似于文献 [9,引理 2.1,引理 2.2,引理 2.3] 的证明方法, 我们可以证明下面三个引理.

引理3.2\left\{\psi _{i,\phi}\right\}_{i=1}^{m+2} 是一个递增的序列, \omega=\left(\omega_{j k}\right) 是一个 (m+1) \times (m+1) 的矩阵, 则有

\begin{align*} &\sum\limits_{i = 1}^m {\frac{{{\psi _{m + 2,\phi}} - {\psi _{1,\phi}}}}{{1 + \sum\limits_{p = i + 1}^{m + 1} {\left( {{\psi _{m + 2,\phi}} - {\psi _{p,\phi}}} \right)} \omega _{ip}^2}}} - \sum\limits_{i = 1}^m {\left( {{\psi _{i + 1,\phi}} - {\psi _{1,\phi}}} \right)} \\ =&\sum\limits_{j = 1}^m {\sum\limits_{i = 1}^{m + 1 - j} {\frac{{\left( {{\psi _{i + j + 1,\phi }} - {\psi _{i + j,\phi }}} \right)\bigg[ {1 - \sum\limits_{p = i + 1}^{i + j} {\left( {{\psi _{p,\phi }} - {\psi _{1,\phi }}} \right)} \omega _{ip}^2} \bigg]}}{{\bigg[ {1 + \sum\limits_{p = i + 1}^{i + j - 1} {\left( {{\psi _{i + j,\phi }} - {\psi _{p,\phi }}} \right)} \omega _{ip}^2} \bigg]\bigg[ {1 + \sum\limits_{p = i + 1}^{i + j} {\left( {{\psi _{i + j + 1,\phi }} - {\psi _{p,\phi }}} \right)} \omega _{ip}^2} \bigg]}}} }. \end{align*}

引理3.3\left\{\psi_{i,\phi}\right\}_{i=1}^{m+2} 是一个递增的序列, \omega=\left(\omega_{j k}\right) 是一个 (m+1) \times (m+1) 的矩阵, 有

\begin{equation*} \begin{aligned} &\sum\limits_{j = 1}^m {\sum\limits_{i = 1}^{m + 1 - j} {\frac{{\left( {{\psi _{i + j + 1,\phi }} - {\psi _{i + j,\phi }}} \right){{\left( {{\psi _{2,\phi }} - {\psi _{1,\phi }}} \right)}^2}}}{{\left( {{\psi _{i + j + 1,\phi }} - {\psi _{1,\phi }}} \right)\left( {{\psi _{i + j,\phi }} - {\lambda _{1,\phi }}} \right)}}} } \bigg[ {1 - \sum\limits_{p = i + 1}^{i + j} {\left( {{\psi _{p,\phi }} - {\psi _{1,\phi }}} \right)} \omega _{ip}^2} \bigg] \\ = &\sum\limits_{j = 2}^{m + 1} {\frac{{\left( {{\psi _{j + 1,\phi }} - {\psi _{j,\phi }}} \right){{\left( {{\psi _{2,\phi }} - {\psi_{1,\phi }}} \right)}^2}}}{{\left( {{\psi _{j + 1,\phi }} - {\psi_{1,\phi }}} \right)\left( {{\psi_{j,\phi }} - {\lambda _{1,\phi }}} \right)}}} \sum\limits_{i = 1}^{j - 1} {\bigg[ {1 - \sum\limits_{p = i + 1}^j {\left( {{\psi _{p,\phi }} - {\psi _{1,\phi }}} \right)} \omega _{ip}^2} \bigg]}. \\ \end{aligned} \end{equation*}

引理3.4\left\{\psi_{i,\phi}\right\}_{i=1}^{m+2} 是一个递增的序列, \omega=\left(\omega_{j k}\right) 是一个 (m+1) \times (m+1) 的矩阵. 则当 1 \leq i \leq m+2, 有

\begin{aligned} & \sum_{j=2}^{i-1} \frac{\psi_{i, \phi}-\psi_{j, \phi}}{\psi_{j, \phi}-\psi_{1, \phi}}\left[1-\left(\psi_{j, \phi}-\psi_{1, \phi}\right) \sum_{k=1}^{j-1} \omega_{k j}^{2}\right] \\ = & \sum_{j=2}^{i-1} \frac{\left(\psi_{i, \phi}-\psi_{1, \phi}\right)\left(\psi_{j+1, \phi}-\psi_{j, \phi}\right)}{\left(\psi_{j+1, \phi}-\psi_{1, \phi}\right)\left(\psi_{j, \phi}-\psi_{1, \phi}\right)} \sum_{k=1}^{j-1}\left[1-\sum_{p=k+1}^{j}\left(\psi_{p, \phi}-\psi_{1, \phi}\right) \omega_{k p}^{2}\right]. \end{aligned}

引理3.5 对于加权 Laplace 算子的 Dirichlet 特征值问题 (1.5), \phi =- C{\left| X \right|^2}, 其中 X\Omega 的位置向量. 则当 C<0 时, 有

\begin{equation*} \sum\limits_{\alpha = 1}^n {\frac{{{\lambda _{k + 1,\phi }} - {\lambda _{1,\phi }}}}{{1 + \sum\limits_{j = \alpha + 1}^k {\left( {{\lambda _{k + 1,\phi }} - {\lambda _{j,\phi }}} \right)} a_{\alpha j}^2}}} \le 4{\lambda _{1,\phi }} - \frac{{{\lambda _{1,\phi }}\left( {{\theta_{l,\phi }} - {\lambda _{1,\phi }}} \right)}}{{{\theta _{l,\phi }}}} - 8nC\left( {1 + \frac{{{\lambda _{1,\phi }}}}{{{\theta _{l,\phi}}}}} \right)\left( {3 - \frac{{{\lambda _{1,\phi }}}}{{{\theta _{l,\phi }}}}} \right). \end{equation*}

C> 0 时, 有

\sum\limits_{\alpha = 1}^n {\frac{{{\lambda _{k + 1,\phi }} - {\lambda _{1,\phi }}}}{{1 + \sum\limits_{j = \alpha + 1}^k {\left( {{\lambda _{k + 1,\phi }} - {\lambda _{j,\phi }}} \right)} a_{\alpha j}^2}}} \le 4{\lambda _{1,\phi }} - \frac{{{\lambda _{1,\phi }}\left( {{\theta _{l,\phi }} - {\lambda _{1,\phi }}} \right)}}{{{\theta _{l,\phi }}}},

其中

\begin{eqnarray*} {\theta _{l,\phi }} - {\lambda _{1,\phi }} = \frac{{{\lambda _{l + 1,\phi }} - {\lambda _{1,\phi }}}}{{1 + \sum\limits_{j = 2}^l {\frac{{{\lambda _{l + 1,\phi }} - {\lambda _{j,\phi }}}}{{{\lambda _{j,\phi }} - {\lambda _{1,\phi }}}}\Big( {1 - \left( {{\lambda _{j,\phi }} - {\lambda _{1,\phi }}} \right)\sum\limits_{\beta = 1}^{j - 1} {a_{\beta j}^2} + \frac{{{M^2}}}{{3{\lambda _{1,\phi }}}}} \Big)} }}. \end{eqnarray*}

\phi = -C{\left| X \right|^2},X \in \mathbb{R}^n,0 \le \left| X \right| \le {\rm diam}\left( \Omega \right), 有

\nabla \phi = - 2CX,\Delta \phi = - 2nC,{\mathbb {L}_\phi }\phi = \Delta \phi - \left\langle {\nabla \phi, \nabla \phi } \right\rangle = - 2nC - 4{C^2}{\left| X \right|^2},

这里 \left| {\nabla \phi } \right| \le M,M = 2\left| C \right|{\rm diam}\left( \Omega \right). 将上式代入到引理 3.1 可得

\begin{equation*} \begin{aligned} &\sum\limits_{\alpha = 1}^n {\frac{{{\lambda _{k + 1,\phi }} - {\lambda _{1,\phi }}}}{{1 + \sum\limits_{j = \alpha + 1}^k {\left( {{\lambda _{k + 1,\phi }} - {\lambda _{j,\phi }}} \right)} a_{\alpha j}^2}}}\\ \leq \,& \frac{(t+1)^2}{2 t-1} \lambda_{ 1,\phi } \frac{{\int_\Omega {u_1^{2t}{\rm d}\mu } }}{{{{(\int_\Omega {u_1^{t + 1}{\rm d}\mu } )}^2}}}+ \frac{{\int_\Omega {u_1^{2t}{{\left( { - 2CX} \right)}^2}} {\rm d}\mu }}{{{{\left( {\int_\Omega {u_1^{t + 1}} {\rm d}\mu } \right)}^2}}} + \frac{{\int_\Omega {\frac{{t + 1}}{t}u_1^{2t}} \left( { - 2nC - 4{C^2}{{\left| X \right|}^2}} \right){\rm d}\mu }}{{{{\left( {\int_\Omega {u_1^{t + 1}} {\rm d}\mu } \right)}^2}}} \\ \le \,& \frac{{{{(t + 1)}^2}}}{{2t - 1}}\frac{{{\lambda _{1,\phi}}}}{{1 - \frac{{{{(t - 1)}^2}}}{{2t - 1}}\frac{{{\lambda _{ 1,\phi }}}}{{{\lambda _{l + 1,\phi}} - {\lambda _{1,\phi}}}}\Big( {1 + \sum\limits_{j = 2}^l {( {{\lambda _{l + 1,\phi}} - {\lambda_{j,\phi}}} )} \beta _j^2} \Big)}} \\ &- \frac{{2nC\left( {t + 1} \right)}}{t}\frac{1}{{1 - \frac{{{{(t - 1)}^2}}}{{2t - 1}}\frac{{{\lambda _{1,\phi}}}}{{{\lambda _{l + 1,\phi}} - {\lambda _{1,\phi}}}}\Big( {1 + \sum\limits_{j = 2}^l {( {{\lambda _{l + 1,\phi}} - {\lambda _{j,\phi}}} )} \beta _j^2} \Big)}}\\ =\,& 4{\lambda _{1,\phi }} - \frac{{{\lambda _{1,\phi }}\left( {{\theta _{l,\phi }} - {\lambda _{1,\phi }}} \right)}}{{{\theta _{l,\phi }}}} - 2nC\Big( {1 + \frac{{{\lambda _{1,\phi }}}}{{{\theta _{l,\phi }}}}} \Big)\Big( {3 - \frac{{{\lambda _{1,\phi }}}}{{{\theta _{l,\phi }}}}} \Big). \Big(\text{选取}~t=\frac{2 \theta_{l,\phi }}{\theta_{l,\phi }+\lambda_{1,\phi}}\Big) \end{aligned} \end{equation*}

由上式可以推出, 当 C>0 时, - 2nC( {1 + \frac{{{\lambda _{1,\phi }}}}{{{\theta _{l,\phi }}}}} )( {3 - \frac{{{\lambda _{1,\phi }}}}{{{\theta _{l,\phi }}}}} ) \le 0; 当 C<0 时, - 2nC( {1 + \frac{{{\lambda _{1,\phi }}}}{{{\theta _{l,\phi }}}}} )( {3 - \frac{{{\lambda _{1,\phi }}}}{{{\theta _{l,\phi }}}}} ) \le -8nC . 代入上式即可完成引理的证明.

定理 1.4 的证明 令引理 3.5 中的 k=n+1, l=i-1, 有

\begin{equation}\label{neq-49} \sum\limits_{\alpha=1}^n\left(\lambda_{\alpha+1,\phi}-\lambda _{1,\phi }\right)+B \leq 4 \lambda _{1,\phi }- \frac{{{\lambda _{1,\phi }}\left( {{\lambda _{i,\phi }} - {\lambda _{1,\phi }}} \right)}}{{{\lambda_{i,\phi }}}} + D - 8nC, \end{equation}(3.15)

其中

\begin{equation*} \begin{aligned} B &= \sum\limits_{\alpha = 1}^n {\frac{{{\lambda _{n + 2,\phi }} - {\lambda _{1,\phi }}}}{{1 + \sum\limits_{j = \alpha + 1}^{n + 1} {\left( {{\lambda _{n + 2,\phi }} - {\lambda _{j,\phi }}} \right)} a_{\alpha j}^2}}} - \sum\limits_{\alpha = 1}^n {\left( {{\lambda _{\alpha + 1,\phi }} - {\lambda _{1,\phi }}} \right)}, \\ D&=\frac{{\left( {{\lambda _{i,\phi }} - {\lambda _{1,\phi }}} \right)\sum\limits_{j = 2}^{i - 1} {\frac{{{\lambda _{i,\phi }} - {\lambda_{j,\phi}}}}{{{\lambda_{j,\phi}} - {\lambda _{1,\phi }}}}} \left[ {1 - \left( {{\lambda_{j,\phi}} - {\lambda _{1,\phi }}} \right)\sum\limits_{\alpha = 1}^{j - 1} {a_{\alpha j}^2} + \frac{{{M^2}}}{{3{\lambda _{1,\phi }}}}} \right]}}{{\frac{{{\lambda _{i,\phi }}}}{{{\lambda _{1,\phi }}}}\bigg\{ {\frac{{{\lambda_{i,\phi}}}}{{{\lambda _{1,\phi }}}} + \sum\limits_{j = 2}^{i - 1} {\frac{{{\lambda _{i,\phi }} - {\lambda_{j,\phi}}}}{{{\lambda_{j,\phi}} - {\lambda_{1,\phi}}}}} \left[ {1 - \left( {{\lambda_{j,\phi}} - {\lambda _{1,\phi }}} \right)\sum\limits_{\alpha = 1}^{j - 1} {a_{\alpha j}^2} + \frac{{{M^2}}}{{3{\lambda _{1,\phi }}}}} \right]} \bigg\}}}. \end{aligned} \end{equation*}

由引理 3.4 可得

\begin{equation}\label{neq-50} B = \sum\limits_{\beta = 1}^n {\sum\limits_{\alpha = 1}^{n + 1 - \beta } {\frac{{\left( {{\lambda _{\alpha + \beta + 1,\phi }} - {\lambda _{\alpha + \beta, \phi }}} \right)\left[ {1 - \sum\limits_{p = \alpha + 1}^{\alpha + \beta } {\left( {{\lambda _{p,\phi }} - {\lambda _{1,\phi }}} \right)} a_{\alpha p}^2} \right]}}{{\left[ {1 + \sum\limits_{p = \alpha + 1}^{\alpha + \beta - 1} {\left( {{\lambda _{\alpha + \beta, \phi }} - {\lambda _{p,\phi }}} \right)} a_{\alpha p}^2} \right]\left[ {1 + \sum\limits_{p = \alpha + 1}^{\alpha + \beta } {\left( {{\lambda _{\alpha + \beta + 1,\phi }} - {\lambda _{p,\phi }}} \right)} a_{\alpha p}^2} \right]}}} }. \end{equation}

对任何正整数 \gamma, 由 (3.2) 式可得 \sum\limits_{p = \alpha + 1}^\gamma {\left( {{\lambda _{p,\phi }} - {\lambda _{1,\phi }}} \right)} a_{\alpha p}^2 \le 1, 则可得递推公式

\begin{aligned} & 1+\sum_{p=\alpha+1}^{\gamma}\left(\lambda_{\gamma+1, \phi}-\lambda_{p, \phi}\right) a_{\alpha p}^{2} \\ = & 1+\frac{\lambda_{\gamma+1, \phi}-\lambda_{\gamma, \phi}}{\lambda_{\gamma, \phi}-\lambda_{1, \phi}} \sum_{p=\alpha+1}^{\gamma}\left(\lambda_{p, \phi}-\lambda_{1, \phi}\right) a_{\alpha p}^{2}-\frac{\lambda_{\gamma+1, \phi}-\lambda_{\gamma, \phi}}{\lambda_{\gamma, \phi}-\lambda_{1, \phi}} \sum_{p=\alpha+1}^{\gamma-1}\left(\lambda_{p, \phi}-\lambda_{1, \phi}\right) a_{\alpha p}^{2} \\ & +\sum_{p=\alpha+1}^{\gamma-1}\left(\lambda_{\gamma+1, \phi}-\lambda_{p, \phi}\right) a_{\alpha p}^{2} \\ \leq & 1+\frac{\lambda_{\gamma+1, \phi}-\lambda_{\gamma, \phi}}{\lambda_{\gamma, \phi}-\lambda_{1, \phi}}-\frac{\lambda_{\gamma+1, \phi}-\lambda_{\gamma, \phi}}{\lambda_{\gamma, \phi}-\lambda_{1, \phi}} \sum_{p=\alpha+1}^{\gamma-1}\left(\lambda_{p, \phi}-\lambda_{1, \phi}\right) a_{\alpha p}^{2}+\sum_{p=\alpha+1}^{\gamma-1}\left(\lambda_{\gamma+1, \phi}-\lambda_{p, \phi}\right) a_{\alpha p}^{2} \\ = & \frac{\lambda_{\gamma+1, \phi}-\lambda_{1, \phi}}{\lambda_{\gamma, \phi}-\lambda_{1, \phi}}\left[1+\sum_{p=\alpha+1}^{\gamma-1}\left(\lambda_{\gamma, \phi}-\lambda_{p, \phi}\right) a_{\alpha p}^{2}\right]. \end{aligned}

因此可得

\begin{equation*} \begin{aligned} &1+\sum\limits_{p=\alpha+1}^{\alpha+\beta}\left(\lambda_{\alpha+\beta+1,\phi}-\lambda_{p,\phi}\right) a_{\alpha p}^2 \\ \leq \,&\frac{\lambda_{\alpha+\beta+1,\phi}-\lambda _{1,\phi }}{\lambda_{\alpha+\beta,\phi}-\lambda _{1,\phi }} \cdot \frac{\lambda_{\alpha+\beta,\phi}-\lambda_{1,\phi}}{\lambda_{\alpha+\beta-1,\phi}-\lambda_{1,\phi}} \cdots \frac{\lambda_{\alpha+2,\phi}-\lambda_{1,\phi} }{\lambda_{\alpha+1,\phi}-\lambda _{1,\phi }} =\frac{\lambda_{\alpha+\beta+1,\phi}-\lambda _{1,\phi }}{\lambda_{\alpha+1,\phi}-\lambda _{1,\phi}}, \end{aligned} \end{equation*}

把类似的参数代入上面的不等式可得

1+\sum\limits_{p=\alpha+1}^{\alpha+\beta-1}\left(\lambda_{\alpha+\beta,\phi}-\lambda_{p,\phi}\right) a_{\alpha p}^2 \leq \frac{\lambda_{\alpha+\beta,\phi}-\lambda _{1,\phi }}{\lambda_{\alpha+1,\phi}-\lambda _{1,\phi }}.

由 (3.16) 、上述不等式和引理 3.5, 可得

\begin{aligned} B & \geq \sum_{\beta=1}^{n} \sum_{\alpha=1}^{n+1-\beta} \frac{\left(\lambda_{\alpha+\beta+1, \phi}-\lambda_{\alpha+\beta, \phi}\right)\left(\lambda_{\alpha+1, \phi}-\lambda_{1, \phi}\right)^{2}}{\left(\lambda_{\alpha+\beta+1, \phi}-\lambda_{1, \phi}\right)\left(\lambda_{\alpha+\beta, \phi}-\lambda_{1, \phi}\right)}\left[1-\sum_{p=\alpha+1}^{\alpha+\beta}\left(\lambda_{p, \phi}-\lambda_{1, \phi}\right) a_{\alpha p}^{2}\right] \\ & \geq \sum_{\beta=1}^{n} \sum_{\alpha=1}^{n+1-\beta} \frac{\left(\lambda_{\alpha+\beta+1, \phi}-\lambda_{\alpha+\beta, \phi}\right)\left(\lambda_{2, \phi}-\lambda_{1, \phi}\right)^{2}}{\left(\lambda_{\alpha+\beta+1, \phi}-\lambda_{1, \phi}\right)\left(\lambda_{\alpha+\beta, \phi}-\lambda_{1, \phi}\right)}\left[1-\sum_{p=\alpha+1}^{\alpha+\beta}\left(\lambda_{p, \phi}-\lambda_{1, \phi}\right) a_{\alpha p}^{2}\right] \\ & =\sum_{\beta=2}^{n+1} \frac{\left(\lambda_{\beta+1, \phi}-\lambda_{\beta, \phi}\right)\left(\lambda_{2, \phi}-\lambda_{1, \phi}\right)^{2}}{\left(\lambda_{\beta+1, \phi}-\lambda_{1, \phi}\right)\left(\lambda_{\beta, \phi}-\lambda_{1, \phi}\right)} \sum_{\alpha=1}^{\beta-1}\left[1-\sum_{p=\alpha+1}^{\beta}\left(\lambda_{p, \phi}-\lambda_{1, \phi}\right) a_{\alpha p}^{2}\right]. \end{aligned}

另一方面, 若 \frac{\lambda_{2,\phi}}{\lambda _{1,\phi }} \geq 2-\frac{\lambda _{1,\phi }}{\lambda_{i,\phi}}, 有 \frac{\lambda_{i,\phi}}{\lambda _{1,\phi }} \geq \frac{\lambda_{i,\phi}-\lambda _{1,\phi }}{\lambda_{2,\phi}-\lambda _{1,\phi }}. 因此, 由引理 3.4 可得

\begin{aligned} D \leq & \frac{\left(\lambda_{i, \phi}-\lambda_{1, \phi}\right)}{\left(\frac{\lambda_{i, \phi}}{\lambda_{1, \phi}}\right)^{2}} \sum_{j=2}^{i-1} \frac{\lambda_{i, \phi}-\lambda_{j, \phi}}{\lambda_{j, \phi}-\lambda_{1, \phi}}\left[1-\left(\lambda_{j, \phi}-\lambda_{1, \phi}\right) \sum_{\alpha=1}^{j-1} a_{\alpha j}^{2}\right] \\ & +\frac{\left(\lambda_{i, \phi}-\lambda_{1, \phi}\right) \sum_{j=2}^{i-1} \frac{\lambda_{i, \phi}-\lambda_{j, \phi}}{\lambda_{j, \phi}-\lambda_{1, \phi}} \frac{M^{2}}{3 \lambda_{1, \phi}}}{\frac{\lambda_{i, \phi}}{\lambda_{1, \phi}}\left\{\frac{\lambda_{i, \phi}}{\lambda_{1, \phi}}+\sum_{j=2}^{i-1} \frac{\lambda_{i, \phi}-\lambda_{j, \phi}}{\lambda_{j, \phi}-\lambda_{1, \phi}}\left[1-\left(\lambda_{j, \phi}-\lambda_{1, \phi}\right) \sum_{\alpha=1}^{j-1} a_{\alpha j}^{2}+\frac{M^{2}}{3 \lambda_{1, \phi}}\right]\right\}} \\ & \leq \sum_{j=2}^{i-1} \frac{\left(\lambda_{2, \phi}-\lambda_{1, \phi}\right)^{2}\left(\lambda_{j+1, \phi}-\lambda_{j, \phi}\right)}{\left(\lambda_{j+1, \phi}-\lambda_{1, \phi}\right)\left(\lambda_{j, \phi}-\lambda_{1, \phi}\right)} \sum_{\alpha=1}^{j-1}\left[1-\sum_{p=\alpha+1}^{j}\left(\lambda_{p, \phi}-\lambda_{1, \phi}\right) a_{\alpha p}^{2}\right]+\frac{\lambda_{i, \phi}-\lambda_{1, \phi}}{\frac{\lambda_{i, \phi}}{\lambda_{1, \phi}}} \\ & \leq \sum_{j=2}^{n+1} \frac{\left(\lambda_{2, \phi}-\lambda_{1, \phi}\right)^{2}\left(\lambda_{j+1, \phi}-\lambda_{j, \phi}\right)}{\left(\lambda_{j+1, \phi}-\lambda_{1, \phi}\right)\left(\lambda_{j, \phi}-\lambda_{1, \phi}\right)} \sum_{\alpha=1}^{j-1}\left[1-\sum_{p=\alpha+1}^{j}\left(\lambda_{p, \phi}-\lambda_{1, \phi}\right) a_{\alpha p}^{2}\right]+\frac{\lambda_{i, \phi}-\lambda_{1, \phi}}{\frac{\lambda_{1, \phi}}{\lambda_{1, \phi}}} \\ & \leq B+\frac{\lambda_{i, \phi}-\lambda_{1, \phi}}{\frac{\lambda_{i, \phi}}{\lambda_{1, \phi}}}. \end{aligned}

最后由 (3.15) 和上述不等式可得

\sum\limits_{\alpha=1}^n\left(\lambda_{\alpha+1,\phi}-\lambda _{1,\phi }\right) \leq 4 \lambda _{1,\phi }- 8nC,

\frac{{{\lambda_{2,\phi}} + \cdots + {\lambda _{n + 1,\phi}}}}{{{\lambda_{1,\phi}}}} \le \left( n + 4 \right) - \frac{{8nC}}{{{\lambda_{1,\phi}}}}.

特别地, 取上述 i=1 即可证明定理 1.4. C>0 时情况类似讨论.

致谢

感谢陈瑞丰博士, 高雅博士在论文写作过程中给予的帮助.

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