In recent years, there is increasing interest in the research of elliptic operators in divergence form (cf. [1-3]). Let $ \Omega $ be a bounded domain in an $ n $-dimensional complete Riemannian manifold $ M $. Let $ A:\Omega\rightarrow \mathrm{End}\left(T\Omega\right) $ be a smooth symmetric and positive definite section of the bundle of all endomorphisms of the tangent bundle $ T\Omega $. Do Carmo, Wang and Xia [4], Sun and Chen [5] researched the Dirichlet eigenvalue problem of elliptic operator in divergence form $ \mathrm{div} \left(A \nabla \right) $ and gave some universal inequalities for its eigenvalues, where $ \nabla $ and $ \rm{div} $ were the gradient operator and the divergence operator on $ M $. The operator $ \mathrm{div} \left(A \nabla \right) $ is an interesting operator. It includes the Laplacian $ \Delta $ as special case. Moreover, it has connections with second order elliptic operators with variable coefficients on $ \Omega \subset \mathbb{R}^n $, the linearized operator $ L_r $ of the $ r $-th mean curvature of a hypersurface and so on.

The purpose of this paper is to investigate the following problem

$ \begin{equation} \left\{\begin{aligned} &\Delta ^2 u - p \mathrm{div} \left(A \nabla u\right) +qu = \lambda \rho u, & \quad \mathrm{in }\quad\Omega , \\ &u |_{\partial \Omega } = \frac{\partial u}{\partial \nu } |_{\partial \Omega } = 0, & \end{aligned}\right. \end{equation} $

(1.1)

where $ \rho $ is a positive continuous function on $ \Omega $, $ \nu $ denotes the outward unit normal to the boundary $ \partial\Omega $ and the constants $ p, q \geq 0 $. It is well known that problem (1.1) has real and discrete spectrum

$ 0 < \lambda_1 \le \lambda_2 \le \lambda_3 \le \cdots \rightarrow +\infty, $

where each eigenvalue is repeated according to its multiplicity. This problem has some connections with some other problems. When $ A $ is the identity, problem (1.1) becomes the following weighted Dirichlet problem of quadratic polynomial operator of the Laplacain.

$ \begin{equation} \left\{\begin{aligned} &\Delta ^2 u - p \Delta u +qu = \lambda \rho u, & \quad \mathrm{in }\quad\Omega , \\ &u |_{\partial \Omega } = \frac{\partial u}{\partial \nu } |_{\partial \Omega } = 0. & \end{aligned}\right. \end{equation} $

(1.2)

Furthermore, when $ p=q=0 $ and $ \rho \equiv 1 $, problem (1.2) becomes the clamped plate problem

$ \begin{equation} \left\{\begin{aligned} &\Delta ^2 u = \lambda u, & \quad \mathrm{in }\quad\Omega , \\ &u |_{\partial \Omega } = \frac{\partial u}{\partial \nu } |_{\partial \Omega } = 0. & \end{aligned}\right. \end{equation} $

(1.3)

Cheng and Yang [6], Cheng, Ichikawa and Mametsuka [7], Wang and Xia [8] established some universal inequalities for problem (1.3). Sun and Chen [9], Sun and Qi [10] obtained some universal inequalities for problem (1.2). Shi [11] gave some inequalities for lower order eigenvalues of problem (1.1) in the case of $ p=1 $ and $ \rho \equiv 1 $.

One of the main goals of this paper is to establish some inequalities for eigenvalues of problem (1.1) on bounded domains of some Riemannian manifolds. Nash's theorem [14] states that any complete Riemannian manifold $ M $ can be isometrically immersed into an Euclidean space. We first give the following result:

Theorem 1.1 Let $ \Omega $ be a connected bounded domain in an $ n $-dimensional complete Riemannian manifold $ M $. Let $ A:\Omega\rightarrow \text{End}(T\Omega) $ be a smooth symmetric and positive definite section of the bundle of all endomorphisms of the tangent bundle $ T\Omega $. Assume that the eigenvalues of $ A $ are bounded below by $ \xi_1 $ and that $ tr\left(A\right)\le n\xi_2 $ throughout $ \Omega $, where $ \xi_1 $ and $ \xi_2 $ are two positive constants. Denote by $ \lambda_i $ the $ i $-th eigenvalue of problem (1.1). Set $ \rho_1=\underset{x\in \Omega}{\min}\rho(x) $ and $ \rho_2=\underset{x\in\Omega}{\max}\rho(x) $. If $ M $ is isometrically immersed in $ \mathbb{R}^N $ with mean curvature vector $ {\bf{H}} $, then we have

$ \begin{equation} \begin{aligned} \sum\limits_{i=1}^k\left(\lambda_{k+1}-\lambda_i\right)^2 \le &\frac{\rho_2}{n\rho_1^{\frac{1}{2}}} \left\{ \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \left[ \left(2n+4\right)B_i +\frac{n^2H_0^2 +pn\xi_2}{\rho_1} \right] \right\}^\frac{1}{2}\\ &\times \left[ \sum\limits_{i=1}^k (\lambda_{k+1}-\lambda_i) \left(4 B_i +\frac{n^2H_0^2}{\rho_1} \right)\right]^\frac{1}{2}, \end{aligned} \end{equation} $

(1.4)

where $ H_0= \underset{x\in\Omega}{\max} \left|{\bf{H}}\right| \left(x\right) $ and $ B_i= \frac{1 }{2\rho_1} \left[ - p\xi_1 + \sqrt {p^2\xi_1^2 + 4\rho_1 (\lambda _i - \frac{ q}{\rho_2} )} \right]. $

As we know, $ H_0 = 0 $ when $ M $ is an $ n $-dimensional complete minimal submanifold in an Euclidean space and $ H_0 = 1 $ when $ M $ is an $ n $-dimensional unit sphere. Therefore, from Theorem 1.1, we can get some results for problem (1.1) on these two kinds of manifolds.

Corollary 1.1 Let $ \Omega $ be a connected bounded domain in an $ n $-dimensional complete minimal submanifold in a Euclidean space. Denote by $ \lambda_i $ the $ i $-th eigenvalue of problem (1.1). Then we have

$ \begin{equation} \begin{aligned} \sum\limits_{i=1}^k\left(\lambda_{k+1}-\lambda_i\right)^2 \le \frac{2\rho_2}{n\rho_1^{\frac{1}{2}}} \left\{ \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \left[ \left(2n+4\right)B_i +\frac{pn\xi_2}{\rho_1} \right] \right\}^\frac{1}{2} \left[ \sum\limits_{i=1}^k (\lambda_{k+1}-\lambda_i) B_i \right]^\frac{1}{2}. \end{aligned} \end{equation} $

(1.5)

Corollary 1.2 Let $ \Omega $ be a connected bounded domain in an $ n $-dimensional unit sphere $ \mathbb{S}^n(1) $. Denote by $ \lambda_i $ the $ i $-th eigenvalue of problem (1.1). Then we have

$ \begin{equation} \begin{aligned} \sum\limits_{i=1}^k\left(\lambda_{k+1}-\lambda_i\right)^2 \le &\frac{\rho_2}{n\rho_1^{\frac{1}{2}}} \left\{ \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \left[ \left(2n+4\right)B_i +\frac{n^2 +pn\xi_2}{\rho_1} \right] \right\}^\frac{1}{2}\\ &\times \left[ \sum\limits_{i=1}^k (\lambda_{k+1}-\lambda_i) \left(4 B_i +\frac{n^2}{\rho_1} \right)\right]^\frac{1}{2}. \end{aligned} \end{equation} $

(1.6)

Moreover, we consider problem (1.1) on Riemannian manifolds admitting some special functions. We obtain the following result:

Theorem 1.2 Let $ \Omega $ be a connected bounded domain in an $ n $-dimensional complete Riemannian manifold $ M $. Let $ A:\Omega\rightarrow \text{End}(T\Omega) $ be a smooth symmetric and positive definite section of the bundle of all endomorphisms of the tangent bundle $ T\Omega $. Assume that $ \xi_1 I \le A \le \xi_2 I $ in the sense that the eigenvalues of $ A $ lie in the interval $ [\xi_1,\xi_2] $ throughout $ \Omega $, where $ \xi_1 $ and $ \xi_2 $ are two positive constants. Denote by $ \lambda_i $ the $ i $-th eigenvalue of problem (1.1). Set $ \rho_1=\underset{x\in \Omega}{\min}\rho(x) $, $ \rho_2=\underset{x\in\Omega}{\max}\rho(x) $ and $ B_i= \frac{1 }{2\rho_1} \left[ - p\xi_1 + \sqrt {p^2\xi_1^2 + 4\rho_1 (\lambda _i - \frac{ q}{\rho_2} )} \right]. $

i) If there exists a function $ \psi:\Omega\rightarrow \mathbb{R} $ and a constant $ C_0 $ such that $ \left|\nabla\psi\right|=1 $ and $ \left|\Delta\psi\right|\le C_0 \ \text{on}\ \Omega $, then

$ \begin{equation} \begin{aligned} \sum\limits_{i=1}^k\left(\lambda_{k+1}-\lambda_i\right)^2\le &\frac{\rho_2}{\rho_1^{\frac{1}{2}}} \left[ \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \left( 6B_i +4C_0\sqrt{\frac{B_i}{\rho_1}} +\frac{ C_0^2 +p\xi_2}{\rho_1} \right)\right]^\frac{1}{2}\\ &\times \left[ \sum\limits_{i=1}^k(\lambda_{k+1}-\lambda_i) \left( 4B_i +4C_0\sqrt{\frac{B_i}{\rho_1}} +\frac{C_0^2}{\rho_1} \right) \right]^\frac{1}{2}; \end{aligned} \end{equation} $

(1.7)

ii) If there exists a function $ \varphi:\Omega\rightarrow \mathbb{R} $ and a constant $ D_0 $ such that $ \left|\nabla\varphi\right|=1 $ and $ \Delta\varphi= D_0 $ on $ \Omega $, then

$ \begin{equation} \begin{aligned} \sum\limits_{i=1}^k\left(\lambda_{k+1}-\lambda_i\right)^2 \le &\frac{\rho_2}{\rho_1^{\frac{1}{2}}} \left[ \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \left( 6B_i -\frac{2D_0^2}{\rho_2} +\frac{ D_0^2 +p\xi_2}{\rho_1} \right) \right]^\frac{1}{2}\\ &\times \left[ \sum\limits_{i=1}^k (\lambda_{k+1}-\lambda_i) \left( 4B_i -\frac{2 D_0^2}{\rho_2} +\frac{D_0^2}{\rho_1} \right) \right]^\frac{1}{2}; \end{aligned} \end{equation} $

(1.8)

iii) If there exist $ l $ functions $ \phi_\alpha:\Omega\rightarrow \mathbb{R} $ such that $ \left<\nabla\phi_\alpha,\nabla\phi_\beta\right>=\delta_{\alpha\beta} $ and $ \Delta\phi_\alpha=0,\,\,\alpha,\beta=1,\cdots,l $ on $ \Omega $, then

$ \begin{equation} \begin{aligned} \sum\limits_{i=1}^k\left(\lambda_{k+1}-\lambda_i\right)^2 \le& \frac{2\rho_2}{l\rho_1^{\frac{1}{2}}} \left\{ \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \left[ \left(2l+4\right)B_i +\frac{pl\xi_2}{\rho_1} \right] \right\}^\frac{1}{2} \left[\sum\limits_{i=1}^k (\lambda_{k+1}-\lambda_i) B_i \right]^\frac{1}{2}; \end{aligned} \end{equation} $

(1.9)

iv) If $ \Omega $ admits an eigenmap $ f=\left(f_1,f_2,\cdots,f_{N+1}\right):\Omega\rightarrow \mathbb{S}^N(1) $ corresponding to an eigenvalue $ \mu $, that is $ \sum_{\alpha=1}^{N+1}f_\alpha^2=1 $ and $ \Delta f_\alpha=-\mu f_\alpha $, $ \alpha=1,2,\cdots,N+1 $, then

$ \begin{equation} \begin{aligned} \sum\limits_{i=1}^k\left(\lambda_{k+1}-\lambda_i\right)^2 \le & \frac{\rho_2}{\mu\rho_1^{\frac{1}{2}}} \left[\sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \left( 6\mu B_i +\frac{\mu^2 +p\mu\xi_2}{\rho_1} \right) \right]^\frac{1}{2}\\ &\times \left[ \sum\limits_{i=1}^k (\lambda_{k+1}-\lambda_i) \left( 4\mu B_i +\frac{\mu^2}{\rho_1}\right) \right]^\frac{1}{2}. \end{aligned} \end{equation} $

(1.10)

As applications of the above results, we can give some results for problem (1.2). For example, when $ A $ is the identity, we can obtain the following corollary for problem (1.2) from Theorem 1.1.

Corollary 1.3 Let $ M $ be an $ n $-dimensional complete Riemannian manifold and $ \Omega $ be a connected bounded domain in $ M $. Denote by $ \lambda_i $ the $ i $-th eigenvalue of problem (1.2). Set $ \rho_1=\underset{x\in \Omega}{\min}\rho(x) $ and $ \rho_2=\underset{x\in\Omega}{\max}\rho(x) $. $ E_i= \frac{1 }{2\rho_1} \left[ - p + \sqrt {p^2 + 4\rho_1 (\lambda _i - \frac{ q}{\rho_2} )} \right]. $ If $ M $ is isometrically immersed in $ \mathbb{R}^N $ with mean curvature vector $ {\bf{H}} $, then

$ \begin{equation} \begin{aligned} \sum\limits_{i=1}^k\left(\lambda_{k+1}-\lambda_i\right)^2 \le &\frac{\rho_2}{n\rho_1^{\frac{1}{2}}} \left\{ \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \left[ \left(2n+4\right)E_i +\frac{n^2H_0^2 +pn}{\rho_1} \right] \right\}^\frac{1}{2}\\ &\times \left[ \sum\limits_{i=1}^k (\lambda_{k+1}-\lambda_i) \left(4 E_i +\frac{n^2H_0^2}{\rho_1} \right)\right]^\frac{1}{2}, \end{aligned} \end{equation} $

(1.11)

where $ H_0=\underset{x\in\Omega}{\max} \left|{\bf{H}}\right| \left(x\right) $.

Remark 1.1 Let $ \{a_i\}_{i=1}^m $, $ \{b_i\}_{i=1}^m $ and $ \{c_i\}_{i=1}^m $ be three sequences of non-negative real numbers with $ \{a_i\}_{i=1}^m $ decreasing and $ \{b_i\}_{i=1}^m $ and $ \{c_i\}_{i=1}^m $ increasing. Then it holds [8]

$ \begin{equation} \left( a_i^2 b_i \right)\left( a_i c_i \right) \leq \left( a_i^2\right)\left( a_i b_i c_i \right). \end{equation} $

(1.12)

Using (1.12), we know that (1.11) becomes (1.11) of Theorem 1.1 in [10].

Moreover, when $ A $ is the identity, we can derive the following corollary for problem (1.2) from Theorem 1.2.

Corollary 1.4 Let $ M $ be an $ n $-dimensional complete Riemannian manifold and $ \Omega $ be a connected bounded domain in $ M $. Denote by $ \lambda_i $ the $ i $-th eigenvalue of problem (1.2). Set $ \rho_1=\underset{x\in \Omega}{\min}\rho(x) $, $ \rho_2=\underset{x\in\Omega}{\max}\rho(x) $ and $ E_i= \frac{1 }{2\rho_1} \left[ - p + \sqrt {p^2 + 4\rho_1 (\lambda _i - \frac{ q}{\rho_2} )} \right]. $

i) If there exists a function $ \psi:\Omega\rightarrow \mathbb{R} $ and a constant $ C_0 $ such that $ \left|\nabla\psi\right|=1 $ and $ \left|\Delta\psi\right|\le C_0 $ on $ \Omega $, then

$ \begin{equation} \begin{aligned} \sum\limits_{i=1}^k\left(\lambda_{k+1}-\lambda_i\right)^2\le & \frac{\rho_2}{\rho_1^{\frac{1}{2}}} \left[ \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \left( 6E_i +4C_0\sqrt{\frac{E_i}{\rho_1}} +\frac{ C_0^2 +p}{\rho_1} \right)\right]^\frac{1}{2}\\ &\times \left[ \sum\limits_{i=1}^k (\lambda_{k+1}-\lambda_i) \left( 4E_i + 4C_0\sqrt{\frac{E_i}{\rho_1}} +\frac{C_0^2}{\rho_1} \right) \right]^\frac{1}{2}; \end{aligned} \end{equation} $

(1.13)

ii) If there exists a function $ \varphi:\Omega\rightarrow \mathbb{R} $ and a constant $ D_0 $ such that $ \left|\nabla\varphi\right|=1 $ and $ \Delta\varphi= D_0 $ on $ \Omega $, then

$ \begin{equation} \begin{aligned} \sum\limits_{i=1}^k\left(\lambda_{k+1}-\lambda_i\right)^2 \le & \frac{\rho_2}{\rho_1^{\frac{1}{2}}} \left[ \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \left( 6E_i -\frac{2D_0^2}{\rho_2} +\frac{ D_0^2 +p}{\rho_1} \right) \right]^\frac{1}{2}\\ &\times \left[ \sum\limits_{i=1}^k (\lambda_{k+1}-\lambda_i) \left( 4 E_i -2\frac{D_0^2}{\rho_2} +\frac{D_0^2}{\rho_1} \right) \right]^\frac{1}{2}; \end{aligned} \end{equation} $

(1.14)

iii) If there exist $ l $ functions $ \phi_\alpha:\Omega\rightarrow \mathbb{R} $ such that $ \left<\nabla\phi_\alpha,\nabla\phi_\beta\right>=\delta_{\alpha\beta} $ and $ \Delta\phi_\alpha=0 $, $ \alpha,\beta=1,\cdots,l $ on $ \Omega $, then

$ \begin{equation} \begin{aligned} \sum\limits_{i=1}^k\left(\lambda_{k+1}-\lambda_i\right)^2 \le& \frac{2\rho_2}{l\rho_1^{\frac{1}{2}}} \left\{ \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \left[ \left(2l+4\right)\rho_1E_i +pl \right] \right\}^\frac{1}{2}\\ &\times \left[ \sum\limits_{i=1}^k (\lambda_{k+1}-\lambda_i) E_i \right]^\frac{1}{2}; \end{aligned} \end{equation} $

(1.15)

iv) If $ \Omega $ admits an eigenmap $ f=\left(f_1,f_2,\cdots,f_{N+1}\right):\Omega\rightarrow \mathbb{S}^N(1) $ corresponding to an eigenvalue $ \mu $, that is $ \sum_{\alpha=1}^{N+1}f_\alpha^2=1 $ and $ \Delta f_\alpha=-\mu f_\alpha $, $ \alpha=1,2,\cdots,N+1 $, then

$ \begin{equation} \begin{aligned} \sum\limits_{i=1}^k\left(\lambda_{k+1}-\lambda_i\right)^2 \le & \frac{\rho_2}{\mu\rho_1^{\frac{1}{2}}} \left[ \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \left( 6\mu E_i +\frac{\mu^2 +p\mu}{\rho_1} \right) \right]^\frac{1}{2}\\ &\times \left[ \sum\limits_{i=1}^k (\lambda_{k+1}-\lambda_i) \left( 4 \mu E_i +\frac{\mu^2}{\rho_1}\right) \right]^\frac{1}{2}. \end{aligned} \end{equation} $

(1.16)

Remark 1.2 It is easy to find that when $ p=q=0 $ and $ \rho \equiv 1 $, (1.13-1.16) for problem (1.2) become the results of Theorem 1.1 for problem (1.3) in [8].

In this section, we first prove a general inequality which plays the key role in the proofs of Theorems 1.1 and 1.2.

Lemma 2.1 Let $ \Omega $ be a bounded domain in an $ n $-dimensional compact Riemannian manifold $ M $. Assume that $ A:\Omega\rightarrow \text{End}(T\Omega) $ is a smooth symmetric and positive definite section of the bundle of all endomorphisms of the tangent bundle $ T\Omega $. Denote by $ \lambda_i $ the $ i $-th eigenvalue of problem (1.1) and $ u_i $ the $ i $-th weighted orthonormal eigenfunctions. For any function $ h\in C^4\left(\bar{\Omega}\right) $ and any integer $ k $, we have

$ \begin{equation} \begin{aligned} &\sum\limits_{i=1}^k\left(\lambda_{k+1}-\lambda_i\right)^2 \int_{\Omega}{u_i^2 \left|\nabla h\right|^2}\\ \le& \frac{1}{\delta} \sum\limits_{i=1}^k ( \lambda_{k+1}-\lambda_i) \int_{\Omega}\frac{1}{\rho} \left[ \left<\nabla h,\nabla u_i\right>^2 +u_i \Delta h \left<\nabla h,\nabla u_i\right> +\frac{u_i^2 \left(\Delta h\right)^2}{4}\right] \\ &+\delta \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \bigg(-2 \int_{\Omega} u_i \Delta u_i \left|\nabla h\right|^2 +4\int_{\Omega} u_i \Delta h \left<\nabla h,\nabla u_i\right> \\ & +4\int_{\Omega} \left<\nabla h,\nabla u_i \right>^2 +\int_{\Omega} u_i^2 \left(\Delta h\right)^2 +p\int_{\Omega} u_i^2 \left<\nabla h,A \nabla h\right>\bigg), \end{aligned} \end{equation} $

(2.1)

where $ \delta $ is any positive constant.

Proof For each $ i=1,\cdots,k $, we define $ \varphi_i:\Omega\rightarrow \mathbb{R} $ by $ \varphi_i = h u_i-\sum_{j=1}^k a_{ij} u_j, $ where $ a_{ij} = \int_{\Omega}{\rho h u_i u_j}=a_{ji}. $ Since $ \int_{\Omega}{\rho \varphi_i u_j}=0,\,\forall i,j=1,\cdots,k, $ we know from Rayleigh-Ritz inequality that

$ \begin{equation} \begin{aligned} \lambda_{k+1}\int_{\Omega}{\rho\varphi_i^2} &\le \int_{\Omega}{\varphi_i \left[\Delta^2 \varphi_i-p \mathrm{div}\left( A \nabla \varphi_i \right) +q\varphi_i\right]}\\ &=\int_{\Omega}{\varphi_i \left[\Delta^2 \left(h u_i\right) -p \mathrm{div}\left(A \nabla\left(h u_i\right)\right) +q h u_i \right]}. \end{aligned} \end{equation} $

(2.2)

By direct computation, we have

$ \begin{equation} \begin{aligned} -\mathrm{div} \left(A \nabla \left(h u_i \right)\right) =-h \mathrm{div}\left(A \nabla u_i\right) -2\left<\nabla h,A \nabla u_i\right> -u_i \mathrm{div}\left(A \nabla h\right) \end{aligned} \end{equation} $

(2.3)

and

$ \begin{equation} \begin{aligned} \Delta^2 \left(h u_i\right) =&\Delta \left( h \Delta u_i +2\left<\nabla h,\nabla u_i\right> +u_i \Delta h \right)\\ =&h \Delta^2 u_i +2\left<\nabla h,\nabla \Delta u_i \right> +2\Delta \left<\nabla h,\nabla u_i\right>+u_i \Delta^2 h\\ & +2\left<\nabla u_i,\nabla \Delta h \right> +2\Delta u_i\Delta h. \end{aligned} \end{equation} $

(2.4)

Substituting (2.3) and (2.4) into (2.2), we get

$ \begin{equation} (\lambda_{k+1} - \lambda_i )\int_{\Omega}{\rho\varphi_i^2} \le \int_\Omega h p_i u_i-\sum\limits_{j=1}^k a_{ij} \int_\Omega p_i u_j, \end{equation} $

(2.5)

where $ p_i =\Delta h \Delta u_i +2\left<\nabla h,\nabla \Delta u_i\right> +2\Delta \left<\nabla h,\nabla u_i\right> +\Delta \left( u_i \Delta h\right)-p u_i \mathrm{div}\left(A \nabla h\right) -2p \left<\nabla h,A \nabla u_i\right>. $

Using the divergence theorem, we deduce

$ \begin{equation} \begin{aligned} \int_{\Omega}{\Delta u_j \left<\nabla h,\nabla u_i\right>} &= -\int_{\Omega} h \mathrm{div} \left(\Delta u_j \nabla u_i\right) = -\int_{\Omega} h \left<\nabla \Delta u_j,\nabla u_i\right> -\int_{\Omega} h \Delta u_i \Delta u_j\\ &= \int_{\Omega} u_i \mathrm{div} \left( h \nabla \Delta u_j \right) -\int_{\Omega} h \Delta u_i \Delta u_j\\ \end{aligned} \end{equation} $

(2.6)

and

$ \begin{equation} \begin{aligned} \int_{\Omega} u_i \mathrm{div} \left( h \nabla \Delta u_j \right) &= \int_{\Omega} h u_i \Delta^2 u_j +\int_{\Omega} u_i \left<\nabla h,\nabla \Delta u_j\right> \\ &= \int_{\Omega} h u_i \Delta^2 u_j -\int_{\Omega} \Delta u_j \mathrm{div} \left(u_i \nabla h\right) \\ &= \int_{\Omega} h u_i \Delta^2 u_j -\int_{\Omega} \Delta u_j \left<\nabla h,\nabla u_i\right> -\int_{\Omega} u_i \Delta u_j \Delta h. \end{aligned} \end{equation} $

(2.7)

Using (2.6) and (2.7), we derive

$ \begin{equation} \begin{aligned} &\int_{\Omega}{\Delta u_j \left<\nabla h,\nabla u_i\right>} -\int_{\Omega}{\Delta u_i \left<\nabla h,\nabla u_j\right>}\\ =& \int_{\Omega} u_i \mathrm{div} \left( h \nabla \Delta u_j \right) -\int_{\Omega} u_j \mathrm{div} \left(h \nabla \Delta u_i\right)\\ =& \int_{\Omega} h u_i \Delta^2 u_j -\int_{\Omega} \Delta u_j \left<\nabla h,\nabla u_i\right> -\int_{\Omega} u_i \Delta u_j \Delta h-\int_{\Omega} h u_j \Delta^2 u_i\\ & +\int_{\Omega} \Delta u_i \left<\nabla h,\nabla u_j\right> +\int_{\Omega} u_j \Delta u_i\Delta h. \end{aligned} \end{equation} $

(2.8)

It yields

$ \begin{equation} \begin{aligned} &2\int_{\Omega}{\Delta u_j \left<\nabla h,\nabla u_i\right>} -2\int_{\Omega}{\Delta u_i \left<\nabla h,\nabla u_j\right>}\\ =& \int_{\Omega} h u_i \Delta^2 u_j -\int_{\Omega} u_i \Delta u_j \Delta h -\int_{\Omega} h u_j \Delta^2 u_i +\int_{\Omega} u_j \Delta u_i \Delta h. \end{aligned} \end{equation} $

(2.9)

Moreover, we have

$ \begin{equation} \begin{aligned} \int_{\Omega}u_iu_j\mathrm{div}\left(A\nabla h\right) +\int_{\Omega}2u_j\left<\nabla h,A\nabla u_i\right> &=\int_{\Omega} h \mathrm{div} \left(A \nabla\left(u_i u_j\right)\right) -\int_{\Omega}2 h \mathrm{div} \left(u_j A \nabla u_i\right) \\ &=\int_{\Omega} h u_i \mathrm{div} \left(A \nabla u_j \right) -\int_{\Omega} h u_j \mathrm{div} \left(A \nabla u_i \right) \end{aligned} \end{equation} $

(2.10)

and

$ \begin{equation} \begin{aligned} \int_{\Omega} u_j \left<\nabla h,\nabla \Delta u_i \right> = -\int_{\Omega}\Delta u_i \mathrm{div} \left(u_j \nabla h \right) = -\int_{\Omega} u_j \Delta u_i \Delta h -\int_{\Omega} \Delta u_i \left<\nabla h,\nabla u_j\right>. \end{aligned} \end{equation} $

(2.11)

Using (2.9)-(2.11), we obtain

$ \begin{equation} \begin{aligned} \int_{\Omega}{ p_i u_j} =&\int_{\Omega} \left[ u_j \Delta u_i \Delta h +2 u_j \left<\nabla h,\nabla \Delta u_i \right> +2u_j \Delta \left<\nabla h,\nabla u_i\right> +u_j \Delta \left(u_i \Delta h\right)\right.\\ &\left. -p u_i u_j \mathrm{div} \left(A \nabla h\right) -2p u_j \left<\nabla h,A \nabla u_i\right>\right]\\ =&\int_{\Omega} \left[ -u_j \Delta u_i \Delta h +h u_i \Delta^2 u_j -u_i \Delta u_j \Delta h -h u_j \Delta^2 u_i +u_j \Delta u_i\Delta h\right.\\ &\left. +u_i \Delta u_j \Delta h -ph u_i \mathrm{div} \left(A \nabla u_j \right) +ph u_j \mathrm{div} \left(A \nabla u_i \right) \right]\\ =&\int_{\Omega} h u_i \left[\Delta^2 u_j -p\mathrm{div} \left(A \nabla u_j \right) +q u_j \right] -\int_{\Omega} h u_j \left[ \Delta^2 u_i -p\mathrm{div} \left(A \nabla u_i \right) +q u_i \right]\\ =&\left(\lambda_j -\lambda_i \right) a_{ij}. \end{aligned} \end{equation} $

(2.12)

Substituting (2.12) into (2.5), we get

$ \begin{equation} \begin{aligned} &\left(\lambda_{k+1}-\lambda_i\right) \int_{\Omega}{\rho \varphi_i^2} \le \int_{\Omega}{h u_i p_i}+\sum\limits_{j=1}^k \left( \lambda_i-\lambda_j\right) a_{ij}^2. \end{aligned} \end{equation} $

(2.13)

Set $ b_{ij}= \int_{\Omega} u_j \left( \left<\nabla h,\nabla u_i\right> +\frac{u_i \Delta h}{2} \right). $ Then one gets from the divergence theorem that

$ \begin{equation} \begin{aligned} b_{ij} +b_{ji} =& \int_{\Omega} \left( u_j \left<\nabla h,\nabla u_i\right> +u_i \left<\nabla h,\nabla u_j\right> +u_i u_j \Delta h\right)\\ =&\int_{\Omega} \left<\nabla h,\nabla \left(u_i u_j\right) \right> + \int_{\Omega} u_i u_j \Delta h\\ =&0. \end{aligned} \end{equation} $

(2.14)

Moreover, since

$ -\int_{\Omega}{ h u_i^2 \Delta h }= \int_{\Omega} \left<\nabla \left(h u_i^2\right),\nabla h\right> = 2\int_{\Omega} h u_i \left<\nabla h,\nabla u_i\right> + \int_{\Omega} u_i^2 \left|\nabla h\right|^2, $

it holds

$ \begin{equation} \begin{aligned} -2\int_{\Omega}{ \varphi_i \left( \left<\nabla h,\nabla u_i\right> +\frac{u_i \Delta h}{2} \right)} =&\int_{\Omega} u_i^2 \left|\nabla h\right|^2 +2\sum\limits_{j=1}^k a_{ij} b_{ij}. \end{aligned} \end{equation} $

(2.15)

Multiplying both sides of (2.15) by $ \left(\lambda_{k+1}-\lambda_i\right)^2 $, using the Schwarz inequality, we get

$ \begin{equation} \begin{aligned} &\left(\lambda_{k+1}-\lambda_i\right)^2 \left( \int_{\Omega}{u_i^2 \left|\nabla h\right|^2} +2\sum\limits_{j=1}^k a_{ij}b_{ij}\right)\\ \le&\frac{\lambda_{k+1}-\lambda_i}{\delta}\int_{\Omega}\left[\frac{1}{\sqrt\rho}\left(\left<\nabla h,\nabla u_i\right>+\frac{u_i\Delta h}{2}\right)-\sum\limits_{j=1}^kb_{ij}\sqrt\rho u_j\right]^2 +\delta \left(\lambda_{k+1}-\lambda_i\right)^3 \int_{\Omega} \rho \varphi_i^2. \end{aligned} \end{equation} $

(2.16)

Substituting (2.13) into (2.16), we obtain

$ \begin{equation} \begin{aligned} &\left(\lambda_{k+1}-\lambda_i\right)^2 \left( \int_{\Omega}{u_i^2 \left|\nabla h\right|^2} +2\sum\limits_{j=1}^k a_{ij}b_{ij}\right)\\ \le&\frac{\lambda_{k+1}-\lambda_i}{\delta}\left[\int_{\Omega}\frac{1}{\rho}\left(\left<\nabla h,\nabla u_i\right>+\frac{u_i\Delta h}{2}\right)^2-\sum\limits_{j=1}^kb_{ij}^2\right]\\ &+ \delta\left(\lambda_{k+1}-\lambda_i\right)^2\left[\int_{\Omega}{hu_ip_i}+\sum\limits_{j=i}^k\left( \lambda_i-\lambda_j\right)a_{ij}^2\right]. \end{aligned} \end{equation} $

(2.17)

Taking the sum on $ i $ from $ 1 $ to $ k $ in (2.17) and noticing

$ a_{ij}=a_{ji}, \quad b_{ij}=-b_{ji}, $

$ \sum\limits_{i,j=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 a_{ij} b_{ij} =-\sum\limits_{i,j=1}^k \left(\lambda_{k+1}-\lambda_i\right) \left(\lambda_i-\lambda_j\right) a_{ij} b_{ij} $

and

$ \sum\limits_{i,j=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \left(\lambda_i-\lambda_j\right) a_{ij}^2 =-\sum\limits_{i,j=1}^k \left(\lambda_{k+1}-\lambda_i\right) \left(\lambda_i-\lambda_j\right)^2 a_{ij}^2, $

we induce

$ \begin{equation} \begin{aligned} \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \int_{\Omega}{u_i^2 \left|\nabla h\right|^2} \le& \frac{1}{\delta}\sum\limits_{i=1}^k (\lambda_{k+1}-\lambda_i) \int_{\Omega} \frac{1}{\rho} \left( \left<\nabla h,\nabla u_i\right> +\frac{u_i \Delta h}{2} \right)^2\\ &+\delta\sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \int_{\Omega} h u_i p_i. \end{aligned} \end{equation} $

(2.18)

Since

$ \begin{equation} \begin{aligned} \int_{\Omega} h u_i \left<\nabla h,\nabla\Delta u_i\right> =& - \int_{\Omega} \Delta u_i \mathrm{div} \left( h u_i \nabla h \right)\\ =& - \int_{\Omega} h u_i \Delta u_i \Delta h -\int_{\Omega} h \Delta u_i\left<\nabla u_i,\nabla h\right> - \int_{\Omega} u_i \Delta u_i \left|\nabla h\right|^2, \end{aligned} \end{equation} $

(2.19)

$ \begin{equation} \begin{aligned} \int_{\Omega} h u_i \Delta \left<\nabla h,\nabla u_i\right> = \int_{\Omega} h \Delta u_i \left<\nabla h,\nabla u_i\right> + \int_{\Omega} u_i \Delta h \left<\nabla h,\nabla u_i\right> +2 \int_{\Omega} \left<\nabla h,\nabla u_i\right>^2, \end{aligned} \end{equation} $

(2.20)

$ \begin{equation} \int_{\Omega} h u_i \Delta \left(u_i \Delta h\right) =\int_{\Omega} u_i^2 \left(\Delta h\right)^2 +\int_{\Omega} h u_i \Delta u_i \Delta h +2\int_{\Omega} u_i \Delta h \left<\nabla h,\nabla u_i\right> \end{equation} $

(2.21)

and

$ \begin{equation} \begin{aligned} \int_{\Omega} 2 h u_i \left<\nabla h,A \nabla u_i\right> =&-\frac{1}{2} \int_{\Omega} u_i^2 \mathrm{div }\left(A \nabla h^2\right)\\ =&-\int_{\Omega} h u_i^2 \mathrm{div} \left(A \nabla h\right) -\int_{\Omega} u_i^2 \left<\nabla h,A \nabla h\right>, \end{aligned} \end{equation} $

(2.22)

we have

$ \begin{equation} \begin{aligned} \int_{\Omega}{h p_i u_i} =&-2\int_{\Omega} u_i \Delta u_i \left|\nabla h\right|^2 +4\int_{\Omega} u_i \Delta h \left<\nabla h,\nabla u_i\right> +4\int_{\Omega} \left<\nabla h,\nabla u_i\right>^2\\ &+\int_{\Omega} u_i^2 \left(\Delta h\right)^2 +p\int_{\Omega} u_i^2 \left<\nabla h,A \nabla h\right>. \end{aligned} \end{equation} $

(2.23)

Substituting (2.23) into (2.18), we obtain (2.1). This completes the proof of Lemma 2.1.

Now we give the proof of Theorems 1.1 and 1.2 by using Lemma 2.1.

Proof of Theorem 1.1 Let $ x_1,x_2,\dots,x_N $ be the standard coordinate functions of $ \mathbb{R}^N $. Then it holds

$ \begin{equation} \sum\limits_{\alpha=1}^N\left|\nabla x_\alpha\right|^2=n, \end{equation} $

(2.24)

$ \begin{equation} \quad\Delta\left(x_1,x_2,\dots,x_N\right)=n{\bf{H}}, \end{equation} $

(2.25)

$ \begin{equation} \sum\limits_{\alpha=1}^N\left<\nabla u_i,\nabla x_\alpha\right>^2=\left|\nabla u_i\right|^2 \end{equation} $

(2.26)

and

$ \begin{equation} \sum\limits_{\alpha=1}^N \Delta x_\alpha \left<\nabla u_i,\nabla x_\alpha\right>=\sum\limits_{\alpha=1}^N\Delta x_\alpha\nabla u_i\left(x_\alpha\right)=\left<n {\bf{H}},\nabla u_i\right>=0. \end{equation} $

(2.27)

Taking $ h=x_\alpha $ in (2.1), taking the sum on $ \alpha $ from $ 1 $ to $ N $, and using (2.24–2.27), we get

$ \begin{equation} \begin{aligned} \frac{n}{\rho_2} \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \le & \delta \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \left[ \left(2n+4\right) \int_{\Omega} \left| \nabla u_i \right|^2 +\int_{\Omega} u_i^2 \left( n {\bf{H}} \right)^2 +p\int_{\Omega} u_i^2 tr(A)\right]\\ &+\frac{1}{\rho_1\delta}\sum\limits_{i=1}^k (\lambda_{k+1}-\lambda_i) \int_{\Omega}{\left( \left| \nabla u_i \right|^2+\frac{\left( n {\bf{H}} \right)^2}{4} u_i^2 \right)}. \end{aligned} \end{equation} $

(2.28)

From

$ \begin{aligned} \lambda_i =\int_{\Omega} u_i \left[ \Delta^2 u_i -p\mathrm{div} \left(A \nabla u_i\right) +q u_i \right] =\int_{\Omega}{\left(\Delta u_i\right)^2} +p\int_{\Omega}{\left<\nabla u_i,A \nabla u_i\right>} +q\int_{\Omega}{u_i^2} \end{aligned} $

and

$ \begin{aligned} \int_\Omega {\left| \nabla u_i \right|}^2 = - \int_\Omega u_1\Delta u_i \le \left( \int_\Omega u_i^2\int_\Omega \left(\Delta u_i\right)^2 \right)^{\frac{1}{2}}, \end{aligned} $

we have

$ \begin{equation} \begin{aligned} \rho_1\rho_2\left( \int_\Omega {\left| \nabla u_i \right|}^2 \right)^2 + p\xi_1\rho_2\int_\Omega {\left| \nabla u_i \right|^2}+q-\rho_2\lambda_i\le 0. \end{aligned} \end{equation} $

(2.29)

This is a quadratic inequality of $ {\int_\Omega {\left| {{\nabla }{u_i}} \right|} ^2} $. Thus we obtain

$ \begin{equation} \begin{aligned} \int_\Omega {\left| \nabla u_i \right|}^2 \leq B_i. \end{aligned} \end{equation} $

(2.30)

Introducing (2.30) into (2.28), we infer

$ \begin{equation} \begin{aligned} \frac{n}{\rho_2} \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 &\le \delta\sum\limits_{i=1}^k \left[(\lambda_{k+1}-\lambda_i\right)^2 \left[ \left(2n+4\right)B_i +\frac{n^2H_0^2+pn\xi_2}{\rho_1}\right]\\ &\quad +\frac{1}{\rho_1\delta}\sum\limits_{i=1}^k (\lambda_{k+1}-\lambda_i ) \left( B_i +\frac{n^2H_0^2}{4\rho_1} \right). \end{aligned} \end{equation} $

(2.31)

Taking

$ \delta=\left\{\frac{\sum_{i=1}^k(\lambda_{k+1}-\lambda_i)\left(B_i+\frac{n^2H_0^2}{4\rho_1}\right)}{\rho_1 \sum_{i=1}^k\left(\lambda_{k+1}-\lambda_i\right)^2\left[\left(2n+4\right)B_i+\frac{n^2H_0^2+pn\xi_2}{\rho_1}\right]}\right\}^\frac{1}{2} $

in (2.31), we can obtain (1.4). This finishes the proof of Theorem 1.1.

Proof of Theorem 1.2 i) Taking $ h=\psi $ in (2.1), we get

$ \begin{equation} \begin{aligned} & \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \int_{\Omega}{u_i^2 \left|\nabla \psi\right|^2}\\ \le& \frac{1}{\delta}\sum\limits_{i=1}^k (\lambda_{k+1}-\lambda_i) \int_{\Omega}{\frac{1}{\rho} \left[\left<\nabla \psi,\nabla u_i\right>^2 +u_i \Delta \psi \left<\nabla \psi,\nabla u_i\right> +\frac{u_i^2 \left(\Delta \psi \right)^2}{4}\right]}\\ &+\delta \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \left( -2 \int_{\Omega} u_i \Delta u_i \left|\nabla \psi \right|^2 +4\int_{\Omega} u_i \Delta \psi \left<\nabla \psi,\nabla u_i\right>\right.\\ &\left.+4\int_{\Omega} \left<\nabla \psi,\nabla u_i \right>^2 +\int_{\Omega} u_i^2 \left(\Delta \psi \right)^2 +p\int_{\Omega} u_i^2 \left<\nabla \psi,A \nabla \psi\right>\right). \end{aligned} \end{equation} $

(2.32)

By using (2.30), we have

$ \begin{equation} \begin{aligned} \int_{\Omega} u_i \Delta \psi \left<\nabla \psi,\nabla u_i\right> \le C_0 \left( \int_{\Omega} u_i^2 \int_{\Omega} \left|\nabla u_i\right|^2 \right)^\frac{1}{2} \le C_0 \sqrt{\frac{B_i}{\rho_1}}. \end{aligned} \end{equation} $

(2.33)

Substituting (2.33) into (2.32), we deduce

$ \begin{equation} \begin{aligned} \frac{1}{\rho_2} \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 &\le \delta\sum\limits_{i=1}^k\left(\lambda_{k+1}-\lambda_i\right)^2 \left( 6B_i +4C_0\sqrt{\frac{B_i}{\rho_1}} +\frac{C_0^2+p\xi_2}{\rho_1} \right)\\ &\quad+\frac{1}{\rho_1\delta}\sum\limits_{i=1}^k (\lambda_{k+1}-\lambda_i) \left( B_i +C_0\sqrt{\frac{B_i}{\rho_1}} +\frac{C_0^2}{4\rho_1} \right). \end{aligned} \end{equation} $

(2.34)

Taking

$ \delta=\left[ \frac{\sum_{i=1}^k (\lambda_{k+1}-\lambda_i) \left( B_i +C_0 \sqrt{\frac{B_i}{\rho_1}} +\frac{C_0^2}{4\rho_1}\right)} {\rho_1\sum_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \left( 6B_i +4C_0 \sqrt{\frac{B_i}{\rho_1}}+\frac{C_0^2 +p\xi_2}{\rho_1} \right)} \right]^\frac{1}{2} $

in (2.34), we get (1.7).

ii) Taking $ h=\varphi $ in (2.1), we have

$ \begin{equation} \begin{aligned} & \sum\limits_{i=1}^k\left(\lambda_{k+1}-\lambda_i\right)^2\int_{\Omega}{u_i^2 \left|\nabla \varphi\right|^2}\\ \le& \frac{1}{\delta} \sum\limits_{i=1}^k (\lambda_{k+1}-\lambda_i) \int_{\Omega}{\frac{1}{\rho} \left[ \left<\nabla \varphi,\nabla u_i\right>^2 +u_i \Delta \varphi \left<\nabla \varphi,\nabla u_i\right> +\frac{u_i^2 \left(\Delta \varphi \right)^2}{4}\right]}\\ & +\delta\sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \left( -2 \int_{\Omega} u_i \Delta u_i \left|\nabla \varphi \right|^2 +4\int_{\Omega} u_i \Delta \varphi \left<\nabla \varphi,\nabla u_i\right>\right.\\ & \left.+4\int_{\Omega} \left<\nabla \varphi,\nabla u_i \right>^2 +\int_{\Omega} u_i^2 \left(\Delta \varphi \right)^2 +p\int_{\Omega} u_i^2 \left<\nabla \varphi,A \nabla \varphi\right>\right). \end{aligned} \end{equation} $

(2.35)

Since

$ \begin{aligned} \int_{\Omega} u_i \Delta \varphi \left<\nabla \varphi,\nabla u_i\right> =\frac{D_0}{2} \int_{\Omega} \left<\nabla\varphi,\nabla u_i^2\right> =-\frac{D_0}{2} \int_{\Omega} u_i^2 \Delta\varphi \le -\frac{D_0^2}{2\rho_2}, \end{aligned} $

we get

$ \begin{equation} \begin{aligned} \frac{1}{\rho_2} \sum\limits_{i=1}^k\left(\lambda_{k+1}-\lambda_i\right)^2 &\le\delta \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2\left(6B_i-\frac{2D_0^2}{\rho_2}+\frac{D_0^2+p\xi_2}{\rho_1}\right)\\ &\quad+ \frac{1}{\rho_1\delta} \sum\limits_{i=1}^k (\lambda_{k+1}-\lambda_i) \left(B_i-\frac{D_0^2}{2\rho_2}+\frac{D_0^2}{4\rho_1}\right). \end{aligned} \end{equation} $

(2.36)

Taking

$ \delta=\left[ \frac{\sum_{i=1}^k (\lambda_{k+1}-\lambda_i) \left(B_i -\frac{D_0^2}{2\rho_2} +\frac{D_0^2}{4\rho_1}\right)} {\rho_1\sum_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \left(6B_i -\frac{2D_0^2}{\rho_2} +\frac{D_0^2+p\xi_2}{\rho_1}\right)} \right]^\frac{1}{2} $

in (2.36), we obtain (1.8).

iii) Because $ \left<\nabla\phi_\alpha,\nabla\phi_\beta\right>=\delta_{\alpha\beta} $ and $ \Delta\phi_\alpha=0 $, $ \alpha,\beta=1,\cdots,l $, we know

$ \sum\limits_{\alpha=1}^l \left<\nabla u_i,\nabla \phi_\alpha \right>^2 \le \left|\nabla u_i\right|^2. $

Thus, taking $ h=\phi_\alpha $ in (2.1), we obtain by summing over $ \alpha $ that

$ \begin{equation} \begin{aligned} & \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \int_{\Omega}{u_i^2 \sum\limits_{\alpha=1}^l \left|\nabla \phi_\alpha\right|^2}\\ \le& \delta \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \left( -2 \int_{\Omega} u_i \Delta u_i \sum\limits_{\alpha=1}^l \left|\nabla \phi_\alpha \right|^2 +4\int_{\Omega} \left|\nabla u_i\right|^2 +p\xi_2 \int_{\Omega} u_i^2 \sum\limits_{\alpha=1}^l \left|\nabla \phi_\alpha\right|^2 \right)\\ &+\frac{1}{\delta}\sum\limits_{i=1}^k (\lambda_{k+1}-\lambda_i) \int_{\Omega}{\frac{1}{\rho} \left|\nabla u_i\right|^2}. \end{aligned} \end{equation} $

(2.37)

It implies

$ \begin{equation} \frac{l}{\rho_2} \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \le \delta \sum\limits_{i=1}^k\left(\lambda_{k+1}-\lambda_i\right)^2\left[(2l+4)B_i+\frac{pl\xi_2}{\rho_1}\right] +\frac{1}{\rho_1\delta} \sum\limits_{i=1}^k(\lambda_{k+1}-\lambda_i)B_i. \end{equation} $

(2.38)

Taking

$ \delta=\left\{\frac{\sum_{i=1}^k(\lambda_{k+1}-\lambda_i)B_i}{\rho_1\sum_{i=1}^k\left(\lambda_{k+1}-\lambda_i\right)^2\left[\left(2l+4\right)B_i+\frac{pl\xi_2}{\rho_1}\right]}\right\}^\frac{1}{2} $

in (2.38), we derive (1.9).

iv) Taking the Laplacian of $ \sum_{\alpha=1}^{N+1} f_\alpha^2 = 1 $ and using the fact that $ \Delta f_\alpha = -\mu f_\alpha, \,\,\alpha=1,2,\cdots,N+1 $, we have

$ \begin{equation} \sum\limits_{\alpha=1}^{N+1} \left|\nabla f_\alpha\right|^2 = \mu \end{equation} $

(2.39)

and

$ \begin{equation} \begin{aligned} \sum\limits_{\alpha=1}^{N+1} \int_{\Omega}u_i\Delta f_\alpha\left<\nabla f_\alpha,\nabla u_i\right> = -\frac{\mu}{4} \sum\limits_{\alpha=1}^{N+1} \int_{\Omega} \left<\nabla f_\alpha^2,\nabla u_i^2\right> = \frac{\mu}{4} \sum\limits_{\alpha=1}^{N+1} \int_{\Omega} u_i^2 \Delta f_\alpha^2 =0. \end{aligned} \end{equation} $

(2.40)

It then follows by taking $ h=f_\alpha $ in (2.1) and summing over $ \alpha $ that

$ \begin{equation} \begin{aligned} &\sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \sum\limits_{\alpha=1}^{N+1} \int_{\Omega}{u_i^2 \left|\nabla f_\alpha\right|^2}\\ \le &\delta \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \left[ -2\int_{\Omega} u_i\Delta u_i \left(\sum\limits_{\alpha=1}^{N+1} \left|\nabla f_\alpha\right|^2\right) + 4\sum\limits_{\alpha=1}^{N+1}\int_{\Omega}u_i\Delta f_\alpha\left<\nabla f_\alpha,\nabla u_i\right>\right.\\ & \left.+4\sum\limits_{\alpha=1}^{N+1} \int_{\Omega}\left<\nabla f_\alpha,\nabla u_i \right>^2 + \int_{\Omega}u_i^2 \sum\limits_{\alpha=1}^{N+1} \left(-\mu f_\alpha\right)^2 +p\int_{\Omega} u_i^2 \sum\limits_{\alpha=1}^{N+1} \left<\nabla f_\alpha,A\nabla f_\alpha\right>\right]\\ & +\frac{1}{\delta} \sum\limits_{i=1}^k(\lambda_{k+1}-\lambda_i) \sum\limits_{\alpha=1}^{N+1} \int_{\Omega} { \frac{1}{\rho}\left[\left<\nabla f_\alpha,\nabla u_i\right>^2+u_i\Delta f_\alpha\left<\nabla f_\alpha,\nabla u_i\right>+\frac{u_i^2\left(-\mu f_\alpha\right)^2}{4}\right]}. \end{aligned} \end{equation} $

(2.41)

It implies

$ \begin{equation} \begin{aligned} \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \frac{\mu}{\rho_2} \le &\delta \sum\limits_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \left( 6\mu B_i +\frac{\mu^2 +p\mu\xi_2}{\rho_1} \right)\\ &+\frac{1}{\rho_1\delta}\sum\limits_{i=1}^k (\lambda_{k+1}-\lambda_i ) \left( \mu B_i +\frac{\mu^2}{4\rho_1}\right). \end{aligned} \end{equation} $

(2.42)

Taking

$ \delta=\left[ \frac{\sum_{i=1}^k (\lambda_{k+1}-\lambda_i) \left( \mu B_i +\frac{\mu^2}{4\rho_1}\right)} {\rho_1\sum_{i=1}^k \left(\lambda_{k+1}-\lambda_i\right)^2 \left( 6\mu B_i +\frac{\mu^2 +p\mu\xi_2}{\rho_1} \right)} \right]^\frac{1}{2} $

in (2.42), we obtain (1.10). This ends the proof of Theorem 1.2.