In 1977, Reilly [19] obtained a sharp lower bound for the first Dirichlet eigenvalue of the Laplacian as following.

Reilly's Theorem   Let $M$ be an $n(\geq 2)$-dimensional compact Riemannian manifold with Ricci curvature bounded below by $(n-1)\kappa > 0$ and boundary. If the mean curvature of $\partial M$ is nonnegative then the first Dirichlet eigenvalue of the Laplacian of $M$ satisfies

$ \begin{eqnarray*} \lambda_1\geq n\kappa, \end{eqnarray*} $

and equality holds if and only $M$ is isometric to an $n$-dimensional Euclidean semi-sphere of radius $\frac{1}{\sqrt{\kappa}}$.}

For the Neumann boundary case a similar result has been proven by Escobar [10] and Xia [25] independently. In [13], Li-Wei generalize these results including Dirichlet boundary and Neumann boundary to the first eigenvalue of the drifting Laplacian ${\Delta _f}:=\Delta-\langle\nabla f, \nabla\rangle$ on compact smooth metric measure spaces $(M, \langle, \rangle, e^{-f}dv)$. Here $(M, \langle, \rangle)$ is $n$-dimensional Riemannian manifold with a metric $\langle, \rangle$, $f$ is a smooth real-valued function on $M$ and $dv$ is the Riemannian volume element related to $\langle, \rangle$ (sometimes, we also call $dv$ the volume density). The drifting Laplacian is also called weighted Laplacian (or Witten Lpalacian), some interesting results concerning eigenvalues of the drifting Laplacian can be found in [7-9, 11, 13, 17, 26]. Compared with Ricci curvature of the Riemannnian manifods, we can define the so-called weighted Ricci curvature $\mathrm{Ric}^{f}$ on smooth metric measure spaces as following

$ \begin{eqnarray*} \mathrm{Ric}^{f}=\mathrm{Ric}_M+\mathrm{Hess}f, \end{eqnarray*} $

which is also called the $\infty$-Bakry-Émery Ricci tensor. The equation $\mathrm{Ric}^f=\kappa\left\langle , \right\rangle $ for some constant $\kappa$ is just the gradient Ricci soliton equation, which plays an important role in the study of Ricci flow. For $\kappa= 0, \kappa> 0, \mathrm{and}~\kappa< 0$, the gradient Ricci soliton $(M, \langle, \rangle, e^{-f}dv, \kappa)$ is called steady, shrinking, and expanding, respectively. Set

$ {\rm{Ri}}{{\rm{c}}^m} = {\rm{Ri}}{{\rm{c}}^f} - \frac{1}{{m - n}}\nabla f \otimes \nabla f, (m > n), $

which is called $m$-weighted Ricci curvature [1] of $M$ (also called the $m$-Bakry-\'Emery Ricci tensor). When $m=n$, let $f$ be a constant and $\mathrm{Ric}^m=\mathrm{Ric}_M$. A smooth metric measure space is not necessarily compact when $\mathrm{Ric}^f\geq\lambda>0$, unlike in the case of Riemannian manifolds where such a complete one is compact if its Ricci curvature is bounded from below uniformly by some positive constant [21], but a interesting fact is that when $\mathrm{Ric}^m\geq(m-1)\kappa>0$, a complete smooth metric measure space $(M, \langle, \rangle, e^{-f}dv)$ is automatically compact and the diameter of $M$ satisfies $diam(M)\leq\pi/\sqrt{\kappa}$ [18].

In [5], Chen-Cheng-Wang-Xia gave some lower bounds for the first eigenvalue of four kinds of eigenvalue problems of the biharmonic operator on compact manifolds with boundary and positive Ricci curvature. In [6], Du-Bezerra generalize these results to the bi-drifting Laplacian on compact manifolds with boundary and positive $m$-weighted Ricci curvature. Recently, Wang-Xia [24] give some new sharp lower bounds for the first eigenvalues of the biharmonic operator on compact manifolds with boundary and positive Ricci curvature. Inspired by these investigations, in this paper, we firstly get

Theorem 1.1   Let $(M, \langle, \rangle, e^{-f}dv)$ be an $n(\geq 2)$-dimensional smooth metric measure space with boundary $\partial M$ and denote by $\nu$ the outward unit normal vector field of $\partial M$. Assume that the $m$-weighted Ricci curvature of $M$ is bounded below by $(m-1)\kappa>0$. Let $\lambda_1$ be the first nonzero eigenvalue with Dirichlet boundary condition of the drifting Laplacian of $M$ and let $\Gamma_1$ be the first eigenvalue of the clamped plate problem on $M$,

$ \begin{eqnarray}\Delta _f^2g =\Gamma_1 g, ~~\mbox{in} ~~M, ~~~~ g=\frac{\partial^2 g}{\partial \nu^2}=0, ~~\mbox{on}~~\partial M. \end{eqnarray} $

(1.2)

Then we have

$ \begin{eqnarray} \Gamma_1\geq\lambda_1(\frac{1}{m}\lambda_1+(m-1)\kappa), \end{eqnarray} $

(1.3)

with equality holding if and only if $M$ is isometric to an n-dimensional Euclidean unit semi-sphere of radius of $1/\sqrt{\kappa}$ and $f$ is constant.

Theorem 1.2   Under the assumption of Theorem 1.2, Let $\Lambda_1$ be the first eigenvalue of the following buckling problem

$ \Delta _f^2\phi = - {\Lambda _1}{\rm{ }}{\Delta _f}\phi , {\rm{in}}M, \phi = \frac{{\partial \phi }}{{\partial \nu }} = 0, {\rm{on}}\partial M.{\rm{ }} $

(1.4)

Then we have

$ \begin{eqnarray}\Lambda_1\geq\frac{1}{m}\lambda_1+(m-1)\kappa, \end{eqnarray} $

(1.5)

with equality holding if and only if $M$ is isometric to an n-dimensional Euclidean unit semi-sphere of radius of $1/\sqrt{\kappa}$ and $f$ is constant.

Remark 1.3   Compared with Theorem 1.7 and 1.8 in [6], we don't need to assume that the weighted mean curvature of $\partial M$ is bounded in Theorem 1.1 and 1.2.

The study of Steklov eigenvalue problem was started by Steklov [20], his motivation came from physics. From then on, for the Steklov eigenvalue problem, many interesting results have been obtained in [2, 3, 15, 16, 22, 23]. Recently, some results for the Steklov eigenvalues of the drifting Laplacian have been given in [4, 12]. In this paper, we consider the following two fourth order Steklov eigenvalue problems for the drifting Laplacian

$ \Delta _f^2u = 0\;{\rm{in}}M, u = {\Delta _f}u - p\frac{{\partial u}}{{\partial \nu }} = 0{\rm{on}}\partial M $

(1.6)

and

$ \Delta _f^2u = 0{\rm{in}}M, u = \frac{{{\partial ^2}u}}{{\partial {\nu ^2}}} - q\frac{{\partial u}}{{\partial \nu }} = 0{\rm{on}}\partial M $

(1.7)

where $\nu$ denotes the outward unit normal vector field of $\partial M$. Then, we obtain

Theorem 1.4   Let $M$ be an $n(\geq2)$-dimensional compact connected smooth metric measure space with boundary $\partial M$ and non-negative $m$-weighted Ricci curvature. Let $p_1, q_1$ be the first eigenvalue of the problem (1.6) and (1.7), respectively. Then we have

$ \begin{eqnarray}\label{1.7}q_1\geq \frac{p_1}{m}, \end{eqnarray} $

(1.8)

the equality holds in (1.8) if and only if $M$ is isometric to an Euclidean ball and $f$ is constant.

Theorem1.5   Let $M$ be an $n(\geq2)$-dimensional compact connected smooth metric measure space with boundary $\partial M$. Assume that $m$-weighted Ricci curvature is bounded below by $-(m-1)\kappa$ for some nonnegative constant $\kappa$, and the weighted mean curvature of $M$ is bounded below by $\frac{(m-1)c}{n-1}$ with a positive constant $c$. Let $\lambda_1$ be the first Dirichlet eigenvalue of the deifting Laplacian, and let $p_1$ is the first eigenvalue of the Steklov problem (1.6), then we have

$ \begin{eqnarray}\label{x7}p_1\geq\frac{mc\lambda_1}{\lambda_1+m\kappa}, \end{eqnarray} $

(1.9)

the equality holds in (1.9) if and only if $M$ is isometric to an Euclidean ball of radius $1/c$ and $f$ is constant.

In this section, we will give the proofs of Theorem 1.1-Theorem 1.2 listed in Section 1. Before proving these results, we first recall some notations. Let $M$ be an $n$-dimensional compact manifold with boundary. We will often write $\left\langle , \right\rangle $ the Riemannian metric on $M$ as well as that induced on $\partial M$. Let $\nabla $ and ${\Delta _f}$ be the connection and the drifting Laplacian on $M$, respectively. Let $\nu$ be the unit outward normal vector of $\partial M$. The shape operator of $\partial M$ is given by $S(X) = \nabla _X\nu$ and the second fundamental form of $\partial M$ is defined as $II(X, Y) = \left\langle S(X), Y\right\rangle$, here $X, Y \in T\partial M$. The eigenvalues of $S$ are called the principal curvatures of $\partial M$ and the mean curvature $H$ of $\partial M$ is given by $H =\frac{1}{n-1} tr S$, here $tr S$ denotes the trace of $S$. We can now state Reilly-type formula (see \cite[Theorem 1]{MD}). For a smooth function $g$ defined on $M$, the following identity holds if $h =\frac{\partial g}{\partial\nu}\big|_{\partial M}, z = g\big|_{\partial M}$

$ \begin{array}{l} \int {_M} ({({\Delta _f}g)^2} - |{\nabla ^2}g{|^2} - {\rm{Ri}}{{\rm{c}}^f}(\nabla g, \nabla g))d\mu = \\ \int {_{\partial M}} ((n - 1){H^f}h + 2\overline {{\Delta _f}} z)hd\mu + \int {_{\partial M}} (II(\overline \nabla z, \overline \nabla z) - \left\langle {\overline \nabla z, \overline \nabla \frac{{\partial z}}{{\partial \nu }}} \right\rangle )d\mu , \end{array} $

(2.1)

Here $ H^f = H+\frac{1}{n-1}\frac{\partial f}{\partial\nu} $ denotes $ f $-mean curvature [21](also called weighted mean curvature), $ \nabla^2g $ is the Hessian of $ g $; $ \bar{\Delta}, \overline{\Delta_{f}} $ and $ \overline{\nabla} $ represent the Laplacian, drifting Laplacian and the gradient on $ \partial M $ with respect to the induced metric on $ \partial M $, respectively. Using the inequalities of $ |\nabla^2 g|^2\geq\frac{1}{n}(\Delta g)^2 $ and $ \frac{1}{n}(a+b)^2\geq\frac{1}{m}a^2-\frac{1}{m-n}b^2 $, then from $ \[{\Delta _f}g = \Delta g - \Lambda \left\langle {\nabla f, \nabla g} \right\rangle \] $, we can get

$ |{\nabla ^2}g{|^2} \ge \frac{1}{n}{(\Delta g)^2} = \frac{1}{n}{({\nabla _f}g + \left\langle {\nabla f, \nabla g} \right\rangle )^2} \ge \frac{{{{({\nabla _f}g)}^2}}}{m} - \frac{{{{\left\langle {\nabla f, \nabla g} \right\rangle }^2}}}{{m - n}}. $

(2.2)

Substituting (2.2) into (2.1), and it then follows from the definition of the $m$-Bakry-\'Emery Ricci tensor that

$ \int {_M} (\frac{{m - 1}}{m}{({\Delta _f}g)^2} - {\rm{Ri}}{{\rm{c}}^m}(\nabla g, \nabla g))d\mu \\ \ge \int {_{\partial M}} ((n - 1){H^\phi }h + 2\overline {{\Delta _f}} z)hd\mu + \int {_{\partial M}} (II(\overline \nabla z, \overline \nabla z) - \left\langle {\overline \nabla z, \overline \nabla \frac{{\partial z}}{{\partial \nu }}} \right\rangle )d\mu . $

(2.3)

Remark 2.6   (i) When $m=n$, we know that $f$ is a constant, then ${\Delta _f}=\Delta$, (2.1) becomes classic Reilly formula [19], the equality holds in (2.3) if and only if ${\nabla ^2}g = \frac{{\Delta g}}{n}\left\langle , \right\rangle $.

(ii) When $m>n$, the equality holds in (2.3) if and only if ${\nabla ^2}g = \frac{{\Delta g}}{n}\left\langle , \right\rangle $ and ${\Delta _f}g + \frac{m}{{m - n}}\left\langle {\nabla f, \nabla g} \right\rangle = 0$.}\end{remark} Using the inequality (2.3), following, we will give the proofs of Theorem 1.1-Theorem 1.2.

Proof of Theorem 1.1   Let $g$ be an eigenfunction of the problem (1.2) corresponding to the first eigenvalue $\Gamma_1$. That is

$ \begin{eqnarray} \Delta_{f}^2 g = \Gamma_1 g\; \; \mathrm{in}\; \; M, \; \; \; \; \; g = \frac{\partial^2 g}{\partial\nu^2} = 0\; \; \; \mathrm{on} \; \; \partial M. \end{eqnarray} $

(2.4)

Multiplying (2.4) by $g$ and integrating on $M$, we infer from the divergence theorem that

$ \begin{eqnarray} \Gamma_1\int_M g^2d\mu = \int_M g\Delta_{f}^2 gd\mu =\\ -\int_M\langle \nabla g, \nabla(\Delta_{f} g)\rangle d\mu = \int_M(\Delta_{f} g)^2d\mu -\int_{\partial M} h\Delta_{f} gd\mu, \end{eqnarray} $

(2.5)

where $ h = \frac{\partial g}{\partial\nu}\Big|_{\partial M} $. On the boundary $ \partial M $, we can write

$ \nabla g = (\nabla g)^{\top}+(\nabla g)^{\bot} = (\nabla g)^{\top}+\langle\nabla g, \nu\rangle\nu, $

where $ (\nabla g)^{\top} $ is tangent to $ \partial M $ and $ (\nabla g)^{\bot} $ is normal to $ \partial M $. Then from $ g\Big|_{\partial M} = \frac{\partial^2 g}{\partial\nu^2}\Big|_{\partial M} = 0 $, we have

$ \begin{eqnarray*} \langle\nabla f, \nabla g\rangle\Big|_{\partial M}& = &\langle(\nabla f)^\top+\langle\nabla f, \nu\rangle\nu, (\nabla g)^{\top}+\langle\nabla g, \nu\rangle\nu\rangle\Big|_{\partial M}\\&& = \langle(\nabla f)^\top, (\nabla g)^\top\rangle\Big|_{\partial M}+\frac{\partial f}{\partial\nu} h = \frac{\partial f}{\partial\nu} h, \end{eqnarray*} $

and

$ \begin{eqnarray*} \Delta g\Big|_{\partial M} = (n-1) H h. \end{eqnarray*} $

So, we infer from above two equalities that

$ \begin{eqnarray} \Delta_{f} g\Big|_{\partial M} = (\Delta g-\langle \nabla f, \nabla g \rangle)\Big|_{\partial M} = (n-1) H h-\frac{\partial f}{\partial\nu} h = (n-1)H^f h. \end{eqnarray} $

(2.6)

Combining (2.5) and (2.6), we have

$ \begin{eqnarray} \Gamma_1 = \frac{\int_M(\Delta_{f} g)^2d\mu -\int_{\partial M} (n-1)H^f h^2d\mu}{\int_M g^2d\mu}. \end{eqnarray} $

(2.7)

Taking $g$ into (2.3), we have

$ \begin{eqnarray*} \nonumber\int_M \frac{m-1}{m}(\Delta_{f} g)^2d\mu&\geq&\int_M\mathrm{Ric}^m(\nabla g, \nabla g)d\mu+\int_{\partial M} (n-1)H^f h^2d\mu \\&\geq&(m-1)\kappa\int_M |\nabla g|^2d\mu+\int_{\partial M} (n-1)H^f h^2d\mu, \end{eqnarray*} $

which implies

$ \begin{eqnarray} -\int_{\partial M} (n-1) H^f h^2d\mu&\geq&-\int_M \frac{m-1}{m}(\Delta_{f} g)^2+(m-1)\kappa \int_M |\nabla g|^2d\mu. \end{eqnarray} $

(2.8)

Combining (2.7) and (2.8), we have

$ \begin{eqnarray} \Gamma_1\geq\frac{\frac{1}{m}\int_M(\Delta_{f} g)^2d\mu +(m-1)\kappa \int_M |\nabla g|^2d\mu}{\int_M g^2d\mu}. \end{eqnarray} $

(2.9)

For any nonzero function which vanishes on $\partial M$, it is well known from the Poincaré inequality and the Schwarz inequality that

$ \begin{eqnarray*} \lambda_1 \int_M g^2d\mu\leq -\int_M g\Delta_{f} gd\mu\leq\left\{\int_M g^2d\mu\int_M(\Delta_{f} g)^2d\mu\right\}^\frac{1}{2}. \end{eqnarray*} $

which implies that

$ \begin{eqnarray} \lambda_1^2 \int_M g^2d\mu\leq\lambda_1\int_M |\nabla g|^2d\mu\leq\int_M(\Delta_{f} g)^2d\mu. \end{eqnarray} $

(2.10)

The equality hold in above inequality if and only if $g$ is the a first Dirichlet eigenfunction of the drifting Laplacian of $M$.

Combining (2.9) and (2.10), we have

$ \begin{eqnarray} \Lambda_1\geq\lambda_1(\frac{1}{m}\lambda_1+(m-1)\kappa). \end{eqnarray} $

(2.11)

If $\Lambda_1=\lambda_1(\frac{1}{m}\lambda_1+(m-1)\kappa)$, we know that equality holds in (2.10), which implies that $g$ is the a first Dirichlet eigenfunction of the drifting Laplacian of $M$, then we have

$ \begin{eqnarray} \Delta_{f} g = -\lambda_1 g, \; \; \mathrm{and} \; \; \; \Delta_{f}^2 g = \lambda_1^2 g, \; \; \; \mathrm{in}\; M. \end{eqnarray} $

(2.12)

Then, we infer from $\Delta_{f}^2 g=\Gamma_1 g$ in $M$ that $\lambda_1=m\kappa>0$.

When $m=n$, then $f$ is a constant and $\Delta_{f}=\Delta$, from [24, Theorem 1.1], we know that $\Gamma_1\geq\lambda_1(\frac{1}{n}\lambda_1+(n-1)\kappa)$ with equality holding if and only if $M$ is isometric to an $n$-dimensional Euclidean unit semi-sphere.

When $m>n$, if $\Gamma_1=\lambda_1(\frac{1}{m}\lambda_1+(m-1)\kappa)$, we know that equality holds in (2.3), which means

$ 0 = \Delta_{f} g+\frac{m}{m-n}\langle\nabla f, \nabla g\rangle = \Delta g+ \frac{n}{m-n}\langle\nabla f, \nabla g\rangle $

holds everywhere on $M$. Multiplying in above inequality with $g$ and integrating on $M$ with respect to $e^{\frac{n}{m-n}f}dv$ give that

$ \begin{eqnarray} 0 = \int_M g(\Delta g+ \frac{n}{m-n}\langle\nabla f, \nabla u\rangle)e^{\frac{n}{m-n}f}d\nu = -\int_M |\nabla g|^2e^{\frac{n}{m-n}f}d\nu. \end{eqnarray} $

(2.13)

From above equality, we know that $g$ is a constant function on $M$, which is a contradiction since $g$ is the first eigenfunction of drifting Laplacian and cannot be a constant. Therefore, we know that $\Lambda_1\geq\lambda_1(\frac{1}{m}\lambda_1+(m-1)\kappa)$ with equality holding if and only if $M$ is isometric to an $n$-dimensional Euclidean unit semi-sphere and $f$ is a constant. This completes the proof of Theorem 1.1.

Proof of Theorem 1.2   Let $\phi$ be an eigenfunction of the problem (1.4) corresponding to the first eigenvalue $\Lambda_1$. That is

$ \begin{eqnarray} \Delta_{f}^2 \phi = -\Lambda_1 \Delta_{f} \phi\; \; \mathrm{in}\; \; M, \; \; \; \; \; \phi = \frac{\partial^2 \phi}{\partial\nu^2} = 0\; \; \; \mathrm{on} \; \; \partial M. \end{eqnarray} $

(2.14)

Multiplying (2.14) by $\phi$ and integrating on $M$, we infer from the divergence theorem that

$ \begin{eqnarray} \Lambda_1\int_M |\nabla \phi|^2d\mu = \int_M(\Delta_{f} \phi)^2d\mu -\int_{\partial M} h\Delta_{f} \phi d\mu, \end{eqnarray} $

(2.15)

where $\psi = \frac{\partial \phi}{\partial\nu}\Big|_{\partial M}$. Since $\phi\Big|_{\partial M} = \frac{\partial^2 \phi}{\partial\nu^2}\Big|_{\partial M} = 0$, we infer from (2.7) that

$ {\Delta _f}\phi {|_{\partial M}} = (\Delta \phi - \left\langle {\nabla f, \nabla \phi } \right\rangle {\rm{ }}){|_{\partial M}} = (n - 1)H\psi - \frac{{\partial f}}{{\partial \nu }}\psi = (n - 1){H^f}\psi . $

(2.16)

Combining (2.15) - (2.16), we have

$ \begin{eqnarray} \Lambda_1 = \frac{\int_M(\Delta_{f} \phi)^2d\mu -\int_{\partial M} (n-1)H^f \psi^2d\mu}{\int_M |\nabla\phi|^2d\mu}. \end{eqnarray} $

(2.17)

Taking $\phi$ into (2.3), we have

$ \begin{eqnarray} -\int_{\partial M} (n-1) H^f \psi^2d\mu&\geq&-\int_M \frac{m-1}{m}(\Delta_{f} \phi)^2d\mu+(m-1)\kappa \int_M |\nabla \phi|^2d\mu. \end{eqnarray} $

(2.18)

Combining (2.17)-(2.18), we have

$ \begin{eqnarray} \Lambda_1\geq\frac{\frac{1}{m}\int_M(\Delta_{f} \phi)^2d\mu +(m-1)\kappa \int_M |\nabla \phi|^2d\mu}{\int_M |\nabla\phi|^2d\mu}. \end{eqnarray} $

(2.19)

From the Poincaré inequality, we have

$ \begin{eqnarray} \lambda_1\int_M |\nabla \phi|^2d\mu\leq\int_M(\Delta_{f} \phi)^2d\mu. \end{eqnarray} $

(2.20)

Taking above inequality into (2.19), we have $\Lambda_1\geq\frac{1}{m}\lambda_1+(m-1)\kappa$, the equality hold in above inequality if and only if $\phi$ is the first Dirichlet eigenfunction of the drifting Laplacian of $M$.

When $m=n$, then $f$ is a constant and $\Delta_{f}=\Delta$, from \cite[Theorem 1.2]{WX}, we know that $\Lambda_1\geq\frac{1}{n}\lambda_1+(n-1)\kappa$ with equality holding if and only if $M$ is isometric to an $n$-dimensional Euclidean unit semi-sphere.

When $m>n$, if $\Lambda_1=\frac{1}{m}\lambda_1+(m-1)\kappa$, we know that equality holds in (2.3), then by similar discussion in the proof of Theorem 1.1, we know that $u$ is a constant function on $M$, which is a contradiction since $\phi$ is the first eigenfunction of drifting Laplacian and cannot be a constant. Therefore, we have $\Lambda_1\geq\frac{1}{m}\lambda_1+(m-1)\kappa$. This completes the proof of Theorem 1.2.

In this section, we will give the proof of Theorem 1.4-Theorem 1.5 listed in the section 1.

Proof of Theorem 1.4   Let $w$ be an eigenfunction corresponding to first eigenvalue $q_1$ of problem (1.7), that is

$ \begin{equation} \Delta_{f}^2 w = 0, \; \; \mbox{in} \; \; M, \; \; \; \; w = \frac{\partial^{2}w}{\partial \nu^{2}}-q_1\frac{\partial w}{\partial \nu} = 0, \; \; \mbox{on}\; \; \partial M, \end{equation} $

(3.1)

We know that $w$ is not a constant since $w\big|_{\partial M} = 0 $. Set $ \eta = \frac{\partial w}{\partial \nu}\big|_{\partial M}$, then $\eta\neq 0$. Otherwise, if $\eta= 0$, we have

$ w = \nabla w = \frac{\partial^2 w}{\partial \nu^2} = 0\; \; \mbox{on}\; \; \partial M, $

combining $ \Delta_{f}^2 w = 0$ in $M$, we can deduce from the maximum principle that $\Delta_{f} w =0$ on $M$, which in turn implies that $w=0.$ This is a contradiction.

Since $w\big|_{\partial M}=0$, we can infer from the divergence theorem that

$ \begin{eqnarray*} \int_M \langle\nabla w, \nabla (\Delta_{f} w)\rangle d\mu = -\int_M w\Delta_{f}^2w d\mu = 0, \end{eqnarray*} $

which implies that

$ \begin{eqnarray} \int_{\partial M} \Delta_{f} w\frac{\partial w}{\partial \nu}d\mu = \int_M \langle\nabla w, \nabla (\Delta_{f} w)\rangle d\mu+\int_M (\Delta_{f} w)^2d\mu = \int_M (\Delta_{f} w)^2d\mu. \end{eqnarray} $

(3.2)

On the other hand, since

$ \begin{eqnarray} \Delta_{f} w\Big|_{\partial M} = (\Delta w-\langle \nabla f, \nabla w \rangle)\Big|_{\partial M} & = &\frac{\partial^2 w}{\partial \nu^2}+(n-1) H \frac{\partial w}{\partial \nu}-\frac{\partial f}{\partial\nu} \frac{\partial w}{\partial \nu} \\& = &\frac{\partial^2 w}{\partial \nu^2}+(n-1)H^f \frac{\partial w}{\partial \nu}\\& = &q_1\frac{\partial w}{\partial \nu}+(n-1)H^f \frac{\partial w}{\partial \nu}, \end{eqnarray} $

(3.3)

we can get

$ {q_1} = \frac{{\int_M {{{({\Delta _f}w)}^2}} d\mu - (n - 1)\int_{\partial M} {{H^f}} {\eta ^2}d\mu }}{{\int_{\partial M} {{\eta ^2}} d\mu }}. $

(3.4)

Substituting $w$ into (2.3) and noticing $w\big|_{\partial M} = 0, \; \mathrm{Ric}^m(\nabla w, \nabla w)\geq 0$, then we have

$ \begin{eqnarray} \int_M \frac{m-1}{m}(\Delta_{f} w)^2d\mu&\geq&\int_M \mathrm{Ric}^m(\nabla w, \nabla w)d\mu+\int_{\partial M} (n-1)H^f \eta^2d\mu\\&\geq&\int_{\partial M} (n-1)H^f \eta^2d\mu. \end{eqnarray} $

(3.5)

Combining (3.4) and (3.5), we have

$ \begin{eqnarray} q_1\geq\frac{1}{m}\frac{\int_M (\Delta_{f} w)^2d\mu}{\int_{\partial M} \eta^2d\mu}. \end{eqnarray} $

(3.6)

Since $p_1$ is given by

$ \begin{eqnarray} p_1 = \mathop{\mathrm{min}}_{w|_{\partial M} = 0, w\neq const.}\frac{\int_M (\Delta_{f} w)^2d\mu}{\int_{\partial M} (\frac{\partial w}{\partial \nu})^2d\mu}, \end{eqnarray} $

(3.7)

we have from the variational characterization that

$ \begin{eqnarray} \frac{\int_M (\Delta_{f} w)^2d\mu}{\int_{\partial M} \eta^2d\mu}\geq p_1. \end{eqnarray} $

(3.8)

Combining (3.6) and (3.8), we have $q_1\geq \frac{1}{m}p_1$.

When $m=n$, then $f$ is constant, so $\Delta_{f}=\Delta$, then by Theorem 1.3 in [24], we know that equality holds in (1.8) if and only if $M$ is isometric to an Euclidean ball.

When $m>n$, if the equality holds in (1.8), the equality holds in (2.3), then $\Delta_{f} w = -\frac{m}{m-n}\langle \nabla f, \nabla w\rangle$, multiplying with $f$ and integrating on $M$ with respect to $e^{\frac{n}{m-n}f}dv$, we can get

$ \int_M w\Delta_{f} w e^{\frac{n}{m-n}f}d\nu = \int_M-w\frac{m}{m-n}\langle \nabla f, \nabla w\rangle e^{\frac{n}{m-n}f}d\nu, $

which implies $\int_M|\nabla w|^2e^{\frac{n}{m-n}f}d\nu = 0$. So $w$ is constant, which is a contradiction. Thus, equality holds in (1.8) if and only if $M$ is isometric to an Euclidean ball of radius $\frac{1}{c}$ and $f$ is constant. This completes the proof of Theorem 1.3.

Proof of Theorem 1.5   Let $\varphi$ be an eigenfunction corresponding to first eigenvalue $p_1$ of problem (1.6), that is

$ \begin{eqnarray} \Delta_{f}^2 \varphi = 0, \; \; \mbox{in} \; \; M, \; \; \; \; \varphi = \Delta_{f} \varphi-q_1\frac{\partial \varphi}{\partial \nu} = 0, \; \; \mbox{on}\; \; \partial M, \end{eqnarray} $

(3.9)

Set $\psi = \frac{\partial \varphi}{\partial \nu}\big|_{\partial M}$, then

$ \begin{eqnarray} q_1 = \frac{\int_M (\Delta_{f} \varphi)^2d\mu}{\int_{\partial M} \psi^2d\mu}. \end{eqnarray} $

(3.10)

Substituting $\varphi$ into (2.3) and noticing $\psi = 0, \; \mathrm{Ric}^m(\nabla \varphi, \nabla \varphi)\geq -(m-1)\kappa $ and $(n-1)H^\varphi\geq (m-1)c$, then we infer from that $-\int_M|\nabla \varphi|^2d\mu\geq\frac{1}{\lambda_1}\int_M (\Delta_{f} \varphi)^2d\mu $ that

$ \begin{eqnarray} \int_M \frac{m-1}{m}(\Delta_{f} \varphi)^2d\mu&\geq&\int_M \mathrm{Ric}^m(\nabla \varphi, \nabla \varphi)d\mu+\int_{\partial M} (n-1)H^\varphi \psi^2d\mu\\&\geq&-(m-1)\kappa\int_M|\nabla\varphi|^2d\mu+(m-1)c\int_{\partial M} \psi^2d\mu\\&\geq&-\frac{(m-1)\kappa}{\lambda_1}\int_M(\Delta_{f} \varphi)^2d\mu+(m-1)c\int_{\partial M} \psi^2d\mu. \end{eqnarray} $

(3.11)

Combining (3.10) and (3.11), we have $q_1\geq \frac{mc\lambda_1}{\lambda_1+m\kappa}.$

When $m=n$, then $f$ is constant, so $\Delta_{f}=\Delta$, then by the Theorem 1.4 in [24], we know that equality holds in (1.8) if and only if $M$ is isometric to an Euclidean ball.

When $m>n$, if the equality holds in (1.8), the equality holds in (2.3), then by the similar discussion, we know that $\varphi$ is constant, which is contradiction. Thus, equality holds in (1.8) if and only if $M$ is isometric to an Euclidean ball of radius $1/c$ and $f$ is constant.