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# Gradient estimates and Liouville-type theorems for a weighted nonlinear elliptic equation

Journal of Inequalities and Applications20182018:112

https://doi.org/10.1186/s13660-018-1705-z

• Received: 1 February 2018
• Accepted: 27 April 2018
• Published:

## Abstract

We consider gradient estimates for positive solutions to the following nonlinear elliptic equation on a smooth metric measure space $$(M, g,e^{-f}\,dv)$$:
$$\Delta_{f} u+au\log u+bu=0,$$
where a, b are two real constants. When the ∞-Bakry–Émery Ricci curvature is bounded from below, we obtain a global gradient estimate which is not dependent on $$|\nabla f|$$. In particular, we find that any bounded positive solution of the above equation must be constant under some suitable assumptions.

## Keywords

• Nonlinear elliptic equation
• Liouville-type theorem

• 58J35
• 35B45

## 1 Introduction

Let $$(M, g)$$ be an n-dimensional complete Riemannian manifold and f be a smooth function defined on M. Then the triple $$(M, g,e^{-f}\, dv)$$ is called a smooth metric measure space, where dv denotes the volume element of the metric g and $$e^{-f}\,dv$$ is called the weighted measure. On the smooth metric measure space $$(M, g,e^{-f}\, dv)$$, the m-Bakry–Émery Ricci curvature (see ) is defined by
$$\mathrm{Ric}_{f}^{m}=\mathrm{Ric}+ \nabla^{2}f-\frac{1}{m-n}\,df\otimes df,$$
(1.1)
where $$m\geq n$$ is a constant, and $$m=n$$ if and only if f is a constant. We define
$$\mathrm{Ric}_{f}=\mathrm{Ric}+\nabla^{2}f.$$
(1.2)
Then $$\mathrm{Ric}_{f}$$ can be seen as the ∞-dimensional Bakry–Émery Ricci curvature. However, there are many differences between the m-Bakry–Émery Ricci curvature and the ∞-Bakry–Émery Ricci curvature. For example, there exist complete noncompact Riemannian manifolds which satisfy $$\mathrm{Ric}_{f}=\lambda g$$ for some positive constant λ (which is called the shrinking gradient Ricci soliton), but not for $$\mathrm{Ric}_{f}^{m}=\lambda g$$. We recall that the f-Laplacian $$\Delta_{f}$$ on $$(M, g,e^{-f}\,dv)$$ is defined by
$$\Delta_{f}=\Delta-\nabla f\nabla.$$
Since we have the Bochner formula with respect to f-Laplacian:
$$\frac{1}{2}\Delta_{f}|\nabla u|^{2}\geq \frac{1}{m}(\Delta_{f}u)^{2}+\nabla u\nabla( \Delta_{f}u)+\mathrm{Ric}_{f}^{m}(\nabla u,\nabla u),$$
which is similar to the Bochner formula associated with the Laplacian, many results with respect to the Laplacian have been generalized to those of the f-Laplacian under the m-dimensional Bakry–Émery Ricci curvature. For example, see  and the references therein. But for elliptic gradient estimates for f-Laplacian under the ∞-Bakry–Émery Ricci curvature, in order to using the weighted comparison theorem, the assumption $$|\nabla f|\leq\theta$$ is forced commonly.
In this paper, under the assumption that the ∞-Bakry–Émery Ricci curvature is bounded from below, we consider the following nonlinear elliptic equation:
$$\Delta_{f}u+au\log u+bu=0,$$
(1.3)
where a, b are two real constants. Inspired by the ideas of Brighton in , we can obtain global gradient estimates for positive solutions to (1.3) without any restriction on $$|\nabla f|$$.

### Theorem 1.1

Let $$(M, g,e^{-f}\,dv)$$ be an n-dimensional complete smooth metric measure space with $$\mathrm{Ric}_{f}(B_{p}(2R))\geq-(n-1)K$$, where $$K\geq0$$ is a constant. Suppose that u is a positive solution to (1.3) with $$u\leq A$$ on $$B_{p}(2R)$$. Then on $$B_{p}(R)$$ with $$R>1$$, the following inequality holds:
\begin{aligned} |\nabla u|^{2}\leq C A^{2} \biggl[\max \biggl\{ \frac{4}{5}b+a \biggl(1+ \frac {4}{5}L \biggr),0 \biggr\} +K+\frac{|\beta|+1}{R} \biggr], \end{aligned}
(1.4)
where C is a positive constant which depends on the dimension n, $$\beta=\max_{\{x|d(x,p)=1\}} \Delta_{f} r(x)$$ and
$$L= \textstyle\begin{cases} \sup_{B_{p}(2R)}(\log u), &\textit{if } a\geq0, \\ \inf_{B_{p}(2R)}(\log u), &\textit{if } a< 0. \end{cases}$$
(1.5)

Letting $$R\rightarrow\infty$$ in (1.4), we obtain the following global estimates on complete noncompact Riemannian manifolds:

### Corollary 1.2

Let $$(M, g,e^{-f}\,dv)$$ be an n-dimensional complete smooth metric measure space with $$\mathrm{Ric}_{f} \geq-(n-1)K$$, where $$K\geq0$$ is a constant. If u is a positive solution to (1.3) with $$u\leq A$$, then we have
\begin{aligned} |\nabla u|^{2}\leq C A^{2} \biggl[\max \biggl\{ \frac{4}{5}b+a \biggl(1+ \frac {4}{5}L \biggr),0 \biggr\} +K \biggr], \end{aligned}
(1.6)
where
$$L= \textstyle\begin{cases} \sup_{M}(\log u), &\textit{if } a\geq0, \\ \inf_{M}(\log u), &\textit{if } a< 0. \end{cases}$$
(1.7)

Using the ideas of the proof of Theorem 1.1, by choosing $$\tilde{h}=\log u$$ a gap develops between the constants, and we also establish the following.

### Theorem 1.3

Let $$(M, g,e^{-f}\,dv)$$ be an n-dimensional complete smooth metric measure space with $$\mathrm{Ric}_{f}(B_{p}(2R))\geq-(n-1)K$$, where $$K\geq0$$ is a constant. Suppose that u is a positive solution to (1.3) on $$B_{p}(2R)$$ such that:
1. (1)

either $$\nabla f \nabla(\log u)-a\log u-b\leq\delta|\nabla (\log u)|^{2}$$ for some $$0\leq\delta<\frac{1}{2}$$;

2. (2)

or $$\nabla f \nabla(\log u)-a\log u-b\geq2 |\nabla(\log u)|^{2}$$.

Then on $$B_{p}(R)$$ with $$R>1$$, the following inequality holds:
$$\big|\nabla(\log u)\big|^{2}\leq\frac{C_{1}(n,\delta,\beta)}{R}+C_{2}(n, \delta)\max\bigl\{ a+(n-1)K,0\bigr\} ,$$
(1.8)
where $$\beta=\max_{\{x|d(x,p)=1\}}\Delta_{f}r(x)$$.

Letting $$R\rightarrow\infty$$ in (1.8), we obtain the following global estimates on complete noncompact Riemannian manifolds:

### Corollary 1.4

Let $$(M, g,e^{-f}\,dv)$$ be an n-dimensional complete smooth metric measure space with $$\mathrm{Ric}_{f} \geq-(n-1)K$$, where $$K\geq0$$ is a constant. Let u be a positive solution to (1.3). Then under the assumption of either (1) or (2) as in Theorem 1.3, we have
\begin{aligned} \big|\nabla(\log u)\big|^{2}\leq C(n, \delta)\max\bigl\{ a+(n-1)K,0\bigr\} . \end{aligned}
(1.9)

Clearly, if either $$u\leq e^{-(\frac{5}{4}+\frac{b}{a})}$$ and $$a>0$$, or $$u\geq e^{-(\frac{5}{4}+\frac{b}{a})}$$ and $$a<0$$, then we have $$\frac {4}{5}b+ a (1+\frac{4}{5}L )\leq0$$. This gives the following result.

### Corollary 1.5

Let $$(M, g,e^{-f}\,dv)$$ be an n-dimensional complete smooth metric measure space with $$\mathrm{Ric}_{f} \geq0$$.
1. (1)

There exists no bounded positive solution to (1.3) with $$a>0$$ and $$u\leq e^{-(\frac{5}{4}+\frac{b}{a})}$$;

2. (2)

if $$a<0$$ and $$u\geq e^{-(\frac{5}{4}+\frac{b}{a})}$$, then any bounded positive solution to (1.3) must be constant $$u=e^{-\frac{b}{a}}$$.

### Remark 1.1

In particular, when $$a=0$$, Eq. (1.3) becomes
$$\Delta_{f}u+bu=0$$
(1.10)
and (1.6) becomes
\begin{aligned} |\nabla u|^{2}\leq C A^{2} \biggl[\max \biggl\{ \frac{4}{5}b,0 \biggr\} +K \biggr]. \end{aligned}
(1.11)
In this case, on a complete smooth metric measure space $$(M, g,e^{-f}\, dv)$$ with $$\mathrm{Ric}_{f} \geq0$$, there exists no bounded positive solution to (1.10) with $$b<0$$. On the other hand, if $$a=b=0$$, our Theorem 1.1 becomes Theorem 1 of Brighton in .

### Remark 1.2

It is easy to see from Corollary 1.4 that if u is a positive solution to (1.3) with $$a\leq-(n-1)K$$ satisfying either (1) or (2) in Theorem 1.3, then $$u=e^{-\frac{b}{a}}$$ is a constant. In particular, if $$a=b=0$$, then our Theorem 1.3 becomes Theorem 3 of Brighton in .

### Remark 1.3

Some related results for gradient estimates of positive solutions to
$$\Delta_{f}u+au\log u=0$$
(1.12)
can be found in . Moreover, Qian in  used a different method to derive similar estimates to (1.12) with constant f. On the other hand, if we assume $$\mathrm{Ric}_{f} \geq -(n-1)K$$ and $$|\nabla f|\leq\theta$$, then from (1.1), we obtain
\begin{aligned}\mathrm{Ric}_{f}^{m}&= \mathrm{Ric}_{f}-\frac{1}{m-n}\, df\otimes df \\ &\geq-(n-1) \biggl(K+\frac{\theta^{2}}{(m-n)(n-1)} \biggr):=-(n-1)\tilde{K}. \end{aligned}
Hence, Theorem 1.5 in  follows from Theorem 1.1 of  immediately. However, our estimates in this paper are not dependent on $$|\nabla f|$$.

## 2 Proof of results

We firstly give the following lemma which plays an important role in the proof of main results.

### Lemma 2.1

Let u be a positive solution to (1.3) with $$u\leq A$$ and $$\mathrm{Ric}_{f}\geq-(n-1)K$$ for some positive constant K. Denote $$\tilde{u}=u/A$$ and $$h=\tilde {u}^{\epsilon}$$ for $$\epsilon\in(0,1)$$. If there exists one positive constant δ satisfying
$$\frac{1}{n}+\frac{2(\epsilon-1)}{n\epsilon\delta}\geq0,$$
(2.1)
then we have
\begin{aligned}[b] \frac{1}{2}\Delta_{f}| \nabla h|^{2}\geq{}& \biggl(\frac{(\epsilon-1)^{2}}{n\epsilon^{2}}-\frac{\epsilon -1}{\epsilon}+ \frac{2\delta(\epsilon-1)}{n\epsilon} \biggr)\frac{|\nabla h|^{4}}{h^{2}}+\frac{\epsilon-1}{\epsilon }\frac{\nabla h}{h} \nabla\bigl(|\nabla h|^{2}\bigr) \\ &-\bigl[a+\tilde{b}\epsilon+(n-1)K+a\tilde{L}\bigr]|\nabla h|^{2}, \end{aligned}
(2.2)
where
$$\tilde{L}= \textstyle\begin{cases} \sup_{M}(\log h), &\textit{if } a\geq0, \\ \inf_{M}(\log h), &\textit{if } a< 0. \end{cases}$$
(2.3)

### Proof

Under the scaling $$u\rightarrow\tilde{u}=u/A$$, it follows from (1.3) that ũ satisfies
$$\Delta_{f} \tilde{u}+a\tilde{u}\log \tilde{u}+\tilde{b} \tilde{u}=0,$$
(2.4)
where the constant is given by $$\tilde{b}=b+a\log A$$. Let $$h=\tilde{u}^{\epsilon}$$, where $$\epsilon\in(0,1)$$ is a constant to be determined. Then we have
$$\log h=\epsilon\log\tilde{u}.$$
(2.5)
Since $$0<\tilde{u}\leq1$$, we have $$\log h\leq0$$ and
\begin{aligned}[b] \Delta_{f} h&= \Delta_{f}\bigl(\tilde{u}^{\epsilon}\bigr)=\epsilon(\epsilon-1) \tilde {u}^{\epsilon-2}|\nabla \tilde{u}|^{2}+\epsilon \tilde{u}^{\epsilon-1}\Delta_{f} \tilde{u} \\ &=\epsilon(\epsilon-1)\tilde{u}^{\epsilon-2}|\nabla\tilde {u}|^{2}-a \epsilon\tilde{u}^{\epsilon}\log\tilde{u}-\tilde{b}\epsilon \tilde{u}^{\epsilon}\\ &=\frac{\epsilon-1}{\epsilon} \frac{|\nabla h|^{2}}{h}-ah\log h-\tilde {b}\epsilon h, \end{aligned}
(2.6)
which implies
\begin{aligned}[b] \nabla h\nabla\Delta_{f} h&= \nabla h\nabla \biggl(\frac{\epsilon-1}{\epsilon} \frac{|\nabla h|^{2}}{h}-ah\log h-\tilde{b} \epsilon h \biggr) \\ &=\frac{\epsilon-1}{\epsilon}\nabla h\nabla\frac{|\nabla h|^{2}}{h}-a\nabla h\nabla(h\log h)- \tilde{b}\epsilon|\nabla h|^{2} \\ &=\frac{\epsilon-1}{\epsilon}\frac{\nabla h}{h}\nabla\bigl(|\nabla h|^{2} \bigr)-\frac{\epsilon-1}{\epsilon}\frac{|\nabla h|^{4}}{h^{2}} \\ &\quad{}-ah\log h\frac{|\nabla h|^{2}}{h}-(a+\tilde{b}\epsilon)|\nabla h|^{2}. \end{aligned}
(2.7)
Thus, under the assumption $$\mathrm{Ric}_{f}\geq-(n-1)K$$, one has
\begin{aligned}[b] \frac{1}{2}\Delta_{f}| \nabla h|^{2}&=|\nabla^{2}h|^{2}+\nabla h\nabla \Delta_{f} h+\mathrm{Ric}_{f} (\nabla h,\nabla h) \\ &\geq\frac{1}{n}(\Delta h)^{2}+\nabla h\nabla \Delta_{f} h-(n-1)K|\nabla h|^{2} \\ &=\frac{1}{n} \biggl(\frac{\epsilon-1}{\epsilon} \frac{|\nabla h|^{2}}{h}+\nabla f\nabla h-ah\log h-\tilde{b}\epsilon h \biggr)^{2}+\frac{\epsilon-1}{\epsilon } \frac{\nabla h}{h}\nabla \bigl(|\nabla h|^{2}\bigr) \\ &\quad{}-\frac{\epsilon-1}{\epsilon}\frac{|\nabla h|^{4}}{h^{2}} -(ah\log h)\frac{|\nabla h|^{2}}{h}-\bigl[a+ \tilde{b}\epsilon+(n-1)K\bigr]|\nabla h|^{2} \\ &= \biggl(\frac{(\epsilon-1)^{2}}{n\epsilon^{2}}-\frac{\epsilon-1}{\epsilon } \biggr)\frac{|\nabla h|^{4}}{h^{2}} + \frac{2(\epsilon-1)}{n\epsilon}\frac{|\nabla h|^{2}}{h} (\nabla f\nabla h-ah\log h-\tilde{b}\epsilon h ) \\ &\quad{}+\frac{1}{n} (\nabla f\nabla h-ah\log h-\tilde{b}\epsilon h )^{2}+\frac{\epsilon-1}{\epsilon }\frac{\nabla h}{h}\nabla\bigl(|\nabla h|^{2}\bigr) \\ &\quad{}-\bigl[a+\tilde{b}\epsilon+(n-1)K+a\log h\bigr]|\nabla h|^{2}. \end{aligned}
(2.8)
For any fixed point p, if there exists a positive constant δ such that $$\nabla f\nabla h-ah\log h-\tilde{b}\epsilon h\leq \delta\frac{|\nabla h|^{2}}{h}$$, then from (2.8), we can deduce
\begin{aligned} \frac{1}{2}\Delta_{f}| \nabla h|^{2}\geq{}& \biggl(\frac{(\epsilon-1)^{2}}{n\epsilon ^{2}}-\frac{\epsilon-1}{\epsilon} \biggr) \frac{|\nabla h|^{4}}{h^{2}} +\frac{2(\epsilon-1)}{n\epsilon}\frac{|\nabla h|^{2}}{h} \biggl(\delta \frac {|\nabla h|^{2}}{h} \biggr) \\ &+\frac{1}{n} (\nabla f\nabla h-ah\log h-\tilde{b}\epsilon h )^{2}+\frac{\epsilon-1}{\epsilon }\frac{\nabla h}{h}\nabla\bigl(|\nabla h|^{2}\bigr) \\ &-\bigl[a+\tilde{b}\epsilon+(n-1)K+a\log h\bigr]|\nabla h|^{2} \\ \geq{}& \biggl(\frac{(\epsilon-1)^{2}}{n\epsilon^{2}}-\frac{\epsilon-1}{\epsilon }+\frac{2\delta(\epsilon-1)}{n\epsilon} \biggr) \frac{|\nabla h|^{4}}{h^{2}}+\frac{\epsilon-1}{\epsilon }\frac{\nabla h}{h}\nabla\bigl(|\nabla h|^{2}\bigr) \\ &-\bigl[a+\tilde{b}\epsilon+(n-1)K+a\tilde{L}\bigr]|\nabla h|^{2}. \end{aligned}
(2.9)
On the contrary, if $$\nabla f\nabla h-ah\log h-\tilde{b}\epsilon h\geq \delta\frac{|\nabla h|^{2}}{h}$$ at the point p, then from (2.8), we can deduce
\begin{aligned}[b] \frac{1}{2}\Delta_{f}| \nabla h|^{2}\geq{}& \biggl(\frac{(\epsilon-1)^{2}}{n\epsilon ^{2}}-\frac{\epsilon-1}{\epsilon} \biggr) \frac{|\nabla h|^{4}}{h^{2}} +\frac{2(\epsilon-1)}{n\epsilon\delta} (\nabla f\nabla h-ah\log h-\tilde{b} \epsilon h )^{2} \\ &+\frac{1}{n} (\nabla f\nabla h-ah\log h-\tilde{b}\epsilon h )^{2}+\frac{\epsilon-1}{\epsilon }\frac{\nabla h}{h}\nabla\bigl(|\nabla h|^{2}\bigr) \\ &-\bigl[a+\tilde{b}\epsilon+(n-1)K+a\log h\bigr]|\nabla h|^{2} \\ ={}& \biggl(\frac{(\epsilon-1)^{2}}{n\epsilon^{2}}-\frac{\epsilon-1}{\epsilon } \biggr)\frac{|\nabla h|^{4}}{h^{2}} + \biggl(\frac{1}{n}+\frac{2(\epsilon-1)}{n\epsilon\delta} \biggr) (\nabla f\nabla h-ah\log h- \tilde{b}\epsilon h )^{2} \\ &+\frac{\epsilon-1}{\epsilon }\frac{\nabla h}{h}\nabla\bigl(|\nabla h|^{2} \bigr)-\bigl[a+\tilde{b}\epsilon +(n-1)K+a\log h\bigr]|\nabla h|^{2} \\ \geq{}& \biggl[\frac{(\epsilon-1)^{2}}{n\epsilon^{2}}-\frac{\epsilon-1}{\epsilon} + \biggl(\frac{1}{n}+ \frac{2(\epsilon-1)}{n\epsilon\delta} \biggr)\delta^{2} \biggr]\frac{|\nabla h|^{4}}{h^{2}}+ \frac{\epsilon-1}{\epsilon }\frac{\nabla h}{h}\nabla\bigl(|\nabla h|^{2}\bigr) \\ &-\bigl[a+\tilde{b}\epsilon+(n-1)K+a\log h\bigr]|\nabla h|^{2} \\ \geq{}& \biggl(\frac{(\epsilon-1)^{2}}{n\epsilon^{2}}-\frac{\epsilon-1}{\epsilon }+\frac{2\delta(\epsilon-1)}{n\epsilon} \biggr) \frac{|\nabla h|^{4}}{h^{2}}+\frac{\epsilon-1}{\epsilon }\frac{\nabla h}{h}\nabla\bigl(|\nabla h|^{2}\bigr) \\ &-\bigl[a+\tilde{b}\epsilon+(n-1)K+a\tilde{L}\bigr]|\nabla h|^{2} \end{aligned}
(2.10)
as long as (2.1) holds.

Therefore, in these two cases the estimate (2.2) holds, which finishes the proof of the Lemma 2.1. □

### 2.1 Proof of Theorem 1.1

In order to obtain the upper bound of $$|\nabla h|$$ by using the maximum principle for (2.2), we need to choose ϵ, δ such that the coefficient of $$\frac{|\nabla h|^{4}}{h^{2}}$$ in (2.2) is positive. That is, we need
$$\frac{(\epsilon-1)^{2}}{n\epsilon^{2}}-\frac{\epsilon-1}{\epsilon}+\frac {2\delta(\epsilon-1)}{n\epsilon}>0.$$
(2.11)
In particular, by choosing $$\epsilon=\frac{4}{5}$$ and letting $$\delta\rightarrow\frac{1}{2}$$, we find that the inequality (2.1) holds and (2.2) becomes
\begin{aligned}[b] \frac{1}{2}\Delta_{f}| \nabla h|^{2}\geq{}&\frac{4n-3}{16n}\frac{|\nabla h|^{4}}{h^{2}}- \frac{1}{4}\frac {\nabla h}{h}\nabla\bigl(|\nabla h|^{2}\bigr) \\ &-\bigl[a+\tilde{b}\epsilon+(n-1)K+a\tilde{L}\bigr]|\nabla h|^{2}. \end{aligned}
(2.12)
As in , we define a cut-off function $$\psi\in C^{2}([0,+\infty))$$ by
$$\psi(t)= \textstyle\begin{cases} 1, & t\in[0,R];\\ 0, & t\in[2R,+\infty], \end{cases}$$
(2.13)
satisfying $$\psi(t)\in[0,1]$$ and
$$-\frac{C}{R}\leq\frac{\psi'(t)}{\sqrt{\psi}}\leq0,\qquad \big| \psi''(t)\big|\leq\frac {C}{R^{2}},$$
(2.14)
where C is a positive constant. Let
$$\phi=\psi\bigl(d(x,p)\bigr).$$
Using Eq. (2.19) in  (see Eq. (4.5) in  or [12, Theorem 3.1]), we obtain
$$\Delta_{f}\phi\geq-\frac{C\beta}{R}- \frac{C(n-1)K(2R-1)}{R}-\frac{C}{R^{2}}$$
(2.15)
and
$$\frac{|\nabla\phi|^{2}}{\phi}\leq\frac{C}{R^{2}}.$$
(2.16)
Denote by $$B_{p}(R)$$ the geodesic ball centered at p with radius R. Let $$G=\phi|\nabla h|^{2}$$. Assume G achieves its maximum at the point $$x_{0}\in B_{p}(2R)$$ and assume $$G(x_{0})>0$$ (otherwise the proof is trivial). Then, at the point $$x_{0}$$,
$$\Delta_{f} G\leq0,\qquad \nabla\bigl(|\nabla h|^{2}\bigr)=- \frac{|\nabla h|^{2}}{\phi} \nabla\phi$$
and
\begin{aligned} [b]0\geq{}& \Delta_{f}G \\ ={}&\phi\Delta_{f}\bigl(|\nabla h|^{2}\bigr)+|\nabla h|^{2}\Delta_{f}\phi+2\nabla\phi\nabla |\nabla h|^{2} \\ ={}&\phi\Delta_{f}\bigl(|\nabla h|^{2}\bigr)+ \frac{\Delta_{f}\phi}{\phi}G -2\frac{|\nabla\phi|^{2}}{\phi^{2}}G \\ \geq{}&\frac{\Delta_{f}\phi}{\phi}G -2\frac{|\nabla\phi|^{2}}{\phi^{2}}G+2\phi \biggl[\frac{4n-3}{16n} \frac{|\nabla h|^{4}}{h^{2}}-\frac{1}{4}\frac{\nabla h}{h}\nabla\bigl(|\nabla h|^{2}\bigr) \\ &-\bigl[a+\tilde{b}\epsilon+(n-1)K+a\tilde{L}\bigr]|\nabla h|^{2} \biggr] \\ ={}&\frac{\Delta_{f}\phi}{\phi}G -2\frac{|\nabla\phi|^{2}}{\phi^{2}}G+\frac {4n-3}{8n} \frac{G^{2}}{\phi h^{2}}+\frac{G}{2\phi}\nabla\phi\frac{\nabla h}{h} \\ &-2\bigl[a+\tilde{b}\epsilon+(n-1)K+a\tilde{L}\bigr]G, \end{aligned}
(2.17)
where in the second inequality, we used (2.12). Multiplying both sides of (2.17) by $$\frac{\phi}{G}$$, we obtain
\begin{aligned}[b] \frac{4n-3}{8n}\frac{G}{h^{2}} \leq{}&{-}\frac{1}{2}\nabla\phi\frac{\nabla h}{h}+2\bigl[a+\tilde{b} \epsilon+(n-1)K+a\tilde{L}\bigr]\phi \\ &-\Delta_{f}\phi+2\frac{|\nabla\phi|^{2}}{\phi}. \end{aligned}
(2.18)
Substituting the Cauchy inequality
\begin{aligned}-\frac{1}{2}\nabla\phi\frac{\nabla h}{h}\leq{}& \frac{1}{2}|\nabla\phi|\frac{|\nabla h|}{h} \\ \leq{}&\frac{n}{4n-3}\frac{|\nabla\phi|^{2}}{\phi} +\frac{4n-3}{16n}\phi \frac{|\nabla h|^{2}}{h^{2}} \\ ={}&\frac{n}{4n-3}\frac{|\nabla\phi|^{2}}{\phi} +\frac{4n-3}{16n}\frac{G}{h^{2}} \end{aligned}
into (2.18) gives
\begin{aligned}[b] \frac{4n-3}{16n}\frac{G}{h^{2}} \leq{}&2\bigl[a+\tilde{b}\epsilon+(n-1)K+a\tilde {L}\bigr]\phi-\Delta_{f} \phi +\frac{9n-6}{4n-3}\frac{|\nabla\phi|^{2}}{\phi} \\ \leq{}&2\bigl[a+\tilde{b}\epsilon+(n-1)K+a\tilde{L}\bigr]+\frac {C_{1}[(n-1)K(2R-1)+\beta]}{R}+ \frac{C_{2}}{R^{2}}, \end{aligned}
(2.19)
where $$C_{1}$$, $$C_{2}$$ are two positive constants depending on n. Hence, on $$B_{p}(R)$$ with $$R>1$$, it follows from (2.19) that
\begin{aligned}[b] \frac{4n-3}{16n}G(x)\leq{}& \frac{4n-3}{16n}G(x_{0}) \\ \leq{}&h^{2}(x_{0}) \biggl[2\bigl[a+\tilde{b} \epsilon+(n-1)K+a\tilde{L}\bigr] \\ &+\frac{C_{1}[(n-1)K(2R-1)+\beta]}{R}+\frac{C_{2}}{R^{2}} \biggr]. \end{aligned}
(2.20)
In particular, the estimate (2.20) gives
\begin{aligned} |\nabla u|^{2}\leq{}&C A^{2} \biggl[\max \biggl\{ \frac{4}{5}b+a \biggl(1+ \frac {4}{5}L \biggr),0 \biggr\} +K+\frac{|\beta|+1}{R} \biggr], \end{aligned}
(2.21)
which finishes the proof of Theorem 1.1.

### 2.2 Proof of Theorem 1.3

We define $$\tilde{h}=\log u$$. Then we have
\begin{aligned}[b] \Delta\tilde{h}-\nabla f\nabla \tilde{h}&=\Delta_{f}\tilde{h} \\ &=\frac{\Delta_{f}u}{u}-\big|\nabla(\log u)\big|^{2} \\ &=-|\nabla\tilde{h}|^{2}-a\tilde{h}-b, \end{aligned}
(2.22)
where, in the last equality of (2.22), we used Eq. (1.3). Using the Bochner formula with respect to the f-Laplacian, we have
\begin{aligned} [b] \frac{1}{2}\Delta_{f}| \nabla\tilde{h}|^{2}={}&\big|\nabla^{2}\tilde{h}\big|^{2}+ \nabla \tilde{h}\nabla\Delta_{f} \tilde{h}+\mathrm{Ric}_{f} (\nabla\tilde {h},\nabla \tilde{h}) \\ \geq{}&\frac{1}{n}(\Delta\tilde{h})^{2}+\nabla \tilde{h}\nabla \Delta_{f} \tilde{h}-(n-1)K|\nabla\tilde{h}|^{2}. \end{aligned}
(2.23)
Moreover, by virtue of (2.22), we have
\begin{aligned}[b] (\Delta\tilde{h})^{2}={}& \bigl({-}|\nabla\tilde{h}|^{2}+\nabla f\nabla\tilde {h}-a\tilde{h}-b \bigr)^{2} \\ ={}&|\nabla\tilde{h}|^{4}-2|\nabla\tilde{h}|^{2}(\nabla f \nabla\tilde {h}-a\tilde{h}-b)+(\nabla f\nabla\tilde{h}-a\tilde{h}-b)^{2}. \end{aligned}
(2.24)
If the assumption (1) holds, then (2.24) yields
\begin{aligned} [b](\Delta\tilde{h})^{2}\geq{}&| \nabla\tilde{h}|^{4}-2\delta|\nabla\tilde {h}|^{4}+(\nabla f \nabla\tilde{h}-a\tilde{h}-b)^{2} \\ \geq{}&(1-2\delta)|\nabla\tilde{h}|^{4}. \end{aligned}
(2.25)
On the other hand, if the assumption (2) holds, then (2.24) shows
\begin{aligned}[b] (\Delta\tilde{h})^{2}\geq{}&| \nabla\tilde{h}|^{4}-(\nabla f\nabla\tilde {h}-a\tilde{h}-b)^{2}+( \nabla f\nabla\tilde{h}-a\tilde{h}-b)^{2} \\ ={}&|\nabla\tilde{h}|^{4} \\ \geq{}&(1-2\delta)|\nabla\tilde{h}|^{4}. \end{aligned}
(2.26)
Therefore, in these two cases, we have
$$(\Delta\tilde{h})^{2}\geq(1-2\delta)|\nabla \tilde{h}|^{4},$$
(2.27)
and (2.23) gives
\begin{aligned} \frac{1}{2}\Delta_{f}| \nabla\tilde{h}|^{2}\geq{}&\frac{1-2\delta}{n}|\nabla \tilde{h}|^{4}- \nabla \tilde{h}\nabla\bigl(|\nabla\tilde{h}|^{2}\bigr)-\bigl[a+(n-1)K \bigr]|\nabla\tilde{h}|^{2}. \end{aligned}
(2.28)
Following the proof of Theorem 1.1 line by line, we obtain on $$B_{p}(R)$$ with $$R>1$$,
$$|\nabla\tilde{h}|^{2}\leq\frac{C_{1}(n,\delta,\beta)}{R}+C_{2}(n, \delta)\max \bigl\{ a+(n-1)K,0\bigr\} ,$$
(2.29)
where δ is taken to zero in the second assumption.

We completed the proof of Theorem 1.3.

## Declarations

### Acknowledgements

The authors want to thank the referee for helpful suggestions, which made the paper more readable. The research of the author is supported by NSFC No. 11401179.

### Authors’ contributions

BM and YD participated in gradient estimates in this paper. All authors read and approved the final manuscript.

### Competing interests

The authors declare that they have no competing interests. 