## Laplace算子特征值和的精细下界

1南京师范大学数学科学学院数学研究所 南京 210023

2莆田学院数学系 福建莆田 351100

3应用数学福建省高校重点实验室(莆田学院) 福建莆田 351100

## Refined Lower Bound for Sums of Eigenvalues of the Laplace Operator

He Yue,1,3,*, Ruan Qihua,2,3

1Institute of Mathematics, School of Mathematics Sciences, Nanjing Normal University, Nanjing 210023

2Department of Mathematics, Putian University, Fujian Putian 351100

3Key Laboratory of Applied Mathematics (Putian University), Fujian Province University, Fujian Putian 351100

 基金资助: 国家自然科学基金.  11871278国家自然科学基金.  11971253应用数学福建省高校重点实验室(莆田学院)开放课题.  SX202101

 Fund supported: The NSFC.  11871278The NSFC.  11971253Key Laboratory of Applied Mathematics of Fujian Province University(Putian University).  SX202101

Abstract

In this paper, we study lower bounds for higher eigenvalues of the Dirichlet eigenvalue problem of the Laplacian on a bounded domain $\Omega$ in $\Bbb R ^n$. It is well known that the $k$-th Dirichlet eigenvalue $\lambda_k(\Omega)$ obeys the Weyl asymptotic formula, that is, $\lambda_k(\Omega)\sim\frac{4\pi^2}{[\omega_nV(\Omega)]^\frac{2}{n}}k^\frac{2}{n} \qquad\hbox{as}\quad k\rightarrow\infty,$ where $\omega_n$ and $V(\Omega)$ are the volume of $n$-dimensional unit ball in $\Bbb R ^n$ and the volume of $\Omega$ respectively. In view of the above formula, Pólya conjectured that $\lambda_k(\Omega)\geq\frac{4\pi^2}{[\omega_nV(\Omega)]^\frac{2}{n}}k^\frac{2}{n} \qquad\hbox{for}\quad k\in{\Bbb N}.$ This is the well-known conjecture of Pólya. Studies on this topic have a long history with much work. In particular, one of the more remarkable achievements in recent tens years has been achieved independently by Berezin[2] and Li and Yau[4], respectively. They solved partially the conjecture of Pólya with a slight difference by a factor $n/(n+2)$. Later, Melas[7] improved Berezin-Li-Yau's estimate by adding an additional positive term of the order of $k$ to the right side. Here, following almost the same argument as Melas, we further refine Melas's estimate.

Keywords： The (fractional) Laplace operator ; The Dirichlet eigenvalue ; Higher eigenvalues ; The Weyl asymptotic formula ; Conjecture of Pólya ; The Berezin-Li-Yau inequality ; The moment of inertia

He Yue, Ruan Qihua. Refined Lower Bound for Sums of Eigenvalues of the Laplace Operator. Acta Mathematica Scientia[J], 2023, 43(1): 14-26 doi:

## 1 引言

$\Omega\subseteq\Bbb R ^n$是一个具有分片光滑边界$\partial\Omega$的有界域.考虑如下Dirichlet特征值问题

$$$\label{dep-s-11-6-8-20-58} \left\{ \begin{array}{ll} -\Delta u=\lambda u,\qquad&\hbox{在}\,\,\Omega\,\,\hbox{中},\\ u=0,\qquad&\hbox{在}\,\,\partial\Omega\,\,\hbox{上}, \end{array} \right.$$$

$$$\label{ineq-21-5-1-21} 0<\lambda_1(\Omega)<\lambda_2(\Omega)\leq\lambda_3(\Omega) \leq\cdots\leq\lambda_k(\Omega)\leq\cdots\nearrow\infty,$$$

$$$\label{Weyl-asymptotic-formula} \lambda_k(\Omega)\sim\frac{4\pi^2}{[\omega_nV(\Omega)]^\frac{2}{n}}k^\frac{2}{n} \qquad\hbox{当}\,\,k\rightarrow\infty\,\,\hbox{时},$$$

$$$\label{Polya-conjecture} \lambda_k(\Omega)\geq\frac{4\pi^2}{[\omega_nV(\Omega)]^\frac{2}{n}}k^\frac{2}{n}.$$$

$$$\label{Lie-1980-estimate} \lambda_k(\Omega)\geq c_n\frac{4\pi^2}{[\omega_nV(\Omega)]^\frac{2}{n}}k^\frac{2}{n}, \quad\forall\,\,k\in{\Bbb N}.$$$

$$$\label{Li-Yau-1983-esti-1} \sum_{j=1}^k\lambda_j(\Omega)\geq \frac{n}{n+2}\cdot\frac{4\pi^2}{[\omega_nV(\Omega)]^\frac{2}{n}}k^{1+\frac{2}{n}}, \quad\forall\,\,k\in{\Bbb N}.$$$

$\sum_j(z-\lambda_j)_+^\sigma\leq L_{\sigma,n}^{cl}V(\Omega)z^{\sigma+\frac{n}{2}}, \quad\forall\,\,\sigma\geq1,$

$$$\label{Li-Yau-1983-esti-2} \lambda_k(\Omega)\geq\frac{n}{n+2}\cdot\frac{4\pi^2}{[\omega_nV(\Omega)]^\frac{2}{n}}k^\frac{2}{n}, \quad\forall\,\,k\in{\Bbb N}.$$$

$$$\label{eqn-14-10-23a-21-38} \sum_{j=1}^k\lambda_j(\Omega)\geq \frac{n}{n+2}\cdot\frac{4\pi^2}{[\omega_nV(\Omega)]^\frac{2}{n}}k^{1+\frac{2}{n}} +\frac{1}{24(n+2)}\cdot\frac{V(\Omega)}{I(\Omega)}k, \quad\forall\,\,k\in{\Bbb N},$$$

$\lambda_k(\Omega)\geq\frac{n}{n+2}\cdot\frac{4\pi^2}{[\omega_nV(\Omega)]^\frac{2}{n}}k^\frac{2}{n} +\frac{1}{24(n+2)}\cdot\frac{V(\Omega)}{I(\Omega)}, \quad\forall\,\,k\in{\Bbb N}.$

$\int_{\tau}^{\tau+1}a(a-b)^{2}{\rm d}a=\int_\frac{1}{3}^\frac{4}{3}a(a-1)^{2}{\rm d}a =\frac{7}{9\times12}<\frac{1}{12}=\frac{1}{2}b^{2}-\frac{2}{3} b+\frac{1}{4}.$

## 2 记号和准备工作

$I(\Omega):=\min_{a\in\Bbb R ^n}\int_\Omega|x-a|^2{\rm d}x=\int_\Omega|x|^2\,{\rm d}x.$

$\{u_j\}_{j=1}^k$是相应于$\{\lambda_j(\Omega)\}_{j=1}^k$的一个标准正交特征函数族.$u_j$的Fourier变换可以表示为

$\hat{u}_j(z):=(2\pi)^{-n/2}\int_\Omega u_j(x){\rm e}^{{\rm i}\langle x,z\rangle}{\rm d}x,$

$$$\label{eq-21-1-1-1} F(z):=\sum_{j=1}^{k}\left|\hat{u}_j(z)\right|^{2}.$$$

$$$\label{eq-21-1-2-16} \int_{\Bbb R ^{n}}F(z){\rm d}z=k,$$$
$$$\label{eq-21-1-8-2} F(z)\leq(2\pi)^{-n}V(\Omega),\quad\forall\,\,z\in\Bbb R ^n,$$$
$$$\label{eq-21-1-1-2} \int_{\Bbb R ^{n}}|z|^{2}F(z){\rm d}z=\sum_{j=1}^{k} \lambda_{j}(\Omega),$$$
$\begin{matrix}\label{eq-21-1-1-3} |\nabla F(z)|\leq2(2\pi)^{-n}\sqrt{V(\Omega)I(\Omega)}:=\rho,\quad\forall\,\,z\in\Bbb R ^n. \end{matrix}$

$$$\label{eq-21-1-1-6} F^*(z)=\phi(|z|)$$$

$$$\label{ineq-21-1-1-5} 0\leq-\phi'(s)\leq\rho$$$

$$$\label{eq(2.13)} k=\int_{\Bbb R ^{n}}F(z){\rm d}z=\int_{\Bbb R ^{n}}F^{*}(z){\rm d}z =n \omega_{n}\int_{0}^{\infty} s^{n-1}\phi(s){\rm d}s.$$$

$$$\label{eq(2.14)} \sum_{j=1}^{k} \lambda_{j}(\Omega)= \int_{\Bbb R ^{n}}|z|^{2}F(z){\rm d}z\geq \int_{\Bbb R ^{n}}|z|^{2}F^{*}(z){\rm d}z =n \omega_{n} \int_{0}^{\infty} s^{n+1} \phi(s){\rm d}s.$$$

$$$\label{eq-21-21-1-2-2} \int_0^{\infty}h(s){\rm d}s=1\qquad\hbox{且}\qquad\int_0^{\infty}s^{n+2}h(s){\rm d}s<\infty.$$$

$$$\label{eq-21-1-2-9} \int_a^{a+1}s^n{\rm d}s=\int_0^{\infty}s^nh(s){\rm d}s\qquad\hbox{且}\qquad \int_a^{a+1}s^{n+2}{\rm d}s\leq\int_0^{\infty}s^{n+2}h(s){\rm d}s.$$$

$$$\label{eq-21-5-13-15} \int_0^{\infty} s^{n-1}\psi(s){\rm d}s=A,\qquad \int_0^{\infty} s^{n+1}\psi(s){\rm d}s=B.$$$

$\begin{matrix}\label{eq-21-1-26-6} B\geq\frac{1}{n+2}\left[(nA)^{1+\frac{2}{n}}+\frac{5}{3}A -\frac{20}{9}n^{-\frac{1}{n}}A^{1-\frac{1}{n}} +\frac{29}{24}n^{-\frac{2}{n}}A^{1-\frac{2}{n}} -\frac{37}{150}n^{-\frac{3}{n}}A^{1-\frac{3}{n}}\right]. \end{matrix}$

$$$\label{eq-20-12-26-19} \int_0^{\infty}h(s){\rm d}s=\psi(0)=1.$$$

$\liminf_{t\to\infty}t^{n+1}\psi(t)=0$和分部积分公式可得

$$$\label{eq-20-12-26-21} \int_{0}^{\infty}s^{n}h(s){\rm d}s =\lim_{t\to\infty}\Big[-t^n\psi(t)+n\int_0^ts^{n-1}\psi(s){\rm d}s\Big] =n\int_{0}^{\infty}s^{n-1}\psi(s){\rm d}s=nA.$$$

$\begin{matrix}\label{ineq-20-12-29-6} \int_0^{\infty}s^{n+2}h(s){\rm d}s &=&\lim_{t\to\infty}\Big[-t^{n+2}\psi(t) +(n+2)\int_0^ts^{n+1}\psi(s){\rm d}s\Big]\nonumber\\ &=&(n+2)\int_0^{\infty}s^{n+1}\psi(s){\rm d}s =(n+2)B.\nonumber \end{matrix}$

$$$\label{eq-20-12-26-8} \int_a^{a+1}s^n{\rm d}s=\int_0^{\infty}s^nh(s){\rm d}s=nA$$$

$$$\label{eq-20-12-22a-20} \int_a^{a+1}s^{n+2}{\rm d}s\leq\int_0^{\infty}s^{n+2}h(s){\rm d}s=(n+2)B.$$$

$\tau:=(nA)^{1/n}$. 显然,使用(3.5)式和Jensen不等式可得

$\tau^n=nA=\int_a^{a+1}s^n{\rm d}s\geq\Big(\int_a^{a+1}s{\rm d}s\Big)^n =\big(a+\frac{1}{2}\big)^n\quad\Longrightarrow\quad \tau\geq a+\frac{1}{2}\quad\Longrightarrow\quad \tau-a\geq\frac{1}{2}.$

$\tau^n=nA=\int_a^{a+1}s^n{\rm d}s\leq(a+1)^n\quad\Longrightarrow\quad\tau\leq a+1 \quad\Longrightarrow\quad\tau-a\leq1.$

$I_i:=\int_a^{a+1}s^i(s-\tau)^2{\rm d}s,\qquad i=0,1,2,3,\cdots.$

$[a,a+1]$上对(2.12)式积分,并使用(3.5)和(3.6)式可得

$\begin{matrix}\label{ineq-22-1-15-9-55} n(n+2)B&\geq&n\int_a^{a+1}s^{n+2}{\rm d}s\nonumber\\ &=&(n+2)\tau^{2}\int_a^{a+1}s^{n}{\rm d}s-2\tau^{n+2} +n\int_a^{a+1}s^{n}(s-\tau)^2{\rm d}s\nonumber\\ &&+2\sum_{i=0}^{n-1}(i+1)\tau^{n-i}\int_a^{a+1}s^i(s-\tau)^2s{\rm d}s\nonumber\\ &=&(n+2)\tau^{2}nA-2\tau^{n+2} +nI_n+2\sum_{i=0}^{n-1}(i+1)I_i\tau^{n-i}\nonumber\\ &=&n\tau^{n+2}+nI_n+2\sum_{i=0}^{n-1}(i+1)I_i\tau^{n-i}. \end{matrix}$

$$$\label{ineq-21-5-13-15} n(n+2)B\geq n\tau^{n+2}+2I_0\tau^n+4I_1\tau^{n-1}+6I_2\tau^{n-2}+8I_3\tau^{n-3},\quad\forall\,\,n\geq4.$$$

$J_j:=\int_a^{a+1}(s-a)^j(s-\tau)^2{\rm d}s,\qquad j=0,1,2,3,\cdots.$

$\begin{matrix}\label{eq-21-6-1-18-1} I_i&=&\int_a^{a+1}\big[(s-a)+a\big]^i(s-\tau)^2{\rm d}s\nonumber\\ &=&\sum_{j=0}^i\bigg(\begin{array}{cc}i\\j\end{array}\bigg)a^{i-j}\int_a^{a+1}(s-a)^j(s-\tau)^2{\rm d}s =\sum_{j=0}^i\bigg(\begin{array}{cc}i\\j\end{array}\bigg)a^{i-j}J_j, \end{matrix}$

$\begin{eqnarray*}\label{} J_j&=&\frac{1}{j+1}\left\{(a+1-\tau)^2-\frac{2}{j+2}(a+1-\tau) +\frac{2}{j+2}\int_a^{a+1}(s-a)^{j+2}{\rm d}s\right\}\nonumber\\ &=&\frac{1}{j+1}\left\{(a+1-\tau)^2-\frac{2}{j+2}(a+1-\tau)+\frac{2}{(j+2)(j+3)}\right\}\nonumber\\ &=&\frac{1}{j+1}\left\{\left[(a+1-\tau)-\frac{1}{j+2}\right]^2+\frac{j+1}{(j+2)^2(j+3)}\right\}\nonumber\\ &\geq&\frac{1}{(j+2)^2(j+3)}. \end{eqnarray*}$

$$$\label{eq-20-12-26-22} I_0=J_0\geq\frac{1}{12},\quad J_1\geq\frac{1}{36}, \quad J_2\geq\frac{1}{80},\quad J_3\geq\frac{1}{150}.$$$

$\begin{matrix}\label{eq-21-1-12-13} I_1=aJ_0+J_1\geq\frac{a}{12}+\frac{1}{36} =\frac{a+1}{12}-\frac{1}{18} \geq\frac{\tau}{12}-\frac{1}{18}. \end{matrix}$

$\begin{matrix}\label{ineq-21-1-26-10} I_2&=&a^2J_0+2aJ_1+J_2 \geq\frac{a^2}{12}+\frac{a}{18}+\frac{1}{80} =\frac{1}{12}\bigg[(a+1)^2-\frac{4}{3}(a+1)+\frac{29}{60}\bigg]. \end{matrix}$

$f(x):=x^2-\frac{4}{3}x+\frac{29}{60}=\big(x-\frac{2}{3}\big)^2+\frac{7}{180}, \quad\forall\,\,x\in\Bbb R.$

$-\frac{1}{6}\leq\tau-\frac{2}{3}\leq a+\frac{1}{3}\qquad\Longrightarrow\qquad \big|\tau-\frac{2}{3}\big|\leq\max\big\{\frac{1}{6},a+\frac{1}{3}\big\}=a+\frac{1}{3}.$

$$$\label{ineq-21-2-2-2} f(a+1)=\big(a+\frac{1}{3}\big)^2+\frac{7}{180} \geq\big(\tau-\frac{2}{3}\big)^2+\frac{7}{180}=f(\tau).$$$

$\begin{matrix}\label{eq-21-1-26-12} I_2\geq\frac{1}{12}f(a+1)\geq\frac{1}{12}f(\tau) =\frac{1}{12}\big(\tau^2-\frac{4}{3}\tau+\frac{29}{60}\big). \end{matrix}$

$\begin{eqnarray*}\label{eq-21-1-26-13} I_3&=&a^3J_0+3a^2J_1+3aJ_2+J_3 \geq\frac{a^3}{12}+\frac{a^2}{12}+\frac{3a}{80}+\frac{1}{150}\\ &=&\frac{1}{12}(a+1)^3-\frac{1}{6}(a+1)^2+\frac{29}{240}(a+1)-\frac{37}{1200}. \end{eqnarray*}$

$g(x):=\frac{1}{12}x^3-\frac{1}{6}x^2+\frac{29}{240}x-\frac{37}{1200}, \quad\forall\,\,x\in[0,\infty).$

$g'(x)=\frac{1}{4}x^2-\frac{1}{3}x+\frac{29}{240} =\frac{1}{4}(x-\frac{2}{3})^2+\frac{7}{720}>0,$

$\begin{eqnarray*}\label{eq-21-3-27-10} \int_0^{\infty} s^{n-1}\psi(s){\rm d}s=A,\qquad \int_0^{\infty} s^{n+1}\psi(s){\rm d}s=B. \end{eqnarray*}$

$\begin{matrix}\label{ineq-21-1-12-6} B&\geq&\frac{1}{n+2}\Big[(nA)^{1+\frac{2}{n}}\psi(0)^{-\frac{2}{n}} +\frac{5}{3}A\rho^{-2}\psi(0)^2 -\frac{20}{9}n^{-\frac{1}{n}}A^{1-\frac{1}{n}}\rho^{-3}\psi(0)^{3+\frac{1}{n}}\nonumber\\ &&+\frac{29}{24}n^{-\frac{2}{n}}A^{1-\frac{2}{n}}\rho^{-4}\psi(0)^{4+\frac{2}{n}} -\frac{37}{150}n^{-\frac{3}{n}}A^{1-\frac{3}{n}}\rho^{-5}\psi(0)^{5+\frac{3}{n}}\Big]. \end{matrix}$

$\widetilde{\psi}(t)=\alpha\psi(\beta t)$,其中$\alpha,\beta>0$是待定常数.显然,令$\alpha=\psi(0)^{-1}$,可得$\widetilde{\psi}(0)=1$. 此外, $\widetilde{\psi}'(t)=\alpha\beta\psi'(\beta t).$ 于是,从$-\rho\leq\psi'(s)\leq0$可推导出 $-\rho\alpha\beta\leq\alpha\beta\psi'(\beta t)\leq0.$ 再令$\beta=\rho^{-1}\psi(0)$可得$\rho\alpha\beta=1$.于是,可得$-1\leq\widetilde{\psi}'(t)\leq0$.$s=\beta t$$t=s/\beta. 由此可得 A=\int_0^{\infty} s^{n-1}\psi(s){\rm d}s =\int_0^{\infty}(\beta t)^{n-1}\psi(\beta t)\beta{\rm d}t =\frac{\beta^n}{\alpha}\int_0^{\infty}t^{n-1}\widetilde{\psi}(t){\rm d}t, 于是,有 \int_0^{\infty}t^{n-1}\widetilde{\psi}(t){\rm d}t=A\alpha\beta^{-n}:=\widetilde{A}. 将(3.2)式应用到\widetilde{\psi}可得 \begin{eqnarray*} \widetilde{B}&:=&\int_0^{\infty}t^{n+1}\widetilde{\psi}(t){\rm d}t\\ &\geq&\frac{1}{n+2}\Big[(n\widetilde{A})^{1+\frac{2}{n}}+\frac{5}{3}\widetilde{A} -\frac{20}{9}n^{-\frac{1}{n}}\widetilde{A}^{1-\frac{1}{n}} +\frac{29}{24}n^{-\frac{2}{n}}\widetilde{A}^{1-\frac{2}{n}} -\frac{37}{150}n^{-\frac{3}{n}}\widetilde{A}^{1-\frac{3}{n}}\Big]. \end{eqnarray*} 于是,将\widetilde{\psi}(t)=\alpha\psi(\beta t)$$\widetilde{A}:=A\alpha\beta^{-n}$代入上面不等式可得

$\begin{matrix}\label{ineq-21-1-2-21} \int_0^{\infty}t^{n+1}\alpha\psi(\beta t){\rm d}t&\geq& \frac{1}{n+2}\Big[(nA\alpha\beta^{-n})^{1+\frac{2}{n}}+\frac{5}{3}A\alpha\beta^{-n} -\frac{20}{9}n^{-\frac{1}{n}}(A\alpha\beta^{-n})^{1-\frac{1}{n}}\nonumber \\ &&+\frac{29}{24}n^{-\frac{2}{n}}(A\alpha\beta^{-n})^{1-\frac{2}{n}} -\frac{37}{150}n^{-\frac{3}{n}}(A\alpha\beta^{-n})^{1-\frac{3}{n}}\Big]. \end{matrix}$

$\begin{matrix}\label{ineq-21-1-2-22} \int_0^{\infty}t^{n+1}\alpha\psi(\beta t){\rm d}t =\alpha\beta^{-(n+2)}\int_0^{\infty}s^{n+1}\psi(s){\rm d}s. \end{matrix}$

$\xi'(t)=-2\omega_{n}^{-\frac{2}{n}}k^{1+\frac{2}{n}}t^{-\frac{2}{n}-1} +\frac{10}{3}k\rho^{-2}t,\quad\forall\,\,t>0.$
$\xi''(t)=2(\frac{2}{n}+1)\omega_{n}^{-\frac{2}{n}}k^{1+\frac{2}{n}}t^{-\frac{1}{n}-2} +\frac{10}{3}k\rho^{-2}>0, \quad\forall\,\,t>0,$

$\Gamma(1+\frac{n}{2})\geq\Gamma(2)=1.$

$k\geq1\geq\big(\frac{12}{5}\big)^{-\frac{n}{2}}\cdot\left[\Gamma(1+\frac{n}{2})\right]^{-2} =\big(\frac{12}{5}\big)^{-\frac{n}{2}}\cdot\frac{\omega_n^2}{\pi^n}.$

$t_*\geq(k\omega_n^{-1}\rho^n)^{\frac{1}{n+1}} \geq\big[\omega_n^{-2}k(2\pi)^{-n^2}V(\Omega)^{n+1}\big]^\frac{1}{n+1} \geq(2\pi)^{-n}V(\Omega).$

$k\geq1\geq2^{-n}\left[\Gamma(1+\frac{n}{2})\right]^{-2}=(2\pi)^{-n}\omega_n^2.$

## 参考文献 原文顺序 文献年度倒序 文中引用次数倒序 被引期刊影响因子

Weyl H.

Das asymptotische verteilungsgesetz der eigenwerte linearer partieller differentialgleichungen

Math Ann, 1912, 71: 441-479

Pólya G.

On the eigenvalues of vibrating membranes

Proc Lond Math Soc, 1961, 11: 419-433

Li P, Yau S T.

On the Schrödinger equation and the eigenvalue problem

Comm Math Phys, 1983, 88: 309-318

Berezin F A.

Covariant and contravariant symbols of operators

Izv Akad Nauk SSSR Ser Mat, 1972, 36: 1134-1167

Lieb E.

The number of bound states of one-body Schroedinger operators and the Weyl problem

Proc Sym Pure Math, 1980, 36: 241-252

Laptev A.

Dirichlet and Neumann eigenvalue problems on domains in Euclidean spaces

J Funct Anal, 1997, 151: 531-545

Melas A D.

A lower bound for sums of eigenvalues of the Laplacian

Proc Amer Math Soc, 2002, 131: 631-636

Cheng Q M, Yang H C.

Estimates for eigenvalues on Riemannian manifolds

J Differ Equations, 2009, 247: 2270-2281

Yolcu S Y, Yolcu T.

Multidimensional lower bounds for the eigenvalues of Stokes and Dirichlet Laplacian operators

J Math Phys, 2012, 53: 043508

Yolcu S Y, Yolcu T.

Refined eigenvalue bounds on the Dirichlet fractional Laplacian

J Math Phys, 2015, 56: 1-12

Davies E B, Safalov Y. Spectral Theory and Geometry. Cambridge: Cambridge University Press, 1999

Wei G X, Sun H J, Zeng L Z.

Lower bounds for fractional Laplacian eigenvalues

Commun Contemp Math, 2014, 16: 1450032

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