Parabolic Gradient Estimates and Harnack Inequalities for a Nonlinear Equation Under The Ricci Flow

Abstract

When the Riemannian metric evolves under the Ricci flow, we investigate parabolic gradient estimates (Li–Yau’s type and J. Li’s type) for positive solutions to the nonlinear parabolic equation $(\Delta -\partial _t)u=(p+1)\frac{|\nabla u|^2}{u}+qu$ on the underlying manifold. Based on these gradient estimates, we derive associated Harnack inequalities, respectively.

1 Introduction

In 1980s, Li and Yau (1986) derived a gradient estimate, which was known as the Li–Yau estimate, for the heat equation on a complete Riemannian manifold. Moreover, they deduced Harnack inequalities. The Harnack inequality also applied to the Ricci flow by Hamilton (1993) and played an important role in solving the Poincaré conjecture Cao and Zhu (2006); Perelman (Perelman). After the fundamental work of Li and Yau (1986), the investigation of Li–Yau estimates for general parabolic partial differential equations of second order has drawn much attentions. To name a few, Yang (2008) proved the Li–Yau estimate for the equation on a Riemannian manifold of $\partial _tu=\Delta u+au\ln u+bu$ with $a,b\in {\mathbb {R}}$, which was introduced by Ma (2016). Moreover, Wu (2010) derived a Li–Yau estimate for the positive solutions to the equation of $\partial _tu=\Delta u-\langle \nabla \phi ,\nabla u\rangle -au\ln u-qu$ on complete Riemannian manifolds. Moreover, Li (1991) proved different type parabolic gradient estimates and Harnack inequalities for positive solutions to a heat-type equation of $(\Delta -\partial _t)u+hu^\alpha =0$ on manifolds.

When the metric evolves along the Ricci flow, Bailesteanu et al. (2010) obtained a series of gradient estimates and Harnack inequalities for positive solutions to the heat equation. Li and Zhu (2016) showed Li–Yau estimates and associated Harnack inequalities for the parabolic partial differential equation of $\partial _tu=\Delta _tu+hu^p$ under the Ricci flow. They also Li and Zhu (2018) derived Li–Yau’s gradient estimates for $(\Delta _t-\partial _t)u=qu+au(\ln u)^\alpha$ coupling with the Ricci flow.

Let $(M^n,g(t))_{t\in [0,T]}$ be a complete solution to the Ricci flow

$$\begin{aligned} \partial _tg(t)=-2Ric_{g(t)}. \end{aligned}$$

(1.1)

Motivating by the works mentioned above, we consider a positive function $u =u(x,t)$ defined on the compact set $Q_{\rho ,T}:=\overline{B({\bar{x}},\rho )}\times [0,T]$ solving the nonlinear parabolic equation of

$$\begin{aligned} (\Delta -\partial _t)u=(p+1)\frac{|\nabla u|^2}{u}+qu \end{aligned}$$

(1.2)

with a nonzero function $p\in C^2(M^n)$ and a time-dependent function $q\in C^{2,1}\big (M^n\times [0,T]\big )$. Here $\Delta$ stands for the Laplacian of g(x, t).

Throughout the paper, we define parabolic cylinders

$$\begin{aligned} Q_{\rho ,t}:=\overline{B({\bar{x}},\rho )}\times [0,t]\subset M\times [0,T], \end{aligned}$$

which is compact for any $0<\rho <+\infty$.

The first result of this paper gives a Li–Yau’s type gradient estimate for positive solutions of the nonlinear parabolic Eq. (1.2) in the case of $p>0$ under the Ricci flow (1.1).

Theorem 1.1

Let $(M^n,g(t))_{t\in [0,T]}$ be a complete solution to the Ricci flow (1.1) on an n-dimensional manifold $M^n$ with $-Kg(t)\le Ric_{g(t)}\le Kg(t)$ on $Q_{\rho ,T}$ for some constant $K>0$ and u(x, t) be a smooth positive solution to the nonlinear parabolic Eq. (1.2) with $p>0$ on $Q_{\rho ,T}$. Then for any $\beta >1$, $0<\epsilon <1$ and $0<a<\frac{1}{\beta }$, there exist positive constants $C_{1/2}$ and ${\bar{C}}$ so that

$$\begin{aligned}&p|\nabla f|^2+\beta \partial _tf+\beta q\nonumber \\&\quad \le \frac{n\beta }{2a(1-\epsilon )\sigma _1}\bigg [A+\frac{C_{1/2}^2\beta n}{4a(\beta -1)\epsilon \rho ^2}+\frac{\sigma _3^2}{(1-a\beta )(\beta -1)\sigma _1^2}\nonumber \\&\qquad +\frac{\theta _1}{2}+\frac{\beta K}{2(\beta -1)}+\frac{\sqrt{n}\sigma _4}{2(\beta -1)\sigma _1}\bigg ]+\bigg (\frac{n\beta }{2a(1-\epsilon )\sigma _1}\bigg )^\frac{1}{2}\nonumber \\&\qquad \cdot \bigg [\frac{\beta ^2 nK^2}{(1-a\beta )\sigma _1}+(\beta -1)\sigma _2\theta _1+\sqrt{n}\beta \theta _2\bigg ]^\frac{1}{2} \end{aligned}$$

(1.3)

on $Q_{\frac{\rho }{2},T}$.

Here $0<\sigma _1\le p\le \sigma _2$, $|\nabla p|\le \sigma _3$, $|Hess\ p|\le \sigma _4$ and $|\nabla q|\le \theta _1$, $|Hess\ q|\le \theta _2$ on $Q_{\rho ,T}$ for some positive constants $\sigma _1$, $\sigma _2$, $\sigma _3$, $\sigma _4$ and $\theta _1$, $\theta _2$. Moreover,

$$\begin{aligned} A:=\frac{C_{1/2}}{\rho }(n-1)\left( \sqrt{K}+\frac{2}{\rho }\right) +\frac{C_{1/2}+2C_{1/2}^2}{\rho ^2}+\frac{(\beta +1){\bar{C}}+1}{t}+(\beta +1)C_{1/2}K. \end{aligned}$$

It is not hard to derive the associated Harnack inequality.

Corollary 1.2

Let $(M^n,g(t))_{t\in [0,T]}$ be a complete solution to the Ricci flow (1.1) on an n-dimensional manifold $M^n$ with $-Kg(t)\le Ric_{g(t)}\le Kg(t)$ on $Q_{\rho ,T}$ for some constant $K>0$ and u(x, t) be a smooth positive solution to the nonlinear parabolic Eq. (1.2) with $p>0$ on $Q_{\rho ,T}$. Then for any $\beta >1$, $0<\epsilon <1$ and $0<a<\frac{1}{\beta }$, there exist positive constants $C_{1/2}$ and ${\bar{C}}$ so that

$$\begin{aligned} \frac{u(y_2,s_2)}{u(y_1,s_1)}\le & {} \bigg (\frac{s_2}{s_1}\bigg )^{\frac{n(\beta {\bar{C}}+\beta +1)}{2a(1-\epsilon )\sigma _1}}\cdot \exp \Bigg \{(s_2-s_1)\bigg \{\theta _0+\frac{n}{2a(1-\epsilon )\sigma _1}\nonumber \\&\quad \times \bigg [\frac{C_{1/2}}{\rho }(n-1)\bigg (\sqrt{K}+\frac{2}{\rho }\bigg )\nonumber \\&+\frac{C_{1/2}+2C_{1/2}^2}{\rho ^2}+(\beta +1)C_{1/2}K+\frac{C_{1/2}^2\beta n}{4a(\beta -1)\epsilon \rho ^2}\nonumber \\&+\frac{\sigma _3^2}{(1-a\beta )(\beta -1)\sigma _1^2}+\frac{\theta _1}{2}\nonumber \\&+\frac{\beta K}{2(\beta -1)}+\frac{\sqrt{n}\sigma _4}{2(\beta -1)\sigma _1}\bigg ]+\sqrt{\frac{n\beta }{2a(1-\epsilon )\sigma _1}}\bigg [\frac{ nK^2}{(1-a\beta )\sigma _1}+\frac{\beta -1}{\beta }\sigma _2\theta _1\nonumber \\&+\frac{\sqrt{n} \theta _2}{\beta }\bigg ]^\frac{1}{2}\bigg \}\Bigg \}. \end{aligned}$$

(1.4)

for any $(y_1,s_1)$, $(y_2,s_2)\in Q_{\frac{\rho }{2},T}$ with $0<s_1<s_2$.

Here, $0<\sigma _1\le p\le \sigma _2$, $|\nabla p|\le \sigma _3$, $|Hess\ p|\le \sigma _4$ and $|q|\le \theta _0$, $|\nabla q|\le \theta _1$, $|Hess\ q|\le \theta _2$ on $Q_{\rho ,T}$ for some positive constants $\sigma _1$, $\sigma _2$, $\sigma _3$, $\sigma _4$ and $\theta _0$, $\theta _1$, $\theta _2$.

Inspired by the works of Li (1991), Li and Zhu (2016), we present a parabolic gradient estimate for positive solutions of the nonlinear parabolic Eq. (1.2) in the case of $p<0$ under the Ricci flow (1.1).

Theorem 1.3

Let $(M^n,g(t))_{t\in [0,T]}$ be a complete solution to the Ricci flow (1.1) on an n-dimensional manifold $M^n$ with $-Kg(t)\le Ric_{g(t)}\le Kg(t)$ on $Q_{\rho ,T}$ for some constant $K>0$ and u(x, t) be a smooth positive solution to the nonlinear parabolic Eq. (1.2) with $p<0$ on $Q_{\rho ,T}$. Then for positive constants b, $k_1$ and $k_2$ with $k_1b\ge 1$ and $\frac{2nb^2}{\min _{Q_{\rho ,T}}(-p)}k_2<1$, there exist positive constants $C_{1/2}$ and ${\bar{C}}$ so that

$$\begin{aligned}&\frac{|\nabla u|^2}{u^2}-k_1q-\frac{k_1\partial _tu}{u}\nonumber \\&\quad \le \frac{nk_1^2}{1-k_2}B+\frac{k_1}{\sqrt{1-k_2}}\left( \sqrt{n}\theta _1+n^\frac{3}{4}\sqrt{\theta _2k_1}\right) \nonumber \\&\qquad +\frac{\sqrt{n}k_1}{b(1-k_2)}\bigg [\frac{1}{n}\bigg (1+\frac{\sigma _5}{b}-\frac{2}{k_1b}\bigg )^2-\frac{1}{k_2}\bigg ]^{-\frac{1}{2}}\nonumber \\&\qquad \cdot \bigg [1+\frac{n\sigma _3^2}{b^2(1-k_2)}\bigg (1+\frac{\sigma _5}{b}-\frac{1}{k_1b}\bigg )^{-2}+\frac{nk_1^2\sigma _6^2C_{1/2}^2}{(1-k_2)\rho ^2}\nonumber \\&\qquad +2(b+\sigma _6)k_1K+2k_1bK\bigg (1+\frac{\sigma _5}{b}-\frac{1}{k_1b}\bigg )\bigg ]\nonumber \\ \end{aligned}$$

(1.5)

on $Q_{\frac{\rho }{2},T}$, where

$$\begin{aligned} B:=\frac{C_{1/2}}{\rho }(n-1)\left( \sqrt{K}+\frac{2}{\rho }\right) +\frac{C_{1/2}+2C_{1/2}^2}{\rho ^2}+\frac{{\bar{C}}}{t}+\left( C_{1/2}+2\right) K. \end{aligned}$$

Here $-\sigma _6\le p\le -\sigma _5<0$, $|\nabla p|\le \sigma _3$, $|Hess\ p|\le \sigma _4$ and $|\nabla q|\le \theta _1$, $|Hess\ q|\le \theta _2$ on $Q_{\rho ,T}$ for some positive constants $\sigma _3$, $\sigma _4$, $\sigma _5$, $\sigma _6$ and $\theta _1$, $\theta _2$.

The Harnack inequality associated to Theorem 1.3 is

Corollary 1.4

Let $(M^n,g(t))_{t\in [0,T]}$ be a complete solution to the Ricci flow (1.1) on an n-dimensional manifold $M^n$ with $-Kg(t)\le Ric_{g(t)}\le Kg(t)$ on $Q_{\rho ,T}$ for some constant $K>0$ and u(x, t) be a smooth positive solution to the nonlinear parabolic Eq. (1.2) with $p<0$ on $Q_{\rho ,T}$. Then for positive constants b, $k_1$ and $k_2$ with $k_1b\ge 1$ and $\frac{2nb^2}{\min _{Q_{\rho ,T}}(-p)}k_2<1$, there exist positive constants $C_{1/2}$ and ${\bar{C}}$ so that

$$\begin{aligned} \frac{u(y_1,s_1)}{u(y_2,s_2)}\le & {} \bigg (\frac{s_2}{s_1}\bigg )^{\frac{nk_1^2{\bar{C}}}{1-k_2}}\cdot \exp \Bigg \{(s_2-s_1)\bigg \{\theta _0+\frac{nk_1^2}{1-k_2}E\nonumber \\&+\frac{k_1}{\sqrt{1-k_2}}(\sqrt{n}\theta _1+n^\frac{3}{4}\sqrt{\theta _2k_1})\nonumber \\&+\frac{\sqrt{n}k_1}{b(1-k_2)}\bigg [\frac{1}{n}\bigg (1+\frac{\sigma _5}{b}-\frac{2}{k_1b}\bigg )^2-\frac{1}{k_2}\bigg ]^{-\frac{1}{2}}\nonumber \\&\cdot \bigg [1++\frac{n\sigma _3^2}{b^2(1-k_2)}\bigg (1+\frac{\sigma _5}{b}-\frac{1}{k_1b}\bigg )^{-2}\nonumber \\&+\frac{nk_1^2\sigma _6^2C_{1/2}^2}{(1-k_2)\rho ^2}+2(b+\sigma _6)k_1K+2k_1bK\bigg (1+\frac{\sigma _5}{b}-\frac{1}{k_1b}\bigg )\bigg ]\bigg \}\Bigg \}.\nonumber \\ \end{aligned}$$

(1.6)

for any $(y_1,s_1)$, $(y_2,s_2)\in Q_{\frac{\rho }{2},T}$ with $0<s_1<s_2$, where

$$\begin{aligned} E:=\frac{C_{1/2}}{\rho }(n-1)\left( \sqrt{K}+\frac{2}{\rho }\right) +\frac{C_{1/2}+2C_{1/2}^2}{\rho ^2}+(C_{1/2}+2)K. \end{aligned}$$

This paper is arranged as follows. In Sect. 2, we introduce the geometric background of (1.2), moreover, we present a fundamental analytical result that we shall need. In Sect. 3, we prove Theorem 1.1 and Corollary 1.2. We finish the proof of Theorem 1.3 in Sect. 4.

2 Preliminaries

In fact, our consideration of the nonlinear parabolic differential Eq. (1.2) is motivated by the understanding of gradient generalized m-quasi-Einstein metrics. First of all, we recall the definition of a gradient generalized m-quasi-Einstein metric (see e.g. Barros and Gomes 2013; Besse 1987; Case et al. 2011).

A gradient generalized m-quasi-Einstein metric on a complete Riemannian manifold $(M^n,g)$ is a choice of a smooth potential function $f:M^n\rightarrow {\mathbb {R}}$ as well as a smooth function $\lambda :M^n\rightarrow {\mathbb {R}}$ such that

$$\begin{aligned} Ric+Hess f-\frac{1}{m}df\otimes df=\lambda g, \end{aligned}$$

(2.1)

where Ric denotes the Ricci tensor of g, while $0<m<+\infty$ is a constant, Hess and $\otimes$ stand for the Hessian and the tensorial product, respectively.

Taking the trace of both sides of (2.1), we have

$$\begin{aligned} R+\Delta f-\frac{|\nabla f|^2}{m}=n\lambda . \end{aligned}$$

(2.2)

Then letting $u=e^f$, we find that

$$\begin{aligned} R+\frac{\Delta u}{u}-\frac{|\nabla u|^2}{u^2}-\frac{|\nabla u|^2}{mu^2}=n\lambda . \end{aligned}$$

Therefore, we obtained a elliptic partial differential equation on Riemannian manifold $(M^n,g)$ that

$$\begin{aligned} \Delta u=\frac{m+1}{m}\frac{|\nabla u|^2}{u}+(n\lambda -R)u. \end{aligned}$$

(2.3)

When the Riemannian metric g evolves along the Ricci flow (1.1), we get a parabolic partial differential equation that

$$\begin{aligned} \partial _tu=\Delta u-\frac{m+1}{m}\frac{|\nabla u|^2}{u}-(n\lambda -R_{g(t)})u. \end{aligned}$$

(2.4)

It is clear that (2.4) is a special case of (1.2).

Next, we present a smooth cut-off function satisfying a basic analytical result stated in the following lemma. (see also Li and Yau 1986; Bailesteanu et al. 2010; Li and Zhu 2018). Note that $r(x,t):=d_{g(t)}(x,{\bar{x}})$ is the distance function from some point ${\bar{x}}\in M^n$ with respect to the metric g(t).

Lemma 2.1

Given $\tau \in (0,T]$, there exists a smooth function ${\overline{\Psi }}:[0,\infty )\times [0,T]\rightarrow {\mathbb {R}}$ satisfying the following requirements:

• 1. The support of ${\overline{\Psi }}$ is a subset of $[0,\rho ]\times [0,T]$, and $0\le {\overline{\Psi }}\le 1$ in $[0,\rho ]\times [0,T]$.

• 2. The equalities ${\overline{\Psi }}=1$ and $\frac{\partial {\overline{\Psi }}}{\partial r}(r,t)=0$ hold in $[0,\frac{\rho }{2}]\times [\tau ,T]$ and $[0,\frac{\rho }{2}]\times [0,T]$, respectively.

• 3. The estimate $|\frac{\partial {\overline{\Psi }}}{\partial t}|\le \frac{{\bar{C}}{\overline{\Psi }}^{\frac{1}{2}}}{\tau }$ is satisfied on $[0,\infty )\times [0,T]$ for some ${\bar{C}}>0$, and ${\overline{\Psi }}(r,0)=0$ for all $r\in [0,\infty )$.

• 4. The inequalities $-\frac{C_\alpha {\overline{\Psi }}^\alpha }{\rho }\le \frac{\partial {\overline{\Psi }}}{\partial r}\le 0$ and $|\frac{\partial ^2{\overline{\Psi }}}{\partial r^2}|\le \frac{C_\alpha {\overline{\Psi }}^\alpha }{\rho ^2}$ hold on $[0,\infty )\times [0,T]$ for every $\alpha \in (0,1)$ with some constant $C_\alpha$ dependent on $\alpha$.

Throughout this paper, we choose the cut-off function ${\bar{\Psi }}$ constructed in Lemma 2.1 for any fixed $\tau \in (0,T]$. Moreover, we define $\Psi :M^n\times [0,T]\rightarrow {\mathbb {R}}$ by

$$\begin{aligned} \Psi (x,t)={\overline{\Psi }}(r(x,t),t). \end{aligned}$$

In the rest of this section, we deal with $(\Delta -\partial _t)\Psi -2\frac{|\nabla \Psi |^2}{\Psi }$ that will be used in the proof of main theorems.

To estimate $\Delta \Psi$, we divide the arguments into two cases:

• Case 1: $r(x,t)<\frac{\rho }{2}$. In this case, it follows from Lemma 2.1 that $\Psi (x,t)\equiv 1$ around (x, t). Therefore, $\Delta \Psi (x,t)=0$.

• Case 2: $r(x,t)\ge \frac{\rho }{2}$. Since $Ric\ge -(n-1)K$ in $B({\bar{x}},\rho )$ for any fixed $t\in [0,T]$, the Laplace comparison theorem (see e.g. Li 1993) implies

  $$\begin{aligned} \Delta r\le (n-1)\sqrt{K}\coth (\sqrt{K}r)\le (n-1)\left( \sqrt{K}+\frac{1}{r}\right) . \end{aligned}$$

  (2.5)

It follows that

$$\begin{aligned} \Delta \Psi (x,t)= & {} \frac{\partial {\overline{\Psi }}}{\partial r}\Delta r+\frac{\partial ^2{\overline{\Psi }}}{\partial r^2}|\nabla r|^2\\\ge & {} -\frac{C_{1/2}\Psi ^{1/2}}{\rho }(n-1)\left( \sqrt{K}+\frac{2}{\rho }\right) -\frac{C_{1/2}\Psi ^{1/2}}{\rho ^2}. \end{aligned}$$

Therefore, we obtain

$$\begin{aligned} \Delta \Psi (x,t)\ge -\frac{C_{1/2}}{\rho }(n-1)\left( \sqrt{K}+\frac{2}{\rho }\right) -\frac{C_{1/2}}{\rho ^2}. \end{aligned}$$

(2.6)

Next, we estimate $\partial _t\Psi$. $\forall$ $x\in B({\bar{x}},\rho )$, let $\gamma :[0,a]\rightarrow M^n$ be a minimal geodesic connecting x and ${\bar{x}}$ at time $t\in [0,T]$. Then we have

$$\begin{aligned} \partial _tr(x,t)= & {} \partial _t\int _0^a|{\dot{\gamma }}(s)|ds\\= & {} -\int _0^aRic({\dot{\gamma }}(s),{\dot{\gamma }}(s))ds\\\le & {} Kr(x,t)\\\le & {} K\rho . \end{aligned}$$

Combining with Lemma 2.1, we know that

$$\begin{aligned} |\partial _t\Psi |\le & {} |\partial _t{\bar{\Psi }}|+|\nabla {\bar{\Psi }}|\cdot |\partial _tr|\nonumber \\\le & {} \frac{{\bar{C}}\Psi ^{\frac{1}{2}}}{\tau }+KC_{1/2}\Psi ^{\frac{1}{2}}\nonumber \\\le & {} \frac{{\bar{C}}}{\tau }+KC_{1/2}. \end{aligned}$$

(2.7)

By Lemma 2.1, we obtain

$$\begin{aligned} \frac{|\nabla \Psi |^2}{\Psi }\le \frac{C_{1/2}^2}{\rho ^2}. \end{aligned}$$

(2.8)

It follows from (2.6), (2.7) and (2.8), we have

$$\begin{aligned} (\Delta -\partial _t)\Psi -2\frac{|\nabla \Psi |^2}{\Psi }\ge & {} -\frac{C_{1/2}}{\rho }(n-1)\left( \sqrt{K}+\frac{2}{\rho }\right) -\frac{C_{1/2}}{\rho ^2}\nonumber \\&-\frac{{\bar{C}}}{\tau }-C_{1/2}K-\frac{2C_{1/2}^2}{\rho ^2}. \end{aligned}$$

(2.9)

For simplicity, we define

$$\begin{aligned} {\tilde{A}}:=\frac{C_{1/2}}{\rho }(n-1)\left( \sqrt{K}+\frac{2}{\rho }\right) +\frac{C_{1/2}+2C_{1/2}^2}{\rho ^2}+\frac{{\bar{C}}}{\tau }+C_{1/2}K. \end{aligned}$$

3 The Case of $p>0$

In this section, we finish the proof of Theorem 1.1 and Corollary 1.2, which presents a Li–Yau estimate for positive solutions of (1.2) coupling with the Ricci flow (1.1). We proceed with following evolution inequality.

Lemma 3.1

Let $(M^n,g(t))_{t\in [0,T]}$ be a complete solution to the Ricci flow (1.1) on an n-dimensional manifold $M^n$ and u(x, t) be a smooth positive solution to the nonlinear parabolic Eq. (1.2) with $p>0$ on $Q_{\rho ,T}$. If $f:=\ln u$, $H:=t\big (p|\nabla f|^2+\beta \partial _tf+\beta q\big )$ for any $\beta >0$ and $0<a<\frac{1}{\beta }$, then we have

$$\begin{aligned}&\Delta H -(1-\beta )\partial _tH\nonumber \\&\quad \ge \frac{2a\beta pt}{n}\bigg (p|\nabla f|^2+\partial _tf+q\bigg )^2-\frac{t}{2(1-a\beta ) p}(2|\nabla p|\cdot |\nabla f|+\beta |Ric|)^2\nonumber \\&\qquad +2p\langle \nabla H,\nabla f\rangle +2(1-\beta )pt\langle \nabla q,\nabla f\rangle +2\beta ptRic(\nabla f,\nabla f)\nonumber \\&\qquad -\sqrt{n}|\nabla f|^2t|Hess\ p|-\sqrt{n}\beta t|Hess\ q|-\frac{H}{t} \end{aligned}$$

(3.1)

on $Q_{\rho ,T}$.

Proof

Since $u=e^{f}$, we have $\partial _tu=e^{f}\partial _tf$, $\nabla u=e^{f}\nabla f$ and $\Delta u=e^{f}(\Delta f+|\nabla f|^2)$. By (1.2), we have

$$\begin{aligned} e^{f}(\Delta f+|\nabla f|^2-\partial _tf)=(p+1)e^{f}|\nabla f|^2+qe^{f}, \end{aligned}$$

i.e.,

$$\begin{aligned} \Delta f=p|\nabla f|^2+\partial _tf+q. \end{aligned}$$

(3.2)

Moreover, we have

$$\begin{aligned} \Delta f=\frac{H}{t}+(1-\beta )(\partial _tf+q). \end{aligned}$$

(3.3)

Note that

$$\begin{aligned} \nabla _iH=t\big (2p\nabla _i\nabla _jf\nabla _jf+\nabla _ip|\nabla f|^2+\beta \nabla _i\partial _tf+\beta \nabla _iq\big ). \end{aligned}$$

Moreover, we can derive that

$$\begin{aligned}&\Delta H\nonumber \\&\quad =2pt|Hess f|^2+2pt\Delta \nabla _jf\nabla _jf+4Hessf(\nabla p,\nabla f) +\Delta p|\nabla f|^2+\beta \Delta \partial _tf+\beta \Delta q\nonumber \\&\quad \ge \frac{2a\beta pt(\Delta f)^2}{n}+2(1-a\beta ) pt|Hess f|^2+2pt\langle \nabla \Delta f,\nabla f\rangle +2ptRic(\nabla f,\nabla f)\nonumber \\&\qquad +4Hessf(\nabla p,\nabla f)+\Delta p|\nabla f|^2+\beta \partial _t\Delta f-2\beta tR_{ij}\nabla _i\nabla _jf+\beta \Delta q\nonumber \\&\quad =\frac{2a\beta pt}{n}(p|\nabla f|^2+\partial _tf+q)^2+2(1-a\beta )pt|Hess\ f|^2+4tHessf(\nabla p,\nabla f)\nonumber \\&\qquad -2\beta tR_{ij}\nabla _i\nabla _jf+2pt\langle \nabla \Bigg [\frac{H}{t}+(1-\beta )(\partial _tf+q)\Bigg ],\nabla f\rangle +2ptRic(\nabla f,\nabla f)\nonumber \\&\qquad +\Delta p|\nabla f|^2t+\beta \Delta qt+\beta t\partial _t\Bigg [\frac{H}{t}+(1-\beta )(\partial _tf+q)\Bigg ]. \end{aligned}$$

(3.4)

By direct computations, we have

$$\begin{aligned}&2(1-a\beta )pt|Hess f|^2+4tHessf(\nabla p,\nabla f)-2\beta tR_{ij}\nabla _i\nabla _jf\nonumber \\&\quad \ge 2t[(1-a\beta ) p|Hess f|^2-|Hess\ f|(2|\nabla p|\cdot |\nabla f|+\beta |Ric|)]\nonumber \\&\quad = 2 t\Bigg [\sqrt{(1-a\beta ) p}|Hess f|-\frac{2|\nabla p|\cdot |\nabla f|+\beta |Ric|}{2\sqrt{(1-a\beta ) p}}\Bigg ]^2\nonumber \\&\qquad -\frac{t}{2(1-a\beta ) p}(2|\nabla p|\cdot |\nabla f|+\beta |Ric|)^2\nonumber \\&\quad \ge -\frac{t}{2(1-a\beta ) p}(2|\nabla p|\cdot |\nabla f|+\beta |Ric|)^2. \end{aligned}$$

(3.5)

Applying (3.5) to (3.4), we obtain

$$\begin{aligned} \Delta H\ge & {} t\frac{2a\beta pt}{n}(p|\nabla f|^2+\partial _tf+q)^2-\frac{t}{2(1-a\beta ) p}(2|\nabla p|\cdot |\nabla f|+\beta |Ric|)^2\nonumber \\&+2p\langle \nabla H,\nabla f\rangle +2(1-\beta )pt\langle \nabla (\partial _tf+q),\nabla f\rangle +2ptRic(\nabla f,\nabla f)\nonumber \\&+\Delta p|\nabla f|^2t+\beta \Delta qt-\beta \frac{H}{t}+\beta \partial _tH+\beta (1-\beta )t\partial _t(\partial _tf+q). \end{aligned}$$

(3.6)

The definition of H implies that

$$\begin{aligned} \partial _tH= & {} \frac{H}{t}+t\partial _t(p|\nabla f|^2+\beta \partial _tf+\beta q)\nonumber \\= & {} \frac{H}{t}+2ptRic(\nabla f,\nabla f)+2pt\langle \nabla \partial _t f,\nabla f\rangle +\beta t\partial _t(\partial _tf+q), \end{aligned}$$

(3.7)

where we used (1.1).

Combining (3.6) with (3.7), we have

$$\begin{aligned}&\Delta H -(1-\beta )\partial _tH\nonumber \\&\quad \ge t\frac{2a\beta pt}{n}(p|\nabla f|^2+\partial _tf+q)^2-\frac{t}{2(1-a\beta ) p}(2|\nabla p|\cdot |\nabla f|+\beta |Ric|)^2\nonumber \\&\qquad +2p\langle \nabla H,\nabla f\rangle +2(1-\beta )pt\langle \nabla (\partial _tf+q),\nabla f\rangle +2ptRic(\nabla f,\nabla f)\nonumber \\&\qquad +\Delta p|\nabla f|^2t+\beta \Delta qt-\beta \frac{H}{t}+\beta (1-\beta )t\partial _t(\partial _tf+q)-(1-\beta )\frac{H}{t}\nonumber \\&\qquad -2(1-\beta )ptRic(\nabla f,\nabla f)-2(1-\beta )pt\langle \nabla \partial _t f,\nabla f\rangle -\beta (1-\beta )t\partial _t(\partial _tf+q)\nonumber \\&\quad \ge \frac{2a\beta pt}{n}\big (p|\nabla f|^2+\partial _tf+q\big )^2-\frac{t}{2(1-a\beta ) p}(2|\nabla p|\cdot |\nabla f|+\beta |Ric|)^2\nonumber \\&\qquad +2p\langle \nabla H,\nabla f\rangle +2(1-\beta )pt\langle \nabla q,\nabla f\rangle +2\beta ptRic(\nabla f,\nabla f)\nonumber \\&\qquad -\sqrt{n}|\nabla f|^2t|Hess\ p|-\sqrt{n}\beta t|Hess\ q|-\frac{H}{t}. \end{aligned}$$

$\square$

Now we are ready to prove Theorem 1.1.

Proof of Theorem 1.1

From Lemma 3.1, we can get

$$\begin{aligned}&\Delta (\Psi H) -(1-\beta )\partial _t(\Psi H)\nonumber \\&\quad =H(\Delta -(1-\beta )\partial _t)\Psi +2\langle \nabla \Psi ,\nabla H\rangle +\Psi (\Delta -(1-\beta )\partial _t)H\nonumber \\&\quad \ge H(\Delta -(1-\beta )\partial _t)\Psi +\frac{2}{\Psi }\langle \nabla \Psi ,\nabla (\Psi H)\rangle -\frac{2|\nabla \Psi |^2}{\Psi }H\nonumber \\&\qquad +\Psi \Bigg [\frac{2a\beta pt}{n}\big (p|\nabla f|^2+\partial _tf+q\big )^2-\frac{t}{2(1-a\beta ) p}(2|\nabla p|\cdot |\nabla f|+\beta |Ric|)^2\nonumber \\&\qquad +2p\langle \nabla H,\nabla f\rangle +2(1-\beta )pt\langle \nabla q,\nabla f\rangle -\sqrt{n}\beta t|Hess\ q|-\frac{H}{t}\nonumber \\&\qquad +2\beta ptRic(\nabla f,\nabla f)-\sqrt{n}|\nabla f|^2t|Hess\ p|\Bigg ]\nonumber \\&\quad \ge H(\Delta -(1-\beta )\partial _t)\Psi +\frac{2}{\Psi }\langle \nabla \Psi ,\nabla (\Psi H)\rangle -\frac{2|\nabla \Psi |^2}{\Psi }H+2p\langle \nabla (\Psi H),\nabla f\rangle \nonumber \\&\qquad -2pH\langle \nabla \Psi ,\nabla f\rangle +\Psi \big [\frac{2a\beta pt}{n}\big (p|\nabla f|^2+\partial _tf+q\big )^2\nonumber \\&\qquad -\frac{t}{2(1-a\beta ) p}(2|\nabla p|\cdot |\nabla f|+\beta \sqrt{n} K)^2-2(\beta -1)pt|\nabla q||\nabla f|\nonumber \\&\qquad -\sqrt{n}\beta t|Hess\ q|-\frac{H}{t}-2\beta ptK|\nabla f|^2-\sqrt{n}|\nabla f|^2t|Hess\ p|\big ]. \end{aligned}$$

(3.8)

For fixed $\tau \in (0,T]$, let $(x_1,t_1)$ be a maximum point for $\Psi H$ in $Q_{\rho ,\tau }:=\overline{B({\bar{x}},\rho )}\times [0,\tau ]\subset Q_{\rho ,T}$. It follows from (3.8) that at such point.

$$\begin{aligned} 0\ge & {} H(\Delta -(1-\beta )\partial _t)\Psi -\frac{2|\nabla \Psi |^2}{\Psi }H-2pH\langle \nabla \Psi ,\nabla f\rangle \nonumber \\&+\frac{2a\beta p\Psi t}{n}\big (p|\nabla f|^2+\partial _tf+q\big )^2\nonumber \\&-\frac{\Psi t}{2(1-a\beta ) p}(2|\nabla p|\cdot |\nabla f|+\beta \sqrt{n}K)^2-2(\beta -1)p\Psi t|\nabla q||\nabla f|\nonumber \\&-\sqrt{n}\beta \Psi t|Hess\ q|-\frac{\Psi H}{t}-2\beta \Psi ptK|\nabla f|^2-\sqrt{n}|\nabla f|^2t|Hess\ p|.\nonumber \\ \end{aligned}$$

(3.9)

Multiplying both sides of (3.9) by $\Psi t$ and define $I=\Psi H$, we will obtain

$$\begin{aligned} 0\ge & {} \left[ (\Delta -\partial _t)\Psi -2\frac{|\nabla \Psi |^2}{\Psi }+\beta \partial _t\Psi \right] It-2Ipt\langle \nabla \Psi ,\nabla f\rangle \nonumber \\&+\frac{2a\beta p\Psi ^2 t^2}{n}\big (p|\nabla f|^2+\partial _tf+q\big )^2\nonumber \\&-\frac{\Psi ^2 t^2}{2(1-a\beta ) p}(2|\nabla p|\cdot |\nabla f|+\beta \sqrt{n}K)^2-2(\beta -1)p\Psi ^2 t^2|\nabla q||\nabla f|\nonumber \\&-\sqrt{n}\beta \Psi ^2 t^2|Hess\ q|\nonumber \\&-\Psi I-2\beta p\Psi ^2 t^2K|\nabla f|^2-\sqrt{n}|\nabla f|^2\Psi t^2|Hess\ p| \end{aligned}$$

(3.10)

at $(x_1,t_1)$.

In the following, we deal with each term of the right-hand side of (3.10).

From Lemma 2.1, we get

$$\begin{aligned} \langle \nabla \Psi ,\nabla f\rangle \le \frac{|\nabla \Psi |}{\Psi ^{1/2}}\Psi ^{1/2}|\nabla f|\le \frac{C_{1/2}}{\rho }\Psi ^\frac{1}{2}|\nabla f|. \end{aligned}$$

(3.11)

Plugging (2.7), (2.9) and (3.11) into (3.10), we have

$$\begin{aligned} 0\ge & {} -\left( {\tilde{A}}+\frac{\beta {\bar{C}}}{\tau }+\beta C_{1/2}K\right) It-\frac{2pC_{1/2}It}{\rho }\Psi ^\frac{1}{2}|\nabla f|\nonumber \\&+\frac{2a\beta p\Psi ^2 t^2}{n}\big (p|\nabla f|^2+\partial _tf+q\big )^2\nonumber \\&-\frac{2|\nabla p|^2|\nabla f|^2\Psi ^2 t^2}{(1-a\beta ) p}-\frac{\beta ^2 nK^2\Psi ^2 t^2}{(1-a\beta )p}-2(\beta -1)p\Psi ^2 t^2|\nabla q||\nabla f|\nonumber \\&-\sqrt{n}\beta \Psi ^2 t^2|Hess\ q|\nonumber \\&-\Psi I-2\beta p\Psi ^2 t^2K|\nabla f|^2-\sqrt{n}|\nabla f|^2\Psi t^2|Hess\ p| \end{aligned}$$

(3.12)

at $(x_1,t_1)$.

As in Chen and Chen (2009), Li and Zhu (2018), Yang (2008), we set $\omega =\frac{|\nabla f|^2(x_1,t_1)}{H(x_1,t_1)}\ge 0$. Therefore, $|\nabla f|=(\omega H)^{\frac{1}{2}}$ and

$$\begin{aligned} p|\nabla f|^2+\partial _tf+q=p\omega H+\frac{1}{\beta }\bigg (\frac{H}{t}-p\omega H\bigg )=H\bigg (p\omega +\frac{1-p\omega t}{\beta t}\bigg ). \end{aligned}$$

We can simplify (3.12) into the following inequality

$$\begin{aligned} 0\ge & {} -\left( {\tilde{A}}+\frac{\beta {\bar{C}}}{\tau }+\beta C_{1/2}K\right) It-\frac{2pC_{1/2}t}{\rho }\omega ^\frac{1}{2}I^\frac{3}{2}+\frac{2ap I^2}{n}[(\beta -1)p\omega t+1]^2\nonumber \\&-\frac{2|\nabla p|^2\omega I\Psi t^2}{(1-a\beta ) p}-\frac{\beta ^2 n K^2 \Psi ^2 t^2}{(1-a\beta )p}-2(\beta -1)p\Psi ^\frac{3}{2} t^2|\nabla q|(\omega I)^\frac{1}{2}\nonumber \\&-\sqrt{n}\beta \Psi ^2 t^2|Hess\ q|-\Psi I-2\beta p\Psi t^2K\omega I-\sqrt{n}\omega It^2|Hess\ p| \end{aligned}$$

(3.13)

at $(x_1,t_1)$.

By Cauchy’s inequality, we get

$$\begin{aligned} 2(\omega I)^{\frac{1}{2}}\le 1+\omega I, \end{aligned}$$

(3.14)

and

$$\begin{aligned} \frac{2pC_{1/2}t}{\rho }\omega ^\frac{1}{2}I^\frac{3}{2}\le \frac{2a \epsilon pI^2 }{n\beta }[(\beta -1)p\omega t+1]^2+\frac{pC_{1/2}^2\beta n\omega It^2}{2a\epsilon \rho ^2[(\beta -1)p\omega t+1]^2} \end{aligned}$$

(3.15)

for any $\epsilon \in (0,1)$.

Applying (3.14) and (3.15) to (3.13), we obtain

$$\begin{aligned} 0\ge & {} -\left( {\tilde{A}}+\frac{\beta {\bar{C}}}{\tau }+\beta C_{1/2}K\right) It+\frac{2a(1-\epsilon )p\beta I^2}{n\beta }[(\beta -1)p\omega t+1]^2\nonumber \\&-\frac{pC_{1/2}^2\beta n\omega It^2}{2a\epsilon \rho ^2[(\beta -1)p\omega t+1]^2}\nonumber \\&-\frac{2|\nabla p|^2\omega I\Psi t^2}{(1-a\beta ) p}-\frac{\beta ^2 nK^2 \Psi ^2 t^2}{(1-a\beta )p}-(\beta -1)p\Psi ^\frac{3}{2} t^2|\nabla q|(1+\omega I)\nonumber \\&-\sqrt{n}\beta \Psi ^2 t^2|Hess\ q|-\Psi I-2\beta p\Psi t^2K\omega I-\sqrt{n}\omega It^2|Hess\ p| \end{aligned}$$

(3.16)

at $(x_1,t_1)$.

Note that $0\le \Psi \le 1$ and $p>0$. It follows from (3.16) that

$$\begin{aligned}&\frac{2a(1-\epsilon )p I^2}{n\beta }\bigg [(\beta -1)p\omega t+1\bigg ]^2\nonumber \\&\quad \le I\bigg [({\tilde{A}}+\frac{\beta {\bar{C}}}{\tau }+\beta C_{1/2}K)t+\frac{pC_{1/2}^2\beta n\omega t^2}{2a\epsilon \rho ^2[(\beta -1)p\omega t+1]^2}+\frac{2|\nabla p|^2\omega t^2}{(1-a\beta ) p}\nonumber \\&\qquad +(\beta -1)p t^2|\nabla q|\omega +1 +\beta p Kt^2\omega +\sqrt{n}\omega t^2|Hess\ p|\bigg ]\nonumber \\&\qquad +\frac{\beta ^2 nK^2 t^2}{(1-a\beta )p}+(\beta -1)p|\nabla q|t^2+\sqrt{n}\beta |Hess\ q|t^2 \end{aligned}$$

(3.17)

at $(x_1,t_1)$.

Since $\beta >1$, we have $\big [(\beta -1)p\omega t+1\big ]^2\ge 1$ and

$$\begin{aligned} \frac{p\omega t}{\big [(\beta -1)p\omega t-1\big ]^2}\le \frac{1}{2(\beta -1)}. \end{aligned}$$

Therefore, (3.17) reduces to

$$\begin{aligned} I^2\le & {} \frac{n\beta }{2a(1-\epsilon )p}\bigg [\bigg ({\tilde{A}}+\frac{\beta {\bar{C}}}{\tau }+\beta C_{1/2}K\bigg )t+\frac{C_{1/2}^2\beta n t}{4a(\beta -1)\epsilon \rho ^2}+\frac{|\nabla p|^2t}{(1-a\beta )(\beta -1)p^2}\nonumber \\&+\frac{|\nabla q|t}{2}+1+\frac{\beta Kt}{2(\beta -1)}+\frac{\sqrt{n}|Hess\ p|t}{2(\beta -1)p}\bigg ]I\nonumber \\&+\frac{n\beta }{2a(1-\epsilon )p}\bigg [\frac{\beta ^2 nK^2 t^2}{(1-a\beta )p}+(\beta -1)p|\nabla q|t^2+\sqrt{n}\beta |Hess\ q|t^2\bigg ] \end{aligned}$$

(3.18)

at $(x_1,t_1)$.

Recall an elementary fact: if $x^2\le a_1x+a_2$ for some $a_1,a_2,x\ge 0$, then

$$\begin{aligned} x\le \frac{a_1}{2}+\sqrt{a_2+\big (\frac{a_2}{2}\big )^2}\le \frac{a_1}{2}+\sqrt{a_2}+\big (\frac{a_2}{2}\big )^2=a_1+\sqrt{a_2}. \end{aligned}$$

Then we get an upper bound for $I(x_1,t_1)$ that

$$\begin{aligned} I(x_1,t_1)\le & {} \frac{n\beta \tau }{2a(1-\epsilon )p}\bigg [{\tilde{A}}+\frac{\beta {\bar{C}}}{\tau }+\beta C_{1/2}K+\frac{C_{1/2}^2\beta n}{4a(\beta -1)\epsilon \rho ^2}\nonumber \\&+\frac{|\nabla p|^2(x_1)}{(1-a\beta )(\beta -1)p^2(x_1)}\nonumber \\&+\frac{|\nabla q|(x_1,t_1)}{2}+\frac{1}{\tau }+\frac{\beta K}{2(\beta -1)}+\frac{\sqrt{n}|Hess\ p|(x_1)}{2(\beta -1)p}\bigg ]\nonumber \\&+\sqrt{\frac{n\beta }{2a(1-\epsilon )p(x_1)}}\bigg [\frac{\beta ^2 nK^2}{(1-a\beta )p(x_1)}+(\beta -1)p(x_1)|\nabla q|(x_1,t_1)\nonumber \\&+\sqrt{n}\beta |Hess\ q|(x_1,t_1)\bigg ]^\frac{1}{2}\tau \nonumber \\\le & {} \frac{n\beta \tau }{2a(1-\epsilon )\sigma _1}\bigg [{\tilde{A}}+\frac{\beta {\bar{C}}}{\tau }+\beta C_{1/2}K+\frac{C_{1/2}^2\beta n}{4a(\beta -1)\epsilon \rho ^2}\nonumber \\&+\frac{\sigma _3^2}{(1-a\beta )(\beta -1)\sigma _1^2}+\frac{\theta _1}{2}+\frac{1}{\tau }+\frac{\beta K}{2(\beta -1)}\nonumber \\&+\frac{\sqrt{n}\sigma _4}{2(\beta -1)\sigma _1}\bigg ]+\sqrt{\frac{n\beta }{2a(1-\epsilon )\sigma _1}}\bigg [\frac{\beta ^2 n K^2}{(1-a\beta )\sigma _1}\nonumber \\&+(\beta -1)\sigma _2\theta _1+\sqrt{n}\beta \theta _2\bigg ]^\frac{1}{2}\tau . \end{aligned}$$

(3.19)

By the construction of $\Psi$, we have

$$\begin{aligned} \sup _{Q_{\frac{\rho }{2},\tau }}H\le \sup _{Q_{\rho ,\tau }}(\Psi H)= I(x_1,t_1) \end{aligned}$$

for all $t\in [0,\tau ]$ with $\tau \le T$ is arbitrary. Therefore, we conclude that

$$\begin{aligned} p\frac{|\nabla u|^2}{u^2}+\beta \frac{\partial _tu}{u}+\beta q= & {} p|\nabla f|^2+\beta \partial _tf+\beta q\nonumber \\\le & {} \frac{n\beta }{2a(1-\epsilon )\sigma _1}\bigg [A+\frac{C_{1/2}^2\beta n}{4a(\beta -1)\epsilon \rho ^2}+\frac{\sigma _3^2}{(1-a\beta )(\beta -1)\sigma _1^2}\nonumber \\&+\frac{\theta _1}{2}+\frac{\beta K}{2(\beta -1)}+\frac{\sqrt{n}\sigma _4}{2(\beta -1)\sigma _1}\bigg ]+\bigg (\frac{n\beta }{2a(1-\epsilon )\sigma _1}\bigg )^\frac{1}{2}\cdot \nonumber \\&\bigg [\frac{\beta ^2 nK^2}{(1-a\beta )\sigma _1}+(\beta -1)\sigma _2\theta _1+\sqrt{n}\beta \theta _2\bigg ]^\frac{1}{2} \end{aligned}$$

on $Q_{\frac{\rho }{2},T}$, where

$$\begin{aligned} A:=\frac{C_{1/2}}{\rho }(n-1)\left( \sqrt{K}+\frac{2}{\rho }\right) +\frac{C_{1/2}+2C_{1/2}^2}{\rho ^2}+\frac{(\beta +1){\bar{C}}+1}{t}+(\beta +1)C_{1/2}K. \end{aligned}$$

$\square$

We apply the Theorem 1.1 above to prove Corollary 1.2 by integrating along a space-time path joining any two points in $M^n$.

Proof of Corollary 1.2

It follows from Theorem 1.1 that

$$\begin{aligned} \frac{\partial _tu}{u}\le & {} \theta _0+\frac{n }{2a(1-\epsilon )\sigma _1}\bigg [\frac{C_{1/2}}{\rho }(n-1)(\sqrt{K}+\frac{2}{\rho })+\frac{C_{1/2}+2C_{1/2}^2}{\rho ^2}\nonumber \\&+\frac{(\beta +1){\bar{C}}+1}{t}+(\beta +1)C_{1/2}K+\frac{C_{1/2}^2\beta n }{4a(\beta -1)\epsilon \rho ^2}+\frac{\sigma _3^2}{(1-a\beta )(\beta -1)\sigma _1^2}\nonumber \\&+\frac{\theta _1}{2}+\frac{\beta K}{\beta -1}\bigg ]\nonumber \\&+\bigg [\frac{n}{2a(1-\epsilon )\sigma _1\beta }\bigg ((\beta -1)\sigma _2\theta _1+\frac{\beta ^2 n}{(1-a\beta )\sigma _1}K^2+ \beta \sqrt{n}\theta _2\bigg )\bigg ]^\frac{1}{2}. \end{aligned}$$

(3.20)

For any $(y_1,s_1)$, $(y_2,s_2)\in Q_{\frac{\rho }{2},T}$ with $0<s_1<s_2$, take the geodesic path $\gamma (t)$ from $y_1$ to $y_2$ at time $s_1$ parametrized proportional to arc length with parameter t starting at $y_1$ at time $s_1$ and ending at $y_2$ at time $s_2$. Now consider the path $(\gamma (t),t)$ in space-time and integrate (3.20) along $\gamma$, we get

$$\begin{aligned} \ln \frac{u(y_2,s_2)}{u(y_1,s_1)}\le & {} \frac{n(\beta {\bar{C}}+\beta +1)}{2a(1-\epsilon )\sigma _1}\ln \frac{s_2}{s_1}+(s_2-s_1)\bigg \{\theta _0+\frac{\sigma _3^2}{(1-a\beta )(\beta -1)\sigma _1^2}\nonumber \\&+\frac{n }{2a(1-\epsilon )\sigma _1}\bigg [\frac{C_{1/2}}{\rho }(n-1)\bigg (\sqrt{K}+\frac{2}{\rho }\bigg )+\frac{C_{1/2}+2C_{1/2}^2}{\rho ^2}\nonumber \\&+\frac{{\bar{C}}+1}{t}+(\beta +1)C_{1/2}K+\frac{C_{1/2}^2\beta n }{4a(\beta -1)\epsilon \rho ^2}+\frac{\theta _1}{2}+\frac{\beta K}{\beta -1}\bigg ]\nonumber \\&+\bigg [\frac{n}{2a(1-\epsilon )\sigma _1\beta }\bigg ((\beta -1)\sigma _2\theta _1+\frac{\beta ^2 n}{(1-a\beta )\sigma _1}K^2+ \beta \sqrt{n}\theta _2\bigg )\bigg ]^\frac{1}{2}\bigg \}. \end{aligned}$$

for any given $(y_1,s_1)$, $(y_2,s_2)\in Q_{\frac{\rho }{2},T}$ with $0<s_1<s_2$.

Now exponentiating and rearranging gives the desired result. $\square$

As corollary of Theorem 1.1, we have a Li–Yau’s type gradient estimate and associated Harnack inequality for (2.4), which has a closer relationship with the gradient generalized m-quasi-Einstein metric, coupling with the Ricci flow (1.1).

Corollary 3.2

Let $(M^n,g(t))_{t\in [0,T]}$ be a complete solution to the Ricci flow (1.1) on an n-dimensional manifold $M^n$ with $-Kg(t)\le Ric_{g(t)}\le Kg(t)$ on $Q_{\rho ,T}$ for some constant $K>0$ and u(x, t) be a smooth positive solution to the nonlinear parabolic Eq. (2.4) on $Q_{\rho ,T}$. Assume $\theta _1=\sup _{Q_{\rho ,T}}|\nabla (n\lambda -R_{g(t)})|$ and $\theta _2=\sup _{Q_{\rho ,T}}|Hess\ (n\lambda -R_{g(t)})|$, then for any $\beta >1$, $0<\epsilon <1$ and $0<a<\frac{1}{\beta }$, there exist positive constants $C_{1/2}$ and ${\bar{C}}$ so that

$$\begin{aligned}&\frac{|\nabla u|^2}{mu^2}+\beta \frac{\partial _tu}{u}+\beta q\le \frac{mn\beta }{2a(1-\epsilon )}\bigg [\frac{C_{1/2}}{\rho }(n-1)\bigg (\sqrt{K}+\frac{2}{\rho }\bigg )+\frac{C_{1/2}+2C_{1/2}^2}{\rho ^2}\nonumber \\&\quad +\frac{\beta {\bar{C}}+\beta +1}{t}+(\beta +1)C_{1/2}K+\frac{C_{1/2}^2\beta n }{4a(\beta -1)\epsilon \rho ^2}+\frac{\theta _1}{2}+\frac{\beta K}{\beta -1}\bigg ]\nonumber \\&\quad +\bigg [\frac{mn\beta }{2a(1-\epsilon )}\bigg (\frac{(\beta -1)\theta _1}{m}+\frac{\beta ^2 nm}{2(1-a\beta )}K^2+ \beta \sqrt{n}\theta _2\bigg )\bigg ]^\frac{1}{2} \end{aligned}$$

(3.21)

on $Q_{\frac{\rho }{2},T}$.

Corollary 3.3

Let $(M^n,g(t))_{t\in [0,T]}$ be a complete solution to the Ricci flow (1.1) on an n-dimensional manifold $M^n$ with $-Kg(t)\le Ric_{g(t)}\le Kg(t)$ on $Q_{\rho ,T}$ for some constant $K>0$ and u(x, t) be a smooth positive solution to the nonlinear parabolic Eq. (2.4) on $Q_{\rho ,T}$. Assume $\theta _1=\sup _{Q_{\rho ,T}}|\nabla (n\lambda -R_{g(t)})|$ and $\theta _2=\sup _{Q_{\rho ,T}}|Hess\ (n\lambda -R_{g(t)})|$, then for any $\beta >1$, $0<\epsilon <1$ and $0<a<\frac{1}{\beta }$, there exist positive constants $C_{1/2}$ and ${\bar{C}}$ so that

$$\begin{aligned} \frac{u(y_2,s_2)}{u(y_1,s_1)}\le & {} \bigg (\frac{s_2}{s_1}\bigg )^{\frac{mn(\beta {\bar{C}}+\beta +1)}{2a(1-\epsilon )}}\cdot \exp \Bigg \{(s_2-s_1)\bigg \{\theta _0\nonumber \\&+\frac{mn }{2a(1-\epsilon )}\bigg [\frac{C_{1/2}}{\rho }(n-1)\bigg (\sqrt{K}+\frac{2}{\rho }\bigg )+\frac{C_{1/2}+2C_{1/2}^2}{\rho ^2}\nonumber \\&+(\beta +1)C_{1/2}K+\frac{C_{1/2}^2\beta n }{4a(\beta -1)\epsilon \rho ^2}+\frac{\theta _1}{2}+\frac{\beta K}{\beta -1}\bigg ]\nonumber \\&+\bigg [\frac{mn}{2a(1-\epsilon )\beta }\bigg (\frac{(\beta -1)\theta _1}{m}+\frac{\beta ^2 nm}{2(1-a\beta )}K^2+\beta \sqrt{n}\theta _2\bigg )\bigg ]^\frac{1}{2}\bigg \}\Bigg \}\nonumber \\ \end{aligned}$$

(3.22)

for any $(y_1,s_1)$, $(y_2,s_2)\in Q_{\frac{\rho }{2},T}$ with $0<s_1<s_2$.

4 The Case of $p<0$

In this section, we give a proof of Theorem 1.3, which presents a Li’s type gradient estimate for positive solutions of (1.2) in the case of $p<0$ under the Ricci flow (1.1).

As in Li (1991), Li and Zhu (2016), we define a new function

$$\begin{aligned} v=u^{-b}, \end{aligned}$$

where b is a small enough positive constant to be determined later. Then we have

$$\begin{aligned} (\Delta -\partial _t)v= & {} -bu^{-b-1}(\Delta -\partial _t)u+b(b+1)u^{-b-2}|\nabla u|^2\nonumber \\= & {} -b(p+1)u^{-b-2}|\nabla u|^2-bqu^{-b}+b(b+1)u^{-b-2}|\nabla u|^2\nonumber \\= & {} \frac{b-p}{b}\frac{|\nabla v|^2}{v}-bqv, \end{aligned}$$

(4.1)

where we used (1.2) in the second equality.

Let $k_1$ be a positive constant such that $k_1b\ge 1$. Similar to Li (1991), Li and Zhu (2016), we consider the evolution of the following three functions.

$$\begin{aligned}&G_0:=|\nabla \ln v|^2-k_1b^2q,\\&G_1:=\partial _t\ln v, \end{aligned}$$

and

$$\begin{aligned} G:=G_0+k_1bG_1. \end{aligned}$$

Proposition 4.1

Let $(M^n,g(t))_{t\in [0,T]}$ be a complete solution to the Ricci flow (1.1) and u be a smooth positive solution to the nonlinear parabolic Eq. (1.2) with $p<0$ on $Q_{\rho ,T}$. Then we have

$$\begin{aligned}&(\Delta -\partial _t) G_0\nonumber \\&\quad =\frac{2|Hess v|^2}{v^2}-\frac{2}{b}|\nabla \ln v|^2\langle \nabla p,\nabla \ln v\rangle -4\frac{Hessv(\nabla \ln v,\nabla \ln v)}{v}+2|\nabla \ln v|^4\nonumber \\&\qquad -2b(1+k_1p)\langle \nabla q,\nabla \ln v\rangle -\frac{2p}{b}\langle \nabla G_0,\nabla \ln v\rangle -k_1b^2(\Delta -\partial _t)q \end{aligned}$$

(4.2)

on $Q_{\rho ,T}$.

Proof

Note that

$$\begin{aligned} \nabla _iG_0=\frac{2\nabla _i\nabla _jv\nabla _jv}{v^2}-\frac{2|\nabla v|^2\nabla _iv}{v^3}-k_1b^2\nabla _iq, \end{aligned}$$

we have

$$\begin{aligned} \Delta G_0= & {} \frac{2|Hess v|^2}{v^2}+\frac{2\Delta \nabla _jv\nabla _jv}{v^2}-\frac{8Hess v(\nabla v,\nabla v)}{v^3}\nonumber \\&-\frac{2|\nabla v|^2\Delta v}{v^3}+\frac{6|\nabla v|^4}{v^4}-k_1b^2\Delta q\nonumber \\= & {} \frac{2|Hess v|^2}{v^2}+\frac{2\langle \nabla \Delta v,\nabla v\rangle }{v^2}+\frac{2Ric(\nabla v,\nabla v)}{v^2}-\frac{8Hess v(\nabla v,\nabla v)}{v^3}\nonumber \\&-\frac{2|\nabla v|^2\Delta v}{v^3}+\frac{6|\nabla v|^4}{v^4}-k_1b^2\Delta q, \end{aligned}$$

(4.3)

where we used Bochner formula.

By (1.1), we obtain

$$\begin{aligned} \partial _tG_0=\frac{2Ric(\nabla v,\nabla v)}{v^2}+\frac{2\langle \nabla \partial _tv,\nabla v\rangle }{v^2}-\frac{2|\nabla v|^2\partial _tv}{v^3}-k_1b^2\partial _tq. \end{aligned}$$

(4.4)

Combining (4.3) and (4.4), we get

$$\begin{aligned} (\Delta -\partial _t)G_0= & {} \frac{2|Hess v|^2}{v^2}+\frac{2\langle \nabla (\Delta -\partial _t)v,\nabla v\rangle }{v^2}-\frac{8Hess v(\nabla v,\nabla v)}{v^3}\nonumber \\&-\frac{2|\nabla v|^2(\Delta -\partial _t) v}{v^3}+\frac{6|\nabla v|^4}{v^4}-k_1b^2(\Delta -\partial _t)q\nonumber \\= & {} \frac{2|Hess v|^2}{v^2}-\frac{2|\nabla v|^2\langle \nabla p,\nabla v\rangle }{bv^3}+\frac{4(b-p)Hessv(\nabla v,\nabla v)}{bv^3}\nonumber \\&-\frac{2(b-p)|\nabla v|^4}{bv^4}-\frac{2b\langle \nabla q,\nabla v\rangle }{v}-\frac{2bq|\nabla v|^2}{v^2}\nonumber \\&-\frac{8Hess v(\nabla v,\nabla v)}{v^3}\nonumber \\&-\frac{2|\nabla v|^2}{v^3}\bigg (\frac{(b-p)|\nabla v|^2}{bv}-bqv\bigg )+\frac{6|\nabla v|^4}{v^4}-k_1b^2(\Delta -\partial _t)q\nonumber \\= & {} \frac{2|Hess v|^2}{v^2}-\frac{2}{b}|\nabla \ln v|^2\langle \nabla p,\nabla \ln v\rangle \nonumber \\&-4\bigg (1+\frac{p}{b}\bigg )\frac{Hessv(\nabla \ln v,\nabla \ln v)}{v}\nonumber \\&+2\bigg (1+\frac{2p}{b}\bigg )|\nabla \ln v|^4-2b\langle \nabla q,\nabla \ln v\rangle -k_1b^2(\Delta -\partial _t)q\nonumber \\= & {} \frac{2|Hess v|^2}{v^2}-\frac{2}{b}|\nabla \ln v|^2\langle \nabla p,\nabla \ln v\rangle \nonumber \\&-4\frac{Hessv(\nabla \ln v,\nabla \ln v)}{v}+2|\nabla \ln v|^4\nonumber \\&-2b(1+k_1p)\langle \nabla q,\nabla \ln v\rangle -\frac{2p}{b}\langle \nabla G_0,\nabla \ln v\rangle \nonumber \\&-k_1b^2(\Delta -\partial _t)q, \end{aligned}$$

(4.5)

on $Q_{\rho ,T}$, where we used (4.1) in the second equality. $\square$

Similarly, we calculate $(\Delta -\partial _t)G_1$.

Proposition 4.2

Let $(M^n,g(t))_{t\in [0,T]}$ be a complete solution to the Ricci flow (1.1) and u be a smooth positive solution to the nonlinear parabolic Eq. (1.2) with $p<0$ on $Q_{\rho ,T}$. Then we have

$$\begin{aligned} (\Delta -\partial _t) G_1=-\frac{2(b-p)}{b}Ric(\nabla \ln v,\nabla \ln v)-\frac{2p}{b}\langle \nabla G_1,\nabla \ln v\rangle -b\partial _t q-\frac{2\langle Ric,\nabla ^2v\rangle }{v} \end{aligned}$$

(4.6)

on $Q_{\rho ,T}$.

Proof

By (1.1), we obtain

$$\begin{aligned} \Delta G_1= & {} \frac{\Delta \partial _tv}{v}-\frac{2\langle \nabla \partial _tv,\nabla v\rangle }{v^2}-\frac{\Delta v\partial _tv}{v^2}+\frac{2|\nabla v|^2\partial _tv}{v^3}\nonumber \\= & {} \frac{\partial _t\Delta v}{v}-\frac{2\langle Ric,\nabla ^2v\rangle }{v}-\frac{2\langle \nabla \partial _tv,\nabla v\rangle }{v^2}-\frac{\Delta v\partial _tv}{v^2}+\frac{2|\nabla v|^2\partial _tv}{v^3}. \end{aligned}$$

(4.7)

Furthermore, we have

$$\begin{aligned} (\Delta -\partial _t)G_1= & {} \frac{\partial _t(\Delta -\partial _t) v}{v}-\frac{2\langle Ric,\nabla ^2v\rangle }{v}-\frac{2\langle \nabla \partial _tv,\nabla v\rangle }{v^2}\nonumber \\&-\frac{(\Delta -\partial _t) v\partial _tv}{v^2}+\frac{2|\nabla v|^2\partial _tv}{v^3}\nonumber \\= & {} \frac{\partial _t\bigg (\frac{(b-p)|\nabla v|^2}{bv}-bqv\bigg )}{v}-\frac{2\langle Ric,\nabla ^2v\rangle }{v}-\frac{2\langle \nabla \partial _tv,\nabla v\rangle }{v^2}\nonumber \\&-\frac{(\frac{(b-p)|\nabla v|^2}{bv}-bqv\bigg )\partial _tv}{v^2}+\frac{2|\nabla v|^2\partial _tv}{v^3}\nonumber \\= & {} \frac{2(b-p)Ric(\nabla v,\nabla v)}{bv^2}+\frac{2(b-p)\langle \nabla \partial _tv,\nabla v\rangle }{bv^2}\nonumber \\&-\frac{(b-p)|\nabla v|^2\partial _tv}{bv^3}-b\partial _tq-bq\partial _t\ln v+\frac{2\langle Ric,\nabla ^2v\rangle }{v}\nonumber \\&-\frac{2\langle \nabla \partial _tv,\nabla v\rangle }{v^2}-\frac{(b-p)|\nabla v|^2\partial _tv}{bv^3}+bq\partial _t\ln v+\frac{2|\nabla v|^2\partial _tv}{v^3}\nonumber \\= & {} -\frac{2(b-p)}{b}Ric(\nabla \ln v,\nabla \ln v)-\frac{2p}{b}\langle \nabla G_1,\nabla \ln v\rangle \nonumber \\&-b\partial _t q-\frac{2\langle Ric,\nabla ^2v\rangle }{v} \end{aligned}$$

(4.8)

on $Q_{\rho ,T}$, where we used (4.1) in the second equality. $\square$

Then we derive the evolution inequality for G.

Lemma 4.3

Let $(M^n,g(t))_{t\in [0,T]}$ be a complete solution to the Ricci flow (1.1) and u be a smooth positive solution to the nonlinear parabolic Eq. (1.2) with $p<0$ on $Q_{\rho ,T}$. Then for some $0<k_2<1$, we have

$$\begin{aligned} (\Delta -\partial _t) G\ge & {} \frac{2(1-k_2)G^2}{nk_1^2b^2}+\frac{4(1-k_2)}{nk_1b}\bigg (1-\frac{p}{b}-\frac{1}{k_1b}\bigg )|\nabla \ln v|^2G\nonumber \\&+2(1-k_2)\bigg [\frac{1}{n}\bigg (1-\frac{p}{b}-\frac{1}{k_1b}\bigg )^2-\frac{1}{k_2}\bigg ]|\nabla \ln v|^4-\frac{2}{b}|\nabla \ln v|^3|\nabla p|\nonumber \\&-2b(1+k_1p)|\nabla q||\nabla \ln v|-\frac{2p}{b}\langle \nabla G,\nabla \ln v\rangle -\sqrt{n}k_1b^2|Hessq|\nonumber \\&-2(b-p)k_1Ric(\nabla \ln v,\nabla \ln v)-\frac{2k_1b\langle Ric,\nabla ^2v\rangle }{v}. \end{aligned}$$

(4.9)

on $Q_{\rho ,T}$.

Proof

Recall that $G=G_0+k_1bG_1$. It follows from Propositions 4.1 and 4.2 that

$$\begin{aligned}&(\Delta -\partial _t) G\nonumber \\&\quad =\frac{2|Hess v|^2}{v^2}-\frac{2}{b}|\nabla \ln v|^2\langle \nabla p,\nabla \ln v\rangle -4\frac{Hessv(\nabla \ln v,\nabla \ln v)}{v}+2|\nabla \ln v|^4\nonumber \\&\qquad -2b(1+k_1p)\langle \nabla q,\nabla \ln v\rangle -\frac{2p}{b}\langle \nabla G_0,\nabla \ln v\rangle -k_1b^2(\Delta -\partial _t)q\nonumber \\&\qquad -2(b-p)k_1Ric(\nabla \ln v,\nabla \ln v)-2k_1p\langle \nabla G_1,\nabla \ln v\rangle -k_1b^2\partial _t q\nonumber \\&\qquad -\frac{2k_1b\langle Ric,\nabla ^2v\rangle }{v}\nonumber \\&\quad \ge 2(1-k_2)\frac{|Hess v|^2}{v^2}+2\bigg (1-\frac{1}{k_2}\bigg )|\nabla \ln v|^4-\frac{2}{b}|\nabla \ln v|^3|\nabla p|\nonumber \\&\qquad -2b(1+k_1p)|\nabla q||\nabla \ln v|-\frac{2p}{b}\langle \nabla G,\nabla \ln v\rangle -\sqrt{n}k_1b^2|Hessq|\nonumber \\&\qquad -2(b-p)k_1Ric(\nabla \ln v,\nabla \ln v)-\frac{2k_1b\langle Ric,\nabla ^2v\rangle }{v}, \end{aligned}$$

(4.10)

for any $0<k_2<1$, where we used Cauchy’s inequality.

It follows from (4.1) that

$$\begin{aligned} \frac{\Delta v}{v}= & {} \partial _t\ln v+\frac{b-q}{b}|\nabla \ln v|^2-bq\nonumber \\= & {} \frac{G}{k_1b}+\bigg (1-\frac{p}{b}-\frac{1}{k_1b}\bigg )|\nabla \ln v|^2 \end{aligned}$$

(4.11)

Furthermore, using Cauchy’s inequality again and then applying (4.11), we have

$$\begin{aligned}&(1-k_2)\frac{|Hess v|^2}{v^2}+\bigg (1-\frac{1}{k_2}\bigg )|\nabla \ln v|^4\nonumber \\&\qquad \ge \frac{1-k_2}{n}\bigg (\frac{\Delta v}{v}\bigg )^2-\frac{1-k_2}{k_2}|\nabla \ln v|^4\nonumber \\&\qquad =\frac{1-k_2}{n}\bigg [\frac{G}{k_1b}+\bigg (1-\frac{p}{b}-\frac{1}{k_1b}\bigg )|\nabla \ln v|^2\bigg ]^2-\frac{1-k_2}{k_2}|\nabla \ln v|^4\nonumber \\&\qquad =\frac{(1-k_2)G^2}{nk_1^2b^2}+\frac{2b(1-k_2)}{nk_1b}\bigg (1-\frac{p}{b}-\frac{1}{k_1b}\bigg )|\nabla \ln v|^2G\nonumber \\&\qquad \quad +(1-k_2)\bigg [\frac{1}{n}\bigg (1-\frac{p}{b}-\frac{1}{k_1b}\bigg )^2-\frac{1}{k_2}\bigg ]|\nabla \ln v|^4. \end{aligned}$$

(4.12)

(4.9) follows immediately by plugging (4.12) into (4.10). $\square$

In the following, we finish the proof of Theorem 1.3.

Proof of Theorem 1.3

We only consider the case of $G\ge 0$.

By Lemma 4.3, we have

$$\begin{aligned}&\Psi (\Delta -\partial _t)(\Psi G)\ge \frac{2(1-k_2)(\Psi G)^2}{nk_1^2b^2}+\frac{4(1-k_2)}{nk_1b}\bigg (1-\frac{p}{b}-\frac{1}{k_1b}\bigg )|\nabla \ln v|^2\Psi ^2G\nonumber \\&\qquad +2(1-k_2)\bigg [\frac{1}{n}\bigg (1-\frac{p}{b}-\frac{1}{k_1b}\bigg )^2-\frac{1}{k_2}\bigg ]\Psi ^2|\nabla \ln v|^4-\frac{2}{b}\Psi ^2|\nabla \ln v|^3|\nabla p|\nonumber \\&\qquad -2b(1+k_1p)\Psi ^2|\nabla q||\nabla \ln v|-\frac{2p}{b}\Psi \langle \nabla (\Psi G),\nabla \ln v\rangle +\frac{2p}{b}\Psi G\langle \nabla \Psi ,\nabla \ln v\rangle \nonumber \\&\qquad -\sqrt{n}k_1b^2\Psi ^2|Hessq|-2(b-p)k_1\Psi ^2Ric(\nabla \ln v,\nabla \ln v)\nonumber \\&\qquad -\frac{2k_1b\Psi ^2\langle Ric,\nabla ^2v\rangle }{v}\nonumber \\&\qquad +2\langle \nabla \Psi ,\nabla (\Psi G)\rangle +\Psi G\bigg [(\Delta -\partial _t)\Psi -2\frac{|\nabla \Psi |^2}{\Psi }\bigg ]\nonumber \\&\quad \ge \frac{2(1-k_2)(\Psi G)^2}{nk_1^2b^2}+\frac{4(1-k_2)}{nk_1b}\bigg (1+\frac{\sigma _5}{b}-\frac{1}{k_1b}\bigg )|\nabla \ln v|^2\Psi ^2G\nonumber \\&\qquad +2(1-k_2)\bigg [\frac{1}{n}\bigg (1+\frac{\sigma _5}{b}-\frac{1}{k_1b}\bigg )^2-\frac{1}{k_2}\bigg ]\Psi ^2|\nabla \ln v|^4-\frac{2\sigma _3}{b}\Psi ^2|\nabla \ln v|^3\nonumber \\&\qquad -2b\theta _1\Psi |\nabla \ln v|-\frac{2p}{b}\Psi \langle \nabla (\Psi G),\nabla \ln v\rangle -\frac{2\sigma _6}{b}\Psi G|\nabla \Psi ||\nabla \ln v|\nonumber \\&\qquad -\sqrt{n}k_1b^2\theta _2-2(b+\sigma _6)k_1K\Psi |\nabla \ln v|^2-2k_1bK\Psi ^2|\frac{\Delta v}{v}|\nonumber \\&\qquad +2\langle \nabla \Psi ,\nabla (\Psi G)\rangle +\Psi G\bigg [(\Delta -\partial _t)\Psi -2\frac{|\nabla \Psi |^2}{\Psi }\bigg ]. \end{aligned}$$

(4.13)

For fixed $\tau \in (0,T]$, let $(x_2,t_2)$ be a maximum point for $\Psi G$ in $Q_{\rho ,\tau }:=\overline{B({\bar{x}},\rho )}\times [0,\tau ]\subset Q_{\rho ,T}$. It follows from (4.13), (4.11) and (2.9) that at such point:

$$\begin{aligned} 0\ge & {} \frac{2(1-k_2)(\Psi G)^2}{nk_1^2b^2}+\frac{4(1-k_2)}{nk_1b}\bigg (1+\frac{\sigma _5}{b}-\frac{1}{k_1b}\bigg )|\nabla \ln v|^2\Psi ^2G\nonumber \\&+2(1-k_2)\bigg [\frac{1}{n}\bigg (1+\frac{\sigma _5}{b}-\frac{1}{k_1b}\bigg )^2-\frac{1}{k_2}\bigg ]\Psi ^2|\nabla \ln v|^4-\frac{2\sigma _3}{b}\Psi ^2|\nabla \ln v|^3\nonumber \\&-2b\theta _1\Psi |\nabla \ln v|-\frac{2\sigma _6}{b}\Psi G|\nabla \Psi ||\nabla \ln v|-\sqrt{n}k_1b^2\theta _2-2(b+\sigma _6)k_1K\Psi |\nabla \ln v|^2\nonumber \\&-2K\Psi G-2k_1bK\bigg (1+\frac{\sigma _5}{b}-\frac{1}{k_1b}\bigg )\Psi |\nabla \ln v|^2-{\tilde{A}}\Psi G. \end{aligned}$$

(4.14)

Since $k_1b\ge 1$ and $\frac{2nb^2}{\sigma _5}k_2<1$, it is easy to verify that

$$\begin{aligned} \frac{1}{n}\bigg (1+\frac{\sigma _5}{b}-\frac{1}{k_1b}\bigg )^2-\frac{2}{k_2}>0, \end{aligned}$$

and

$$\begin{aligned} \frac{1-k_2}{nk_1b}\big (1+\frac{\sigma _5}{b}-\frac{1}{k_1b}\big )>0, \end{aligned}$$

(4.15)

By Cauchy’s inequality and the fact of $0\le \Psi \le 1$, we have

$$\begin{aligned} -\frac{2\sigma _3}{b}\Psi ^2|\nabla \ln v|^3\ge & {} -\frac{1-k_2}{n}\bigg (1+\frac{\sigma _5}{b}-\frac{1}{k_1b}\bigg )^2\Psi ^2|\nabla \ln v|^4\nonumber \\&-\frac{n\sigma _3^2}{b^2(1-k_2)}\bigg (1+\frac{\sigma _5}{b}-\frac{1}{k_1b}\bigg )^{-2}\Psi |\nabla \ln v|^2, \end{aligned}$$

(4.16)

and

$$\begin{aligned} -2b\theta _1\Psi |\nabla \ln v|\ge -\Psi |\nabla \ln v|^2-b^2\theta _1^2. \end{aligned}$$

(4.17)

Using Cauchy’s inequality and Lemma 2.1, we obtain

$$\begin{aligned}&-\frac{2\sigma _6}{b}\Psi G|\nabla \Psi ||\nabla \ln v|\nonumber \\&\quad \ge -\frac{(1-k_2)(\Psi G)^2}{nk_1^2b^2}-\frac{nk_1^2\sigma _6^2}{1-k_2}\frac{|\nabla \Psi |^2}{\Psi }\Psi |\nabla \ln v|^2\nonumber \\&\quad \ge -\frac{(1-k_2)(\Psi G)^2}{nk_1^2b^2}-\frac{nk_1^2\sigma _6^2C_{1/2}^2}{(1-k_2)\rho ^2}\Psi |\nabla \ln v|^2. \end{aligned}$$

(4.18)

Plugging (4.15) to (4.18) into (4.14), we get

$$\begin{aligned} 0\ge & {} \frac{(1-k_2)(\Psi G)^2}{nk_1^2b^2}-(2K+{\tilde{A}})\Psi G-b^2\theta _1^2-\sqrt{n}k_1b^2\theta _2\nonumber \\&+(1-k_2)\bigg [\frac{1}{n}\bigg (1+\frac{\sigma _5}{b}-\frac{2}{k_1b}\bigg )^2-\frac{1}{k_2}\bigg ]\Psi ^2|\nabla \ln v|^4\nonumber \\&-\bigg [\frac{n\sigma _3^2}{b^2(1-k_2)}\bigg (1+\frac{\sigma _5}{b}-\frac{1}{k_1b}\bigg )^{-2}+1+\frac{nk_1^2\sigma _6^2C_{1/2}^2}{(1-k_2)\rho ^2}\nonumber \\&+2(b+\sigma _6)k_1K+2k_1bK\bigg (1+\frac{\sigma _5}{b}-\frac{1}{k_1b}\bigg )\bigg ]\Psi |\nabla \ln v|^2\nonumber \\\ge & {} \frac{(1-k_2)(\Psi G)^2}{nk_1^2b^2}-(2K+{\tilde{A}})\Psi G-b^2\theta _1^2-\sqrt{n}k_1b^2\theta _2\nonumber \\&-\frac{1}{1-k_2}\bigg [\frac{1}{n}\bigg (1+\frac{\sigma _5}{b}-\frac{2}{k_1b}\bigg )^2-\frac{1}{k_2}\bigg ]^{-1}\nonumber \\&\cdot \bigg [1+\frac{n\sigma _3^2}{b^2(1-k_2)}\bigg (1+\frac{\sigma _5}{b}-\frac{1}{k_1b}\bigg )^{-2}+\frac{nk_1^2\sigma _6^2C_{1/2}^2}{(1-k_2)\rho ^2}\nonumber \\&+2(b+\sigma _6)k_1K+2k_1bK\bigg (1+\frac{\sigma _5}{b}-\frac{1}{k_1b}\bigg )\bigg ]^2, \end{aligned}$$

(4.19)

at $(x_2,t_2)$, i.e.,

$$\begin{aligned} (\Psi G)^2(x_2,t_2)\le & {} \frac{nk_1^2b^2}{1-k_2}(2K+{\tilde{A}})\Psi G+\frac{nk_1^2b^4\theta _1^2}{1-k_2}+\frac{n^\frac{3}{2}k_1^3b^4\theta _2}{1-k_2}\nonumber \\&+\frac{nk_1^2b^2}{(1-k_2)^2}\bigg [\frac{1}{n}\bigg (1+\frac{\sigma _5}{b}-\frac{2}{k_1b}\bigg )^2-\frac{1}{k_2}\bigg ]^{-1}\nonumber \\&\cdot \bigg [1+\frac{n\sigma _3^2}{b^2(1-k_2)}\bigg (1+\frac{\sigma _5}{b}-\frac{1}{k_1b}\bigg )^{-2}+\frac{nk_1^2\sigma _6^2C_{1/2}^2}{(1-k_2)\rho ^2}\nonumber \\&+2(b+\sigma _6)k_1K+2k_1bK\bigg (1+\frac{\sigma _5}{b}-\frac{1}{k_1b}\bigg )\bigg ]^2. \end{aligned}$$

(4.20)

Recall an elementary fact: if $x^2\le a_1x+a_2$ for some $a_1,a_2,x\ge 0$, then

$$\begin{aligned} x\le \frac{a_1}{2}+\sqrt{a_2+\bigg (\frac{a_2}{2}\bigg )^2}\le \frac{a_1}{2}+\sqrt{a_2}+\bigg (\frac{a_2}{2}\bigg )^2=a_1+\sqrt{a_2}. \end{aligned}$$

Then we get an upper bound for $(\Psi G)(x_2,t_3)$ that

$$\begin{aligned} (\Psi G)(x_2,t_2)\le & {} \frac{nk_1^2b^2}{1-k_2}(2K+{\tilde{A}})+\frac{k_1b^2}{\sqrt{1-k_2}}(\sqrt{n}\theta _1+n^\frac{3}{4}\sqrt{\theta _2k_1})\nonumber \\&+\frac{\sqrt{n}k_1b}{1-k_2}\bigg [\frac{1}{n}\bigg (1+\frac{\sigma _5}{b}-\frac{2}{k_1b}\bigg )^2-\frac{1}{k_2}\bigg ]^{-\frac{1}{2}}\nonumber \\&\cdot \bigg [1+\frac{n\sigma _3^2}{b^2(1-k_2)}\bigg (1+\frac{\sigma _5}{b}-\frac{1}{k_1b}\bigg )^{-2}+\frac{nk_1^2\sigma _6^2C_{1/2}^2}{(1-k_2)\rho ^2}\nonumber \\&+2(b+\sigma _6)k_1K+2k_1bK\bigg (1+\frac{\sigma _5}{b}-\frac{1}{k_1b}\bigg )\bigg ]. \end{aligned}$$

(4.21)

By the construction of $\Psi$, we have

$$\begin{aligned} \sup _{Q_{\frac{\rho }{2},\tau }}G\le \sup _{Q_{\rho ,\tau }}(\Psi G)=(\Psi G)(x_2,t_2) \end{aligned}$$

for all $t\in [0,\tau ]$ with $\tau \le T$ is arbitrary. Therefore, we conclude that

$$\begin{aligned} G= & {} |\nabla \ln v|^2-k_1b^2q+k_1b\partial _t\ln v\nonumber \\= & {} \frac{b^2|\nabla u|^2}{u^2}-k_1b^2q-\frac{k_1b^2\partial _tu}{u}\nonumber \\\le & {} \frac{nk_1^2b^2}{1-k_2}B+\frac{k_1b^2}{\sqrt{1-k_2}}(\sqrt{n}\theta _1+n^\frac{3}{4}\sqrt{\theta _2k_1})\nonumber \\&+\frac{\sqrt{n}k_1b}{1-k_2}\bigg [\frac{1}{n}\bigg (1+\frac{\sigma _5}{b}-\frac{2}{k_1b}\bigg )^2-\frac{1}{k_2}\bigg ]^{-\frac{1}{2}}\nonumber \\&\cdot \bigg [1+\frac{n\sigma _3^2}{b^2(1-k_2)}\bigg (1+\frac{\sigma _5}{b}-\frac{1}{k_1b}\bigg )^{-2}+\frac{nk_1^2\sigma _6^2C_{1/2}^2}{(1-k_2)\rho ^2}\nonumber \\&+2(b+\sigma _6)k_1K+2k_1bK\bigg (1+\frac{\sigma _5}{b}-\frac{1}{k_1b}\bigg )\bigg ] \end{aligned}$$

(4.22)

on $Q_{\frac{\rho }{2},T}$, where

$$\begin{aligned} B:=\frac{C_{1/2}}{\rho }(n-1)\left( \sqrt{K}+\frac{2}{\rho }\right) +\frac{C_{1/2}+2C_{1/2}^2}{\rho ^2}+\frac{{\bar{C}}}{t}+(C_{1/2}+2)K. \end{aligned}$$

$\square$

Remark 4.4

As we proved Theorem 1.1, it is not hard to derive Corollary 1.4 by similar arguments as in the proof of Corollary 1.2.