\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 258, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/258\hfil Liouville theorem and gradient estimates]
{Liouville theorem and gradient estimates for nonlinear elliptic equations
on Riemannian manifolds}

\author[W. Wang, H. Zhou, X. Zhang \hfil EJDE-2017/258\hfilneg]
{Wen Wang, Hui Zhou, Xinquan Zhang}

\address{Wen Wang (corresponding author)\newline
School of Mathematics and Statistics,
Hefei Normal University,
Hefei 230601,  China. \newline
School of mathematical Science,
University of Science and Technology of China,
Hefei 230026,  China}
\email{wwen2014@mail.ustc.edu.cn}

\address{Hui Zhou  (corresponding author)\newline
School of Mathematics and Statistics,
Hefei Normal University,
Hefei 230601,  China. \newline
School of mathematical Science,
University of Science and Technology of China,
Hefei 230026,  China}
\email{zhouhui0309@126.com}

\address{Xinquan Zhang \newline
School of Mathematics and Statistics,
Hefei Normal University,
Hefei 230601,  China}
\email{285171107@qq.com}

\dedicatory{Communicated by Giovanni Molica Bisci}

\thanks{Submitted October 28, 2016. Published October 16, 2017.}
\subjclass[2010]{58J35, 35K05, 53C21}
\keywords{Gradient estimate; nonlinear elliptic equation;
 Liouville theorem; 
\hfill\break\indent Harnack inequality}

\begin{abstract}
 In this article we study a nonlinear elliptic equation by using
 the maximum principle and cutoff functions,
 We establish related gradient estimates, the Liouville theorem,
 and the Harnack inequality.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\allowdisplaybreaks

\section{Introduction and statement of main results}

In 1981, Gidas-Spruck \cite{3} derived the following result.

\begin{theorem} \label{thmA}
Let $M^n$ be a complete manifold with nonnegative Ricci curvature.
Assume that $h(x)\in C^{2}(M^n)$ and $\alpha>0$ satisfy
the following conditions:
\begin{itemize}
\item[(1)] $h(x)\geq0$ on $M^n$;
\item[(2)] $\Delta h(x)\geq0$ on $M^n$;
\item[(3)] for $r(x)$ large, $|\nabla\log h(x)|\leq C/r(x)$ and if
$n\geq 4$, $h(x)\geq C(r(x))^{\sigma}$ with $\sigma\geq -\frac{2}{n-3}$,
where $r(x)$ is the geodesic distance between $x$ and some fixed point $p$;
\item[(4)] $1\leq \alpha\leq\frac{n+2}{n-2}$.
\end{itemize}
If $u(x)$ is a nonnegative solution of
\begin{equation*}
\Delta u +hu^{\alpha}=0,
\end{equation*}
then $u(x)\equiv 0$.
\end{theorem}

For $\alpha=1$, Li-Yau \cite{9} demonstrated the same result under the
condition that $|\nabla h(x)|=o(r(x))$ as $r(x)\to\infty$.
Later, Li \cite{6} proved that
as $1\leq \alpha\leq \frac{n}{n-2}$  $(n\geq 4)$, the condition (3)
of Theorem \ref{thmA} is unnecessary. On these conditions were further weakened,
see \cite{1,5,7}.
 In 2010, Yang \cite{11} studied the  equation
 \begin{equation*}
\Delta u +cu^{-\alpha}=0
\end{equation*}
on a noncompact complete Riemannian manifold, where $\alpha>0$ and $c$ are
two real constants. The corresponding gradient estimates and Liouville type
theorem are also derived.

Recently, Wang \cite{10} deduced  gradient estimates and Liouville type
theorem for positive solutions to the equation
 \begin{equation*}
\Delta_{f} u^{m} +cu=0
\end{equation*}
on smooth measure space with $m$-Bakry-\'{E}mery curvature bounded
by $Ric_{f,m}\geq -(m-1)K$, where $K\geq 0$.

Inspired by the  works \cite{3,8,10,11}, we investigate the nonlinear
elliptic equation
\begin{equation}\label{1.1}
\Delta u^{m}+\lambda(x) u^{l}=0,\quad m>1
\end{equation}
on a complete Riemannian manifold with  Ricci
curvature  bounded below, where $m>1$ and $l$ are real numbers, and
$\lambda(x)\in C^{2}(M^n)$.
If $M^{n}=\Omega$ is  a bounded smooth domain in $\mathbb{R}^n$ and
$\lambda(x)\leq 0$ is a constant, the equation \eqref{1.1}
is regarded as the thin film equation, which depict a steady state of
the thin film. Concrete content can be seen \cite{4}.
Our main results reads as follows.

\begin{theorem}\label{thm1.1}
 Let $(M^{n}, g)$ be a complete
Riemannian manifold without boundary. Suppose that $B_{2R}$ is a geodesic
ball of radius $2R$ around
$p\in M$ and $\operatorname{Ric}(B_{2R})\geq -K$ with $K\geq 0$.
Also suppose that there exist two positive numbers $\delta$ and $\tau$ such that
$|\lambda(x,t)|\leq \delta$ and $|\nabla\lambda|^{2}\leq \tau|\lambda|^2$.
Let $u(x)$ is a positive solution to the equation  \eqref{1.1} and
$v=\frac{m}{m-1}u^{m-1}$.

(a) Assume that $l\geq 1$, then
\begin{equation} \label{1.2}
\sup_{B_{p}(R)}\frac{|\nabla v|^2}{v}
\leq\frac{C_{4}(m-1)}{m+1}[\frac{1}{R^2}(1+\sqrt{K}R)
+2K+\tau]\sup_{x\in M^n}v+H_1.
\end{equation}

(b) Assume that $l< 1$, then
\begin{equation}\label{1.3}
\sup_{B_{p}(R)}\frac{|\nabla v|^2}{v}
\leq\frac{C_{4}(m-1)}{m+1}[\frac{1}{R^2}(1+\sqrt{K}R)
+2K+\tau]\sup_{x\in M^n}v+H_2.
\end{equation}
Where $C_4$ is a constant depending only on $n$, and
\begin{gather*}
H_1=\frac{(m-1)(n-1)}{m(m+1)}|\frac{2(m+1)}{n-1}+(m-2l+1)|
 \delta\Big(\frac{m-1}{m}\sup_{x\in M^n}v\Big)^{\frac{l-1}{m-1}},\\
H_2=\frac{(m-1)(n-1)}{m(m+1)}|\frac{2(m+1)}{n-1}+(m-2l+1)|
\delta\Big(\frac{m-1}{m}\inf_{x\in M^n}v\Big)^{\frac{l-1}{m-1}}.
\end{gather*}
Moreover, if $(M^{n}, g)$ has nonnegative Ricci curvature, letting
$R\to\infty$, we have
following estimate for $l\geq 1$,
\begin{equation}\label{1.4}
\frac{|\nabla v|^2}{v} \leq C(m,n,l,K,\delta,\tau,\sup_{x\in M^n}v),
\end{equation}
 and for $l< 1$,
\begin{equation}\label{1.5}
\frac{|\nabla v|^2}{v} \leq C(m,n,l,K,\delta,\tau,\sup_{x\in M^n}v).
\end{equation}
 Where \begin{gather*}
C(m,n,l,K,\delta,\tau,\sup_{x\in M^n}v)
=\frac{C_{4}(m-1)}{m+1}(2K+\tau)\sup_{x\in M^n}v+H_1,\\
C'(m,n,l,K,\delta,\tau,\sup_{x\in M^n}v,\inf_{x\in M^n}v)
=\frac{C_{4}(m-1)}{m+1}(2K+\tau)\sup_{x\in M^n}v+H_2.
\end{gather*}
\end{theorem}

By using  \eqref{1.4} and  \eqref{1.5}, we derive the related Harnack inequalities.

\begin{corollary}\label{coro1.2}
 Let $(M^{n}, g)$ be a noncompact complete
Riemannian manifold without boundary. Suppose that $\operatorname{Ric}(M^{n})\geq  0$.
Let $u(x)$ is a positive solution of the  equation  \eqref{1.1}, and
$v=\frac{m}{m-1}u^{m-1}$.

If $l\geq 1$, then
\begin{equation}\label{1.6}
 v(x)\leq v(y)\exp{\Big[r(x,y)\sqrt{\frac{C(m,n,l,K,\delta,\tau,
\sup_{x\in M^n}v)}{\inf_{x\in M^n}v}}\Big]}.
\end{equation}

If  $l<1$, then
\begin{equation}\label{1.7}
 v(x)\leq v(y)\exp{\Big[r(x,y)\sqrt{\frac{C'(m,n,l,K,\delta,\tau,
\sup_{x\in M^n}v,\inf_{x\in M^n}v)}{\inf_{x\in M^n}v}}\Big]}.
\end{equation}
\end{corollary}

\begin{theorem}\label{thm1.3}
Let $(M^{n}, g)$ be a complete
Riemannian manifold without boundary. Suppose that $B_{2R}$ is a geodesic
 ball of radius $2R$ around $p\in M$
and $\operatorname{Ric}(B_{2R})\geq -K$ with $K\geq 0$.
Let $u(x)$ is a positive solution of the  equation  \eqref{1.1}.
Let $v=\frac{m}{m-1}u^{m-1}$ and
 $|\nabla\lambda|^{2}\leq \tau|\lambda|^2$  for some positive constant $\tau$.
 If $\lambda\geq0$ and $l\leq \frac{(n+1)(m+1)}{2(n-1)}$ or $\lambda\leq0$ and
$l\geq\frac{(n+1)(m+1)}{2(n-1)}$, then we have
\begin{equation}\label{1.8}
\sup_{B_{p}(R)}\frac{|\nabla v|^2}{v}
\leq C_{4}\big[\frac{1}{R^2}(1+\sqrt{K}R)+2K+\tau\big]\sup_{x\in M^n}v,
\end{equation}
where $C_4$ is a constant depending only on $n$.

Letting $R\to\infty$ , then we infer on a complete noncompact Riemannian manifold,
\begin{equation}\label{1.9}
\frac{|\nabla v|^2}{v} \leq C_{4}(2K+\tau)\sup_{x\in M^n}v.
\end{equation}
\end{theorem}

Applying  \eqref{1.9}, we can derive the following Liouville type theorem
as $\lambda(x)$ is a constant.

\begin{corollary} \label{coro1.4}
Let $(M^{n}, g)$ be a noncompact complete
Riemannian manifold without boundary. Suppose that  $\operatorname{Ric}(M^n)\geq 0$
and $u(x)$ is a positive solution of the equation  \eqref{1.1}, where
$\lambda(x)$ is a constant. If $\lambda\geq0$ and $l\leq \frac{(n+1)(m+1)}{2(n-1)}$
or $\lambda\leq0$ and $l\geq\frac{(n+1)(m+1)}{2(n-1)}$,  then $u$ is a constant.
\end{corollary}

\begin{theorem}\label{thm1.5}
 Let $(M^{n}, g)$ be a complete noncompact
Riemannian manifold with $\operatorname{Ric}(M^n)\geq -K$ with $K\geq 0$.
Let $u(x)$ is a positive  solution to the equation
\begin{equation}\label{1.10}
\Delta u^{m}+\lambda u^{l}=0,
\end{equation}
where $\lambda>0$ is a constant. Let $v=\frac{m}{m-1}u^{m-1}$
and $1\leq l\leq \frac{(n+1)(m+1)}{2(n-1)}$.
If
\begin{equation}\label{1.11}
\lambda\leq\frac{2(m-1)(n-1)K}{\frac{2(m+1)}{n-1}+(m-2l+1)}
\big(\frac{m-1}{m}\big)^{\frac{l-1}{m-1}}\big(\sup_{x\in M^n}v\big)^{\frac{m-l}{m-1}},
\end{equation}
then for any $x\in M^n$,
\begin{align*}
\frac{|\nabla v|^2}{v}
&\leq\frac{2(m-1)^{2}(n-1)^{2}}{m(m+1)}K\sup_{x\in M^n}v\\
&\quad -\frac{(m-1)(n-1)}{m(m+1)}\big[\frac{2(m+1)}{n-1}+(m-2l+1)\big]
 \lambda\big(\frac{m-1}{m}\sup_{x\in M^n}v\big)^{\frac{l-1}{m-1}}.
\end{align*}
 If
$$
\lambda\geq\frac{2(m-1)(n-1)K}{\frac{2(m+1)}{n-1}+(m-2l+1)}
\big(\frac{m-1}{m}\big)^{\frac{l-1}{m-1}}\big(\sup_{x\in M^n}v\big)^{\frac{m-l}{m-1}},
$$
then $v$ must be a constant.
\end{theorem}

Note that by  taking $l=1$ in Theorem \ref{thm1.5}, our result partially generalize Wang's
result in [10].
By  \eqref{1.11}, we can find the lower bound estimate as $m\geq l$,
and the upper bound estimate as $m\leq l$ for positive solutions of
 \eqref{1.10}.

\section{Preliminaries}

To prove our main results, we need the lemma below.
Let $v=\frac{m}{m-1}u^{m-1}$, then
\begin{equation}\label{2.1}
(m-1)v\Delta v+|\nabla v|^{2}
=-\lambda (m-1)\big(\frac{m-1}{m}\big)^{\frac{l-1}{m-1}}v^{1+\frac{l-1}{m-1}}.
\end{equation}

\begin{lemma}\label{lem2.1}
 Let $(M^{n}, g)$ be a complete
Riemannian manifold without boundary. Suppose that $B_{2R}$ is a geodesic
 ball of radius $2R$ around $p\in M$ and $Ricci(B_{2R})\geq -K$ with $K\geq 0$.
Let $u(x)$ is a positive solution to the equation  \eqref{1.1} and
$v=\frac{m}{m-1}u^{m-1}$. Let $w=\frac{|\nabla v|^2}{v}$ and $G=\varphi w$,
where $\varphi(x)$ is a smooth cutoff function (see the proof of Theorem \ref{thm1.1}).
 Suppose that $G(x)$ reaches the maximum value at $x_{0}$ and $\varphi(x_0)>0$.
Then at $x_0$,
\begin{equation}\label{2.2}
\begin{aligned}
\varphi\Delta w
&\geq \big[\frac{n}{2(n-1)}+\frac{2m}{(n-1)(m-1)^2}+\frac{1}{m-1}\big]
\frac{G^2}{v\varphi}\\
&\quad +\big[\frac{2m}{m-1}-\frac{n}{n-1}-\frac{2}{(n-1)(m-1)}\big]
 \frac{\nabla v\nabla\varphi}{v\varphi}G+\frac{n}{2(n-1)}
 \frac{|\nabla \varphi|^2}{\varphi^2}G\\
&\quad +\big[\frac{2(m+1)}{(n-1)(m-1)}+\frac{m-2l+1}{m-1}\big]
\lambda\big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}\frac{G}{v}-2(n-1)KG\\
&\quad -\frac{2}{v}\varphi|\nabla v||\nabla\lambda|
 \big(\frac{m-1}{m}v \big)^{\frac{l-1}{m-1}}-\frac{2}{n-1}\lambda
 \big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}\frac{\nabla v\nabla\varphi}{v}\\
&\quad +\frac{2}{n-1}\lambda^{2}\big(\frac{m-1}{m}v\big)^{2\frac{l-1}{m-1}}
 \frac{\varphi}{v}.
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
After calculations we obtain that
\begin{equation}\label{2.3}
\begin{aligned}
\Delta w&= \Delta\Big(\frac{|\nabla v|^2}{v}\Big)\\
&= \frac{\Delta|\nabla v|^{2}}{v}-\frac{2\nabla|\nabla v|^{2}\Delta v}{v^2}
 -\frac{|\nabla v|^{2}\Delta v}{v^2}+\frac{2|\nabla v|^4}{v^3}\\
&= \frac2{v}\big[|\operatorname{Hess} v|^{2}+\nabla v\cdot \nabla\Delta v
+\operatorname{Ric}((\nabla v,\nabla v))\big]\\
&\quad -\frac{2\nabla|\nabla v|^{2}\cdot \nabla v}{v^2}
-\frac{|\nabla v|^{2}\Delta v}{v^2}+\frac{2|\nabla v|^4}{v^3}.
\end{aligned}
\end{equation}
 Since $G$ reaches at the maximum at $x_0$, so we have  $\nabla G=0$. Then at $x_0$,
\begin{gather}\label{2.4}
\nabla w=-\frac{G\nabla \varphi}{\varphi^2}, \\
\label{2.5}
\nabla |\nabla v|^{2}=-\frac{vG}{\varphi^2}\nabla \varphi+\frac{G}{\varphi}\nabla v.
\end{gather}
Choose an orthonormal frame $\{e_1, e_2,\cdots,e_{n}\}$ around $x_0$, such that
$|\nabla v|e_1=\nabla v$. Then
\begin{equation}\label{2.6}
\frac{|\nabla|\nabla v|^2|^2}{4|\nabla v|^2}=\sum_{j=1}^{n}v^{2}_{1j},
\end{equation}
\begin{equation}\label{2.7}
\frac{\nabla v\cdot\nabla|\nabla v|^2}{2|\nabla v|^2}=v_{11}.
\end{equation}
On the other hand, we have
\begin{equation}\label{2.8}
\begin{aligned}
|\operatorname{Hess} v|^{2}
&\geq v^{2}_{11}+2\sum_{\alpha=2}^{n}v^{2}_{1\alpha}
 +\sum_{\alpha=2}^{n}v^{2}_{\alpha\alpha}\\
&\geq v^{2}_{11}+2\sum_{\alpha=2}^{n}v^{2}_{1\alpha}
 +\frac{1}{n-1}\Big(\sum_{\alpha=2}^{n}v_{\alpha\alpha}\Big)^{2}\\
&= v^{2}_{11}+2\sum_{\alpha=2}^{n}v^{2}_{1\alpha}
 +\frac{1}{n-1}(\Delta v-v_{11})^{2}\\
&= \frac{n}{n-1}v^{2}_{11}+2\sum_{\alpha=2}^{n}v^{2}_{1\alpha}
 -\frac{2}{n-1}\Delta v v_{11}+\frac{1}{n-1}(\Delta v)^{2}\\
&= \frac{n}{n-1}v^{2}_{11}+2\sum_{\alpha=2}^{n}v^{2}_{1\alpha}
 +\frac{2v_{11}}{n-1}\left[\frac{1}{m-1}w
 +\lambda\big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}\right]\\
&\quad +\frac{1}{n-1}\left[\frac{1}{m-1}w
 +\lambda\big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}\right]^{2}.
\end{aligned}
\end{equation}
Putting  \eqref{2.6} and  \eqref{2.7} into  \eqref{2.8}, and using  \eqref{2.5},
we have
\begin{align*}
&|\operatorname{Hess} v|^{2}+\operatorname{Ric}(\nabla v,\nabla v)\\
&\geq \frac{n}{n-1}v^{2}_{11}+2\sum_{\alpha=2}^{n}v^{2}_{1\alpha}
 +\frac{2v_{11}}{n-1}\big[\frac{1}{m-1}w
 +\lambda\big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}\big]\\
&\quad +\frac{1}{n-1}\big[\frac{1}{m-1}w+\lambda
 \big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}\big]^{2}-(n-1)K|\nabla v|^{2}\\
&\geq \frac{n}{n-1}\sum_{j=1}^{n}v^{2}_{1j}+\frac{2v_{11}}{n-1}
\big[\frac{1}{m-1}w+\lambda\big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}\big]\\
&\quad  +\frac{1}{n-1}\big[\frac{1}{m-1}w+\lambda
 \big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}\big]^{2}-(n-1)K|\nabla v|^{2}\\
&= \frac{n}{n-1}\frac{|\nabla|\nabla v|^2|^2}{4|\nabla v|^2}
 +\frac{1}{n-1}\cdot\frac{\nabla v\cdot\nabla|\nabla v|^2}{|\nabla v|^2}
 \big[\frac{1}{m-1}w+\lambda\big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}\big]\\
&\quad +\frac{1}{n-1}\big[\frac{1}{m-1}w+\lambda
 \big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}\big]^{2}-(n-1)K|\nabla v|^{2}\\
&= \frac{n}{4(n-1)|\nabla v|^2}\big[\frac{G}{\varphi}\nabla v
 -\frac{vG}{\varphi^2}\nabla\varphi\big]^{2}
 +\frac{1}{n-1}\big[\frac{1}{m-1}w+\lambda\big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}
 \big]^{2}\\
&\quad +\frac{1}{(n-1)|\nabla v|^2}\big[\frac{G}{\varphi}|\nabla v|^{2}
 -\frac{vG}{\varphi^2}\nabla v\nabla\varphi\big]
\big[\frac{1}{m-1}w+\lambda\big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}\big]\\
&\quad -(n-1)K|\nabla v|^{2}\\
&= \frac{n}{4(n-1)}\big(\frac{G}{\varphi}\big)^{2}
 +\frac{nv^2}{4(n-1)|\nabla v|^2}\frac{G^2}{\varphi^4}|\nabla\varphi|^{2}
 -\frac{n}{2(n-1)|\nabla v|^2}\cdot\frac{vG^{2}\nabla v\nabla \varphi}{\varphi^3}\\
&\quad +\frac{w^2}{(n-1)(m-1)^2}+\frac{2w}{(n-1)(m-1)}\lambda
\big(\frac{m}{m-1}v\big)^{\frac{l-1}{m-1}}\\
&\quad +\frac{1}{n-1}\lambda^{2}\big(\frac{m-1}{m}v\big)^{2\frac{l-1}{m-1}}
 +\frac{1}{(n-1)(m-1)}\frac{G}{\varphi}w
 +\frac{1}{n-1}\frac{G}{\varphi}\lambda\big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}\\
& -\frac{1}{(n-1)|\nabla v|^{2}}\frac{vG}{\varphi^2}\nabla v\nabla\varphi\frac{w}{m-1}
-\frac{1}{(n-1)|\nabla v|^2}\frac{vG}{\varphi^2}\nabla v\nabla\varphi\lambda
 \big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}\\
&\quad -(n-1)K|\nabla v|^{2}\\
&= \frac{n}{4(n-1)}\big(\frac{G}{\varphi}\big)^{2}
 +\frac{nv}{4(n-1)}\frac{|\nabla\varphi|^2}{\varphi^2}\frac{G}{\varphi}
 -\frac{n}{2(n-1)}\frac{G}{\varphi}\frac{\nabla v\nabla\varphi}{\varphi}\\
&\quad +\frac{1}{(n-1)(m-1)^2}\big(\frac{G}{\varphi}\big)^{2}
+\frac{2\lambda}{(n-1)(m-1)}\big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}
\frac{G}{\varphi}\\
&\quad +\frac{\lambda^{2}}{n-1}\big(\frac{m-1}{m}v\big)^{2\frac{l-1}{m-1}}
+\frac{1}{(n-1)(m-1)}\big(\frac{G}{\varphi}\big)^{2}\\
&\quad +\frac{1}{n-1}\lambda\big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}
 \frac{G}{\varphi}
-\frac{1}{(n-1)(m-1)}\frac{G}{\varphi}\frac{\nabla v\nabla\varphi}{\varphi}\\
&\quad -\frac{1}{n-1}\lambda\big(\frac{m}{m-1}v\big)^{\frac{l-1}{m-1}}
 \frac{\nabla v\nabla\varphi}{\varphi}-(n-1)K|\nabla v|^{2}\\
&= \big[\frac{n}{4(n-1)}+\frac{m}{(n-1)(m-1)^2}\big]
 \frac{G^2}{\varphi^2}-\big[\frac{n}{2(n-1)}+\frac{1}{(n-1)(m-1)}\big]
 \frac{G}{\varphi^2}\nabla v\nabla\varphi\\
&\quad +\frac{n}{4(n-1)}\frac{|\nabla\varphi|^2}{\varphi^3}vG
 +\frac{m+1}{(n-1)(m-1)}\lambda\big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}
 \frac{G}{\varphi}\\
&\quad -\frac{1}{n-1}\lambda\big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}
\frac{\nabla v\nabla\varphi}{\varphi}+\frac{\lambda^2}{n-1}
 \big(\frac{m-1}{m}v\big)^{2\frac{l-1}{m-1}}\\
&= -(n-1)K|\nabla v|^{2}.
\end{align*} %\label{2.9}
Noting that $v>0$ when $m>1$. Using the above inequality in  \eqref{2.3} and
applying  \eqref{2.1} and  \eqref{2.5}, then  \eqref{2.2} can be inferred.
\end{proof}

\section{The proof of main results}

\begin{proof}[Proof of Theorem \ref{thm1.1}]
Construct a smooth function $\theta(t): [0,+\infty)\to[0,1]$
\[
\theta(t)=\begin{cases}
1,&0\leq t\leq 1\\
0,&t>2
\end{cases}
\]
such that
\begin{equation}\label{3.1}
-C_1\sqrt{\theta}\leq \theta'\leq 0,\quad |\theta''|\leq C_2\theta.
\end{equation}
Define the smooth cutoff function $\varphi:M\to\mathbb{R}$ by
$\varphi(x)=\theta(\frac{r(x)}{R})$. We suppose that
$G=\varphi w=\varphi\frac{|\nabla v|^2}{v}$ attains its maximal value  at
$x_{0}\in B_{2R}$. We can
suppose that $G(x_0)>0$, because otherwise the proof is trivial.
 Then  at $x_0$, we have
\begin{align*}
\Delta G
&= \Delta\varphi\cdot w+2\nabla\varphi\nabla w+\varphi\Delta w\\
&= \Delta\varphi\cdot w-2G\frac{|\nabla\varphi|^2}{\varphi^2}+\varphi\Delta w\\
&= \frac{\Delta\varphi}{\varphi}G-2G\frac{|\nabla\varphi|^2}{\varphi^2}
+\varphi\Delta w.
\end{align*}
Note that
\begin{gather*}
\nabla \varphi=\frac{\theta'\nabla r}{R},\\
\Delta\varphi=\frac{\theta''}{R^2}
+\frac{\theta'\Delta r}{R}\geq \frac{\theta''}{R^2}
+\frac{(n-1)(1+\sqrt{K}R)\theta'}{R^2}.
\end{gather*}
Since $\Delta G\leq 0$ is valid at $x_0$, we have
\begin{equation}\label{3.2}
\begin{aligned}
0&\geq \big[ \frac{\theta''}{\theta R^2}+\frac{(n-1)
(1+\sqrt{K}R)\theta'}{\theta R^2}\big]G
-\frac{3n-4}{2(n-1)}\frac{(\theta')^2}{R^{2}\theta^2}G\\
&\quad +\frac{(m-1)(mn+n-2)+4m}{2(m-1)^{2}(n-1)}\frac{G^2}{v\theta}
 -\frac{(m+1)(n-2)}{(n-1)(m-1)}\frac{G\sqrt{G}|\theta'|}{R\theta\sqrt{v\theta}}\\
&\quad +\big[\frac{2(m+1)}{(m-1)(n-1)}+\frac{m-2l+1}{m-1}\big]\lambda
 \big(\frac{m}{m-1}v\big)^{\frac{l-1}{m-1}}\frac{G}{v}\\
&\quad -2(n-1)KG-\frac{2}{v}\theta|\nabla v||\nabla\lambda|
 \big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}\\
&\quad -\frac{2|\lambda|}{n-1}\big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}
 \frac{\sqrt{G}|\theta'|}{R\sqrt{v\theta}}
+\frac{2\lambda^{2}}{n-1}\big(\frac{m-1}{m}v\big)^{2\frac{l-1}{m-1}}
 \frac{\theta}{v}.
\end{aligned}
\end{equation}
Applying the inequality $ax^{2}+bx\geq -\frac{b^2}{4a}$ with $a>0$,
we have
\begin{equation}\label{3.3}
\begin{gathered}
\begin{aligned}
&\frac{\lambda^{2}}{n-1}\big(\frac{m-1}{m}v\big)^{2\frac{l-1}{m-1}}
\frac{\theta}{v}-\frac{2|\lambda|}{n-1}
\big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}
\frac{\sqrt{G}|\theta'|}{R\sqrt{v\theta}}\\
&\geq -\frac{G(\theta')^2}{(n-1)R^{2}\theta^2},
\end{aligned}\\
\begin{aligned}
&\frac{\lambda^{2}}{n-1}\big(\frac{m-1}{m}v\big)^{2\frac{l-1}{m-1}}
 \frac{\theta}{v}-\frac{2}{v}\varphi|\nabla v||\nabla\lambda|
 \big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}\\
&\geq -(n-1)\frac{|\nabla\lambda|^2}{\lambda^2}G.
\end{aligned}
\end{gathered}
\end{equation}
By the Cauchy inequality, it follows that
\begin{equation}\label{3.4}
-\frac{G\sqrt{G}|\theta'|}{R\theta\sqrt{v\theta}}
\geq -\frac{G^2}{2v\theta}-\frac{G(\theta')^2}{2R^{2}\theta^2}.
\end{equation}
Substituting   \eqref{3.3} and  \eqref{3.4} into  \eqref{3.2}, we obtain
\begin{equation}\label{3.5}
\begin{aligned}
0&\geq \Big[ \frac{\theta''}{\theta R^2}+\frac{(n-1)
(1+\sqrt{K}R)\theta'}{\theta R^2}\Big]G-\frac{3n-2}{2(n-1)}
\frac{(\theta')^2}{R^{2}\theta^2}G\\
&\quad +\frac{m(m+1)}{(m-1)^{2}(n-1)}\frac{G^2}{v\theta}
 -\frac{(m+1)(n-2)}{2(m-1)(n-1)}\frac{G(\theta')^2}{R^{2}\theta^2}\\
&\quad -2(n-1)KG-(n-1)\frac{|\nabla\lambda|^2}{\lambda^2}G\\
&\quad+\big[\frac{2(m+1)}{(m-1)(n-1)}+\frac{m-2l+1}{m-1}\big]
\lambda\big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}\frac{G}{v}.
\end{aligned}
\end{equation}
From \eqref{3.1} and $|\nabla\lambda|^{2}\leq \tau\lambda^{2}$, we have
\begin{equation}\label{3.6}
\begin{aligned}
0&\geq -\Big[ \frac{C_2}{R^2}+\frac{(n-1)(1+\sqrt{K}R)C_1}
{\sqrt{\theta} R^2}\Big]G+\frac{m(m+1)}{(m-1)^{2}(n-1)}\frac{G^2}{v\theta}\\
&\quad -\Big[\frac{2m}{m-1}-\frac{n}{(m-1)(n-1)}\Big]
 \frac{C^{2}_1}{R^{2}\theta}G-2(n-1)KG-(n-1)\tau G\\
&\quad +\Big[\frac{2(m+1)}{(m-1)(n-1)}+\frac{m-2l+1}{m-1}\Big]
\lambda\big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}\frac{G}{v}\\
&\geq -\Big[ \frac{C_2}{R^2}+\frac{(n-1)(1+\sqrt{K}R)C_1}{\sqrt{\theta} R^2}
\Big]G+\frac{m(m+1)}{(m-1)^{2}(n-1)}\frac{G^2}{v\theta}\\
&\quad -\frac{2m}{m-1}\frac{C^{2}_1}{R^{2}\theta}G-2(n-1)KG-(n-1)\tau G\\
&\quad +\Big[\frac{2(m+1)}{(m-1)(n-1)}+\frac{m-2l+1}{m-1}\Big]
\lambda\big(\frac{m-1}{m}v\big)^{\frac{l-1}{m-1}}\frac{G}{v}.
\end{aligned}
\end{equation}
Multiply by $v\theta$ to both side of  \eqref{3.6}, and using
 $0\leq \theta\leq 1$  we obtain for $l\geq1$
\begin{equation}\label{3.7}
\begin{aligned}
0&\geq \frac{m(m+1)}{(m-1)^{2}(n-1)}G^{2}
 -\frac{2m}{m-1}\frac{C^{2}_1}{R^{2}}\sup_{x\in M^n}vG\\
&\quad -\Big[ \frac{C_2}{R^2}+\frac{(n-1)(1+\sqrt{K}R)C_1}{R^2}\Big]
 G\sup_{x\in M^n}v\\
&\quad -\Big[2(n-1)K+(n-1)\tau\Big]G\sup_{x\in M^n}v\\
&\quad -|\frac{2(m+1)}{(m-1)(n-1)}+\frac{m-2l+1}{m-1}|
 |\lambda|\big(\frac{m-1}{m}\sup_{x\in M^n}v\big)^{\frac{l-1}{m-1}}G.
\end{aligned}
\end{equation}
Meanwhile,  for $l<1$ we obtain
\begin{equation}\label{3.8}
\begin{aligned}
0&\geq \frac{m(m+1)}{(m-1)^{2}(n-1)}G^{2}
 -\frac{2m}{m-1}\frac{C^{2}_1}{R^{2}}G\sup_{x\in M^n}v\\
&\quad -\Big[ \frac{C_2}{R^2}+\frac{(n-1)(1+\sqrt{K}R)C_1}{R^2}\Big]
 G\sup_{x\in M^n}v\\
&\quad -\Big[2(n-1)K+(n-1)\tau\Big]G\sup_{x\in M}v\\
&\quad -|\frac{2(m+1)}{(m-1)(n-1)}+\frac{m-2l+1}{m-1}| \,
|\lambda|\Big(\frac{m-1}{m}\inf_{x\in M^n}v\Big)^{\frac{l-1}{m-1}}G.
\end{aligned}
\end{equation}
We observe that
\begin{equation}\label{3.9}
\Big[ \frac{C_2}{R^2}+\frac{(n-1)(1+\sqrt{K}R)C_1}{R^2}\Big]
\sup_{x\in M^n}v\leq \frac{C_3}{R^2}(1+\sqrt{K}R)\sup_{x\in M^n}v,
\end{equation}
for some constant $C_3$ depending only on $n$.

On the other hand, for the equation $Ax^{2}-Bx\leq 0$ with $A>0,B>0$,
we have $x\leq \frac{B}{A}$. By utilize the equation to
 \eqref{3.7} and  \eqref{3.8}, and noting   \eqref{3.9} we obtain at
the the maximum point $x_0$ for $l\geq 1$
\begin{align*}
\sup_{B_{p}(R)}w(x)
&\leq \varphi w(x_0)=G(x_0)\\
&\leq\frac{2(m-1)(n-1)}{m+1}\frac{C^{2}_1}{R^{2}}\sup_{x\in M^n}v+\frac{(m-1)^{2}(n-1)}{m(m+1)}\frac{C_3}{R^2}(1+\sqrt{K}R)\sup_{x\in M^n}v\\
&\quad +\frac{(m-1)^{2}(n-1)^{2}}{m(m+1)}(2K+\tau)\sup_{x\in M^n}v\\
&\quad +\frac{(m-1)(n-1)}{m(m+1)}|\frac{2(m+1)}{n-1}+(m-2l+1)|\,
 |\lambda|\Big(\frac{m-1}{m}\sup_{x\in M^n}v\Big)^{\frac{l-1}{m-1}},
\end{align*}
and for $l< 1$,
\begin{align*}
\sup_{B_{p}(R)}w(x)
&\leq \varphi w(x_0)=G(x_0)\\
&\leq\frac{2(m-1)(n-1)}{m+1}\frac{C^{2}_1}{R^{2}}\sup_{x\in M^n}v+\frac{(m-1)^{2}(n-1)}{m(m+1)}\frac{C_3}{R^2}(1+\sqrt{K}R)\sup_{x\in M^n}v\\
&\quad +\frac{(m-1)^{2}(n-1)^{2}}{m(m+1)}(2K+\tau)\sup_{x\in M^n}v\\
&\quad +\frac{(m-1)(n-1)}{m(m+1)}|\frac{2(m+1)}{n-1}+(m-2l+1)|\,
 |\lambda|\Big(\frac{m-1}{m}\inf_{x\in M^n}v\Big)^{\frac{l-1}{m-1}}.
\end{align*}
The proof is complete.
\end{proof}

\begin{proof}[The proof of Theorem \ref{thm1.3}]
Simple calculations show that
$\frac{2(m+1)}{(m-1)(n-1)}+\frac{m-2l+1}{m-1}\geq 0$ as
$\lambda\geq 0$ and $l\leq \frac{(n+1)(m+1)}{2(n-1)}$ or $\lambda\leq 0$
and $l\geq\frac{(n+1)(m+1)}{2(n-1)}$.
Hence, dropping the last term in  \eqref{3.6} which is nonnegative, we have
\begin{align*}
0&\geq -\Big[\frac{C_2}{R^2}+\frac{(n-1)(1+\sqrt{K}R)C_1}{\sqrt{\theta} R^2}\Big]G
 +\frac{m(m+1)}{(m-1)^{2}(n-1)}\frac{G^2}{v\theta}\\
&\quad -\frac{2m}{m-1}\frac{C^{2}_1}{R^{2}\theta}G-(n-1)(2K+\tau) G.
\end{align*}
Multiplying by $v\theta$ on  both sides, and using
 $0\leq \theta\leq 1$,  we obtain
\begin{align*}
0&\geq \frac{m(m+1)}{(m-1)^{2}(n-1)}G^{2}
 -\frac{2m}{m-1}\frac{C^{2}_1}{R^{2}}G\sup_{x\in M^n}v\\
&\quad -\Big[ \frac{C_2}{R^2}+\frac{(n-1)(1+\sqrt{K}R)C_1}{R^2}\Big]G
 \sup_{x\in M^n}v
 -(n-1)(2K+\tau)G\sup_{x\in M}v.
\end{align*}
Therefore, at the the maximum point $x_0$ we obtain
\begin{align*}
\sup_{B_{p}(R)}w(x)
&\leq \varphi w(x_0)=G(x_0)\\
&\leq \frac{2(m-1)(n-1)}{m+1)}\frac{C^{2}_1}{R^{2}}\sup_{x\in M^n}v
 +\frac{(m-1)^{2}(n-1)}{m(m+1)}\frac{C_3}{R^2}(1+\sqrt{K}R)\sup_{x\in M^n}v\\
&\quad +\frac{(m-1)^{2}(n-1)^{2}}{m(m+1)}(2K+\tau)\sup_{x\in M^n}v,
\end{align*}
where we used  \eqref{3.9}.
The proof is complete.
\end{proof}

\begin{proof}[The proof of Theorem \ref{thm1.5}]
It is not difficult to find that $\frac{2(m+1)}{(m-1)(n-1)}+\frac{m-2l+1}{m-1}\geq 0$
for $1\leq l\leq \frac{(n+1)(m+1)}{2(n-1)}$.
Then we have form \eqref{3.6},
\begin{equation}\label{3.10}
\begin{aligned}
0&\geq \frac{m(m+1)}{(m-1)^{2}(n-1)}G^{2}-\frac{2m}{m-1}
 \frac{C^{2}_1}{R^{2}}\sup_{x\in M^n}vG\\
&\quad -\Big[ \frac{C_2}{R^2}+\frac{(n-1)(1+\sqrt{K}R)C_1}{R^2}\Big]G\sup_{x\in M^n}v\\
&\quad -\big[2(n-1)K+(n-1)\tau\big]G\sup_{x\in M^n}v\\
&\quad +\Big[\frac{2(m+1)}{(m-1)(n-1)}+\frac{m-2l+1}{m-1}\Big]
\lambda\Big(\frac{m-1}{m}\sup_{x\in M^n}v\Big)^{\frac{l-1}{m-1}}G.
\end{aligned}
\end{equation}
By \eqref{3.10}, and   \eqref{3.9} we obtain at the the maximum point $x_0$,
\begin{align*}
\sup_{B_{p}(R)}w(x)
&\leq \varphi w(x_0)=G(x_0)\\
&\leq\frac{2(m-1)(n-1)}{m+1}\frac{C^{2}_1}{R^{2}}\sup_{x\in M^n}v+\frac{(m-1)^{2}(n-1)}{m(m+1)}\frac{C_3}{R^2}(1+\sqrt{K}R)\sup_{x\in M^n}v\\
&\quad +\frac{2(m-1)^{2}(n-1)^{2}}{m(m+1)}K\sup_{x\in M^n}v\\
&\quad -\frac{(m-1)(n-1)}{m(m+1)}\big[\frac{2(m+1)}{n-1}+(m-2l+1)\big]
 \lambda\Big(\frac{m-1}{m}\sup_{x\in M^n}v\Big)^{\frac{l-1}{m-1}}.
\end{align*}
Letting $R\to \infty$, we infer as
$$
\lambda\leq\frac{2(m-1)(n-1)K}{\frac{2(m+1)}{n-1}+(m-2l+1)}
\big(\frac{m-1}{m}\big)^{\frac{l-1}{m-1}}(\sup_{x\in M^n}v)^{\frac{m-l}{m-1}},
$$
and
\begin{align*}
\frac{|\nabla v|^2}{v}
&\leq\frac{2(m-1)^{2}(n-1)^{2}}{m(m+1)}K\sup_{x\in M^n}v\\
&\quad -\frac{(m-1)(n-1)}{m(m+1)}\big[\frac{2(m+1)}{n-1}+(m-2l+1)\big]
\lambda\Big(\frac{m-1}{m}\sup_{x\in M^n}v\Big)^{\frac{l-1}{m-1}}.
\end{align*}
On the other hand, as
$$
\lambda\geq\frac{2(m-1)(n-1)K}{\frac{2(m+1)}{n-1}+(m-2l+1)}
\big(\frac{m-1}{m}\big)^{\frac{l-1}{m-1}}(\sup_{x\in M^n}v)^{\frac{m-l}{m-1}},
$$
we derive that $v$ must be constant.
\end{proof}

\begin{proof}[Proof of Corollary \ref{coro1.2}]
Let minimal geodesic $\gamma(s):[0,1]\to M^n$, so that $\gamma(0)=y$, $\gamma(1)=x$,
 then
\begin{align*}
\ln\frac{v(x)}{v(y)}
&=\int_{0}^{1}\frac{d\ln(v(\gamma(s)))}{ds}
 =\int_{0}^{1}\frac{\nabla v\cdot\gamma'}{v(\gamma(s))}ds\\
&\leq\int_{0}^{1}\frac{|\nabla v|\cdot|\gamma'|}{|v(\gamma(s))|}ds
 =r(x,y)\int_{0}^{1}\frac{|\nabla v|}{|v(\gamma(s))|}ds\\
&\leq r(x,y)\int_{0}^{1}\sqrt{\frac{C(m,n,l,K,\delta,\tau,
\sup_{x\in M^n}v)}{\inf_{x\in M^n}v}}ds\\
&=r(x,y)\sqrt{\frac{C(m,n,l,K,\delta,\tau,\sup_{x\in M}v)}{\inf_{x\in M^n}v}}.
\end{align*}
\end{proof}

\subsection*{Acknowledgments}
This work was supported by the Higher School Natural Science Foundation of Anhui
Province (KJ2016A310, KJ2017A937), by the Higher School outstanding young talent
 support project of Anhui province in 2017 (gxyq 2017048), by
the Academic Research Project of Hefei Normal University
(2017QN41,2017QN44),
and by the Natural Science Foundation of Anhui Province (1708085MA16).

We are grateful to Professor Jiayu Li for his encouragement.
We also thank Professor Qi S Zhang for introduction of this
subject in the summer course. We thank the anonymous referee
for suggestions and references.

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\end{document}