自1956年, Payne-Póyla和Weinberger^[14]开始研究欧氏空间拉普拉斯算子的特征值之后, 许多数学家开始关注该问题.其中杨洪苍教授^[19]证明了著名的第一杨型不等式

$ \sum\limits_{r=1}^{k}(\lambda_{k+1}-\lambda_{r})^{2}\leq \frac{4}{n}\sum\limits_{r=1}^{k}(\lambda_{k+1}-\lambda_{r})\lambda_{r} $

和第二杨型不等式

$ \lambda_{k+1}\leq \Big(1+\frac{4}{n}\Big)\frac{1}{k}\sum\limits_{r=1}^{k} \lambda_{r}. $

关于该算子特征值的研究进展, 可以参考文献[2-4, 7]等.近来, 一些数学家研究了权重黎曼流形 $(M, g, {\rm e}^{-\phi}{\rm d}v)$ 上的Witten -拉普拉斯算子的特征值问题^[10-12]

$ \begin{equation} \left\{ \begin{array}{ll} (-\Delta+\langle\nabla \phi, \nabla(\cdot)\rangle) u=-\lambda u, x\in\Omega, \\[1mm] u=0, x\in\partial\Omega, \\[1mm] \end{array} \right. \end{equation} $

(1.1)

这里 $\phi$ 是 $\overline{\Omega}$ 上的光滑函数.在2013年, 夏昌玉和许洪伟教授^[18]考虑了黎曼流形上该算子的特征值估计.

2003年, 钮鹏程教授^[13]研究了Hesienberg群 $H^{n}$ 上次拉普拉斯算子 $\Delta_{H^{n}}$ 的特征值问题

$ \begin{equation} \left\{ \begin{array}{ll} -\Delta_{H^{n}} u=\lambda u, x\in\Omega, \\[1mm] u=0, x\in\partial\Omega, \\[1mm] \end{array} \right. \end{equation} $

(1.2)

得到了下面的不等式

$ \begin{equation} \lambda_{k+1}-\lambda_{k}\leq \frac{2}{nk}\sum\limits_{r=1}^{k}\lambda_{r}. \end{equation} $

(1.3)

针对问题(1.2), Soufi, Harrel和Ilias^[15]利用交换子的方法得到了下面的结果

$ \begin{equation} \sum\limits_{r=1}^{k}(\lambda_{k+1}-\lambda_{r})^{2}\leq \frac{2}{n}\sum\limits_{r=1}^{k}(\lambda_{k+1}-\lambda_{r})\lambda_{r}. \end{equation} $

(1.4)

2010年, 黄广月^[8]考虑了 $H^{n}$ 上双拉普拉斯算子 $\Delta_{H}^{2}$ 的特征值估计问题.孙和军^[16]考虑了 $H^{n}$ 上薛定谔算子的特征值估计问题, 把文献[15]中的结果推广到加权特征值上. 2012年, 杜锋、吴传喜、李光汉等人^[5]考虑了 $H^{n}$ 上散度形算子 $- {\rm div}_{H}(A\nabla_{H})+V$ 的特征值估计, 得到如下结果

$ \begin{equation} \sum\limits_{i=1}^{k}(\lambda_{k+1}-\lambda_{i})^{2}\leq\frac{2\sigma^{2}\xi_{2}}{n\xi_{1}\tau^{2}}\sum\limits_{i=1}^{k}(\lambda_{k+1}-\lambda_{i})(\lambda_{i}-V_{0}\tau). \end{equation} $

(1.5)

本文中, 我们将主要研究具有概率测度 ${\rm d}\mu:={\rm e}^{-\phi(w)}{\rm d}v$ 的Heisenberg型群(简称H型群) G上散度形算子

$ \begin{eqnarray*} Lu=-{\rm div}_{G}(A\nabla_{G} u)+\langle A\nabla_{G}\phi, \nabla_{G}u\rangle+Vu \end{eqnarray*} $

的特征值估计, 其中 $\phi$ 在d $(w)\rightarrow \infty$ 时, $\frac{1}{4}|\nabla_{G}\phi|-\frac{1}{2}\Delta_{G}\phi\rightarrow \infty$ (d $(w)$ 表示 $G$ 上任一点 $w$ 和原点的距离), $ V$ 为非负位势函数, $ A$ 为对称正定矩阵.经分部积分运算可知

$ \begin{eqnarray} &&\int_{\Omega}f(-{\rm div}_{G}(A\nabla_{G} g)+\langle A\nabla_{G}\phi, \nabla_{G} g\rangle+Vg){\rm d}\mu \\ &=&\int_{\Omega}g(-{\rm div}_{G}(A\nabla_{G} f)+\langle A\nabla_{G}\phi, \nabla_{G} f\rangle+Vf){\rm d}\mu, \end{eqnarray} $

(1.6)

$ \begin{equation} \int_{\Omega}f(-{\rm div}_{G}(A\nabla_{G} g)+\langle A\nabla_{G}\phi, \nabla_{G} g\rangle ){\rm d}\mu=\int_{\Omega}\langle A\nabla_{G} g, \nabla_{G} f \rangle {\rm d}\mu , \end{equation} $

(1.7)

其中 $f$ 和 $g$ 均为 $\Omega$ 上的光滑函数, 并且满足 $f|_{\partial \Omega}=g|_{\partial \Omega}=0$ .由文献[9]可知, 该算子是 $L^{2}({\rm d}\mu)$ 上的正定自伴算子, 我们将考虑有界区域 $\Omega$ 上该算子的Dirichlet特征值问题

$ \begin{equation} \left\{ \begin{array}{ll} -{\rm div}_{G}(A\nabla_{G} u)+\langle A\nabla_{G}\phi, \nabla_{G} u\rangle +Vu=\lambda \rho u, \quad \hbox{in}\ \Omega, \\[1mm] u|_{\partial \Omega }=0. \end{array} \right. \end{equation} $

(1.8)

由文献[6, 9]可知, 该算子有离散的特征值

$ \begin{eqnarray*} 0<\lambda_{1}\leq\lambda_{2}\leq \cdots \leq\lambda_{r}\leq\cdots\nearrow. \end{eqnarray*} $

本文利用经典的试验函数法得到了如下主要结果：

定理1.1  设 $\lambda_{i}$ 为问题(1.8)的第 $i$ 个特征值, $i=1, 2, \cdots, n$ , $u_{i}$ 为 $\lambda_{i}$ 相应的特征函数, 且对应于不同特征值的特征函数相互正交, 令 $\tau=(\sup_{\Omega}\rho)^{-1}, \sigma=(\inf_{\Omega}\rho)^{-1}, $ $V_{0}=\min_{\Omega}V, \quad |\nabla\phi|\leq C_{0}, \xi_{1}I\leq A\leq\xi_{2}I$ , 则

$ \begin{eqnarray} \sum\limits_{i=1}^{k}(\lambda_{k+1}-\lambda_{i})^{2} \leq\frac{\xi_{2}\sigma^{2}}{2n\tau^{2}\xi_{1}^{2}}\sum\limits_{i=1}^{n}\left(\lambda_{k+1}-\lambda_{i}\right)\left[\xi_{2}C_{0}\sigma+(\xi_{2}^{2}C_{0}^{2}\sigma-4\xi_{1}V_{0}\tau +4\xi_{1}\lambda_{i})^{\frac{1}{2}}\right]^{2}, \end{eqnarray} $

(1.9)

其中 $\rho$ 为 $\overline{\Omega}$ 上的正连续函数, $\xi_{1}, \xi_{2}$ 为正的常数.

在2n+m维欧氏空间 $\mathbb{R}^n \times \mathbb{R}^n \times \mathbb{R}^{m}$ , 定义下列群运算

$ \begin{equation} (z, t)\circ (z', t')=(x, y, t)\circ (x', y', t')=\left(z_{i}+z_{i}', t_{j}+t_{j}'+\frac{1}{2}\langle z, U^{j}z' \rangle\right), \end{equation} $

(2.1)

$i=1, 2\cdots, 2n;j=1, 2, \cdots, m$ , 其中 $z=(x, y)\in \mathbb{R}^{2n}, t\in \mathbb{R}^{m}$ , $\langle\cdot, \cdot\rangle$ 表示欧氏内积, 矩阵 $U^{1}, U^{2}, \cdots, U^{2n}$ 满足下列条件:

(1) $U^{j}$ 是 $2n\times 2n$ 阶反对称正交矩阵, $\forall j=1, 2, \cdots, m$ ;

(2) $U^{i}U^{j}+U^{j}U^{i}=0$ , $ \forall i, j\in \{ 1, 2, \cdots, m\}$ , $i\neq j$ .满足这种群运算的 $2n+m$ 维欧氏空间称为Heisenberg型群, 简称为H型群.李代数 $g$ 的基底为

$ \begin{equation} X_{j}=\frac{\partial}{\partial x_{j}}+\frac{1}{2}\sum\limits_{k=1}^{m} \bigg(\sum\limits_{l=1}^{2n}z_{l}U_{l, j}^{(k)}\bigg)\frac{\partial}{\partial t_{k}}, \quad Y_{j}=\frac{\partial}{\partial y_{j}}+\frac{1}{2}\sum\limits_{k=1}^{m} \bigg(\sum\limits_{l=1}^{2n}z_{l}U_{l, j+n}^{(k)}\bigg)\frac{\partial}{\partial t_{k}}, \quad T_{k}=\frac{\partial}{\partial t_{k}}, \end{equation} $

(2.2)

其中

$ \left(U_{l, j}^{(k)}\right )_{2n\times 2n}=U^{(k)}. $

在H型群G上

$ \Delta_{G}=\sum\limits_{j=1}^{n}(X_{j}^{2}+Y_{j}^{2}), \ \ \ \ \nabla_{G}=(X_{1}, X_{2}, \cdots, X_{n}, Y_{1}, Y_{2}, \cdots, Y_{n}), \\ \ \ \ \ {\rm div}_{G}\varphi=\sum\limits_{j=1}^{n}X_{j}\varphi_{j}+Y_{j}\varphi_{n+j}, \forall\varphi=(\varphi_{1}, \varphi_{2}, \cdots, \varphi_{2n})\in C^{2}(\Omega). $

G上任一点 $w=(z, t)$ 与原点的距离定义为

$ {\rm d}(w)=(|z|^{4}+16|t|^{2})^{\frac{1}{4}}. $

H型群的最简单例子就是著名的Heisenberg群. $(2n+1)$ 维的Heisenberg群 $H^{n}$ 满足下列非交换群运算

$ (x, y, t)(x', y', t')=\left(x+x', y+y', t+t'+\frac{1}{2}(\langle x', y\rangle-\langle x, y'\rangle)\right), $

其中 $x, y, x', y'\in \mathbb{R}^n $ , $t, t'\in \mathbb{R}$ , $\langle\cdot, \cdot\rangle$ 表示 $\mathbb{R}^n $ 中的欧氏内积. $H^{n}$ 的李代数的基底 $\{X_{j}, Y_{j}, T\}, $ $j=1, 2, \cdots, n.$

$ \begin{eqnarray*} X_{j}=\frac{\partial}{\partial x_{j}}+\frac{y_{j}}{2}\frac{\partial}{\partial t}, \quad Y_{j}=\frac{\partial}{\partial y_{j}}-\frac{x_{j}}{2}\frac{\partial}{\partial t}, \quad T=\frac{\partial}{\partial t}. \end{eqnarray*} $

$H^{n}$ 上的拉普拉斯算子为 $\Delta_{H}=\sum\limits_{j=1}^{n}(X_{j}^{2}+Y_{j}^{2})$ .

关于上述算子, 有下列命题成立.

引理2.1^[17]  假设 $\Omega$ 是G上的有界区域, 则对 $\forall f, g \in C^{k}(\Omega), k\geq2, $

$ \begin{equation} \nabla_{G}(fg)=f\nabla_{G}(g)+g\nabla_{G}(f); \end{equation} $

(2.3)

$ \begin{equation} \Delta_{G}fg=f\Delta_{G}g+2\langle\nabla_{G}(f), \nabla_{G}(g) \rangle+g\Delta_{G}f . \end{equation} $

(2.4)

当 $f|_{\partial{\Omega}}=g|_{\partial{\Omega}}=0$ ,

$ \ \ \ \ \int_{\Omega}(X_{i}f)( X_{j}g){\rm d}\mu=-\int_{\Omega}f(X_{i}X_{j}g){\rm d}\mu;\\ \int_{\Omega}f{\rm div}_{G}(\nabla_{G}g) {\rm d}\mu=-\int_{\Omega}\langle\nabla_{G}(f), \nabla_{G}(g) \rangle {\rm d}\mu;\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \int_{\Omega}f\Delta_{G}g{\rm d}\mu=\int_{\Omega}g\Delta_{G}f{\rm d}\mu. $

引理3.1  设 $\lambda_{i}$ 为问题(1.8)的第 $i$ 个特征值, $i=1, 2, \cdots, n$ , $u_{i}$ 为 $\lambda_{i}$ 相应的特征函数, 且对应于不同特征值的特征函数相互正交, 即

$ \begin{equation} \left\{ \begin{array}{lll} -{\rm div}_{G}(A\nabla_{G} u_{i})+\langle A\nabla_{G}\phi, \nabla_{G} u_{i}\rangle+Vu_{i}=\lambda_{i} \rho u_{i}, \quad \hbox{in}\ \Omega, \\[2mm] \int_{\Omega}\rho u_{i}u_{j}{\rm d}\mu=\delta_{ij}, \\[2mm] u_{i}|_{\partial \Omega}=0, \end{array} \right. \end{equation} $

(3.1)

对任意的 $h\in C^{3}(\Omega)\cap C^{2}(\partial \Omega)$ 及任意整数 $k$ , 有

$ \begin{eqnarray} &&\sum\limits_{i=1}^{k}(\lambda_{k+1}-\lambda_{i})^{2} \int_{\Omega}u_{i}^{2}\langle\nabla_{G} h, \nabla_{G} h\rangle {\rm d}\mu \\ &\leq& \epsilon\sum\limits_{i=1}^{k}(\lambda_{k+1}-\lambda_{i})^{2} \int_{\Omega}u_{i}^{2}\langle A\nabla_{G} h, \nabla_{G} h\rangle {\rm d}\mu \\ &&+\sum\limits_{i=1}^{k}\frac{\lambda_{k+1}-\lambda_{i}} {\epsilon}\int_{\Omega}\frac{1}{\rho}(\langle\nabla_{G} u_{i}, \nabla_{G} h\rangle)+\frac{1}{2}u_{i}\Delta_{G} h)^{2}{\rm d}\mu, \end{eqnarray} $

(3.2)

其中 $\epsilon$ 为任意正数.

证  定义试验函数

$ \begin{equation} \varphi_{i}=hu_{i}-\sum\limits_{j=1}^{k}a_{ij}u_{j}, \end{equation} $

(3.3)

其中 $a_{ij}=\int_{\Omega}\rho hu_{i}u_{j}{\rm d}\mu=a_{ji}$ , 则

$ \begin{equation} \int_{\Omega}\rho\varphi_{i}u_{j}{\rm d}\mu=0, \varphi_{i}|_{\partial \Omega}=0, \forall i, j=1, \cdots, k. \end{equation} $

(3.4)

直接计算可知

$ \begin{eqnarray} L\varphi_{i}&=&(-{\rm div}_{G}(A\nabla_{G} (\cdot))+\langle A\nabla_{G}\phi, \nabla_{G} (\cdot) \rangle+V)\varphi_{i} \\ &=&hLu_{i}+u_{i}Lh-Vhu_{i}-2\langle A\nabla_{G} h, \nabla_{G} u_{i}\rangle-\sum\limits_{j=1}^{k}a_{ij}\lambda_{j}u_{j}. \end{eqnarray} $

(3.5)

将上式代入Rayleigh-Ritz^[1]不等式

$ \begin{equation} \lambda_{k+1}\leq \frac {\int_{\Omega}\varphi_{i} L\varphi_{i}}{\int_{\Omega}\rho \varphi_{i}^{2}{\rm d}\mu}, \end{equation} $

(3.6)

可得

$ \begin{equation} (\lambda_{k+1}-\lambda_{i})\int_{\Omega}\rho\varphi_{i}^{2}{\rm d}\mu\leq\int_{\Omega}\varphi_{i}(u_{i}Lh-Vhu_{i}+2\langle A\nabla_{G} h, \nabla_{G} u_{i}\rangle){\rm d}\mu. \end{equation} $

(3.7)

令

$ \begin{equation} b_{ij}=\int_{\Omega}(u_{i}Lh-Vhu_{i}-2\langle A\nabla_{G} h, \nabla_{G} u_{i}\rangle)u_{j}{\rm d}\mu, \end{equation} $

(3.8)

直接计算可知

$ \begin{equation} b_{ij}=-b_{ji}, b_{ij}=(\lambda_{i}-\lambda_{j})a_{ij}. \end{equation} $

(3.9)

结合(3.7)式, 可得

$ \begin{eqnarray} (\lambda_{k+1}-\lambda_{i})\int_{\Omega}\rho\varphi_{i}^{2}{\rm d}\mu &\leq&-\int_{\Omega}h u_{i}(u_{i}({\rm div}_{G}(A\nabla_{G} h)-\langle A\nabla_{G} \phi, \nabla_{G} h\rangle \\ &&-2\langle\nabla_{G} u_{i}, A\nabla_{G} h\rangle)){\rm d}\mu \\ &&+\sum\limits_{j=1}^{k}(\lambda_{i}-\lambda_{j})a_{ij}^{2}. \end{eqnarray} $

(3.10)

令

$ c_{ij}=\int_{\Omega}u_{j}(\langle \nabla_{G} u_{i}, \nabla h \rangle+\frac{1}{2}u_{i}\Delta_{G} h){\rm d}\mu, $

则 $c_{ij}=-c_{ji}, $ 并且

$ \begin{eqnarray} &&\int_{\Omega}(-2)\varphi_{i}(\langle \nabla_{G} u_{i}, \nabla_{G} h \rangle+\frac{1}{2}u_{i}\Delta_{G} h){\rm d}\mu \\ &=&-2\int_{\Omega}h u_{i}(\langle \nabla_{G} u_{i}, \nabla_{G} h \rangle +\frac{1}{2}u_{i}\Delta_{G} h){\rm d}\mu+2\sum\limits_{j=1}^{k}a_{ij}c_{ij} \\ &=&\int_{\Omega}u_{i}^{2}\langle \nabla_{G} h, \nabla_{G} h \rangle {\rm d}\mu+2\sum\limits_{j=1}^{k}a_{ij}c_{ij}. \end{eqnarray} $

(3.11)

将(3.11)式两端同乘 $(\lambda_{k+1}-\lambda_{i})^{2}$ , 则

$ \begin{align} & \ \ \ \ \ \ {{({{\lambda }_{k+1}}-{{\lambda }_{i}})}^{2}}(\int_{\Omega }{u_{i}^{2}}\langle {{\nabla }_{G}}h,{{\nabla }_{G}}h\rangle {\rm{d}}\mu +2\sum\limits_{j=1}^{k}{{{a}_{ij}}}{{c}_{ij}}) \\ & ={{({{\lambda }_{k+1}}-{{\lambda }_{i}})}^{2}}\int_{\Omega }{(-2)}\sqrt{\rho }{{\phi }_{i}}(\frac{1}{\sqrt{\rho }}(\langle {{\nabla }_{G}}{{u}_{i}},{{\nabla }_{G}}h\rangle +\frac{1}{2}{{u}_{i}}{{\Delta }_{G}}h)-\sum\limits_{j=1}^{k}{{{c}_{ij}}}\sqrt{\rho }{{u}_{j}}){\rm{d}}\mu \\ & \ \ \ \le \epsilon {{({{\lambda }_{k+1}}-{{\lambda }_{i}})}^{3}}\int_{\Omega }{\rho }\phi _{i}^{2}{\rm{d}}\mu \\ & \ \ \ \ \ \ +\frac{{{\lambda }_{k+1}}-{{\lambda }_{i}}}{\epsilon }\int_{\Omega }{(}\frac{1}{\sqrt{\rho }}(\langle {{\nabla }_{G}}{{u}_{i}},{{\nabla }_{G}}h\rangle +\frac{1}{2}{{u}_{i}}{{\Delta }_{G}}h)-\sum\limits_{j=1}^{k}{{{c}_{ij}}}\sqrt{\rho }{{u}_{j}}{{)}^{2}}{\rm{d}}\mu \\ & =\epsilon {{({{\lambda }_{k+1}}-{{\lambda }_{i}})}^{3}}\int_{\Omega }{\rho }\phi _{i}^{2}{\rm{d}}\mu +\frac{{{\lambda }_{k+1}}-{{\lambda }_{i}}}{\epsilon }(\int_{\Omega }{\frac{1}{\rho }}{{(\langle {{\nabla }_{G}}{{u}_{i}},{{\nabla }_{G}}h\rangle +\frac{1}{2}{{u}_{i}}{{\Delta }_{G}}h)}^{2}}{\rm{d}}\mu -\sum\limits_{j=1}^{k}{c_{ij}^{2}}) \\ & \le \epsilon {{({{\lambda }_{k+1}}-{{\lambda }_{i}})}^{2}}(-\int_{\Omega }{h}{{u}_{i}}({{u}_{i}}{\rm{di}}{{\rm{v}}_{G}}(A{{\nabla }_{G}}h)-\langle A{{\nabla }_{G}}\phi ,{{\nabla }_{G}}h\rangle +2\langle {{\nabla }_{G}}{{u}_{i}},A{{\nabla }_{G}}h\rangle ){\rm{d}}\mu \\ & \ \ \ \ \ +\sum\limits_{j=1}^{k}{({{\lambda }_{i}}-{{\lambda }_{j}})}a_{ij}^{2})+\frac{{{\lambda }_{k+1}}-{{\lambda }_{i}}}{\epsilon }(\int_{\Omega }{\frac{1}{\rho }}{{(\langle {{\nabla }_{G}}{{u}_{i}},{{\nabla }_{G}}h\rangle +\frac{1}{2}{{u}_{i}}{{\Delta }_{G}}h)}^{2}}{\rm{d}}\mu -\sum\limits_{j=1}^{k}{c_{ij}^{2}}), \\ \end{align} $

(3.12)

这里 $\epsilon$ 为任意正常数.关于 $i$ 从1加到 $k$ , 可得

$ \ \ \ \sum\limits_{j=1}^{k}(\lambda_{k+1}-\lambda_{i})^{2}\int_{\Omega}u_{i}^{2}\langle \nabla_{G} h, \nabla_{G} h \rangle {\rm d}\mu-2\sum\limits_{i, j=1}^{k}(\lambda_{k+1}-\lambda_{i})(\lambda_{i}-\lambda_{j})a_{ij}c_{ij} \\ \leq \epsilon\sum\limits_{i=1}^{k}(\lambda_{k+1}-\lambda_{i})^{2} \bigg(-\int_{\Omega}h u_{i}(u_{i}{\rm div}_{G}(A\nabla_{G} h)-\langle A\nabla_{G}\phi, \nabla_{G} h\rangle+2\langle\nabla_{G} u_{i}, A\nabla_{G} h\rangle){\rm d}\mu\bigg) \\ \ \ \ \ \ \ +\sum\limits_{i=1}^{k}\frac{\lambda_{k+1}-\lambda_{i}}{\epsilon} \bigg(\int_{\Omega}\Big(\frac{1}{\rho}(\langle \nabla_{G} u_{i}, \nabla_{G} h \rangle+\frac{1}{2}u_{i}\Delta_{G} h)^{2}{\rm d}\mu \\ \ \ \ \ \ \ -\sum\limits_{i, j=1}^{k} (\lambda_{k+1}-\lambda_{i})\epsilon (\lambda_{i}-\lambda_{j})^{2} a_{ij}^{2}\Big)\bigg) -\sum\limits_{i, j=1}^{k}\frac{(\lambda_{k+1}-\lambda_{i})}{\epsilon}c_{ij}^{2}. $

(3.13)

因为 $a_{ij}=a_{ji}, c_{ij}=-c_{ji}, $ 从而

$ \begin{eqnarray} &&\sum\limits_{j=1}^{k}(\lambda_{k+1}-\lambda_{i})^{2}\int_{\Omega}u_{i}^{2}\langle \nabla_{G} h, \nabla_{G} h \rangle {\rm d}\mu \\ &\leq&\epsilon\sum\limits_{i=1}^{k}(\lambda_{k+1}-\lambda_{i})^{2} \bigg(-\int_{\Omega}h u_{i}(u_{i}({\rm div}_{G}(A\nabla_{G} h) -\langle A\nabla_{G}\phi, \nabla_{G} h\rangle \\ &&+2\langle\nabla_{G} u_{i}, A\nabla_{G} h\rangle)){\rm d}\mu\bigg) \\ &&+\sum\limits_{i=1}^{k}\frac{\lambda_{k+1}-\lambda_{i}}{\epsilon} \bigg(\int_{\Omega}\Big(\frac{1}{\rho}(\langle \nabla_{G} u_{i}, \nabla_{G} h \rangle+\frac{1}{2}u_{i}\Delta_{G} h)\Big)^{2}{\rm d}\mu\bigg). \end{eqnarray} $

(3.14)

考虑到 ${\rm d}\mu={\rm e}^{-\phi}{\rm d}v$ , 可得

$ \int_{\Omega}hu_{i}^{2}{\rm div}_{G}(A\nabla_{G} h){\rm d}\mu= \int_{\Omega}hu_{i}^{2}{\rm div}_{G}(A\nabla_{G} h){\rm e}^{-\phi}{\rm d}v \\ =-\int_{\Omega}\langle A\nabla_{G} h, \nabla_{G} h \rangle u_{i}^{2} {\rm e}^{-\phi}{\rm d}v-\int_{\Omega}\langle A\nabla_{G} h, \nabla_{G} u_{i} \rangle 2hu_{i}{\rm e}^{-\phi}{\rm d}v+\int_{\Omega}\langle A\nabla_{G} h, \nabla_{G} \phi \rangle hu_{i}^{2}\\ {\rm e}^{-\phi}{\rm d}v \\ =-\int_{\Omega}\langle A\nabla_{G} h, \nabla_{G} h \rangle u_{i}^{2}{\rm d}\mu -\int_{\Omega}\langle A\nabla_{G} h, \nabla_{G} u_{i} \rangle 2hu_{i}{\rm d}\mu+\int_{\Omega}\langle A\nabla_{G} h, \nabla_{G} \phi \rangle hu_{i}^{2}{\rm d}\mu. $

(3.15)

将(3.15)式代入(3.14)式, 引理3.1得证.

定理3.2  在引理3.1的条件下, 令 $\tau=(\sup_{\Omega}\rho)^{-1}, \sigma=(\inf_{\Omega}\rho)^{-1}, $ $ V_{0}=\min_{\Omega}V, $ $ |\nabla\phi|\leq C_{0}, $ $\xi_{1}I\leq A\leq\xi_{2}I$ , 则

$ \begin{eqnarray} \sum\limits_{i=1}^{k}(\lambda_{k+1}-\lambda_{i})^{2}\leq\frac{\xi_{2}\sigma^{2}}{2n\tau^{2}\xi_{1}^{2}}\sum\limits_{i=1}^{n}(\lambda_{k+1}-\lambda_{i})\left[\xi_{2}C_{0}\sigma+(\xi_{2}^{2}C_{0}^{2}\sigma-4\xi_{1}V_{0}\tau +4\xi_{1}\lambda_{i})^{\frac{1}{2}}\right]^{2}. \end{eqnarray} $

(3.16)

证  将 $h=x_{\alpha}, $ 或 $h=y_{\alpha}$ 代入(3.2)式, 并考虑到

$ \begin{equation} \Delta_{G}x_{\alpha}=\Delta_{G}y_{\alpha}=0, \end{equation} $

(3.17)

$ \begin{equation} \sum\limits_{\alpha=1}^{n}(|\nabla_{G}x_{\alpha}|^{2}+|\nabla_{G}y_{\alpha}|^{2})=2n, \end{equation} $

(3.18)

$ \begin{equation} \sum\limits_{\alpha=1}^{n}(\langle\nabla_{G}u_{i}, \nabla_{G}x_{\alpha}\rangle^{2}+\langle\nabla_{G}u_{i}, \nabla_{G}y_{\alpha}\rangle^{2})=|\nabla_{G}u_{i}|^{2}, \end{equation} $

(3.19)

以及

$ \xi_{1}I\leq A\leq\xi_{2}I, $

可以得到

$ \begin{equation} 2n(1-\varepsilon\xi_{2})\sum\limits_{i=1}^{k}(\lambda_{k+1}-\lambda_{i})^{2}\int_{\Omega}u_{i}^{2}{\rm d}\mu\leq\frac{\sigma}{\varepsilon}\sum\limits_{i=1}^{k}(\lambda_{k+1}-\lambda_{i})\int_{\Omega}|\nabla_{G}u_{i}|^{2}{\rm d}\mu. \end{equation} $

(3.20)

很显然

$ \begin{equation} \sigma= \sigma\int_{\Omega}\rho u_{i}^{2}{\rm d}\mu\geq\int_{\Omega}u_{i}^{2}{\rm d}\mu\geq \tau \int_{\Omega}\rho u_{i}^{2}{\rm d}\mu=\tau. \end{equation} $

(3.21)

并且

$ \begin{eqnarray} \lambda_{i}&=&\int_{\Omega}\langle A\nabla_{G} u_{i}, \nabla_{G} u_{i}\rangle {\rm d}\mu+\int_{\Omega}u_{i}\langle A\nabla_{G}\phi, \nabla_{G} u_{i}\rangle {\rm d}\mu+\int_{\Omega}Vu_{i}^{2}{\rm d}\mu \\ &\geq &\xi_{1}\int_{\Omega}|\nabla_{G} u_{i}|^{2}{\rm d}\mu+\int_{\Omega}u_{i}\langle A\nabla_{G}\phi, \nabla_{G} u_{i}\rangle {\rm d}\mu+\int_{\Omega}Vu_{i}^{2}{\rm d}\mu. \end{eqnarray} $

(3.22)

其中

$ \begin{eqnarray} \left|\int_{\Omega}u_{i}\langle A\nabla_{G} \phi, \nabla_{G} u_{i}\rangle {\rm d}\mu\right|&\leq& \xi_{2}\int_{\Omega}|u_{i}||\nabla_{G}\phi||\nabla_{G} u_{i}|{\rm d}\mu \\ &\leq&\xi_{2}C_{0}\left(\int_{\Omega}u_{i}^{2}{\rm d}\mu\right)^{\frac{1}{2}}\left(\int_{\Omega}|\nabla_{G} u_{i} |^{2}{\rm d}\mu\right)^{\frac{1}{2}} \\ &\leq& \xi_{2}C_{0}\sigma\left(\int_{\Omega}|\nabla_{G} u_{i}|^{2}{\rm d}\mu\right)^{\frac{1}{2}}, \end{eqnarray} $

(3.23)

从而

$ \begin{eqnarray} \lambda_{i}&\geq &\xi_{1}\int_{\Omega}|\nabla_{G} u_{i}|^{2}{\rm d}\mu+\int_{\Omega}u_{i}\langle A\nabla_{G}\phi, \nabla_{G}u_{i}\rangle {\rm d}\mu+\int_{\Omega}Vu_{i}^{2}{\rm d}\mu \\ &\geq& \xi_{1}\int_{\Omega}|\nabla_{G} u_{i}|^{2}{\rm d}\mu- \xi_{2}C_{0}\sigma\bigg(\int_{\Omega}|\nabla_{G} u_{i}|^{2}{\rm d}\mu \bigg)^{\frac{1}{2}}+V_{0}\tau. \end{eqnarray} $

(3.24)

解上面的不等式, 可得

$ \begin{equation} \int_{\Omega}|\nabla_{G} u_{i}|^{2}{\rm d}\mu\leq \frac{1}{4\xi_{1}^{2}}\left\{\xi_{2}C_{0}\sigma+(\xi_{2}^{2}C_{0}^{2}\sigma^{2}-4\xi_{1}V_{0}\tau+4\xi_{1}\lambda_{i})^{\frac{1}{2}}\right\}^{2}. \end{equation} $

(3.25)

将(3.18)和(3.15)式代入(3.15)式, 可得

$ \begin{equation} \epsilon=\left(\frac{A}{B}\right)^{\frac{1}{2}}, \end{equation} $

(3.26)

其中

$ A=\frac{\sigma}{4\xi_{1}^{2}}\sum\limits_{i=1}^{n}(\lambda_{k+1}-\lambda_{i})\left[\xi_{2}C_{0}\sigma+(\xi_{2}^{2}C_{0}^{2}\sigma-4\xi_{1}V_{0}\tau +4\xi_{1}\lambda_{i})^{\frac{1}{2}}\right]^{2}, \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ B=2n\xi_{2}\sigma\sum\limits_{i=1}^{n}(\lambda_{k+1}-\lambda_{i})^{2}, $

证毕.

推论3.3  在引理3.1的条件下, 假设 $\phi$ =常数, 则

$ \begin{equation} \sum\limits_{i=1}^{k}(\lambda_{k+1}-\lambda_{i})^{2}\leq\frac{2\sigma^{2}\xi_{2}}{n\xi_{1}\tau^{2}}\sum\limits_{i=1}^{k}(\lambda_{k+1}-\lambda_{i})(\lambda_{i}-V_{0}\tau). \end{equation} $

(3.27)

我们可以发现推论3.3的结果恰好就是文献[5]中的结果(5).通过求解定理3.2中的不等式, 可以得到一个第二杨型不等式.

推论3.4  在引理3.1的条件下, 有

$ \begin{equation} \lambda_{k+1}\leq \frac{1}{2k}\left\{2\sum\limits_{i=1}^{k}\lambda_{i}+nEF +\left[\bigg(2\sum\limits_{i=1}^{k}\lambda_{i}+nEF\bigg)^{2} -4k\bigg(\sum\limits_{i=1}^{k}\lambda_{i}^{2}+EF\sum\limits_{i=1}^{k}\lambda_{i} \bigg)\right]^{\frac{1}{2}}\right\}, \end{equation} $

(3.28)

其中

$ E=\frac{\xi_{2}\sigma^{2}}{2n\tau^{2}\xi_{1}^{2}}, \quad F=\left[\xi_{2}C_{0}\sigma+\left(\xi_{2}^{2}C_{0}^{2}\sigma-4\xi_{1}V_{0}\tau +4\xi_{1}\lambda_{i}\right)^{\frac{1}{2}}\right]^{2}. $

通过该推论, 我们可以利用前 $k$ 个特征值来估计第 $k+1$ 个特征值.