Lower bounds for eigenvalues of Laplacian operator and the clamped plate problem

Abstract

We first investigate the lower bound for higher eigenvalues $\lambda _i$ of the Laplace operator on a bounded domain and obtain a sharp lower bound. Then, we extent this estimate of the eigenvalues to general cases. Finally, we study the eigenvalues $\Gamma _i$ for the clamped plate problem and deliver a sharp bound for the clamped plate problem for arbitrary dimension.

1 Introduction

Let $\Omega$ be a bounded domain with piecewise smooth boundary $\partial \Omega$ in an n-dimensional Euclidean space $\textbf{R}^n$. First of all, we focus on the following Dirichlet eigenvalue problem of Laplacian

$$\begin{aligned} {\left\{ \begin{array}{ll} \Delta u=-\lambda u \ {} &{} \textrm{in}\ \Omega , \\ u=0 \ {} &{} \textrm{on}\ \partial \Omega . \end{array}\right. } \end{aligned}$$

(1.1)

It is well known that the spectrum of eigenvalue problem (1.1) is real and discrete (cf. [2, 6, 12, 15, 21])

$$\begin{aligned} 0<\lambda _1<\lambda _2\le \lambda _3\le \cdots \rightarrow \infty , \end{aligned}$$

where each $\lambda _i$ has finite multiplicity which is counted by its multiplicity.

Let $V(\Omega )$ be the volume of $\Omega$, and $\omega _n$ the volume of the unit ball in ${R}^n$. Then the following well-known Weyl’s asymptotic formula holds

$$\begin{aligned} \lambda _k\sim \frac{4\pi ^2}{(\omega _nV(\Omega ))^\frac{2}{n}}k^{\frac{2}{n}},\ k\rightarrow \infty , \end{aligned}$$

which implies that

$$\begin{aligned} \frac{1}{k}\sum _{i=1}^k \lambda _i \sim \frac{n}{n+2}\frac{4\pi ^2}{(\omega _nV(\Omega ))^\frac{2}{n}}k^{\frac{2}{n}},\ k\rightarrow \infty . \end{aligned}$$

(1.2)

In 1961, Pólya [23] proved that, if $n=2$ and $\Omega$ is a tiling domain in ${R}^2$, then

$$\begin{aligned} \lambda _k \ge \frac{4\pi ^2}{(\omega _nV(\Omega ))^\frac{2}{n}}k^{\frac{2}{n}},\ \ \textrm{for}\ k=1,2,\ldots , \end{aligned}$$

Based on the result above, he proposed the famous conjecture:

Conjecture of Pólya. If$\Omega$ is a bounded domain in ${R}^n$, then k-th eigenvalue $\lambda _k$ of the eigenvalue problem (1.1) satisfies

$$\begin{aligned} \lambda _k \ge \frac{4\pi ^2}{(\omega _nV(\Omega ))^\frac{2}{n}}k^{\frac{2}{n}},\ \ \textrm{for}\ k=1,2,\ldots . \end{aligned}$$

During the past six decades, many mathematicians have focused on this problem and the related topics, there are a lot of important results on this aspect (cf. [4, 5, 7, 10, 11, 13, 14, 16, 18]) and we suggest that readers refer [25, 29] for more details. In 1983, Li and Yau [17] verified the famous Li-Yau inequality

$$\begin{aligned} \frac{1}{k}\sum _{i=1}^k \lambda _i \ge \frac{n}{n+2}\frac{4\pi ^2}{(\omega _nV(\Omega ))^\frac{2}{n}}k^{\frac{2}{n}},\ k=1,2,\ldots . \end{aligned}$$

(1.3)

It’s seen from the asymptotic formula (1.2), that Li-Yau’s inequality is the best possible in the sense of the average of eigenvalues. From (1.3), one can derive

$$\begin{aligned} \lambda _k \ge \frac{n}{n+2}\frac{4\pi ^2}{(\omega _nV(\Omega ))^\frac{2}{n}}k^{\frac{2}{n}},\ \ \textrm{for}\ k=1,2,\ldots , \end{aligned}$$

which gives a partial solution to the Pólya conjecture with a factor $\frac{n}{n+2}$. This conjecture is still open up to now.

In [20], Melas obtained the following beautiful estimate which improves (1.3) for $n\ge 1$ and $k\ge 1$

$$\begin{aligned} \sum _{i=1}^k \lambda _i \ge \frac{n}{n+2}\frac{4\pi ^2}{(\omega _nV(\Omega ))^\frac{2}{n}}k^{\frac{n+2}{n}}+c_n\frac{V(\Omega )}{I(\Omega )}k,\ \ \textrm{for}\ k=1,2,\ldots , \end{aligned}$$

(1.4)

where $c_n$ is a positive constant depending only on n and

$$\begin{aligned} I(\Omega )=\min _{a\in \textbf{R}^n} \int _{\Omega } |x-a|^2dx \end{aligned}$$

is called the moment of inertia of $\Omega$. In fact $c_n\le \frac{1}{24(n+2)}$. Obviously,

$$\begin{aligned} I(\Omega )\ge \frac{n}{n+2}V(\Omega )\left( \frac{V(\Omega )}{\omega _n} \right) ^{\frac{2}{n}}. \end{aligned}$$

In the formula (2.27) of [20], Males requires $c\le \min \{\frac{1}{6},\frac{(2\pi )^2}{{\omega _n}^{\frac{4}{n}}}\}$. According to $\frac{\omega _n^{\frac{4}{n}}}{(2\pi )^2}\le \frac{1}{2}$, we get $c\le \frac{1}{6}$. Putting $c\le \frac{1}{6}$ into the formula (2.27) of [20], we get $c_n\le \frac{1}{24(n+2)}$ in (1.4).

Afterwards, Kovařík, Vugalter and Weidl [13] improved this results when $n=2$. They proved that

$$\begin{aligned} \sum _{i=1}^k \lambda _i \ge \frac{2\pi }{V(\Omega )}k^2+C(a_0)V(\Omega )^{-\frac{3}{2}}k^{\frac{3}{2}-\varepsilon (k)}+(1-a_0)\frac{V(\Omega )}{32I(\Omega )}k, \end{aligned}$$

(1.5)

where $C(a_0)$ is a positive constant depending on $a_0\in [0,1]$ and the length of the smooth parts of $\partial \Omega$, $\varepsilon (k)=\frac{2}{\sqrt{\log _2(\frac{2\pi k}{c})}}$ and $c=\sqrt{\frac{3\pi }{14}}10^{-11}$.

The first purpose of this paper is to improve Melas’s estimate (1.4) by giving a sharper polynomial inequality, see Corollary 2.4. For more general cases, where $n\ge m\ge 2$ and $k\ge 1$, we obtain a lower bound for eigenvalues in Sect. 3, and we should mention that our result gives a sharp lower bounds by comparing Lemma 2.2 with the polynomial inequality in [20]. As a consequence of our result, we prove the Theorem 3.1. An interesting problem is to investigate the similar problem in a Cartan-Hadamard manifold and we recommend readers to refer to [27, 28] for details.

The second purpose of this paper is to estimate eigenvalues of the following clamped plate problem. Let $\Omega$ be a bounded domain in ${R}^n$. We consider the following clamped plate problem, which describes characteristic vibrations of a clamped plate:

$$\begin{aligned} {\left\{ \begin{array}{ll} \Delta ^2 u=\Gamma u, \ {} &{} \textrm{in}\ \Omega , \\ u=\frac{\partial u}{\partial \nu }=0, \ {} &{} \textrm{on}\ \partial \Omega , \end{array}\right. } \end{aligned}$$

where $\Delta$ is the Laplacian operator and $\nu$ denotes the outward unit normal to the boundary $\partial \Omega$. As is known, this problem has a real and discrete spectrum (cf. [1])

$$\begin{aligned} 0<\Gamma _1\le \Gamma _2\le \Gamma _3\le \cdots \rightarrow \infty , \end{aligned}$$

where each $\Gamma _i$ has finite multiplicity which is repeated according to its multiplicity.

For the eigenvalues of the clamped plate problem, Agmon [1] and Pleijel [22] gave the following asymptotic formula

$$\begin{aligned} \Gamma _k \sim \frac{16\pi ^2}{(\omega _n V(\Omega ))^{\frac{4}{n}}}k^{\frac{4}{n}},\ k\rightarrow \infty . \end{aligned}$$

This implies that

$$\begin{aligned} \frac{1}{k}\sum _{i=1}^k\Gamma _i \sim \frac{n}{n+4} \frac{16\pi ^2}{(\omega _n V(\Omega ))^{\frac{4}{n}}}k^{\frac{4}{n}},\ k\rightarrow \infty . \end{aligned}$$

(1.6)

Furthermore, Levine and Protter [16] proved that the eigenvalues of the clamped plate problem satisfy

$$\begin{aligned} \frac{1}{k}\sum _{i=1}^k\Gamma _i \ge \frac{n}{n+4} \frac{16\pi ^4}{(\omega _n V(\Omega ))^{\frac{4}{n}}}k^{\frac{4}{n}}. \end{aligned}$$

The formula (1.6) shows that the coefficient of $k^{\frac{4}{n}}$ is the best possible in the sense of the average of eigenvalues. Later, Cheng and Wei [8] improved the above estimate as follows:

$$\begin{aligned} \frac{1}{k}\sum _{i=1}^k\Gamma _i \ge&\frac{n}{n+4} \frac{16\pi ^4}{(\omega _nV(\Omega ))^{\frac{4}{n}}}k^{\frac{4}{n}}\\&+\left( \frac{n+2}{12n(n+4)}-\frac{1}{1152n^2(n+4)} \right) \frac{V(\Omega )}{I(\Omega )}\frac{n}{n+2} \frac{4\pi ^2}{(\omega _nV(\Omega ))^{\frac{2}{n}}}k^{\frac{2}{n}}\\&+\left( \frac{1}{576n(n+4)}-\frac{1}{27648n^2(n+2)(n+4)} \right) \left( \frac{V(\Omega )}{I(\Omega )}\right) ^2,\\ \end{aligned}$$

where $n\ge 1$ and $k\ge 1$.

Recently, by using a different method, Cheng and Wei [9] got better lower bounds for eigenvalues of the clamped plate problem and proved that

$$\begin{aligned} \begin{aligned} \frac{1}{k}\sum _{i=1}^k\Gamma _i \ge&\frac{n}{n+4} \frac{16\pi ^4}{(\omega _nV(\Omega ))^{\frac{4}{n}}}k^{\frac{4}{n}}+\frac{n+2}{12n(n+4)}\frac{V(\Omega )}{I(\Omega )}\frac{n}{n+2} \frac{4\pi ^2}{(\omega _nV(\Omega ))^{\frac{2}{n}}}k^{\frac{2}{n}}\\&+\frac{(n+2)^2}{1152n(n+4)^2}\left( \frac{V(\Omega )}{I(\Omega )}\right) ^2, \end{aligned} \end{aligned}$$

(1.7)

where $n\ge 2$ and $k\ge 1$.

Furthermore, they gave upper bounds for the sum of $\Gamma _i$,

$$\begin{aligned} \frac{1}{k}\sum _{i=1}^k\Gamma _i \le \frac{1+\frac{4(n+4)(n^2+2n+6)}{n+2} \frac{V(\Omega _{r_0})}{V(\Omega )}}{\left( 1-\frac{V(\Omega _{r_0})}{V(\Omega )} \right) ^{\frac{n+4}{n}}} \frac{n}{n+4} \frac{16\pi ^4}{(\omega _nV(\Omega ))^{\frac{4}{n}}}k^{\frac{4}{n}}, \end{aligned}$$

where $k\ge V(\Omega )r_0^n$, and

$$\begin{aligned} \Omega _r=\left\{ x\in \Omega \ \Big | \ \textrm{dist}(x,\partial \Omega )<\frac{1}{r} \right\} . \end{aligned}$$

In [30], Yildirim and Yolcu improved Cheng and Wei’s estimates by replacing the last term in the right hand of (1.7) by a positive term of $k^{\frac{1}{n}}$. For any bounded open set $\Omega \subseteq R^n$, where $n\ge 2$ and $k \ge 1$, Yildirim and Yolcu got the following inequality

$$\begin{aligned} \begin{aligned} \sum _{i=1}^k \Gamma _i \ge&\frac{n}{n+4}(\omega _n)^{-\frac{4}{n}}\alpha ^{-\frac{4}{n}}k^{\frac{4+n}{n}} +\frac{1}{3(n+4)}\frac{(\omega _n)^{-\frac{2}{n}}\alpha ^{\frac{2n-2}{n}}k^{\frac{n+2}{n}}}{\rho ^2}\\&+\frac{2}{9(n+4)}\frac{(\omega _n)^{-\frac{1}{n}}\alpha ^{\frac{3n-1}{n}}k^{\frac{n+1}{n}}}{\rho ^3}, \end{aligned} \end{aligned}$$

(1.8)

where

$$\begin{aligned} \alpha =\frac{V(\Omega )}{(2\pi )^n},\,\,\,\rho =2(2\pi )^{-n}\sqrt{V(\Omega )I(\Omega )}. \end{aligned}$$

(1.9)

In Sect. 4, we will improve Yildirim and Yolcu’s [30] estimate (1.8) by giving a shaper polynomial inequality when $n\ge 3$, see Corollary 4.4.

2 Lower bounds for sums of Dirichlet eigenvalues

In this section we prove the following theorem.

Theorem 2.1

For any bounded domain $\Omega \subseteq R^n$, $n\ge 2$ we have

$$\begin{aligned}&\sum _{j=1}^k\lambda _j(\Omega )\ge \omega _n^{-\frac{2}{n}}{\alpha }^{-\frac{2}{n}}k^{\frac{n+2}{n}}-\frac{s_3^3\alpha ^2}{(n+2)\rho ^2}k\\&\quad +c_1 \omega _n^{\frac{1}{n}}\frac{s_4^4\alpha ^{\frac{3n+1}{n}}k^{\frac{n-1}{n}}}{(n+2)\rho ^3}, \end{aligned}$$

where

$$\begin{aligned} c_1\le&\min \left\{ 1,\max \left\{ \frac{4\sqrt{2} n s_3^3k^{\frac{1}{n}}}{(3n+1)s_4^4},\frac{4\sqrt{2}(n+2)k^{\frac{3}{n}}}{(3n+1)s_4^4} \right\} \right\} ,\\ s_l^l=&(a+1)^{l}-a^l, \end{aligned}$$

$\alpha$, $\rho$ are defined by (1.9) and a is defined by (2.16).

Firstly, we introduce some notations and definitions. For a bounded domain $\Omega$, the moment of inertia of $\Omega$ is defined by

$$\begin{aligned} I(\Omega )=\min _{a\in R^n}\int _{\Omega } |x-a|^2 dx. \end{aligned}$$

By a translation of the origin and a suitable rotation of axes, we can assume that the center of mass is the origin and

$$\begin{aligned} I(\Omega )=\int _{\Omega } |x|^2 dx. \end{aligned}$$

We now fix a $k \ge 1$ and let $u_1,\ldots ,u_k$ denote an orthonormal set of eigenfunctions of (1.1) corresponding to the set of eigenvalues $\lambda _1(\Omega ),\ldots ,\lambda _k(\Omega )$. We consider the Fourier transform of each eigenfunction

$$\begin{aligned} f_j(\xi )={\hat{u}}_j(\xi )=(2\pi )^{-n/2}\int _{\Omega } u_j(x)e^{ix\xi }dx. \end{aligned}$$

It seems from Plancherel’s Theorem that $f_1,.\ldots , f_k$ is an orthonormal set in $R^n$. Since these eigenfunctions $u_1,\ldots ,u_k$ are also orthonormal in $L_2(\Omega )$, Bessel’s inequality implies that for every $\xi \in R^n$

$$\begin{aligned} \sum _{j=1}^k |f_j(\xi )|^2\le (2\pi )^{-n}\int _{\Omega } |e^{ix\xi }|^2dx=(2\pi )^{-n}V(\Omega ). \end{aligned}$$

(2.1)

Since

$$\begin{aligned} \nabla f_j(\xi )=(2\pi )^{-n/2}\int _{\Omega }ixu_j(x)e^{ix\xi }dx, \end{aligned}$$

we have

$$\begin{aligned} \sum _{j=1}^k |\nabla f_j(\xi )|^2\le (2\pi )^{-n/2}\int _{\Omega }|ixe^{ix\xi }|^2dx=(2\pi )^{-n}I(\Omega ). \end{aligned}$$

By the boundary condition, we get

$$\begin{aligned} \int _{R^n}|\xi |^2|f_j(\xi )|^2d\xi =\int _{\Omega }|\nabla u_j (x)|^2dx=\lambda _j(\Omega ) \end{aligned}$$

for each $1\le j\le k$. Set

$$\begin{aligned} F(\xi )=\sum _{j=1}^k|f_j(\xi )|^2. \end{aligned}$$

From (2.1), we have

$$\begin{aligned}&0\le F(\xi )\le (2\pi )^{-n}V(\Omega ),\end{aligned}$$

(2.2)

$$\begin{aligned} |\nabla F(\xi )|\le 2\left( \sum _{j=1}^k|f_j(\xi )|^2 \right) ^{1/2}&\left( \sum _{j=1}^k|\nabla f_j(\xi )|^2 \right) ^{1/2}\le 2(2\pi )^{-n}\sqrt{V(\Omega )I(\Omega )} \end{aligned}$$

(2.3)

for each $\xi \in R^n$. We also get

$$\begin{aligned} \int _{R^n}F(\xi )d\xi&=k, \end{aligned}$$

(2.4)

$$\begin{aligned} \int _{R^n} |\xi |^2F(\xi )d\xi&=\sum _{j=1}^k \lambda _j(\Omega ). \end{aligned}$$

(2.5)

Assume (by approximating F) that the decreasing function $\phi : [0,+\infty )\rightarrow [0,(2\pi )^{-n}V(\Omega )]$ is absolutely continuous. Let $F^*(\xi ) = \phi (|\xi |)$ denote the decreasing radial rearrangement of F. Put $\mu (t)=|\{F^*>t \}|=|\{F>t\}|$. It follows from the coarea formula that

$$\begin{aligned} \mu (t)=\int _{t}^{(2\pi )^{-n}V(\Omega )} \int _{\{F=s \}}\frac{1}{|\nabla F|}d\sigma _sds. \end{aligned}$$

Since $F^*$ is radial, we have $\mu (\phi (s))=|\{F^*>\phi (s) \}|=\omega _n s^n$. Differentiating both side of the above equality, we have $n\omega _n s^{n-1}={\mu }'(\phi (s))\phi '(s)$ for almost all s. This together with (2.3), $\rho =2(2\pi )^{-n}\sqrt{V(\Omega )I(\Omega )}$ and the isoperimetric inequality implies

$$\begin{aligned} -\mu '(\phi (s))&=\int _{\{F=\phi (s) \}} |\nabla F|^{-1} d\sigma _{\phi (s)}\\&\ge \rho ^{-1}\textrm{Vol}_{n-1}(\{F=\phi (s)\})\\&\ge \rho ^{-1}n\omega _n s^{n-1}. \end{aligned}$$

For almost all s, we have

$$\begin{aligned} -\rho \le \phi '(s)\le 0. \end{aligned}$$

(2.6)

Since the map $\xi \mapsto |\xi |^2$ is radial and increasing, applying (2.5), we get

$$\begin{aligned} k=\int _{R^n}F(\xi )d\xi =\int _{R^n}F^*(\xi )d\xi =n\omega _n\int _0^{\infty }s^{n-1}\phi (s)ds \end{aligned}$$

(2.7)

and

$$\begin{aligned} \sum _{j=1}^k\lambda _j(\Omega )=\int _{R^n}|\xi |^2F(\xi )d\xi \ge \int _{R^n}|\xi |^2F^*(\xi )d\xi =n\omega _n\int _0^\infty s^{n+1}\phi (s)ds. \end{aligned}$$

(2.8)

The following lemma will be used in the proof of Theorem 2.1.

Lemma 2.2

Let $n\ge 2$, $\rho >0$, $A>0$. If $\psi : [0,+\infty )\rightarrow [0,+\infty )$ is a decreasing function (and absolutely continuous) satisfying

$$\begin{aligned} -\rho \le -\psi '(s)\le 0 \end{aligned}$$

(2.9)

and

$$\begin{aligned} \int _0^\infty s^{n-1}\psi (s)ds=A. \end{aligned}$$

Then

$$\begin{aligned} \int _0^{\infty } s^{n+1}\psi (s)ds \ge&\frac{(nA)^{\frac{n+2}{n}}{\psi (0)}^{-\frac{2}{n}}}{n}-\frac{s_3^3(nA)\psi (0)^2}{n(n+2)\rho ^2} +\frac{s_4^4(nA)^{\frac{n-1}{n}}\psi (0)^{\frac{3n+1}{n}}}{n(n+2)\rho ^3}, \end{aligned}$$

where

$$\begin{aligned} s_l^l=(a+1)^l-a^l\ge 1. \end{aligned}$$

Proof

We choose the function $\alpha \psi (\beta t)$ for appropriate $\alpha , \beta >0$, such that $\rho = 1$ and $\psi (0) = 1$. By [20] we can also assume that $B=\int _0^\infty s^{n+1}\psi (s)ds <\infty$. If we let $q(s)=-\psi ^{'}(s)$ for $s\ge 0$, we have $0\le q(s)\le 1$ and $\int _0^\infty q(s)=\psi (0)=1.$ Moreover, integration by parts implies that

$$\begin{aligned} \int _0^\infty s^{n}q(s)ds=n\int _0^\infty s^{n-1}\psi (s)ds=nA \end{aligned}$$

and

$$\begin{aligned} \int _0^\infty s^{n+2}q(s)ds\le (n+2)B. \end{aligned}$$

Next, let $0\le a < +\infty$ satisfies that

$$\begin{aligned} \int _a^{a+1} s^{n}ds=\int _0^\infty s^{n}q(s)ds=nA. \end{aligned}$$

(2.10)

By the same argument as in Lemma 1 of [17], such real number a exists. From [20], we have

$$\begin{aligned} (n+2)B\ge \int _0^{\infty }s^{n+2}q(s)ds\ge \int _a^{a+1}s^{n+2}ds. \end{aligned}$$

(2.11)

To estimate the last integral we take $\tau > 0$ to be chosen later. Applying (2.11) and integrating the both sides of the following inequality

$$\begin{aligned} ns^{n+2}-(n+2)\tau ^2s^n+2\tau ^{n+2}\ge 2\tau ^n(s-\tau )^2+4s\tau ^{n-1}(s-\tau )^2, \,s\in [a,a+1], \end{aligned}$$

(2.12)

we get

$$\begin{aligned} \begin{aligned}&n(n+2)B-(n+2)\tau ^2nA+2\tau ^{n+2}\\ \quad&\ge 2\tau ^n \int _a^{a+1}(s-\tau )^2 +4\tau ^{n-1}\int _a^{a+1}s(s-\tau )^2ds\\ \quad&\ge 2\tau ^n\left( \frac{s^3}{3}-s^2\tau +s\tau ^2 \right) \bigg |_{a}^{a+1}\\&\qquad +4\tau ^{n-1}\left( \frac{s^4}{4}-\frac{2s^3\tau }{3}+\frac{s^2\tau ^2}{2} \right) \bigg |_{a}^{a+1}\\ \quad&=2s\tau ^{n+2}+2s^2\tau ^{n+1}-2s^3\tau ^{n}-2s^2\tau ^{n+1}+s^4\tau ^{n-1}\bigg |_{a}^{a+1}\\ \quad&=2\tau ^{n+2}-2s_3^3\tau ^{n}+s_4^4\tau ^{n-1}, \end{aligned} \end{aligned}$$

(2.13)

where

$$\begin{aligned} s_l^l=(a+1)^l-a^l\ge 1. \end{aligned}$$

Putting, $\tau = (nA)^{1/n}$ we get

$$\begin{aligned} B\ge&\frac{1}{n}(nA)^{\frac{n+2}{n}}-\frac{s_3^3}{n(n+2)}(nA)+\frac{s_4^4}{n(n+2)}(nA)^{\frac{n-1}{n}}. \end{aligned}$$

This proves Lemma 2.2.

To prove (2.12), we need to show that for any $\tau >0$ we have

$$\begin{aligned} ns^{n+2}-(n+2)\tau ^2s^n+2\tau ^{n+2}- 2\tau ^n(s-\tau )^2 -4s\tau ^{n-1}(\tau -s)^2\ge 0. \end{aligned}$$

(2.14)

Taking $t=\frac{s}{\tau }$, we define f(t) (for $t>0$) by

$$\begin{aligned} f(t)=nt^{n+2}-(n+2)t^n+2-2(t-1)^2-4t(t-1)^2. \end{aligned}$$

Differentiating, f(t) we have

$$\begin{aligned} f'(t)&=n(n+2)t^{n+1}-(n+2)nt^{n-1}-4(t-1)-4(t-1)^2-8t(t-1)\\&=\left[ n(n+2)t^{n-2}(t+1)-12 \right] t(t-1). \end{aligned}$$

It follows from the above formula that if $n\ge 2$, then $t=1$ is the minimum point of f and $f\ge \min \{f(1)=0,f(0)=0\}$. This implies

$$\begin{aligned} f(t)\tau ^{n+2}=ns^{n+2}-(n+2)\tau ^2s^n+2\tau ^{n+2}- 2\tau ^n(s-\tau )^2 -4s\tau ^{n-1}(\tau -s)^2 \ge 0. \end{aligned}$$

$\square$

Next we will give the proof of Theorem 2.1.

Proof of Theorem 2.1

Applying Lemma 2.2 to the function $\phi$ with $A=(n\omega _n)^{-1}k, \rho =2(2\pi )^{-n}\sqrt{V(\Omega )I(\Omega )}$ and submitting it to (2.8), we obtain

$$\begin{aligned} \begin{aligned} \sum _{j=1}^k\lambda _j(\Omega )\ge&\omega _n^{-\frac{2}{n}}{\psi (0)}^{-\frac{2}{n}}k^{\frac{n+2}{n}}-\frac{s_3^3\psi (0)^2}{(n+2)\rho ^2}k\\&+c_1 \omega _n^{\frac{1}{n}}\frac{s_4^4\psi (0)^{\frac{3n+1}{n}}k^{\frac{n-1}{n}}}{(n+2)\rho ^3}, \end{aligned} \end{aligned}$$

(2.15)

where $0<c_1\le 1$ is a constant and a is defined by

$$\begin{aligned} \int _a^{a+1} \xi ^{n}d\xi =\int _0^\infty -\xi ^{n}\phi '(\xi )d\xi . \end{aligned}$$

(2.16)

We observe the following facts

• (i) $0<\psi (0)\le (2\pi )^{-n}V(\Omega )$,

• (ii) if R is a positive constant such that $\omega _n R^n=V(\Omega )$, then

$$\begin{aligned} I(\Omega )\ge \int _{B(R)}|x|^2dx=\frac{n\omega _nR^{n+2}}{n+2}. \end{aligned}$$

(2.17)

It follows from the above properties

$$\begin{aligned} \rho \ge (2\pi )^{-n}\omega _n^{-\frac{1}{n}}V(\Omega )^{\frac{n+1}{n}}. \end{aligned}$$

(2.18)

On the other hand, we consider the following function

$$\begin{aligned} g(t)=g_1(t)+g_2(t), \end{aligned}$$

for $t\in (0, (2\pi )^{-n}V(\Omega )]$, where

$$\begin{aligned} g_1(t)&=\omega _n^{-\frac{2}{n}}t^{-\frac{2}{n}}k^{\frac{n+2}{n}} \end{aligned}$$

and

$$\begin{aligned} g_2(t)=-\frac{s_3^3t^2}{(n+2)\rho ^2}k+c_1 \omega _n^{\frac{1}{n}}\frac{s_4^4t^{\frac{3n+1}{n}}k^{\frac{n-1}{n}}}{(n+2)\rho ^3}. \end{aligned}$$

Then we have

$$\begin{aligned} (n+2)\rho ^2g_2'(t) =-{2s_3^3k t}+c_1\omega _n^{-\frac{1}{n}}\frac{s_4^4k^{\frac{n-1}{n}}}{\rho }\frac{3n+1}{n}t^{\frac{2n+1}{n}}. \end{aligned}$$

By a direct calculation, we see from $\omega _n=\frac{2\pi ^{\frac{n}{2}}}{n\Gamma (\frac{n}{2})}$ that

$$\begin{aligned} \frac{\omega _n^{\frac{4}{n}}}{(2\pi )^2}\le \frac{1}{2}, \end{aligned}$$

where $\Gamma (t)$ is the Gamma function.

Therefore, in view of (2.18), if

$$\begin{aligned} c_1\le \min \left\{ 1, \frac{4\sqrt{2} n s_3^3k^{\frac{1}{n}}}{(3n+1)s_4^4} \right\} , \end{aligned}$$

then $g_2(t)$ is decreasing on $(0, (2\pi )^{-n}V(\Omega )]$. Now we consider another estimate. Setting

$$\begin{aligned} G(t)=G_1(t)+G_2(t), \end{aligned}$$

where

$$\begin{aligned} G_1(t)&=\omega _n^{-\frac{2}{n}}{\psi (0)}^{-\frac{2}{n}}k^{\frac{n+2}{n}}+c_1 \omega _n^{\frac{1}{n}}\frac{s_4^4\psi (0)^{\frac{3n+1}{n}}k^{\frac{n-1}{n}}}{(n+2)\rho ^3} \end{aligned}$$

and

$$\begin{aligned} G_2(t)=-\frac{s_3^3\psi (0)^2}{(n+2)\rho ^2}k, \end{aligned}$$

we have

$$\begin{aligned} G_1'(t)\rho ^2 =-\frac{2}{n}\omega _n^{-\frac{2}{n}}t^{-\frac{n+2}{n}}k^{\frac{n+2}{n}}+\frac{c_1(3n+1)\omega _n^{-\frac{1}{n}}}{n}\frac{s_4^4t^{\frac{2n+1}{n}}}{(n+2)\rho ^2}k^{\frac{n-1}{n}}. \end{aligned}$$

Therefore, we conclude that if

$$\begin{aligned} c_1\le \frac{4\sqrt{2}(n+2)k^{\frac{3}{n}}}{(3n+1)s_4^4}, \end{aligned}$$

then G(t) is decreasing on $(0, (2\pi )^{-n}V(\Omega )]$. Finally, we obtain

$$\begin{aligned} \begin{aligned} \sum _{j=1}^k\lambda _j(\Omega )\ge&\omega _n^{-\frac{2}{n}}{\alpha }^{-\frac{2}{n}}k^{\frac{n+2}{n}}-\frac{s_3^3\alpha ^2}{(n+2)\rho ^2}k\\&+c_1 \omega _n^{\frac{1}{n}}\frac{s_4^4\alpha ^{\frac{3n+1}{n}}k^{\frac{n-1}{n}}}{(n+2)\rho ^3}, \end{aligned} \end{aligned}$$

(2.19)

where $\alpha$, $\rho$ are defined in the (1.9) and

$$\begin{aligned} c_1\le&\min \left\{ 1,\max \left\{ \frac{4\sqrt{2} n s_3^3k^{\frac{1}{n}}}{(3n+1)s_4^4},\frac{4\sqrt{2}(n+2)k^{\frac{3}{n}}}{(3n+1)s_4^4} \right\} \right\} . \end{aligned}$$

$\square$

Note that $\lambda _1<\lambda _2\le \lambda _3\le \cdots$. This together with the above lemma implies the following estimate for higher eigenvalues.

Corollary 2.3

For any bounded domain $\Omega \subseteq R^n$, $n\ge 2$ and any $k \ge 1$ we have

$$\begin{aligned} \lambda _k(\Omega )\ge&\omega _n^{-\frac{2}{n}}{\alpha }^{-\frac{2}{n}}k^{\frac{2}{n}}-\frac{s_3^3\alpha ^2}{(n+2)\rho ^2}\\&+c_1 \omega _n^{\frac{1}{n}}\frac{s_4^4\alpha ^{\frac{3n+1}{n}}k^{\frac{-1}{n}}}{(n+2)\rho ^3}, \end{aligned}$$

where

$$\begin{aligned} c_1\le&\min \left\{ 1,\max \left\{ \frac{4\sqrt{2} n s_3^3k^{\frac{1}{n}}}{(3n+1)s_4^4},\frac{4\sqrt{2}(n+2)k^{\frac{3}{n}}}{(3n+1)s_4^4} \right\} \right\} ,\\ s_l^l=&(a+1)^{l}-a^l, \end{aligned}$$

$\alpha$, $\rho$ are defined by (1.9).

In fact, if we choose special a in (2.13), we also have the following result.

Corollary 2.4

For any bounded domain $\Omega \subseteq R^n$, $n\ge 2$ and any $k \ge 1$ we have

$$\begin{aligned} \begin{aligned} \sum _{i=1}^k\lambda _i\ge&\frac{n{\omega _n}^{-\frac{2}{n}}(2\pi )^2{V(\Omega )}^{-\frac{2}{n}}}{n+2}k^{\frac{n+2}{n}} +\frac{1}{24(n+2)}\left( \frac{V(\Omega )}{I(\Omega )}\right) k\\&+\frac{\omega _n^{-\frac{1}{n}}\alpha ^{\frac{3n+1}{n}}}{9(n+2)\rho ^3}k^{\frac{n-1}{n}}. \end{aligned} \end{aligned}$$

(2.20)

Proof

Combining with the formula (2.25) in [20] and (57) in [30], we have

$$\begin{aligned} n(n+2)B-(n+2)\tau ^2nA+2\tau ^{n+2}&\ge 2\tau ^n \int _a^{a+1}(s-\tau )^2 +4\tau ^{n-1}\int _a^{a+1}s(s-\tau )^2ds\\&\ge \frac{\tau ^n}{6}+\frac{\tau ^{n-1}}{9}. \end{aligned}$$

By using similar discussion in the proof of Theorem 2.1, we get

$$\begin{aligned} \sum _{i=1}^k\lambda _i\ge \frac{n{\omega _n}^{-\frac{2}{n}}\phi (0)^{-\frac{2}{n}}}{n+2}k^{\frac{n+2}{n}}+\frac{C_1\phi (0)^2}{(n+2)\rho ^2}k +\frac{C_2\omega _n^{-\frac{1}{n}}\phi (0)^{\frac{3n+1}{n}}}{(n+2)\rho ^3}k^{\frac{n-1}{n}}, \end{aligned}$$

(2.21)

where $0<C_1\le \frac{1}{6}$ and $0<C_2\le \frac{1}{9}$ are two constants which will be determined later. We consider the following function

$$\begin{aligned} g(t)=\frac{n{\omega _n}^{-\frac{2}{n}}t^{-\frac{2}{n}}}{n+2}k^{\frac{n+2}{n}}+\frac{C_1 t^2}{(n+2)\rho ^2}k +\frac{C_2\omega _n^{-\frac{1}{n}} t^{\frac{3n+1}{n}}}{(n+2)\rho ^3}k^{\frac{n-1}{n}}, \end{aligned}$$

which would be decreasing on $(0, (2\pi )^{-n}V(\Omega )]$ if $g'((2\pi )^{-n}V(\Omega ))\le 0$. In view of (2.18), the degression of g(t) is equal to the following inequality

$$\begin{aligned} 2k^{\frac{2}{n}}\ge 2C_1\frac{\omega _n^{\frac{2}{n}}}{(2\pi )^2}+C_2\frac{3n+1}{n}\frac{\omega _n^{\frac{2}{n}}}{(2\pi )^3} \end{aligned}$$

Since $k\ge 1$ and $\frac{\omega _n^{\frac{4}{n}}}{(2\pi )^2}\le \frac{1}{2}$, we can choose $C_1=\frac{1}{6}$. Therefore, $C_2$ satisfies

$$\begin{aligned} C_2\le \min \{ \frac{1}{9}, {\tilde{C}}_2 \}, \end{aligned}$$

where

$$\begin{aligned} {\tilde{C}}_2=\frac{\sqrt{2}(2\pi )^4}{3.5}\left( 2-\frac{1}{6} \right) . \end{aligned}$$

Obviously, ${\tilde{C}}_2\ge \frac{1}{9}$. Hence, we complete our proof. $\square$

3 Lower bounds for Dirichlet eigenvalues in higher dimensions

In this section we will give a universal lower bound on the sum of eigenvalues for $n\ge m+1$, where $m\ge 2$.

Theorem 3.1

For any bounded domain $\Omega \subseteq R^n$, $n\ge m+1\ge 3$ and $k \ge 1$, we have

$$\begin{aligned} \sum _{i=1}^k \lambda _i \ge&\omega _n^{-\frac{2}{n}} {\alpha ^{-\frac{2}{n}}}k^{\frac{n+2}{n}}-\frac{2\omega _n^{\frac{m-1}{n}}S_{m+2}\alpha ^{\frac{(m+1)n+m-1}{n}}}{(n+2)\rho ^{m+1}}k^{\frac{n-m+1}{n}}\\&+c_2\frac{2\omega _n^{\frac{m}{n}}(m+1)S_{m+3}\alpha ^{\frac{(m+2)n+m}{n}}}{(n+2)(m+3)\rho ^{m+2}}k^{\frac{n-m}{n}}, \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} c_2\le&\min \left\{ 1, \frac{(m+1)n+m-1 }{(m+2)n+m}\frac{\sqrt{2}S_{m+2}}{S_{m+3}}\frac{m+3}{m+1}k^{\frac{1}{n}} \right\} ,\\ S_{l}=&(a+1)^l-a^l, \end{aligned} \end{aligned}$$

$\alpha =\frac{V(\Omega )}{(2\pi )^n},\,\,\,\,\rho =2(2\pi )^{-n}\sqrt{V(\Omega )I(\Omega )}$ and a is defined by (2.16).

The following lemma will be used in the proof of Theorem 3.1.

Lemma 3.2

For an integer $n\ge m+1\ge 0$ and positive real numbers s and $\tau$ we have the following inequality:

$$\begin{aligned} ns^{n+2}-(n+2)\tau ^2s^n+2\tau ^{n+2}-\sum _{k=1}^{m+1}2ks^{k-1}\tau ^{n-k+1}(\tau -s)^2\ge 0. \end{aligned}$$

Proof

Setting $t=\frac{s}{\tau }$, and putting

$$\begin{aligned} f(t)=nt^{n+2}-(n+2)t^n+2-\sum _{k=1}^{m+1}2kt^{k-1}(t-1)^2, \end{aligned}$$

for $t\ge 0$, we get

$$\begin{aligned} \begin{aligned} f'(t)=&n(n+2)t^{n+1}-n(n+2)t^{n-1}\\&\quad -\left[ 4(t-1)+\sum _{k=1}^m (2k(k+1)t^{k-1}(t-1)^2 + 4(k+1)t^k(t-1) ) \right] \\ =&n(n+2)t^{n+1}-n(n+2)t^{n-1}\\&\quad -(t-1)\left[ 4+\sum _{k=1}^m [2k(k+1)t^{k-1}(t-1)+4(k+1)t^k ] \right] \\ =&n(n+2)t^{n+1}-n(n+2)t^{n-1}\\&\quad -(t-1)\left[ 2(m+2)(m+1)t^m +\sum _{k=1}^{m-1}( 2k(k+1)t^k-2k(k+1)t^{k-1}+ 4(k+1)t^k ) \right] \\ =&n(n+2)t^{n+1}-n(n+2)t^{n-1}-2(m+2)(m+1)t^m(t-1) \\ =&t^m(t-1)\left[ n(n+2)t^{n-m-1}(t+1)-2(m+2)(m+1) \right] . \end{aligned} \end{aligned}$$

(3.1)

It follows from the above formula that if $n \ge m+1$, then $t = 1$ is the minimum point of f(t) and $f \ge \min \{f(1) = 0,f(0) = 0\}$. So, we get

$$\begin{aligned} \tau ^{n+2}f(t)=ns^{n+2}-(n+2)\tau ^2s^n-\sum _{k=1}^{m+1}2ks^{k-1}\tau ^{n-k+1}(\tau -s)^2\ge 0. \end{aligned}$$

$\square$

Next we will give the proof of Theorem 3.1.

Proof of Theorem 3.1

For $l\ge 0$, $\tau \ge \frac{1}{2}$ and $a \ge 0$, we have

$$\begin{aligned} \begin{aligned} \int _a^{a+1}s^l(\tau -s)^2ds&=\frac{s^{l+3}}{l+3}-\frac{2s^{l+2}}{l+2}\tau +\frac{s^{l+1}}{l+1}\tau ^{2}\bigg |_a^{a+1}\\&=\frac{S_{l+3}}{l+3}-\frac{2S_{l+2}}{l+2}\tau +\frac{S_{l+1}}{l+1}\tau ^{2}, \end{aligned} \end{aligned}$$

(3.2)

where

$$\begin{aligned} S_j=(a+1)^j-a^j\ge 1. \end{aligned}$$

Therefore, we get

$$\begin{aligned} n(n+2)B-(n+2)\tau ^2nA+2\tau ^{n+2} \ge&\sum _{k=1}^{m+1}2k\tau ^{n-k+1}\left( \frac{S_{k+2}}{k+2}-\frac{2S_{k+1}}{k+1}\tau +\frac{S_{k}}{k}\tau ^{2}\right) . \end{aligned}$$

From

$$\begin{aligned} \sum _{k=1}^{m+1}&2k\tau ^{n-k+1}\left( \frac{S_{k}}{k}\tau ^{2}-\frac{2S_{k+1}}{k+1}\tau +\frac{S_{k+2}}{k+2}\right) \\ =&2\tau ^{n+2}+2\sum _{k=1}^{m}S_{k+1}\tau ^{n-k+2}-2\sum _{k=1}^{m+1}\frac{2kS_{k+1}}{k+1}\tau ^{n-k+2}+2\sum _{k=1}^{m+1}\frac{kS_{k+2}}{k+2}\tau ^{n-k+1}\\ =&2\tau ^{n+2}+2S_2\tau ^{n+1}+\frac{2mS_{m+2}}{m+2}\tau ^{n-m+1}+\frac{2(m+1)S_{m+3}}{m+3}\tau ^{n-m}\\ {}&-2S_2\tau ^{n+1}- \frac{4(m+1)S_{m+2}}{m+2}\tau ^{n-m+1}\\&+2\sum _{k=2}^{m }\left( 1+\frac{k-1}{k+1}-\frac{2k}{k+1} \right) S_{k+1}\tau ^{n-k+2}\\ =&2\tau ^{n+2}+2S_2\tau ^{n+1}+\frac{2mS_{m+2}}{m+2}\tau ^{n-m+1}+\frac{2(m+1)S_{m+3}}{m+3}\tau ^{n-m}\\&-2S_2\tau ^{n+1}- \frac{4(m+1)S_{m+2}}{m+2}\tau ^{n-m+1}\\ =&2\tau ^{n+2}-2S_{m+2}\tau ^{n-m+1}+\frac{2(m+1)S_{m+3}}{m+3}\tau ^{n-m}, \end{aligned}$$

and

$$\begin{aligned} \sum _{k=2}^{m }\left( 1+\frac{k-1}{k+1}-\frac{2k}{k+1} \right) S_{k+1}\tau ^{n-k+2}=0, \end{aligned}$$

we obtain

$$\begin{aligned} n(n+2)B-(n+2)\tau ^2nA+2\tau ^{n+2} \ge&2\tau ^{n+2}-2S_{m+2}\tau ^{n-m+1}\\&+\frac{2(m+1)S_{m+3}}{m+3}\tau ^{n-m}. \end{aligned}$$

Choosing $\tau =(nA)^{\frac{1}{n}}$, we get

$$\begin{aligned} \begin{aligned} B\ge&\frac{(nA)^{\frac{n+2}{n}}}{n}-\frac{2S_{m+2}(nA)^{\frac{n-m+1}{n}}}{n(n+2)}+\frac{2(m+1)S_{m+3}(nA)^{\frac{n-m}{n}}}{n(n+2)(m+3)}. \end{aligned} \end{aligned}$$

(3.3)

It follows from (3.3) that

$$\begin{aligned} \begin{aligned} \int _0^\infty s^{n+1}\psi (s)ds\ge&\frac{(nA)^{\frac{n+2}{n}}\psi (0)^{-\frac{2}{n}}}{n}-\frac{2S_{m+2}(nA)^{\frac{n-m+1}{n}}\psi (0)^{\frac{(m+1)n+m-1}{n}}}{n(n+2)\rho ^{m+1}}\\&+\frac{2(m+1)S_{m+3}(nA)^{\frac{n-m}{n}}\psi (0)^{\frac{(m+2)n+m}{n}}}{n(n+2)(m+3)\rho ^{m+2}}. \end{aligned} \end{aligned}$$

(3.4)

From (2.8), we know

$$\begin{aligned} \begin{aligned} \sum _{i=1}^k \lambda _i \ge&n\omega _n \int _0^\infty s^{n+1}\psi (s)ds\\ \ge&\omega _n {(nA)^{\frac{n+2}{n}}\psi (0)^{-\frac{2}{n}}}-\frac{2\omega _nS_{m+2}(nA)^{\frac{n-m+1}{n}}\psi (0)^{\frac{(m+1)n+m-1}{n}}}{(n+2)\rho ^{m+1}}\\&+\frac{2\omega _n(m+1)S_{m+3}(nA)^{\frac{n-m}{n}}\psi (0)^{\frac{(m+2)n+m}{n}}}{(n+2)(m+3)\rho ^{m+2}}. \end{aligned} \end{aligned}$$

In view of $A=\frac{k}{n\omega _n}$, we have

$$\begin{aligned} \begin{aligned} \sum _{i=1}^k \lambda _i \ge&\omega _n^{-\frac{2}{n}} {\psi (0)^{-\frac{2}{n}}}k^{\frac{n+2}{n}}-\frac{2\omega _n^{\frac{m-1}{n}}S_{m+2}\psi (0)^{\frac{(m+1)n+m-1}{n}}}{(n+2)\rho ^{m+1}}k^{\frac{n-m+1}{n}}\\&+c_2\frac{2\omega _n^{\frac{m}{n}}(m+1)S_{m+3}\psi (0)^{\frac{(m+2)n+m}{n}}}{(n+2)(m+3)\rho ^{m+2}}k^{\frac{n-m}{n}}, \end{aligned} \end{aligned}$$

(3.5)

where $0<c_2\le 1$ is a constant.

When $m=1$, we complete the proof of Theorem 2.1 in Sect. 2. We assume that $m\ge 2$. Putting

$$\begin{aligned} g(t)=g_1(t)+g_2(t), \end{aligned}$$

where

$$\begin{aligned} g_1(t)&= \omega _n^{-\frac{2}{n}} {t^{-\frac{2}{n}}}k^{\frac{n+2}{n}} \end{aligned}$$

and

$$\begin{aligned} g_2(t)=&-\frac{2\omega _n^{\frac{m-1}{n}}S_{m+2}t^{\frac{(m+1)n+m-1}{n}}}{(n+2)\rho ^{m+1}}k^{\frac{n-m+1}{n}}\\&+c_2\frac{2\omega _n^{\frac{m}{n}}(m+1)S_{m+3}t^{\frac{(m+2)n+m}{n}}}{(n+2)(m+3)\rho ^{m+2}}k^{\frac{n-m}{n}}, \end{aligned}$$

we have

$$\begin{aligned} \frac{(n+2)\rho ^{m+1}\omega _n^{\frac{m}{n}}g_2'(t)}{2k^{\frac{n-m}{n}}}=&-\frac{(m+1)n+m-1}{n}\omega _n^{-\frac{1}{n}}S_{m+2}t^{\frac{mn+m-1}{n}}k^{\frac{1}{n}}\\&+c_2\frac{(m+2)n+m}{n} \frac{(m+1)S_{m+3}}{(m+3)\rho }t^{\frac{(m+1)n+m}{n} }. \end{aligned}$$

When

$$\begin{aligned} \begin{aligned} c_2\le \frac{(m+1)n+m-1 }{(m+2)n+m}\frac{\sqrt{2}S_{m+2}}{S_{m+3}}\frac{m+3}{m+1}k^{\frac{1}{n}}, \end{aligned} \end{aligned}$$

(3.6)

we get that $g_2(t)$ is decreasing on $(0, (2\pi )^{-n}V(\Omega )]$ by using the following formulas

$$\begin{aligned} nA&=\frac{k}{\omega _n},\\ \rho \ge (2\pi )^{-n}&\omega _n^{-\frac{1}{n}}V(\Omega )^{\frac{n+1}{n}}. \end{aligned}$$

Hence g(t) is also decreasing on $(0, (2\pi )^{-n}V(\Omega )]$. This implies

$$\begin{aligned} \begin{aligned} \sum _{i=1}^k \lambda _i \ge&\omega _n^{-\frac{2}{n}} {\psi (0)^{-\frac{2}{n}}}k^{\frac{n+2}{n}}-\frac{2\omega _n^{\frac{m-1}{n}}S_{m+2}\psi (0)^{\frac{(m+1)n+m-1}{n}}}{(n+2)\rho ^{m+1}}k^{\frac{n-m+1}{n}}\\&+c_2\frac{2\omega _n^{\frac{m}{n}}(m+1)S_{m+3}\psi (0)^{\frac{(m+2)n+m}{n}}}{(n+2)(m+3)\rho ^{m+2}}k^{\frac{n-m}{n}}, \end{aligned} \end{aligned}$$

(3.7)

where

$$\begin{aligned} \psi (0)&=\frac{V(\Omega )}{(2\pi )^n},\\ \end{aligned}$$

and

$$\begin{aligned} \rho&=\frac{V(\Omega )^{\frac{n+1}{n}}}{(2\pi )^n\omega _n^{\frac{1}{n}}}. \end{aligned}$$

$\square$

From the above lemma, we have the following universal lower bounds for higher eigenvalues.

Corollary 3.3

For any bounded domain $\Omega \subseteq R^n$, $n\ge m+1\ge 3$ and $k \ge 1$ we have

$$\begin{aligned} \lambda _k \ge&\omega _n^{-\frac{2}{n}} {\alpha ^{-\frac{2}{n}}}k^{\frac{2}{n}}-\frac{2\omega _n^{\frac{m-1}{n}}S_{m+2}\alpha ^{\frac{(m+1)n+m-1}{n}}}{(n+2)\rho ^{m+1}}k^{\frac{-m+1}{n}}\\&+c_2\frac{2\omega _n^{\frac{m}{n}}(m+1)S_{m+3}\alpha ^{\frac{(m+2)n+m}{n}}}{(n+2)(m+3)\rho ^{m+2}}k^{\frac{-m}{n}}, \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} c_2\le&\min \left\{ 1, \frac{(m+1)n+m-1 }{(m+2)n+m}\frac{\sqrt{2}S_{m+2}}{S_{m+3}}\frac{m+3}{m+1}k^{\frac{1}{n}} \right\} ,\\ S_{l}=&(a+1)^l-a^l, \end{aligned} \end{aligned}$$

$\alpha =\frac{V(\Omega )}{(2\pi )^n},\,\,\,\,\rho =2(2\pi )^{-n}\sqrt{V(\Omega )I(\Omega )}$ and a is defined by (2.16).

Due to the similar discussion to Corollary 2.4, we have

Corollary 3.4

For any bounded domain $\Omega \subseteq R^n$, $n\ge 3$ and any $k \ge 1$ we have

$$\begin{aligned} \sum _{i=1}^k\lambda _i\ge&\frac{n{\omega _n}^{-\frac{2}{n}}(2\pi )^2{V(\Omega )}^{-\frac{2}{n}}}{n+2}k^{\frac{n+2}{n}} +\frac{1}{24(n+2)}\left( \frac{V(\Omega )}{I(\Omega )}\right) k\\&+\frac{\omega _n^{-\frac{1}{n}}\alpha ^{\frac{3n+1}{n}}}{9(n+2)\rho ^3}k^{\frac{n-1}{n}}+ \frac{3\omega _n^{-\frac{1}{n}}\alpha ^{\frac{4n+2}{n}}}{80(n+2)\rho ^4}k^{\frac{n-2}{n}}. \end{aligned}$$

Proof

According to Lemma 3.2, we have

$$\begin{aligned} n(n+2)B-(n+2)\tau ^2nA+2\tau ^{n+2}&\ge \frac{\tau ^n}{6}+\frac{\tau ^{n-1}}{9}+\frac{3\tau ^{n-2}}{80}. \end{aligned}$$

By using similar discussion in the proof of Theorem 2.1, we get

$$\begin{aligned} \sum _{i=1}^k\lambda _i\ge&\frac{n{\omega _n}^{-\frac{2}{n}}\phi (0)^{-\frac{2}{n}}}{n+2}k^{\frac{n+2}{n}}+\frac{\phi (0)^2}{6(n+2)\rho ^2}k +\frac{\omega _n^{-\frac{1}{n}}\phi (0)^{\frac{3n+1}{n}}}{9(n+2)\rho ^3}k^{\frac{n-1}{n}}\end{aligned}$$

(3.8)

$$\begin{aligned}&+\frac{C_3\omega _n^{-\frac{1}{n}}\phi (0)^{\frac{4n+2}{n}}}{(n+2)\rho ^4}k^{\frac{n-2}{n}}, \end{aligned}$$

(3.9)

where $0<C_3\le \frac{3}{80}$ is a constant which will be chosen. By using the similar discussion in the proof of Corollary 4.4, one can choose $C_3=\frac{3}{80}$. Hence, we complete our proof. $\square$

4 A universal lower bound on eigenvalues of the clamped plate problem

In this section, let $\phi (z)$ be the decreasing radial rearrangement of h(z) where h(z) is defined as (4.9). Then, a is defined by

$$\begin{aligned} \int _a^{a+1} z^{n+3}dz=\int _0^\infty -z^{n+3}\phi '(z)dz. \end{aligned}$$

(4.1)

We will give a universal lower bounds on the sum of eigenvalues for $n\ge m$, where $m\ge 1$.

Theorem 4.1

For any bounded domain $\Omega \subseteq R^n$, $n\ge m\ge 1$ and $k \ge 1$ we have

• (1) When $n=1$ and

  $$\begin{aligned} \frac{2\sqrt{2}S_3}{5} \le k, \end{aligned}$$

  we have

  $$\begin{aligned} \begin{aligned} \sum _{i=1}^k\Gamma _i \ge&\omega _n^{-\frac{4}{n}} {\alpha ^{-\frac{4}{n}}k^{1+\frac{4}{n}}} -\omega _n^{\frac{m-4}{n}} \frac{4S_{m+2}}{(n+4)\rho ^{m}}\alpha ^{\frac{mn+m-4}{n}}k^{\frac{n-m+4}{n}}\\&+\omega _n^{\frac{m-3}{n}} \frac{4mS_{m+2}}{(n+4)(m+2)\rho ^{m+1}}\alpha ^{\frac{(m+1)n+m-3}{n}}k^{\frac{n-m+3}{n}}, \end{aligned} \end{aligned}$$

  (4.2)

  where $\alpha$, $\rho$ are defined by (1.9) and

  $$\begin{aligned} S_l=(a+1)^l-a^l. \end{aligned}$$

• (2) When $m\ge 2$, we have

  $$\begin{aligned} \begin{aligned} \sum _{i=1}^k\Gamma _i \ge&\omega _n^{-\frac{4}{n}} {\alpha ^{-\frac{4}{n}}k^{1+\frac{4}{n}}} -\omega _n^{\frac{m-4}{n}} \frac{4S_{m+2}}{(n+4)\rho ^{m}}\alpha ^{\frac{mn+m-4}{n}}k^{\frac{n-m+4}{n}}\\&+c_3\omega _n^{\frac{m-3}{n}} \frac{4mS_{m+2}}{(n+4)(m+2)\rho ^{m+1}}\alpha ^{\frac{(m+1)n+m-3}{n}}k^{\frac{n-m+3}{n}}, \end{aligned} \end{aligned}$$

  where

  $$\begin{aligned} c_3\le \min \left\{ 1, \frac{2^{\frac{m+1}{2}}(n+2)(m+2)}{S_{m+2}[(m+1)n+m-3]}k^{\frac{m+1}{n}} \right\} . \end{aligned}$$

Next, we recall the definition and serval properties of the symmetric decreasing rearrangements. Let $\Omega \subset R^n$ be a bounded domain. Its symmetric rearrangement $\Omega ^*$ is the open ball with the same volume as $\Omega$,

$$\begin{aligned} \Omega ^*=\left\{ x\in R^n| \,\, |x|<\left( \frac{V(\Omega )}{\omega _n} \right) \right\} . \end{aligned}$$

By using a symmetric rearrangement of $\Omega$, we have

$$\begin{aligned} I(\Omega )=\int _{\Omega } |x|^2 dx\ge \int _{\Omega ^*} |x|^2 dx=\frac{n}{n+2}V(\Omega )\left( \frac{V(\Omega )}{\omega _n} \right) ^{\frac{2}{n}}. \end{aligned}$$

(4.3)

Then we have

$$\begin{aligned} \int _{R^n}|x|^4F(x)dx\ge \int _{R^n}|x|^4F^{*}(x)dx=n\omega _n\int _0^{\infty } s^{n+3}\phi (s)ds. \end{aligned}$$

(4.4)

The following lemma is useful in the proof of Theorem 4.1.

Lemma 4.2

For integers $n\ge m\ge 1$ and positive real numbers s and $\tau$, we have the following inequality:

$$\begin{aligned} ns^{n+4}-(n+4)\tau ^4s^n+4\tau ^{n+4}-\sum _{k=1}^{m}4ks^{k-1}\tau ^{n-k+3}(\tau -s)^2\ge 0. \end{aligned}$$

(4.5)

Proof

Taking $t=\frac{s}{\tau }$, and putting f(t)

$$\begin{aligned} f(t)=nt^{n+4}-(n+4)t^n+4-4(t-1)^2 - \sum _{k=2}^{m}4kt^{k-1}(t-1)^2, \end{aligned}$$

for $t\ge 0$, we get

$$\begin{aligned} \begin{aligned} f'(t)=&n(n+4)t^{n+3}-n(n+4)t^{n-1}\\&-\left[ 8(t-1)+ \sum _{k=2}^{m}4k(k-1)t^{k-2}(t-1)^2+\sum _{k=2}^{m}8kt^{k-1}(t-1) \right] \\ =&n(n+4)t^{n+3}-n(n+4)t^{n-1}\\&-(t-1)\left[ 8+ \sum _{k=2}^{m}4k(k-1)t^{k-2}(t-1)+\sum _{k=2}^{m}8kt^{k-1} \right] \\ =&n(n+4)t^{n+3}-n(n+4)t^{n-1}\\&-(t-1)\left[ 8+ \sum _{k=2}^{m}4k(k-1)t^{k-1} -\sum _{k=2}^{m}4k(k-1)t^{k-2} +\sum _{k=2}^{m}8kt^{k-1} \right] \\ =&n(n+4)t^{n+3}-n(n+4)t^{n-1}\\&-(t-1)\left[ 4m(m+1)t^{m-1} + \sum _{k=3}^{m}(4(k-1)(k-2) -4k(k-1)+8(k-1) )t^{k-2} \right] \\ =&n(n+4)t^{n+3}-n(n+4)t^{n-1}-4m(m+1)t^{m-1}(t-1)\\ =&\left[ n(n+4)t^{n-m}(t^2+1)(t+1)-4m(m+1) \right] t^{n-m}(t-1). \end{aligned} \end{aligned}$$

From the above formula, it is clear that when $n \ge m$, we have $t = 1$ is the minimum point of f(t) and then $f \ge \min \{f(1) = 0,f(0) = 0\}$. We get

$$\begin{aligned} \tau ^{n+4}f(t)=ns^{n+4}-(n+4)\tau ^4s^n-\sum _{k=1}^{m}4ks^{k-1}\tau ^{n-k+3}(\tau -s)^2\ge 0. \end{aligned}$$

$\square$

Now, we will give the proof of Theorem 4.1.

Proof of Theorem 4.1

Let $\{u_j\}_{j=1}^{\infty }$ be the eigenfunction corresponding to the eigenvalue $\Gamma _j$, $j=1,2.\ldots$ which satisfy

$$\begin{aligned} {\left\{ \begin{array}{ll} \Delta ^2 u_j=\Gamma _j u_j,\,\,&{}\textrm{in}\,\,\Omega ,\\ u_j=\frac{\partial u_j}{\partial \nu }=0,\,\,&{}\textrm{on}\,\,\partial \Omega ,\\ \int _{\Omega }u_i(x) u_j(x)dx=\delta _{ij},\,\,&{}\textrm{for}\,\,\textrm{any}\,\,i,j. \end{array}\right. } \end{aligned}$$

Thus, $\{ u_j \}_{j=1}^{\infty }$ forms an orthonormal basis of $L^2(\Omega )$. We define a function $\varphi _j$ by

$$\begin{aligned} \varphi _j(x)={\left\{ \begin{array}{ll} u_j(x), \ {} &{}x \in \Omega , \\ 0, \ {} &{}x \in \textbf{R}^n \backslash \Omega . \end{array}\right. } \end{aligned}$$

Denote by ${\widehat{\varphi }}_j(z)$ the Fourier transform of $\varphi _j (x)$. For any $z \in \textbf{R}^n$, we have

$$\begin{aligned} {\widehat{\varphi }}_j(z)=(2\pi )^{-\frac{n}{2}}\int _{\textbf{R}^n}\varphi _j(x)e^{i\langle x,z\rangle }dx=(2\pi )^{-\frac{n}{2}}\int _{\Omega }u_j(x)e^{i\langle x,z\rangle }dx. \end{aligned}$$

By the Plancherel formula, we have

$$\begin{aligned} \int _{\textbf{R}^n}{\widehat{\varphi }}_i(z){\widehat{\varphi }}_j(z)=\delta _{ij} \end{aligned}$$

(4.6)

for any i, j. Since $\{ u_j \}_{j=1}^{\infty }$ is an orthonormal basis in $L^2(\Omega )$, the Bessel inequality implies that

$$\begin{aligned} \sum _{j=1}^k |{\widehat{\varphi }}_j(z) |^2\le (2\pi )^{-n}\int _{\Omega }|e^{i\langle x,z \rangle } |^2dx=(2\pi )^{-n}V(\Omega ). \end{aligned}$$

For each $j = 1,\ldots ,k$, we deduce from the divergence theorem and $u_j|_{\partial \Omega }=\frac{\partial u_j}{\partial \nu }|_{\partial \Omega }=0$ that

$$\begin{aligned} z_p^2{\widehat{\varphi }}_j(z)&=(2\pi )^{-\frac{n}{2}}\int _{\textbf{R}^n}\varphi _j(x)(-i)^2 \frac{\partial ^2e^{i\langle x,z \rangle } }{\partial x_p^2}dx\\&=-(2\pi )^{-\frac{n}{2}}\int _{\textbf{R}^n}\frac{\partial ^2\varphi _j(x) }{\partial x_p^2} e^{i\langle x,z \rangle }dx\\&=-\frac{\widehat{\partial ^2\varphi _j}}{\partial x_p^2}(z). \end{aligned}$$

It follows from the Parseval’s identity that

$$\begin{aligned} \begin{aligned} \int _{\textbf{R}^n}|z|^4|{\widehat{\varphi }}_j(z) |^2dz&=\int _{\textbf{R}^n}(|z|^2|{\widehat{\varphi }}_j(z) |)^2dz\\&=\int _{\Omega }|\Delta u_j(x)|^2dx\\&=\Gamma _j. \end{aligned} \end{aligned}$$

(4.7)

Since

$$\begin{aligned} \nabla {\widehat{\varphi }}_j(z) =(2\pi )^{-\frac{n}{2}}\int _{\Omega } ixu_j(x)e^{i\langle x,z \rangle }dx, \end{aligned}$$

we obtain

$$\begin{aligned} \sum _{j=1}^k |\nabla {\widehat{\varphi }}_j(z) |^2\le (2\pi )^{-n}\int _{\Omega } |ixe^{i\langle x,z \rangle }|^2dx=(2\pi )^{-n}I(\Omega ). \end{aligned}$$

(4.8)

Putting

$$\begin{aligned} h(z):=\sum _{j=1}^k | {\widehat{\varphi }}_j(z) |^2, \end{aligned}$$

(4.9)

one derives from (4.6) that $0 \le h(z) \le (2\pi )^{-n}V(\Omega )$. It follows from (4.8) and the Cauchy-Schwarz inequality that

$$\begin{aligned} |\nabla h(z)|&\le 2\left( \sum _{j=1}^k | {\widehat{\varphi }}_j(z) |^2 \right) ^{\frac{1}{2}}\left( \sum _{j=1}^k |\nabla {\widehat{\varphi }}_j(z) |^2 \right) ^{\frac{1}{2}}\\&\le 2(2\pi )^{-n}\sqrt{V(\Omega )I(\Omega )} \end{aligned}$$

for every $z\in \textbf{R }^n$. From the Parseval’s identity, we derive

$$\begin{aligned} \int _{\textbf{R }^n}h(z)dz=\sum _{j=1}^k \int _{\Omega }|u_j(x)|^2dx=k. \end{aligned}$$

(4.10)

Applying the symmetric decreasing rearrangement to h(z) and noting that $\zeta =\sup |\nabla h|\le 2(2\pi )^{-n}\sqrt{V(\Omega )I(\Omega )}:=\eta ,$ we see from (2.6)

$$\begin{aligned} -\eta \le -\zeta \le \phi {'}(s)\le 0 \end{aligned}$$

for almost every s. According to (4.4) and (4.7), we infer

$$\begin{aligned} \begin{aligned} \sum _{i=1}^k\Gamma _i&=\int _{\textbf{R}^n}|z|^4h(z)dz\\&\ge \int _{\textbf{R}^n}|z|^4h^{*}(z)dz\\&=n\omega _n \int _{0}^{\infty } s^{n+3}\phi (s)ds. \end{aligned} \end{aligned}$$

(4.11)

In order to apply Lemma 4.2, from (4.4) and the definition of A, we take

$$\begin{aligned} \psi (s)=\phi (s),\,\, A=\frac{k}{n\omega _n},\,\,\eta =2(2\pi )^{-n}\sqrt{V(\Omega )I(\Omega )}. \end{aligned}$$

(4.12)

From (4.3), we deduce that

$$\begin{aligned} \rho \ge 2(2\pi )^{-n}\left( \frac{n}{n+2} \right) ^{\frac{1}{2}}\omega _n^{-\frac{1}{n}}V(\Omega )^{\frac{n+1}{n}}. \end{aligned}$$

(4.13)

On the other hand, $0 < \phi (0) \le \sup h^*(z) = \sup h(z) \le (2\pi )^{-n} V(\Omega )$.

For any $k\ge 1$ and $a \ge 0$, we have

$$\begin{aligned} \begin{aligned} \int _a^{a+1}s^{k-1}(\tau -s)^2ds&=\frac{s^{k+2}}{k+2}-\frac{2s^{k+1}}{k+1}\tau +\frac{s^{k}}{k}\tau ^{2}\bigg |_a^{a+1}\\&=\frac{S_{k+2}}{k+2}-\frac{2S_{k+1}}{k+1}\tau +\frac{S_{k}}{k}\tau ^{2}, \end{aligned} \end{aligned}$$

(4.14)

where

$$\begin{aligned} S_l=(a+1)^l-a^l. \end{aligned}$$

Let $D'=\int ^{a+1}_a s^{n+4}ds$, from the above lemma, integrating the both sides of (4.5) over $[a,a+1]$, we get

$$\begin{aligned} n(n+4)D'-(n+4)\tau ^4 nA+4\tau ^{n+4}\ge \sum _{k=1}^m4k\tau ^{n-k+3}\left( \frac{S_{k}}{k}\tau ^{2}-\frac{2S_{k+1}}{k+1}\tau +\frac{S_{k+2}}{k+2}\right) . \end{aligned}$$

(4.15)

From

$$\begin{aligned} \sum _{k=1}^{m}&4k\tau ^{n-k+3}\left( \frac{S_{k}}{k}\tau ^{2}-\frac{2S_{k+1}}{k+1}\tau +\frac{S_{k+2}}{k+2}\right) \\ =&4\tau ^{n+4}+4\sum _{k=1}^{m-1}S_{k+1}\tau ^{n-k+4}-4\sum _{k=1}^{m}\frac{2kS_{k+1}}{k+1}\tau ^{n-k+4}+4\sum _{k=1}^{m }\frac{kS_{k+2}}{k+2}\tau ^{n-k+3}\\ =&4\tau ^{n+4}+4S_2\tau ^{n+3}+\frac{4mS_{m+2}}{m+2}\tau ^{n-m+3}+\frac{4(m-1)S_{m+1}}{m+1}\tau ^{n-m+4}\\ {}&-4S_2\tau ^{n+3}- \frac{8mS_{m+2}}{m+1}\tau ^{n-m+4}\\&+4\sum _{k=2}^{m-1 }\left( 1+\frac{k-1}{k+1}-\frac{2k}{k+1} \right) S_{k+1}\tau ^{n-k+4}\\ =&4\tau ^{n+4}+4S_2\tau ^{n+3}+\frac{4mS_{m+2}}{m+2}\tau ^{n-m+3}+\frac{4(m-1)S_{m+1}}{m+1}\tau ^{n-m+4}\\&-4S_2\tau ^{n+3}- \frac{8mS_{m+2}}{m+1}\tau ^{n-m+4}\\ =&4\tau ^{n+4}-4S_{m+2}\tau ^{n-m+4}+\frac{4mS_{m+2}}{m+2}\tau ^{n-m+3}, \end{aligned}$$

and

$$\begin{aligned} 4\sum _{k=2}^{m-1 }\left( 1+\frac{k-1}{k+1}-\frac{2k}{k+1} \right) S_{k+1}\tau ^{n-k+4}=0, \end{aligned}$$

we get

$$\begin{aligned} n(n+4)D'-(n+4)\tau ^4 nA+4\tau ^{n+4}\ge&4\tau ^{n+4}-4S_{m+2}\tau ^{n-m+4}\\&+\frac{4mS_{m+2}}{m+2}\tau ^{n-m+3}. \end{aligned}$$

This implies that

$$\begin{aligned} n(n+4)D'\ge&(n+4)\tau ^4(nA) -4S_{m+2}\tau ^{n-m+4}\\&+\frac{4mS_{m+2}}{m+2}\tau ^{n-m+3}. \end{aligned}$$

Taking $\tau =(nA)^{\frac{1}{n}}$, we get

$$\begin{aligned} D'\ge&\frac{(nA)}{n}\tau ^4 -\frac{4S_{m+2}}{n(n+4)}\tau ^{n-m+4}\\&+\frac{4mS_{m+2}}{n(n+4)(m+2)}\tau ^{n-m+3}\\ \ge&\frac{(nA)^{\frac{n+4}{n}}}{n}-\frac{4S_{m+2}(nA)^{\frac{n-m+4}{n}}}{n(n+4)}\\&+\frac{4mS_{m+2}(nA)^{\frac{n-m+3}{n}}}{n(n+4)(m+2)}. \end{aligned}$$

Then, we get

$$\begin{aligned} \begin{aligned} \int _0^{\infty }s^{n+3}\psi (s)ds\ge&\frac{(nA)^{1+\frac{4}{n}}}{n}\psi (0)^{-\frac{4}{n}}-\frac{4S_{m+2}(nA)^{\frac{n-m+4}{n}}}{n(n+4)\rho ^{m}}\psi (0)^{\frac{mn+m-4}{n}}\\&+\frac{4mS_{m+2}(nA)^{\frac{n-m+3}{n}}}{n(n+4)(m+2)\rho ^{m+1}}\rho ^{\frac{(m+1)n+m-3}{n}}. \end{aligned} \end{aligned}$$

According to (4.4), (4.7) and the above inequality, we conclude

$$\begin{aligned} \begin{aligned} \sum _{i=1}^k\Gamma _i =&\int _{\textbf{R}^n}|z|^4h(z)dz\\ \ge&\int _{\textbf{R}^n}|z|^4h^{*}(z)dz\\ =&n\omega _n \int _{0}^{\infty } s^{n+3}\phi (s)ds\\ \ge&n\omega _n \frac{(nA)^{1+\frac{4}{n}}}{n}\psi (0)^{-\frac{4}{n}}-n\omega _n \frac{4S_{m+2}(nA)^{\frac{n-m+4}{n}}}{n(n+4)\rho ^{m}}\psi (0)^{\frac{mn+m-4}{n}}\\&+n\omega _n \frac{4mS_{m+2}(nA)^{\frac{n-m+3}{n}}}{n(n+4)(m+2)\rho ^{m+1}}\psi (0)^{\frac{(m+1)n+m-3}{n}}\\ =&\omega _n {(nA)^{1+\frac{4}{n}}}\psi (0)^{-\frac{4}{n}}-\omega _n \frac{4S_{m+2}(nA)^{\frac{n-m+4}{n}}}{(n+4)\rho ^{m}}\psi (0)^{\frac{mn+m-4}{n}}\\&+\omega _n \frac{4mS_{m+2}(nA)^{\frac{n-m+3}{n}}}{(n+4)(m+2)\rho ^{m+1}}\psi (0)^{\frac{(m+1)n+m-3}{n}}. \end{aligned} \end{aligned}$$

For $m=1$ and $n=1$, we define f(t) as follows

$$\begin{aligned} f(t)=&f_1(t)+f_2(t), \end{aligned}$$

on $(0,(2\pi )^{-n}V(\Omega )]$, where

$$\begin{aligned} f_1(t)&=\xi \omega _n {(nA)^{1+\frac{4}{n}}}t^{-\frac{4}{n}}-\omega _n \frac{4S_{m+2}(nA)^{\frac{n-m+4}{n}}}{(n+4)\rho ^{m}}t^{\frac{-2}{n}}, \end{aligned}$$

and

$$\begin{aligned} f_2(t)&=(1-\xi )\omega _n {(nA)^{1+\frac{4}{n}}}t^{-\frac{4}{n}}+\omega _n \frac{4mS_{m+2}(nA)^{\frac{n-m+3}{n}}}{(n+4)(m+2)\rho ^{m+1}}t^{\frac{2n-2}{n}}\\&=(1-\xi )\omega _n {(nA)^{1+\frac{4}{n}}}t^{-\frac{4}{n}}+\omega _n \frac{4mS_{m+2}(nA)^{\frac{n-m+3}{n}}}{(n+4)(m+2)\rho ^{m+1}}t^{\frac{2n-2}{n}}, \end{aligned}$$

for $0<\xi \le 1$. Then

$$\begin{aligned} \frac{nf_1'(t)}{4\omega _n(nA)^{\frac{4}{n}}}=-\xi (nA)t^{-\frac{n+4}{n}}+\frac{2S_{n+2}}{(n+4)\rho }t^{-\frac{n+2}{n}}. \end{aligned}$$

When

$$\begin{aligned} \frac{2\sqrt{2}S_3}{5k}\le \xi \le 1, \end{aligned}$$

we prove that f(t) decreases on $(0,(2\pi )^{-n}V(\Omega )]$ by using

$$\begin{aligned} \frac{\omega _n^{\frac{4}{n}}}{(2\pi )^2} \le&\frac{1}{2}, \end{aligned}$$

and

$$\begin{aligned} \rho \ge (2\pi )^{-n}&\omega _n^{-\frac{1}{n}}V(\Omega )^{\frac{n+1}{n}}. \end{aligned}$$

Therefore, if

$$\begin{aligned} \frac{2\sqrt{2}S_3}{5} \le k, \end{aligned}$$

we get

$$\begin{aligned} \begin{aligned} \sum _{i=1}^k\Gamma _i \ge&\omega _n {(nA)^{1+\frac{4}{n}}}\alpha ^{-\frac{4}{n}}-\omega _n \frac{4S_{m+2}(nA)^{\frac{n-m+4}{n}}}{(n+4)\rho ^{m}}\alpha ^{\frac{mn+m-4}{n}}\\&+\omega _n \frac{4mS_{m+2}(nA)^{\frac{n-m+3}{n}}}{(n+4)(m+2)\rho ^{m+1}}\alpha ^{\frac{(m+1)n+m-3}{n}}, \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \alpha =\frac{V(\Omega )}{(2\pi )^n}, \end{aligned}$$

and

$$\begin{aligned} \rho&=\frac{V(\Omega )^{\frac{n+1}{n}}}{(2\pi )^n\omega _n^{\frac{1}{n}}}. \end{aligned}$$

Noting that $A=\frac{k}{n\omega _n}$, we obtain the following inequality

$$\begin{aligned} \begin{aligned} \sum _{i=1}^k\Gamma _i \ge&\omega _n^{-\frac{4}{n}} {\alpha ^{-\frac{4}{n}}k^{1+\frac{4}{n}}} -\omega _n^{\frac{m-4}{n}} \frac{4S_{m+2}}{(n+4)\rho ^{m}}\alpha ^{\frac{mn+m-4}{n}}k^{\frac{n-m+4}{n}}\\&+\omega _n^{\frac{m-3}{n}} \frac{4mS_{m+2}}{(n+4)(m+2)\rho ^{m+1}}\alpha ^{\frac{(m+1)n+m-3}{n}}k^{\frac{n-m+3}{n}}. \end{aligned} \end{aligned}$$

(4.16)

When $m\ge 2$, F(t) is defined by

$$\begin{aligned} F(t)=&F_1(t)+F_2(t) \end{aligned}$$

for $t\in (0,(2\pi )^{-n}V(\Omega )]$, where

$$\begin{aligned} F_1(t)&=\omega _n {(nA)^{1+\frac{4}{n}}}t^{-\frac{4}{n}}+c_3 \omega _n \frac{4mS_{m+2}(nA)^{\frac{n-m+3}{n}}}{(n+4)(m+2)\rho ^{m+1}}t^{\frac{(m+1)n+m-3}{n}}, \end{aligned}$$

for $0<c_3\le 1$ and

$$\begin{aligned} F_2(t)&=-\omega _n \frac{4S_{m+2}(nA)^{\frac{n-m+4}{n}}}{(n+4)\rho ^{m}}t^{\frac{mn+m-4}{n}}. \end{aligned}$$

This implies

$$\begin{aligned} \frac{F_1'(t)}{4\omega _n}=&-\frac{1}{n}(nA)^{1+\frac{4}{n}}t^{-\frac{n+4}{n}}\\&+c_3\frac{(m+1)n+m-3}{n}\frac{mS_{m+2}(nA)^{\frac{n-m+3}{n}}}{(n+2)(m+2)\rho ^{m+1}}t^{\frac{mn+m-3}{n}}. \end{aligned}$$

So, if

$$\begin{aligned} c_3\le \frac{2^{\frac{m+1}{2}}(n+2)(m+2)}{S_{m+2}[(m+1)n+m-3]}k^{\frac{m+1}{n}}, \end{aligned}$$

we obtain that F(t) decreases on $(0,(2\pi )^{-n}V(\Omega )]$, which yields that

$$\begin{aligned} \begin{aligned} \sum _{i=1}^k\Gamma _i \ge&\omega _n^{-\frac{4}{n}} {\alpha ^{-\frac{4}{n}}k^{1+\frac{4}{n}}} -\omega _n^{\frac{m-4}{n}} \frac{4S_{m+2}}{(n+4)\rho ^{m}}\alpha ^{\frac{mn+m-4}{n}}k^{\frac{n-m+4}{n}}\\&+c_3\omega _n^{\frac{m-3}{n}} \frac{4mS_{m+2}}{(n+4)(m+2)\rho ^{m+1}}\alpha ^{\frac{(m+1)n+m-3}{n}}k^{\frac{n-m+3}{n}}, \end{aligned} \end{aligned}$$

where

$$\begin{aligned} c_3\le \min \left\{ 1, \frac{2^{\frac{m+1}{2}}(n+2)(m+2)}{S_{m+2}[(m+1)n+m-3]}k^{\frac{m+1}{n}} \right\} . \end{aligned}$$

$\square$

For higher eigenvalues, we have the following universal lower bounds

Corollary 4.3

For any bounded domain $\Omega \subseteq R^n$, $n\ge m\ge 1$ and any $k \ge 1$ we have

• (1) When $n=1$ and

  $$\begin{aligned} \frac{2\sqrt{2}S_3}{5} \le k, \end{aligned}$$

  we have

  $$\begin{aligned} \begin{aligned} \Gamma _k \ge&\omega _n^{-\frac{4}{n}} {\alpha ^{-\frac{4}{n}}k^{\frac{4}{n}}} -\omega _n^{\frac{m-4}{n}} \frac{4S_{m+2}}{(n+4)\rho ^{m}}\alpha ^{\frac{mn+m-4}{n}}k^{\frac{-m+4}{n}}\\&+\omega _n^{\frac{m-3}{n}} \frac{4mS_{m+2}}{(n+4)(m+2)\rho ^{m+1}}\alpha ^{\frac{(m+1)n+m-3}{n}}k^{\frac{-m+3}{n}}, \end{aligned} \end{aligned}$$

  (4.17)

  where $\alpha$, $\rho$ are defined by (1.9) and

  $$\begin{aligned} S_l=(a+1)^l-a^l. \end{aligned}$$

• (2) When $m\ge 2$, we have

  $$\begin{aligned} \begin{aligned} \Gamma _k \ge&\omega _n^{-\frac{4}{n}} {\alpha ^{-\frac{4}{n}}k^{\frac{4}{n}}} -\omega _n^{\frac{m-4}{n}} \frac{4S_{m+2}}{(n+4)\rho ^{m}}\alpha ^{\frac{mn+m-4}{n}}k^{\frac{-m+4}{n}}\\&+c_3\omega _n^{\frac{m-3}{n}} \frac{4mS_{m+2}}{(n+4)(m+2)\rho ^{m+1}}\alpha ^{\frac{(m+1)n+m-3}{n}}k^{\frac{-m+3}{n}}, \end{aligned} \end{aligned}$$

  (4.18)

  where

  $$\begin{aligned} c_3\le \min \left\{ 1, \frac{2^{\frac{m+1}{2}}(n+2)(m+2)}{S_{m+2}[(m+1)n+m-3]}k^{\frac{m+1}{n}} \right\} . \end{aligned}$$

According to Lemma 4.2 and the proof of Theorem 4.1, we also have the following result.

Corollary 4.4

For any bounded domain $\Omega \subseteq R^n$, $n\ge 3$ and any $k \ge 1$ we have

$$\begin{aligned} \begin{aligned} \sum _{i=1}^k \Gamma _i \ge&\frac{n}{n+4}(\omega _n)^{-\frac{4}{n}}\alpha ^{-\frac{4}{n}}k^{\frac{4+n}{n}} +\frac{1}{3(n+4)}\frac{(\omega _n)^{-\frac{2}{n}}\alpha ^{\frac{2n-2}{n}}k^{\frac{n+2}{n}}}{\rho ^2}\\&+\frac{2}{9(n+4)}\frac{(\omega _n)^{-\frac{1}{n}}\alpha ^{\frac{3n-1}{n}}k^{\frac{n+1}{n}}}{\rho ^3}+\frac{3\alpha ^{4}}{40(n+4)\rho ^4}k. \end{aligned} \end{aligned}$$

(4.19)

Proof

By using Lemma 4.2, we get

$$\begin{aligned} ns^{n+4}-(n+4)\tau ^4s^n+4\tau ^{n+4}\ge&4\tau ^{n+2}(\tau -s)^2+8s\tau ^{n+1}(\tau -s)^2\\&+12s^2\tau ^n(\tau -s)^2. \end{aligned}$$

In view of (4.15) and (57) in [30], integrating the both sides of the above inequality over $[a,a+1]$, we have

$$\begin{aligned} \begin{aligned} n(n+4)D'-(n+4)\tau ^4 nA+4\tau ^{n+4}\ge&\frac{\tau ^{n+2}}{3}+\frac{2\tau ^{n+1}}{9}+12\tau ^n\int ^{\frac{1}{2}}_0s^2(\tau -s)^2\\ \ge&\frac{\tau ^{n+2}}{3}+\frac{2\tau ^{n+1}}{9}+12\tau ^n\min _{\tau \ge \frac{1}{2}}\int ^{\frac{1}{2}}_0s^2(\tau -s)^2\\ \ge&\frac{\tau ^{n+2}}{3}+\frac{2\tau ^{n+1}}{9}+\frac{3\tau ^n}{40}. \end{aligned} \end{aligned}$$

By using similar discussion in the proof of Theorem 4.1 and taking $\tau =(nA)^{\frac{1}{n}}$, we get

$$\begin{aligned} \begin{aligned} \int _0^{\infty }s^{n+3}\psi (s)ds\ge&\frac{1}{n+4}\tau ^{n+4}+\frac{\tau ^{n+2}}{3n(n+4)}+\frac{2\tau ^{n+1}}{9n(n+4)}+\frac{3\tau ^n}{40n(n+4)}. \end{aligned} \end{aligned}$$

Hence,we arrive at

$$\begin{aligned} \begin{aligned} \sum _{i=1}^k\Gamma _i \ge&\frac{n\omega _n}{n+4}(nA)^{\frac{n+4}{n}}\psi (0)^{-\frac{4}{n}}+\frac{\omega _n(nA)^{\frac{n+2}{n}}}{3(n+4)\rho ^{2}}\psi (0)^{\frac{2n-2}{n}}\\&+\frac{2\omega _n(nA)^{\frac{n+1}{n}}}{9(n+4)\rho ^3}\psi (0)^{\frac{3n-1}{n}}+d_1\frac{3\omega _n(nA)}{40(n+4)\rho ^4}\psi (0)^{4}\\ =&\frac{n\omega _n}{n+4}\left( \frac{k}{\omega _n} \right) ^{\frac{n+4}{n}}\psi (0)^{-\frac{4}{n}}+ \frac{\omega _n\left( \frac{k}{\omega _n} \right) ^{\frac{n+2}{n}}}{3(n+4)\rho ^{2}}\psi (0)^{\frac{2n-2}{n}}\\&+ \frac{2\omega _n\left( \frac{k}{\omega _n} \right) ^{\frac{n+1}{n}}}{9(n+4)\rho ^3}\psi (0)^{\frac{3n-1}{n}}+d_1\frac{3\omega _n\left( \frac{k}{\omega _n} \right) }{40(n+4)\rho ^4}\psi (0)^{4}, \end{aligned} \end{aligned}$$

where $0<d_1\le 1$ is a constant to be determined. Let $t\in (0,(2\pi )^{-n}V(\Omega )]$, we define

$$\begin{aligned} \begin{aligned} Q(t)=&\frac{n}{n+4}\left( \frac{k}{\omega _n} \right) ^{\frac{n+4}{n}}t^{-\frac{4}{n}}+ \frac{\left( \frac{k}{\omega _n} \right) ^{\frac{n+2}{n}}}{3(n+4)\rho ^{2}}t^{\frac{2n-2}{n}}\\&+ \frac{2\left( \frac{k}{\omega _n} \right) ^{\frac{n+1}{n}}}{9(n+4)\rho ^3}t^{\frac{3n-1}{n}}+d_1\frac{3\left( \frac{k}{\omega _n} \right) }{40(n+4)\rho ^4}t^{4}, \end{aligned} \end{aligned}$$

which would be decreasing on $(0, (2\pi )^{-n}V(\Omega )]$ if $Q'((2\pi )^{-n}V(\Omega ))\le 0$. Obviously, $Q'((2\pi )^{-n}V(\Omega ))\le 0$ is equal to

$$\begin{aligned} \begin{aligned} 4\left( \frac{k}{\omega _n} \right) ^{\frac{4}{n}}\left( \frac{(2\pi )^n}{V(\Omega )} \right) ^{1+\frac{4}{n}}\ge&\frac{(2n-2)}{n}\left( \frac{k}{\omega _n} \right) ^{\frac{2}{n}}\frac{1}{\rho ^2}\left( \frac{V(\Omega )}{(2\pi )^n} \right) ^{\frac{n-2}{n}}\\&+\frac{3n-1}{n}\frac{2\left( \frac{k}{\omega _n} \right) ^{\frac{1}{n}}}{9(n+4)\rho ^3}\left( \frac{V(\Omega )}{(2\pi )^n} \right) ^{\frac{2n-1}{n}}\\&+4d_1\frac{3}{40(n+4)\rho ^4}\left( \frac{V(\Omega )}{(2\pi )^n} \right) ^{3}. \end{aligned} \end{aligned}$$

Due to (2.18) and $\frac{\omega _n^{\frac{4}{n}}}{(2\pi )^2}\le \frac{1}{2}$, if

$$\begin{aligned} d_1\le \min \{1,d_0\}, \end{aligned}$$

we have $Q'((2\pi )^{-n}V(\Omega ))\le 0$, where

$$\begin{aligned} d_0=\frac{140(2\pi )^2}{3}\Bigg (4\left( \frac{k}{\omega _n} \right) ^{\frac{4}{n}}-\frac{1}{\sqrt{2}\pi } \left( \frac{k}{\omega _n} \right) ^{\frac{2}{n}}-\frac{2}{21(2\pi )^{\frac{3}{2}}}\left( \frac{1}{2}\right) ^{\frac{3}{4}}\left( \frac{k}{\omega _n} \right) ^{\frac{1}{n}} \Bigg ). \end{aligned}$$

By direct computation, one has $d_0>1$. Therefore, we obtain the following eigenvalue inequality

$$\begin{aligned} \begin{aligned} \sum _{i=1}^k\Gamma _i \ge&\frac{n\omega _n}{n+4}\left( \frac{k}{\omega _n} \right) ^{\frac{n+4}{n}}\alpha ^{-\frac{4}{n}}+ \frac{\omega _n\left( \frac{k}{\omega _n} \right) ^{\frac{n+2}{n}}}{3(n+4)\rho ^{2}}\alpha ^{\frac{2n-2}{n}}\\&+ \frac{2\omega _n\left( \frac{k}{\omega _n} \right) ^{\frac{n+1}{n}}}{9(n+4)\rho ^3}\alpha ^{\frac{3n-1}{n}}+\frac{3\omega _n\left( \frac{k}{\omega _n} \right) }{20(n+4)\rho ^4}\alpha ^{4}. \end{aligned} \end{aligned}$$

$\square$