# Nash-Moser iteration approach to gradient estimate and Liouville Property of
quasilinear elliptic equations on complete Riemannian manifolds

Jie He  School of Mathematics and Physics, Beijing University of Chemical
Technology, Chaoyang District, Beijing 100029, China<EMAIL_ADDRESS>,
Jingchen Hu  Hua Loo-Keng Center for Mathematical Sciences, Academy of
Mathematics and Systems Science, Chinese Academy of Sciences, Beijing 100190,
China<EMAIL_ADDRESS>and  Youde Wang  1\. School of Mathematics and
Information Sciences, Guangzhou University; 2. Hua Loo-Keng Key Laboratory of
Mathematics, Institute of Mathematics, Academy of Mathematics and Systems
Science, Chinese Academy of Sciences, Beijing 100190, China; 3. School of
Mathematical Sciences, University of Chinese Academy of Sciences, Beijing
100049, China<EMAIL_ADDRESS>
###### Abstract.

In this paper, we employ the Nash-Moser iteration technique to analyze the
local and global properties of positive solutions to the equation

$\Delta_{p}u+a|\nabla u|^{q}u^{r}=0$

on a complete Riemannian manifold with Ricci curvature bounded from below,
where $p>1$, $q$, $r$ and $a$ are some real constants. Assuming certain
conditions on $a,\,p,\,q$ and $r$, we derive a succinct Cheng-Yau type
gradient estimate for such solutions. This gradient estimate allows us to
obtain a Liouville-type theorem and a Harnack inequality. Our Liouville-type
result is novel even in Euclidean space.

*Corresponding author

Key words: gradient estimate; Nash-Moser iteration; Liouville type theorem

MSC 2020: 58J05; 35B45; 35J92

##  1\. Introduction

One trend in Riemannian geometry since the 1950’s has been the study of how
curvature affects global properties of partial differential equations and
global quantities like the eigenvalues of the Laplacian. It is well-known that
gradient estimates are a fundamental and powerful tool in the analysis of
partial differential equations on Riemannian manifolds.

The most classical results about gradient estimates can be traced back to
Cheng-Yau’s gradient estimate for positive harmonic functions(see [9, 39]).
Let $(M,g)$ be a complete non-compact Riemannian manifold $(M,g)$ with
$\mathrm{Ric}_{g}\geq-(n-1)\kappa g$, Cheng and Yau deduced for any positive
harmonic function $v$ in a geodesic ball $B_{R}(o)\subset M$,

$\displaystyle\sup_{B_{R/2}(o)}\frac{|\nabla v|}{v}\leq
C_{n}\frac{1+\sqrt{\kappa}R}{R}.$ (1.1)

An important feature of Cheng-Yau’s estimate is that the RHS of (1.1) depends
only on $n,\kappa$ and $R$, it does not depend on some other geometric
quantities such as the injective radius and the Hessian or Laplacian of
distance function etc.

Cheng-Yau’s estimate turned out to be very useful, for instance, Harnack
inequality follows immediately from Cheng-Yau’s estimate and Liouville theorem
for global positive harmonic functions on non-compact manifolds with
$\mathrm{Ric}_{g}\geq 0$ is also direct consequence of (1.1). It is well-known
that the Liouville theorem has had a huge impact across many fields, such as
complex analysis, partial differential equations, geometry, probability,
discrete mathematics and complex and algebraic geometry. The impact of the
Liouville theorem has been even larger as the starting point of many further
developments. For more details on the Liouville properties of harmonic
functions and some related theory of function on a manifold we refer to an
expository paper [10] written by T. H. Colding (also see [11, 12, 13, 29,
39]).

On the other hand, Cheng-Yau’s approach can also be used to derive the
estimate of the spectrum of manifolds ([29, 33]), and investigate the geometry
of manifolds ([22, 36, 41]). After Cheng-Yau’s work, gradient estimates for
many other equations defined on Riemannian manifolds are established (see for
example ([20, 21, 23, 37, 39])). This topic has attracted the attention of
many mathematicians.

In this article, we are concerned with the equation

$\begin{cases}\Delta_{p}v+a|\nabla v|^{q}v^{r}=0,&\quad\text{in}\,M;\\\
v>0,&\quad\text{in}\,M;\\\ p>1,~{}a,~{}q,~{}r\in\mathbb{R},&\end{cases}$ (1.2)

where $(M,g)$ is a complete Riemannian manifold with Ricci curvature bounded
from below. Equation (1.2) arises from many classical equations and there are
many questions related to equation (1.2).

The primary objectives of the present paper are twofold: Firstly, we want to
establish an exact Cheng-Yau type gradient estimates for positive solutions to
equation (1.2) on a Riemannian manifold with Ricci curvature bounded below.
Secondly, as application we would like to show some global properties of
solutions to (1.2), for instance, Liouville-type theorems. In addition, we try
to extend the ranges of values of $p,\,q$ and $r$, obtained in the previous
work, such that the Liouville results on solutions to equation (1.2) on an
Euclidean space hold true. To achieve these goals, we employ the Nash-Moser
iteration method and develop an adaptive pointwise estimate for the linearized
operator of the $p$-Laplace operator.

Now, let’s recall some relevant work with the above equation (1.2). In the
case $a=0$, (1.2) reduces to the $p$-Laplace equation. In 2011, Wang and Zhang
[36] proved any positive $p$-harmonic function $v$ on $(M,g)$ with Ricci
curvature $\mathrm{Ric}_{g}\geq-(n-1)\kappa g$ satisfies

$\displaystyle\sup_{B_{R/2}(o)}\frac{|\nabla v|}{v}\leq
C_{n,p}\frac{1+\sqrt{\kappa}R}{R}.$

Later on, Sung and Wang [33] in 2014 proved that the optimal constant
$C_{n,p}$ in the above estimate is $(n-1)/(p-1)$. Wang and Zhang extended
Cheng-Yau’s result for the case $p=2$ ([39]) to any $p>1$ and improved
Kotschwar-Ni’s results [22] by weakening sectional curvature condition to the
Ricci curvature condition.

In the case $a=1$, $q=0$ and $M=\mathbb{R}^{n}$, the equation (1.2) reduces to
the classical Lane-Emden-Fowler equation

$\displaystyle\begin{cases}\Delta_{p}v+v^{r}=0\quad\text{in}\quad\mathbb{R}^{n};\\\
v>0\quad\text{in}\quad\mathbb{R}^{n},\end{cases}$ (1.3)

where $r>0$ and $1<p<n$. The equation (1.3) has already been the subject of
countless publications (see [1, 2, 6, 8, 19, 26, 27, 30, 31]). One of the
questions solved (see [30]) is that the Liouville property holds (i.e.
equation (1.3) admits no positive solution) if and only if

$r\in\left(0,\,\,p^{*}-1\right),$

where $p^{*}=np/(n-p)$ is the Sobolev duality of $p$.

Here we recall some previous work on the gradient estimates of positive
solutions to Lane-Emden-Fowler equation on a complete manifold $(M,g)$ with
$\mathrm{Ric}_{g}\geq-(n-1)\kappa$, i.e.

$\displaystyle\begin{cases}\Delta_{p}v+av^{r}=0,\quad&\text{in}\,M;\\\
v>0,\quad&\text{in}\,M.\end{cases}$ (1.4)

In the case $p=2$ and $a>0$, Wang and Wei [37] derived Cheng-Yau type gradient
estimates for positive solutions to (1.4) under the assumption

$r\in\left(-\infty,\quad\frac{n+1}{n-1}+\frac{2}{\sqrt{n(n-1)}}\right).$

For any $p>1$ and $a>0$, He, Wang and Wei [21] proved that the Cheng-Yau type
gradient estimate holds for positive solutions of (1.4) when

$r\in\left(-\infty,\quad\frac{n+3}{n-1}(p-1)\right).$

We refer to [40, 41] when the $p$-Laplace operator in the equation (1.4) is
replaced by weighted $p$-Laplace Lichnerowicz operator.

On the other hand, if $r=0$ and $a=-1$, then (1.2) reduces to the so called
quasilinear Hamilton-Jacobi equation

$\displaystyle\Delta_{p}v-|\nabla v|^{q}=0.$ (1.5)

Lions [24] proved that any $C^{2}$ solution to (1.5) on $\mathbb{R}^{n}$ with
$q>1$ and $p=2$ must be a constant. Later, Bidaut-Véron, Garcia-Huidobro and
Véron [3] proved that when $n\geq p>1$ and $q>p-1$, the gradient of the
solutions to (1.5) on $\Omega\subset\mathbb{R}^{n}$ can be controlled by the
distance fuction to the boundary of $\Omega$

$\displaystyle|\nabla v(x)|\leq
c(n,p,q)(d(x,\partial\Omega))^{-\frac{1}{q+1-p}},\quad\forall
x\in\Omega\subset\mathbb{R}^{n},$ (1.6)

and obtained some Liouville-type theorems. They also extended their results to
Riemannian manifolds with sectional curvature bounded from below. Recently,
Han, He and Wang in [20] proved any solution $u\in C^{1}(B_{R}(o))$ of
equation (1.5) with $q>p-1$ defined on a geodesic ball $B_{R}(o)$ of a
complete Riemannian manifold $(M,g)$ with $\mathrm{Ric}_{g}\geq-(n-1)\kappa g$
satisfies

$\displaystyle\sup_{B_{R/2}(o)}|\nabla u|\leq
C(n,p,q)\left(\frac{1+\sqrt{\kappa}R}{R}\right)^{q-p+1}.$ (1.7)

If $a$, $q$ and $r$ are all non-zero, one also discussed a more general case
where $p>1$, i.e., the equation

$\displaystyle\begin{cases}\Delta_{p}v+|\nabla
v|^{q}v^{r}=0,\quad&\text{in}\,\mathbb{R}^{n};\\\ v>0,&\text{in
}\,\mathbb{R}^{n}.\end{cases}$ (1.8)

Provided $n>p>1$, Mitidieri and Pohozaev (Theorem 15.1 in [25]) proved that if
$r>0$, $q\geq 0$ and $q+r>p-1$ such that

$\displaystyle r(n-p)+q(n-1)<n(p-1)$ (1.9)

holds, then all nonnegative functions $u\in C^{1}(\mathbb{R}^{n})$ satisfying

$-\Delta_{p}u\geq|\nabla u|^{q}u^{r}\geq 0$

are constant.

Recently, Chang, Hu and Zhang [7] proved that if $n\geq 2$, $\,p,\,q$ and $r$
satisfy one of the following two conditions,

  * •

$0\leq q<p,\,r>0\quad$ and $\quad
1<\frac{q}{p-1}+\frac{r}{p-1}<\frac{n+4}{n}$;

  * •

$1<p<n,\,r\geq 0,\,q\geq 0\quad$ and $\quad r(n-p)+q(n-1)<n(p-1)$,

then positive solutions to the equation (1.8) are constant, and in some sense
their results can be viewed as a complement of Theorem 15.1 in [25]. Bidaut-
Véron [5] obtained some Liouville-type results for (1.8) if $n>p>1,\,r\geq 0$
and $q\geq p$ without the assumption of boundedness on the solution. For more
details on equation (1.2) on $\mathbb{R}^{n}$, we refer to [15, 16, 17, 25]
and the references therein.

In particular, as $p=2$, in many cases one would like to assumes that the
superlinearity of the right-hand side (i.e., $q+r-1>0$ and $0\leq q<2$) of the
equation (1.2) on an Euclidean space. It is easy to see that (1.2) on
$\mathbb{R}^{n}$ is invariant under the action of the transformations
$T_{\sigma}$ defined for $\sigma>0$ by

$T_{\sigma}(u)(x)=\sigma^{\frac{2-q}{q+r-1}}u(\sigma x).$

If we looks for radial positive solutions under the form
$u(x)=\Lambda|x|^{-\gamma}$, we find that, if $q<2$ and $q+r-1>0$, then
$\gamma=\gamma_{q,r}=\frac{2-q}{q+r-1}$ and

$\Lambda=\Lambda_{n,q,r}=\gamma_{q,r}^{\frac{1-q}{q+r-1}}\left(n-\frac{q+2r}{q+r-1}\right)^{\frac{1}{q+r-1}}.$

However, this last quantity exists if and only if the exponents belong to the
supercritical range, that is, when

$q(n-1)+r(n-2)>n.$

In the subcritical range of exponents, that is, when

$q(n-1)+r(n-2)<n,$

we have mentioned in the above that Mitidieri and Pohozaev (Theorem 15.1 in
[25]) proved that if $r>0$ and $q\geq 0$ satisfy $q+r>1$ and

$\displaystyle r(n-2)+q(n-1)<n,$ (1.10)

then all nonnegative functions $u\in C^{1}(\mathbb{R}^{n})$ satisfying

$-\Delta u\geq|\nabla u|^{q}u^{r}\geq 0$

are constant.

Recently, Bidaut-Véron, Garcia-Huidobro and Véron [4] also adopted the well-
known Bernstein method and the Keller-Osserman comparison method to discuss
this equation. Concretely, they consider the following equation

$\displaystyle\begin{cases}\Delta v+|\nabla v|^{q}v^{r}=0,&\text{in
}\,\mathbb{R}^{n};\\\ v>0,&\text{in }\,\mathbb{R}^{n},\end{cases}$ (1.11)

where $0\leq q<2$, $r\geq 0$ and $q+r>1$. The following key gradient
estimates, which are one of the main results of their paper (see Theorem B,
page 1489 in [4]), were derived

######  Theorem.

Let $n\geq 2$, $0\leq q<2$, and $r\geq 0$ be such that $r+q-1>0$. If $u$ is a
positive solution of (1.11) in $B_{R}(0)\subset\mathbb{R}^{n}$ and either of
the assumptions

  1. (1)

$1\leq r<\frac{n+3}{n-1}\quad\text{and}\quad q+r-1<\frac{4}{n-1}$,

  2. (2)

$0\leq r<1\quad\text{and}\quad q+r-1<\frac{(r+1)^{2}}{r(n-1)}$,

is fulfilled, then there exist positive constants $a^{*}=a^{*}(n,q,r)$ and
$c_{1}=c_{1}(n,q,r)$ such that

$|\nabla u^{a^{*}}(0)|\leq c_{1}R^{-1-a^{*}\frac{2-q}{q+r-1}}.$

Note that $\frac{4}{n-1}\leq\frac{(r+1)^{2}}{r(n-1)}$ always holds. The
authors of [4] also pointed out that the value of the exponent $a^{*}$ is not
easy to compute.

As a consequence of the above theorem, they also derived some Liouville
properties for (1.2) on $\mathbb{R}^{n}$ or (1.11), which can be stated as
follows: under the assumptions on $n$, $q$ and $r$ of the above Theorem, any
positive solution of (1.11) in $\mathbb{R}^{n}$ is constant.

Besides, they also gave some other regions of $(q,r)$ for which the Liouville
property holds in the case $0\leq q<2$, $r\geq 0$ and $q+r>1$.

Filippucci, Pucci and Souplet [18] also proved that any positive bounded
classical solution of (1.11) is identically equal to a constant provided
$q>2$.

As for the equation (1.2) defined on a Riemannian manifold, to our knowledge,
the only known Liouville type result is due to Sun, Xiao and Xu [32]. They
showed that, under some additional conditions on the volume growth of geodesic
ball on $M$, there exists no nontrivial non-negative solution of equation
(1.2) with $a>0$. In fact, they discussed the nonexistence of non-negative
solutions to the following differential inequality $\Delta_{p}v+|\nabla
v|^{q}v^{r}\leq 0$ in $M$.

A natural problem remaining in [4] is for us how to obtain the concise
gradient estimates for the positive solutions to (1.2) on $\mathbb{R}^{n}$ or
(1.11) since it is hard to compute $a^{*}$ appeared in the above estimates
obtained in [4]. Furthermore, for the positive solutions to equation (1.2)
with $p>1$ can one also establish a priori estimates of gradients of its
solutions?

On the other hand, one wants to know whether or not the conditions supposed in
the above theorem can be weaken? For instance, whether or not the condition
(1) in the above theorem

$0\leq q<2,\quad r\geq 0,\quad q+r>1,\quad 1\leq
r<\frac{n+3}{n-1}\quad\text{and}\quad q+r<\frac{n+3}{n-1}$

can be weaken to

$\displaystyle q+r<\frac{n+3}{n-1}$ (1.12)

where $q$ and $r$ are any real constants.

The motivation of this article is to answer the above problems and deal with
the the same problems on equation (1.2) defined on a complete Riemannian
manifold. The method we employ is Nash-Moser iteration method, which is
completely different from that adopted in [4]. Now we are in the position to
state our main results.

######  Theorem 1.1.

Let $(M,g)$ be an $n$-dimensional complete Riemannian manifold with Ricci
curvature satisfying ${\rm Ric}_{g}\geq-(n-1)\kappa g$, where $\kappa$ is a
nonnegative constant. Suppose that $p>1$ and $v$ is a $C^{1}$-positive
solution of (1.2) on a geodesic ball $B_{R}(o)\subset M$. If $n$, $p$, $q$,
$r$ and $a$ satisfy

$\displaystyle a\left(\frac{n+1}{n-1}-\frac{q+r}{p-1}\right)\geq 0,$ (1.13)

or

$\displaystyle 1<\frac{q+r}{p-1}<\frac{n+3}{n-1},\quad\forall a\in\mathbb{R},$
(1.14)

then there holds true

$\displaystyle\sup_{B_{R/2}(o)}\frac{|\nabla v|}{v}\leq
C(n,p,q,r)\frac{1+\sqrt{\kappa}R}{R},$ (1.15)

where the constant $C(n,p,q,r)$ depends only on $n$, $p$, $q$ and $r$.

Theorem 1.1 can cover some previous results and give some new conclusions for
some related equations.

######  Remark 1.

Let $(M,g)$ be an $n$-dimensional complete Riemannian manifold with Ricci
curvature ${\rm Ric}_{g}\geq-(n-1)\kappa g$, where $\kappa$ is a non-negative
constant.

  1. (1)

In the case where $a=0$, Wang-Zhang’s gradient estimate (see [36]) can be
recovered by the above Theorem 1.1.

  2. (2)

In the case $q=0$, suppose that $n$, $p$, $r$ and $a$ satisfy

$\displaystyle a\left(\frac{n+1}{n-1}-\frac{r}{p-1}\right)\geq 0,$ (1.16)

or

$\displaystyle 1<\frac{r}{p-1}<\frac{n+3}{n-1},\quad\forall a\in\mathbb{R}.$
(1.17)

Then, the above estimate (1.15) can cover the gradient estimate obtained in
[21].

  3. (3)

In the case $r=0$ and $a=-1$, Theorem 1.1 implies that any positive solution
$v$ of the quasi-linear Hamilton-Jacobi equation (1.5) defined on a geodesic
ball $B_{R}(o)\subset M$ satisfies (1.15) if $q>p-1$.

The results due to Han, He and Wang [20] demonstrate that any solution $u$ of
the quasi-linear Hamilton-Jacobi equation defined on a geodesic ball
$B_{R}(o)\subset M$ satisfies (1.7). For positive solution $v$ of (1.2), $\ln
v$ satisfies a different equation (see (2.1)), so the corresponding gradient
estimates for $\ln v$ and $v$ are different from each other. However, the
necessary condition for both gradient estimates are the same, i.e., $q>p-1$.

By carefully analyzing the conditions (1.13) and (1.14) in Theorem 1.1, it is
easy to see that the following results holds.

######  Corollary 1.2.

Let $p>1$ and $(M,g)$ be an $n$-dim complete Riemannian manifold with ${\rm
Ric}_{g}\geq-(n-1)\kappa g$, where $\kappa$ is a non-negative constant. If

$\displaystyle a\geq 0\quad\text{ and }\quad q+r<\frac{n+3}{n-1}(p-1),$ (1.18)

or

$\displaystyle a\leq 0\quad\text{ and }\quad q+r>p-1,$ (1.19)

then, for any $C^{1}$-positive solution $v$ of (1.2) on a geodesic ball
$B_{R}(o)\subset M$, we have

$\displaystyle\sup_{B_{R/2}(o)}\frac{|\nabla v|}{v}\leq
C(n,p,q,r)\frac{1+\sqrt{\kappa}R}{R}.$

######  Remark 2.

Firstly Theorem 1.1 is the Cheng-Yau type estimate for positive solutions of
equation (1.2). In the case $p=2$ and $a=1$, the assumption (1.18) is just
(1.12) under which the Liouville theorem has been established in [4].

By the gradient estimations obtained in the above, we can get Liouville-type
results for such solutions on a Riemannian manifold with non-negative Ricci
curvature.

######  Theorem 1.3.

Let $p>1$ and $(M,g)$ be an $n$-dimensional complete non-compact Riemannian
manifold with non-negative Ricci curvature. If

$\displaystyle a\geq 0\quad\text{ and }\quad q+r<\frac{n+3}{n-1}(p-1),$

or

$\displaystyle a\leq 0\quad\text{ and }\quad q+r>p-1,$

then any $C^{1}$ smooth positive solution of equation (1.2) on $M$ is a
constant.

In fact, we can also establish a Liouville theorem for equation (1.2) on a
non-compact complete Riemannian manifold with non-negative Ricci curvature on
the same range of values of $p$, $q$ and $r$ where [25, Theorem 15.1]
(Liouville theorem) holds (also see [7]).

######  Theorem 1.4.

Let $(M,g)$ be complete non-compact manifold with non-negative Ricci
curvature. Suppose that $1<p<n$ and $r>0,\,q\geq 0,\,q+r>p-1$, and $a>0$. If

$\displaystyle r+\frac{n-1}{n-p}q<\frac{n(p-1)}{n-p},$

then the problem

$\displaystyle\begin{cases}\Delta_{p}u+au^{r}|\nabla u|^{q}\leq
0,&\quad\text{in}\,\,M;\\\ u>0,&\quad\text{in}\,\,M\end{cases}$ (1.20)

admits no solution $u\in C^{1}(M)$.

######  Remark 3.

To our best knowledge, Theorem 1.3 is the first Liouville type result for
equation (1.2) on Riemannian manifolds with non-negative Ricci curvature. Sun-
Xiao-Xu [32] also obtained some Liouville-type results for (1.2) under the
assumption on volume growth rate of geodesic ball. The classical Cheng-Yau’s
results on harmonic functions on complete Riemannian manifolds show that the
condition on the Ricci tensor in Theorem 1.3 is natural.

######  Remark 4.

When $a=1$ and the equation (1.2) is defined on $\mathbb{R}^{n}$, the previous
results (see [5, 4, 7, 18]) are derived by the Bernstein method. The essential
difference between the Liouville-type theorems of this paper and the previous
Liouville-type theorems is that $q$ and $r$ may be negative in present
situation, however, one needs to assume that $q$ and $r$ are nonnegative in
the previous results. Moreover, we also consider the case where $a<0$ in
Theorem 1.1.

$q=p$$(0,0)$$r$$q$

(a) [5]

$\frac{r}{p-1}+\frac{q}{p-1}=\frac{n+4}{n}$$\frac{q}{p-1}+\frac{r}{p-1}=1$$r\frac{n-p}{p-1}+q\frac{n-1}{p-1}=n$$(0,0)$$r$$q$

(b) [7]

$\frac{q}{p-1}+\frac{r}{p-1}=\frac{n+3}{n-1}$$\frac{q}{p-1}+\frac{r}{p-1}=1$$r\frac{n-p}{p-1}+q\frac{n-1}{p-1}=n$$(0,0)$$r$$q$

(c) ours

Figure 1.  The region of $(q,r)$ where Liouville result holds when $n=4$ and
$p=3$. Our results and previous results can’t cover each other.

######  Remark 5.

In the case where $a=1$, $q\geq 0$ and $r\geq 0$, even for the equation (1.2)
which is defined on $\mathbb{R}^{n}$, the results obtained here are new. For
the Liouville-type results established in the present paper, there is no upper
bound restriction on $p$ while one always need to assume that $p<n$ in the
previous all results. In the case $p=2$ and $a>0$, the range of $(q,r)$ in
Theorem 1.3 is just one of the conditions given in [4], we extend one of the
Liouville results proved for the case $p=2$ in [4] to the case $p>1$. Our
results (Theorem 1.3 Theorem 1.4) and the Liouville type results in [5] can
not cover each other, but our results can completely cover the Liouville
results in [7, 18]. In the case $n=4$ and $p=3$, Figure 1 shows the regions of
$(q,r)$ where the Liouville result holds by different methods.

######  Theorem 1.5.

Let $(M,g)$ be a complete non-compact Riemannian manifold with Ricci curvature
${\rm Ric}_{g}\geq-(n-1)\kappa g$, where $\kappa$ is a non-negative constant.
Suppose $u\in C^{1}(B_{R}(o))$ is a positive solution to equation (1.2),
defined on a geodesic ball $B_{R}(o)\subset M$, with $a$, $p$ and $q$
satisfying (1.18) or (1.19). Then, for any $x,y\in B_{R/2}(o)$ there holds

$u(x)/u(y)\leq e^{c(n,p,q)(1+\sqrt{\kappa}R)}.$

Moreover, if $u\in C^{1}(M)$ is an entire positive solution to equation (1.2)
with $a$, $p$ and $q$ satisfying (1.18) or (1.19) in $M$, then, for any
$x,\,y\in M$ there holds

$u(x)/u(y)\leq e^{c(n,p,q)\sqrt{\kappa}d(x,y)},$

where $d(x,y)$ is the geodesic distance between $x$ and $y$.

The remainder of our paper is organized as follows. In Section 2, we will give
a meticulous estimate of $\mathcal{L}\left(|\nabla\log v|^{2\alpha}\right)$
(see (2.4) for the explicit definition of the operator $\mathcal{L}$) and
recall Saloff-Coste’s Sobolev embedding theorem. In section 3, we use
delicately the Nash-Moser iteration to provide the proofs of the main results
in this paper. The proof of Theorem 1.4 is provided in Section 4.

##  2\. preliminary

In this paper, $(M,g)$ is an $n$-dimensional Riemann manifold and $\nabla$ is
the associated Levi-Civita connection. For any function $\varphi\in C^{1}(M)$,
we denote $\nabla\varphi\in\Gamma(T^{*}M)$ by
$\nabla\varphi(X)=\nabla_{X}\varphi$. The volume form is denoted as
${\rm{vol}}=\sqrt{\det(g_{ij})}dx_{1}\wedge\ldots\wedge dx_{n}$ where
$(x_{1},\ldots,x_{n})$ is a local coordinate. For simplicity, we omit the
volume form in integrations.

For $p>1$, the $p$-Laplace operator is defined by

$\Delta_{p}u:={\rm div}\left(|\nabla u|^{p-2}\nabla u\right).$

The solution of $p$-Laplace equation $\Delta_{p}u=0$ is the critical point of
the energy functional

$E(u)=\int_{M}|\nabla u|^{p}.$

######  Definition 2.1.

A function $v$ is said to be a positive (weak) solution of equation (1.2) on a
region $\Omega$ if $v\in C^{1}(\Omega),v>0$ and for all $\psi\in
C_{0}^{\infty}(\Omega)$, we have

$\displaystyle-\int_{\Omega}|\nabla v|^{p-2}\bm{\langle}\nabla
v,\nabla\psi\bm{\rangle}+a\int_{\Omega}|\nabla u|^{q}v^{r}\psi=0.$

It is worth mentioning that any solution $v$ of equation (1.2) satisfies $v\in
W^{2,2}_{loc}(\Omega)$ and $v\in C^{1,\alpha}(\Omega)$ for some
$\alpha\in(0,1)$(for example, see [14, 34, 35]). Moreover, $v$ is in fact
smooth away from $\\{\nabla v=0\\}$.

Next, we recall Saloff-Coste’s Sobolev inequalities (see [28, Theorem 3.1])
which shall play an important role in our proof of the main theorem.

######  Lemma 2.2 ([28]).

Let $(M,g)$ be a complete manifold with $\mathrm{Ric}_{g}\geq-(n-1)\kappa$.
For $n>2$, there exists a positive constant $c_{0}$ depending only on $n$,
such that for all $B\subset M$ of radius R and volume $V$ we have for $f\in
C^{\infty}_{0}(B)$

$\|f\|_{L^{\frac{2n}{n-2}}(B)}^{2}\leq
e^{c_{0}(1+\sqrt{\kappa}R)}V^{-\frac{2}{n}}R^{2}\left(\int_{B}|\nabla
f|^{2}+R^{-2}f^{2}\right).$

For $n=2$, the above inequality holds if we replace $n$ by some fixed
$n^{\prime}>2$.

For any positive solution $v$ to the equation (1.2), if we take a logarithmic
transformation $v=e^{-\frac{u}{p-1}}$, then we have

$\displaystyle\Delta_{p}u-|\nabla u|^{p}-be^{cu}|\nabla u|^{q}=0,$ (2.1)

where

$\displaystyle b=a(p-1)^{p-1-q}\quad\text{and}\quad c=1-\frac{q+r}{p-1}.$
(2.2)

Denote $f=|\nabla u|^{2}$. It is easy to see that (2.1) becomes

$\displaystyle\Delta_{p}u-f^{\frac{p}{2}}-be^{cu}f^{\frac{q}{2}}=0.$ (2.3)

Now we consider the linearization operator $\mathcal{L}$ of $p$-Laplace
operator:

$\displaystyle\mathcal{L}(\psi)={\rm
div}\left(f^{p/2-1}A\left(\nabla\psi\right)\right),$ (2.4)

where

$\displaystyle f=|\nabla u|^{2}$ (2.5)

and

$\displaystyle
A(\nabla\psi)=\nabla\psi+(p-2)f^{-1}\bm{\langle}\nabla\psi,\nabla
u\bm{\rangle}\nabla u.$ (2.6)

We first derive a useful expression of $\mathcal{L}(f^{\alpha})$ for any
$\alpha>0$.

######  Lemma 2.3.

For any $\alpha>0$, the following identity holds point-wisely in
$\\{x:f(x)>0\\}$,

$\displaystyle\begin{split}\mathcal{L}(f^{\alpha})=&\alpha(\alpha+\frac{p}{2}-2)f^{\alpha+\frac{p}{2}-3}|\nabla
f|^{2}+2\alpha f^{\alpha+\frac{p}{2}-2}\left(|\nabla\nabla u|^{2}+{\rm
Ric}(\nabla u,\nabla u)\right)\\\
&+\alpha(p-2)(\alpha-1)f^{\alpha+\frac{p}{2}-4}\langle\nabla f,\nabla
u\rangle^{2}+2\alpha f^{\alpha-1}\langle\nabla\Delta_{p}u,\nabla
u\rangle.\end{split}$ (2.7)

###### Proof.

By the definition of $A$ in (2.6), we have

$A\Big{(}\nabla(f^{\alpha})\Big{)}=\alpha f^{\alpha-1}\nabla
f+\alpha(p-2)f^{\alpha-2}\langle\nabla f,\nabla u\rangle\nabla u=\alpha
f^{\alpha-1}A(\nabla f).$

Hence

$\displaystyle\mathcal{L}(f^{\alpha})$
$\displaystyle=\alpha\text{div}\Big{(}f^{\alpha-1}f^{\frac{p}{2}-1}A(\nabla
f)\Big{)}=\alpha\Big{\langle}\nabla(f^{\alpha-1}),f^{\frac{p}{2}-1}A(\nabla
f)\Big{\rangle}+\alpha f^{\alpha-1}\mathcal{L}(f).$

Direction computation shows that

$\displaystyle\alpha\Big{\langle}\nabla(f^{\alpha-1}),f^{\frac{p}{2}-1}A(\nabla
f)\Big{\rangle}=$
$\displaystyle\Big{\langle}\alpha(\alpha-1)f^{\alpha-2}\nabla
f,f^{\frac{p}{2}-1}\nabla f+(p-2)f^{\frac{p}{2}-2}\langle\nabla f,\nabla
u\rangle\nabla u\Big{\rangle},$ (2.8)

and

$\displaystyle\begin{split}\alpha f^{\alpha-1}\mathcal{L}(f)=&\alpha
f^{\alpha-1}\Big{(}\left(\frac{p}{2}-1\right)f^{\frac{p}{2}-2}|\nabla
f|^{2}+f^{\frac{p}{2}-1}\Delta
f+(p-2)\left(\frac{p}{2}-2\right)f^{\frac{p}{2}-3}\langle\nabla f,\nabla
u\rangle^{2}\\\ &+(p-2)f^{\frac{p}{2}-2}\langle\nabla\langle\nabla f,\nabla
u\rangle,\nabla u\rangle+(p-2)f^{\frac{p}{2}-2}\langle\nabla f,\nabla
u\rangle\Delta u\Big{)}.\end{split}$ (2.9)

Combining (2.8) and (2.9) yields

$\displaystyle\begin{split}\mathcal{L}(f^{\alpha})=&\alpha\left(\alpha+\frac{p}{2}-2\right)f^{\alpha+\frac{p}{2}-3}|\nabla
f|^{2}+\alpha f^{\alpha+\frac{p}{2}-2}\Delta f\\\
&+\alpha(p-2)\left(\alpha+\frac{p}{2}-3\right)f^{\alpha+\frac{p}{2}-4}\langle\nabla
f,\nabla u\rangle^{2}\\\
&+\alpha(p-2)f^{\alpha+\frac{p}{2}-3}\langle\nabla\langle\nabla f,\nabla
u\rangle,\nabla u\rangle+\alpha(p-2)f^{\alpha+\frac{p}{2}-3}\langle\nabla
f,\nabla u\rangle\Delta u.\end{split}$ (2.10)

On the other hand, by the definition of the $p$-Laplacian we have

$\displaystyle\langle\nabla\Delta_{p}u,\nabla u\rangle$
$\displaystyle=\left(\frac{p}{2}-1\right)\left(\frac{p}{2}-2\right)f^{\frac{p}{2}-3}\langle\nabla
f,\nabla
u\rangle^{2}+\left(\frac{p}{2}-1\right)f^{\frac{p}{2}-2}\langle\nabla\langle\nabla
f,\nabla u\rangle,\nabla u\rangle$
$\displaystyle\quad+\left(\frac{p}{2}-1\right)f^{\frac{p}{2}-2}\langle\nabla
f,\nabla u\rangle\Delta u+f^{\frac{p}{2}-1}\langle\nabla\Delta u,\nabla
u\rangle.$

Thus, the last term of (2.10) can be rewritten as

$\displaystyle\begin{split}\alpha(p-2)f^{\alpha+\frac{p}{2}-3}\langle\nabla
f,\nabla u\rangle\Delta u&=2\alpha f^{\alpha-1}\langle\nabla\Delta_{p}u,\nabla
u\rangle-2\alpha f^{\alpha+\frac{p}{2}-2}\langle\nabla\Delta u,\nabla
u\rangle\\\
&\quad-\alpha(p-2)\left(\frac{p}{2}-2\right)f^{\alpha+\frac{p}{2}-4}\langle\nabla
f,\nabla u\rangle^{2}\\\
&\quad-\alpha(p-2)f^{\alpha+\frac{p}{2}-3}\langle\nabla\langle\nabla f,\nabla
u\rangle,\nabla u\rangle.\end{split}$ (2.11)

By (2.11) and the following Bochner formula

$\frac{1}{2}\Delta f=|\nabla\nabla u|^{2}+{\rm Ric}(\nabla u,\nabla
u)+\langle\nabla\Delta u,\nabla u\rangle,$

we have

$\displaystyle\mathcal{L}(f^{\alpha})=$
$\displaystyle\alpha\left(\alpha+\frac{p}{2}-2\right)f^{\alpha+\frac{p}{2}-3}|\nabla
f|^{2}+2\alpha f^{\alpha+\frac{p}{2}-2}\left(|\nabla\nabla u|^{2}+{\rm
Ric}(\nabla u,\nabla u)\right)$
$\displaystyle+\alpha(p-2)(\alpha-1)f^{\alpha+\frac{p}{2}-4}\langle\nabla
f,\nabla u\rangle^{2}+2\alpha f^{\alpha-1}\langle\nabla\Delta_{p}u,\nabla
u\rangle.$

∎

######  Remark 6.

In the above lemma, we have used the following two facts:

  1. (1)

By elliptic regularity theory, one know that $u$ is smooth away from
$\\{x:f(x)=0\\}$. So the above equality holds true in the smooth sense.

  2. (2)

The parameter $\alpha$ is employed to get a larger range of $(q,r)$ where the
Cheng-Yau type gradient estimate holds for (1.2). Obviously, the larger
$\alpha$ is, the better the estimate of $\mathcal{L}(f^{\alpha})/\alpha$ is.
Taking $\alpha\to\infty$ leads to the critical range of $(q,r)$ where the
Liouville theorem is establishes by our method. When $\alpha=1$ and
$\alpha=p/2$, Lemma 2.3 has been established in [22] and [38].

##  3\. Proof of Main Theorem

We divide the proof of Theorem 1.1 into three parts. In the first part, we
derive a fundamental integral inequality on $f=|\nabla u|^{2}$, which will be
used in the second and third parts. In the second part, we give an
$L^{\beta}$-estimate of $f$ on a geodesic ball with radius $3R/4$, where
$L^{\beta}$ norm of $f$ determines the initial state of the Nash-Moser
iteration. Finally, we give a complete proof of our theorem by an intensive
use of Nash-Moser iteration method.

###  3.1. Integral inequality

In order to establish the main theorem, we need to show some lemmas on
integral estimates. We first introduce the following point-wise estimate for
$\mathcal{L}(f^{\alpha})$.

######  Lemma 3.1.

Let $(M,g)$ be an $n$-dim complete Riemannian manifold with ${\rm
Ric}_{g}\geq-(n-1)\kappa g$ where $\kappa$ is a non-negative constant. Assume
that $v$ is a positive solution to the equation (1.2), $u=-(p-1)\log v$ and
$f=|\nabla u|^{2}$. If $n,\,p,\,q,\,r$ and $a$ satisfy

$\displaystyle a\left(\frac{n+1}{n-1}-\frac{q+r}{p-1}\right)\geq 0.$

or

$\displaystyle 1<\frac{q+r}{p-1}<\frac{n+3}{n-1},\quad\forall a\in\mathbb{R}.$

Then there exists constant $\alpha_{0}=\alpha_{0}(n,p,q,r)>0$ such that the
following inequality

$\displaystyle\begin{split}\mathcal{L}(f^{\alpha_{0}})\geq&2\alpha_{0}\beta_{n,p,q,r}f^{\alpha_{0}+\frac{p}{2}}-2\alpha_{0}(n-1)\kappa
f^{\alpha_{0}+\frac{p}{2}-1}-\alpha_{0}a_{1}|\nabla
f|f^{\alpha_{0}+\frac{p-3}{2}}\end{split}$

holds point-wisely in $\\{x:f(x)>0\\}$, where
$a_{1}=\left|p-\frac{2(p-1)}{n-1}\right|$.

###### Proof.

Let $\\{e_{1},e_{2},\ldots,e_{n}\\}$ be an orthonormal frame of $TM$ on a
domain with $f\neq 0$ such that $e_{1}=\frac{\nabla u}{|\nabla u|}$. We have
$u_{1}=f^{1/2}$. Direct computation shows

$\displaystyle
u_{11}=\frac{1}{2}f^{-1/2}f_{1}=\frac{1}{2}f^{-1}\bm{\langle}\nabla u,\nabla
f\bm{\rangle},$ (3.1)

and

$\displaystyle|\nabla f|^{2}/f=4\sum_{i=1}^{n}u_{1i}^{2}\geq 4u^{2}_{11}.$
(3.2)

It follows from (2.3) and (3.1) that

$\displaystyle\begin{split}\langle\nabla\Delta_{p}u,\nabla u\rangle=&\
\frac{p}{2}f^{\frac{p}{2}-1}\bm{\langle}\nabla f,\nabla
u\bm{\rangle}+be^{cu}\frac{q}{2}f^{\frac{q}{2}-1}\bm{\langle}\nabla f,\nabla
u\bm{\rangle}+bce^{cu}f^{\frac{q}{2}+1}\\\ =&\
pf^{\frac{p}{2}}u_{11}+bqe^{cu}f^{\frac{q}{2}}u_{11}+bce^{cu}f^{\frac{q}{2}+1}.\end{split}$
(3.3)

By Cauchy inequality, we arrive at

$\displaystyle|\nabla\nabla u|^{2}\geq$ $\displaystyle
u_{11}^{2}+\sum_{i=2}u_{ii}^{2}\geq
u_{11}^{2}+\frac{1}{n-1}\left(\sum_{i=2}u_{ii}\right)^{2}.$ (3.4)

From now only, we always assume that $\alpha\geq 3/2$, which can guarantee
$\alpha+p/2-1>0$. Substituting (3.1), (3.2), (3.3) and (3.4) into (2.7) and
collecting terms about $u_{11}^{2}$, we obtained

$\displaystyle\begin{split}\frac{f^{2-\alpha-\frac{p}{2}}}{2\alpha}\mathcal{L}(f^{\alpha})\geq&\
(2\alpha-1)(p-1)u_{11}^{2}+{\rm Ric}(\nabla u,\nabla
u)+\frac{1}{n-1}\left(\sum_{i=2}u_{ii}\right)^{2}\\\
&+pfu_{11}+bqe^{cu}f^{1+\frac{p-q}{2}}u_{11}+bce^{cu}f^{\frac{q-p}{2}+2}.\end{split}$
(3.5)

If we express the $p$-Lapalce in terms of $f$, we have

$\displaystyle\Delta_{p}u=$ $\displaystyle
f^{\frac{p}{2}-1}\left((p-1)u_{11}+\sum_{i=2}^{n}u_{ii}\right).$ (3.6)

Substituting (3.6) into equation (2.3), we obtain

$\displaystyle(p-1)u_{11}+\sum_{i=2}^{n}u_{ii}^{2}=f+be^{cu}f^{\frac{q-p}{2}+1}.$
(3.7)

It follows from (3.7) that

$\displaystyle\begin{split}\frac{1}{n-1}(\sum_{i=2}^{n}u_{ii}^{2})=\frac{1}{n-1}&\left(f+be^{cu}f^{\frac{q-p}{2}+1}-(p-1)u_{11}\right)^{2}\\\
=\frac{1}{n-1}&\Big{(}f^{2}+b^{2}e^{2cu}f^{q-p+2}+(p-1)^{2}u^{2}_{11}+2be^{cu}f^{\frac{q-p}{2}+2}\\\
&\ -2(p-1)fu_{11}-2b(p-1)e^{cu}f^{\frac{q-p}{2}+1}u_{11}\Big{)}.\end{split}$
(3.8)

It follows from (3.5) and (3.8) that

$\displaystyle\begin{split}\frac{f^{2-\alpha-\frac{p}{2}}}{2\alpha}\mathcal{L}(f^{\alpha})\geq&\left((2\alpha-1)(p-1)+\frac{(p-1)^{2}}{n-1}\right)u_{11}^{2}+\frac{f^{2}}{n-1}+\frac{b^{2}e^{2cu}f^{q-p+2}}{n-1}\\\
&+\left(p-\frac{2(p-1)}{n-1}\right)fu_{11}+{\rm Ric}(\nabla u,\nabla u)\\\
&+b\left(q-\frac{2(p-1)}{n-1}\right)e^{cu}f^{1+\frac{q-p}{2}}u_{11}+b\left(c+\frac{2}{n-1}\right)e^{cu}f^{\frac{q-p}{2}+2}.\end{split}$
(3.9)

Making use of $a^{2}+2ab\geq-b^{2}$ , we have

$\displaystyle\begin{split}&\left((2\alpha-1)(p-1)+\frac{(p-1)^{2}}{n-1}\right)u_{11}^{2}+b\left(q-\frac{2(p-1)}{n-1}\right)e^{cu}f^{1+\frac{q-p}{2}}u_{11}\\\
\geq&-\frac{\left(q(n-1)-2(p-1)\right)^{2}}{4(n-1)(p-1)((2\alpha-1)(n-1)+p-1)}b^{2}e^{2cu}f^{q-p+2}.\end{split}$
(3.10)

Substituting (3.10) into (3.9) yields

$\displaystyle\begin{split}\frac{f^{2-\alpha-\frac{p}{2}}}{2\alpha}\mathcal{L}(f^{\alpha})\geq&\
\frac{f^{2}}{n-1}+B_{n,p,q,\alpha}b^{2}e^{2cu}f^{q-p+2}+{\rm Ric}(\nabla
u,\nabla u)\\\
&+\left(p-\frac{2(p-1)}{n-1}\right)fu_{11}+b\left(c+\frac{2}{n-1}\right)e^{cu}f^{\frac{q-p}{2}+2},\end{split}$
(3.11)

where

$\displaystyle
B_{n,p,q,\alpha}=\frac{1}{n-1}-\frac{\left(q(n-1)-2(p-1)\right)^{2}}{(n-1)(p-1)((2\alpha-1)(n-1)+p-1)}.$
(3.12)

For any $n,p,q$ with $n>1,p>1$, we can always choose $\alpha$ large enough
such that $B_{n,p,q,\alpha}$ is positive.

Case 1.

$a\left(\frac{n+1}{n-1}-\frac{q+r}{p-1}\right)\geq 0.$

In this case, by the definition of $b,c$ in (2.2) we have

$b\left(c+\frac{2}{n-1}\right)=(p-1)^{p-q-1}a\left(\frac{n+1}{n-1}-\frac{q+r}{p-1}\right)\geq
0.$

Hence,

$\displaystyle\frac{f^{2-\alpha-\frac{p}{2}}}{2\alpha}\mathcal{L}(f^{\alpha})\geq$
$\displaystyle\ \frac{f^{2}}{n-1}+B_{n,p,q,\alpha}b^{2}e^{2cu}f^{q-p+2}+{\rm
Ric}(\nabla u,\nabla u)$
$\displaystyle+\left(p-\frac{2(p-1)}{n-1}\right)fu_{11}.$

By (3.12), we have

$\lim_{\alpha\to\infty}B_{n,p,q,\alpha}=\frac{1}{n-1}>0.$

So there exists a constant $\alpha_{1}=\alpha_{1}(n,p,q)\geq 3/2$, for any
$\alpha\geq\alpha_{1}$, we have $B_{n,p,q,\alpha}>0$ and

$\displaystyle\frac{f^{2-\alpha-\frac{p}{2}}}{2\alpha}\mathcal{L}(f^{\alpha})\geq$
$\displaystyle\frac{f^{2}}{n-1}+{\rm Ric}(\nabla u,\nabla
u)+\left(p-\frac{2(p-1)}{n-1}\right)fu_{11}.$

Case 2.

$1<\frac{q+r}{p-1}<\frac{n+3}{n-1}.$

This condition is equivalent to

$\displaystyle\begin{split}\frac{1}{n-1}-\frac{(n-1)}{4}\left(\frac{q+r}{p-1}-\frac{n+1}{n-1}\right)^{2}>0.\end{split}$
(3.13)

Since

$B_{n,p,q}b^{2}e^{2cu}f^{q-p+2}+b\left(c+\frac{2}{n-1}\right)e^{cu}f^{\frac{q-p}{2}+2}\geq-\frac{1}{4B_{n,p,q,\alpha}}\left(c+\frac{2}{n-1}\right)^{2}f^{2},$

we can conclude from (3.11) that

$\displaystyle\frac{f^{2-\alpha-\frac{p}{2}}}{2\alpha}\mathcal{L}(f^{\alpha})\geq$
$\displaystyle\left(\frac{1}{n-1}-\frac{1}{4B_{n,p,q,\alpha}}\left(c+\frac{2}{n-1}\right)^{2}\right)f^{2}$
$\displaystyle+{\rm Ric}\left(\nabla u,\nabla
u\right)+\left(p-\frac{2(p-1)}{n-1}\right)fu_{11}.$

It follows from (3.12) and (3.13) that

$\lim_{\alpha\to\infty}\frac{1}{n-1}-\frac{1}{4B_{n,p,q,\alpha}}\left(c+\frac{2}{n-1}\right)^{2}=\frac{1}{n-1}-\frac{(n-1)}{4}\left(\frac{q+r}{p-1}-\frac{n+1}{n-1}\right)^{2}>0.$

Then there exists a constant $\alpha_{2}=\alpha_{2}(n,p,q,r)\geq 3/2$, if we
choose $\alpha\geq\alpha_{2}$, we have

$\beta_{n,p,q,r,\alpha}:=\frac{1}{n-1}-\frac{1}{4B_{n,p,q,\alpha}}\left(c+\frac{2}{n-1}\right)^{2}>0,$

then we have

$\displaystyle\begin{split}\frac{f^{2-\alpha-\frac{p}{2}}}{2\alpha}\mathcal{L}(f^{\alpha})\geq&\
\beta_{n,p,q,r,\alpha}f^{2}+{\rm Ric}(\nabla u,\nabla
u)+\left(p-\frac{2(p-1)}{n-1}\right)fu_{11}\\\ \geq&\
\beta_{n,p,q,r,\alpha}f^{2}-(n-1)\kappa
f-\frac{a_{1}}{2}f^{\frac{1}{2}}|\nabla f|,\end{split}$ (3.14)

where

$a_{1}=\left|p-\frac{2(p-1)}{n-1}\right|.$

From now on, we choose $\alpha_{0}=\max\\{\alpha_{1},\alpha_{2}\\}$ and denote
$\beta_{n,p,q,r}=\beta_{n,p,q,r,\alpha_{0}}$. We summarize the above two cases
to obtain the following point-wise estimate

$\displaystyle\begin{split}\mathcal{L}(f^{\alpha_{0}})\geq&2\alpha_{0}\beta_{n,p,q,r}f^{\alpha_{0}+\frac{p}{2}}-2\alpha_{0}(n-1)\kappa
f^{\alpha_{0}+\frac{p}{2}-1}-\alpha_{0}a_{1}|\nabla
f|f^{\alpha_{0}+\frac{p-3}{2}},\end{split}$ (3.15)

since

$\beta_{n,p,q,r}\leq\frac{1}{n-1}.$

Thus, we complete the proof of this lemma. ∎

######  Lemma 3.2.

Let $(M,g)$ be an $n$-dim complete Riemannian manifold with ${\rm
Ric}_{g}\geq-(n-1)\kappa g$ where $\kappa$ is a non-negative constant and
$\Omega=B_{R}(o)\subset M$ be a geodesic ball. Assume that $v$ is a positive
solution to the equation (1.2), $u=-(p-1)\ln v$ and $f=|\nabla u|^{2}$. Then
there exist constants $\beta_{n,p,q,r}$, $a_{3},\,a_{4}$ and $a_{5}$ depending
only on $n,p,q$ and $r$ such that for any $t\geq t_{0}$, where $t_{0}$ is
defined in (3.29), and we have

$\displaystyle\begin{split}&\beta_{n,p,q,r}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t}\eta^{2}+\frac{a_{3}}{t}e^{-t_{0}}V^{\frac{2}{n}}R^{-2}\left\|f^{\frac{\alpha_{0}+t-1}{2}+\frac{p}{4}}\eta\right\|_{L^{\frac{2n}{n-2}}}^{2}\\\
\leq\
&a_{5}t_{0}^{2}R^{-2}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t-1}\eta^{2}+\frac{a_{4}}{t}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t-1}|\nabla\eta|^{2},\end{split}$

where $\eta\in C^{\infty}_{0}(\Omega,\mathbb{R})$.

###### Proof.

By the regularity theory on elliptic equations, $u$ is smooth away from
$\\{f=0\\}$. So the both sides of (3.15) are in fact smooth. Let $\epsilon>0$
and $\psi=f_{\epsilon}^{\alpha}\eta^{2}$, where
$f_{\epsilon}=(f-\epsilon)^{+}$, $\eta\in C^{\infty}_{0}(B_{R}(o))$ is non-
negative and $\alpha>1$ which will be determined later. Now, we multiply the
both sides of (3.15) by $\psi$, integrate over $\Omega$ to get

$\displaystyle\begin{split}&-\int_{\Omega}\alpha_{0}tf^{\alpha_{0}+\frac{p}{2}-2}f_{\epsilon}^{t-1}|\nabla
f|^{2}\eta^{2}+t\alpha_{0}(p-2)f^{\alpha_{0}+\frac{p}{2}-3}f_{\epsilon}^{t-1}\bm{\langle}\nabla
f,\nabla u\bm{\rangle}^{2}\eta^{2}\\\
&-\int_{\Omega}2\eta\alpha_{0}f^{\alpha+\frac{p}{2}-2}f_{\epsilon}^{t}\bm{\langle}\nabla
f,\nabla\eta\bm{\rangle}+2\alpha_{0}\eta(p-2)f^{\alpha+\frac{p}{2}-3}f_{\epsilon}^{t}\bm{\langle}\nabla
f,\nabla u\bm{\rangle}\bm{\langle}\nabla u,\nabla\eta\bm{\rangle}\\\
\geq&2\beta_{n,p,q,r}\alpha_{0}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}}f_{\epsilon}^{t}\eta^{2}-2(n-1)\alpha_{0}\kappa\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}-1}f_{\epsilon}^{t}\eta^{2}-a_{1}\alpha_{0}\int_{\Omega}f^{\alpha_{0}+\frac{p-3}{2}}f_{\epsilon}^{t}|\nabla
f|\eta^{2}.\end{split}$ (3.16)

It is easy to see that

$\displaystyle f_{\epsilon}^{t-1}|\nabla
f|^{2}+(p-2)f_{\epsilon}^{t-1}f^{-1}\bm{\langle}\nabla f,\nabla
u\bm{\rangle}^{2}\geq a_{2}f_{\epsilon}^{t-1}|\nabla f|^{2},$ (3.17)

where $a_{2}=\min\\{1,p-1\\}$, and

$\displaystyle f_{\epsilon}^{t}\bm{\langle}\nabla
f,\nabla\eta\bm{\rangle}+(p-2)f_{\epsilon}^{t}f^{-1}\bm{\langle}\nabla
f,\nabla u\bm{\rangle}\bm{\langle}\nabla
u,\nabla\eta\bm{\rangle}\geq-(p+1)f_{\epsilon}^{t}|\nabla f||\nabla\eta|.$
(3.18)

Substituting (3.17)) and (3.18) into (3.16) and letting $\epsilon\to 0$, we
obtain

$\displaystyle\begin{split}&2\beta_{n,p,q,r}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t}\eta^{2}+\int_{\Omega}a_{2}tf^{\alpha_{0}+\frac{p}{2}+t-3}|\nabla
f|^{2}\eta^{2}\\\
\leq&~{}2(n-1)\kappa\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t-1}\eta^{2}+a_{1}\int_{\Omega}f^{\alpha_{0}+\frac{p-3}{2}+t}|\nabla
f|\eta^{2}\\\ &+2(p-1)\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t-2}|\nabla
f||\nabla\eta|\eta,\end{split}$ (3.19)

where we have divided the both sides of the inequality by $\alpha_{0}$.

Using Cauchy-inequality, we have

$\displaystyle\begin{split}a_{1}f^{\alpha_{0}+\frac{p-3}{2}+t}|\nabla
f|\eta^{2}\leq&\ \frac{a_{2}t}{4}f^{\alpha_{0}+\frac{p}{2}+t-3}|\nabla
f|^{2}\eta^{2}+\frac{a_{1}^{2}}{a_{2}t}f^{\alpha_{0}+\frac{p}{2}+t}\eta^{2},\\\
\end{split}$ (3.20)

and

$\displaystyle\begin{split}2(p+1)f^{\alpha_{0}+\frac{p}{2}+t-2}|\nabla
f||\nabla\eta|\eta\leq&\ \frac{a_{2}t}{4}f^{\alpha_{0}+\frac{p}{2}+t-3}|\nabla
f|^{2}\eta^{2}+\frac{4(p+1)^{2}}{a_{2}t}f^{\alpha_{0}+\frac{p}{2}+t-1}|\nabla\eta|^{2}.\end{split}$
(3.21)

Now we choose $t$ large enough such that

$\displaystyle\frac{a_{1}^{2}}{a_{2}t}\leq\beta_{n,p,q,r}.$ (3.22)

It follows from (3.19), (3.20), (3.21) and (3.22) that

$\displaystyle\begin{split}&\beta_{n,p,q,r}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t}\eta^{2}+\frac{a_{2}t}{2}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t-3}|\nabla
f|^{2}\eta^{2}\\\ \leq&\
2(n-1)\kappa\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t-1}\eta^{2}+\frac{4(p+1)^{2}}{a_{2}t}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t-1}|\nabla\eta|^{2}.\end{split}$
(3.23)

On the other hand, we have

$\displaystyle\begin{split}\left|\nabla\left(f^{\frac{\alpha_{0}+t-1}{2}+\frac{p}{4}}\eta\right)\right|^{2}\leq&\
2\left|\nabla
f^{\frac{\alpha_{0}+t-1}{2}+\frac{p}{4}}\right|^{2}\eta^{2}+2f^{\alpha_{0}+t-1+\frac{p}{2}}|\nabla\eta|^{2}\\\
=&\
\frac{(\alpha_{0}+t+\frac{p}{2}-1)^{2}}{2}f^{\alpha_{0}+t+\frac{p}{2}-3}|\nabla
f|^{2}\eta^{2}+2f^{\alpha_{0}+t-1+\frac{p}{2}}|\nabla\eta|^{2}.\end{split}$
(3.24)

Now substituting (3.24) into (3.23) leads to

$\displaystyle\begin{split}&\beta_{n,p,q,r}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t}\eta^{2}+\frac{4a_{2}t}{(2\alpha_{0}+2t+p-2)^{2}}\int_{\Omega}\left|\nabla\left(f^{\frac{\alpha_{0}+t-1}{2}+\frac{p}{4}}\eta\right)\right|^{2}\\\
\leq&\
2(n-1)\kappa\int_{\Omega}f^{\alpha_{0}+t+\frac{p}{2}-1}\eta^{2}+\frac{4(p-1)^{2}}{a_{2}t}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t-1}|\nabla\eta|^{2}\\\
&\
+\frac{8a_{2}t}{(2\alpha_{0}+2t+p-2)^{2}}\int_{\Omega}f^{\alpha_{0}+t+\frac{p}{2}-1}|\nabla\eta|^{2}.\end{split}$
(3.25)

From now on we use $a_{1},\,a_{2},\,a_{3}\,\cdots$ to denote constants
depending only on $n,\,p,\,q$ and $r$. We choose $a_{3}$ and $a_{4}$ such that

$\displaystyle\frac{a_{3}}{t}\leq\frac{4a_{2}t}{(2\alpha_{0}+2t+p-2)^{2}}\quad\text{and}\quad\frac{8a_{2}t}{(2\alpha_{0}+2t+p-2)^{2}}+\frac{4(p-1)^{2}}{a_{2}t}\leq\frac{a_{4}}{t},$
(3.26)

since $t$ satisfies (3.22).

It follows from (3.25) and (3.26) that

$\displaystyle\begin{split}&\beta_{n,p,q,r}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t}\eta^{2}+\frac{a_{3}}{t}\int_{\Omega}\left|\nabla\left(f^{\frac{\alpha_{0}+t-1}{2}+\frac{p}{4}}\eta\right)\right|^{2}\\\
\leq&\
2(n-1)\kappa\int_{\Omega}f^{\alpha_{0}+t+\frac{p}{2}-1}\eta^{2}+\frac{a_{4}}{t}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t-1}\left|\nabla\eta\right|^{2}.\end{split}$
(3.27)

Noting that Saloff-Coste’s Sobolev inequality implies

$e^{-c_{0}(1+\sqrt{\kappa}R)}V^{\frac{2}{n}}R^{-2}\left\|f^{\frac{\alpha_{0}+t-1}{2}+\frac{p}{4}}\eta\right\|_{L^{\frac{2n}{n-2}}(\Omega)}^{2}\leq\int_{\Omega}\left|\nabla\left(f^{\frac{\alpha_{0}+t-1}{2}+\frac{p}{4}}\eta\right)\right|^{2}+R^{-2}\int_{\Omega}f^{\alpha_{0}+t+\frac{p}{2}-1}\eta^{2},$

furthermore we can infer from (3.27) that

$\displaystyle\begin{split}&\beta_{n,p,q,r}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t}\eta^{2}+\frac{a_{3}}{t}e^{-c_{0}(1+\sqrt{\kappa}R)}V^{\frac{2}{n}}R^{-2}\left\|f^{\frac{\alpha_{0}+t-1}{2}+\frac{p}{4}}\eta\right\|_{L^{\frac{2n}{n-2}}}^{2}\\\
\leq&\
2(n-1)\kappa\int_{\Omega}f^{\alpha_{0}+t+\frac{p}{2}-1}\eta^{2}+\frac{a_{4}}{t}\int_{\Omega}f^{\alpha_{0}+t+\frac{p}{2}-1}|\nabla\eta|^{2}+\frac{a_{3}}{t}\int_{\Omega}R^{-2}f^{\alpha_{0}+\frac{p}{2}+t-1}\eta^{2}.\end{split}$
(3.28)

Now we choose

$\displaystyle
t_{0}=c_{n,p,q,r}\left(1+\sqrt{\kappa}R\right)\quad\text{and}\quad
c_{n,p,q,r}=\max\left\\{c_{0}+1,\frac{a_{1}^{2}}{a_{2}\beta_{n,p,q,r}}\right\\},$
(3.29)

and pick $t$ such that $t\geq t_{0}$. Since

$\displaystyle 2(n-1)\kappa
R^{2}\leq\frac{2(n-1)}{c_{n,p,q,r}^{2}}t_{0}^{2}\quad\text{
and}\quad\frac{a_{3}}{t}\leq\frac{a_{3}}{c_{n,p,q,r}},$

there exists $a_{5}=a_{5}(n,p,q,r)>0$ such that

$\displaystyle 2(n-1)\kappa R^{2}+\frac{a_{3}}{t}\leq
a_{5}t_{0}^{2}=a_{5}c^{2}_{n,p,q,r}\left(1+\sqrt{\kappa}R\right)^{2}.$ (3.30)

Immediately, it follows from (3.28) and (3.30) that

$\displaystyle\begin{split}&\beta_{n,p,q,r}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t}\eta^{2}+\frac{a_{3}}{t}e^{-t_{0}}V^{\frac{2}{n}}R^{-2}\left\|f^{\frac{\alpha_{0}+t-1}{2}+\frac{p}{4}}\eta\right\|_{L^{\frac{2n}{n-2}}}^{2}\\\
\leq\
&a_{5}t_{0}^{2}R^{-2}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t-1}\eta^{2}+\frac{a_{4}}{t}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t-1}|\nabla\eta|^{2}.\end{split}$
(3.31)

Thus, we finish the proof. ∎

###  3.2. $L^{\beta}$ bound of gradient in a geodesic ball with radius $3R/4$

######  Lemma 3.3.

Let $(M,g)$ be an $n$-dimensional complete Riemannian manifold with Ricci
curvature ${\rm Ric}_{g}\geq-(n-1)\kappa g$ where $\kappa$ is a non-negative
constant and $\Omega=B_{R}(o)\subset M$ is a geodesic ball. Assume that $f$ is
the same as in the above lemma. For a given positive number

$\beta=\left(\alpha_{0}+t_{0}+\frac{p}{2}-1\right)\frac{n}{n-2},$

there exists a constant $a_{8}=a_{8}(n,p,q)>0$ such that

$\displaystyle\|f\|_{L^{\beta}(B_{3R/4}(o))}\leq
a_{8}V^{\frac{1}{\beta}}\frac{t_{0}^{2}}{R^{2}},$ (3.32)

where $V$ is the volume of geodesic ball $B_{R}(o)$.

###### Proof.

We set $t=t_{0}$ in (3.31) and obtain

$\displaystyle\begin{split}&\beta_{n,p,q,r}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t_{0}}\eta^{2}+\frac{a_{3}}{t_{0}}e^{-t_{0}}V^{\frac{2}{n}}R^{-2}\left\|f^{\frac{\alpha_{0}+t_{0}-1}{2}+\frac{p}{4}}\eta\right\|_{L^{\frac{2n}{n-2}}}^{2}\\\
\leq&\
a_{5}t_{0}^{2}R^{-2}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t_{0}-1}\eta^{2}+\frac{a_{4}}{t_{0}}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t_{0}-1}|\nabla\eta|^{2}.\end{split}$
(3.33)

From (3.33) we know that, if

$f\geq\frac{2a_{5}t_{0}^{2}}{\beta_{n,p,q,r}R^{2}},$

then there holds true

$a_{5}t_{0}^{2}R^{-2}\int_{\Omega}f^{\alpha+\frac{p}{2}+t_{0}-1}\eta^{2}\leq\frac{\beta_{n,p,q,r}}{2}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t_{0}}\eta^{2}.$

Now we denote

$\Omega_{1}=\left\\{f<\frac{2a_{5}t_{0}^{2}}{\beta_{n,p,q,r}R^{2}}\right\\},$

then $\Omega$ can be decomposed into two parts $\Omega_{1}$ and $\Omega_{2}$,
i.e., $\Omega=\Omega_{1}\cup\Omega_{2}$ where $\Omega_{2}$ is the complement
of $\Omega_{1}$. Hence, we have

$\displaystyle\begin{split}a_{5}t_{0}^{2}R^{-2}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t_{0}-1}\eta^{2}=&\
a_{5}t_{0}^{2}R^{-2}\int_{\Omega_{1}}f^{\alpha_{0}+\frac{p}{2}+t_{0}-1}\eta^{2}+a_{5}t_{0}^{2}R^{-2}\int_{\Omega_{2}}f^{\alpha_{0}+\frac{p}{2}+t_{0}-1}\eta^{2}\\\
\leq&\
\frac{a_{5}t_{0}^{2}}{R^{2}}\left(\frac{2a_{5}t_{0}^{2}}{\beta_{n,p,q,r}R^{2}}\right)^{\alpha_{0}+\frac{p}{2}+t_{0}-1}V+\frac{\beta_{n,p,q,r}}{2}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t_{0}}\eta^{2},\end{split}$
(3.34)

where $V$ is the volume of $B_{R}(o)$. It follows from (3.31) and (3.34)

$\displaystyle\begin{split}&\frac{\beta_{n,p,q,r}}{2}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t_{0}}\eta^{2}+\frac{a_{3}}{t_{0}}e^{-t_{0}}V^{\frac{2}{n}}R^{-2}\left\|f^{\frac{\alpha_{0}+t_{0}-1}{2}+\frac{p}{4}}\eta\right\|_{L^{\frac{2n}{n-2}}}^{2}\\\
\leq\
&\frac{a_{5}t_{0}^{2}}{R^{2}}\left(\frac{2a_{5}t_{0}^{2}}{\beta_{n,p,q,r}R^{2}}\right)^{\alpha_{0}+\frac{p}{2}+t_{0}-1}V+\frac{a_{4}}{t_{0}}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t_{0}-1}|\nabla\eta|^{2}.\end{split}$
(3.35)

Now we choose $\gamma\in C^{\infty}_{0}(B_{R}(o))$ satisfying

$\begin{cases}0\leq\gamma(x)\leq
1,\quad|\nabla\gamma(x)|\leq\frac{C}{R},&\forall x\in B_{R}(o);\\\
\gamma(x)\equiv 1,&\forall x\in B_{{3R}/{4}}(o),\end{cases}$

and let $\eta=\gamma^{\alpha_{0}+\frac{p}{2}+t_{0}}$. Then, we have

$\displaystyle a_{4}R^{2}|\nabla\eta|^{2}\leq
a_{4}C^{2}\left(t_{0}+\frac{p}{2}+\alpha_{0}\right)^{2}\eta^{\frac{2\alpha_{0}+2t_{0}+p-2}{\alpha_{0}+p/2+t_{0}}}\leq
a_{6}t^{2}_{0}\eta^{\frac{2\alpha_{0}+2t_{0}+p-2}{\alpha_{0}+p/2+t_{0}}}.$
(3.36)

By Hölder inequality and Young inequality, we have

$\displaystyle\begin{split}\frac{a_{4}}{t_{0}}\int_{\Omega}f^{\frac{p}{2}+\alpha_{0}+t_{0}-1}|\nabla\eta|^{2}\leq\
&\frac{a_{6}t_{0}}{R^{2}}\int_{\Omega}f^{\frac{p}{2}+\alpha_{0}+t_{0}-1}\eta^{\frac{2\alpha_{0}+p+2t_{0}-2}{\alpha_{0}+p/2+t_{0}}}\\\
\leq\
&\frac{a_{6}t_{0}}{R^{2}}\left(\int_{\Omega}f^{\alpha_{0}+t_{0}+\frac{p}{2}}\eta^{2}\right)^{\frac{\alpha_{0}+p/2+t_{0}-1}{\alpha+p/2+t_{0}}}V^{\frac{1}{\alpha_{0}+t_{0}+p/2}}\\\
\leq\
&\frac{\beta_{n,p,q,r}}{2}\left[\int_{\Omega}f^{\alpha_{0}+t_{0}+\frac{p}{2}}\eta^{2}+\left(\frac{2a_{6}t_{0}}{\beta_{n,p,q,r}R^{2}}\right)^{\alpha_{0}+t_{0}+\frac{p}{2}}V\right].\end{split}$
(3.37)

We conclude from (3.35) and (3.37) that

$\displaystyle\begin{split}&\left(\int_{\Omega}f^{\frac{n(p/2+\alpha_{0}+t_{0}-1)}{n-2}}\eta^{\frac{2n}{n-2}}\right)^{\frac{n-2}{n}}\\\
\leq\
&\frac{t_{0}}{a_{3}}e^{t_{0}}V^{1-\frac{2}{n}}R^{2}\left[\frac{2a_{5}t_{0}^{2}}{R^{2}}\left(\frac{2a_{5}t_{0}^{2}}{\beta_{n,p,q,r}R^{2}}\right)^{\alpha_{0}+t_{0}+\frac{p}{2}-1}+\frac{a_{6}t^{2}_{0}}{R^{2}}\left(\frac{2a_{6}t_{0}}{\beta_{n,p,q,r}R^{2}}\right)^{\alpha_{0}+t_{0}+\frac{p}{2}-1}\right]\\\
\leq\
&a_{7}^{t_{0}}e^{t_{0}}V^{1-\frac{2}{n}}t_{0}^{3}\left(\frac{t_{0}^{2}}{R^{2}}\right)^{\alpha_{0}+t_{0}+\frac{p}{2}-1},\end{split}$
(3.38)

where the constant $a_{7}$ depends only on $n,\,p,\,q$ and $r$, and satisfies

$a_{7}^{t_{0}}\geq\frac{2a_{5}}{a_{3}}\left(\frac{4a_{5}}{\beta_{n,p,q,r}}\right)^{\alpha_{0}+t_{0}+\frac{p}{2}-1}+\frac{a_{6}}{a_{3}}\left(\frac{4a_{6}}{\beta_{n,p,q,r}t_{0}}\right)^{\alpha_{0}+t_{0}+\frac{p}{2}-1}.$

Here we have used the fact $t_{0}\geq 1$.

Taking $\frac{1}{\alpha_{0}+t_{0}+p/2-1}$ power of the both sides of (3.38)
gives

$\displaystyle\left\|f\eta^{\frac{2}{\alpha_{0}+t_{0}+p/2-1}}\right\|_{L^{\beta}(\Omega)}\leq
a_{7}^{\frac{t_{0}}{\alpha_{0}+t_{0}+p/2-1}}V^{1/\beta}t_{0}^{\frac{3}{\alpha_{0}+t_{0}+p/2-1}}\frac{t_{0}^{2}}{R^{2}}\leq
a_{8}V^{\frac{1}{\beta}}\frac{t_{0}^{2}}{R^{2}},$ (3.39)

where $a_{8}$ depends only on $n,\ p,\ q,\ r$ and satisfies

$a_{8}\geq
a_{7}^{\frac{t_{0}}{\alpha_{0}+t_{0}+p/2-1}}t_{0}^{\frac{3}{\alpha_{0}+t_{0}+p/2-1}}.$

Since $\eta\equiv 1$ in $B_{3R/4}$, we obtain that

$\|f\|_{L^{\beta}(B_{3R/4}(o))}\leq
a_{8}V^{\frac{1}{\beta}}\frac{t_{0}^{2}}{R^{2}}.$

Thus, the proof of this lemma is finished. ∎

###  3.3. Moser iteration

Now we turn to providing the proof of Theorem 1.1.

###### Proof.

We discard the first term in (3.31) to obtain

$\displaystyle\frac{a_{3}}{t}e^{-t_{0}}V^{\frac{2}{n}}R^{-2}\left\|f^{\frac{\alpha_{0}+t-1}{2}+\frac{p}{4}}\eta\right\|_{L^{\frac{2n}{n-2}}}^{2}\leq\frac{a_{5}\alpha_{0}^{2}}{R^{2}}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t-1}\eta^{2}+\frac{a_{4}}{t}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t-1}|\nabla\eta|^{2}.$
(3.40)

For $k=1,\,2,\,\cdots$, let

$r_{k}=\frac{R}{2}+\frac{R}{4^{k}}$

and $\Omega_{k}=B_{r_{k}}(o)$. Choosing a sequence of cut-off function
$\eta_{k}\in C^{\infty}(\Omega_{k})$ which satisfy

$\displaystyle\begin{cases}0\leq\eta_{k}(x)\leq
1,\quad|\nabla\eta_{k}(x)|\leq\frac{C4^{k}}{R},&\forall x\in
B_{r_{k+1}}(o);\\\ \eta_{k}(x)\equiv 1,&\forall x\in
B_{r_{k+1}}(o),\end{cases}$ (3.41)

where $C$ is a constant that does not depend on $(M,g)$ and equation (1.2),
and substituting $\eta_{k}$ into (3.40) instead of $\eta$, we arrive at

$\displaystyle
a_{3}e^{-t_{0}}V^{\frac{2}{n}}\left\|f^{\frac{\alpha_{0}+t-1}{2}+\frac{p}{4}}\eta_{k}\right\|_{L^{\frac{2n}{n-2}}(\Omega_{k})}^{2}\leq\
$ $\displaystyle
a_{5}t_{0}^{2}t\int_{\Omega_{k}}f^{\alpha_{0}+\frac{p}{2}+t-1}\eta_{k}^{2}+a_{4}R^{2}\int_{\Omega_{k}}f^{\alpha_{0}+\frac{p}{2}+t-1}\left|\nabla\eta_{k}\right|^{2}$
$\displaystyle\leq\ $
$\displaystyle\left(a_{5}t_{0}^{2}t+C^{2}16^{k}\right)\int_{\Omega_{k}}f^{\alpha_{0}+\frac{p}{2}+t-1}.$

Now we choose $\beta_{1}=\beta$, $\beta_{k+1}=n\beta_{k}/(n-2)$,
$k=1,\,2,\,\cdots$, and let $t=t_{k}$ such that

$t_{k}+\frac{p}{2}+\alpha_{0}-1=\beta_{k},$

then we have

$\displaystyle
a_{3}\left(\int_{\Omega_{k}}f^{\beta_{k+1}}\eta_{k}^{\frac{2n}{n-2}}\right)^{\frac{n-2}{n}}\leq\
e^{t_{0}}V^{-\frac{2}{n}}\left(a_{5}t_{0}^{2}\left(t_{0}+\alpha_{0}+\frac{p}{2}-1\right)\left(\frac{n}{n-2}\right)^{k}+C^{2}16^{k}\right)\int_{\Omega_{k}}f^{\beta_{k}}.$

Since $n/(n-2)<16,~{}\forall n>2$, there exists some constant $a_{9}$ such
that

$\displaystyle\left(\int_{\Omega_{k}}f^{\beta_{k+1}}\eta_{k}^{\frac{2n}{n-2}}\right)^{\frac{n-2}{n}}\leq\
a_{9}t_{0}^{3}16^{k}e^{t_{0}}V^{-\frac{2}{n}}\int_{\Omega_{k}}f^{\beta_{k}}.$
(3.42)

Taking $1/\beta_{k}$ power of the both sides of (3.42) respectively and using
the fact $\eta_{k}\equiv 1\in\Omega_{k+1}$, we obtain

$\displaystyle\begin{split}\|f\|_{L^{\beta_{k+1}}(\Omega_{k+1})}\leq&\left(a_{9}t_{0}^{3}16^{k}e^{t_{0}}V^{-\frac{2}{n}}\right)^{\frac{1}{\beta_{k}}}16^{\frac{k}{\beta_{k}}}\|f\|_{L^{\beta_{k}}(\Omega_{k})}.\end{split}$
(3.43)

Noting

$\sum_{k=1}^{\infty}\frac{1}{\beta_{k}}=\frac{n}{2\beta_{1}}\quad\text{and}\quad\sum_{k=1}^{\infty}\frac{k}{\beta_{k}}=\frac{n^{2}}{4\beta_{1}},$

we have

$\displaystyle\|f\|_{L^{\infty}(B_{R/2}(o))}\leq\
a_{10}V^{-\frac{1}{\beta}}\|f\|_{L^{\beta_{1}}(B_{3R/4}(o))},$ (3.44)

where

$a_{10}\geq\left(a_{9}t_{0}^{3}16^{k}e^{t_{0}}\right)^{\frac{n}{2\beta_{1}}}16^{\frac{n^{2}}{4\beta_{1}}}.$

By (3.32), we obtain

$\displaystyle\|f\|_{L^{\infty}(B_{R/2}(o))}\leq\
a_{11}\frac{(1+\sqrt{\kappa}R)^{2}}{R^{2}},$ (3.45)

where $a_{11}=a_{10}a_{8}c_{n,p,q,r}$. Reminding $f=|\nabla u|^{2}$ and
$u=-(p-1)\log v$, we obtain the desired estimate. Thus, we complete the proof
of Theorem 1.1. ∎

###  3.4. Proof of Theorem 1.3

In this case, we have $\kappa=0$. By Corollory 1.2, we have

$\displaystyle\frac{|\nabla
v(x_{0})|}{v(x_{0})}\leq\sup_{B(x_{0},\frac{R}{2})}\frac{|\nabla
v|}{v}\leq\frac{C(n,p,q,r)}{R},\quad\forall x_{0}\in M.$ (3.46)

Letting $R\to\infty$ in (3.46), we obtain

$\nabla v(x_{0})=0,\quad\forall x_{0}\in M.$

thus $v$ is a constant. ∎

###  3.5. Proof of Theorem 1.5

For any $x\in B_{R/2}(o)\subset M$, by Corollory 1.2, we have

$\displaystyle|\nabla\ln v(x)|\leq c(n,p,q)\frac{1+\sqrt{\kappa}R}{R}.$ (3.47)

Choosing a minimizing geodesic $\gamma(t)$ with arc length parameter
connecting $o$ and $x$:

$\gamma:[0,d]\to M,\quad\gamma(0)=o,\quad\gamma(d)=x.$

where $d=d(x,o)$ is the geodesic distance from $o$ to $x$, we have

$\displaystyle\ln v(x)-\ln v(o)=\int_{0}^{d}\frac{d}{dt}\ln
v\circ\gamma(t)dt.$ (3.48)

Since $|\gamma^{\prime}|=1$, we have

$\displaystyle\left|\frac{d}{dt}\ln v\circ\gamma(t)\right|\leq|\nabla\ln
v||\gamma^{\prime}(t)|\leq c(n,p,q)\frac{1+\sqrt{\kappa}R}{R}.$ (3.49)

Thus it follows from $d\leq R/2$ and the above inequality that

$e^{-c(n,p,q)(1+\sqrt{\kappa}R)/2}\leq v(x)/v(o)\leq
e^{c(n,p,q)(1+\sqrt{\kappa}R)/2}.$

So, for any $x,y\in B_{R/2}(o)$ we have

$v(x)/v(y)\leq e^{c(n,p,q)(1+\sqrt{\kappa}R)}.$

Now suppose $v$ is a global positive solution of equation (1.2) on $M$.
Letting $R\to\infty$ in (3.47), we obtain that

$|\nabla\ln v(x)|\leq c(n,p,q)\sqrt{\kappa},\quad\forall x\in M.$

For any $y\in M$, choose a minimizing geodesic $\gamma(t)$ parameterized by
arc length which connects $x$ and $y$:

$\gamma:[0,d]\to M,\quad\gamma(0)=x,\quad\gamma(d)=y.$

where $d=d(x,y)$ is the distance from $x$ to $y$. Then, we have

$\displaystyle\ln v(y)-\ln v(x)=\int_{0}^{d}\frac{d}{dt}\ln
v\circ\gamma(t)dt.$ (3.50)

Since $|\gamma^{\prime}(t)|=1$, we have

$\displaystyle\left|\frac{d}{dt}\ln v\circ\gamma(t)\right|\leq|\nabla\ln
v||\gamma^{\prime}(t)|=c(n,p,q)\sqrt{\kappa}.$ (3.51)

It follows from (3.50) and (3.51) that

$\displaystyle v(y)/v(x)\leq e^{c(n,p,q)\sqrt{\kappa}d(x,y)}.$ (3.52)

Thus we finish the proof by (3.52).

###  3.6. The case $\dim(M)=2$.

In the proof of Theorem 1.1 above, we have used Saloff-Coste’s Sobolev
inequality on the embedding $W^{1,2}(B)\hookrightarrow L^{\frac{2n}{n-2}}(B)$
on a manifold. Since the Sobolev exponent $2^{*}=2n/(n-2)$ requires $n>2$, we
did not consider the case $n=2$ in Theorem 1.1. In this subsection, we will
explain briefly that Theorem 1.1 can also be established when $n=2$.

When $\dim M=n=2$, we need the special case of Saloff-Coste’s Sobolev theorem,
i.e., Lemma 2.2. For any $n^{\prime}>2$, there holds

$\|f\|_{L^{\frac{2n^{\prime}}{n^{\prime}-2}}(B)}^{2}\leq
e^{c_{0}(1+\sqrt{\kappa}R)}V^{-\frac{2}{n^{\prime}}}R^{2}\left(\int_{B}|\nabla
f|^{2}+R^{-2}f^{2}\right)$

for any $f\in C^{\infty}_{0}(B)$.

For example, we can choose $n^{\prime}=4$, then we have

$\displaystyle\left(\int_{\Omega}f^{2\alpha+p}\eta^{4}\right)^{2}\leq
e^{c_{0}(1+\sqrt{\kappa}R)}V^{-\frac{1}{2}}\left(R^{2}\int_{\Omega}\left|\nabla\left(f^{\frac{p}{4}+\frac{\alpha}{2}}\eta\right)\right|^{2}+\int_{\Omega}f^{\frac{p}{2}+\alpha}\eta^{2}\right).$

By the above inequality and (3.33), we can deduce the following integral
inequality by almost the same method as in Lemma 3.2.

$\displaystyle\begin{split}&\beta_{n,p,q,r}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t}\eta^{2}+\frac{a_{3}}{t}e^{-t_{0}}V^{\frac{1}{2}}R^{-2}\left\|f^{\frac{\alpha_{0}+t-1}{2}+\frac{p}{4}}\eta\right\|_{L^{4}}^{2}\\\
\leq\
&a_{5}t_{0}^{2}R^{-2}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t-1}\eta^{2}+\frac{a_{4}}{t}\int_{\Omega}f^{\alpha_{0}+\frac{p}{2}+t-1}|\nabla\eta|^{2},\end{split}$
(3.53)

where $\alpha_{0}$ is the same as $\alpha_{0}$ defined in Section 3.1, but the
constants $a_{i}$, $i=3,\,4,\,5$, may differ from those defined in Section
3.1.

By repeating the same procedure as in Section 3.2, we can deduce from (3.53)
the $L^{\beta}$-bound of $f$ in a geodesic ball with radius $3R/4$

$\displaystyle\|f\|_{L^{\beta}(B_{3R/4}(o))}\leq
a_{8}V^{\frac{1}{\beta}}\frac{t_{0}^{2}}{R^{2}},$ (3.54)

where $\beta=p+2\alpha_{0}+2t_{0}-2$.

For the Nash-Moser iteration, we set

$\beta_{l}=2^{l}(\alpha_{0}+t_{0}+p/2-1)$

and $\Omega_{l}$ by the similar way with that in Section 3.3, and can obtain
the following inequality

$\displaystyle\|f\|_{L^{\infty}(B_{\frac{R}{2}}(o))}\leq\ $ $\displaystyle
a_{11}V^{-\frac{1}{\beta}}\|f\|_{L^{\beta}(B_{\frac{3R}{4}}(o))}.$ (3.55)

Combining (3.54) and (3.55), we finally obtain the Cheng-Yau type gradient
estimate. Harnack inequality and Liouville type results follow from the Cheng-
Yau type gradient estimate, here we omit the details.

##  4\. Proof of Theorem 1.4

In this section, we generalized [25, Theorem 15.1] in $\mathbb{R}^{n}$ to non-
compact Riemannian manifold with non-negative Ricci curvature.

######  Theorem 4.1 (Theorem 1.4).

Let $(M,g)$ be complete non-compact manifold with non-negative Ricci
curvature. Suppose that $1<p<n$, $r>0,\,q\geq 0,\,q+r>p-1$, and $a>0$. If

$\displaystyle r+\frac{n-1}{n-p}q<\frac{n(p-1)}{n-p},$

then the problem

$\displaystyle\begin{cases}\Delta_{p}u+au^{r}|\nabla u|^{q}\leq
0,&\quad\text{in}\,\,M;\\\ u>0,&\quad\text{in}\,\,M\end{cases}$ (4.1)

admits no solution $u\in C^{1}(M)$.

###### Proof.

We show this theorem by almost the same arguments as in the proof of Theorem
15.1 in [25] due to Mitidieri and Pohozaev. Let $\varphi\in C_{0}^{1}(M)$ be a
non-negative cut-off function and $\alpha<0$ be small. Multiplying (4.1) by
$\varphi u^{\alpha}$ and integrating the inequality obtained over $M$, we
arrive at

$\displaystyle\int_{M}au^{r+\alpha}|\nabla
u|^{q}\varphi+|\alpha|\int_{M}|\nabla
u|^{p}u^{\alpha-1}\varphi\leq\int_{M}|\nabla
u|^{p-1}u^{\alpha}|\nabla\varphi|.$ (4.2)

It follows from (4.2) that

$\displaystyle\int_{M}au^{r+\alpha}|\nabla
u|^{q}\varphi+|\alpha|\int_{M}|\nabla u|^{p}u^{\alpha-1}\varphi\leq$
$\displaystyle\int_{M}|\nabla
u|^{\frac{p}{s}}u^{\frac{\alpha-1}{s}}\varphi^{\frac{1}{s}}u^{\frac{\alpha
s+1-\alpha}{s}}|\nabla\varphi||\nabla
u|^{\frac{s(p-1)-p}{s}}\varphi^{-\frac{1}{s}},$

where $s>1$ is a constant which will be determined later. Now, applying Young
inequality with exponents $s,\,t,\,\frac{1}{s}+\frac{1}{t}=1$ and $\epsilon>0$
for the integral function on the right hand side of the above inequality we
have

$\displaystyle\begin{split}&\int_{M}au^{r+\alpha}|\nabla
u|^{q}\varphi+|\alpha|\int_{M}|\nabla u|^{p}u^{\alpha-1}\varphi\\\
\leq&\frac{\epsilon}{s}\int_{M}|\nabla
u|^{p}u^{\alpha-1}\varphi+\frac{\epsilon^{-\frac{1}{s-1}}}{t}\int_{M}u^{\frac{\alpha
s+1-\alpha}{s-1}}|\nabla\varphi|^{\frac{s}{s-1}}|\nabla
u|^{\frac{s(p-1)-p}{s-1}}\varphi^{-\frac{1}{s-1}},\end{split}$ (4.3)

where $\epsilon$ is small enough such that $|\alpha|\geq\frac{\epsilon}{s}$.

Hence, by Hölder inequality we can infer from (4.3) the following

$\displaystyle\begin{split}a\int_{M}u^{r+\alpha}|\nabla
u|^{q}\varphi\leq&\frac{\epsilon^{-\frac{1}{s-1}}}{t}\int_{M}u^{\frac{\alpha
s+1-\alpha}{s-1}}|\nabla\varphi|^{\frac{s}{s-1}}|\nabla
u|^{\frac{s(p-1)-p}{s-1}}\varphi^{-\frac{1}{s-1}}\\\
\leq&\frac{\epsilon^{-\frac{1}{s-1}}}{t}\left(\int_{M}u^{\frac{\alpha
s+1-\alpha}{s-1}\mu}|\nabla
u|^{\frac{s(p-1)-p}{s-1}\mu}\varphi\right)^{\frac{1}{\mu}}\left(\int_{M}|\nabla\varphi|^{\frac{s\nu}{s-1}}\varphi^{1-\frac{s\nu}{s-1}}\right)^{\frac{1}{\nu}}.\end{split}$
(4.4)

Now we set $s$ and $\mu$ by

$\begin{cases}\frac{\alpha s+1-\alpha}{s-1}\mu=r+\alpha,\\\
\frac{s(p-1)-p}{s-1}\mu=q.\end{cases}$

It is easy to check that

$\displaystyle s=\frac{pr+\alpha(p-q)+q}{r(p-1)+\alpha(p-1-q)}.$ (4.5)

For $\alpha>1-p$, we can verify that

$\begin{cases}pr+\alpha(p-q)+q>0,\\\
r(p-1)+\alpha(p-1-q)>0\end{cases}\quad\text{and}\quad s>\frac{p}{p-1}>1.$

It follows from (4.4) that

$\displaystyle\int_{M}u^{r+\alpha}|\nabla
u|^{q}\varphi\leq\left(\frac{\epsilon^{-\frac{1}{s-1}}}{at}\right)^{\nu}\int_{M}|\nabla\varphi|^{\frac{s\nu}{s-1}}\varphi^{1-\frac{s\nu}{s-1}}.$
(4.6)

Now we can choose $\gamma\in C^{\infty}_{0}(B_{2R}(o))$ such that

$\begin{cases}\gamma(x)\equiv 1,&\quad\text{in}\quad B_{R}(0),\\\
1\geq\gamma(x)\geq 0,\quad|\gamma(x)|\leq\frac{C}{R},&\quad\text{in}\quad
B_{2R}(0)\end{cases}$

and let $\varphi=\gamma^{l}$ with $l$ large enough, then we have

$|\nabla\varphi|^{\frac{s\nu}{s-1}}\varphi^{1-\frac{s\nu}{s-1}}=l^{\frac{s\nu}{s-1}}\gamma^{l-\frac{s\nu}{s-1}}|\nabla\gamma|^{\frac{s\nu}{s-1}}\leq
CR^{-\frac{s\nu}{s-1}}.$

Since $\mathrm{Ric}_{g}\geq 0$, by the volume comparison theorem we conclude
from (4.6) and the last inequality that

$\displaystyle\int_{M}u^{r+\alpha}|\nabla u|^{q}\varphi\leq
CR^{n-\frac{s\nu}{s-1}}=CR^{n-\frac{s}{s-1}\frac{\mu}{\mu-1}}.$

Direct computation shows that we can always choose $\alpha<0$ with $|\alpha|$
small enough such that

$n-\frac{s}{s-1}\frac{\mu}{\mu-1}<0,$

if

$r(n-p)+q(n-1)<n(p-1).$

Taking $R\to\infty$ leads to that

$\int_{M}u^{r+\alpha}|\nabla u|^{q}\varphi=0$

which implies $u\equiv 0$. Thus, we complete the proof of the theorem. ∎

Let $r>0,\,q\geq 0$, and $p>1$. For small enough $\alpha<0$ we define

$S_{\alpha}=\\{u:M\to\mathbb{R}:u\geq 0,\,u^{\alpha+r}|\nabla
u|^{q},\,u^{\alpha-1}|\nabla u|^{p},\,u^{\alpha}|\nabla u|^{p-1}\in
L^{1}_{loc}(M)\\},$

where $\nabla u$ is understood in the sense of distribution. Actually, we can
also prove the following more general conclusions:

######  Theorem 4.2.

Let $(M,g)$ be complete non-compact manifold with non-negative Ricci
curvature. Suppose that $1<p<n$, $r>0,\,q\geq 0,\,q+r>p-1$, and $a>0$. If

$\displaystyle r+\frac{n-1}{n-p}q<\frac{n(p-1)}{n-p},$

then the problem

$\displaystyle\begin{cases}\Delta_{p}u+au^{r}|\nabla u|^{q}\leq
0,&\quad\text{in}\,\,M;\\\ u\geq 0,&\quad\text{in}\,\,M\end{cases}$

admits no solution in $S_{\alpha}$ unless $u\equiv 0$.

The proof of this theorem goes almost the same as the proof of Theorem 15.1 in
[25] except for we need to use the volume comparison theorem. So, we omit it.

Acknowledgements: The author Y. Wang is supported partially by NSFC (Grant
No.11971400) and National Key Research and Development projects of China
(Grant No. 2020YFA0712500). The author J. Hu is supported by National Natural
Science Foundation of China(Grant No. 12288201) and the Project of Stable
Support for Youth Team in Basic Research Fields, CAS(Grant No. YSBR-001).

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