In recent years, with the increasing attention to analysis and geometry in metric space, the investigations of Carnot-Carathéodory (CC) spaces as a special kind of metric space were carried out with prosperous results. In this case, the Carnot group plays a fundamental role. As we all know, it is a graded Lie group whose Lie algebra is nilpotent. Roughly speaking, the Carnot group can be regarded as a local model of CC spaces since they can be seen as the natural tangent space of the CC space, just as the Euclidean space is tangent to the manifold (see[1,2]). Carnot groups occupy a central position in the study of harmonic analysis, partial differential equation, sub-Riemannian geometry, mechanical engineering and so on (cf.[3-5]).

As we know, the Heisenberg group $ \mathbb{H}^n $ is the classical example of a non-Abelian Carnot group. Niu and Zhang[6], Sun[7, 8] established some universal inequalities of eigenvalues for the Kohn-Laplacian, the Folland-Stein operator on the Heisenberg group.

The horizontal Laplacian $ \Delta_H $ on a Carnot group $ G $ is a hypoelliptic operator of Hörmander type. In 2013, Aribi and El Soufi[9] gave some universal bounds for the eigenvalues of the horizontal Laplacian on Carnot groups. In 2017, Du, Wu, Li and Xia[10] investigated the following eigenvalue problem of the biharmonic horizontal Laplacian on a bounded domain $ \Omega $ in a Carnot group $ G $ with an $ d_1 $-dimensional sub-bundle

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

(1.1)

where $ v $ is the outwards unit normal vector field of $ \partial \Omega $. They obtained the following inequalities for eigenvalues of this problem

$ $ \begin{equation} \sum\limits_{i = 1}^k(\lambda_{k+1}-\lambda_i)^2 \leq (\frac{8d_{1}+16}{d^2_{1}})^\frac{1}{2}\{\sum\limits_{i = 1}^k (\lambda_{k+1}-\lambda_i)^2 \lambda_i^{\frac{1}{2}}\}^\frac{1}{2}\{\sum\limits_{i = i}^k (\lambda_{k+1}-\lambda_i)\lambda_i^{\frac{1}{2}}\}^\frac{1}{2} \end{equation} $ $

(1.2)

and

$ $ \begin{equation} \sum\limits_{i = 1}^k (\lambda _{i + 1} - \lambda _1)^{\frac{1}{2}} \le 2\sqrt {2d_{1} + 4} \lambda _1^{\frac{1}{2}}. \end{equation} $ $

(1.3)

Quadratic polynomial operator of the Laplacian is one important kind of differential operator in the research of differential geometry and partial differential equation (see[11]). In this paper, we consider the following Dirichlet weighted eigenvalue problem of quadratic polynomial operator of the horizontal Laplacian

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

(1.4)

where $ \rho $ is a positive function continuous on $ \bar \Omega $ and the constants $ a, b\geq 0 $. It is known that this eigenvalue problem has a discrete spectrum $ 0 < {\lambda _1} \le \cdots \le {\lambda _k} \le \cdots \nearrow $, where each eigenvalue is repeated with its multiplicity (see[12]). In particular, when $ \rho \equiv 1 $ and $ b = 0 $, problem (1.4) becomes the following eigenvalue problem

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

(1.5)

Furthermore, when $ a = 0 $, problem becomes (1.1).

We derive the following results for problem (1.4).

Theorem 1.1 Let $ \Omega $ be a bounded domain in a Carnot group $ G $ with an $ d_1 $-dimensional sub-bundle. Denote by $ \lambda_i $ the $ i $-th eigenvalue of problem (1.4). Set $ {\sigma} = {( {\mathop {\inf }\limits_{\bar \Omega } \rho } )^{ - 1}} $ and $ {\tau} = {( {\mathop {\max }\limits_{\bar \Omega } \rho } )^{ - 1}} $. Then we have

$ $ \begin{equation} \begin{aligned} \sum\limits_{i = 1}^k(\lambda_{k+1}-\lambda_i)^2 \leq \frac{2 \sigma^{\frac{1}{2}} }{\tau d_1}\{\sum\limits_{i = 1}^k (\lambda_{k+1}-\lambda_i)^2 \big[2(d_1+2)\xi_i+ad_1\sigma\big]\}^\frac{1}{2}\{\sum\limits_{i = 1}^k (\lambda_{k+1}-\lambda_i) \xi_i\}^\frac{1}{2}, \end{aligned} \end{equation} $ $

(1.6)

where

$ \xi_i = \frac{ - a\sigma + \sqrt {a^2\sigma^2 + 4\sigma( \lambda _i - b\tau )} }{2}. $

Theorem 1.2 Under the assumptions of Theorem 1.1, we have

$ $ \begin{equation} \sum\limits_{j = 1}^{d_1} ( \lambda _{i + 1} - \lambda _1 )^{\frac{1}{2}} \leq \frac{2\sigma^{\frac{1}{2}} }{\tau} [( 4 + 2d_1) \xi_1 +a \sigma d_1 ]^{\frac{1}{2}} \xi_1^{\frac{1}{2}}. \end{equation} $ $

(1.7)

From Theorem 1.1 and Theorem 1.2, we can get the following results for problem (1.5).

Corollary 1.1 Let $ \Omega $ be a bounded domain in a Carnot group $ G $ with an $ d_1 $-dimensional sub-bundle. Denote by $ \lambda_i $ the $ i $-th eigenvalue of problem (1.5). Then we have

$ $ \begin{equation} \begin{aligned} \sum\limits_{i = 1}^k(\lambda_{k+1}-\lambda_i)^2 \leq \frac{2 }{ d_1}\{\sum\limits_{i = 1}^k (\lambda_{k+1}-\lambda_i)^2 \big[2(d_1+2)\zeta_i+ad_1 \big]\}^\frac{1}{2}\{\sum\limits_{i = 1}^k (\lambda_{k+1}-\lambda_i) \zeta_i\}^\frac{1}{2} \end{aligned} \end{equation} $ $

(1.8)

and

$ $ \begin{equation} \sum\limits_{j = 1}^{d_1} ( \lambda _{i + 1} - \lambda _1 )^{\frac{1}{2}} \leq 2 [( 4 + 2d_1) \zeta_1 +a d_1 ]^{\frac{1}{2}} \zeta_1^{\frac{1}{2}}. \end{equation} $ $

(1.9)

where $ \zeta_i = \frac{ - a + \sqrt {a^2 + 4\lambda _i } }{2}. $

Remark It is easy to find that $ \xi_i = \lambda_i^{\frac{1}{2}} $ when $ \rho\equiv1 $ and $ a = b = 0 $. Thus (1.6) and (1.7) respectively become (1.2) and (1.3) when $ \rho\equiv1 $ and $ a = b = 0 $. Therefore our results cover the results of Du, Wu, Li and Xia[10] for the biharmonic horizontal Laplacian.

In this section, we give the proofs of Theorem 1.1 and Theorem 1.2.

For the convenience of readers, we fisrt give some basic knowledge about the Carnot group. A Carnot group $ G $ of step $ r $ is a connected, simply connected Lie group whose Lie algebra $ \mathfrak{g} $ admits a stratification $ \mathfrak{g} = {V_1} \oplus {V_2} \oplus \cdots \oplus {V_r} $. It is $ r $-nilpotent, i.e., $ [V_1, V_j] = V_{j+1}, j = 1, \cdots, r-1, [V_j, V_r] = {0}, j = 1, \cdots, r $. We also assume that there exists a scalar product $ \langle, \rangle_\mathfrak{g} $ on $ \mathfrak{g} $, such that the $ V'_js $ are mutually orthogonal. The whole $ \mathfrak{g} $ is generated by the layer $ V_1 $ which induces a subbundle $ HG $ of $ TG $ of rank $ d_1 = $dim$ V_1 $. We call $ HG $ the horizontal bundle of the Carnot group. Let $ \{ {e_1^i, \cdots , e_{d_i}^i, i \le r} \} $ be an orthonormal basis of $ V_i $ and $ \{ {X_1^i, \cdots , X_{d_i}^i} \} $ denotes the system of left invariant vector fields that coincides with $ \{ {e_1^i, \cdots , e_{d_i}^i} \} $ at the identity element of $ G $. We consider $ G $ is endowed with a left-invariant Riemannian metric $ \mathfrak{g}_G $ with respect to which the family $ \{ {X_1^1, \cdots , X_{d_r}^r} \} $ constitute an orthonormal frame for $ TG $. The corresponding Levi-Civita connection $ \nabla $ induces a connection $ \nabla^H $ on $ HG $ that we call "horizontal connection": if X is a vector field and Y is a horizontal vector field on $ G $, then $ \nabla _X^HY = {\pi _H}{\nabla _X}Y $, where $ {\pi _H}:TG \to HG $ is the orthogonal projection. The horizontal Laplacian $ \Delta_H $ is defined by

$ $ \begin{equation} \Delta _Hu: = \text{div}_H \nabla^H u = \sum\limits_{j = 1}^{d_1} \left( X_j^1 \right)^2u , \end{equation} $ $

(2.1)

where $ u \in C^2 $.

Define $ {S^{2, 2}}( \Omega ) $ by

$ $ \begin{equation} {S^{2, 2}}(\Omega ): = \{ {f \in {L^2}( \Omega )| {| {{\nabla ^H}f} |, {{| {{\nabla ^H}f} |}^2} \in {L^2}( \Omega )} .} \}, \end{equation} $ $

(2.2)

where $ {| {{\nabla ^H}f} |^2} = \sum\limits_{j = 1}^{d_1} {{{( {{X_j^1}f} )}^2}} $. Then $ {S^{2, 2}}( \Omega ) $ is a Hilbert space with a Sobolev norm $ {\| f \|^2} = \int_\Omega {( {\sum\limits_{p = 1}^2 {{{| {(\nabla ^H)^p f} |}^2} + {f^2}} } )}. $ Furthermore, we consider the subspace $ S_0^{2, 2}( \Omega ) $ defined by

$ $ \begin{equation} S_0^{2, 2}(\Omega ): = \{ {f \in {S^{2, 2}}(\Omega )\left| {f\left| {_{\partial \Omega }} \right. = \frac{{\partial f}}{{\partial \upsilon }}\left| {_{\partial \Omega }} \right. = 0} \right.} \}. \end{equation} $ $

(2.3)

For every $ f, g \in S_0^{2, 2}\left( \Omega \right) $, we have

$ $ \begin{equation} \int_\Omega fX_j^1g = - \int_\Omega gX_j^1f . \end{equation} $ $

(2.4)

Proof of Theorem 1.1 In order to construct a good trial function, we use some functions introduced by Danielli, Garofalo and Nhieu[4]. Since the Carnot group $ G $ of step $ r $ is nilpotent, the exponential exp$ :\mathfrak{g}\rightarrow G $ is a global diffeomorphism. Setting $ \{ e_1, \cdots, e_{d_1} \} $ and $ \{ X_1^1, \cdots, X_{d_1}^1 \} $ be an orthonormal basis of $ V_1 $ and the system of left invariant vector fields, respectively. We can define a smooth map $ x_i:G\rightarrow R $ by

$ $ \begin{equation} {x_i}(g): = {\left\langle {{{\exp }^{ - 1}}(g), e_i} \right\rangle _\mathfrak{g}}, \quad i = 1, \cdots, d_1. \end{equation} $ $

(2.5)

These functions satisfy

$ $ \begin{equation} X_j^1 ( x_i ) = \delta _{ij} , \quad \Delta _H x_i = 0 \quad \mbox{for} \quad i, j = 1, \cdots , d_1. \end{equation} $ $

(2.6)

Denote by $ u_i $ the $ i $-th weighted orthonormal eigenfunction of problem (1.5) corresponding to the eigenvalue $ \lambda_i $, namely, we have

$ $ \begin{equation} \left\{ \begin{aligned} &\Delta _H^2 u_i - a \Delta _H u_i + b u_i = \lambda _i \rho u_i, \quad u_i \in \Omega , \\ &u_i |_{\partial \Omega } = \frac{\partial u_i }{ \partial \nu } |_{\partial \Omega } = 0, \\ &\int_\Omega \rho u_i u_j = \delta _{ij}. \end{aligned} \right. \end{equation} $ $

(2.7)

Set $ \varphi_{ij} = x_i u_j-\sum\limits_{l = 1}^k {q_{ij}^l u_l} $, where $ q_{ij}^l = \int_\Omega \rho x_i u_j u_l $. It is easy to find

$ $ \begin{equation} {\varphi_{ij}}\left| {_{\partial \Omega }} \right. = \frac{{\partial {\varphi_{ij}}}}{{\partial \nu }}\left| {_{\partial \Omega }} \right. = 0, \quad \int_\Omega {{\rho\varphi_{ij}}{u_l}} = 0 \quad \mbox{for} \quad l = 1, \cdots , k. \end{equation} $ $

(2.8)

Using the Rayleigh-Ritz inequality, we have

$ $ \begin{equation} \begin{split} \lambda _{k + 1} \int_\Omega {\rho \varphi_{ij}^2} &\le \int_\Omega \varphi _{ij} \left[ \Delta _H^2 \varphi_{ij} - a \Delta _H \varphi_{ij} + b \varphi _{ij} \right] \\ & = {\lambda _j}\int_\Omega {\rho \varphi _{ij}^2} + \int_\Omega {\varphi _{ij}}\left[ \big( \Delta _H^2 \varphi_{ij} - a \Delta _H \varphi_{ij} + b \varphi _{ij} \big) - {\lambda _j}\rho x_i u_j \right]\\ & = {\lambda _j}\int_\Omega {\rho \varphi _{ij}^2} + \int_\Omega {\varphi _{ij}} ( \Delta _H^2 - a \Delta _H + b - {\lambda _j}\rho ) (x_i u_j)\\ & = {\lambda _j}\int_\Omega {\rho \varphi _{ij}^2} + \int_\Omega {x_i} u_j ( \Delta _H^2 - a \Delta _H + b - {\lambda _j}\rho ) (x_i u_j) - \sum\limits_{l = 1}^k q_{ij}^l s_{ij}^l. \end{split} \end{equation} $ $

(2.9)

Substituting

$ $ \begin{equation} \begin{aligned} s_{ij}^l = &\int_\Omega u_l ( \Delta _H^2 - a \Delta _H + b - {\lambda _j}\rho ) (x_i u_j)\\ = &\int_\Omega {x_i}{u_j} \big( \Delta _H^2 - a \Delta _H + b \big) u_l -\lambda _j \int_\Omega \rho x_i u_j u_l \\ = & (\lambda_l-\lambda_j)q_{ij}^l \end{aligned} \end{equation} $ $

(2.10)

into (2.9), we derive

$ $ \begin{equation} (\lambda _{k + 1}- \lambda _j)\int_\Omega \rho \varphi _{ij}^2 \leq \int_\Omega {x_i}{u_j} (\Delta _H^2 - a\Delta _H + b -\lambda_j \rho ) ( x_i u_j) + \sum\limits_{l = 1}^k ( {{\lambda _j} - {\lambda _l}})(q_{ij}^l)^2. \end{equation} $ $

(2.11)

By direct calculations, we have

$ $ \begin{equation} {\Delta _H}( {{x_i}{u_j}} ) = 2 X_i^1 u_j + {x_i}{\Delta _H}{u_j} \end{equation} $ $

(2.12)

and

$ $ \begin{equation} \Delta_H^2(x_i u_j) = 2\Delta_H X_i^1 u_j +2X_i^1(\Delta_H u_j)+x_i\Delta_H^2 u_j. \end{equation} $ $

(2.13)

Using (2.12) and (2.13), we obtain

$ $ \begin{equation} (\Delta _H^2 - a\Delta _H + b )( x_i u_j) = 2\left[ \Delta_H X_i^1 u_j +X_i^1(\Delta_H u_j)-aX_i^1 u_j \right]+\lambda_j \rho x_iu_j. \end{equation} $ $

(2.14)

Substituting (2.14) into (2.11), we get

$ $ \begin{equation} \begin{aligned} (\lambda_{k+1}-\lambda_j)\int_\Omega \rho \varphi_{ij}^2 \leq& 2\int_\Omega x_iu_j[\Delta_H X_i^1 u_j+X_i^1(\Delta_H u_j)-aX_i^1u_j ] + \sum\limits_{l = 1}^k ( {{\lambda _j} - {\lambda _l}}) (q_{ij}^l)^2 \\ = &2\int_\Omega [ X_i^1u_j\Delta_H(x_i u_j)-X_i^1(x_i u_j)\Delta_H u_j ]-2a\int_\Omega x_i u_j X_i^1 u_j \\ &+\sum\limits_{l = 1}^k ( {{\lambda _j} - {\lambda _l}}) (q_{ij}^l)^2 \\ = &\int_\Omega [4(X_i^1u_j )^2-2u_j\Delta_H u_j+a u_j^2 ] +\sum\limits_{l = 1}^k ( {{\lambda _j} - {\lambda _l}}) (q_{ij}^l)^2. \end{aligned} \end{equation} $ $

(2.15)

From

$ $ \begin{equation} -2\int_\Omega x_iu_jX_i^1 u_j = 2\int_\Omega u_jX_i^1( x_iu_j) = 2\int_\Omega u_j^2 + 2\int_\Omega x_iu_jX_i^1 u_j \end{equation} $ $

(2.16)

and

$ $ \begin{equation} \tau \le \int_\Omega u_j^2 = \int_\Omega \frac{1}{\rho } \rho u_j^2 \le \sigma, \end{equation} $ $

(2.17)

we can get

$ $ \begin{equation} -2\int_\Omega {x_i}{u_j} X_i^1 u_j = \int_\Omega u_j^2\geq \tau. \end{equation} $ $

(2.18)

Set $ p_{ij}^l = \int_\Omega u_l X_i^1 u_j $. Then we have

$ $ \begin{equation} -2\int_\Omega \varphi_{ij} X_i^1 u_j = -2\int_\Omega {x_i}{u_j}X_i^1 u_j + 2\sum\limits_{l = 1}^k q_{ij}^lp_{ij}^l \geq \tau+2\sum\limits_{l = 1}^k q_{ij}^lp_{ij}^l. \end{equation} $ $

(2.19)

Multiplying the both sides of (2.19) by $ (\lambda _{k + 1} - \lambda _j)^2 $, using the Cauchy-Schwarz inequality and (2.15), we have

$ $ \begin{equation} \begin{aligned} &(\lambda _{k + 1} - \lambda _j)^2(\tau+2\sum\limits_{l = 1}^k q_{ij}^l p_{ij}^l)\\ \leq& -2 (\lambda _{k + 1} - \lambda _j)^2 \int_\Omega \sqrt{\rho}\varphi_{ij} ( \frac{1}{\sqrt{\rho} } X_i^1 u_j-\sum\limits_{l = 1}^k p_{ij}^l \sqrt{\rho} u_l )\\ \leq& \delta(\lambda _{k + 1} - \lambda _j)^3\int_\Omega \rho \varphi_{ij}^2 +\frac{(\lambda _{k + 1} - \lambda _j)}{\delta}\int_\Omega (\frac{1}{\sqrt{\rho}} X_i^1 u_j -\sum\limits_{l = 1}^k p_{ij}^l \sqrt{\rho} u_l )^2\\ \leq& \delta(\lambda _{k + 1} - \lambda _j)^2\{\int_\Omega [4(X_i^1 u_j)^2-2u_j\Delta_H u_j+au_j^2 ] +\sum\limits_{l = 1}^k ( {{\lambda _j} - {\lambda _l}})(q_{ij}^l)^2 \}\\ & +\frac{\sigma}{\delta}(\lambda _{k + 1} - \lambda _j)\int_\Omega (X_i^1u_j)^2-\frac{1}{\delta}(\lambda _{k + 1} - \lambda _j)\sum\limits_{l = 1}^k (p_{ij}^l)^2, \end{aligned} \end{equation} $ $

(2.20)

where $ \delta $ is any positive number. Taking sum on $ j $ from $ 1 $ to $ k $ in (2.20), and noticing $ q_{ij}^l = q_{il}^j, p_{ij}^l = -p_{il}^j $, we get

$ $ \begin{equation} \begin{aligned} &\tau\sum\limits_{j = 1}^k (\lambda_{k+1}-\lambda_j)^2-2\sum\limits_{j, l = 1}^k (\lambda_{k+1}-\lambda_j)(\lambda_j-\lambda_l)q_{ij}^lp_{ij}^l\\ \leq &\sum\limits_{j = 1}^k \delta(\lambda _{k + 1} - \lambda _j)^2 \int_\Omega \left[4(X_i^1 u_j )^2-2u_j\Delta_H u_j+a u_j^2 \right] +\frac{\sigma}{\delta}\sum\limits_{j = 1}^k (\lambda _{k + 1} - \lambda _j)\int_\Omega (X_i^1 u_j)^2\\ & -\delta\sum\limits_{j, l = 1}^k(\lambda_{k+1}-\lambda_j)(\lambda_j-\lambda_l)^2 (q_{ij}^l)^2-\frac{1}{\delta}\sum\limits_{j, l = 1}^k (\lambda_{k+1}-\lambda_j) (p_{ij}^l)^2. \end{aligned} \end{equation} $ $

(2.21)

It implies that

$ $ \begin{equation} \begin{split} \tau \sum\limits_{j = 1}^k (\lambda_{k+1}-\lambda_j)^2 &\leq \delta\sum\limits_{j = 1}^k(\lambda_{k+1}-\lambda_i)^2 \int_\Omega [4(X_i^1 u_j)^2- 2u_j\Delta_H u_j+ a u_j^2 ] \\ &\quad +\frac{\sigma}{\delta}\sum\limits_{j = 1}^k (\lambda _{k + 1} - \lambda _j)\int_\Omega (X_i^1 u_j )^2. \end{split} \end{equation} $ $

(2.22)

Taking sum on $ i $ from $ 1 $ to $ d_1 $ in (2.22), we get

$ $ \begin{equation} \begin{split} \tau d_1\sum\limits_{j = 1}^k (\lambda_{k+1}-\lambda_j)^2 &\leq \delta \sum\limits_{j = 1}^k(\lambda_{k+1}-\lambda_i)^2 \int_\Omega \left[2(d_1+2) |\nabla^H u_j |^2+ a d_1 u_j^2 \right] \\ &\quad +\frac{\sigma}{\delta} \sum\limits_{j = 1}^k (\lambda _{k + 1} - \lambda _j)\int_\Omega |\nabla^H u_j|^2. \end{split} \end{equation} $ $

(2.23)

Using the Schwarz inequality,

$ $ \begin{equation} \begin{split} \int_\Omega {| \nabla ^H u_j |}^2 & = - \int_\Omega u_j\Delta_H u_j \le ( \int_\Omega u_j^2 )^{\frac{1}{2}} ( \int_\Omega (\Delta _H u_j)^2 )^{\frac{1}{2}} \\ &\le \sigma^{\frac{1}{2}} ( \lambda _j - a\int_\Omega {| \nabla ^H u_j |}^2 - b\tau )^{\frac{1}{2}}. \end{split} \end{equation} $ $

(2.24)

Then we have

$ $ \begin{equation} ( \int_\Omega {| \nabla ^H u_j |}^2 )^2 + a\sigma\int_\Omega {| \nabla ^H u_j |}^2 - \sigma( \lambda _j-b\tau )\le 0. \end{equation} $ $

(2.25)

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

$ $ \begin{equation} \int_\Omega {| \nabla ^H u_j |}^2 \leq \xi_j. \end{equation} $ $

(2.26)

Substituting (2.26) into (2.23), we get

$ $ \begin{equation} \begin{aligned} \tau d_1\sum\limits_{j = 1}^k(\lambda_{k+1}-\lambda_j)^2 \leq\delta\sum\limits_{j = 1}^k (\lambda_{k+1}-\lambda_j)^2 [2(d_1+2)\xi_j+ad_1\sigma] +\frac{\sigma }{\delta} \sum\limits_{j = 1}^k (\lambda_{k+1}-\lambda_j)\xi_j. \end{aligned} \end{equation} $ $

(2.27)

Taking

$ \delta = \left\{\frac{\sigma\sum\limits_{j = 1}^k(\lambda_{k+1}-\lambda_j)\xi_j}{\sum\limits_{j = 1}^k(\lambda_{k+1}-\lambda_j)^2[2(d_1+2)\xi_j+ad_1\sigma]}\right\}^\frac{1}{2} $

in (2.27), we can obtain (1.6). This completes the proof of Theorem 1.1.

Proof of Theorem 1.2 Similar to the proof of Theorem 1.1, let $ x_i $'s be the functions satisfying (2.5) and (2.6). Denote by $ u_i $ the $ i $-th weighted orthonormal eigenfunction of problem (1.5) corresponding to the eigenvalue $ \lambda_i $. Define a $ {d_1} \times {d_1} $-matrix $ Q = (q_{ij} )_{d_1 \times d_1} $, where $ q_{ij} = \int_\Omega \rho x_i u_1 u_{j + 1} $. According to the QR-factorization theorem, we know that there exists an orthogonal matrix $ T = (t_{ij})_{d_1 \times d_1} $ such that $ R = TQ $ is an upper triangle matrix, namely, we have

$ $ \begin{equation} R_{ij} = \sum\limits_{k = 1}^{d_1} t_{ik}c_{kj} = \sum\limits_{k = 1}^{d_1} \int_\Omega t_{ik} \rho x_k u_1 u_{j + 1} = 0 \quad \mbox{for} \quad 0 < j < i \le d_1. \end{equation} $ $

(2.28)

Set $ {y_i} = \sum\limits_{k = 1}^{d_1} t_{ik} x_k $. Then (2.28) can be written as

$ $ \begin{equation} \int_\Omega \rho {y_i}{u_1} u_{j + 1} = 0 \quad \mbox{for} \quad 0 < j < i \le {d_1}. \end{equation} $ $

(2.29)

Moreover, from (2.6), we get

$ $ \begin{equation} X_j^1 (y_i ) = \sum\limits_{k = 1}^{d_1} t_{ik}\delta _{kj} = t_{ij}, \quad \Delta _H y_i = 0 \quad \mbox{for} \quad i, j = 1, \cdots, d_1. \end{equation} $ $

(2.30)

Consider the function $ \phi _i = (y_i - a_i) u_1 $, where $ {a_i} = \int_\Omega {\rho {y_i}u_1^2} $. It is easy to find that $ \int_\Omega {\rho {\phi _i}{u_{j + 1}}} = 0 \quad \mbox{for} \quad 0 \le j < i \le {d_1}. $ Using the Rayleigh-Ritz inequality, we have

$ $ \begin{equation} \begin{split} \lambda _{i + 1} \int_\Omega {\rho \phi _i^2} &\le \int_\Omega \phi _i \big( \Delta _H^2 \phi_i - a \Delta _H \phi_i + b \phi _i \big) \\ & = {\lambda _1}\int_\Omega {\rho \phi _i^2} + \int_\Omega {\phi _i}\left[ \big( \Delta _H^2 \phi_i - a \Delta _H \phi_i + b \phi _i \big) - {\lambda _1}\rho {y_i}{u_1} \right]\\ & = {\lambda _1}\int_\Omega {\rho \phi _i^2} + \int_\Omega {\phi _i}\left[ \Delta _H^2 (y_i u_1) - a \Delta _H (y_i u_1) + b y_i u_1 - {\lambda _1}\rho {y_i}{u_1} \right]\\ & = {\lambda _1}\int_\Omega {\rho \phi _i^2} + \int_\Omega {y_i} u_1 \left[ \Delta _H^2 (y_i u_1) - a \Delta _H (y_i u_1) + b y_i u_1 - {\lambda _1}\rho {y_i}{u_1} \right] - {a_i}{s_i}. \end{split} \end{equation} $ $

(2.31)

Since

$ $ \begin{equation} \begin{aligned} s_i = &\int_\Omega u_1 \left[ \big( \Delta _H^2 (y_i u_1) - a \Delta _H (y_i u_1) + b y_i u_1 \big) - {\lambda _1}\rho y_i u_1 \right] \\ = &\int_\Omega {y_i}{u_1} \big( \Delta _H^2 u_1 - a \Delta _H u_1 + b u_1 \big) -\lambda _1 \int_\Omega \rho y_i u_1^2 \\ = & 0, \end{aligned} \end{equation} $ $

(2.32)

we get

$ $ \begin{equation} (\lambda _{i + 1} - \lambda _1) \int_\Omega \rho \phi _i^2 \le \int_\Omega y_i u_1 [ \Delta _H^2 (y_i u_1) - a \Delta _H (y_i u_1) + b y_i u_1 - \lambda _1 \rho y_i u_1 ] . \end{equation} $ $

(2.33)

On the one hand, using (2.30), we obtain

$ $ \begin{equation} \begin{split} \Delta _H \left( y_i u_1 \right) & = \sum\limits_{j = 1}^{d_1} (X_j^1)^2\left( y_i u_1 \right) = \sum\limits_{j = 1}^{d_1} {X_j^1}\left( t_{ij} u_1+ y_i X_j^1 u_1 \right)\\ & = 2\sum\limits_{j = 1}^{d_1} t_{ij} X_j^1 u_1 + y_i\Delta _H u_1 \end{split} \end{equation} $ $

(2.34)

and

$ $ \begin{equation} \begin{split} \Delta _H^2 ( y_i u_1 ) & = \Delta _H ( 2\sum\limits_{j = 1}^{d_1} t_{ij} X_j^1 u_1 + y_i\Delta _H u_1 )\\ & = 2\sum\limits_{j = 1}^{d_1} t_{ij} \Delta _H X_j^1 u_1 + 2\sum\limits_{j = 1}^{d_1} t_{ij} X_j^1 ( \Delta _H u_1 ) + y_i\Delta _H^2 u_1. \end{split} \end{equation} $ $

(2.35)

It follows from (2.34) and (2.35) that

$ $ \begin{equation} \begin{aligned} &(\Delta _H^2 - a\Delta _H + b ) ( y_i u_1 )\\ = &2[ \sum\limits_{j = 1}^{d_1} t_{ij} \Delta _H ( X_j^1 u_1 ) + \sum\limits_{j = 1}^{d_1} t_{ij} X_j^1 ( \Delta _H u_1 ) - a\sum\limits_{j = 1}^{d_1} t_{ij} X_j^1 u_1 ] + \lambda _1 \rho y_i u_1. \end{aligned} \end{equation} $ $

(2.36)

Substituting (2.36) into (2.33), we obtain

$ $ \begin{equation} \begin{split} &(\lambda _{i + 1}-\lambda _1)\int_\Omega \rho \phi _i^2 \\ \leq& 2\int_\Omega y_i u_1 [ \sum\limits_{j = 1}^{d_1} t_{ij} \Delta _H ( X_j^1 u_1 ) + \sum\limits_{j = 1}^{d_1} t_{ij} X_j^1 (\Delta _H u_1 ) - a\sum\limits_{j = 1}^{d_1} t_{ij} X_j^1 u_1 ] \\ = & 2\int_\Omega [ \sum\limits_{j = 1}^{d_1} t_{ij} X_j^1 u_1 \Delta _H ( y_i u_1 ) - \sum\limits_{j = 1}^{d_1} t_{ij} \Delta _H u_1 X_j^1 ( y_i u_1 ) + a\sum\limits_{j = 1}^{d_1} t_{ij} u_1 X_j^1 ( y_i u_1 ) ] \\ = & \int_\Omega [ 4 (\sum\limits_{j = 1}^{d_1} t_{ij} X_j^1 u_1 )^2 - 2u_1\Delta _H u_1 + a\sum\limits_{j = 1}^{d_1} t_{ij}^2 u_1^2 ]. \end{split} \end{equation} $ $

(2.37)

On the other hand, we have

$ $ \begin{equation*} \label{eqn3.14} \begin{split} \int_\Omega \phi _i \sum\limits_{j = 1}^{d_1} t_{ij} X_j^1 u_1 & = -\sum\limits_{j = 1}^{d_1} t_{ij}^2 \int_\Omega u_1^2 -\sum\limits_{j = 1}^{d_1} \int_{\Omega} ( y_i - a_i )u_1 t_{ij} X_j^1 u_1 \\ & = - \int_\Omega u_1^2 - \int_\Omega \phi _i\sum\limits_{j = 1}^{d_1} t_{ij} X_j^1 u_1. \end{split} \end{equation*} $ $

It implies

$ $ \begin{equation} 2\int_\Omega \phi _i\sum\limits_{j = 1}^{d_1} t_{ij} X_j^1 u_1 = - \int_\Omega u_1^2 . \end{equation} $ $

(2.38)

Multiplying the both sides of (2.38) by $ ( \lambda _{i + 1} - \lambda _1 )^{\frac{1}{2}} $, and using the Cauchy-Schwarz inequality, we derive

$ $ \begin{equation} \begin{aligned} (\lambda _{i + 1} - \lambda _1 )^{\frac{1}{2}} \int_\Omega u_1^2 & = -2( \lambda _{i + 1}-\lambda _1)^{\frac{1}{2}}\int_\Omega \phi _i\sum\limits_{j = 1}^{d_1} t_{ij} X_j^1 u_1 \\ &\le \delta( \lambda _{i + 1} - \lambda _1 )\int_\Omega \rho \phi _i^2 + \frac{1}{\delta }\int_\Omega \frac{1}{\rho } ( \sum\limits_{j = 1}^{d_1} t_{ij} X_j^1 u_1 )^2, \end{aligned} \end{equation} $ $

(2.39)

where $ \delta $ is any positive number. Substituting (2.37) into (2.39), and taking sum on $ i $ from 1 to $ d_1 $, we have

$ $ \begin{equation} \begin{split} &\sum\limits_{i = 1}^{d_1} ( \lambda _{i + 1} - \lambda _1 )^{\frac{1}{2}} \int_\Omega u_1^2\\ \le& \delta \int_\Omega \sum\limits_{i = 1}^{d_1} [ 4 (\sum\limits_{j = 1}^{d_1} t_{ij} X_j^1 u_1 )^2 - 2u_1\Delta _H u_1 + a\sum\limits_{j = 1}^{d_1} t_{ij}^2 u_1^2 ] + \frac{1}{\delta } \int_\Omega \frac{1}{\rho } \sum\limits_{i = 1}^{d_1}(\sum\limits_{j = 1}^{d_1} t_{ij} X_j^1 u_1 )^2 \\ = & 4\delta \int_\Omega \sum\limits_{j = 1}^{d_1} (X_j^1 u_1)^2 - 2d_1\delta \int_\Omega u_1\Delta _H u_1 + ad_1\delta \int_\Omega u_1^2 + \frac{1}{\delta }\int_\Omega \frac{1}{\rho } \sum\limits_{j = 1}^{d_1} ( X_j^1 u_1)^2 \\ = & ( 4 + 2d_1 ) \delta \int_\Omega {| \nabla ^H u_1 |} ^2 + \frac{1}{\delta }\int_\Omega \frac{1}{\rho } {| \nabla ^H u_1 |}^2 + ad_1\delta \int_\Omega u_1^2. \end{split} \end{equation} $ $

(2.40)

Using (2.17) and (2.26) and (2.40), we get

$ $ \begin{equation} \tau\sum\limits_{j = 1}^{d_1} ( \lambda _{i + 1} - \lambda _1 )^{\frac{1}{2}} \leq [( 4 + 2d_1) \xi_1 +a \sigma d_1 ] \delta + \frac{\sigma}{\delta } \xi_1. \end{equation} $ $

(2.41)

Taking

$ \delta = [\frac{\sigma\xi_1}{(4 + 2d_1) \xi_1 +a \sigma d_1}]^{\frac{1}{2}} $

in (2.41), we obtain (1.7). This concludes the proof of Theorem 1.2.