Let $ M $ be an $ n $-dimensional complete Riemannian manifold. The Dirichlet eigenvalue problem of the Laplacian $ \Delta $ on a bounded domain $ \Omega $ of $ M $ is described by

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

(1.1)

Many mathematicians have obtained some universal inequalities for eigenvalues of problem (1.1) (cf. [1–4]).

Let $ A:\Omega\rightarrow \rm{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 $ of $ M $. Define the following elliptic operator in divergence form

$ \begin{equation} L_{A}=-\operatorname{div}(A\nabla), \end{equation} $

(1.2)

where $ \mathrm{div} $ is the divergence operator and $ \nabla $ is the gradient operator. It is easy to see that the operator $ L_{A} $ defined in (1.2) includes the Laplacian operator as a special case. In 2010, Do Carmo, Wang and Xia [5] considered the eigenvalue problem of $ L_{A} $ as follows

$ \begin{equation} \left\{ \begin{aligned} &L_{A} u+Vu = \lambda\rho u, & &\text{in}\ M, \\ &u=0, & &\text{on}\ \partial M, \end{aligned} \right. \end{equation} $

(1.3)

where $ V $ is a non-negative continuous function and $ \rho $ is a positive continuous function on $ M $. They obtained the following Yang-type inequality

$ \begin{equation} \sum\limits_{i=1}^{k}(\lambda_{k+1}-\lambda_i)^2\leq\dfrac{4\xi_2\rho_2^2}{n\rho_1^2}\sum\limits_{i=1}^k(\lambda_{k+1} - \lambda_i)\left[\dfrac{1}{\xi_1}\left(\lambda_i - \dfrac{V_0}{\rho_2}\right)+\dfrac{n^2H_0^2}{4\rho_1}\right], \end{equation} $

(1.4)

where $ \xi_{1} $, $ \xi_{2} $, $ \rho_{1} $, $ \rho_{2} $ are positive constants, $ H_0=\max\limits_{x\in M}|\mathbf{H}(x)| $, $ V_0=\min\limits_{x\in M}V(x) $ and $ \mathbf{H} $ is the mean curvature vector of $ M $ in $ \mathbb{R}^m $. For more reference about $ L_{A} $, we refer to [6].

In recent years, metric measure spaces have received a lot of attention in geometry and analysis. For some significant results about metric measure spaces, we refer to [7, 8] and the references therein. A smooth metric measure space is actually a Riemannian manifold equipped with some measures which is absolutely continuous with respect to the usual Riemannian measure. More precisely, for a given $ n $-dimensional complete Riemannian manifold $ (M, g) $ with a smooth metric $ g $, we say that the triple $ (M, g, d\mu) $ is a smooth metric measure space, where $ d\mu=e^{-f}d\upsilon $, $ f $ is a smooth real-valued function on $ M $ and $ d\upsilon $ is the Riemannian volume element related to $ g $.

Let $ \Omega $ be a bounded domain in a smooth metric measure space $ (M, g, e^{-f}d\upsilon) $. Define the elliptic operator in weighted divergence form $ \mathfrak{L}_{A, f} $ as

$ \begin{equation} \mathfrak{L}_{A, f}=-\operatorname{div}_{f}A\nabla, \end{equation} $

(1.5)

where $ \operatorname{div}_{f}X=e^f\operatorname{div}\left(e^{-f}X\right) $ is the weighted divergence of the vector field $ X $ on $ M $. When $ A $ is an identity map, $ -\mathfrak{L}_{A, f} $ becomes the drifting Laplacian $ \Delta_f=\operatorname{div}_{f} \nabla $. Moreover, when $ f $ is a constant, $ \mathfrak{L}_{A, f} $ becomes $ L_{A} $ defined in (1.2). There have been some interesting results for $ \mathfrak{L}_{A, f} $ (see [9–11]).

As an important example of complete metric measure spaces, we consider Ricci solitons introduced by Hamilton [12, 13]. They are corresponding to self-similar solutions of Hamilton's Ricci flow. We say that $ (M, g, f) $ is a gradient Ricci soliton if there is a constant $ K $, such that

$ \begin{equation} \mathrm{Ric}+\mathrm{Hess}f=Kg. \end{equation} $

(1.6)

The function $ f $ is called a potential function of the gradient Ricci soliton. For $ K>0 $, $ K=0 $ and $ K<0 $, the Ricci soliton is called shrinking, steady or expanding respectively. When the dimension is two, Hamilton discovered the first complete non-compact example of a steady Ricci soliton on $ \mathbb{R}^2 $, called the cigar soliton. The metric and potential function of the cigar soliton $ (\mathbb{R}^2, g, f) $ is given by

$ \begin{equation} \notag g=\frac{d(x^1)^2+d(x^2)^2}{1+|x|^2} \end{equation} $

and

$ \begin{equation} \notag f=-{\rm log}(1+|x|^2), \end{equation} $

where $ |x|^2=(x^1)^2+(x^2)^2 $. In physics, the cigar soliton $ (\mathbb{R}^2, g, f) $ is regarded as the Euclidean-Witten black hole under first-order Ricci flow of the world-sheet sigma model. As an important tractable model for understanding black hole physics (cf. [14]), it is of great significance in both geometry and physics. In 2018, Zeng [15] considered the following Dirichlet eigenvalue problem of the drifting Laplacian $ \Delta_f $ on a bounded domain $ \Omega $ in the cigar soliton $ (\mathbb{R}^2, g, f) $

$ \begin{equation} \left\{ \begin{aligned} &\Delta_fu = -\lambda u, & &in\ \Omega, \\ &u=0, & &on\ \partial\Omega \end{aligned} \right. \end{equation} $

(1.7)

and derived

$ \begin{equation} \begin{aligned} \sum\limits_{i=1}^k(\lambda_{k+1}-\lambda_i)^2\leq&\sum\limits_{i=1}^k(\lambda_{{k+1}}-\lambda_i)\left\{\left[2\left(1+\max\limits_{x\in\Omega}|x|^2\right)+\min\limits_{x\in\Omega}|x|^2\right]\lambda_i\right.\\ &\left.-\min\limits_{x\in\Omega}|x|^2\lambda_{k+1}-2\left(2+3\min\limits_{x\in\Omega}|x|^2\right)\right\}. \end{aligned} \end{equation} $

(1.8)

In this paper, on a bounded domain $ \Omega $ of the cigar soliton $ (\mathbb{R}^2, g, f) $, we consider the Dirichlet eigenvalue problem of $ \mathfrak{L}_{A, f} $ as follows

$ \begin{equation} \left\{ \begin{aligned} &\mathfrak{L}_{A, f}u+V u = \lambda\rho u, & &in\ \Omega, \\ &u=0, & &on\ \partial\Omega, \end{aligned} \right. \end{equation} $

(1.9)

where $ V $ is a non-negative continuous function and $ \rho $ is a positive continuous function on $ M $. We obtain the following results.

Theorem 1.1 Let $ \Omega $ be a bounded domain in the cigar soliton $ (\mathbb{R}^2, g, f) $. Let $ \lambda_{i} $ be the $ i $-th eigenvalue of problem (1.9). Assume that $ \xi_{1}I\leq A\leq\xi_{2}I $ throughout $ \Omega $, and $ \rho_{1}\leq \rho{(x)}\leq\rho_{2} $, $ \forall x\in\Omega $, where $ I $ is the identity map, $ \xi_{1} $, $ \xi_{2} $, $ \rho_{1} $, $ \rho_{2} $ are positive constants. Then we have

$ \begin{equation} \sum\limits_{i=1}^{k}(\lambda_{k+1}-\lambda_{i})^{2}\leq\frac{2\rho_2\xi_2^2}{\rho_{1}\xi_1(1+C_1)}\sum\limits_{i=1}^{k}(\lambda_{k+1}-\lambda_{i})\left[\frac{(1+C_0)(\lambda_{i}-\rho_{2}^{-1}V_{0})}{\xi_{1}}-\frac{2+3C_1}{\rho_{2}}\right], \end{equation} $

(1.10)

where $ C_0=\max\limits_{x\in\Omega}\left\{|x|^2\right\} $, $ C_1=\min\limits_{x\in\Omega}\left\{|x|^2\right\} $ and $ V_{0}=\min\limits_{x\in\Omega}\left\{V(x)\right\} $.

Remark 1.1 If $ A $ is an identity map, $ \rho(x)\equiv1 $ and $ V(x)\equiv0 $, then $ \xi_{1}=\xi_{2}=1 $, $ \rho_{1}=\rho_{2}=1 $ and $ V_{0}=0 $. Thus (1.10) becomes (1.8). Therefore, our result generalizes (1.8) of [15].

Moreover, we derive the following result for lower order eigenvalues of problem (1.9).

Theorem 1.2 Let $ \Omega $ be a bounded domain in the cigar soliton $ (\mathbb{R}^2, g, f) $. Let $ \lambda_{i} $ be the $ i $-th eigenvalue of problem (1.9). Assume that $ \xi_{1}I\leq A\leq\xi_{2}I $ throughout $ \Omega $, and $ \rho_{1}\leq \rho{(x)}\leq\rho_{2} $, $ \forall x\in\Omega $, where $ I $ is the identity map, $ \xi_{1} $, $ \xi_{2} $, $ \rho_{1} $, $ \rho_{2} $ are positive constants. Then we have

$ \begin{equation} \sum\limits_{p=1}^{2}(\lambda_{p+1}-\lambda_{1})^{\frac{1}{2}}\leq\frac{2\rho_2}{\rho_1(1+C_1)}\left\{2\xi_2(1+C_0)\left[\frac{(1+C_0)\left(\lambda_1-\rho_2^{-1}V_0\right)}{\xi_1}-\frac{2+3C_1}{\rho_2}\right]\right\}^{\frac{1}{2}}, \end{equation} $

(1.11)

where $ C_0=\max\limits_{x\in\Omega}\left\{|x|^2\right\} $, $ C_1=\min\limits_{x\in\Omega}\left\{|x|^2\right\} $ and $ V_{0}=\min\limits_{x\in\Omega}\left\{V(x)\right\} $.

Corollary 1.1 Let $ \Omega $ be a bounded domain in the cigar soliton $ (\mathbb{R}^2, g, f) $. Denote by $ \lambda_i $ the $ i $-th eigenvalue of problem (1.7). Then we have

$ \begin{equation} \sum\limits_{p=1}^{2}(\lambda_{p+1}-\lambda_{1})^{\frac{1}{2}}\leq\frac{2}{1+C_1}\bigg\{2\left(1+C_0\right)\bigg[\left(1+C_0\right)\lambda_1-\left(2+3C_1\right)\bigg]\bigg\}^{\frac{1}{2}}, \end{equation} $

(1.12)

where $ C_0=\max\limits_{x\in\Omega}\left\{|x|^2\right\} $ and $ C_1=\min\limits_{x\in\Omega}\left\{|x|^2\right\} $.

In this section, we give the proofs of the main results.

Proof of Theorem 1.1 Suppose that $ x^p $ is the $ p $-th local coordinate of $ x_{0}\in \Omega\subset\mathbb{R}^2 $, where $ p=1, 2 $. Consider the test functions

$ \begin{equation} \varphi_{i}=x^pu_{i}-\sum\limits_{j=1}^ka_{ij}u_{j}, \ \mbox{for} \ i=1, \ldots, k. \end{equation} $

(2.1)

where

$ \begin{equation} \notag a_{ij}=\int_{\Omega}\rho x^p u_iu_jd\mu. \end{equation} $

It is easy to find that

$ \begin{equation} \varphi_{i}|_{\partial\Omega}=0, \ \int_\Omega\rho\varphi_i u_jd\mu=0, \ \forall i, j=1, \ldots, k. \end{equation} $

(2.2)

Hence the Rayleigh-Ritz inequality reads as

$ \begin{equation} \lambda_{k+1}\int_\Omega\rho\varphi_i^2d\mu\leq\int_\Omega\varphi_i\big(\mathfrak{L}_{A, f}+V\big)\varphi_id\mu. \end{equation} $

(2.3)

According to the definition of $ \mathfrak{L}_{A, f} $, we have

$ \begin{equation} \begin{aligned} \mathfrak{L}_{A, f}(x^pu_{i})&=-\operatorname{div}_f\left(A\nabla(x^p u_{i})\right)\\ &=-e^f\operatorname{div}\left(e^{-f}\left(A\left(x^p\nabla u_{i}+u_{i}\nabla x^p\right)\right)\right)\\ &=-\operatorname{div}\left(A\left( x^p\nabla u_{i}+u_{i}\nabla x^p\right)\right)-\langle\nabla f, A\left(x^p\nabla u_{i}+u_{i}\nabla x^p\right)\rangle\\ &=-x^p\operatorname{div}_{f}\left(A\nabla u_{i}\right)-\langle\nabla x^p, A\nabla u_{i}\rangle-u_{i}\operatorname{div}_{f}(A\nabla x^p)-\left\langle\nabla u_{i}, A\nabla x^p\right)\\ &=x^p\mathfrak{L}_{A, f}u_i+u_i \mathfrak{L}_{A, f}x^p-2\langle\nabla x^p, A\nabla u_{i}\rangle. \end{aligned} \end{equation} $

(2.4)

Hence, we derive

$ \begin{equation} \begin{aligned} \int_\Omega\varphi_i\big(\mathfrak{L}_{A, f}+V\big)\varphi_id\mu &=\int_\Omega\varphi_i\left[\left(\mathfrak{L}_{A, f}+V\right)(x^pu_{i})-\rho\sum\limits_{j=1}^ka_{ij}\lambda_{j} u_{j}\right]d\mu\\ &=\int_\Omega\varphi_i\left(x^p\lambda_{i}\rho u_i+u_i \mathfrak{L}_{A, f}x^p-2\langle\nabla x^p, A\nabla u_{i}\rangle\right)d\mu\\ &=\lambda_{i}\int_\Omega\rho\varphi_i^2d\mu+P_i, \end{aligned} \end{equation} $

(2.5)

where $ P_i=\int_\Omega\varphi_i\left(u_i\mathfrak{L}_{A, f}x^p-2\langle\nabla x^p, A\nabla u_{i}\rangle\right)d\mu $. Substituting (2.5) into (2.3), we can get

$ \begin{equation} (\lambda_{k+1}-\lambda_i)\int_\Omega\rho\varphi_i^2d\mu\leq P_i. \end{equation} $

(2.6)

Set

$ \begin{equation} \notag b_{ij}=\int_\Omega\left(u_i \mathfrak{L}_{A, f}x^p-2\langle\nabla x^p, A\nabla u_{i}\rangle\right)u_jd\mu, \ \forall i, j=1, \ldots, k. \end{equation} $

Then we have

$ \begin{equation} \begin{aligned} b_{ij}&=\int_\Omega 2x^p\langle\nabla u_j, A\nabla u_{i}\rangle d\mu-\int_\Omega 2x^pu_j\mathfrak{L}_{A, f}u_id\mu+\int_\Omega x^p\mathfrak{L}_{A, f}\left(u_iu_j\right)d\mu\\ &=\int_\Omega x^p u_{i}\mathfrak{L}_{A, f}u_{j}d\mu-\int_\Omega x^p u_{j}\mathfrak{L}_{A, f}u_{i}d\mu\\ &=\int_\Omega x^p u_{i}\left(\mathfrak{L}_{A, f}+V\right)u_{j}d\mu-\int_\Omega x^p u_{j}\left(\mathfrak{L}_{A, f}+V\right)u_{i}d\mu\\ &=(\lambda_{j}-\lambda_{i})a_{ij}. \end{aligned} \end{equation} $

(2.7)

Using the Cauchy-Schwarz inequality, we obtain

$ \begin{equation} \begin{aligned} P_i^2& =\left[\int_\Omega\varphi_i\left(u_i \mathfrak{L}_{A, f}x^p-2\langle\nabla x^p, A\nabla u_{i}\rangle-\rho \sum\limits_{j=1}^k b_{ij}u_j\right)d\mu\right]^2\\ &\leq\int_\Omega\rho\varphi_i^2d\mu\cdot\int_\Omega\frac{1}{\rho}\left(u_i \mathfrak{L}_{A, f}x^p-2\langle\nabla x^p, A\nabla u_{i}\rangle-\rho \sum\limits_{j=1}^k b_{ij}u_j\right)^2d\mu. \end{aligned} \end{equation} $

(2.8)

Combining (2.6) and (2.8), we infer that

$ \begin{equation} \begin{aligned} (\lambda_{k+1}-\lambda_i)P_i^2&\leq(\lambda_{k+1}-\lambda_i)\int_\Omega\rho\varphi_i^2d\mu\cdot\int_\Omega\frac{1}{\rho}\left(u_i \mathfrak{L}_{A, f}x^p-2\langle\nabla x^p, A\nabla u_{i}\rangle-\rho \sum\limits_{j=1}^k b_{ij}u_j\right)^2d\mu\\ &\leq P_i\cdot\int_\Omega\frac{1}{\rho}\left(u_i \mathfrak{L}_{A, f}x^p-2\langle\nabla x^p, A\nabla u_{i}\rangle-\rho \sum\limits_{j=1}^k b_{ij}u_j\right)^2d\mu. \end{aligned} \end{equation} $

(2.9)

It implies that

$ \begin{equation} (\lambda_{k+1}-\lambda_i)P_i\leq\int_\Omega\frac{1}{\rho}\left(u_i \mathfrak{L}_{A, f}x^p-2\langle\nabla x^p, A\nabla u_{i}\rangle-\rho \sum\limits_{j=1}^k b_{ij}u_j\right)^2d\mu. \end{equation} $

(2.10)

Multiplying both sides of (2.10) by $ (\lambda_{k+1}-\lambda_i) $, taking the sum over $ i $ from 1 to $ k $ and $ p $ from 1 to 2, we get

$ \begin{equation} \begin{aligned} &\sum\limits_{i=1}^k(\lambda_{k+1}-\lambda_i)^2\sum\limits_{p=1}^2P_i \\ \leq&\sum\limits_{i=1}^k(\lambda_{k+1}-\lambda_i)\sum\limits_{p=1}^2\int_\Omega\frac{1}{\rho}\left(u_i \mathfrak{L}_{A, f}x^p-2\langle\nabla x^p, A\nabla u_{i}\rangle-\rho \sum\limits_{j=1}^k b_{ij}u_j\right)^2d\mu. \end{aligned} \end{equation} $

(2.11)

Using (2.7), we deduce

$ \begin{equation} \begin{aligned} &\sum\limits_{p=1}^2\int_\Omega\frac{1}{\rho}\left(u_i \mathfrak{L}_{A, f}x^p-2\langle\nabla x^p, A\nabla u_{i}\rangle-\rho\sum\limits_{j=1}^k b_{ij}u_j\right)^2d\mu\\ =&\sum\limits_{p=1}^2\int_\Omega\frac{1}{\rho}\left[(u_i \mathfrak{L}_{A, f}x^p)^2+4\langle\nabla x^p, A\nabla u_{i}\rangle^2-4u_i \mathfrak{L}_{A, f}x^p\langle\nabla x^p, A\nabla u_{i}\rangle\right]d\mu-2\sum\limits_{j=1}^k b_{ij}^2\\ =&\sum\limits_{p=1}^2\int_\Omega\frac{1}{\rho}\left[(u_i \mathfrak{L}_{A, f}x^p)^2+4\langle\nabla x^p, A\nabla u_{i}\rangle^2-4u_i \mathfrak{L}_{A, f}x^p\langle\nabla x^p, A\nabla u_{i}\rangle\right]d\mu-2\sum\limits_{j=1}^k(\lambda_i-\lambda_j)^2a_{ij}^2 \end{aligned} \end{equation} $

(2.12)

and

$ \begin{equation} \begin{aligned} \sum\limits_{p=1}^2P_i &=\sum\limits_{p=1}^2\int_\Omega\left(x^p u_i^2 \mathfrak{L}_{A, f}x^p-2x^p u_i\langle\nabla x^p, A\nabla u_{i}\rangle\right)d\mu-2\sum\limits_{j=1}^ka_{ij}b_{ij}\\ &=\sum\limits_{p=1}^2\int_\Omega u_i^2\langle\nabla x^p, A\nabla x^p\rangle d\mu+2\sum\limits_{j=1}^k(\lambda_i-\lambda_j)a_{ij}^2. \end{aligned} \end{equation} $

(2.13)

Using (2.12) and (2.13), we obtain

$ \begin{equation} \begin{aligned} &\sum\limits_{i=1}^k(\lambda_{k+1}-\lambda_i)^2 \sum\limits_{p=1}^2\int_\Omega u_i^2\langle\nabla x^p, A\nabla x^p\rangle d\mu+ 2 \sum\limits_{i, j=1}^k(\lambda_{k+1}-\lambda_i)^2 (\lambda_i-\lambda_j)a_{ij}^2\\ \leq&\sum\limits_{i=1}^k(\lambda_{k+1}-\lambda_i)\sum\limits_{p=1}^2\int_\Omega\frac{1}{\rho}\left[(u_i\mathfrak{L}_{A, f}x^p)^2+4\langle\nabla x^p, A\nabla u_{i}\rangle^2-4u_i \mathfrak{L}_{A, f}x^p\langle\nabla x^p, A\nabla u_{i}\rangle\right]d\mu\\ &-2\sum\limits_{i, j=1}^k(\lambda_{k+1}-\lambda_i)(\lambda_i-\lambda_j)^2a_{ij}^2. \end{aligned} \end{equation} $

(2.14)

Moreover, observe that

$ \begin{equation} \begin{aligned} &\sum\limits_{i, j=1}^k(\lambda_{k+1}-\lambda_i)^2(\lambda_i-\lambda_j)a_{ij}^2\\ =&\sum\limits_{i, j=1}^k(\lambda_{k+1}-\lambda_i)(\lambda_{k+1}-\lambda_j)(\lambda_i-\lambda_j)a_{ij}^2-\sum\limits_{i, j=1}^k(\lambda_{k+1}-\lambda_i)(\lambda_i-\lambda_j)^2a_{ij}^2\\ =&-\sum\limits_{i, j=1}^k(\lambda_{k+1}-\lambda_i)(\lambda_i-\lambda_j)^2a_{ij}^2. \end{aligned} \end{equation} $

(2.15)

Therefore, combining (2.14) with (2.15), we have

$ \begin{equation} \begin{aligned} &\sum\limits_{i=1}^k(\lambda_{k+1}-\lambda_i)^2 \sum\limits_{p=1}^2\int_\Omega u_i^2\langle\nabla x^p, A\nabla x^p\rangle d\mu\\ \leq&\sum\limits_{i=1}^k(\lambda_{k+1}-\lambda_i)\sum\limits_{p=1}^2\int_\Omega\frac{1}{\rho}\left[(u_i \mathfrak{L}_{A, f}x^p)^2+4\langle\nabla x^p, A\nabla u_{i}\rangle^2-4u_i \mathfrak{L}_{A, f}x^p\langle\nabla x^p, A\nabla u_{i}\rangle\right]d\mu. \end{aligned} \end{equation} $

(2.16)

Now it is necessary to calculate and estimate some terms in (2.16). It is not difficult to obtain

$ \begin{equation} \langle\nabla x^1, \nabla x^1\rangle=\langle\nabla x^2, \nabla x^2\rangle=1+|x|^2, \end{equation} $

(2.17)

$ \begin{equation} \langle\nabla x^1, \nabla x^2\rangle=0 \end{equation} $

(2.18)

and

$ \begin{equation} \Delta x^1=\Delta x^2=0. \end{equation} $

(2.19)

Using (2.17) and (2.18), and noticing

$ \begin{equation} \nabla(-\log(1+|x|^2)) =-\frac{1}{1+|x|^2} \nabla|x|^2 =-\frac{2}{1+|x|^2}\sum\limits_{q=1}^2 x^q \nabla x^q, \end{equation} $

(2.20)

we infer that

$ \begin{equation} \langle\nabla f, \nabla x^p\rangle =-\frac{2}{1+|x|^2} \sum\limits_{q=1}^2 x^q\langle\nabla x^q, \nabla x^p\rangle =-2x^p. \end{equation} $

(2.21)

It implies

$ \begin{equation} \langle\nabla f, \nabla |x|^2\rangle =-\frac{1}{1+|x|^2}\sum\limits_{p, q=1}^24x^px^q\langle\nabla x^p, \nabla x^q\rangle =-4|x|^2. \end{equation} $

(2.22)

Hence, (2.19) and (2.21) yield

$ \begin{equation} \Delta_fx^p=\Delta x^p-\langle\nabla f, \nabla x^p\rangle=2x^p. \end{equation} $

(2.23)

Furthermore, utilizing (2.17) and (2.19), we have

$ \begin{equation} \Delta|x|^2=\sum\limits_{p=1}^2\Delta(x^p)^2=\sum\limits_{p=1}^2\left(2x^p\Delta x^p+2\langle\nabla x^p, \nabla x^p\rangle\right)=4(1+|x|^2). \end{equation} $

(2.24)

Combining (2.22) and (2.24), we derive

$ \begin{equation} \Delta_f|x|^2=\Delta|x|^2-\langle\nabla f, \nabla |x|^2\rangle=4(1+2|x|^2). \end{equation} $

(2.25)

According to the assumptions of the theorem, we have

$ \begin{equation} \rho_2^{-1}\leq\int_\Omega u_i^2d\mu=\int_\Omega\frac{1}{\rho}\rho u_i^2d\mu\leq\rho_1^{-1}. \end{equation} $

(2.26)

Now we calculate the righthand side of (2.16). Since $ A\leq\xi_{2}I $, we can infer from (2.23) and (2.25) that

$ \begin{equation} \begin{aligned} \sum\limits_{p=1}^2\int_\Omega(u_i \mathfrak{L}_{A, f}x^p)^2d\mu \leq \xi_2^2\sum\limits_{p=1}^2\int_\Omega u_i^2(\Delta_fx^p)^2d\mu =4\xi_2^2\int_\Omega u_i^2|x|^2d\mu \end{aligned} \end{equation} $

(2.27)

and

$ \begin{equation} \begin{aligned} \sum\limits_{p=1}^2\int_\Omega\langle\nabla x^p, A\nabla u_{i}\rangle^2d\mu\leq \xi_2^2\sum\limits_{p=1}^2\int_\Omega\langle\nabla x^p, \nabla u_{i}\rangle^2d\mu =\xi_2^2\int_\Omega \left(1+|x|^2\right)|\nabla u_i|^2d\mu. \end{aligned} \end{equation} $

(2.28)

Moreover, from

$ \begin{equation} \notag \begin{aligned} \sum\limits_{p=1}^2\int_\Omega u_i\Delta_fx^p\langle\nabla x^p, \nabla u_{i}\rangle d\mu &=\frac{1}{2}\int_\Omega\langle\nabla |x|^2, \nabla u_i^2\rangle d\mu =-\frac{1}{2}\int_\Omega u_i^2 \Delta_f |x|^2d\mu, \end{aligned} \end{equation} $

we obtain

$ \begin{equation} \begin{aligned} \sum\limits_{p=1}^2\int_\Omega\left(-u_i \mathfrak{L}_{A, f}x^p\langle\nabla x^p, A\nabla u_{i}\rangle\right)d\mu &\leq-2\xi_2^2\int_\Omega u_i^2 \left(1+2|x|^2\right)d\mu. \end{aligned} \end{equation} $

(2.29)

Using (2.26) and noticing that $ A\geq\xi_1I $, we have

$ \begin{equation} \notag \lambda_i=\int_\Omega u_i \left(\mathfrak{L}_{A, f}+V\right)u_id\mu =\int_\Omega\langle\nabla u_i, A\nabla u_i\rangle d\mu+\int_\Omega Vu_i^2d\mu \geq\xi_1\int_\Omega |\nabla u_i|^2d\mu+\rho_2^{-1}V_0. \end{equation} $

It implies that

$ \begin{equation} \int_\Omega |\nabla u_i|^2d\mu\leq \frac{\lambda_i-\rho_2^{-1}V_0}{\xi_1}. \end{equation} $

(2.30)

Then it follows from (2.27–2.30) that

$ \begin{equation} \begin{aligned} &\sum\limits_{p=1}^2\int_\Omega\frac{1}{\rho}\left[(u_i \mathfrak{L}_{A, f}x^p)^2+4\langle\nabla x^p, A\nabla u_{i}\rangle^2-4u_i \mathfrak{L}_{A, f}x^p\langle\nabla x^p, A\nabla u_{i}\rangle\right]d\mu\\ \leq&4\xi_2^2\int_\Omega\frac{1}{\rho}\left[\left(1+|x|^2\right)|\nabla u_i|^2-\left(2+3|x|^2\right)u_i^2\right]d\mu\\ \leq&\frac{4\xi_2^2}{\rho_1}\left[\frac{(1+C_0)(\lambda_{i}-\rho_{2}^{-1}V_{0})}{\xi_{1}}-\frac{2+3C_1}{\rho_{2}}\right]. \end{aligned} \end{equation} $

(2.31)

Moreover, we acquire

$ \begin{equation} \sum\limits_{p=1}^2\int_\Omega u_i^2\langle\nabla x^p, A\nabla x^p\rangle d\mu \geq 2\xi_1\int_\Omega u_i^2\left(1+|x|^2\right)d\mu \geq2\xi_1 \frac{1+C_1}{\rho_2}. \end{equation} $

(2.32)

It follows from (2.16), (2.31) and (2.32) that

$ \begin{equation} \begin{aligned} &2\xi_1 \frac{1+C_1}{\rho_2} \sum\limits_{i=1}^k(\lambda_{k+1}-\lambda_i)^2 \\ \leq&\frac{4\xi_2^2}{\rho_1}\sum\limits_{i=1}^k(\lambda_{k+1}-\lambda_i)\left[\frac{(1+C_0)(\lambda_{i}-\rho_{2}^{-1}V_{0})}{\xi_{1}}-\frac{2+3C_1}{\rho_{2}}\right]. \end{aligned} \end{equation} $

(2.33)

Hence we can know that (1.10) holds. This completes the proof of Theorem 1.1.

Now we give the proof of Theorem 1.2.

Proof of Theorem 1.2 Define a $ 2\times2 $ matrix $ C=(C_{ps}) $, where $ C_{ps}=\int_\Omega \rho x^pu_1u_{s+1}d\mu $. Using the orthogonalization of Gram-Schmidt, we know that there exist an upper triangle matrix $ R=(R_{ps}) $ and an orthogonal matrix $ T=(T_{ps}) $ such that $ R=TC $. That is to say, for $ 1\leq s<p\leq2 $, we have

$ \begin{equation} \notag R_{ps}=\sum\limits_{k=1}^2T_{pk}C_{ks}=\int_\Omega\sum\limits_{k=1}^2T_{pk}\rho x^ku_1u_{s+1}d\mu=0. \end{equation} $

Setting $ y^p=\sum\limits_{k=1}^2T_{pk}x^k $, we get

$ \begin{equation} \int_\Omega \rho y^pu_1u_{s+1}d\mu=0, \ \ \mbox{for} \ 1\leq s<p \leq2. \end{equation} $

(2.34)

For $ p=1, 2 $, define the test functions $ \varphi_p $ by

$ \begin{equation} \varphi_p=y^pu_1-a_pu_1, \end{equation} $

(2.35)

where

$ \begin{equation} \notag a_p=\int_\Omega\rho y^pu_1^2d\mu. \end{equation} $

Since (2.34) holds, it yields

$ \begin{equation} \int_\Omega\rho\varphi_p u_{s+1}d\mu=0, \ \ \mbox{for} \ 0\leq s<p \leq2. \end{equation} $

(2.36)

According to the Rayleigh-Ritz inequality, we have

$ \begin{equation} \lambda_{p+1}\int_\Omega\rho\varphi_p^2d\mu\leq\int_\Omega\varphi_p\big(\mathfrak{L}_{A, f}+V\big)\varphi_pd\mu. \end{equation} $

(2.37)

It follows from (2.36) that

$ \begin{equation} \int_\Omega\rho\varphi_p^2d\mu =\int_\Omega\rho\varphi_py^pu_1d\mu-a_p\int_\Omega\rho\varphi_pu_1d\mu =\int_\Omega\rho\varphi_py^pu_1d\mu. \end{equation} $

(2.38)

Similar to the proof of (2.4), we acquire

$ \begin{equation} \mathfrak{L}_{A, f}(y^pu_1)=y^p\mathfrak{L}_{A, f}u_1+u_1 \mathfrak{L}_{A, f}y^p-2\langle\nabla y^p, A\nabla u_{1}\rangle. \end{equation} $

(2.39)

Combining (2.38) and (2.39), we have

$ \begin{equation} \begin{aligned} \int_\Omega\varphi_p\big(\mathfrak{L}_{A, f}+V\big)\varphi_pd\mu &=\lambda_1\int_\Omega\rho\varphi_p^2d\mu+\int_\Omega y^pu_1\left(u_1 \mathfrak{L}_{A, f}y^p-2\langle\nabla y^p, A\nabla u_{1}\rangle\right)d\mu. \end{aligned} \end{equation} $

(2.40)

At the same time, using

$ \begin{equation} \notag -2\int_\Omega y^pu_1\langle\nabla y^p, A\nabla u_1\rangle d\mu=\int_\Omega u_1^2\langle\nabla y^p, A\nabla y^p\rangle d\mu-\int_\Omega y^pu_1^2\mathfrak{L}_{A, f}y^pd\mu, \end{equation} $

we obtain

$ \begin{equation} \int_\Omega y^pu_1\left(u_1 \mathfrak{L}_{A, f}y^p-2\langle\nabla y^p, A\nabla u_{1}\rangle\right)d\mu=\int_\Omega u_1^2\langle\nabla y^p, A\nabla y^p\rangle d\mu. \end{equation} $

(2.41)

Substituting (2.40) and (2.41) into (2.37), we deduce

$ \begin{equation} \left(\lambda_{p+1}-\lambda_1\right)\int_\Omega\rho\varphi_p^2d\mu\leq\int_\Omega u_1^2\langle\nabla y^p, A\nabla y^p\rangle d\mu. \end{equation} $

(2.42)

Observing that

$ \begin{equation} \notag \int_\Omega u_1\left(\langle\nabla u_1, \nabla y^p\rangle+\frac{1}{2}u_1\Delta_fy^p\right)d\mu=0, \end{equation} $

we infer

$ \begin{equation} \begin{aligned} -2\int_\Omega\varphi_p\left(\langle\nabla u_1, \nabla y^p\rangle+\frac{1}{2}u_1\Delta_fy^p\right)d\mu =&-2\int_\Omega y^pu_1\left(\langle\nabla u_1, \nabla y^p\rangle+\frac{1}{2}u_1\Delta_fy^p\right)d\mu\\ =&\int_\Omega u_1^2|\nabla y^p|^2d\mu. \end{aligned} \end{equation} $

(2.43)

Therefore, using (2.42) and (2.43), and summing over $ p $ from 1 to 2, we derive

$ \begin{equation} \begin{aligned} &\sum\limits_{p=1}^2(\lambda_{p+1}-\lambda_1)^{\frac{1}{2}}\int_\Omega u_1^2|\nabla y^p|^2d\mu \\ =&-2\sum\limits_{p=1}^2(\lambda_{p+1}-\lambda_1)^{\frac{1}{2}}\int_\Omega\varphi_p\left(\langle\nabla u_1, \nabla y^p\rangle+\frac{1}{2}u_1\Delta_fy^p\right)d\mu \\ \leq& \delta \sum\limits_{p=1}^2(\lambda_{p+1}-\lambda_1)\int_\Omega\rho\varphi_p^2d\mu +\frac{1}{\delta}\sum\limits_{p=1}^2\int_\Omega\frac{1}{\rho}\left(\langle\nabla u_1, \nabla y^p\rangle+\frac{1}{2}u_1\Delta_fy^p\right)^2d\mu \\ \leq&\delta\sum\limits_{p=1}^2\int_\Omega u_1^2\langle\nabla y^p, A\nabla y^p\rangle d\mu +\frac{1}{\delta} \sum\limits_{p=1}^2\int_\Omega\frac{1}{\rho}\left(\langle\nabla u_1, \nabla y^p\rangle+\frac{1}{2}u_1\Delta_fy^p\right)^2d\mu, \end{aligned} \end{equation} $

(2.44)

where $ \delta $ is any positive constant.

Since $ y^p=\sum\limits_{k=1}^2T_{pk}x^k $ and $ T $ is an orthogonal matrix, we know that $ y^1 $ and $ y^2 $ are the standard coordinate functions of $ \mathbb{R}^2 $. It is not difficult to check that

$ \begin{equation} \notag |y|^2=|x|^2, \end{equation} $

$ \begin{equation} |\nabla y^p|^2=1+|x|^2, \end{equation} $

(2.45)

$ \begin{equation} \Delta_fy^p=2y^p \end{equation} $

(2.46)

and

$ \begin{equation} \Delta_f|y|^2=4(1+2|x|^2). \end{equation} $

(2.47)

Noticing that $ \rho_2^{-1}\leq\int_\Omega u_1^2d\mu\leq\rho_1^{-1} $, and using (2.45), we obtain

$ \begin{equation} \begin{aligned} \sum\limits_{p=1}^2(\lambda_{p+1}-\lambda_1)^{\frac{1}{2}}\int_\Omega u_1^2|\nabla y^p|^2d\mu&=\sum\limits_{p=1}^2(\lambda_{p+1}-\lambda_1)^{\frac{1}{2}}\int_\Omega u_1^2\left(1+|x|^2\right)d\mu\\ &\geq\frac{1+C_1}{\rho_2}\sum\limits_{p=1}^2(\lambda_{p+1}-\lambda_1)^{\frac{1}{2}} \end{aligned} \end{equation} $

(2.48)

and

$ \begin{equation} \begin{aligned} \sum\limits_{p=1}^2\int_\Omega u_1^2\langle\nabla y^p, A\nabla y^p\rangle d\mu\leq2\xi_2\int_\Omega u_1^2\left(1+|x|^2\right)d\mu \leq2\xi_2 \frac{1+C_0}{\rho_1}. \end{aligned} \end{equation} $

(2.49)

Similar to the proof of (2.30), we have

$ \begin{equation} \int_\Omega |\nabla u_1|^2d\mu\leq \frac{\lambda_1-\rho_2^{-1}V_0}{\xi_1}. \end{equation} $

(2.50)

Then it follows from (2.46), (2.47) and (2.50) that

$ \begin{equation} \begin{aligned} &\sum\limits_{p=1}^2\int_\Omega\frac{1}{\rho}\left(\langle\nabla u_1, \nabla y^p\rangle+\frac{1}{2}u_1\Delta_fy^p\right)^2d\mu\\ =&\int_\Omega\frac{1}{\rho}\left[\left(1+|x|^2\right)|\nabla u_1|^2-\frac{1}{2}u_1^{2}\Delta_f|y|^2+u_1^{2}|y|^2\right]d\mu \\ =&\int_\Omega\frac{1}{\rho}\left[\left(1+|x|^2\right)|\nabla u_1|^2-\left(2+3|x|^2\right)u_1^{2}\right]d\mu \\ \leq&\frac{1}{\rho_1}\left[\frac{(1+C_0)(\lambda_1-\rho_2^{-1}V_0)}{\xi_1}-\frac{2+3C_1}{\rho_2}\right]. \end{aligned} \end{equation} $

(2.51)

Substituting (2.48), (2.49) and (2.51) into (2.44), we get

$ \begin{equation} \frac{1+C_1}{\rho_2}\sum\limits_{p=1}^2(\lambda_{p+1}-\lambda_1)^{\frac{1}{2}} \leq\delta\left[\frac{2\xi_2\left(1+C_0\right)}{\rho_1}\right]+\frac{1}{\delta}\left\{\frac{1}{\rho_1}\left[\frac{(1+C_0)(\lambda_1-\rho_2^{-1}V_0)}{\xi_1}-\frac{2+3C_1}{\rho_2}\right]\right\}. \end{equation} $

(2.52)

Taking

$ \begin{equation} \notag \delta=\dfrac{\left\{\dfrac{1}{\rho_1}\left[\dfrac{(1+C_0)(\lambda_1-\rho_2^{-1}V_0)}{\xi_1}-\dfrac{2+3C_1}{\rho_2}\right]\right\}^{\frac{1}{2}}}{\left[\dfrac{2\xi_2\left(1+C_0\right)}{\rho_1}\right]^{\frac{1}{2}}} \end{equation} $

in (2.52), we obtain (1.11). This finishes the proof of Theorem 1.2.