In recent years, there are some researches on eigenvalue estimates for quadratic polynomial operator of Laplacian:

$ \begin{equation}\label{1} \left\{ \begin{array}{l} \Delta^{2}u-p\Delta{u}+qu=\lambda{u}\ \ \ \ \ \hbox{in}\, \, \Omega, \\ u|_{\partial \Omega}=\frac{\partial u}{\partial v}|_{\partial \Omega}=0, \end{array} \right. \end{equation} $

(1.1)

where $\Omega$ is a bounded domain in an $n$-dimensional complete Riemannian manifold $M, v$ denotes the outwards unit normal vector field of $\partial \Omega$, the constants $p, q \geq0$, $\Delta(=-{\rm div}\nabla)$ is the Laplacian and $\Delta^{2}$ is the biharmonic operator on $M$. In 2011, Sun and Qi [3] obtained universal eigenvalue inequalities of problem (1.1),

$\begin{eqnarray*} \sum\limits_{i=1}^{k}(\lambda_{k+1}-\lambda_{i})^{2}\leq\frac{1}{n^{2}}\sum\limits_{i=1}^{k}(\lambda_{k+1}-\lambda_{i}) \left((nH_{0})^{2}+(2n+4)E_{i}+np\right)\left((nH_{0})^{2}+4E_{i}\right), \end{eqnarray*} $

where $H_{0}$ is a constant which only depends on the mean curvature of $M$ and

$ E_{i}=\frac{1}{2}(-p+\sqrt{p^{2}+4(\lambda_{i}-q)}). $

For other related results, one can see [2, 4, 5, 7, 8].

Let $(M^{n}, \langle, \rangle)$ be an $n(\geq2)$-dimensional complete Riemannian manifold isometrically immersed in the Euclidean space $\mathbb{R}^{N}$ by $\Psi$, $\Omega$ is a bounded domain in $M^{n}$. In this paper, we are interested in extrinsic upper bounds for the lower order eigenvalues of following operator defined on $\Omega$:

$ \begin{equation}\label{2} \left\{ \begin{array}{l} \Delta^{2}u-{\rm div} A\nabla{u}+au=\lambda{u}\ \ \hbox{in}\, \, \Omega, \\ u|_{\partial \Omega}=\frac{\partial u}{\partial v}|_{\partial \Omega}=0, \end{array} \right. \end{equation} $

(1.2)

where $A$ is a positive definite symmetric (1, 1)-tensor on $M^{n}$, the eigenvalues of $A$ on $\Omega$ are bounded below by a positive constant $b$, ${\rm tr}(A)|_{\Omega}\leq c$ (i.e., $A$ can also be viewed as a smooth symmetric and positive defined section of the bundle of all endomorphisms of $T(M^{n})$, $\Delta(={\rm div}\nabla)$ is the Laplacian, $\nabla$ is the gradient operator on $M$ and $a$ is a nonnegative constant.

It is well known that problem (1.2) has a real and discrete spectrum

$ 0<\lambda_{1}\leq \lambda_{2}\leq \lambda_{3}\leq \cdots \rightarrow +\infty, $

where each eigenvalue is repeated according to its multiplicity. Our results are stated as follows.

Theorem 1.1  Let $(M^{n}, \langle, \rangle)$ be an $n(\geq 2)$-dimensional complete Riemannian manifold isometrically immersed in the Euclidean space $\mathbb{R}^{N}$ by $\Psi$, $\Omega$ is a bounded domain in $M^{n}$. $A$ is a positive definite symmetric (1, 1)-tensor on $M^{n}$. Assume that the eigenvalues of $A$ on $\Omega$ are bounded below by a positive constant $b$ (that is, $\bar{\lambda}(A)|_{\Omega}\geq b)$ and that ${\rm tr}(\hbox{A})|_{\Omega}\leq c$. Then for the eigenvalues of problem (1.2),

$\begin{eqnarray*} \sum\limits_{\alpha=1}^{n}(\lambda_{\alpha+1}-\lambda_{1})^{\frac{1}{2}}\leq \sqrt{\left(n^{2}\sup\limits_{\Omega}|H|^2+(n+2)B+c\right)\left(n^{2}\sup\limits_{\Omega}|H|^2+2B\right)}, \end{eqnarray*} $

(1.3)

where $B=-b+\sqrt{b^2+4(\lambda_{1}-a)}$ and $H$ is the mean curvature vector of the immersion $\Psi$.

First, we give the following lemma.

Lemma 2.1  Let $\lambda_{i}$ be the $i$-th eigenvalue of problem (1.2) and $u_{i}$ the orthonormal eigenfunction corresponding to $\lambda_{i}$ $({\rm i.e.} \displaystyle\int_{\Omega}u_{i}u_{j}=\delta_{ij})$. For any function $m_{\alpha}\in C^2(\bar{\Omega})$ satisfying

$ \begin{equation} \int_{\Omega}m_{\alpha}u_{1}u_{\beta}=0 \quad {\rm for} \ \beta=2, \cdots, \alpha, \end{equation} $

(2.1)

we have for any positive constant $\delta$,

$ \begin{equation} \begin{aligned} &(\lambda_{\alpha+1}-\lambda_{1})^{\frac{1}{2}}\int_{\Omega}u_{1}^2|\nabla m_{\alpha}|^2\\ \leq&\frac{\delta}{2}\left(||u_{1}\Delta m_{\alpha}+2\langle\nabla m_{\alpha}, \nabla u_{1}\rangle||^2 -2\int_{\Omega}|\nabla m_{\alpha}|^2 u_{1}\Delta u_{1} +\int_{\Omega}u_{1}^2 \langle\nabla m_{\alpha}, A\nabla m_{\alpha}\rangle\right)\\ &+\frac{1}{2\delta}||u_{1}\Delta m_{\alpha}+2\langle\nabla m_{\alpha}, \nabla u_{1}\rangle||^2, \end{aligned} \end{equation} $

(2.2)

where $||f||^2=\int_{\Omega}f^2$.

Proof  We consider the function $T_{\alpha}=m_{\alpha}u_{1}-u_{1}\displaystyle\int_{\Omega} m_{\alpha}u_{1}^2$. Then, it is easy to check that $T_{\alpha}|_{\partial \Omega}=0$ and

$ \begin{equation}\label{s1} \int_{\Omega}T_{\alpha}u_{1}=0. \end{equation} $

(2.3)

Hence

$ \begin{equation} \int_{\Omega}T_{\alpha}u_{l}=0 \ \ {\rm for} \ l=1, \cdots, \alpha. \end{equation} $

(2.4)

It follows from the Rayleigh-Ritz inequality(or min-max principle) that

$ \begin{equation}\label{s2} \lambda_{\alpha+1}||T_{\alpha}||^2\leq\int_{\Omega}T_{\alpha}[\Delta^{2}-{\rm div} A\nabla+a]T_{\alpha}. \end{equation} $

(2.5)

By virtue of (2.3), a direct calculation yields

$ \begin{equation}\label{s3} \begin{aligned} \lambda_{\alpha+1}||T_{\alpha}||^2 \leq&\int_{\Omega}T_{\alpha}[\Delta^{2}-{\rm div} A\nabla+a]T_{\alpha} =\int_{\Omega}T_{\alpha}[\Delta^{2}-{\rm div} A\nabla+a](m_{\alpha}u_{1})\\ =&\int_{\Omega}T_{\alpha}[u_{1}\Delta^{2}m_{\alpha}+2\Delta m_{\alpha}\Delta u_{1}+2\Delta\langle\nabla m_{\alpha}, \nabla u_{1}\rangle\\ &+2\langle\nabla m_{\alpha}, \nabla\Delta u_{1}\rangle+2\langle\nabla u_{1}, \nabla\Delta m_{\alpha}\rangle\\ &+m_{\alpha}\Delta^{2}u_{1}-m_{\alpha}{\rm div} A\nabla u_{1}+a m_{\alpha}u_{1} -u_{1}{\rm div} A\nabla m_{\alpha}\\ &-2\langle\nabla m_{\alpha}, A\nabla u_{1}\rangle]\\ =&\int_{\Omega}T_{\alpha}[u_{1}\Delta^{2}m_{\alpha}+2\Delta m_{\alpha}\Delta u_{1}+2\Delta\langle\nabla m_{\alpha}, \nabla u_{1}\rangle\\ &+2\langle\nabla m_{\alpha}, \nabla\Delta u_{1}\rangle+2\langle\nabla u_{1}, \nabla\Delta m_{\alpha}\rangle\\ &-u_{1}{\rm div} A\nabla m_{\alpha}-2\langle\nabla m_{\alpha}, A\nabla u_{1}\rangle]+\lambda_{1}||T_{\alpha}||^2. \end{aligned} \end{equation} $

(2.6)

Substituting (2.6) into inequality (2.5) and utilizing the Green's formula, we have

$ \begin{equation}\label{s4} \begin{aligned} & (\lambda_{\alpha+1}-\lambda_{1})||T_{\alpha}||^2 \leq \int_{\Omega}T_{\alpha}[u_{1}\Delta^{2}m_{\alpha}+2\Delta m_{\alpha}\Delta u_{1} +2\Delta\langle\nabla m_{\alpha}, \nabla u_{1}\rangle\\ &+2\langle\nabla m_{\alpha}, \nabla\Delta u_{1}\rangle+2\langle\nabla u_{1}, \nabla\Delta m_{\alpha}\rangle -u_{1}{\rm div} A\nabla m_{\alpha}-2\langle\nabla m_{\alpha}, A\nabla u_{1}\rangle]\\ \leq&\int_{\Omega}(m_{\alpha}u_{1})[u_{1}\Delta^{2}m_{\alpha}+2\Delta m_{\alpha}\Delta u_{1} +2\Delta\langle\nabla m_{\alpha}, \nabla u_{1}\rangle\\ &+2\langle\nabla m_{\alpha}, \nabla\Delta u_{1}\rangle+2\langle\nabla u_{1}, \nabla\Delta m_{\alpha}\rangle\\ &-u_{1}{\rm div} A\nabla m_{\alpha}-2\langle\nabla m_{\alpha}, A\nabla u_{1}\rangle]\\ &-\int_{\Omega}m_{\alpha}u_{1}^2\int_{\Omega}u_{1}[u_{1}\Delta^{2}m_{\alpha}+2\Delta m_{\alpha}\Delta u_{1} +2\Delta\langle\nabla m_{\alpha}, \nabla u_{1}\rangle\\ &+2\langle\nabla m_{\alpha}, \nabla\Delta u_{1}\rangle+2\langle\nabla u_{1}, \nabla\Delta m_{\alpha}\rangle\\ &-u_{1}{\rm div} A\nabla m_{\alpha}-2\langle\nabla m_{\alpha}, A\nabla u_{1}\rangle]\\ =&\int_{\Omega}(m_{\alpha}u_{1})[u_{1}\Delta^{2}m_{\alpha}+2\Delta m_{\alpha}\Delta u_{1} +2\Delta\langle\nabla m_{\alpha}, \nabla u_{1}\rangle\\ &+2\langle\nabla m_{\alpha}, \nabla\Delta u_{1}\rangle+2\langle\nabla u_{1}, \nabla\Delta m_{\alpha}\rangle\\ &-u_{1}{\rm div} A\nabla m_{\alpha}-2\langle\nabla m_{\alpha}, A\nabla u_{1}\rangle]. \end{aligned} \end{equation} $

(2.7)

From the divergence theorem, we get

$ \begin{eqnarray}\label{al1} &&\int_{\Omega}2m_{\alpha}u_{1}\Delta\langle\nabla m_{\alpha}, \nabla u_{1}\rangle\nonumber\\ &=&2\int_{\Omega}(u_{1}\langle\nabla m_{\alpha}, \nabla u_{1}\rangle \Delta m_{\alpha}+m_{\alpha}\langle\nabla m_{\alpha}, \nabla u_{1}\rangle\Delta u_{1} +2\langle\nabla m_{\alpha}, \nabla u_{1}\rangle^2), \end{eqnarray} $

(2.8)

$\begin{eqnarray} \label{al2} &&\int_{\Omega}2m_{\alpha}u_{1}\langle\nabla m_{\alpha}, \nabla\Delta u_{1}\rangle\nonumber\\ &=&-2\int_{\Omega}(m_{\alpha}\langle\nabla m_{\alpha}, \nabla u_{1}\rangle \Delta u_{1}+\langle\nabla m_{\alpha}, \nabla m_{\alpha}\rangle u_{1}\Delta u_{1} +m_{\alpha}u_{1}\Delta u_{1}\Delta m_{\alpha}), \end{eqnarray} $

(2.9)

$\begin{eqnarray} &&\int_{\Omega}2m_{\alpha}u_{1}\langle\nabla u_{1}, \Delta\nabla m_{\alpha}\rangle\nonumber\\ &=&\int_{\Omega}(\Delta m_{\alpha})^2 u_{1}^2 +\langle\nabla u_{1}^2, \nabla m_{\alpha}\rangle \Delta m_{\alpha}-m_{\alpha}u_{1}^2 \Delta^2m_{\alpha}. \end{eqnarray} $

(2.10)

Then, putting (2.8)-(2.10) into (2.7), one get

$ \begin{equation}\label{a14} \begin{aligned} &(\lambda_{\alpha+1}-\lambda_{1})||T_{\alpha}||^2\\ \leq&\int_{\Omega}[u_{1}^2(\Delta m_{\alpha})^2+4\langle\nabla m_{\alpha}, \nabla u_{1}\rangle^2 -2\langle\nabla m_{\alpha}, \nabla m_{\alpha}\rangle u_{1}\Delta u_{1}\\ &+4u_{1}\langle\nabla m_{\alpha}, \nabla u_{1}\rangle\Delta m_{\alpha}-m_{\alpha}u_{1}^2 ({\rm div} A\nabla m_{\alpha}) -\frac{1}{2}\langle\nabla (m_{\alpha}^2), A\nabla (u_{1}^2)\rangle]\\ =&||u_{1}\Delta m_{\alpha}+2\langle\nabla m_{\alpha}, \nabla u_{1}\rangle||^2-2\int_{\Omega}|\nabla m_{\alpha}|^2 u_{1}\Delta u_{1} +\int_{\Omega}u_{1}^2\langle\nabla m_{\alpha}, A\nabla m_{\alpha}\rangle. \end{aligned} \end{equation} $

(2.11)

Utilizing the divergence theorem again, it holds

$ \begin{equation} \begin{aligned} 2\int_{\Omega}u_{1}m_{\alpha}\langle\nabla m_{\alpha}, \nabla u_{1}\rangle &=\frac{1}{2}\int_{\Omega}\langle\nabla (m_{\alpha}^2), \nabla (u_{1}^2)\rangle =-\frac{1}{2}\int_{\Omega}u_{1}^2 \Delta(m_{\alpha}^2)\\ &=-\int_{\Omega}u_{1}^2 m_{\alpha}\Delta m_{\alpha}-\int_{\Omega}u_{1}^2 |\nabla m_{\alpha}|^2. \end{aligned} \end{equation} $

(2.12)

Hence

$ \begin{equation} \begin{aligned}\label{a15} &\int_{\Omega}T_{\alpha}(u_{1}\Delta m_{\alpha}+2\langle\nabla m_{\alpha}, \nabla u_{1}\rangle)\\ =&\int_{\Omega}u_{1}^2 m_{\alpha}\Delta m_{\alpha}+2\int_{\Omega}u_{1}m_{\alpha}\langle\nabla m_{\alpha}, \nabla u_{1}\rangle\\ &-2\int_{\Omega}m_{\alpha}u_{1}^2(\frac{1}{2}\int_{\Omega}u_{1}^2 \Delta m_{\alpha}+\int_{\Omega}u_{1}\langle\nabla m_{\alpha}, \nabla u_{1}\rangle)\\ =&-\int_{\Omega}u_{1}^2 \left|\nabla m_{\alpha}\right|^2. \end{aligned} \end{equation} $

(2.13)

By virtue of (2.11), (2.13) and the Cauchy's inequality, it is not difficult to get

$ \begin{equation} \begin{aligned} &(\lambda_{\alpha+1}-\lambda_{1})^{\frac{1}{2}}\int_{\Omega}u_{1}^2|\nabla m_{\alpha}|^2\\ =&-(\lambda_{\alpha+1}-\lambda_{1})^{\frac{1}{2}} \int_{\Omega}T_{\alpha}(u_{1}\Delta m_{\alpha}+2\langle\nabla m_{\alpha}, \nabla u_{1}\rangle)\\ \leq&\frac{\delta}{2}(\lambda_{\alpha+1}-\lambda_{1})\|T_{\alpha}\|^2+\frac{1}{2\delta} \|u_{1}\Delta m_{\alpha}+2\langle\nabla m_{\alpha}, \nabla u_{1}\rangle\|^2\\ =&\frac{\delta}{2}(\|u_{1}\Delta m_{\alpha}+2\langle\nabla m_{\alpha}, \nabla u_{1}\rangle\|^2 -2\int_{\Omega}|\nabla m_{\alpha}|^2 u_{1}\Delta u_{1} +\int_{\Omega}u_{1}^2 \langle\nabla m_{\alpha}, A\nabla m_{\alpha}\rangle)\\ &+\frac{1}{2\delta}\|u_{1}\Delta m_{\alpha}+2\langle\nabla m_{\alpha}, \nabla u_{1}\rangle\|^2. \end{aligned} \end{equation} $

(2.14)

Now we give the proof of Theorem 1.1:

Proof of Theorem 1.1  Let $x^1, x^2, \cdots, x^{N}$ be the standard coordinate functions on $\mathbb{R}^{N}$. We consider the $N\times N$ matrix

$ D\triangleq(d_{\alpha\beta}), ~~ d_{\alpha\beta}=\displaystyle\int_{\Omega}x^{\alpha}u_{1}u_{\beta+1}~~(\alpha, \beta=1, \cdots N). $

From the orthogonalization of Gram-Schmidt (QR-factorization theorem), we know that $D$ can be written by $ R=QD, $ where $Q=(q_{\alpha\beta})$ is an orthogonal $N\times N$ matrix and $R=(r_{\alpha\beta})$ is an upper triangular matrix. Hence, we have

$\begin{eqnarray*} r_{\alpha\beta}=\sum\limits_{\gamma=1}^N q_{\alpha\gamma}d_{\gamma\beta}=\int_{\Omega}\sum\limits_{\gamma=1}^N q_{\alpha\gamma}x^{\gamma}u_{1}u_{\beta+1}=0 \ \ {\rm for} \ \beta=1, \cdots, \alpha-1.\end{eqnarray*} $

Taking $m_{\alpha}=\sum\limits_{\gamma=1}^N q_{\alpha\gamma}x^{\gamma}$, thus $\displaystyle\int_{\Omega}m_{\alpha}u_{1}u_{l}=0 \quad {\rm for} \ l=2, \cdots, \alpha$. Applying Lemma 2.1 to $m_{\alpha}=\sum\limits_{\gamma=1}^N q_{\alpha\gamma}x^{\gamma}$ and summing on $\alpha$ from $1$ to $N$, we get

$ \begin{eqnarray} &&\sum\limits_{\alpha=1}^N(\lambda_{\alpha+1}-\lambda_{1})^{\frac{1}{2}}\int_{\Omega}u_{1}^2|\nabla m_{\alpha}|^2\nonumber\\ &\leq&\frac{\delta}{2}\sum\limits_{\alpha=1}^N\left(||u_{1}\Delta m_{\alpha}+2\langle\nabla m_{\alpha}, \nabla u_{1}\rangle||^2 -2\int_{\Omega}|\nabla m_{\alpha}|^2 u_{1}\Delta u_{1}\right.\nonumber\\ &&\left.+\int_{\Omega}u_{1}^2 \langle\nabla m_{\alpha}, A\nabla m_{\alpha}\rangle\right) +\frac{1}{2\delta}\sum\limits_{\alpha=1}^N||u_{1}\Delta m_{\alpha}+2\langle\nabla m_{\alpha}, \nabla u_{1}\rangle||^2\nonumber\\ &\leq&\frac{\delta}{2}\sum\limits_{\gamma=1}^N\left(||u_{1}\Delta x^{\gamma}+2\langle\nabla x^{\gamma}, \nabla u_{1}\rangle||^2 -2\int_{\Omega}|\nabla x^{\gamma}|^2 u_{1}\Delta u_{1}\right.\nonumber\\ &&\left.+\int_{\Omega}u_{1}^2 \langle\nabla x^{\gamma}, A\nabla x^{\gamma}\rangle\right) +\frac{1}{2\delta}\sum\limits_{\gamma=1}^N||u_{1}\Delta x^{\gamma}+2\langle\nabla x^{\gamma}, \nabla u_{1}\rangle||^2\nonumber\\ &\leq&\frac{\delta}{2}\left(n^2\int_{\Omega}u_{1}^2|H|^2+(2n+4)\int_{\Omega}|\nabla u_{1}|^2+\int_{\Omega}u_{1}^2({\rm tr} A)\right)\nonumber\\ &&+\frac{1}{2\delta}\int_{\Omega}\left(n^2 u_{1}^2|H|^2+4|\nabla u_{1}|^2\right)\nonumber\\ &\leq&\frac{\delta}{2}\left(n^2\int_{\Omega}u_{1}^2|H|^2+(2n+4)\int_{\Omega}|\nabla u_{1}|^2+c\right)\nonumber\\ &&+\frac{1}{2\delta}\int_{\Omega}\left(n^2 u_{1}^2|H|^2+4|\nabla u_{1}|^2\right), \end{eqnarray} $

(2.15)

where

$ \begin{eqnarray*}&& \sum\limits_{\gamma=1}^N|\nabla x^{\gamma}|^2=n, \ \sum\limits_{\gamma=1}^N\langle\nabla x^{\gamma}, \ \nabla u_{1}\rangle^2=|\nabla u_{1}|^2, \\ && \sum\limits_{\gamma=1}^N(\Delta x^{\gamma})^2=n^2|H|^2\, \ \sum\limits_{\gamma=1}^N(\langle\nabla x^{\gamma}, \nabla u_{1}\rangle\Delta x^{\gamma})=0, \end{eqnarray*} $

see [1, Lemma 2.1]. Since

$ \begin{equation} \begin{aligned} \int_{\Omega}|\nabla u_{1}|^2=&\int_{\Omega}-u_{1}\Delta u_{1}\leq(\int_{\Omega}u_{1}^2)^\frac{1}{2} (\int_{\Omega}(\Delta u_{1})^2)^\frac{1}{2} =(\int_{\Omega}u_{1}\Delta^2 u_{1})^\frac{1}{2}\\ &=(\int_{\Omega}u_{1}({\rm div} A\nabla u_{1}-a u_{1}+\lambda_1 u_{1}))^\frac{1}{2}\\ &=(-a+\lambda_1-\int_{\Omega}\langle\nabla u_{1}, A\nabla u_{1}\rangle)^\frac{1}{2}\\ &\leq(-a+\lambda_1-b\int_{\Omega}|\nabla u_{1}|^2)^\frac{1}{2}. \end{aligned} \end{equation} $

(2.16)

Hence

$ \int_{\Omega}|\nabla u_{1}|^2\leq\frac{-b+\sqrt{b^2+4(\lambda_1-a)}}{2}. $

Then, we obtain

$ \begin{equation}\label{ad1} \begin{aligned} &\sum\limits_{\alpha=1}^N(\lambda_{\alpha+1}-\lambda_{1})^{\frac{1}{2}}\int_{\Omega}u_{1}^2|\nabla m_{\alpha}|^2\\ \leq&\frac{\delta}{2}(n^{2}\sup\limits_{\Omega}|H|^2+(n+2)B+c) +\frac{1}{2\delta}(n^{2}\sup\limits_{\Omega}|H|^2+2B), \end{aligned} \end{equation} $

(2.17)

where $B=-b+\sqrt{b^2+4(\lambda_{1}-a)}$ and $H$ is the mean curvature vector of its immersion $\Psi$.

Minimizing the right hand side of (2.17), one get

$ \begin{equation}\label{ad2} \begin{aligned} &\sum\limits_{\alpha=1}^{N}(\lambda_{\alpha+1}-\lambda_{1})^{\frac{1}{2}}\int_{\Omega}u_{1}^2|\nabla m_{\alpha}|^2\\ \leq &\sqrt{(n^{2}\sup\limits_{\Omega}|H|^2+(n+2)B+c)(n^{2}\sup\limits_{\Omega}|H|^2+2B)}. \end{aligned} \end{equation} $

(2.18)

It is easy to get $\sum\limits_{\alpha=1}^{N}|\nabla m_{\alpha}|^2=n$ and $|\nabla m_{\alpha}|^2\leq 1 \ {\rm for}\ \alpha=1, \cdots, n$, see [6]. Therefore,

$ \begin{equation}\label{ad3} \begin{aligned} &\sum\limits_{\alpha=1}^{N}(\lambda_{\alpha+1}-\lambda_{1})^{\frac{1}{2}}|\nabla m_{\alpha}|^2\\ \geq&\sum\limits_{\alpha=1}^{n}(\lambda_{\alpha+1}-\lambda_{1})^{\frac{1}{2}}|\nabla m_{\alpha}|^2+(\lambda_{n+1}-\lambda_{1})^{\frac{1}{2}}\sum\limits_{t=n+1}^{N} |\nabla m_{t}|^2\\ =&\sum\limits_{\alpha=1}^{n}(\lambda_{\alpha+1}-\lambda_{1})^{\frac{1}{2}}|\nabla m_{\alpha}|^2 +(\lambda_{n+1}-\lambda_{1})^{\frac{1}{2}}(n-\sum\limits_{\beta=1}^{n}|\nabla m_{\beta}|^2)\\ =&\sum\limits_{\alpha=1}^{n}(\lambda_{\alpha+1}-\lambda_{1})^{\frac{1}{2}}|\nabla m_{\alpha}|^2 +(\lambda_{n+1}-\lambda_{1})^{\frac{1}{2}}\sum\limits_{\beta=1}^{n}(1-|\nabla m_{\beta}|^2)\\ \geq&\sum\limits_{\alpha=1}^{n}(\lambda_{\alpha+1}-\lambda_{1})^{\frac{1}{2}}. \end{aligned} \end{equation} $

(2.19)

From (2.18) and (2.19), we have

$ \begin{eqnarray*} \sum\limits_{\alpha=1}^{n}(\lambda_{\alpha+1}-\lambda_{1})^{\frac{1}{2}}\leq \sqrt{(n^{2}\sup\limits_{\Omega}|H|^2+(n+2)B+c)(n^{2}\sup\limits_{\Omega}|H|^2+2B)}. \end{eqnarray*} $

This completes the proof of Theorem 1.1.

Corollary 2.2  Let $(M^{n}, \langle, \rangle)$ be an $n(\geq 2)$-dimensional complete Riemannian manifold isometrically immersed in the Euclidean space $\mathbb{R}^{N}$ by $\Psi$, $\Omega$ is a bounded domain in $M^{n}$. For the lower order eigenvalues of problem (1.1), where $p>0$, it holds that

$ \begin{eqnarray*} \sum\limits_{\alpha=1}^{n}(\lambda_{\alpha+1}-\lambda_{1})^{\frac{1}{2}}\leq \sqrt{(n^{2}\sup\limits_{\Omega}|H|^2+(n+2)C-2p)(n^{2}\sup\limits_{\Omega}|H|^2+2C-2p)} \end{eqnarray*} $

(2.20)

where $C=\sqrt{p^2+4(\lambda_{1}-q)}$ and $H$ is the mean curvature vector of its immersion $\Psi$.

Corollary 2.3  Let $\Omega$ be a bounded domain in an $n(\geq 2)$-dimensional unit sphere $\mathbb{S}^{n}(1)$. For the lower order eigenvalues of problem (1.1), where $p>0$, It holds that

$ \begin{eqnarray*}\label{3} \sum\limits_{\alpha=1}^{n}(\lambda_{\alpha+1}-\lambda_{1})^{\frac{1}{2}}\leq \sqrt{\left(n^{2}+(n+2)C-2p\right)\left(n^{2}+2C-2p\right)}, \end{eqnarray*} $

(2.21)

where $C=\sqrt{p^2+4(\lambda_{1}-q)}$.

Remark  For the unit sphere $\mathbb{S}^{n}(1)$, by taking $\Omega=\mathbb{S}^{n}(1)$, $\partial \Omega=\emptyset$, we know that eigenvalues of problem (1.1) satisfy $\lambda_{1}=q$ and $\lambda_{2}=\cdots=\lambda_{n+1}=n^2+pn+q$. Then inequality in (2.21) become equality. So inequality (2.21) is sharp.