Li–Yau-Type Gradient Estimate along Geometric Flow

Abstract

In this article we derive a Li–Yau-type gradient estimate for a generalized weighted parabolic heat equation with potential on a weighted Riemannian manifold evolving by a geometric flow. As an application, a Harnack-type inequality is also derived in the end.

1. Introduction

The gradient estimation for both elliptic and parabolic equations plays a significant role in geometric analysis. Harnack estimation is also one of the powerful tools in heat kernel analysis. The local and global behavior of positive solutions of nonlinear elliptic equations on ${\mathbb{R}}^n(n>2)$ near an isolated singularity were studied by Gidas and Spruck [1]. In [2], Hamilton proved a Harnack estimate on the Riemannian manifold for Ricci flow with a weakly positive curvature operator, which was later used in solving the Poincaré conjecture. Li and Yau [3] established parabolic gradient estimates on solutions to the linear heat equation

$$\begin{array}{ccc}\hfill (\Delta -{\partial }_t)u& =& q(x,t)u\hfill \end{array}$$

(1)

on Riemannian manifold having Ricci curvature bounded from below, where $q(x,t)$ is $C^2$ in first variable x and $C^1$ in second variable t, where $C^2$ and $C^1$ denote the space of all twice differentiable and one-time differentiable functions, respectively. After a remarkable work by Perelman [4,5,6] in Ricci flow, this topic gained massive importance. Thus, this topic becomes one of the important tools in geometric analysis and modern PDE theory. In [7], Jiyau Li considered the heat-type equation

$$\begin{array}{ccc}\hfill (\Delta -{\partial }_t)u(x,t)+h(x,t)u^{\alpha }(x,t)& =& 0\hfill \end{array}$$

(2)

on $M\times [0,\infty )$, where $h(x,t)$, is a function on $M\times [0,\infty )$, which is $C^2$ in the first variable and $C^1$ in the second variable, $\alpha \in \mathbb{R}$ and derived the gradient estimates and Harnack inequalities for a positive solution to the above nonlinear parabolic equation. This equation represents a simple ecological model for population dynamics, where $u(x,t)$ is the population density at time t.

Wu [8] studied gradient estimates for the nonlinear parabolic equation

$$\begin{array}{ccc}\hfill ({\Delta }_{\varphi }-{\partial }_t)u+\mu (x,t)u+p(x,t)u^{\alpha }+q(x,t)u^{\beta }& =& 0,\hfill \end{array}$$

(3)

where ${\Delta }_{\varphi }$ is the weighted Laplacian, $p(x,t),q(x,t)$ are $C^2$ in x and $C^1$ in t. Abolarinwa et al. [9,10,11,12] studied gradient and Harnack estimates for various nonlinear parabolic equations. In [13], Dung et al. studied various gradient estimations for solutions of the following f-heat type equations

$$\begin{array}{ccc}\hfill u_t& =& {\Delta }_{f}u+aulogu+bu+C{u}^p+D{u}^{-q}\hfill \end{array}$$

(4)

$$\begin{array}{ccc}\hfill \mathrm{and}{u}_t& =& {\Delta }_{f}u+C{e}^{pu}+D{e}^{-pu}+E,\hfill \end{array}$$

(5)

where $a,b\in \mathbb{R}$ and $C,D,E$ are smooth functions, on a complete smooth metric measure space $(M,g,e^{-f}d\nu )$ with Bakry–Émery Ricci curvature bounded from below. In [14], Azami studied gradient estimates for a weighted parabolic equation

$$\begin{array}{ccc}\hfill ({\Delta }_{\varphi }-{\partial }_t)u(x,t)& =& q(x,t)u^{a+1}(x,t)+p(x,t)A\left(u(x,t)\right)\hfill \end{array}$$

(6)

evolving under the geometric flow, where $p(x,t),q(x,t),A\left(u\right(x,t\left)\right)$ are $C^2$ in x and $C^1$ in t. Thereafter many authors studied the geometric aspect of analysis on the Riemannian manifold, see [15,16,17,18,19,20,21,22,23] and the references therein. Recently, Hui et al. studied Hamilton-Souplet-Zhang type gradient estimation for nonlinear weighted parabolic equation in [24], the same estimation for a system of equations in [25] and Saha et al. [26] studied first eigenvalue of weighted p-Laplacian along the Cotton flow.

Motivated by the above works in this paper we consider a generalized non-linear parabolic equation with potential by

$${\Delta }_{\varphi }u=\frac{\partial u}{\partial t}+A\left(u\right)p(x,t)+B\left(u\right)q(x,t)+\xi (x,t)u(x,t),$$

(7)

where $p(x,t),q(x,t)$ and $\xi (x,t)$ are $C^2$ functions of $x,t$. We derive a Li–Yau-type gradient estimate for a positive solution of (7) on a weighted Riemannian manifold which evolves under an abstract geometric flow.

In particular, if we consider $A\left(u\right)=u^{\alpha },B\left(u\right)=u^{\beta },\xi =\mu (x,t)$ then (7) reduces to (3), which was studied by Wu [8]. If we take $A\left(u\right)=ulogu,B\left(u\right)=u,\xi =C{u}^p+D{u}^q$ then (7) reduces to (4) and if $A\left(u\right)=C{e}^{pu},B\left(u\right)=D{e}^{-pu},\xi =\frac{E}{u}$ then (7) reduces to (5), both of which were studied by Dung et al. [13]. The generalized Lichnerowicz type equation studied by Dung [13] comes from our Equation (7) by considering $A\left(u\right)=u^{\alpha }logu,B\left(u\right)=u^{\beta }$ and $p,q,\xi$ are suitable constants. Finally for $B\left(u\right)=u^{a+1}$ and $\xi =0$ we have (6), which was studied by Azami [14]. Thus, our Equation (7) generalizes all the cases.

2. Preliminaries

Let us consider an n-dimensional closed weighted Riemannian manifold $(M^n,g,e^{-\varphi }d\mu )$, where $e^{-\varphi }d\mu$ is the weighted volume measure, g is Riemannian metric and $\varphi \in C^2\left(M\right)$. Choose $\{e_1,e_2,\cdots ,e_n\}$ as an orthonormal frame on M. Let $g\left(t\right)$ be a one-parameter family of Riemannian metrics evolving along the following abstract geometric flow

$$\frac{\partial }{\partial t}{g}_{ij}\left(t\right)=2S_{ij}\left(t\right),$$

(8)

where $S_{ij}\left(t\right):=S(e_i,e_j)\left(t\right)$ is smooth symmetric (0, 2)-type tensor on $(M,g(t\left)\right)$. Let us define one parameter family of functions $S\left(t\right)=trace\left(S\right)\left(t\right)=g^{ij}\left(t\right)S_{ij}\left(t\right)$ on M. The weighted Laplacian operator is defined by

$${\Delta }_{\varphi }=\Delta -\nabla \varphi \nabla ,$$

where $\Delta$ is the Laplace operator and ∇ is the gradient operator. Let $u=e^f$ be a positive solution of (7), then Equation (7) transforms to

$${\Delta }_{\varphi }f={\partial }_{t}f-{|\nabla f|}^2+\widehat{A}\left(f\right)p+\widehat{B}\left(f\right)q+\xi ,$$

(9)

where $\widehat{A}\left(f\right)=\frac{A\left(u\right)}{u},\widehat{B}\left(f\right)=\frac{B\left(u\right)}{u}$. We define

$$\begin{array}{c}\hfill {\widehat{A}}_f=A^{\prime }\left(u\right)-\frac{A\left(u\right)}{u},{\widehat{A}}_{ff}=u{A}^{″}\left(u\right)-A^{\prime }\left(u\right)+\frac{A\left(u\right)}{u}.\end{array}$$

(10)

Example 1.

Let $u=e^f$ and $A\left(u\right)={\left|u\right|}^{\alpha -1}u$. Therefore $\widehat{A}\left(f\right)=\frac{A\left(u\right)}{u}=e^{(\alpha -1)f}$, which gives

1. 

${\widehat{A}}_f=(\alpha -1)e^{(\alpha -1)f}$

2. 

${\widehat{A}}_{ff}={(\alpha -1)}^2{e}^{(\alpha -1)f}$

3. 

$\nabla \widehat{A}=(\alpha -1)e^{(\alpha -1)f}\nabla f={\widehat{A}}_f\nabla f$

4. 

$\Delta \widehat{A}={(\alpha -1)}^2{e}^{(\alpha -1)f}{|\nabla f|}^2+(\alpha -1)e^{(\alpha -1)f}\Delta f={\widehat{A}}_{ff}{|\nabla f|}^2+{\widehat{A}}_f\Delta f$.

Let $\overline{f}=\widehat{A}p+\widehat{B}q+\xi$ so that Equation (9) reduces to

$${\Delta }_{\varphi }f=-{|\nabla f|}^2+f_t+\overline{f}.$$

(11)

Definition 1.

([27] Bakry–Émery Ricci tensor). For any integer $m>n$, an $(m-n)-$ Bakry–Émery tensor is defined by

$$Ri{c}_{\varphi }^{m-n}:=Ric+Hess\varphi -\frac{\nabla \varphi \otimes \nabla \varphi }{m-n},$$

where Hess is the Hessian operator. The case when $m=n$ occurs if and only if ϕ is a constant function. Furthermore, when $m\to \infty$ the $\infty -$ Bakry–Émery Ricci tensor is defined by

$$Ri{c}_{\varphi }:=Ric+Hess\varphi .$$

Lemma 1

([14] Weighted Bochner Formula). For any smooth function u on a weighted Riemannian manifold $(M,g,e^{-\varphi }d\mu )$, we have the weighted version of Bochner formula

$$\frac{1}{2}{\Delta }_{\varphi }{|\nabla u|}^2={\left|Hessu\right|}^2+\langle \nabla {\Delta }_{\varphi }u,\nabla u\rangle +Ri{c}_{\varphi }(\nabla u,\nabla u),$$

where $\langle ·,·\rangle$ is the induced inner product by the Riemannian metric g.

Lemma 2

([14]). Under the geometric flow Equation (8) and for any smooth function u on a weighted Riemannian manifold $(M,g,e^{-\varphi }d\mu )$ we have the following evolution formulas

1. 

$\frac{\partial }{\partial t}{|\nabla u|}^2=-2S(\nabla u,\nabla u)+2\langle \nabla u,\nabla u_t\rangle$,

2. 

$\frac{\partial }{\partial t}\left({\Delta }_{\varphi }u\right)={\Delta }_{\varphi }{u}_t-2S^{ij}{\nabla }_i{\nabla }_{j}u-\langle 2divS-\nabla S,\nabla u\rangle +2S(\nabla \varphi ,\nabla u)-\langle \nabla u,\nabla {\varphi }_t\rangle$, where $divS$ denotes the divergence of $S$ and $S^{ij}=g^{ik}{g}^{jl}{S}_{kl}$.

Let $T>0$ be any real number. For any two points $x,y\in M$ and for any $t\in [0,T]$, the quantity $d(x,y,t)$ denotes the geodesic distance between x and y under the metric $g\left(t\right)$. For any fixed $x_0\in M$ and $R>0$ we define a compact set

$$\begin{array}{c}\hfill Q_{2R,T}=\{(x,t):d(x,x_0,t)\le 2R,0\le t\le T\}\subset M^n\times (-\infty ,+\infty ).\end{array}$$

(12)

Now for $u>0$ we define some non-negative real numbers

${\lambda }_1:=\underset{Q_{2R,T}}{\sup }\left|\widehat{A}\right|$	${\lambda }_2:=\underset{Q_{2R,T}}{\sup }\left|{\widehat{A}}_f\right|$	${\lambda }_3:=\underset{Q_{2R,T}}{\sup }\left|{\widehat{A}}_{ff}\right|$
${\mathsf{\Lambda }}_1:=\underset{M\times [0,T]}{\sup }\left|\widehat{A}\right|$	${\mathsf{\Lambda }}_2:=\underset{M\times [0,T]}{\sup }\left|{\widehat{A}}_f\right|$	${\mathsf{\Lambda }}_3:=\underset{M\times [0,T]}{\sup }\left|{\widehat{A}}_{ff}\right|$
$b_1:=\underset{Q_{2R,T}}{\sup }\left|\widehat{B}\right|$	$b_2:=\underset{Q_{2R,T}}{\sup }\left|{\widehat{B}}_f\right|$	$b_3:=\underset{Q_{2R,T}}{\sup }\left|{\widehat{B}}_{ff}\right|$
$B_1:=\underset{M\times [0,T]}{\sup }\left|\widehat{B}\right|$	$B_2:=\underset{M\times [0,T]}{\sup }\left|{\widehat{B}}_f\right|$	$B_3:=\underset{M\times [0,T]}{\sup }\left|{\widehat{B}}_{ff}\right|$
${\sigma }_1:=\underset{Q_{2R,T}}{\sup }\left|q\right|$	${\sigma }_2:=\underset{Q_{2R,T}}{\sup }|\nabla q|$	${\sigma }_3:=\underset{Q_{2R,T}}{\sup }\left|{\Delta }_{\varphi }q\right|$
${\mathsf{\Sigma }}_1:=\underset{M\times [0,T]}{\sup }\left|q\right|$	${\mathsf{\Sigma }}_2:=\underset{M\times [0,T]}{\sup }|\nabla q|$	${\mathsf{\Sigma }}_3:=\underset{M\times [0,T]}{\sup }\left|{\Delta }_{\varphi }q\right|$
${\gamma }_1:=\underset{Q_{2R,T}}{\sup }\left|p\right|$	${\gamma }_2:=\underset{Q_{2R,T}}{\sup }|\nabla p|$	${\gamma }_3:=\underset{Q_{2R,T}}{\sup }\left|{\Delta }_{\varphi }p\right|$
${\mathsf{\Gamma }}_1:=\underset{M\times [0,T]}{\sup }\left|p\right|$	${\mathsf{\Gamma }}_2:=\underset{M\times [0,T]}{\sup }|\nabla p|$	${\mathsf{\Gamma }}_3:=\underset{M\times [0,T]}{\sup }\left|{\Delta }_{\varphi }p\right|$
${\theta }_1:=\underset{Q_{2R,T}}{\sup }|\nabla \varphi |$	${\theta }_2:=\underset{Q_{2R,T}}{\sup }|\nabla {\varphi }_t|$	${\mathsf{\Theta }}_1:=\underset{M\times [0,T]}{\sup }|\nabla \varphi |$
${\mathsf{\Theta }}_2:=\underset{M\times [0,T]}{\sup }|\nabla {\varphi }_t|$	$m_1:=\underset{Q_{2R,T}}{\sup }|\nabla \xi |$	$m_2:=\underset{Q_{2R,T}}{\sup }\left|{\Delta }_{\varphi }\xi \right|$
$m_3:=\underset{Q_{2R,T}}{\sup }\left|\xi \right|$	$M_1:=\underset{M\times [0,T]}{\sup }|\nabla \xi |$	$M_2:=\underset{M\times [0,T]}{\sup }\left|{\Delta }_{\varphi }\xi \right|$
$M_3:=\underset{M\times [0,T]}{\sup }\left|\xi \right|$

Lemma 3

([14]). For any smooth function f on an n-dimensional Riemannian manifold $(M^n,g,e^{-\varphi }$$d\mu )$ and $m>n$ we have the following relation connecting Hessian and weighted Laplacian

$${\left|Hessf\right|}^2\ge \frac{{\left({\Delta }_{\varphi }f\right)}^2}{m}-\frac{1}{m-n}{\langle \nabla f,\nabla \varphi \rangle }^2.$$

(13)

Proof. 

Let $m>n$. Then we see that

$$\begin{array}{ccc}\hfill 0& \le & {\left(\sqrt{\frac{m-n}{mn}}\Delta f+\sqrt{\frac{n}{m(m-n)}}\langle \nabla f,\nabla \varphi \rangle \right)}^2\hfill \\ & =& (\frac{1}{n}-\frac{1}{m}){(\Delta f)}^2+\frac{2}{m}\Delta f\langle \nabla f,\nabla \varphi \rangle +(\frac{1}{m-n}-\frac{1}{m}){\langle \nabla f,\nabla \varphi \rangle }^2\hfill \\ & \le & {\left|Hessf\right|}^2-\frac{1}{m}\left({(\Delta f)}^2-2\Delta f\langle \nabla f,\nabla \varphi \rangle +{\langle \nabla f,\nabla \varphi \rangle }^2\right)+\frac{1}{m-n}{\langle \nabla f,\nabla \varphi \rangle }^2\hfill \\ & =& {\left|Hessf\right|}^2-\frac{{\left({\Delta }_{\varphi }f\right)}^2}{m}+\frac{1}{m-n}{\langle \nabla f,\nabla \varphi \rangle }^2.\hfill \end{array}$$

Thus ${\left|\mathrm{Hess}f\right|}^2\ge \frac{{\left({\Delta }_{\varphi }f\right)}^2}{m}-\frac{1}{m-n}{\langle \nabla f,\nabla \varphi \rangle }^2$. □

Lemma 4

([28] Young’s inequality). If $a,b$ are nonnegative real numbers and $p>1,q>1$ are real numbers such that $\frac{1}{p}+\frac{1}{q}=1$ then

$$ab\le \frac{a^p}{p}+\frac{b^q}{q}.$$

Let $\alpha >0$ be any real number. Put $a=\alpha a$ and $b=\frac{b}{\alpha }$ in the above expression we get Peter-Paul type inequality

$$\begin{array}{c}\hfill ab\le {\alpha }^p\frac{a^p}{p}+\frac{b^q}{{\alpha }^{q}q}.\end{array}$$

(14)

If we put $a=a\sqrt{2\alpha }$, $b=\frac{b}{\sqrt{2\alpha }}$, $p=q=2$ in Young’s inequality then we have the well known Peter-Paul inequality

$$\begin{array}{ccc}\hfill ab& \le & \alpha a^2+\frac{b^2}{4\alpha }.\hfill \end{array}$$

(15)

In this paper we use these inequalities with a suitable choice of $\alpha$.

3. Li-iYau-Type Gradient Estimation

In this section, we are going to derive a bound for the quantity $\frac{{|\nabla u|}^2}{u^2}$ on a compact domain $Q_{2R,T}$ of M, where u satisfies (7). This estimation is known as local Li–Yau-type estimation. After that, we derive global Li–Yau-type estimation on the whole of M. This method enables us to find the heat ratio between two points on a manifold by deriving a Harnack-type inequality. For this, we fix a point $x_0\in M$ and let $R>0$ be a real number. Let u be a positive solution to (7) in $Q_{2R,T}$.

Theorem 1.

If $k_1,k_2,k_3,k_4$ are positive constants such that

$$Ri{c}_{\varphi }^{m-n}\ge -(m-1)k_{1}g,-k_{2}g\le S\le k_{3}g,|\nabla S|\le k_4$$

on $Q_{2R,T}$, then for any solution u of (7), any $\lambda >1$ and $\delta \in (0,1)$ we have

$$\begin{array}{c}\hfill \frac{{|\nabla u|}^2}{u^2}-\lambda \left(\frac{u_t}{u}+\frac{A\left(u\right)}{u}p+\frac{B\left(u\right)}{u}q+\xi \right)\le \frac{m{\lambda }^2}{2t(1-\lambda ϵ)}+\frac{m{\lambda }^2}{2(1-\lambda ϵ)}\widetilde{D_1}+\widetilde{E_1},\end{array}$$

(16)

where

$$\begin{array}{ccc}\hfill \widetilde{D_1}& =& \frac{c_0}{R}(m-1)(\sqrt{k_1}+\frac{2}{R})+\frac{3c_1}{R^2}+c_2{k}_2+\frac{m{\lambda }^2{c}_1}{4(1-\lambda ϵ)(\lambda -1)R^2}+\frac{1}{\lambda },\hfill \\ \hfill \widetilde{E_1}& =& {\left(\frac{m{\lambda }^2}{2(1-\lambda ϵ)}{E}_1\right)}^{\frac{1}{2}},\hfill \\ \hfill E_1& =& \frac{m{\lambda }^2}{4(1-\lambda ϵ)(1-\delta ){(\lambda -1)}^2}{\overline{C_1}}^2+2\lambda k_2ϵ{\theta }_1^2+\frac{n\lambda }{2ϵ}{(k_2+k_3)}^2\hfill \\ & +& \frac{9}{8}n{\lambda }^2{k}_4+({\lambda }_1{\gamma }_3+b_1{\sigma }_3)+\frac{3}{4}{\left(\frac{2m{\lambda }^2}{(1-\lambda ϵ)(\lambda -1)\delta }\right)}^{\frac{1}{3}}(2{\lambda }_2{\gamma }_2\hfill \\ & +& 2b_2{\sigma }_2{)}^{\frac{4}{3}}+m_2+\frac{3}{4}{\left(\frac{m{\lambda }^2}{2(1-\lambda ϵ)(1-\delta ){(\lambda -1)}^2}\right)}^{\frac{1}{3}}{\lambda }^{\frac{4}{3}}{\theta }_2^{\frac{4}{3}}\hfill \\ & +& \frac{3}{4}{\left(\frac{m{\lambda }^2}{2(1-\lambda ϵ)(1-\delta ){(\lambda -1)}^2}\right)}^{\frac{1}{3}}{({\lambda }_1{\gamma }_2+b_1{\sigma }_2+m_1)}^{\frac{4}{3}},\hfill \\ \hfill \overline{C_1}& =& \frac{\lambda k_2}{2ϵ}+2k_4+{\lambda }_3{\gamma }_1+b_3{\sigma }_1+2(\lambda -1)k_3+\frac{\lambda -1}{\lambda }+{\gamma }_1{\lambda }_2+{\sigma }_1{b}_2\hfill \\ & & +2(1-\lambda ϵ)(m-1)k_1.\hfill \end{array}$$

To prove the theorem we need the following lemma.

Lemma 5.

If $u=e^f$ is a positive solution to (7) and $F:=t\left(\right|\nabla f{|}^2-\lambda (f_t+\overline{f}))$, where $\overline{f}=\widehat{A}p+\widehat{B}q+\xi$ then for any $ϵ\in (0,\frac{1}{\lambda })$ and assuming conditions of Theorem 1 we have

$$\begin{array}{cc}\hfill \left({\Delta }_{\varphi }-{\partial }_t\right)F& \ge 2t(1-\lambda ϵ)\frac{{\left({\Delta }_{\varphi }f\right)}^2}{m}-\frac{\lambda t{k}_2}{2ϵ}{|\nabla f|}^2-2\lambda t{k}_2ϵ{|\nabla \varphi |}^2\hfill \\ \hfill & -2t(1-\lambda ϵ)(m-1)k_1{|\nabla f|}^2-2\nabla F\nabla f-\frac{F}{t}-2t(\lambda -1)k_3{|\nabla f|}^2\hfill \\ \hfill & -\frac{n\lambda t}{2ϵ}{\left(k_2+k_3\right)}^2-3\lambda t\sqrt{n}{k}_4{|\nabla f|}^2+\mathcal{H},\hfill \end{array}$$

(17)

where $\mathcal{H}=-2t(\lambda -1)\nabla \overline{f}\nabla f-\lambda t\nabla f\nabla {\varphi }_t-\lambda t{\Delta }_{\varphi }\overline{f}$.

Proof. 

Let u be a solution of (7) and consider $F:=t\left(\right|\nabla f{|}^2-\lambda (f_t+\overline{f}))$, where $\overline{f}=\widehat{A}p+\widehat{B}q+\xi$. Hence

$$\begin{array}{ccc}\hfill \frac{F}{t}& =& {|\nabla f|}^2-\lambda (f_t+\overline{f})\hfill \end{array}$$

(18)

and applying Lemma 1 (Weighted Bochner formula) we have

$$\begin{array}{cc}\hfill {\Delta }_{\varphi }F& ={2t|Hessf|}^2+2t〈\nabla {\Delta }_{\varphi }f,\nabla f〉+2t{\mathit{Ric}}_{\varphi }(\nabla f,\nabla f)-\lambda t{\Delta }_{\varphi }{f}_t\hfill \\ \hfill & -\lambda t{\Delta }_{\varphi }\overline{f}.\hfill \end{array}$$

(19)

Now ${\Delta }_{\varphi }f=-\frac{F}{t}-(\lambda -1)(f_t+\overline{f})$, so $\nabla {\Delta }_{\varphi }f=-\frac{\nabla F}{t}-(\lambda -1)(\nabla f_t+\nabla \overline{f})$.

Hence

$$\begin{array}{cc}\hfill {\Delta }_{\varphi }F& =2t\mid Hess{\left.f\right|}^2-2\nabla F\nabla f-2t(\lambda -1)\left(\nabla f_t+\nabla \overline{f}\right)\nabla f+2t{\mathit{Ric}}_{\varphi }(\nabla f,\nabla f)\hfill \\ \hfill & -\lambda t{\Delta }_{\varphi }{f}_t-\lambda t{\Delta }_{\varphi }\overline{f}\hfill \end{array}$$

(20)

Furthermore,

$${\partial }_t\left({\Delta }_{\varphi }f\right)=\frac{F}{t^2}-\frac{F_t}{t}-(\lambda -1)(f_{tt}+{\overline{f}}_t).$$

(21)

Using (21) on (20) we get

$$\begin{array}{cc}\hfill {\Delta }_{\varphi }F& ={2t|Hessf|}^2-2\nabla F\nabla f-2t(\lambda -1)\left(\nabla f_t+\nabla \overline{f}\right)\nabla f+2t{\mathit{Ric}}_{\varphi }(\nabla f,\nabla f)\hfill \\ \hfill & -\frac{\lambda F}{t}+\lambda F_t+\lambda (\lambda -1)t\left(f_{tt}+{\overline{f}}_t\right)-2\lambda t\langle S,Hessf\rangle -2\lambda t\langle divS-\frac{1}{2}\nabla S,\nabla f\rangle \hfill \\ \hfill & +2\lambda tS(\nabla \varphi ,\nabla f)-\lambda t〈\nabla f,\nabla {\varphi }_t〉-\lambda t{\Delta }_{\varphi }\overline{f},\hfill \end{array}$$

(22)

and

$${\partial }_{t}F=\frac{F}{t}+t({\partial }_t{|\nabla f|}^2-\lambda (f_{tt}+{\overline{f}}_t)).$$

(23)

From (22) and (23) we get

$$\begin{array}{cc}\hfill \left({\Delta }_{\varphi }-{\partial }_t\right)F& ={2t|Hessf|}^2-2\nabla F\nabla f-2t(\lambda -1)\left(\nabla f_t+\nabla \overline{f}\right)\nabla f\hfill \\ \hfill & +2t{\mathit{Ric}}_{\varphi }(\nabla f,\nabla f)-2t(\lambda -1)S(\nabla f,\nabla f)\hfill \\ \hfill & +2t(\lambda -1)\nabla f\nabla f_t-2\lambda t\langle S,Hessf\rangle \hfill \\ \hfill & -2\lambda t\langle divS-\frac{1}{2}\nabla S,\nabla f\rangle +2\lambda tS(\nabla \varphi ,\nabla f)-\lambda t\nabla f\nabla {\varphi }_t\hfill \\ \hfill & -\lambda t{\Delta }_{\varphi }\overline{f}-\frac{F}{t}.\hfill \end{array}$$

(24)

$$\begin{array}{cc}\hfill \mathrm{or},\left({\Delta }_{\varphi }-{\partial }_t\right)F& ={2t|Hessf|}^2+2t{\mathit{Ric}}_{\varphi }(\nabla f,\nabla f)-2\nabla f\nabla f-\frac{F}{t}\hfill \\ \hfill & -2t(\lambda -1)S(\nabla f,\nabla f)+2\lambda tS(\nabla \varphi ,\nabla f)-2\lambda t\langle S,Hessf\rangle \hfill \\ \hfill & -2\lambda t\langle divS-\frac{1}{2}\nabla S,\nabla f\rangle +\mathcal{H},\hfill \end{array}$$

(25)

where $\mathcal{H}=-2t(\lambda -1)\nabla \overline{f}\nabla f-\lambda t\nabla f\nabla {\varphi }_t-\lambda t{\Delta }_{\varphi }\overline{f}$.

Given that

$$-(k_2+k_3)g_{ij}\le S_{ij}\le (k_2+k_3)g_{ij},$$

(26)

which implies

$${\left|S\right|}^2\le n{(k_2+k_3)}^2,$$

(27)

as $S_{ij}$ is a symmetric tensor.

Following [14], for any $ϵ\in (0,\frac{1}{\lambda })$ using Young’s inequality, we have

$$\langle S,Hessf\rangle \le {ϵ|Hessf|}^2+\frac{n}{4ϵ}{(k_2+k_3)}^2,$$

(28)

$$2\lambda tS(\nabla \varphi ,\nabla f)\ge -\frac{\lambda t{k}_2}{2ϵ}{|\nabla f|}^2-2\lambda t{k}_2ϵ{|\nabla \varphi |}^2.$$

(29)

Also

$$|div{S}_{ij}-\frac{1}{2}\nabla S|\le \frac{3}{2}\sqrt{n}{k}_4.$$

(30)

Using Lemma 3, Equations (26)–(30) and bounds of $Ri{c}_{\varphi }^{m-n}$, $S$ in (25) we have (17). □

Proof of Theorem 1.

Consider a $C^2$-function $\psi$ on $[0,\infty )$,

$$\psi \left(s\right)=\left\{\begin{array}{c}1,s\in [0,1],\hfill \\ 0,s\in [2,\infty ),\hfill \end{array}\right.$$

and it satisfies $\psi \left(s\right)\in [0,1]$, $-c_0\le {\psi }^{\prime }\left(s\right)\le 0$, ${\psi }^{\prime \prime }\left(s\right)\ge -c_1$ and $\frac{|{\psi }^{″}{\left(s\right)|}^2}{\psi \left(s\right)}\le c_1$, where $c_1$ is a constant and for $R\ge 1$ we defined a function

$$\eta (x,t)=\psi \left(\frac{r(x,t)}{R}\right),$$

where $r(x,t)=d(x,x_0,t)$. Applying the same argument as in [3] we can apply a maximum principle and use Calabi’s trick [29] to assume everywhere smoothness of $\eta (x,t)$, as $\psi \left(s\right)$ is Lipschitz.

By generalized Laplacian comparison theorem [14], we have

• ${\Delta }_{\varphi }r\left(x\right)\le (m-1)\sqrt{k_1}coth\left(\sqrt{k_1}r\left(x\right)\right)$,
• ${\Delta }_{\varphi }\eta \ge -\frac{c_0}{R}(m-1)(\sqrt{k_1}+\frac{2}{R})-\frac{c_1}{R^2}$,
• $\frac{{|\nabla \eta |}^2}{\eta }\le \frac{c_1}{R^2}$.

Let $G=\eta F$. Fix any $T_1\in (0,T]$ and assume G achieves maximum at $(x_0,t_0)\in Q_{2R,T_1}$. If $G(x_0,t_0)\le 0$ then the result is trivial and hence nothing to be proved, so assume that $G(x_0,t_0)\ge 0$.

Thus, at $(x_0,t_0)$ we have

$$\nabla G=0,\Delta G\le 0,{\partial }_{t}G\ge 0.$$

Therefore

$$\nabla F=-\frac{F}{\eta }\nabla \eta$$

(31)

and

$$0\ge ({\Delta }_{\varphi }-{\partial }_t)G=F({\Delta }_{\varphi }-{\partial }_t)\eta +\eta ({\Delta }_{\varphi }-{\partial }_t)F+2\langle \nabla \eta ,\nabla F\rangle .$$

(32)

By [16], there is a constant $c_2$ such that

$$\begin{array}{c}\hfill -F{\eta }_t\ge -c_2{k}_{2}F.\end{array}$$

(33)

Using (31) and (33) in (32) we get

$$\begin{array}{c}\hfill 0\ge -\left(\frac{c_0}{R}(m-1)(\sqrt{k_1}+\frac{2}{R})+\frac{3c_1}{R^2}+c_2{k}_2\right)F+\eta ({\Delta }_{\varphi }-{\partial }_t)F.\end{array}$$

(34)

Following [14,20,23], we set

$$\xi =\frac{{|\nabla f|}^2}{F}{|}_{(x_0,t_0)}\ge 0,$$

then at $(x_0,t_0)$ we have

$$\begin{array}{ccc}\hfill |\nabla f|& =& \sqrt{\xi F},\hfill \end{array}$$

(35)

$$\begin{array}{ccc}\hfill (\xi -\frac{t_0\xi -1}{\lambda t_0})F& =& {|\nabla f|}^2-(f_t+\overline{f}),\hfill \end{array}$$

(36)

$$\begin{array}{ccc}\hfill \eta \langle \nabla f,\nabla F\rangle & \le & \frac{\sqrt{c_1}}{R}{\eta }^{\frac{1}{2}}F|\nabla f|,\hfill \end{array}$$

(37)

$$\begin{array}{ccc}\hfill 3\lambda \sqrt{n}{k}_4|\nabla f|& \le & 2k_4{|\nabla f|}^2+\frac{9}{8}n{\lambda }^2{k}_4.\hfill \end{array}$$

(38)

Using Lemma 5 in (34) we have

$$\begin{array}{ccc}\hfill 0& \ge & -\left(\frac{c_0}{R}(m-1)(\sqrt{k_1}+\frac{2}{R})+\frac{3c_1}{R^2}+c_2{k}_2\right)F+\frac{2\eta t_0(1-\lambda ϵ)}{m}{\left({\Delta }_{\varphi }f\right)}^2-\frac{\lambda \eta t_0{k}_2}{2ϵ}|\nabla f^2|\hfill \\ & -& 2\lambda t_0\eta k_2{ϵ|\nabla \varphi |}^2-2\eta t_0(1-\lambda ϵ)(m-1)k_1{|\nabla f|}^2-2\eta \nabla F\nabla f-\frac{\eta F}{t_0}\hfill \\ \hfill & -& 2t_0(\lambda -1)\eta k_3{|\nabla f|}^2-\frac{n\eta \lambda t_0}{2ϵ}{(k_2+k_3)}^2-3\eta \lambda t_0\sqrt{n}{k}_4|\nabla f|+\eta \mathcal{H}.\hfill \end{array}$$

(39)

Multiplying (39) with $\eta t_0$ and using results from (35)–(38) we get

$$\begin{array}{ccc}\hfill 0& \ge & \hfill -\left(\frac{c_0}{R}(m-1)({\sqrt{k}}_1+\frac{2}{R})+\frac{3c_1}{R^2}+c_2{k}_2\right)G{t}_0+\frac{2{\eta }^2{t}_0^2(1-\lambda ϵ)}{m}{\left({\Delta }_{\varphi }f\right)}^2\\ \hfill & -& \frac{\lambda \eta t_0^2{k}_2\xi }{2ϵ}G-2\lambda {\eta }^2{t}_0^2{k}_2ϵ{|\nabla \varphi |}^2-2\eta \xi t_0^2(1-\lambda ϵ)(m-1)k_{1}G-2t_0\frac{{\sqrt{c}}_1}{R}{G}^{\frac{3}{2}}{\xi }^{\frac{1}{2}}\hfill \\ & -& \eta G-2{\eta }^2{t}_0^2(\lambda -1)k_3{|\nabla f|}^2-\frac{n{\eta }^2{t}_0^2\lambda }{2ϵ}{(k_2+k_3)}^2-2k_4{t}_0^2\xi G\eta \hfill \\ \hfill & -& \frac{9}{8}n{\lambda }^2{\eta }^2{k}_4{t}_0^2+{\eta }^2{t}_0\mathcal{H}.\hfill \end{array}$$

(40)

Now we use Young’s inequality by choosing suitable values for $a,b,\alpha ,p,q$ as in Lemma 4.

Set $a=\frac{2\sqrt{c_1}}{R}{G}^{\frac{1}{2}}$, $b=G{\xi }^{\frac{1}{2}}$, $p=2$, $q=2$, $\alpha =\frac{m{\lambda }^2}{4(1-\lambda ϵ)(\lambda -1)}$ and apply Lemma 4 (Young’s inequality) we get

$$\begin{array}{ccc}\hfill 2t_0\frac{{\sqrt{c}}_1}{R}{G}^{\frac{3}{2}}{\xi }^{\frac{1}{2}}& \le & \frac{4(1-\lambda ϵ)}{m{\lambda }^2}(\lambda -1)\xi G^2{t}_0+\frac{m{\lambda }^2{c}_1{t}_{0}G}{4(1-\lambda ϵ)(\lambda -1)R^2}.\hfill \end{array}$$

(41)

Cauchy–Schwarz inequality gives

$$\begin{array}{ccc}\hfill {\eta }^2\lambda \langle \nabla f,\nabla {\varphi }_t\rangle & \le & {\eta }^2\lambda |\nabla f||\nabla {\varphi }_t|\hfill \\ \hfill & \le & \lambda {\theta }_2{G}^{\frac{1}{2}}{\xi }^{\frac{1}{2}}.\hfill \end{array}$$

Set $a=\lambda {\theta }_2$, $b={\xi }^{\frac{1}{2}}{G}^{\frac{1}{2}}$, $p=\frac{4}{3}$, $q=4$, $\alpha ={\left(\frac{m{\lambda }^2}{2(1-\lambda ϵ)(1-\delta ){(\lambda -1)}^2}\right)}^{\frac{1}{4}}$ and apply Lemma 4 we get

$$\begin{array}{ccc}\hfill {\eta }^2\lambda \langle \nabla f,\nabla {\varphi }_t\rangle & \le & \frac{(1-\lambda ϵ)(1-\delta )}{2m{\lambda }^2}{(\lambda -1)}^2{\xi }^2{G}^2+\frac{3}{4}{\left(\frac{m{\lambda }^2}{2(1-ϵ\lambda )(1-\delta ){(\lambda -1)}^2}\right)}^{\frac{1}{3}}{\lambda }^{\frac{4}{3}}{\theta }_2^{\frac{4}{3}},\hfill \\ \hfill & & \mathrm{for}\mathrm{all}\delta \in (0,1).\hfill \end{array}$$

(42)

We have $\overline{f}=\widehat{A}p+\widehat{B}q+\xi$. Hence

$$\begin{array}{ccc}\hfill \nabla \overline{f}& =& p{\widehat{A}}_f\nabla f+q{\widehat{B}}_f\nabla f+\widehat{A}\nabla p+\widehat{B}\nabla q+\nabla \xi ,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill \Delta \overline{f}& =& \nabla \nabla \overline{f}\hfill \\ \hfill & =& (\widehat{A}\Delta p+\widehat{B}\Delta q)+p({\widehat{A}}_{ff}{|\nabla f|}^2+{\widehat{A}}_f\Delta f)+q({\widehat{B}}_{ff}{|\nabla f|}^2+{\widehat{B}}_f\Delta f)\hfill \\ \hfill & +& 2({\widehat{A}}_f\langle \nabla f,\nabla p\rangle +{\widehat{B}}_f\langle \nabla f,\nabla q\rangle )+\Delta \xi .\hfill \end{array}$$

Hence

$$\begin{array}{ccc}\hfill {\Delta }_{\varphi }\overline{f}& =& {\eta }^2(\Delta \overline{f}-\nabla \varphi \nabla \overline{f})\hfill \\ & =& (\widehat{A}{\Delta }_{\varphi }p+\widehat{B}{\Delta }_{\varphi }q)+(p{\widehat{A}}_f+q{\widehat{B}}_f){\Delta }_{\varphi }f+(p{\widehat{A}}_{ff}+q{\widehat{B}}_{ff}){|\nabla f|}^2\hfill \\ \hfill & +& 2({\widehat{A}}_f\langle \nabla p,\nabla f\rangle +{\widehat{B}}_f\langle \nabla q,\nabla f\rangle )+{\Delta }_{\varphi }\xi .\hfill \end{array}$$

(43)

Again

$$\begin{array}{ccc}\hfill 2{\eta }^2{\widehat{A}}_f\langle \nabla p,\nabla f\rangle & \le & 2{\lambda }_2{\eta }^2|\nabla p||\nabla f|,\mathrm{using}\mathrm{Cauchy-Schwarz}\mathrm{inequality}\hfill \\ & \le & 2{\lambda }_2{\gamma }_2{\eta }^2|\nabla f|\hfill \\ \hfill & \le & 2{\lambda }_2{\gamma }_2{\xi }^{\frac{1}{2}}{G}^{\frac{1}{2}}.\hfill \end{array}$$

(44)

Similarly

$$\begin{array}{ccc}\hfill 2{\eta }^2{\widehat{B}}_f\langle \nabla q,\nabla f\rangle & \le & 2b_2{\sigma }_2{\xi }^{\frac{1}{2}}{G}^{\frac{1}{2}}.\hfill \end{array}$$

(45)

Adding (44) and (45) gives

$$\begin{array}{ccc}\hfill 2{\eta }^2({\widehat{A}}_f\langle \nabla p,\nabla f\rangle +{\widehat{B}}_f\langle \nabla q,\nabla f\rangle )& \le & 2({\gamma }_2{\lambda }_2+b_2{\sigma }_2){\xi }^{\frac{1}{2}}{G}^{\frac{1}{2}}.\hfill \end{array}$$

(46)

Using (46) in (43) and applying Young’s inequality with $a=2({\gamma }_2{\lambda }_2+b_2{\sigma }_2)$, $b={\xi }^{\frac{1}{2}}{G}^{\frac{1}{2}}$, $p=\frac{4}{3}$, $q=4$ and $\alpha ={\left(\frac{2m{\lambda }^2}{(1-\lambda ϵ)\delta (\lambda -1)}\right)}^{\frac{1}{4}}$ we obtain

$$\begin{array}{ccc}\hfill {\eta }^2{\Delta }_{\varphi }\overline{f}& \le & ({\lambda }_1{\gamma }_3+b_1{\sigma }_3)+({\gamma }_1{\lambda }_2+{\sigma }_1{b}_2){\eta }^2{\Delta }_{\varphi }f\hfill \\ & +& ({\lambda }_3{\gamma }_1+b_3{\sigma }_1)\xi G+\frac{2(1-\lambda ϵ)\delta }{m{\lambda }^2}{(\lambda -1)}^2{\xi }^2{G}^2\hfill \\ \hfill & +& \frac{3}{4}{\left(\frac{2m{\lambda }^2}{(1-\lambda ϵ)\delta (\lambda -1)}\right)}^{\frac{1}{3}}{(2{\lambda }_2{\gamma }_2+2b_2{\sigma }_2)}^{\frac{4}{3}}+m_2.\hfill \end{array}$$

(47)

Similarly we get

$$\begin{array}{ccc}\hfill {\eta }^2\langle \nabla \overline{f},\nabla f\rangle & \le & \frac{(1-\lambda ϵ)(1-\delta ){(\lambda -1)}^2}{2m{\lambda }^2}{\xi }^2{G}^2\hfill \\ & +& \frac{3}{4}{\left(\frac{m{\lambda }^2}{2(1-\lambda ϵ)(1-\delta ){(\lambda -1)}^2}\right)}^{\frac{1}{3}}{({\lambda }_1{\gamma }_2+b_1{\sigma }_2+m_1)}^{\frac{4}{3}}\hfill \\ \hfill & +& ({\gamma }_1{\lambda }_2+{\sigma }_1{b}_2)\xi G.\hfill \end{array}$$

(48)

Equations (47) and (48) are the quantities that estimates $\mathcal{H}$.

From (36) we have

$$\begin{array}{ccc}\hfill {\Delta }_{\varphi }f& =& {|\nabla f|}^2-(f_t+\overline{f})\hfill \\ \hfill & =& \left(\xi -\frac{t_0\xi -1}{\lambda t_0}\right)F.\hfill \end{array}$$

Thus

$$\begin{array}{ccc}\hfill \frac{2{\eta }^2{t}_0^2(1-\lambda ϵ)}{m}{\left({\Delta }_{\varphi }f\right)}^2& =& \frac{2(1-\lambda ϵ)}{m{\lambda }^2}{G}^2-\frac{4\xi t_0(1-\lambda ϵ)(\lambda -1)}{m{\lambda }^2}{G}^2\hfill \\ & +& \frac{2(1-\lambda ϵ)}{m{\lambda }^2}{\xi }^2{t}_0^2{(\lambda -1)}^2{G}^2\hfill \end{array}$$

(49)

and

$$\begin{array}{ccc}\hfill {\eta }^2{\Delta }_{\varphi }f& =& -\frac{1}{\lambda t_0}G-\frac{t_0(\lambda -1)}{\lambda t_0}\xi G.\hfill \end{array}$$

(50)

Set

$$\overline{C_1}:=\left\{\frac{\lambda k_2}{2ϵ}+2k_4+{\lambda }_3{\gamma }_1+b_3{\sigma }_1+2(\lambda -1)k_3+\frac{\lambda -1}{\lambda }+{\gamma }_1{\lambda }_2+{\sigma }_1{b}_2+2(1-\lambda ϵ)(m-1)k_1\right\}$$

and apply Peter-Paul inequality with $a=\xi G$, $b=\overline{C_1}$, $\alpha =\frac{m{\lambda }^2}{(1-ϵ\lambda )(1-\delta ){(\lambda -1)}^2}$ we get

$$\begin{array}{ccc}\hfill \overline{C_1}\xi G& \le & \frac{(1-ϵ\lambda )(1-\delta ){(\lambda -1)}^2}{m{\lambda }^2}{\xi }^2{G}^2+\frac{m{\lambda }^2}{4(1-\lambda ϵ)(1-\delta ){(\lambda -1)}^2}{\overline{C_1}}^2.\hfill \end{array}$$

(51)

Set

$$\begin{array}{ccc}\hfill D_1& :=& 1+t_0\left(\frac{c_0}{R}(m-1)(\sqrt{k_1}+\frac{2}{R})+\frac{3c_1}{R^2}+c_2{k}_2+\frac{m{\lambda }^2{c}_1}{4(1-\lambda ϵ)(\lambda -1)R^2}+\frac{1}{\lambda }\right),\hfill \end{array}$$

(52)

$$\begin{array}{ccc}\hfill E_1& :=& \frac{m{\lambda }^2}{4(1-\lambda ϵ)(1-\delta ){(\lambda -1)}^2}{\overline{C_1}}^2+2\lambda k_2ϵ{\theta }_1^2\hfill \\ \hfill & +& \frac{n\lambda }{2ϵ}{(k_2+k_3)}^2+\frac{9}{8}n{\lambda }^2{k}_4+({\lambda }_1{\gamma }_3+b_1{\sigma }_3)+\frac{3}{4}{\left(\frac{2m{\lambda }^2}{(1-\lambda ϵ)(\lambda -1)\delta }\right)}^{\frac{1}{3}}(2{\lambda }_2{\gamma }_2\hfill \\ & +& 2b_2{\sigma }_2{)}^{\frac{4}{3}}+m_2+\frac{3}{4}{\left(\frac{m{\lambda }^2}{2(1-\lambda ϵ)(1-\delta ){(\lambda -1)}^2}\right)}^{\frac{1}{3}}{\lambda }^{\frac{4}{3}}{\theta }_2^{\frac{4}{3}}\hfill \\ \hfill & +& \frac{3}{4}{\left(\frac{m{\lambda }^2}{2(1-\lambda ϵ)(1-\delta ){(\lambda -1)}^2}\right)}^{\frac{1}{3}}{({\lambda }_1{\gamma }_2+b_1{\sigma }_2+m_1)}^{\frac{4}{3}}.\hfill \end{array}$$

(53)

Using (41) to (52) in (40) we obtain

$$\begin{array}{ccc}\hfill 0& \ge & \frac{2(1-\lambda ϵ)}{m{\lambda }^2}{G}^2-D_{1}G-t_0^2{E}_1.\hfill \end{array}$$

(54)

For a positive number p and two non-negative numbers $q,r$, the quadratic inequality of the form $p{x}^2-qx-r\le 0$ implies that $x\le \frac{q}{p}+\sqrt{\frac{r}{p}}$.

So at $(x_0,t_0)$ we have

$$\begin{array}{ccc}\hfill G& \le & D_1\frac{m{\lambda }^2}{2(1-\lambda ϵ)}+t_0\sqrt{\frac{m{\lambda }^2{E}_1}{2(1-\lambda ϵ)}}.\hfill \end{array}$$

(55)

Since $\eta (x,t)=1$ whenever $d(x,x_0,T_1)\le R$, hence

$$\frac{F(x,T_1)}{T_1}={\left(\right|\nabla f|}^2-\lambda (f_t+\overline{f})){|}_{(x,T_1)}\le \frac{G(x_0,t_0)}{T_1}\le \frac{1}{T_1}\left(D_1\frac{m{\lambda }^2}{2(1-\lambda ϵ)}+t_0\sqrt{\frac{m{\lambda }^2{E}_1}{2(1-\lambda ϵ)}}\right).$$

Since $t_0\le T_1$, so

$$\frac{1}{T_1}\left(D_1\frac{m{\lambda }^2}{2(1-\lambda ϵ)}+t_0\sqrt{\frac{m{\lambda }^2{E}_1}{2(1-\lambda ϵ)}}\right)\le \frac{m{\lambda }^2}{2T_1(1-\lambda ϵ)}+\frac{\tilde{D}}{T_1}\frac{m{\lambda }^2}{2(1-\lambda ϵ)}+\sqrt{\frac{m{\lambda }^2{E}_1}{2(1-\lambda ϵ)}},$$

where $\tilde{D}=t_0\widetilde{D_1}$ and $\widetilde{D_1}=\left(\frac{c_0}{R}(m-1)(\sqrt{k_1}+\frac{2}{R})+\frac{3c_1}{R^2}+c_2{k}_2\right)+\frac{m{\lambda }^2{c}_1}{4(1-\lambda ϵ)(\lambda -1)R^2}+\frac{1}{\lambda }$ satisfying $\frac{\tilde{D}}{T_1}\le \widetilde{D_1}$. Since $T_1$ is arbitrary so

$$\begin{array}{c}\hfill {|\nabla f|}^2-\lambda (f_t+\widehat{A}p+\widehat{B}q+\xi )\le \frac{m{\lambda }^2}{2t(1-\lambda ϵ)}+\frac{m{\lambda }^2}{2(1-\lambda ϵ)}\widetilde{D_1}+\widetilde{E_1},\end{array}$$

(56)

where $\widetilde{E_1}={\left(\frac{m{\lambda }^2}{2(1-\lambda ϵ)}{E}_1\right)}^{\frac{1}{2}}$.

Substituting $f=logu$ on (56) and using the definition of $\widehat{A},\widehat{B}$, we get (16). This completes the proof. □

Corollary 1.

If $k_1,k_2,k_3,k_4$ are positive constants such that

$$Ri{c}_{\varphi }^{m-n}\ge -(m-1)k_{1}g,-k_{2}g\le S\le k_{3}g,|\nabla S|\le k_4$$

on M, then for any $\lambda >1$ and $\delta \in (0,1)$ we have

$$\begin{array}{c}\hfill \frac{{|\nabla u|}^2}{u^2}-\lambda \left(\frac{u_t}{u}+\frac{A\left(u\right)}{u}p+\frac{B\left(u\right)}{u}q+\xi \right)\le \frac{m{\lambda }^2}{2t(1-\lambda ϵ)}+\frac{m{\lambda }^2}{2(1-\lambda ϵ)}\widetilde{D_2}+\widetilde{E_2},\end{array}$$

(57)

where

$$\begin{array}{ccc}\hfill \widetilde{D_2}& =& c_2{k}_2+\frac{1}{\lambda },\hfill \\ \hfill \widetilde{E_2}& =& {\left(\frac{m{\lambda }^2}{2(1-\lambda ϵ)}{E}_2\right)}^{\frac{1}{2}},\hfill \\ \hfill E_2& =& \frac{m{\lambda }^2}{4(1-\lambda ϵ)(1-\delta ){(\lambda -1)}^2}{\overline{C_2}}^2+2\lambda k_2ϵ{\mathsf{\Theta }}_1^2+\frac{n\lambda }{2ϵ}{(k_2+k_3)}^2\hfill \\ & +& \frac{9}{8}n{\lambda }^2{k}_4+({\mathsf{\Lambda }}_1{\mathsf{\Gamma }}_3+B_1{\mathsf{\Sigma }}_3)+\frac{3}{4}{\left(\frac{2m{\lambda }^2}{(1-\lambda ϵ)(\lambda -1)\delta }\right)}^{\frac{1}{3}}{(2{\mathsf{\Lambda }}_2{\mathsf{\Gamma }}_2+2B_2{\mathsf{\Sigma }}_2)}^{\frac{4}{3}}\hfill \\ & +& M_2+\frac{3}{4}{\left(\frac{m{\lambda }^2}{2(1-\lambda ϵ)(1-\delta ){(\lambda -1)}^2}\right)}^{\frac{1}{3}}{\lambda }^{\frac{4}{3}}{\mathsf{\Theta }}_2^{\frac{4}{3}}\hfill \\ & +& \frac{3}{4}{\left(\frac{m{\lambda }^2}{2(1-\lambda ϵ)(1-\delta ){(\lambda -1)}^2}\right)}^{\frac{1}{3}}{({\mathsf{\Lambda }}_1{\mathsf{\Gamma }}_2+B_1{\mathsf{\Sigma }}_2+M_1)}^{\frac{4}{3}},\hfill \\ \hfill \overline{C_2}& =& \left(\frac{\lambda k_2}{2ϵ}+2k_4+{\mathsf{\Lambda }}_3{\mathsf{\Gamma }}_1+B_3{\mathsf{\Sigma }}_1+2(\lambda -1)k_3+\frac{\lambda -1}{\lambda }+2(1-\lambda ϵ)(m-1)k_1\right).\hfill \end{array}$$

Proof. 

We know $g\left(t\right)$ is uniformly equivalent to the initial metric $g\left(0\right)$. For a fixed $\delta \in (0,1)$ if we let R tend to $+\infty$ then we obtain our result. □

Theorem 2.

If $k_1,k_2,k_3,k_4$ are positive constants such that

$$Ri{c}_{\varphi }^{m-n}\ge -(m-1)k_{1}g,-k_{2}g\le S\le k_{3}g,|\nabla S|\le k_4$$

on M and let u be a positive solution to (7) under the flow (8) then we have the Harnack inequality

$$\begin{array}{cc}\hfill u(y_1,s_1)& \le u(y_2,s_2){(\frac{s_2}{s_1})}^{\frac{m\lambda }{2(1-\lambda ϵ)}}exp\left\{\frac{\lambda }{4}\mathcal{I}(s_1,s_2)+(s_2-s_1)({\mathsf{\Lambda }}_1{\mathsf{\Gamma }}_1+B_1{\mathsf{\Sigma }}_1+M_3+\frac{1}{\lambda }\widetilde{F_2})\right\},\hfill \end{array}$$

(58)

where $\mathcal{I}(s_1,s_2)=\underset{\zeta }{inf}{\int }_{s_1}^{s_2}{\left|{\zeta }^{\prime }\left(t\right)\right|}^{2}dt$ and $\zeta :[s_1,s_2]\to M$ is a path joining the points $(y_1,s_1)$, $(y_2,$$s_2)$ in $M\times [0,T]$ and $\widetilde{F_2}=\frac{m{\lambda }^2}{2(1-\lambda ϵ)}\widetilde{D_2}+\widetilde{E_2}$.

Proof. 

Set $\widetilde{F_2}=\frac{m{\lambda }^2}{2(1-\lambda ϵ)}\widetilde{D_2}+\widetilde{E_2}$ then (57) becomes

$$\begin{array}{ccc}\hfill \frac{{|\nabla u|}^2}{u^2}-\lambda \left(\frac{u_t}{u}+\frac{A\left(u\right)}{u}p+\frac{B\left(u\right)}{u}q+\xi \right)& \le & \frac{m{\lambda }^2}{2t(1-\lambda ϵ)}+\widetilde{F_2}.\hfill \end{array}$$

(59)

For $u=e^f$ we have

$$\begin{array}{ccc}\hfill {|\nabla f|}^2-\lambda \left(f_t+\widehat{A}p+\widehat{B}q+\xi \right)& \le & \frac{m{\lambda }^2}{2t(1-\lambda ϵ)}+\widetilde{F_2}.\hfill \end{array}$$

(60)

Let $(y_1,s_1),(y_2,s_2)\in M\times [0,T]$ be such that $s_1<s_2$. Take a geodesic path $\zeta :[s_1,s_2]\to M$ satisfying $\zeta \left(s_1\right)=y_1,\zeta \left(s_2\right)=y_2$. Using (60) we obtain

$$\begin{array}{ccc}\hfill f(y_1,s_1)-f(y_2,s_2)& =& -{\int }_{s_1}^{s_2}\frac{d}{dt}f(\zeta \left(t\right),t)dt\hfill \\ \hfill & =& -{\int }_{s_1}^{s_2}{\partial }_{t}fdt-{\int }_{s_1}^{s_2}\langle \nabla f,{\zeta }^{\prime }\left(t\right)\rangle dt\hfill \\ & \le & \frac{m\lambda }{2(1-\lambda ϵ)}ln(\frac{s_2}{s_1})+(s_2-s_1)({\mathsf{\Lambda }}_1{\mathsf{\Gamma }}_1+B_1{\mathsf{\Sigma }}_1+M_3+\frac{1}{\lambda }\widetilde{F_2})\hfill \\ \hfill & -& {\int }_{s_1}^{s_2}\frac{1}{\lambda }{|\nabla f|}^{2}dt-{\int }_{s_1}^{s_2}\langle \nabla f,{\zeta }^{\prime }\left(t\right)\rangle dt.\hfill \end{array}$$

(61)

Now using the relation $-a{x}^2-bx\le \frac{b^2}{4a}$, we set $x=\nabla f$, $a=\frac{1}{\lambda }$ and $b={\zeta }^{\prime }\left(t\right)$ we get

$$\begin{array}{ccc}\hfill f(y_1,s_1)-f(y_2,s_2)& \le & \frac{m\lambda }{2(1-\lambda ϵ)}ln\left(\frac{s_2}{s_1}\right)-{\int }_{s_1}^{s_2}\frac{\lambda |{\zeta }^{\prime }{\left(t\right)|}^2}{4}dt\hfill \\ & +& (s_2-s_1)({\mathsf{\Lambda }}_1{\mathsf{\Gamma }}_1+B_1{\mathsf{\Sigma }}_1+M_3+\frac{1}{\lambda }\widetilde{F_2}).\hfill \end{array}$$

(62)

Take infimum of (62) over all possible curves $\zeta$ on M and put $f=lnu$ to obtain (58). □

4. Conclusions

In this paper, we have established Li–Yau-type estimate for a positive solution of the equation

$$\begin{array}{c}\hfill {\Delta }_{\varphi }u=\frac{\partial u}{\partial t}+A\left(u\right)p(x,t)+B\left(u\right)q(x,t)+\xi (x,t)u(x,t),\end{array}$$

along the flow ${\partial }_t{g}_{ij}=2S_{ij}$ and related Harnack type inequality. In particular if $\xi (x,t)=0$, $B\left(u\right)=u^{a+1}$ then the results are same as in Section 2 of [14]. Thus, our paper generalizes some results of [14].

Further $A\left(u\right)=B\left(u\right)=\xi \left(u\right)=0$ gives the classical Li–Yau-type estimate for positive solution of the weighted heat equation

$${\Delta }_{\varphi }u={\partial }_{t}u$$

(63)

under the geometric flow ${\partial }_t{g}_{ij}=2S_{ij}$. To obtain this estimate we put

• $A\left(u\right)=B\left(u\right)=\xi (x,t)=0$
• ${\lambda }_1={\lambda }_2={\lambda }_3=0$
• $b_1=b_2=b_3=0$
• $p(x,t)=q(x,t)=0$

in (16) and get

$$\begin{array}{ccc}\hfill \frac{{|\nabla u|}^2}{u^2}-\lambda \frac{u_t}{u}& \le & \frac{m{\lambda }^2}{2t(1-\lambda ϵ)}+\frac{m{\lambda }^2}{2(1-\lambda ϵ)}\widetilde{D_3}+\widetilde{E_3},\hfill \end{array}$$

(64)

where

$$\begin{array}{ccc}\hfill \widetilde{D_3}& =& \frac{c_0}{R}(m-1)(\sqrt{k_1}+\frac{2}{R})+\frac{3c_1}{R^2}+c_2{k}_2+\frac{m{\lambda }^2{c}_1}{4(a-\lambda ϵ)(\lambda -1)R^2}+\frac{1}{\lambda },\hfill \\ \hfill \widetilde{E_3}& =& {\left(\frac{m{\lambda }^2}{2(1-\lambda ϵ)}{E}_3\right)}^{\frac{1}{2}},\hfill \end{array}$$

$$\begin{array}{ccc}\hfill E_3& =& \frac{m{\lambda }^2}{4(1-\lambda ϵ)(1-\delta ){(\lambda -1)}^2}{\overline{C_3}}^2+2\lambda k_2ϵ{\theta }_1^2+\frac{n\lambda }{2ϵ}{(k_2+k_3)}^2+\frac{9}{8}n{\lambda }^2{k}_4\hfill \\ \hfill & & +\frac{3}{4}{\left(\frac{m{\lambda }^2}{2(1-\lambda ϵ)(1-\delta ){(\lambda -1)}^2}\right)}^{\frac{1}{3}}{\lambda }^{\frac{4}{3}}{\theta }_2^{\frac{4}{3}},\hfill \\ \hfill \overline{C_3}& =& \frac{\lambda k_2}{2ϵ}+2k_4+2(\lambda -1)k_3+\frac{\lambda -1}{\lambda }+2(1-\lambda ϵ)(m-1)k_1.\hfill \end{array}$$

Here if we let $R\to +\infty$ then we get the classical Li–Yau-type global gradient estimate for (63) along the flow ${\partial }_t{g}_{ij}=2S_{ij}$. The key ingredient in this estimation is the assumption of bounds for the weight function $\varphi$ and its derivative $|\nabla \varphi |$ (see Preliminaries section), it would be interesting if one can derive Li–Yau-type estimation for a positive solution u of (7) without assuming bounds for $\varphi ,|\nabla \varphi |$. One can consider this problem as a future work for this article.