\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2015 (2015), No. 12, pp. 1--11.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2015 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2015/12\hfil Gradient estimates] {Gradient estimates for a nonlinear parabolic equation with potential under geometric flow} \author[A. Abolarinwa \hfil EJDE-2015/12\hfilneg] {Abimbola Abolarinwa} % in alphabetical order \address{Abimbola Abolarinwa \newline Department of Mathematics, University of Sussex, Brighton, BN1 9QH, United Kingdom} \email{A.Abolarinwa@sussex.ac.uk} \thanks{Submitted December 8, 2014. Published January 8, 2015.} \subjclass[2000]{35K55, 53C21, 53C44, 58J35} \keywords{Gradient estimates; Harnack inequalities; parabolic equations; \hfill\break\indent geometric flows} \begin{abstract} Let $(M, g)$ be an $n$ dimensional complete Riemannian manifold. In this article we prove local Li-Yau type gradient estimates for all positive solutions to the nonlinear parabolic equation \[ (\partial_t - \Delta_g + \mathcal{R}) u( x, t) = - a u( x, t) \log u( x, t) \] along the generalised geometric flow on $M$. Here $\mathcal{R} = \mathcal{R} (x, t)$ is a smooth potential function and $a$ is an arbitrary constant. As an application we derive a global estimate and a space-time Harnack inequality. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{corollary}[theorem]{Corollary} \allowdisplaybreaks \section{Preliminaries and main results} Gradient and Harnack estimates are fundamental tools to tackle classical and modern problems in geometric analysis. These methods applied to parabolic equations were first studied by Li and Yau in their celebrated paper \cite{LY86}. They have been applied successfully to the setting of various geometric flows, for more details see %\cite{Ab1}--\cite{CZ11}, \cite{HHL12}--\cite{KZh08}, \cite{Ni07} \cite{Ab1,Ab2,Ba13,BCP,CaH09,CZ11,HHL12,HM10,KZh08,Ni07} and the references therein. In this article we derive various gradient estimates for the following nonlinear parabolic equation with potential \begin{equation}\label{eq52} \big(\frac{\partial }{ \partial t} - \Delta + \mathcal{R}\big) u( x, t) = - a u( x, t) \log u( x, t), \end{equation} in a more general setting of geometric flow. Here the symbol $\Delta = \Delta_g $ is the Laplace-Beltrami operator acting on functions in space with respect to metric $g(t)$ in time, $a$ is a constant and $\mathcal{R}: M \times [0, T] \to \mathbb{R}$ is a $C^\infty$-function on $M$. For instance if we take $\mathcal{R}$ to be the scalar curvature of the manifold and we allow $g$ to evolve by the Ricci flow, $ \partial_t g = - 2Rc$, where $Rc$ is the Ricci curvature tensor, it then reduces to the study of gradient Ricci soliton. Taking $ f = \log u$, a standard calculation yields \begin{align}\label{eqn2} \big(\frac{\partial }{ \partial t} - \Delta \big) f = |\nabla f|^2 - af - \mathcal{R}. \end{align} The study of gradient estimates on $M$ can be reduced to the study of the properties of the solution $f$ of \eqref{eqn2} and it is related to gradient soliton equation \cite{CaH09,CZ11}. Let $(M, g(t)), 0 \leq t \leq T$, be an $n$-dimensional complete Riemannian manifold whose metric $g(t)$ evolves by the geometric flow \begin{equation}\label{eq51} \frac{\partial }{\partial t} g_{ij}(x,t) = 2 h_{ij}(x, t), \quad (x, t) \in M \times [0, T], \end{equation} where $h_{ij}$ is a general time-dependent symmetric $(0, 2)$-tensor and $0 0$ and some constants $C_8$ and $C_9$ depending only on $n, \alpha$ and uniform bounds for $ | \nabla h |$, $|\nabla \mathcal{R}| $ and $ | \Delta \mathcal{R} |$, where $ f = \log u$ and $\alpha > 1$ are given such that $ \frac{1}{p} + \frac{1}{q} = \frac{1}{\alpha}$ for any real numbers $p, q > 0$. \end{theorem} As an application of the above result we obtain global gradient estimates (Cf. Remark \ref{rmk31}, equation \eqref{eq512}). We then apply the global estimates obtained to derive classical Harnack inequalities by integrating along a space-time path joining any two points in $M$. The rest of the paper is as follows; in the next section we state and prove an important lemma that will be applied to prove the theorem above. The last section is devoted to the descriptions of the cut-off function needed in the proof and the detail of the proof of Theorem \ref{thm54} itself and its application to Harnack inequality (Cf. Corollary \ref{corN5}). \section{Important Lemma} We first prove the following technical lemma which is a generalization of \cite[Lemma 3.1]{Ab1}. It is originally proved for heat equation on static metric by Li and Yau \cite{LY86}. This is very crucial to derivation of both local and global estimate of Li-Yau type. \begin{lemma} \label{lem53} Let $( M, g(t))$ be a complete solution to the generalized flow \eqref{eq51} in some time interval $[0, T]$. Suppose there exist some nonnegative constants $k_1, k_2, k_3, $ and $k_4$ such that $ R_{ij}(g) \geq - k_1 g $, $- k_2 g \leq h \leq k_3 g$ and $| \nabla h | \leq k_4$ for all $ t \in [0,T]$. For any smooth positive solution $ u \in C^{2,1 } ( M \times [0,T])$ to equation \eqref{eq52} in the geodesic ball $ \mathcal{B}_{ 2\rho, T}$, it holds that \begin{equation}\label{eq55} \begin{aligned} ( \Delta - \partial_t) F &\geq - 2 \langle \nabla f, \nabla F \rangle - \frac{2 \alpha t }{ np} (\Delta f )^2 - \frac{F}{t} - 3 \alpha n^{1/2} k_4 t |\nabla f| \\ &\quad - \big( ( \alpha - 1)(2 k_3 + a) + 2k_1 \big) t| \nabla f |^2 - \alpha t \Delta \mathcal{R} \\ &\quad - 2 t ( \alpha -1) \langle \nabla f , \nabla R \rangle - \frac{\alpha n q }{2} t ( k_2 + k_3)^2 + aF, \end{aligned} \end{equation} where $ f = \log u, \ F = t (| \nabla f |^2 - \alpha \partial_t f - \alpha \mathcal{R} - \alpha a f )$ and $\alpha \geq 1$ are given such that $ \frac{1}{p} + \frac{1}{q} = \frac{1}{\alpha}$ for any real numbers $p, q > 0$. \end{lemma} \begin{proof} Recall from \cite[Lemma 2.1]{Ab1} the following evolutions under the flow \begin{gather}\label{eqn34} ( | \nabla f|^2)_t = - 2 h_{ij} f_i f_j + 2 f_i f_{ti},\\ \label{eqn35} ( \Delta f )_t = \Delta (f_t) - 2 h_{ij}f_{ij} - 2 \langle \operatorname{div}h , \nabla f \rangle + \langle \nabla H , \nabla f \rangle, \end{gather} where $\operatorname{div}$ is the divergence operator, i.e., $(\operatorname{div}h)_k = g^{ij} \nabla_i h_{jk}$. Notice also that $ f_t = \Delta f + | \nabla f |^2 - \mathcal{R} - a f $. Taking covariant derivative of $F$ we have $$ F_i = t ( 2 f_j f_{ji} - \alpha f_{ti} - \alpha \mathcal{R}_i - \alpha a f_i) $$ and with Bochner-Weitzenb\"ock's formula \begin{equation} \Delta | \nabla f|^2 = 2 |f_{ij}|^2 + 2 f_j f_{jji} + 2 R_{ij} f_i f_j \end{equation} we have $$ \Delta F = \sum_{i =1}^n F_{ii} = t \Big( 2 f_{ij}^2 + 2 f_j f_{jji} + 2 R_{ij} f_{ij} - \alpha \Delta (f_t) - \alpha \Delta \mathcal{R} - \alpha a \Delta f \Big). $$ Using \eqref{eqn35} we obtain \begin{align*} \Delta F &= t \Big[ 2 f_{ij}^2 + 2 f_j f_{jji} + 2 R_{ij} f_i f_j - \alpha( \Delta f)_t - 2 \alpha h_{ij} f_{ij} \\ &\quad - 2 \alpha (\operatorname{div}h)_i f_j + \alpha H_i f_j - \alpha \Delta \mathcal{R} - \alpha a \Delta f \Big] \\ & = t \big( 2 f_{ij}^2 - 2 \alpha h_{ij} f_{ij} \big) + 2t \langle \nabla f , \nabla ( f_t + af + \mathcal{R} - | \nabla f |^2 ) \rangle \\ &\quad - \alpha t ( f_t + af + \mathcal{R} - | \nabla f |^2 )_t - 2 \alpha t (\operatorname{div}h)_i f_j \\ &\quad + \alpha t H_i f_j - \alpha t \Delta \mathcal{R} - \alpha a t \Delta f + 2 t R_{ij} f_{ij}. \end{align*} Notice that \begin{gather}\label{eq56} \begin{aligned} - \alpha t ( f_t + af + \mathcal{R} - | \nabla f |^2 )_t &= t ( \alpha | \nabla f |^2 - \alpha f_t - \alpha a f - \alpha t\mathcal{R} )_t \\ &= t \Big( | \nabla f |^2 - \alpha f_t - \alpha a f - \alpha \mathcal{R} + ( \alpha - 1) | \nabla f |^2 \Big)_t \\ &= t \Big(\frac{F}{t} + ( \alpha - 1) | \nabla f |^2 \Big)_t \\ &= F_t - \frac{F}{t} + t ( \alpha - 1) ( | \nabla f |^2 )_t , \end{aligned} \\ \label{eq57} \begin{aligned} &2t \langle \nabla f , \nabla ( f_t + af + \mathcal{R} - | \nabla f |^2 ) \rangle + t( \alpha - 1) ( | \nabla f |^2 )_t \\ &= 2t \langle \nabla f , \nabla ( f_t + af + \mathcal{R} - | \nabla f |^2 ) \rangle + 2 t ( \alpha - 1) \langle \nabla f, \nabla (f_t) \rangle - 2 t ( \alpha - 1) h_{ij} f_i f_j \\ &= 2t \langle \nabla f , \nabla ( \alpha f_t + af + \mathcal{R} - | \nabla f |^2 ) \rangle - 2 t ( \alpha - 1) h_{ij} f_i f_j \\ &= - 2t \langle \nabla f , \nabla \Big( \frac{F}{t} +( \alpha - 1) ( af + \mathcal{R}) \Big) \rangle - 2 t ( \alpha - 1) h_{ij} f_i f_j \\ &= - 2 \langle \nabla f , \nabla F \rangle - 2 t ( \alpha - 1) \langle \nabla f , \nabla \mathcal{R} \rangle - 2 t ( \alpha - 1) a | \nabla f|^2 - 2 t ( \alpha - 1) h_{ij} f_i f_j, \end{aligned} \\ \label{eqn38} \begin{aligned} - \alpha a t \Delta f &= a t ( \alpha | \nabla f |^2 - \alpha f_t - \alpha a f - \alpha \mathcal{R} ) \\ &= aF + t ( \alpha - 1) ( | \nabla f |^2 ). \end{aligned} \end{gather} From \eqref{eq56}--\eqref{eqn38} we obtain \begin{equation}\label{eqn39} \begin{aligned} \Delta F - F_t &= t \Big( 2 f_{ij}^2 - 2 \alpha h_{ij} f_{ij} \Big) - 2 \langle \nabla f , \nabla F \rangle - \frac{F}{t} - 2 t ( \alpha - 1) h_{ij} f_i f_j \\ &\quad - \alpha t ( 2 (\operatorname{div}h)_i f_j - H_i f_j ) - 2 t ( \alpha - 1) \langle \nabla f , \nabla \mathcal{R} \rangle - \alpha t \Delta \mathcal{R} \\ &\quad + 2 t R_{ij} f_{ij} - 2 t ( \alpha - 1) a | \nabla f|^2 + t ( \alpha - 1) a | \nabla f|^2 + a F. \end{aligned} \end{equation} We now choose any two real numbers $ p, q > 0$ such that $ \frac{1}{p} + \frac{1}{q} = \frac{1}{\alpha}$ so that we can write \begin{align*} 2 f_{ij}^2 - 2 \alpha h_{ij} f_{ij} & = \frac{2 \alpha }{p} f_{ij}^2 + 2 \alpha \Big(\frac{1}{q} f_{ij}^2- h_{ij} f_{ij} \Big) \\ & \geq \frac{2 \alpha }{p} f_{ij}^2 - \frac { \alpha q}{2} h_{ij}^2, \end{align*} where we have used completing the square method to arrive at the last inequality. Also by Cauchy-Schwarz inequality we have $ f_{ij}^2 \geq \frac{1}{n} ( \Delta f)^2$. We can also write the boundedness condition on $h_{ij}$ as $ -( k_2 + k_3) g \leq h_{ij} \leq ( k_2 + k_3) g$ so that $$ \sup_M |h_{ij}|^2 \leq n ( k_2 + k_3)^2 $$ since $h_{ij}$ is a symmetric tensor. Therefore, \begin{equation}\label{eq58} t \Big( 2 f_{ij}^2 - 2 \alpha h_{ij} f_{ij} \Big) \geq \frac{2 \alpha t }{ np} ( \Delta f )^2 - \frac{\alpha n q}{2}t (k_2 + k_3)^2. \end{equation} Notice also that \begin{align*} \alpha t \Big( 2 (\operatorname{div}h)_i f_j - H_i f_j \Big) &= 2 \alpha t \Big( div \ h - \frac{1}{2} \nabla H \Big) f_j \\ &= 2 \alpha t \Big( g^{kl} \nabla_k h_{li} - \frac{1}{2} g^{kl} \nabla_i h_{kl} \Big) \nabla_j f \\ &\leq 2 \alpha t \Big( \frac{3}{2} |g| | \nabla h| \Big) | \nabla f| \\ &\leq 3 \alpha t n^{\frac{1}{2} } k_4 | \nabla f| . \end{align*} Putting together the last inequality, \eqref{eq58} and \eqref{eqn39} with the assumption that $R_{ij} \geq -k_1g$, we arrive at \begin{align*} ( \Delta - \partial_t ) F & \geq - 2 \langle \nabla f , \nabla F \rangle - \frac{2 \alpha t }{ np} (\Delta f )^2 - \frac{F}{t} + aF - 2 t ( \alpha - 1) k_3 | \nabla f |^2 \\ & - 2 t k_1 | \nabla f|^2 - 3 \alpha t n^{\frac{1}{2} } k_4 | \nabla f| - \frac{\alpha n q}{2} t (k_2 + k_3)^2 - \alpha t \Delta \mathcal{R}\\ & - 2 t ( \alpha - 1) \langle \nabla f , \nabla \mathcal{R} \rangle - t ( \alpha - 1) a |\nabla f|^2. \end{align*} Our calculation is valid in the ball $ \mathcal{B}_{2 \rho, T}$. Hence the desired claim follows. \end{proof} \section{Proof of Theorem \ref{thm54}} To prove Theorem \ref{thm54} we use the lemma above and the assumptions that the sectional curvature, $ \| \nabla h \|$, $| \mathcal{R} |$, $|\nabla \mathcal{R}| $ and $ | \Delta \mathcal{R} |$ are uniformly bounded on $M \times [0, T]$. Then we write equation \eqref{eq55} as \begin{equation}\label{eqp 1} \begin{aligned} ( \Delta - \partial_t) F &\geq - 2 \langle \nabla f, \nabla F \rangle - \frac{2 \alpha t }{ np} (\Delta f )^2 - \frac{F}{t} + aF - C_1 t | \nabla f |^2 \\ &\quad - C_2 t | \nabla f | - 2k_1 t | \nabla f |^2 - \frac{\alpha n q }{2} t ( k_2 + k_3)^2 , \end{aligned} \end{equation} where constants $C_1 >0$ depends on $\alpha$, \ $ \max \{a, 0 \}$, $\sup| h|$ and $ \|\nabla h\|$, and $C_2 >0$ depends on $\alpha, \ n$ and the space-time bounds of $ \|\nabla h\|$, $|\nabla \mathcal{R}|$, $| \Delta \mathcal{R} |$. We have used the following inequality $$ 3 \alpha n^{1/2} k_4 t | \nabla f | \leq 2t k_4 | \nabla f |^2 + 2 \alpha^2 nt k_4. $$ Furthermore, by using $$ - C_2 t | \nabla f | \geq - \delta^{-1} t C_2^2 - \delta t | \nabla f |^2 $$ for any number $\delta > 0$, we have \begin{equation}\label{eqp2} \begin{aligned} ( \Delta - \partial_t) F & \geq - 2 \langle \nabla f, \nabla F \rangle - \frac{2 \alpha t }{ np} (\Delta f )^2 - \frac{F}{t} + aF - C_3 t | \nabla f |^2 \\ &\quad - C_4 t - 2k_1 t | \nabla f |^2 - \frac{\alpha n q }{2} t ( k_2 + k_3)^2 , \end{aligned} \end{equation} where $C_3 >0$ depends on $C_1$ and $\delta$ and $C_4$ depends on $C_2$ and $\delta$. \subsection*{Estimating the cut-off function} A natural function that will be defined on $M$ is the distance function from a given point. Namely, let $ y \in M$ and define $d(x, y)$ for all $ x\in M$, where $d(\cdot, \cdot )$ is the geodesic distance. Note that $d$ is everywhere continuous except on the cut locus of $y$ and on the point where $x$ and $y$ coincide. It is then easy to see that $ | \nabla d | = g^{ij} \partial_i d \partial_j d = 1$ on $M \setminus \{ \{ y \} \cup cut(y) \} $. Let $d(x, y, t)$ be the geodesic distance between $x$ and $y$ with respect to the metric $g(t)$, we define a smooth cut-off function $ \varphi(x, t)$ with support in the geodesic ball $$ \mathcal{B}_{ 2\rho, T} := \big\{ ( x, t) \in M \times (0, T] : d(x, y, t) \leq 2\rho \big\}. $$ For any $C^2$-function $ \psi( s)$ on $[0, + \infty )$ with $ \psi(s) = 1$ on $ 0 \leq s \leq 1$ and $ \psi(s) = 0$ on $ 2 \leq s \leq + \infty$ such that $ - C_5 \leq \psi'(s) \leq 0$, $- C_6 \leq \psi''(s) \leq C_6$ and $- C_6 \psi \leq | \psi'|^2 \leq C_6 \psi$, where $C_5, C_6$ are absolute constants. Let $\rho \geq 1$ and define a smooth function $$ \varphi(x, t) = \psi \Big( \frac{d( x, p, t)}{\rho } \Big) \quad\text{and}\quad \varphi \big|_{ \mathcal{B}_{ 2\rho, T} } =1 . $$ We will apply maximum principle and invoke Calabi's trick to assume everywhere smoothness of $ \varphi(x, t)$ since $ \psi(s)$ is in general Lipschitz (see the argument of Li-Yau in \cite{LY86}). We need Laplacian comparison theorem \cite{SY94} to do some calculation on $\varphi(x, t)$. Let $M$ be a complete $n$-dimensional Riemannian manifold whose Ricci curvature is bounded from below by $ Rc \geq -(n-1) k_1 $ for some constant $ k_1 \in \mathbb{R}$, then the Laplacian of the distance function satisfies $$ \Delta d(x, y) \leq (n-1)\sqrt{|k_1|} \coth ( \sqrt{|k_1|} \rho), \quad \forall x \in M\; d(x, y) \geq \rho. $$ We need the following calculation \[ \frac{| \nabla \varphi |^2}{ \varphi} = \frac{| \psi' |^2 \cdot | \nabla d|^2 }{\rho^2 \psi} \leq \frac{C_6}{\rho^2} \] and by the Laplacian comparison theorem we have \begin{align*} \Delta \varphi &= \frac{\psi' \Delta d}{\rho} + \frac{\psi'' | \nabla d |^2}{\rho^2} \\ & \geq -\frac{C_5}{\rho}(n-1) \sqrt{k_1} \coth( \sqrt{k_1} \rho ) - \frac{C_6}{\rho^2}\\ & \geq -\frac{C_5\sqrt{k_1}}{\rho} - \frac{C_6}{\rho^2}. \end{align*} Next is to estimate time derivative of $\varphi$: consider a fixed smooth path $\gamma :[a, b] \to M$ whose length at time $t$ is given by $d(\gamma) = \int_a^b |\gamma'(s)|_{g(t)} ds$, where $s$ is the arc length along the path. Differentiating we get $$ \frac{\partial}{\partial t} (d(\gamma)) = \frac{1}{2} \int_a^b \big|\gamma'(s) \big|^{-1}_{g(t)} \frac{\partial g}{\partial t} \Big(\gamma'(s), \gamma'(s)\Big) ds = \int_\gamma h_{ij}(X, X) ds, $$ where $X$ is the unit tangent vector to the path $\gamma$. Now \begin{align*} \frac{\partial}{\partial t} \varphi & = \psi ' \frac{1}{\rho} \frac{d}{dt} (d(t)) = \psi' \frac{1}{\rho} \int_\gamma h_{ij}(X, X) ds \\ & \leq \frac{\sqrt{C_6} \psi^{1/2}}{\rho} (k_2 +k_3)^2 \int_\gamma dr \le \sqrt{C_6} (k_2 +k_3)^2 \end{align*} by choosing $0\le s\le 1$, $\rho \ge1$ and fixing the path to be of length not more than unit so that it always stays inside the geodesic ball $ \mathcal{B}_{ 2\rho, T}$. Hence we denote $$ ( \Delta - \partial_t ) \varphi \geq \Big( - \frac{C_5 \sqrt{k_1}}{\rho} - \frac{C_6 }{\rho^2} - \sqrt{C_6} (k_2 +k_3)^2\Big) =: C_7. $$ which will be used in the proof of our result. \begin{proof}[Proof of Theorem \ref{thm54}] Using the same notation as in the previous lemma, we write $ \widetilde{K} = ( k_2 + k_3)^2$. For a fixed $ \tau \in (0, T]$ and a smooth cut-off function $ \varphi(x,t)$ (chosen as before), we now estimate the inequality \eqref{eqp2} at the point $( x_0, t_0) \in \mathcal{B}_{2\rho, T } \subset ( M \times [0,T])$ such that $ d( x, x_0, t) < 2 \rho$. The argument follows from the identity \begin{equation}\label{eq38} (\Delta - \partial_t )( \varphi F) = 2 \nabla \varphi \nabla F + \varphi (\Delta - \partial_t )F+ F (\Delta - \partial_t ) \varphi. \end{equation} Suppose $ ( \varphi F)$ attains its maximum value at $( x_0, t_0) \in M \times [0, T]$, for $t_0 >0$. If $ ( \varphi F)(x_0, t_0) \leq 0$ for any $\rho \geq 1$, then the result holds trivially in $M \times [0, T]$ and we are done. Hence we may assume without loss of generality that there exists $ ( \varphi F)(x_0, t_0) > 0$. Then since $ ( \varphi F)(x,0) = 0$ for all $x \in M$, we have by the maximum principle that \begin{equation}\label{eq39} \nabla ( \varphi F) ( x_0, t_0) = 0, \quad \frac{\partial}{\partial t}( \varphi F) ( x_0, t_0) \geq 0, \quad \Delta ( \varphi F) ( x_0, t_0) \leq 0, \end{equation} where the function $( \varphi F)$ is being considered with support on $ \mathcal{B}_{2\rho} \times [0, T]$ and we have assumed that $ ( \varphi F) ( x_0, t_0) >0$ for $ t_0 >0$. By \eqref{eq39} we notice that $$ (\Delta - \partial_t )( \varphi F)( x_0, t_0) \leq 0. $$ Hence we have by using the inequality \eqref{eqp2} and equation \eqref{eq38}: \begin{equation} \label{eq3.10} \begin{aligned} 0 &\geq ( \Delta - \partial_t ) ( \varphi F) \\ &\geq 2 \nabla \varphi \nabla F + C_7 F + \varphi \Big\{ \frac{2 \alpha}{np} t_0 (\Delta f )^2 - 2 \langle \nabla f, \nabla F \rangle - \frac{F}{t_0} + a F\\ &\quad - C_3 t_0 | \nabla f |^2 - C_4t_0 - 2k_1 t_0 | \nabla f |^2 - \frac{\alpha n q }{2} t_0 ( k_2 + k_3)^2 \Big\}. \end{aligned} \end{equation} The above inequality holds in the part of $ \mathcal{B}_{2\rho, T }$ where $ \varphi(x, t)$ is strictly positive ($ 0 < \varphi(x, t) \leq 1$ ). Notice that since $\nabla (\varphi F) = 0$, the product rule tells us that we can always replace $-F\nabla \varphi$ with $\varphi \nabla F$ at the maximum point $(x_0, t_0)$. Indeed, the following equalities hold \begin{gather*} 2 \nabla \varphi \nabla F = 2 \varphi \frac{\nabla \varphi}{\varphi} \nabla F = 2 \frac{\nabla \varphi}{\varphi} (-F \nabla \varphi) = -2 F \frac{C_6}{\rho^2}, \\ - 2 \varphi \nabla F \cdot \nabla f = 2 F \nabla \varphi \cdot \nabla f = 2 F | \nabla f| \varphi \frac{| \nabla \varphi |}{\varphi } \geq -2 \frac{\sqrt{C_6}}{\rho} | \nabla f| \varphi^{1/2} F. \end{gather*} Multiplying \eqref{eq3.10} by $( t_0 \varphi)$, after some simple calculations involving the last two identities at the maximum point we obtain \begin{align*} 0 &\geq -2 t_0 \frac{C_6}{\rho^2} \varphi F - \varphi^2 F - 2 t_0 \frac{\sqrt{C_6}}{\rho} | \nabla f| \varphi^{\frac{3}{2}} F + C_7 t_0 \varphi F + a \varphi^2 t_0 F \\ &\quad + \varphi \frac{2 t_0^2}{n} \Big( \frac{\alpha}{p} ( \varphi | \nabla f |^2 - \varphi ( f_t + a f + \varphi \mathcal{R} ) \Big)^2 - \frac{\alpha n q }{2} t^2_0 \widetilde{K} \varphi^2 \\ &\quad - C_3 t^2_0 \varphi^2| \nabla f |^2 - C_4 \varphi^2t^2_0 - 2k_1 t^2_0 \varphi^2| \nabla f |^2 \\ &\geq -2 t_0 \frac{C_6}{\rho^2} \varphi F - \varphi^2 F - 2 t_0 \frac{\sqrt{C_6}}{\rho} | \nabla f| \varphi^{\frac{3}{2}} F + C_7 t_0 \varphi F + a \varphi^2 t_0 F \\ &\quad + \varphi \frac{2 t_0^2}{n} \Big( \frac{\alpha}{p} ( \varphi | \nabla f |^2 - \varphi ( f_t + a f + \varphi \mathcal{R} ) \Big)^2 - \frac{\alpha n q }{2} t^2_0 \widetilde{K} \varphi^2 - C_8 t^2_0 \varphi^2| \nabla f |^2, \end{align*} where $C_8$ depends on $C_3$, $C_4$ and $k_1$. Using a similar technique as in Li-Yau paper \cite{LY86}, when $ t_0 > 0$, let $ y = \varphi | \nabla f |^2 $ and $ z = \varphi ( f_t + af + \mathcal{R})$ to obtain $ \varphi^2 | \nabla f |^2 = \varphi y \leq y $, $ y^{1/2}( y - \alpha z) = \frac{1}{t_0} | \nabla f | \varphi^{\frac{3}{2}} F $ and $ \varphi F = t_0 ( y - \alpha z )$. We obtain \begin{equation} \label{eqp6} \begin{aligned} 0 &\geq \frac{2 t_0^2}{n} \Big( \frac{\alpha}{p} ( y-z)^2 -\frac{C_8}{2}n y - \frac{n \sqrt{C_6}}{\rho} y^{1/2}(y- \alpha z) \Big) \\ &\quad - \frac{\alpha n q }{2} t^2_0 \widetilde{K} \varphi^2 + \Big( C_7 t_0 - 2 t_0 \frac{C_6}{\rho^2} - 1 + a \varphi t_0 \Big)( \varphi F). \end{aligned} \end{equation} Notice that by direct calculations, \begin{align*} ( y- z)^2 &= [ \frac{1}{\alpha}( y- \alpha z) + \frac{\alpha-1}{\alpha}y]^2 \\ &= \frac{1}{\alpha^2}( y- \alpha z)^2 + \frac{(\alpha-1)^2}{\alpha^2} y^2 + \frac{2(\alpha-1)}{\alpha^2} y( y- \alpha z). \end{align*} Then, the first term in the right hand side of inequality \eqref{eqp6} can be simplified as follows: \begin{align*} &\frac{2 t_0^2}{n} \Big \{ \frac{\alpha}{p} \Big[ ( y-z)^2 - \frac{C_8 np}{ 2\alpha} y - \frac{n p}{\alpha} \frac{\sqrt{C_6}}{\rho}y( y - \alpha z) \Big] \Big \} \\ &= \frac{2 t_0^2}{n} \Big \{ \frac{\alpha}{p} \Big[ \frac{1}{\alpha^2} ( y - \alpha z)^2 + \Big( \frac{( \alpha-1)^2}{\alpha^2} y^2 -\frac{C_8 np}{ 2\alpha} y \Big) \\ &\quad + \Big( \frac{2(\alpha-1)}{\alpha^2} y - \frac{np}{\alpha} \frac{\sqrt{C_6}}{\rho} y^{1/2} \Big)( y- \alpha z) \Big] \Big \} \\ &\geq \frac{2 t_0^2}{n} \Big \{ \frac{1}{\alpha p}( y- \alpha z)^2 - \frac{C^4_8 \alpha n^2 p }{ 16( \alpha-1)^2 } - \frac{C_6 \alpha n^2 p}{8 \rho^2 (\alpha-1)} ( y- \alpha z ) \Big \} \\ &= \frac{2}{\alpha np} ( \varphi F)^2 - \frac{C_8^2 \alpha n p}{8 (\alpha-1)^2 } t_0^2 - \frac{C_6 \alpha n p}{4 \rho^2 (\alpha-1)} t_0 ( \varphi F ). \end{align*} We have used the inequality of the form $ ax^2 - b x \geq -\frac{b^2}{ 4a}$, $( a, b > 0)$, to compute \begin{gather*} \frac{(\alpha-1)^2}{\alpha^2} y^2 - \frac{C^2_8np}{2 \alpha} y \geq - \frac{C_8^4 n^2 p^2}{ 16 (\alpha-1)^2 } ,\\ \frac{2(\alpha-1 )}{\alpha^2} y - \frac{np}{\alpha} \frac{\sqrt{C_6}}{\rho} y^{1/2} \geq - \frac{C_6 n^2 p^2}{8(\alpha-1) \rho^2}. \end{gather*} Therefore, putting all these together into \eqref{eqp6}, we get a quadratic polynomial in $( \varphi F)$ \begin{align*} 0 &\geq \frac{2}{\alpha np} ( \varphi F)^2 + \Big( C_7 t_0 - 2 t_0 \frac{C_6}{\rho^2} - 1 + a t_0 - \frac{C_6 \alpha n p}{4 \rho^2 (\alpha-1)} t_0 \Big) ( \varphi F ) \\ &\quad - \Big( \frac{C_8^2 \alpha n p}{ 8 (\alpha-1)^2 } t_0^2 + \frac{\alpha n q }{2} t^2_0 \widetilde{K} \varphi^2 \Big). \end{align*} Then we develop a formula for quadratic inequality of the form $ ax^2 + b x + c \leq 0$, for $x \in \mathbb{R}$. Note that when $ a > 0$ and $ c < 0$, then $ b^2 - 4ac > 0$ and we have an upper bound \begin{align}\label{eqp9} x \leq \frac{-b + \sqrt{b^2 - 4ac}}{ 2a} \leq \frac{1}{a} \Big \{ - b + \sqrt{-ac} \Big \} . \end{align} The next is to make more explicit the term \begin{align*} b &:= \Big(C_7 t_0 - 2 t_0 \frac{C_6}{\rho^2} - 1 + a t_0 - \frac{C_6 \alpha n p}{4 \rho^2 (\alpha-1)} t_0\Big)\\ & = \Big( - \frac{C_5 \sqrt{k_1}}{\rho}t_0 - \frac{C_6 }{\rho^2}t_0 - \sqrt{C_6} \widetilde{K}t_0 - 2 t_0 \frac{C_6}{\rho^2} -1 + a t_0 - \frac{C_6 \alpha n p}{4 \rho^2 (\alpha-1)} t_0\Big) \\ &= - \Big(\frac{C_9}{\rho^2} t_0 \Big( \frac{\alpha p}{\alpha - 1} + \rho \sqrt{k_1} + \rho^2 \widetilde{K} \Big) -a t_0 + 1 \Big), \end{align*} where $C_9 >0$ depends on $C_6$ and $n$. Hence, we have by applying \eqref{eqp9} \begin{align*} \varphi F & \leq \frac{\alpha n p}{2} + \frac{\alpha n p}{2 } \Big \{ \frac{C_9}{\rho^2} t_0 \Big( \frac{\alpha p}{\alpha - 1} + \rho \sqrt{k_1} +\rho^2( k_2 + k_3 )^2 \Big) - a t_0 \Big \} \\ &\quad + \frac{\alpha n p }{ 4(\alpha -1) } C_8 t_0 +\frac{\alpha n }{2} ( k_2 + k_3) t_0 \varphi \sqrt{pq}. \end{align*} To obtain the required bound on $F(x, \tau)$ for an appropriate range of $ x \in M$, we take $ \varphi(x, \tau) \equiv 1$ whenever $d( x , x_0, \tau ) < 2 \rho$ and since $( x_0, t_0)$ is the maximum point for $( \varphi F)$ in $ \mathcal{B}_{2\rho, T}$, we have $$ F(x, \tau) = ( \varphi F)( x, \tau) \leq ( \varphi F)(x_0, t_0) $$ for all $ x \in M$, such that $d( x , x_0, \tau ) < \rho$ and $ \tau \in ( 0, T]$ was arbitrarily chosen, then we have the conclusion in a more compact way, that \begin{align}\label{eq511} \sup_{ x \in \mathcal{B}_{2\rho}} \Big \{ | \nabla f |^2 - \alpha f_t - \alpha a f - \alpha \mathcal{R} \Big \} \leq \frac{\alpha np}{2 t} (1 - a t) + C_{10}, \end{align} where $C_{10}$ depends on $ \alpha, \tau, \rho, k_1, k_2, k_3, n, p$ and $q$. This completes the proof. . \end{proof} \begin{remark}\label{rmk31} \rm Global estimate follows by letting $ \rho \rightarrow \infty$ for all $ t > 0$. For instance, if we set $ p = 2 \alpha = q $ and allow $ \rho$ goes to infinity, we have the estimate \begin{align}\label{eq512} \frac{| \nabla u|^2}{u^2} - \alpha \frac{u_t}{u} - \alpha a \log u - \alpha \mathcal{R} \leq \frac{\alpha^2 n }{t} + C_{11} \end{align} where $C_{11}$ is an absolute constant depending on $n, \tau , \alpha$ and the upper bounds of $|Rc|, | \nabla \mathcal{R}|$, $| \Delta \mathcal{R}|$, $|h|$, $| \nabla h|$ and $ - \min \{a, 0 \}$. \end{remark} As an application of the global gradient estimates derived in Theorem \ref{thm54}, we obtain the following result for the corresponding Harnack estimates. \begin{corollary}[Hanarck estimates]\label{corN5} With the same assumption as in Theorem \ref{thm54}. The estimate \begin{align}\label{eq514} \frac{u(x_1, t_1)}{u(x_2, t_2)^{e^{-a(t_2 -t_1)}}} \leq \Big( \frac{t_2}{t_1}\Big)^{ \alpha n} \exp \Big \{ \frac{d^2(x_1, x_2)}{4(t_2 - t_t)} + C_{12}(t_2 - t_1) \Big \} \end{align} holds for all $ (x, t) \in M \times (0, T]$, where $C_{12}$ is an absolute constant depending on $n, \tau , \alpha$ and the upper bounds of $|Rc|$, $|\mathcal{R}|$, $| \nabla \mathcal{R}|$, $| \Delta \mathcal{R}|$, $|h|$, $| \nabla h|$ and $ - \min \{a, 0 \}$. Here $d(x_1, x_2)$ is the geodesic distance between points $x_1$ and $x_2$. The space-time path $ \gamma :[t_1, t_2] \to M$ connects points $x_1= \gamma(t_1)$ and $x_2 = \gamma(t_2)$ in $M$. Denote $|\dot{\gamma}(t) | = d(x_1, x_2)/(t_2 - t_1)$, where the norm $| \cdot|$ depends on $t$. \end{corollary} \begin{proof} Equation \eqref{eq512} implies \begin{align}\label{eq515} f_t \geq \frac{1}{\alpha} |\nabla f|^2 - \frac{\alpha n }{t} - a f - \mathcal{R} - \frac{1}{\alpha} C_{11}. \end{align} Straight computation yields \begin{align*} &e^{at_2} f(x_2, t_2) - e^{a t_1} f(x_1, t_1)\\ &= \int_{t_1}^{t_2} \frac{d}{dt} \Big( e^{at} f( \gamma(t), t) \Big) dt\\ &= \int_{t_1}^{t_2} \Big \{ e^{at} (f_t + \langle \nabla f( \gamma(t), t), \dot{\gamma}(t) \rangle) + a e^{at} f \Big \} dt \\ & \geq \int_{t_1}^{t_2} \Big \{ e^{at} \Big( \frac{1}{\alpha} |\nabla f|^2 - \frac{\alpha n }{t} - \mathcal{R} - \frac{1}{\alpha} C_{11} + \langle \nabla f( \gamma(t), t), \dot{\gamma}(t) \rangle \Big)\Big\} dt \\ & \geq - e^{at_1} \Big( \int_{t_1}^{t_2} \Big \{ \frac{\alpha | \dot{\gamma}(t) |^2}{4} +\frac{\alpha n }{t} +\frac{1}{\alpha} C_{11} \Big\} dt \Big) \\ & = - e^{at_1} \Big( \int_{t_1}^{t_2} \frac{\alpha d^2(x_1, x_2)}{4(t_2 - t_t)^2} dt + \log \Big(\frac{t_2}{t_1}\Big)^{\alpha n} + C_{12}(t_2 - t_1) \Big). \end{align*} In the computation above we have used \eqref{eq515} to arrive at the inequality in the third line, whereas the quantity $e^{at}$ coming up in the first line helped to get rid of the term $af$ in \eqref{eq515}. We also used an inequality of the form $Ay^2 + B y \geq - B^2 / 4A$ to obtain the inequality in the fourth line and denoted $|\dot{\gamma}(t) | = d(x_1, x_2)/(t_2 - t_1)$ by defining a curve $\eta$ in $M \times (0, T_\varepsilon)$, $\eta :[t_1, t_2] \to M\times (0, T_\varepsilon)$, by $\eta(s) = (\gamma(s), s)$. Positive constant $C_{12}$ depends on $\alpha, C_{11}$ and the uniform bound for $|\mathcal{R}|$. Now, multiplying both sides by $e^{-a t_1}$ the expression in the left hand side becomes \[ f(x_1, t_1) - e^{a(t_2-t_1)} f(x_2, t_2) = \log \Big(\frac{u(x_1, t_1)}{u(x_2, t_2)^{e^{a(t_2-t_1)}}} \Big). \] By exponentiation we arrive at \[ u(x_1, t_1) \leq u(x_2, t_2)^{e^{a(t_2-t_1)}} \Big(\frac{t_2}{t_1} \Big)^{\alpha n} \exp \Big \{\int_{t_1}^{t_2} \frac{d^2(x_1, x_2)}{4(t_2 - t_t)^2} dt + C_{12}(t_2 - t_1) \Big\}, \] which concludes the proof of the corollary. \end{proof} \subsection*{Acknowledgements} The author wishes to thank the anonymous referees for their useful comments and for informing him that reference \cite{Ba13} will appear in Advances in Geometry. His research is supported by the TETFund of Federal Government of Nigeria and University of Sussex, United Kingdom. \begin{thebibliography}{00} \bibitem{Ab1} A. Abolarinwa; \emph{Gradient estimates for heat-type equations on evolving manifolds}, Journal of Nonlinear Evolution Equation and Applications. To appear. \bibitem{Ab2} A. Abolarinwa; \emph{Differential Harnack inequalities for nonlinear parabolic equation on time-dependent metrics}, Adv. Th. Appl. Math., \textbf{9}(2) (2014), 155-166. \bibitem{Ba13} M. B\v{a}ile\c{s}teanu; \emph{Gradient estimates for the heat equation under the Ricci-harmonic map flow}, arXiv:1309.0139. \bibitem{BCP} M. B\v{a}ile\c{s}teanu, X. Cao, A. Pulemotov; \emph{Gradient estimates for the heat equation under the Ricci flow}, J. Funct. Anal., \textbf{258} (2010), 3517--3542. \bibitem{CaH09} X. Cao, R. S. Hamilton; \emph{Differential Harnack estimates for time-dependent heat equations with potentials}, Geom. Funct. Anal., \textbf{19}(4)(2009), 989--1000. \bibitem{CZ11} X. Cao, Z. Zhang; \emph{Differential Harnack estimates for parabolic equations}, Com. and Diff. Geom. Springer Proceedings in Mathematics \textbf{8}, (2011), 87--98. \bibitem{CCG} B. Chow, S.-c Chu, D. Glickenstein, C. Guenther, J. Isenberg, T. Ivey, D. Knopf, L. Peng; Luo, N. Feng; L. Nei; \emph{The Ricci Flow: Techniques and Applications. Parts I--III, Geometric-analytic Aspects}, Math. Surveys and Monographs, 135, 144, 163, AMS, Providence, RI, (2007, 2008, 2010). \bibitem{CLN06} B. Chow, P. Lu, L. Ni; \emph{Hamilton's Ricci Flow: An Introduction}. American Mathematics Society, (2006). \bibitem{HHL12} G. Huang, Z. Huang, H. Li; \emph{Gradient estimates and differential Harnack inequalities for a nonlinear parabolic equation on Riemannian manifolds}, Annals of Global Analysis and Geometry, \textbf{23}(3) (1993), 209--232. \bibitem{HM10} G. Huang, B. Ma; \emph{Gradient estimates for a nonlinear parabolic equation on Riemannian manifolds}, Arch. Math. 94 (2010), 265--275. \bibitem{KZh08} S. Kuang, Q. S. Zhang; \emph{A gradient estimate for all positive solutions of the conjugate heat equation under Ricci flow}, J. Funct. Anal., \textbf{255}(4)( 2008), 1008--1023. \bibitem{LY86} P. Li, S-T. Yau; \emph{On the parabolic kernel of the Schr\"odinger operator}, Acta Mathematica \textbf{156}(1) (1986), 153--201. \bibitem{Ni07} L. Ni; \emph{A matrix Li-Yau-Hamilton estimate for K\"ahler-Ricci flow}, J. Differential Geom., \textbf{75}(2) (2007), 303--358. \bibitem{SY94} R. Schoen, S.-T. Yau; \emph{Lectures on differential geometry}. International Press, Cambridge, MA, (1994). \end{thebibliography} \end{document}