Geometric flows and differential Harnack estimates for heat equations with potentials

Abstract

Let $M$ be a closed Riemannian manifold with a Riemannian metric $g_{ij}(t)$ evolving by a geometric flow $\partial _{t}g_{ij} = -2{S}_{ij}$, where $S_{ij}(t)$ is a symmetric two-tensor on $(M, g(t))$. Suppose that $S_{ij}$ satisfies the tensor inequality $2{\mathcal H}(S, X)+{\mathcal E}(S,X) \ge 0$ for all vector fields $X$ on $M$, where ${\mathcal H}(S, X)$ and ${\mathcal E}(S,X)$ are introduced in Definition 1 below. Then, we shall prove differential Harnack estimates for positive solutions to time-dependent forward heat equations with potentials. In the case where $S_{ij} = R_{ij}$, the Ricci tensor of $M$, our results correspond to the results proved by Cao and Hamilton (Geom Funct Anal 19:983–989, 2009). Moreover, in the case where the Ricci flow coupled with harmonic map heat flow introduced by Müller (Ann Sci Ec Norm Super 45(4):101–142, 2012), our results derive new differential Harnack estimates. We shall also find new entropies which are monotone under the above geometric flow.

1 Introduction

The main purpose of the current article is to study time-dependent heat equations with potentials on closed Riemannian $n$-manifolds $M$ evolving by the geometric flow

$$\begin{aligned} \frac{\partial }{\partial t}g_{ij} = - 2S_{ij}, \end{aligned}$$

(1)

where $S_{ij}(t)$ is a symmetric two-tensor on $(M, g(t))$. A typical example would be the case where $S_{ij} = R_{ij}$ is the Ricci tensor and $g(t)$ is a solution of the Ricci flow introduced by Hamilton [8]. We shall derive Li–Yau type differential Harnack inequalities [13] for positive solutions to the following heat equation with a potential term:

$$\begin{aligned} \frac{\partial f}{\partial t} = {\Delta }f + Sf, \end{aligned}$$

(2)

where the symbol $\Delta$ stands for the Laplacian of the evolving metric $g(t)$ and $S ={g}^{ij}S_{ij}$ is the trace of $S_{ij}$. In the current article, we shall use the Einstein summation convention. For simplicity, we omit $g(t)$ in the above notation. All geometric operators are with respect to the evolving metric $g(t)$. Notice also that we have

$$\begin{aligned} \frac{\partial }{\partial t}\left( \,\,\int _{M}f{\mathrm{d}}{\mu }_{g} \right) =\int _{M}\left( \frac{\partial f}{\partial t} -S \right) {\mathrm{d}}{\mu }_{g} = \int _{M}{\Delta }_{}f{\mathrm{d}}{\mu }_{g} = 0. \end{aligned}$$

The main results of the current article are Theorems A, B, C, D and E stated below, which can be seen as natural generalizations of results proved by Cao and Hamilton [4]. See also Sect. 2 below.

The study of differential Harnack inequalities for parabolic equations originated with the work of Li and Yau [13]. They first proved a gradient estimate for the heat equation using the maximal principle. By integrating the gradient estimate along a space-time path, a classical Harnack inequality was derived. Therefore, Li–Yau type gradient estimate is often called differential Harnack inequality. Hamilton adapted similar techniques to prove Harnack inequalities for the Ricci flow [10] and the mean curvature flow [11]. Many authors used similar techniques to prove Harnack inequalities for geometric flows. For instance, see [1–3, 5–7, 12, 17–19].

To state the main results of the current article, we shall introduce evolving tensor quantities associated with the tensor ${S}_{ij}$.

Definition 1

Suppose that $g(t)$ evolves by the geometric flow (1) and let $X=X^{i}\frac{\partial }{\partial x^{i}} \in \Gamma (TX)$ be a vector field on $M$. We define

$$\begin{aligned} {\mathcal I}(S, X)&= (R^{ij} - S^{ij})X_{i}X_{j}, \\ {\mathcal H}(S, {X})&= \frac{\partial S}{\partial t} + \frac{S}{t} + 2{\nabla }_{i}SX^{i} + 2{S}^{ij}X_{i}{X}_{j}, \\ {\mathcal E}(S, {X})&= \left( \frac{\partial S}{\partial t} - \Delta S - 2|S_{ij}|^{2} \right) - 2 (2{\nabla }^{i}S_{i\ell } - {\nabla }_{\ell }S ){X}^{\ell } + 2 {\mathcal I}(S, X), \end{aligned}$$

where $R^{ij} = {g}^{ik}g^{j \ell }R_{k \ell }$, $S^{ij} = {g}^{ik}g^{j \ell }S_{k \ell }$, $S = g^{ij}S_{ij}$, ${\nabla }^{i} = g^{ij}{\nabla }_{j}$ and $X_{k} = {g}_{ik}X^{i}$.

We notice that these quantities are also introduced by Müller [15] to prove the monotonicity of Perelman type reduced volume under (1).

The first main result of the current article is as follows:

Theorem A

Suppose that $g(t)$ evolves by the geometric flow (1) on a closed oriented smooth $n$-manifold $M$ and

$$\begin{aligned} 2{\mathcal H}(S, X)+{\mathcal E}(S,X) \ge 0 \end{aligned}$$

(3)

holds for all vector fields $X$ and all time $t \in [0, T)$ for which the flow exists. Let $f$ be a positive solution to the heat equation (2), $u = -\log f$, and

$$\begin{aligned} H_{S} = 2{\Delta }u - |\nabla u|^{2}-3{S} - \frac{2n}{t}. \end{aligned}$$

Then, $H_{S} \le 0$ for all time $t \in (0, T)$.

In the case of the heat equation without the potential term, we shall also prove

Theorem B

Suppose that $g(t)$ evolves by the geometric flow (1) on a closed oriented smooth $n$-manifold $M$ and

$$\begin{aligned} {\mathcal I}(S, X) \ge 0 \end{aligned}$$

(4)

holds for all vector fields $X$ and all time $t \in [0, T)$ for which the flow exists. Let $f \ (< 1)$ be a positive solution to the heat equation $\frac{\partial f}{\partial t} = \Delta f$, $u = -\log f$ and

$$\begin{aligned} H_{S} = |\nabla u|^{2} - \frac{u}{t}. \end{aligned}$$

Then, $H_{S} \le 0$ holds for all time $t \in (0, T)$. Hence we have the following on $(0, T)$,

$$\begin{aligned} |\nabla f|^{2} \le -\frac{f^{2}}{t}\log f. \end{aligned}$$

On the other hand, a similar technique with the proof of Theorem A also enables us to prove

Theorem C

Suppose that $g(t)$ evolves by the geometric flow (1) on a closed oriented smooth $n$-manifold $M$ and $2{\mathcal H}(S, X) + {\mathcal E}(S,X) \ge 0$ holds for all vector fields $X$ and all time $t \in [0, T)$ for which the flow exists. Let $f$ be a positive solution to the heat equation (2), $v = -\log f- \frac{n}{2}\log (4{\pi }t)$ and

$$\begin{aligned} P_{S} = 2{\Delta }v - |\nabla v|^{2}-3{S} + \frac{v}{t} - d\frac{n}{t}, \end{aligned}$$

where $d$ is any constant. Then for all time $t \in (0, T)$,

$$\begin{aligned} \frac{\partial }{\partial t}(tP_{S})&= {\Delta }(t{P}_{S}) - 2{\nabla }^{i}(tP_{S}){\nabla }_{i}v -2t\left| \nabla _{i}\nabla _{j}v - S_{ij} - \frac{1}{2t}g_{ij}\right| ^{2} \\&-t\left( 2{\mathcal H}(S, {\nabla }v)+{\mathcal E}(S, {\nabla }v) \right) \end{aligned}$$

and $\max (t{P}_{S})$ is non-increasing.

Moreover, inspired by the works of Perelman [17] and Cao and Hamilton [4], we shall introduce two entropies which are associated with the above Harnack quantities. The first one is associated with $H_{S}$.

Theorem D

Under the same assumption with Theorem A, we define

$$\begin{aligned} {\mathcal F}_{S} = \int _{M}t^{2}\mathrm{e}^{-u}H_{S}{\mathrm{d}}{\mu }_{g}. \end{aligned}$$

Then for all time $t \in (0, T)$, we have ${\mathcal F}_{S} \le 0$ and

$$\begin{aligned} \frac{\mathrm{d}}{{\mathrm{d}}t} {\mathcal {F}}_{S} \le 0. \end{aligned}$$

Moreover, suppose that ${\mathcal H}(S, X) \ge 0$ and ${\mathcal E}(S, X) \ge 0$ holds for all vector fields $X$ and all time $t \in [0, T)$ for which the flow exists. If $\frac{\mathrm{d}}{{\mathrm{d}}t} {\mathcal F}_{S} = 0$ holds for some time $t$, then the following holds:

$$\begin{aligned} S_{ij} = -\frac{1}{t}g_{ij}, \quad {\nabla }u = 0, \, {\mathcal H}(S, {\nabla }u) = 0, \, {\mathcal E}(S, {\nabla v}) = 0 \end{aligned}$$

(5)

The second one is associated with $P_{S}$ as follows.

Theorem E

Under the same assumption with Theorem C, we define

$$\begin{aligned} {\mathcal W}_{S} = \int _{M}tP_{S}(4{\pi }t)^{-\frac{n}{2}}\mathrm{e}^{-v}{\mathrm{d}}{\mu }_{g}. \end{aligned}$$

Then for all time $t \in (0, T)$, we have

$$\begin{aligned} \frac{\mathrm{d}}{{\mathrm{d}}t} {\mathcal W}_{S} \le 0. \end{aligned}$$

Moreover, suppose that ${\mathcal H}(S, X) \ge 0$ and ${\mathcal E}(S,X) \ge 0$ holds for all vector fields $X$ and all time $t \in [0, T)$ for which the flow exists. If $\frac{\mathrm{d}}{{\mathrm{d}}t} {\mathcal W}_{S} = 0$ holds for some time $t$, then the following holds:

$$\begin{aligned} S_{ij} - {\nabla }_{i}{\nabla }_{j}v + \frac{1}{2t}g_{ij}=0, \quad {\mathcal H}(S, {\nabla }v)=0, \, {\mathcal E}(S, {\nabla v}) = 0. \end{aligned}$$

(6)

2 Examples

2.1 The Ricci flow

Let $g(t)$ be a solution to the Ricci flow:

$$\begin{aligned} \frac{\partial }{\partial t}g_{ij} = - 2R_{ij}. \end{aligned}$$

Namely, we have $S_{ij} = {R}_{ij}$ and $S = R$ the scalar curvature. Notice that it is known that the scalar curvature $R$ evolves by $\frac{\partial R}{\partial t} - \Delta R - 2|R_{ij}|^{2} = 0$. Moreover, we have the twice contracted second Bianchi identity $2{\nabla }^{i}R_{i\ell } - {\nabla }_{\ell }R=0$. Hence, we have ${\mathcal E}(S, X) =0$ in this case. Therefore, (3) is equivalent to

$$\begin{aligned} {\mathcal H}(S, X) = \frac{\partial R}{\partial t} + \frac{R}{t} + 2{\nabla }_{i}RX^{i} + 2{R}^{ij}X_{i}X_{j} \ge 0. \end{aligned}$$

This tells us that $g(t)$ has weakly positive curvature operator (see also [9]). Moreover, we have ${\mathcal I}(S, X) =0$. Hence, (4) holds. Therefore, Theorems A, B, C, D and E in the case where $S_{ij} = {R}_{ij}$ just correspond to the results proved by Cao and Hamilton [4]. Notice also that (5) particularly tells us that $R_{ij} = -\frac{1}{t}g_{ij}$, i.e., $g(t)$ is Einstein. Similarly, (6) implies $R_{ij} +{\nabla }_{i}{\nabla }_{j}(-v) + \frac{1}{2t}g_{ij}=0$. This tells us that $g(t)$ is an expanding gradient Ricci soliton. Since it is known [17] that any expanding Ricci soliton on a closed manifold must be Einstein, $g(t)$ is Einstein.

2.2 Bernhard List’s flow

List [14] introduced a geometric flow closely related to the Ricci flow:

$$\begin{aligned} \frac{\partial }{\partial t}g_{ij}&= - 2R_{ij} + 4{\nabla }_i \psi {\nabla }_j \psi , \\ \frac{\partial \psi }{\partial t}&= {\Delta } \psi , \end{aligned}$$

where $\psi : M \rightarrow {\mathbb R}$ is a smooth function. If we set $S_{ij} = {R}_{ij}-2{\nabla }_i \psi {\nabla }_j \psi$, it is clear that the first of List’s flow has the form (1). Notice also that we have $S =R -2|\nabla \psi |^{2}$. List [14] pointed out that $S$ satisfies the following evolution equation:

$$\begin{aligned} \frac{\partial S}{\partial t} -\Delta S - 2|S_{ij}|^{2} = 4|\Delta \psi |^{2}. \end{aligned}$$

On the other hand, we get $2{\nabla }^{i}S_{i\ell } - {\nabla }_{\ell }S = 2{\nabla }^{i}({R}_{i\ell }-2{\nabla }_i \psi {\nabla }_\ell \psi )-{\nabla }_{\ell }(R -2|\nabla \psi |^{2}) = -4{\nabla }^{i}({\nabla }_i \psi {\nabla }_\ell \psi ) - 2{\nabla }_{\ell }(\nabla ^{k} \psi \nabla _{k} \psi ) = -4{\Delta }\psi \nabla _{\ell } \psi$, where notice the twice contracted second Bianchi identity $2{\nabla }^{i}R_{i\ell } = {\nabla }_{\ell }R$. Therefore, we have

$$\begin{aligned} {\mathcal E}(S, X) = 4|\Delta \psi |^{2} + 8{\Delta }\psi \nabla _{\ell } \psi X^{\ell } + 4{\nabla }^i \psi {\nabla }^j \psi X_{i}X_{j} = 4|\Delta \psi + {\nabla }_{X}\psi |^{2} \ge 0. \end{aligned}$$

(7)

In particular, (3) is particularly satisfied if

$$\begin{aligned} {\mathcal H}(S, X) \ge 0. \end{aligned}$$

On the other hand, we have ${\mathcal I}(S, X) = {\nabla }^i \psi {\nabla }^j \psi X_{i}X_{j} = ({\nabla }_{X}\psi )^{2} \ge 0$. Hence, (4) holds. Therefore, Theorems A, B, C, D and E just correspond to the result proved by Fang [6]. Notice also that, under the situation on Theorem D, we particularly have the following by (5) and (7):

$$\begin{aligned} {R}_{ij}-2{\nabla }_i \psi {\nabla }_j \psi = -\frac{1}{t}g_{ij}, \quad {\nabla }u = 0, \, \Delta \psi + {\nabla }^{i}\psi {\nabla }_{i}u = 0. \end{aligned}$$

Since it follows that $\Delta \psi =0$, $\psi$ must be a harmonic function on the closed manifold $M$. This implies that $\psi$ is a constant for the time $t$. Therefore, we have ${R}_{ij}= -\frac{1}{t}g_{ij}$, i.e., $M$ is Einstein.

2.3 Rent Müller’s flow

Let $(Y, h)$ be a fixed Riemannian manifold. Let $(g(t), \phi (t))$ be the couple consisting of a family of metric $g(t)$ on $M$ and a family of maps $\phi (t)$ from $M$ to $Y$. We call $(g(t), \phi (t))$ a solution of Rent Müller’s flow [16] (also known as the Ricci flow coupled with harmonic map heat flow) with coupling function $\alpha (t) \ge 0$ if

$$\begin{aligned} \frac{\partial }{\partial t}g_{ij}&= - 2R_{ij} + 2\alpha (t){\nabla }_i \phi {\nabla }_j \phi , \\ \frac{\partial \phi }{\partial t}&= {\tau } _g \phi , \end{aligned}$$

where ${\tau } _g \phi$ is the tension field of the map $\phi$ with respect to the metric $g(t)$. List’s flow is a special case of this flow. If we set $S_{ij} = {R}_{ij}-\alpha (t){\nabla }_i \phi {\nabla }_j \phi$, the first of Müller’s flow has the form (1). Notice that $S =R -2\alpha (t)|\nabla \phi |^{2}$ holds. Müller [16] proved that $S$ satisfies

$$\begin{aligned} \frac{\partial S}{\partial t} -\Delta S - 2|S_{ij}|^{2} = 2\alpha (t)|\tau _{g} \phi |^{2} -\left( \frac{\partial \alpha (t)}{\partial t}\right) |\nabla \phi |^{2}. \end{aligned}$$

Since we are able to get $2{\nabla }^{i}S_{i\ell } - {\nabla }_{\ell }S = -2\alpha (t){\tau }_{g}\phi \nabla _{\ell } \phi$, the following holds:

$$\begin{aligned} {\mathcal E}(S, X)&= 2\alpha (t)|\tau _{g} \phi |^{2} -\left( \frac{\partial \alpha (t)}{\partial t}\right) |\nabla \phi |^{2}+4\alpha (t){\tau }_{g}\phi \nabla _{\ell } \phi X^{\ell }+ 2\alpha (t){\nabla }^i \phi {\nabla }^j \phi X_{i}X_{j} \\&= 2{\alpha (t)}\left| \tau _{g} \phi + {\nabla }_{X}\phi \right| ^{2} -\left( \frac{\partial \alpha (t)}{\partial t}\right) |\nabla \phi |^{2}. \end{aligned}$$

Therefore, ${\mathcal E}(S, X) \ge 0$ holds if $\alpha (t) \ge 0$ is non-increasing. In this case, (3) is particularly satisfied if

$$\begin{aligned} {\mathcal H}(S, X) \ge 0. \end{aligned}$$

Notice also that ${\mathcal I}(S, X) = \alpha (t){\nabla }^i \phi {\nabla }^j \phi X_{i}X_{j} = \alpha (t)({\nabla }_{X}\phi )^{2} \ge 0$. Hence, (4) holds. To the best of our knowledge, Theorems A, B, C, D and E in the case where $S_{ij} = {R}_{ij}-\alpha (t){\nabla }_i \phi {\nabla }_j \phi$ are new.

On the other hand, under the situation on Theorem D, we have the following by (5) and the above computation if $\alpha (t)$ is constant:

$$\begin{aligned} {R}_{ij}-2{\nabla }_i \phi {\nabla }_j \phi = -\frac{1}{t}g_{ij}, \quad {\nabla }u = 0, \quad \tau _{g} \phi + {\nabla }^{i}\phi {\nabla }_{i}u = 0. \end{aligned}$$

Therefore, we have ${R}_{ij}-2{\nabla }_i \phi {\nabla }_j \phi = -\frac{1}{t}g_{ij}$ and $\tau _{g} \phi = 0$. In particular, $\phi$ must be a harmonic map.

3 Proofs of Theorems A and B

Let $f$ be a positive solution of the following heat equation with potential:

$$\begin{aligned} \frac{\partial f}{\partial t} = {\Delta }_{}f - cSf, \end{aligned}$$

(8)

where $c$ is a constant. In what follows, let $u = - \log f$. By a direct computation, we are able to see that $u$ satisfies

$$\begin{aligned} \frac{\partial u}{\partial t} = \Delta u - |\nabla u|^{2} + cS. \end{aligned}$$

(9)

Let us introduce the following:

Definition 2

Suppose that $g(t)$ evolves by (1) and let $S$ be the trace of $S_{ij}$. Let $X=X^{i}\frac{\partial }{\partial x^{i}} \in \Gamma (TX)$ be a vector field on $M$. We define

$$\begin{aligned} {\mathcal D}_{(a, \alpha , \beta )}(S, X)&= a \left( \frac{\partial S}{\partial t} - \Delta S - 2|S_{ij}|^{2} \right) + \alpha (2{\nabla }^{i}S_{i\ell } - {\nabla }_{\ell }S){X}^{\ell } \\&+ 2{\beta } (R^{ij} -S^{ij}){X}_{i}{X}_{j}, \end{aligned}$$

where $a, \alpha$ and $\beta$ are constants.

Notice that we have ${\mathcal E}(S, {X}) = {\mathcal D}_{(1, -2, 1)}(S, X)$.

Lemma 1

Let $g(t)$ be a solution to the geometric flow (1) and $u$ satisfies (9). Let

$$\begin{aligned} H_{S} = \alpha \Delta u - \beta |\nabla u|^{2} + a{S} - b\frac{u}{t}-d\frac{n}{t}, \end{aligned}$$

(10)

where $\alpha , \beta , a, b$ and $d$ are constants. Then, $H_{S}$ satisfies

$$\begin{aligned} \frac{\partial H_{S}}{\partial t}&= {\Delta }H_{S} - 2{\nabla }^{i}H_{S}{\nabla }_{i}u + 2(a-\beta c){\nabla }^{i}S{\nabla }_{i}u-2(\alpha -\beta )|\nabla \nabla u|^{2} \\&-2{\alpha }R^{ij}{\nabla }_{i}u{\nabla }_{j}u + 2{\alpha }S^{ij}{\nabla }_{i}{\nabla }_{j}u + {\alpha }c{\Delta }S -\frac{b}{t}|\nabla u|^{2} - \frac{b}{t}cS + \frac{b}{t^{2}}u + d\frac{n}{t^{2}} \\&+ 2a|S_{ij}|^{2} + {\mathcal D}_{(a, \alpha , \beta )}(S, \nabla u). \end{aligned}$$

Proof

First of all, notice that we have the following three evolution equations, which follow from standard computation:

$$\begin{aligned} \frac{\partial }{\partial t} (\Delta u)&= 2{S}^{ij}{\nabla }_{i}{\nabla }_{j}u + \Delta \left( \frac{\partial u}{\partial t}\right) - g^{ij}\left( \frac{\partial }{\partial t}\Gamma ^{k}_{ij} \right) \nabla _{k}u, \\ \frac{\partial }{\partial t}(|\nabla u|^{2})&= 2{S}^{ij}{\nabla }_{i}u{\nabla }_{j}u + 2 \nabla ^{i}\left( \frac{\partial u}{\partial t}\right) \nabla _{i} u, \\ g^{ij}\left( \frac{\partial }{\partial t}\Gamma ^{k}_{ij} \right)&= -g^{k \ell } (2\nabla ^{i}S_{i\ell } - \nabla _{\ell }S). \end{aligned}$$

By (9), (10) and these equations, we are able to obtain

$$\begin{aligned} \frac{\partial H_{S}}{\partial t}&= \alpha \frac{\partial }{\partial t} (\Delta u) - \beta \frac{\partial }{\partial t}(|\nabla u|^{2}) + a\frac{\partial S}{\partial t}- \frac{b}{t}\frac{\partial u}{\partial t} + \frac{b}{t^{2}}u + d\frac{n}{t^{2}} \\&= \alpha \left( 2{S}^{ij}{\nabla }_{i}{\nabla }_{j}u + \Delta \left( \frac{\partial u}{\partial t}\right) - g^{ij}\left( \frac{\partial }{\partial t}\Gamma ^{k}_{ij} \right) \nabla _{k}u \right) \\&\quad - \beta \left( 2{S}^{ij}{\nabla }_{i}u{\nabla }_{j}u + 2 \nabla ^{i}\left( \frac{\partial u}{\partial t}\right) \nabla _{i} u \right) + a\frac{\partial S}{\partial t} - \frac{b}{t}\frac{\partial u}{\partial t} + \frac{b}{t^{2}}u + d\frac{n}{t^{2}} \\&= \alpha (2{S}^{ij}{\nabla }_{i}{\nabla }_{j}u + \Delta (\Delta u - |\nabla u|^{2} + cS) + g^{k \ell } (2\nabla ^{i}S_{i\ell } - \nabla _{\ell }S)\nabla _{k}u) \\&\quad - \beta (2{S}^{ij}{\nabla }_{i}u{\nabla }_{j}u + 2 \nabla ^{i}(\Delta u - |\nabla u|^{2} + cS) \nabla _{i} u) + a\frac{\partial S}{\partial t} \\&\quad - \frac{b}{t}(\Delta u - |\nabla u|^{2} + cS) + \frac{b}{t^{2}}u + d\frac{n}{t^{2}} \\&= 2 \alpha {S}^{ij}{\nabla }_{i}{\nabla }_{j}u + \alpha \Delta (\Delta u) - \alpha \Delta (|\nabla u|^{2}) + \alpha c \Delta S+ \alpha (2\nabla ^{i}S_{i\ell } - \nabla _{\ell }S)\nabla ^{\ell }u \\&\quad - 2\beta {S}^{ij}{\nabla }_{i}u{\nabla }_{j}u - 2{\beta }\nabla ^{i}(\Delta u)\nabla _{i}u + 2{\beta }\nabla ^{i}(|\nabla u|^{2})\nabla _{i}u-2{\beta } c\nabla ^{i}S \nabla _{i}u \\&\quad + \frac{b}{t}|\nabla u|^{2} - \frac{b}{t}cS + \frac{b}{t^{2}}u + d\frac{n}{t^{2}} + a\frac{\partial S}{\partial t} - \frac{b}{t}\Delta u. \end{aligned}$$

On the other hand, we also have the following by (10):

$$\begin{aligned} {\Delta }H_{S}&= \alpha \Delta (\Delta u) - \beta \Delta (|\nabla u|^{2}) + a\Delta {S} - \frac{b}{t}\Delta u. \\ {\nabla }^{i}H_{S}&= \alpha \nabla ^{i}(\Delta u) - \beta \nabla ^{i}(|\nabla u|^{2}) + a\nabla ^{i}{S} - \frac{b}{t}\nabla ^{i}u \end{aligned}$$

Therefore, we get

$$\begin{aligned} {\Delta }H_{S}-2{\nabla }^{i}H_{S}{\nabla }_{i}u&= \alpha \Delta (\Delta u) - \beta \Delta (|\nabla u|^{2}) + a\Delta {S} - \frac{b}{t}\Delta u \\&- 2{\alpha }{\nabla }^{i}(\Delta u){\nabla }_{i}u + 2\beta {\nabla }^{i}(|\nabla u|^{2}){\nabla }_{i}u - 2a{\nabla }^{i}S{\nabla }_{i}u + \frac{2b}{t}|\nabla u|^{2}. \end{aligned}$$

Using this, we are able to obtain

$$\begin{aligned} \frac{\partial H_{S}}{\partial t}&= {\Delta }H_{S} - 2{\nabla }^{i}H{\nabla }_{i}u + 2 \alpha {S}^{ij}{\nabla }_{i}{\nabla }_{j}u - 2\beta {S}^{ij}{\nabla }_{i}u{\nabla }_{j}u - (\alpha - \beta ) \Delta (|\nabla u|^{2}) \\&+ (\alpha c-a)\Delta S + \alpha (2{\nabla }^{i}S_{i \ell } - {\nabla }_\ell S){\nabla }^{\ell }u + 2(\alpha -\beta ){\nabla }^{i}(\Delta u){\nabla }_{i}u \\&+ 2(a-\beta c){\nabla }^{i}S{\nabla }_{i}u -\frac{b}{t}|\nabla u|^{2} + a\frac{\partial S}{\partial t} - \frac{b}{t}cS + \frac{b}{t^{2}}u + d\frac{n}{t^{2}}. \end{aligned}$$

On the other hand, we also have the following Bochner–Weitzenbock type formula:

$$\begin{aligned} \Delta (|\nabla u|^{2}) = 2|\nabla \nabla u|^{2} + 2{\nabla }^{i}(\Delta u){\nabla }_{i}u + 2{R}^{ij}{\nabla }_{i}u{\nabla }_{j}u. \end{aligned}$$

Using this formula, we get

$$\begin{aligned} \frac{\partial H_{S}}{\partial t}&= {\Delta }H_{S} - 2{\nabla }^{i}H_{S}{\nabla }_{i}u + 2(a-\beta c){\nabla }^{i}S{\nabla }_{i}u-2(\alpha -\beta )|\nabla \nabla u|^{2} \\&-2{\alpha }R^{ij}{\nabla }_{i}u{\nabla }_{j}u + 2{\alpha }S^{ij}{\nabla }_{i}{\nabla }_{j}u + {\alpha }c{\Delta }S -\frac{b}{t}|\nabla u|^{2} - \frac{b}{t}cS + \frac{b}{t^{2}}u + d\frac{n}{t^{2}} \\&+ 2a|S_{ij}|^{2} + a \left( \frac{\partial S}{\partial t} -{\Delta }S - 2|S_{ij}|^{2} \right) + \alpha (2{\nabla }^{i}S_{i \ell } - {\nabla }_\ell S){\nabla }^{\ell }u \\&+ 2{\beta }(R^{ij} - S^{ij}){\nabla }_{i}u{\nabla }_{j}u \\&= {\Delta }H_{S} - 2{\nabla }^{i}H_{S}{\nabla }_{i}u + 2(a-\beta c){\nabla }^{i}S{\nabla }_{i}u-2(\alpha -\beta )|\nabla \nabla u|^{2} \\&-2{\alpha }R^{ij}{\nabla }_{i}u{\nabla }_{j}u + 2{\alpha }S^{ij}{\nabla }_{i}{\nabla }_{j}u + {\alpha }c{\Delta }S -\frac{b}{t}|\nabla u|^{2} - \frac{b}{t}cS + \frac{b}{t^{2}}u + d\frac{n}{t^{2}} \\&+ 2a|S_{ij}|^{2} + {\mathcal D}_{(a, \alpha , \beta )}(S, \nabla u), \end{aligned}$$

where notice that Definition 2. $\square$

In particular, we shall use Lemma 1 to prove Theorem B. The following result is used to prove Theorem A.

Proposition 1

The evolution equation in Lemma 1 can be rewritten as follows:

$$\begin{aligned} \frac{\partial H_{S}}{\partial t}&= \Delta {H_{S}} -2{\nabla }^{i}H_{S}{\nabla }_{i}u - 2(\alpha - \beta )\left| \nabla _{i}\nabla _{j}u - \frac{\alpha }{2(\alpha -\beta )}S_{ij}-\frac{\lambda }{2t}g_{ij} \right| ^{2} \\&+ 2(a-\beta c){\nabla }^{i}u{\nabla }_{i}S - \frac{2(\alpha -\beta )}{\alpha }\frac{\lambda }{t}H_{S} + \frac{(\alpha -\beta )n{\lambda }^{2}}{2t^{2}}- \left( b + \frac{2(\alpha -\beta )\lambda {\beta }}{\alpha } \right) \frac{|\nabla u|^{2}}{t} \\&+ \left( 2a + \frac{\alpha ^{2}}{2(\alpha -\beta )} \right) |S_{ij}|^{2} + \left( \alpha {\lambda } - bc + \frac{2(\alpha -\beta )\lambda a}{\alpha } \right) \frac{S}{t} + \left( 1-\frac{2(\alpha -\beta )\lambda }{\alpha } \right) \frac{b}{t^{2}}u \\&+ \left( 1-\frac{2(\alpha -\beta )\lambda }{\alpha } \right) \frac{d}{t^{2}}n + \alpha c {\Delta }S - 2{\alpha }R^{ij}{\nabla }_{i}u{\nabla }_{j}u + {\mathcal D}_{(a, \alpha , \beta )}(S, {\nabla }u), \end{aligned}$$

where $\lambda$ is a constant, $\alpha \not = 0$ and $\alpha \not = \beta$.

Proof

A direct computation implies

$$\begin{aligned}&-2(\alpha -\beta )\left| \nabla _{i}\nabla _{j}u - \frac{\alpha }{2(\alpha -\beta )}S_{ij}-\frac{\lambda }{2t}g_{ij} \right| ^{2} = -2(\alpha -\beta )|\nabla \nabla u|^{2} + 2{\alpha }S^{ij}{\nabla }_{i}{\nabla }_{j}u \\&\quad + 2(\alpha -\beta )\frac{\lambda }{t} \Delta u - \frac{\lambda }{t}S - \frac{\alpha ^{2}}{2(\alpha -\beta )} |S_{ij}|^{2} - \frac{(\alpha - \beta )\lambda ^{2} n}{2t^{2}}. \end{aligned}$$

Therefore, we get

$$\begin{aligned}&-2(\alpha -\beta )|\nabla \nabla u|^{2} + 2{\alpha }S^{ij}{\nabla }_{i}{\nabla }_{j}u + 2a|S_{ij}|^{2}\\&\quad = -2(\alpha -\beta )\left| \nabla _{i}\nabla _{j}u - \frac{\alpha }{2(\alpha -\beta )}S_{ij}-\frac{\lambda }{2t}g_{ij} \right| ^{2} - 2(\alpha -\beta )\frac{\lambda }{t} \left( \Delta u - \frac{\alpha S}{2(\alpha -\beta )} \right) \\&\qquad + \frac{(\alpha - \beta )\lambda ^{2} n}{2t^{2}} + \left( 2a + \frac{\alpha ^{2}}{2(\alpha -\beta )} \right) |S_{ij}|^{2}. \end{aligned}$$

By this and Lemma 1, we obtain

$$\begin{aligned} \frac{\partial H_{S}}{\partial t}&= {\Delta }H_{S} - 2{\nabla }^{i}H_{S}{\nabla }_{i}u - 2(\alpha -\beta )\left| \nabla _{i}\nabla _{j}u - \frac{\alpha }{2(\alpha -\beta )}S_{ij} - \frac{\lambda }{2t}g_{ij} \right| ^{2} \\&+ 2(a-\beta c){\nabla }^{i}S{\nabla }_{i}u - 2(\alpha -\beta )\frac{\lambda }{t} \left( \Delta u - \frac{\alpha S}{2(\alpha -\beta )} \right) + \frac{(\alpha - \beta )\lambda ^{2} n}{2t^{2}} \\&+ \left( 2a + \frac{\alpha ^{2}}{2(\alpha -\beta )} \right) |S_{ij}|^{2} + \alpha c {\Delta }S - 2\alpha {R}^{ij}{\nabla }_{i}u{\nabla }_{j}u - \frac{b}{t}|\nabla u|^{2} - \frac{b}{t}cS \\&+ \frac{b}{t^{2}}u + d\frac{n}{t^{2}} + {\mathcal D}_{(a, \alpha , \beta )}(S, \nabla u). \end{aligned}$$

The desired result now follows from the above equation and the following:

$$\begin{aligned}&-2(\alpha -\beta )\frac{\lambda }{t} \left( \Delta u - \frac{\alpha S}{2(\alpha -\beta )} \right) - \frac{b}{t}|\nabla u|^{2} - \frac{b}{t}cS + \frac{b}{t^{2}}u + d\frac{n}{t^{2}} \\&\quad = - \frac{2(\alpha - \beta )}{\alpha }\frac{\lambda }{t}H_{S} - \left( b + \frac{2(\alpha -\beta )\lambda \beta }{\alpha } \right) \frac{|\nabla u|^{2}}{t} + \left( 1- \frac{2(\alpha - \beta )\lambda }{\alpha } \right) \frac{b}{t^{2}}u \\&\quad \quad + \left( \alpha {\lambda } -bc + \frac{2(\alpha -\beta )\lambda a}{\alpha } \right) \frac{S}{t} + \left( 1-\frac{2(\alpha -\beta )\lambda }{\alpha } \right) \frac{d}{t^{2}}n. \end{aligned}$$

This equation also follows from a direct computation. $\square$

As a corollary of the above proposition, we obtain the following result which is a key to prove Theorem A:

Corollary 1

Suppose that $g(t)$ evolves by the geometric flow (1) on a closed oriented smooth $n$-manifold $M$. Let $f$ be a positive solution to the heat equation (8) with $c=-1$, $u = -\log f$ and

$$\begin{aligned} H_{S} = 2{\Delta }u - |\nabla u|^{2}-3{S} - \frac{2n}{t}. \end{aligned}$$

Then,

$$\begin{aligned} \frac{\partial H_{S}}{\partial t}&= \Delta {H_{S}} -2{\nabla }^{i}H_{S}{\nabla }_{i}u - 2\left| \nabla _{i}\nabla _{j}u - S_{ij}-\frac{1}{t}g_{ij} \right| ^{2} - \frac{2}{t}H_{S} - \frac{2}{t}{|\nabla u|^{2}} \\&\quad - \left( 2{\mathcal H}(S, {\nabla }u) + {\mathcal E}(S, {\nabla }u) \right) . \end{aligned}$$

Proof

By Proposition 1 in the case where $\alpha = 2$, $\beta = 1$, $a = -3$, $c=-1$, $\lambda = 2$, $b = 0$ and $d= 2$, we get the desired result as follows:

$$\begin{aligned} \frac{\partial H_{S}}{\partial t}&= \Delta {H}_{S} -2{\nabla }^{i}H_{S}{\nabla }_{i}u - 2\left| \nabla _{i}\nabla _{j}u - S_{ij}-\frac{1}{t}g_{ij} \right| ^{2} - \frac{2}{t}H_{S} - \frac{2}{t}{|\nabla u|^{2}} \\&- 4{\nabla }_{i}S{\nabla }^{i}u - 4|S_{ij}|^{2} -2 \frac{S}{t} - 2{\Delta }S - 4R^{ij}{\nabla }_{i}u{\nabla }_{j}u + {\mathcal D}_{(-3, 2, 1)}(S, {\nabla }u) \\&= \Delta {H_{S}} -2{\nabla }^{i}H_{S}{\nabla }_{i}u - 2\left| \nabla _{i}\nabla _{j}u - S_{ij}-\frac{1}{t}g_{ij} \right| ^{2} - \frac{2}{t}H_{S} - \frac{2}{t}{|\nabla u|^{2}} \\&- 2\left( \frac{\partial S}{\partial t} + \frac{S}{t} + 2{\nabla }_{i}S{\nabla }^{i}u + 2{S}^{ij}{\nabla }_{i}u{\nabla }_{j}u\right) + 2\frac{\partial S}{\partial t} + 4({S}^{ij}-R^{ij}){\nabla }_{i}u{\nabla }_{j}u \\&- 4|S_{ij}|^{2}- 2{\Delta }S + {\mathcal D}_{(-3, 2, 1)}(S, {\nabla }u) \\&= \Delta {H_{S}} -2{\nabla }^{i}H_{S}{\nabla }_{i}u - 2\left| \nabla _{i}\nabla _{j}u - S_{ij}-\frac{1}{t}g_{ij} \right| ^{2} - \frac{2}{t}H_{S} - \frac{2}{t}{|\nabla u|^{2}} \\&- 2\left( \frac{\partial S}{\partial t} + \frac{S}{t} + 2{\nabla }_{i}S{\nabla }^{i}u + 2{S}^{ij}{\nabla }_{i}u{\nabla }_{j}u\right) + 2\left( \frac{\partial S}{\partial t} - \Delta S - 2|S_{ij}|^{2}\right) \\&- 4(R^{ij}-{S}^{ij}){\nabla }_{i}u{\nabla }_{j}u - 3 \left( \frac{\partial S}{\partial t} - \Delta S - 2|S_{ij}|^{2} \right) + 2 (2{\nabla }^{i}S_{i\ell } - {\nabla }_{\ell }S){\nabla }^{\ell }u \\&+ 2 (R^{ij} -S^{ij}){\nabla }_{i}u{\nabla }_{j}u \\&= \Delta {H_{S}} -2{\nabla }^{i}H_{S}{\nabla }_{i}u - 2\left| \nabla _{i}\nabla _{j}u - S_{ij}-\frac{1}{t}g_{ij} \right| ^{2} - \frac{2}{t}H - \frac{2}{t}{|\nabla u|^{2}} \\&- 2\left( \frac{\partial S}{\partial t} + \frac{S}{t} + 2{\nabla }_{i}S{\nabla }^{i}u + 2{S}^{ij}{\nabla }_{i}u{\nabla }_{j}u\right) \\&- \left( \frac{\partial S}{\partial t} - \Delta S - 2|S_{ij}|^{2} \right) + 2 (2{\nabla }^{i}S_{i\ell } - {\nabla }_{\ell }S){\nabla }^{\ell }u -2 (R^{ij} -S^{ij}){\nabla }_{i}u{\nabla }_{j}u \\&= \Delta {H_{S}} -2{\nabla }^{i}H_{S}{\nabla }_{i}u - 2\left| \nabla _{i}\nabla _{j}u - S_{ij}-\frac{1}{t}g_{ij} \right| ^{2} - \frac{2}{t}H_{S} - \frac{2}{t}{|\nabla u|^{2}} \\&- \left( 2{\mathcal H}(S, {\nabla }u) + {\mathcal E}(S, {\nabla }u) \right) \!, \end{aligned}$$

where we used Definition 2. $\square$

We are now in a position to prove Theorem A. First of all, notice that, for $t$ small enough, we get $H_{S} < 0$. Since we assumed that (3) holds, the maximal principle and Corollary 1 tell us that

$$\begin{aligned} H_{S} \le 0 \end{aligned}$$

for all time $t \in (0, T)$. Hence, we have proved Theorem A.

By Theorem A and integrating along a space-time path, we are able to get a classical Harnack inequality as follows:

Corollary 2

Suppose that $g(t)$ evolves by the geometric flow (1) on a closed oriented smooth $n$-manifold $M$ and $2{\mathcal H}(S, X)+{\mathcal E}(S,X) \ge 0$ holds for all vector fields $X$ and all time $t \in [0, T)$ for which the flow exists. Let $f$ be a positive solution to the heat equation (2). Assume that $(x_{1}, t_{1})$ and $(x_{2}, t_{2})$ are two points in $M \times (0, T)$, where $0 < t_{1} < {t}_{2}$. Let

$$\begin{aligned} L = \inf _{\ell }\int ^{t_{2}}_{t_{1}}\left( |\dot{\ell }|^{2} + S \right) {\mathrm{d}}t, \end{aligned}$$

where $\ell$ is any space-time path joining $(x_{1}, t_{1})$ and $(x_{2}, t_{2})$. Then,

$$\begin{aligned} f(x_{1}, t_{1}) \le f(x_{2}, t_{2}) \left( \frac{t_{2}}{t_{1}} \right) ^{n}\exp \left( {\frac{L}{2}}\right) \!. \end{aligned}$$

(11)

Proof

The strategy of the proof is now standard. For the reader, let us include the proof. First of all, we have $H_{S} \le 0$ by Theorem A. And $u = -\log f$ satisfies (9) with $c=-1$, i.e.,

$$\begin{aligned} \frac{\partial u}{\partial t} = \Delta u - |\nabla u|^{2} -S. \end{aligned}$$

Therefore, we get

$$\begin{aligned} 2\frac{\partial u}{\partial t} + |\nabla u|^{2} - S - 2\frac{n}{t} = H_{S} \le 0. \end{aligned}$$

(12)

Pick a space-time path $\ell (x, t)$ joining $(x_{1}, t_{1})$ and $(x_{2}, t_{2})$. Then, we obtain the following along the path $\ell (x, t)$ using (12):

$$\begin{aligned} \frac{{\mathrm{d}}u}{{\mathrm{d}}t}&= \frac{\partial u}{\partial t} + {\nabla u} \cdot \dot{\ell } \\&\le \frac{S}{2} + \frac{n}{t} - \frac{|\nabla u|^{2}}{2} + {\nabla u} \cdot \dot{\ell } \\&\le \frac{n}{t} + \frac{1}{2}\left( |\dot{\ell }|^{2} + S \right) \!. \end{aligned}$$

This implies

$$\begin{aligned} u(x_{2}, {t}_{1}) -u(x_{1}, {t}_{1}) \le \frac{L}{2} + n \log \left( \frac{t_{2}}{t_{1}}\right) \!. \end{aligned}$$

This tells us that (11) holds. $\square$

Let us close this section with the proof of Theorem B. Let $f$ be a positive solution to linear heat equation $\frac{\partial f}{\partial t} = \Delta f$. Then, we may assume that $f< 1$ by the linearity. Then, $u = -\log f$ satisfies (9) with $c=0$. Therefore, by taking $\alpha = 0$, $\beta = -1$, $a = c =0$, $\lambda = 2$, $b = 1$ and $d= 0$ in Lemma 1, we have

$$\begin{aligned} H_{S} = |\nabla u|^{2} - \frac{u}{t} \end{aligned}$$

and

$$\begin{aligned} \frac{\partial H_{S}}{\partial t}&= {\Delta }H_{S} - 2{\nabla }^{i}{H}_{S}{\nabla }_{i}u - 2|\nabla \nabla u|^{2} - \frac{1}{t}|\nabla u|^{2} + \frac{1}{t^{2}}u +{\mathcal D}_{(0,0,-1)} \\&= {\Delta }H_{S} - 2{\nabla }^{i}{H}_{S}{\nabla }_{i}u - \frac{1}{t}H_{S} - 2|\nabla \nabla u|^{2} - 2{\mathcal I}(S, {\nabla }u). \end{aligned}$$

Notice that as $t$ small enough, $H_{S}<0$. By the maximal principle and this evolution equation, Theorem B follows as desired.

4 Proof of Theorem C

Let $f$ be a positive solution of (8). In what follows, let $v = - \log f-\frac{n}{2}\log (4{\pi }t)$. By a direct computation, we see that $v$ satisfies

$$\begin{aligned} \frac{\partial v}{\partial t} = \Delta v - |\nabla v|^{2} + cS - \frac{n}{2t}. \end{aligned}$$

(13)

Then, we have

Proposition 2

Let $g(t)$ be a solution to the geometric flow (1) and $v$ satisfies (13). Let

$$\begin{aligned} P_{S} = \alpha \Delta v - \beta |\nabla v|^{2} + a{S} - b\frac{v}{t}-d\frac{n}{t}, \end{aligned}$$

(14)

where $\alpha , \beta , a, b$ and $d$ are constants. Then, $P_{S}$ satisfies

$$\begin{aligned} \frac{\partial P_{S}}{\partial t}&= \Delta {P_{S}} -2{\nabla }^{i}P_{S}{\nabla }_{i}v - 2(\alpha - \beta )\left| \nabla _{i}\nabla _{j}v - \frac{\alpha }{2(\alpha -\beta )}S_{ij}\!-\!\frac{\lambda }{2t}g_{ij} \right| ^{2} \\&+ 2(a-\beta c){\nabla }^{i}v{\nabla }_{i}S \!-\! \frac{2(\alpha -\beta )}{\alpha }\frac{\lambda }{t}P_{S} \!+\! \frac{(\alpha -\beta )n{\lambda }^{2}}{2t^{2}}- \left( b + \frac{2(\alpha -\beta )\lambda {\beta }}{\alpha } \right) \frac{|\nabla v|^{2}}{t} \\&+ \left( 2a \!+\! \frac{\alpha ^{2}}{2(\alpha -\beta )} \right) |S_{ij}|^{2} \!+\! \left( \alpha {\lambda } - bc \!+\! \frac{2(\alpha -\beta )\lambda a}{\alpha } \right) \frac{S}{t} \!+\! \left( 1\!-\!\frac{2(\alpha \!-\!\beta )\lambda }{\alpha } \right) \frac{b}{t^{2}}v \\&+ \left( 1-\frac{2(\alpha -\beta )\lambda }{\alpha } \right) \frac{d}{t^{2}}n + \alpha c {\Delta }S - 2{\alpha }R^{ij}{\nabla }_{i}v{\nabla }_{j}v + \frac{bn}{2{t}^{2}} + {\mathcal D}_{(a, \alpha , \beta )}(S, {\nabla }v), \end{aligned}$$

where $\lambda$ is a constant, $\alpha \not =0$ and $\alpha \not = \beta$.

Proof

A similar computation with Proposition 1 enables us to prove this result. In fact, notice that we have $v = u - \frac{n}{2}\log (4{\pi }t)$. Therefore, we get $\nabla u = \nabla v$ and $\Delta u = \Delta v$. We also have $P_{S} =H_{S} + \frac{bn}{2t}\log (4{\pi }t)$. Then, Proposition 1 and a direct computation imply

$$\begin{aligned} \frac{\partial P_{S}}{\partial t}&= \frac{\partial {H_S}}{\partial t} - \frac{bn}{2{t}^{2}}\log (4{\pi }t) + \frac{bn}{2t^{2}} \\&= \Delta {P_{S}} -2{\nabla }^{i}P_{S}{\nabla }_{i}v - 2(\alpha - \beta )\left| \nabla _{i}\nabla _{j}v - \frac{\alpha }{2(\alpha -\beta )}S_{ij}-\frac{\lambda }{2t}g_{ij} \right| ^{2} \\&\quad + 2(a-\beta c){\nabla }^{i}v{\nabla }_{i}S - \frac{2(\alpha -\beta )}{\alpha }\frac{\lambda }{t}P_{S} + \frac{(\alpha -\beta )n{\lambda }^{2}}{2t^{2}}- \left( b + \frac{2(\alpha -\beta )\lambda {\beta }}{\alpha } \right) \frac{|\nabla v|^{2}}{t} \\&\quad + \left( 2a + \frac{\alpha ^{2}}{2(\alpha -\beta )} \right) |S_{ij}|^{2} \!+\! \left( \alpha {\lambda } - bc \!+\! \frac{2(\alpha -\beta )\lambda a}{\alpha } \right) \frac{S}{t} \!+\! \left( 1-\frac{2(\alpha -\beta )\lambda }{\alpha } \right) \frac{b}{t^{2}}v \\&\quad + \left( 1-\frac{2(\alpha -\beta )\lambda }{\alpha } \right) \frac{d}{t^{2}}n + \alpha c {\Delta }S - 2{\alpha }R^{ij}{\nabla }_{i}v{\nabla }_{j}v + \frac{bn}{2{t}^{2}} + {\mathcal D}_{(a, \alpha , \beta )}(S, {\nabla }v). \end{aligned}$$

Hence we obtained the desired result. $\square$

As a special case of Proposition 2, we get

Corollary 3

Suppose that $g(t)$ evolves by the geometric flow (1) on a closed oriented smooth $n$-manifold $M$. Let $f$ be a positive solution to the heat equation (8) with $c= -1$, $v = - \log f-\frac{n}{2}\log (4{\pi }t)$ and

$$\begin{aligned} P_{S} = 2{\Delta }v - |\nabla v|^{2}-3{S} + \frac{v}{t} - d\frac{n}{t}. \end{aligned}$$

Then,

$$\begin{aligned} \frac{\partial P_{S}}{\partial t} + \frac{1}{t}P_{S}&= \Delta {P_{S}} -2{\nabla }^{i}P_{S}{\nabla }_{i}v - 2\left| \nabla _{i}\nabla _{j}v - S_{ij}-\frac{1}{2t}g_{ij} \right| ^{2} \\&\quad - \left( 2{\mathcal H}(S, {\nabla }v) + {\mathcal E}(S, {\nabla }v) \right) . \end{aligned}$$

$\square$

Proof

By Proposition 2 in the case where $\alpha = 2$, $\beta = 1$, $a = -3$, $b = -1$, $c=-1$, $\lambda = 1$, we obtain

$$\begin{aligned} \frac{\partial P_{S}}{\partial t}&= \Delta {P_{S}} -2{\nabla }^{i}P_{S}{\nabla }_{i}v - 2\left| \nabla _{i}\nabla _{j}v - S_{ij}-\frac{1}{2t}g_{ij} \right| ^{2} - \frac{1}{t}P_{S} \\&\quad - 4{\nabla }_{i}S{\nabla }^{i}u - 4|S_{ij}|^{2} -2 \frac{S}{t} - 2{\Delta }S - 4R^{ij}{\nabla }_{i}u{\nabla }_{j}u \\&\quad + {\mathcal D}_{(-3, 2, 1)}(S, {\nabla }u) \\&= \Delta {P}_{S} -2{\nabla }^{i}P_{S}{\nabla }_{i}v - 2\left| \nabla _{i}\nabla _{j}v - S_{ij}-\frac{1}{2t}g_{ij} \right| ^{2} - \frac{1}{t}P_{S} \\&\quad - 2\left( \frac{\partial S}{\partial t} + \frac{S}{t} + 2{\nabla }_{i}S{\nabla }^{i}u + 2{S}^{ij}{\nabla }_{i}u{\nabla }_{j}u\right) + 2\left( \frac{\partial S}{\partial t} - \Delta S - 2|S_{ij}|^{2}\right) \\&\quad + 4({S}^{ij}-R^{ij}){\nabla }_{i}u{\nabla }_{j}u + {\mathcal D}_{(-3, 2, 1)}(S, {\nabla }u) \\&= \Delta {P_{S}} -2{\nabla }^{i}P_{S}{\nabla }_{i}v - 2\left| \nabla _{i}\nabla _{j}v - S_{ij}-\frac{1}{t}g_{ij} \right| ^{2} - \frac{1}{t}P_{S} \\&\quad - 2\left( \frac{\partial S}{\partial t} + \frac{S}{t} + 2{\nabla }_{i}S{\nabla }^{i}v + 2{S}^{ij}{\nabla }_{i}v{\nabla }_{j}v\right) + 2\left( \frac{\partial S}{\partial t} - \Delta S - 2|S_{ij}|^{2}\right) \\&\quad - 4(R^{ij}-{S}^{ij}){\nabla }_{i}u{\nabla }_{j}v - 3 \left( \frac{\partial S}{\partial t} - \Delta S - 2|S_{ij}|^{2} \right) + 2 (2{\nabla }^{i}S_{i\ell } - {\nabla }_{\ell }S){\nabla }^{\ell }u \\&\quad + 2 (R^{ij} -S^{ij}){\nabla }_{i}u{\nabla }_{j}u \\&= \Delta {P_{S}} -2{\nabla }^{i}P_{S}{\nabla }_{i}v - 2\left| \nabla _{i}\nabla _{j}v - S_{ij}-\frac{1}{t}g_{ij} \right| ^{2} - \frac{1}{t}P_{S} \\&\quad - 2\left( \frac{\partial S}{\partial t} + \frac{S}{t} + 2{\nabla }_{i}S{\nabla }^{i}v + 2{S}^{ij}{\nabla }_{i}v{\nabla }_{j}v\right) \\&\quad - \left( \frac{\partial S}{\partial t} - \Delta S - 2|S_{ij}|^{2} \right) + 2 (2{\nabla }^{i}S_{i\ell } - {\nabla }_{\ell }S ){\nabla }^{\ell }u -2 (R^{ij} -S^{ij}){\nabla }_{i}u{\nabla }_{j}u \\&= \Delta {P_{S}} -2{\nabla }^{i}P_{S}{\nabla }_{i}v - 2\left| \nabla _{i}\nabla _{j}v - S_{ij}-\frac{1}{2t}g_{ij} \right| ^{2} - \frac{1}{t}P_{S} \\&\quad - \left( 2{\mathcal H}(S, {\nabla }v) + {\mathcal E}(S, {\nabla }v) \right) . \end{aligned}$$

Therefore, the desired result follows. $\square$

We shall prove Theorem C as follows. In fact, we are able to obtain the following by Corollary 3:

$$\begin{aligned} \frac{\partial }{\partial t}(tP_{S})&= t\frac{\partial P_{S}}{\partial t} + P_{S} = {\Delta }(t{P}_{S}) - 2{\nabla }^{i}(tP_{S}){\nabla }_{i}v -2t\left| \nabla _{i}\nabla _{j}v - S_{ij} - \frac{1}{2t}g_{ij}\right| ^{2} \\&\quad - t\left( 2{\mathcal H}(S, {\nabla }v)+{\mathcal E}(S, {\nabla }v) \right) . \end{aligned}$$

Furthermore, the monotonicity of $\max (t{P}_{S})$ follows from this equation and the maximal principle.

5 Proof of Theorem D

First of all, notice that ${\mathcal F}_{S} \le 0$ follows from the definition of ${\mathcal F}_{S}$ and $H_{S} \le 0$, where we used Theorem A. Let us consider the following quantity:

$$\begin{aligned} A = 2t\mathrm{e}^{-u}{H}_{S} - t^{2}\mathrm{e}^{-u}\frac{\partial u}{\partial t}H_{S} + t^{2}\mathrm{e}^{-u}\frac{\partial H_{S}}{\partial t} -St^{2}\mathrm{e}^{-u}H_{S}. \end{aligned}$$

On the other hand, a direct computation tells us that the following holds:

$$\begin{aligned} {\Delta }(t^{2}\mathrm{e}^{-u}H_{S}) = t^{2}\mathrm{e}^{-u} ({\Delta } H_{S} - 2{\nabla }^{i}H_{S}{\nabla }_{i}u - H_{S}{\Delta }u + H_{S}|\nabla u|^{2}). \end{aligned}$$

(15)

By (9) with $c=-1$, Corollary 1 and (15), we get

$$\begin{aligned} A&= 2t\mathrm{e}^{-u}{H}_{S} - t^{2}\mathrm{e}^{-u} \left( \Delta u - |\nabla u|^{2} -S \right) H_{S} \\&+ t^{2}\mathrm{e}^{-u}\left( \Delta {H_{S}} -2{\nabla }^{i}H_{S}{\nabla }_{i}u - 2\left| \nabla _{i}\nabla _{j}u - S_{ij}-\frac{1}{t}g_{ij} \right| ^{2} - \frac{2}{t}H_{S} - \frac{2}{t}{|\nabla u|^{2}}- (2{\mathcal H}(S, {\nabla }u) + {\mathcal E}(S, {\nabla }u) ) \right) \\&- St^{2}\mathrm{e}^{-u}H_{S} \\&= {\Delta }(t^{2}\mathrm{e}^{-u}H_{S}) - 2t^{2}\mathrm{e}^{-u}\left| \nabla _{i}\nabla _{j}u - S_{ij}-\frac{1}{t}g_{ij} \right| ^{2} -2t{e}^{-u}|\nabla u|^{2} \\&- t^{2}\mathrm{e}^{-u}\left( 2{\mathcal H}(S, {\nabla }u) + {\mathcal E}(S, {\nabla }u) \right) \!. \end{aligned}$$

On the other hand, notice that we have

$$\begin{aligned} \frac{\mathrm{d}}{{\mathrm{d}}t} {\mathcal F}_{S} = \frac{\mathrm{d}}{{\mathrm{d}}t} \left( \int _{M}t^{2}\mathrm{e}^{-u}H_{S}{\mathrm{d}}{\mu }_{g}\right) = \int _{M}A{\mathrm{d}}{\mu }_{g}. \end{aligned}$$

Therefore, the following holds:

$$\begin{aligned} \frac{\mathrm{d}}{{\mathrm{d}}t} {\mathcal F}_{S}&= \int _{M}\bigg ({\Delta }(t^{2}\mathrm{e}^{-u}H_{S}) - 2t^{2}\mathrm{e}^{-u}\left| \nabla _{i}\nabla _{j}u - S_{ij}-\frac{1}{t}g_{ij} \right| ^{2} -2t{e}^{-u}|\nabla u|^{2} \\&-\, t^{2}\mathrm{e}^{-u}(2{\mathcal H}(S, {\nabla }u) + {\mathcal E}(S, {\nabla }u))\bigg ) {\mathrm{d}}{\mu }_{g} \\&= -\int _{M}\bigg (2t^{2}\mathrm{e}^{-u}\left| \nabla _{i}\nabla _{j}u - S_{ij}-\frac{1}{t}g_{ij} \right| ^{2} + 2t\mathrm{e}^{-u}|\nabla u|^{2}\\&+\, t^{2}\mathrm{e}^{-u}(2{\mathcal H}(S, {\nabla }u) + {\mathcal E}(S, {\nabla }u))\bigg ) {\mathrm{d}}{\mu }_{g} \le 0. \end{aligned}$$

Assume moreover that ${\mathcal H}(S, X) \ge 0$ and ${\mathcal E}(S, X) \ge 0$ holds. Suppose also that $\frac{\mathrm{d}}{{\mathrm{d}}t} {\mathcal F}_{S} = 0$ holds for some time $t$. Then, we obtain

$$\begin{aligned} \nabla _{i}\nabla _{j}u - S_{ij}-\frac{1}{t}g_{ij} = 0, \ \nabla u = 0, \ {\mathcal H}(S, {\nabla }u) = 0, \ {\mathcal E}(S, {\nabla }u) = 0. \end{aligned}$$

These imply (5) as desired. We proved Theorem D.

6 Proof of Theorem E

Let us consider the following quantity:

$$\begin{aligned} B = \mathrm{e}^{-v}{P}_{S} + t\mathrm{e}^{-v}\frac{\partial P_{S}}{\partial t} -\frac{n}{2}\mathrm{e}^{-v}P_{S} - t\mathrm{e}^{-v}\frac{\partial v}{\partial t}P_{S} -St\mathrm{e}^{-v}P_{S}. \end{aligned}$$

A direct computation tells us that the following holds:

$$\begin{aligned} {\Delta }(t\mathrm{e}^{-v}P_{S}) = t\mathrm{e}^{-v} ({\Delta } P_{S} - 2{\nabla }^{i}P_{S}{\nabla }_{i}v - P_{S}{\Delta }v + P_{S}|\nabla v|^{2}) \end{aligned}$$

(16)

By (13) with $c=-1$, Corollary 3 and (16), we obtain the following:

$$\begin{aligned} B&= \mathrm{e}^{-v}{P}_{S} + t\mathrm{e}^{-v}(\Delta {P_{S}} -2{\nabla }^{i}P_{S}{\nabla }_{i}v - 2\left| \nabla _{i}\nabla _{j}v - S_{ij}-\frac{1}{2t}g_{ij} \right| ^{2} - \frac{1}{t}P_{S} \\&-\, (2{\mathcal H}(S, {\nabla }v) + {\mathcal E}(S, {\nabla }v)) -\frac{n}{2}\mathrm{e}^{-v}P_{S}- t\mathrm{e}^{-v}P_{S}\left( \Delta v - |\nabla v|^{2} -S - \frac{n}{2t} \right) \\&-\, St\mathrm{e}^{-v}P_{S} \\&= {\Delta }(t\mathrm{e}^{-v}P_{S}) -2t\mathrm{e}^{-v}\left| \nabla _{i}\nabla _{j}v - S_{ij}-\frac{1}{2t}g_{ij} \right| ^{2} - t\mathrm{e}^{-v}\left( 2{\mathcal H}(S, {\nabla }v) + {\mathcal E}(S, {\nabla }v) \right) \!. \end{aligned}$$

By a direct computation, we also have

$$\begin{aligned} \frac{\mathrm{d}}{{\mathrm{d}}t} {\mathcal W}_{S} = \frac{\mathrm{d}}{{\mathrm{d}}t}\left( \,\,\int _{M}tP_{S}(4{\pi }t)^{-\frac{n}{2}}\mathrm{e}^{-v}{\mathrm{d}}{\mu }_{g} \right) = \int _{M}B(4{\pi }t)^{-\frac{n}{2}}{\mathrm{d}}{\mu }_{g}. \end{aligned}$$

Therefore, we obtain

$$\begin{aligned} \frac{\mathrm{d}}{{\mathrm{d}}t} {\mathcal W}_{S}&= \int _{M}({\Delta }(t\mathrm{e}^{-v}P_{S}) -2t\mathrm{e}^{-v}\left| \nabla _{i}\nabla _{j}v - S_{ij}-\frac{1}{2t}g_{ij} \right| ^{2} \\&-\, t\mathrm{e}^{-v}( 2{\mathcal H}(S, {\nabla }v) + {\mathcal E}(S, {\nabla }v)))(4{\pi }t)^{-\frac{n}{2}}{\mathrm{d}}{\mu }_{g} \\&= - \int _{M}\bigg (2t\mathrm{e}^{-v}\left| \nabla _{i}\nabla _{j}v - S_{ij}-\frac{1}{2t}g_{ij} \right| ^{2}\\&+\, t\mathrm{e}^{-v}( 2{\mathcal H}(S, {\nabla }v) + {\mathcal E}(S, {\nabla }v)) \bigg )(4{\pi }t)^{-\frac{n}{2}}{\mathrm{d}}{\mu }_{g} \le 0. \end{aligned}$$

Assume moreover that ${\mathcal H}(S, X) \ge 0$ and ${\mathcal E}(S, X) \ge 0$. Suppose also that $\frac{\mathrm{d}}{{\mathrm{d}}t} {\mathcal W}_{S} =0$ for some time $t$. Then, we obtain

$$\begin{aligned} S_{ij}-\nabla _{i}\nabla _{j}v +\frac{1}{2t}g_{ij} = 0,\quad \ {\mathcal H}(S, {\nabla }v) = 0,\quad \ {\mathcal E}(S, {\nabla }v) = 0. \end{aligned}$$

Hence, we have proved Theorem E.