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eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2021; 61(3): 631-644

Published online September 30, 2021

Copyright © Kyungpook Mathematical Journal.

### Harnack Estimate for Positive Solutions to a Nonlinear Equation Under Geometric Flow

Ghodratallah Fasihi-Ramandi, Shahroud Azami*

Department of Pure Mathematics, Faculty of Science Imam Khomeini International University, Qazvin, Iran
e-mail : fasihi@sci.ikiu.ac.ir and azami@sci.ikiu.ac.ir

Received: August 28, 2019; Revised: June 16, 2020; Accepted: August 18, 2020

### Abstract

In the present paper, we obtain gradient estimates for positive solutions to the following nonlinear parabolic equation under general geometric flow on complete noncompact manifolds $∂u∂t=△u+a(x,t)up+b(x,t)uq$ where, 0 < p, q < 1 are real constants and a(x, t) and b(x, t) are functions which are C2 in the x-variable and C1 in the t-variable. We shall get an interesting Harnack inequality as an application.

Keywords: Geometric Flow, Harnack Estimate, Nonlinear Parabolic Equations.

### 1. Introduction and Main Results

Gradient estimates for nonlinear partial differential equations are of classical interest, and have been extensively studied, leading to many important results, especially in the area of geometric analysis. They were developed by Li and Yau [6] as a method to study the heat equation. Hamilton applied this method {to Ricci flow on manifolds with scalar curvature} [4]. Since then, there has been a lot of work on gradient estimates for solutions of differential equations under geometric flows, see, for instance [5, 7]. Extending some of this work, Sun [9] studied gradient estimates for positive solutions of the heat equation under the geometric flow. Also, the differential Harnack estimates plays an important role in solving the Poincaré conjecture and the geometrization conjecture [8].

In the present paper, we study the following nonlinear parabolic equation under general geometric flow on complete noncompact manifolds M,

$∂u∂t=△u+a(x,t)up+b(x,t)uq$

where, $0 are real constants and a(x,t) and b(x,t) are functions which are C2 in the x-variable and C1 in the t-variable. Before presenting our main results about the equation, {we motivate its consideration as a topic of study.} If a(x,t) and b(x,t) are identically zero, then (1.1) is the heat equation. In bio-mathematics, the following equation

$∂u∂t=△u+a(x,t)up, p>0,$

could be used to model population dynamics. Similar equations arise in the study of the conformal deformation of scalar curvature on a manifold (See [10], equation (1.4)).

Let (M,g(t)) be a smooth 1-parameter family of complete Riemannian metrics on a manifold M evolving by equation

$∂gij∂t=2sij$

for t in some time interval [0,T], where sij are componnents of a symmmetric (0,2)-tensor s. Notice that

• if $sij=−Rij$ then geometric flow (1.2) called Ricci flow,

• if $sij=−12Rgij$ then geometric flow (1.2) called Yamabe flow,

• if $sij=−(Rij+ρRgij)$ then geometric flow (1.2) called Ricci-Bourguignon flow,

• if $sij=−Rij+α∇ϕ⊗∇ϕ$ where $∂ϕ∂t=τgϕ$ then geometric flow (1.2) called harmonic-Ricci flow.

Now we present our main results about the equation (1.1) as follows.

Theorem 1.1. Suppose (M,g(t)) is the family of complete Riemannian manifolds evolving by (1.2). Let M be complete under the initial metric g(0). Given x0∈ M, and $M1,R>0$, let u be a positive solution to the nonlinear equation (1.1) with $u≥M1$ in the cube $Q2R,T={(x,t)|d(x,x0,t)≤2R,0≤t≤T}$. Suppose that there exist constants $K1,K2,K3,K4≥0$ such that

$Ric≥−K1g, −K2g≤s≤K3g, |∇s|≤K4$

on $Q2R,T$. Moreover, assume that there exist positive constants $θa,θb,γa,γb$ such that $△a≤θa$, $|∇a|≤γa$, $△b≤θb$ and $|∇b|≤γb$ in $Q2R,T$. Then for any constant $0<β<1$ and $(x,t)∈Q2R,T$ if $β we have

$β|∇u|2u2+aup−1+buq−1−utu≤H1+H2+nβ1t,$

where,

$H1=nβ((n−1)(1+K1R)c12+c2+2c12R2+c3K2+|a|(1−p)M1(p−1)+|b|(1−q)M1(q−1)+nc122R2(β−β2)),$
$H2=[n24β2(1−β)2(2(1−β)K3+2βK1+32K4)2+nβ{M1(p−1)θa+M1(q−1)θb+n(1β(K2+K3)2+32K4)}−nβ{[(p−β)M1(p−1)γa+(q−β)M1(q−1)γb]2|a|(p−β)(p−1)M1(p−1)+|b|(q−β)(q−1)M1(q−1)}]12.$

When R approaches infinity, we get the global Li-Yau type gradient estimates (see [6]) for equation (1.1) as follows.

Corollary 1.2. Let (M, g(0)) be a complete noncompact Riemannian manifold without boundary, and suppose that g(t) evolves by $∂gij∂t=2sij$ for $t∈[0,T]$ and satisfies

$Ric≥−K1g, −K2g≤s≤K3g, |∇s|≤K4.$

Also, assume that $△a≤θa$, $△b≤θb$, $|∇a|≤γa$ and $|∇b|≤γb$ in $M×[0,T)$ for some constants $θa,θb,γa$ and γb. Let u be a positive solution of (1.1) with u ≥ M1. Then for any constant $0<β<1$, if $β, we have

$β|∇u|2u2+aup−1+buq−1−utu≤H1¯+H2+nβ1t,$

where

$H1¯=nβ(c3K2+|a|(1−p)M1(p−1)+|b|(1−q)M1(q−1)).$

As an application, we get the following Harnack inequality.

Corollary 1.3. Let (M, g(0)) be a complete noncompact Riemannian manifold without boundary, and suppose that g(t) evolves by $∂gij∂t=2sij$ for $t∈[0,T]$ and satisfies

$Ric≥−K1g, −K2g≤s≤K3g, |∇s|≤K4.$

Also, assume that $△a≤θa$, $△b≤θb$, $|∇a|≤γa$ and $|∇b|≤γb$ in $M×[0,T)$ for some constants $θa,θb,γa$ and γb. Let u(x,t) be a positive solution of (1.1) in $M×[0,T)$ with u ≥ M1 where, a and b are positive constants. Then for any constant $0<β<1$, if $β, for any points (x1,t1) and (x2,t2) on $M×[0,T)$ with $0, we have the following Harnack inequality,

$u(x1,t1)≤u(x2,t2)(t2 t1 )nβeΨ(x1,x2,t1,t2)+(H1 ¯+H2)(t2−t1),$

where $Ψ(x1,x2,t1,t2)=infγ∫t1 t2 1 4β|γ′|2dt$, and $H1¯=nβ(c3K2+a(1−p)M1(p−1)+b(1−q)M1(q−1))$, and

$H2=[n24β2(1−β)2(2(1−β)K3+2βK1+32K4)2+nβ{M1(p−1)θa+M1(q−1)θb+n(1β(K2+K3)2+32K4)}−nβ{[(p−β)M1(p−1)γa+(q−β)M1(q−1)γb]2a(p−β)(p−1)M1(p−1)+b(q−β)(q−1)M1(q−1)}]12.$

### 2. Methods and Proofs

Let u be a positive solution to (1.1). Let $w=lnu$, then a simple computation shows that w satisfies the following equation

$wt=△w+|∇w|2+ae(p−1)w+be(q−1)w.$

We need the following lemmas of [3, 9] to prove our main theorem.

Lemma 2.1. If the metric evolves by (1.2) then for any smooth function w, we have

$∂∂t|∇w|2=−2s(∇w,∇w)+2∇w∇wt$

and

$∂∂t△w=△wt−2s∇2w−2∇w(divs−12∇(trgs)),$

where, $divs$ denotes the divergence of s.

Lemma 2.2. Assume that (M, g(t)) satisfies the hypotheses of Proposition 1.1. Then for any constant $0<β<1$ and $(x,t)∈QR,T$, if $β, we have

$(△−∂∂t)F≥−2∇w∇F+t{βn(wt−|∇w|2−ae(p−1)w−be(q−1)w)2 +[a(p−β)(p−1)e(p−1)w+b(q−β)(q−1)e(q−1)w+2(β−1)K3 −2βK1−32K4]|∇w|2 +2(p−β)e(p−1)w∇w∇a+2(q−β)e(q−1)w∇w∇b +e(p−1)w△a+e(q−1)w△b−n(1β(K2+K3)2+32K4)} −a(p−1)e(p−1)wF−b(q−1)e(q−1)wF−Ft,$

where

$F=t(β|∇w|2+ae(p−1)w+be(q−1)w−wt).$

Proof. Define

$F=t(β|∇w|2+ae(p−1)w+be(q−1)w−wt).$

By the Bochner formula, we can write

$△|∇w|2≥2|∇2w|2+2∇w∇(△w)−2K1|∇w|2.$

Note that

$△wt=(△w)t+2s∇2w+2∇w(divs−12∇(trgs)) =wtt−(|∇w|2)t−ate(p−1)w−ae(p−1)w−bte(q−1)w−be(q−1)w +2s∇2w+2∇w(divs−12∇(trgs)) =2s(∇w,∇w)−2∇w∇wt−ate(p−1)w−ae(p−1)w −bte(q−1)w−be(q−1)w+wtt+2s∇2w+2∇w(divs−12∇(trgs)),$

and

$△w=−|∇w|2−ae(p−1)w−be(q−1)w+wt =(1β−1)(ae(p−1)w+be(q−1)w−wt)−Ftβ =(β−1)|∇w|2−Ft.$

We can write,

$△F=t(β△|∇w|2+△(ae(p−1)w)+△(be(q−1)w)−△wt).$

According to the above computations, we obtain

$β△|∇w|2≥2β|∇2w|2+2β∇w∇(△w)−2βK1|∇w|2 =2β|∇2w|2+2β∇w∇([(1β−1)(ae(p−1)w+be(q−1)w−wt)−Ftβ]) −2βK1|∇w|2 =2β|∇2w|2−2t∇w∇F+2(1−β)e(p−1)w∇w∇a +2(1−β)e(q−1)w∇w∇b +2a(1−β)(p−1)e(p−1)w|∇w|2 +2b(1−β)(q−1)e(q−1)w|∇w|2 +2(1−β)∇w∇wt−2K1β|∇w|2,$

and, we know

$△(ae(p−1)w)=e(p−1)w△a+2(p−1)e(p−1)w∇w∇a+a(p−1)2e(p−1)w|∇w|2 +a(p−1)e(p−1)w△w =e(p−1)w△a+2(p−1)e(p−1)w∇w∇a+a(p−1)2e(p−1)w|∇w|2 +a(p−1)e(p−1)w[(β−1)|∇w|2−Ft].$

So we have

$△F≥t{2β|∇2w|2−2t∇w∇F+2(1−β)e(p−1)w∇w∇a+2(1−β)e(q−1)w∇w∇b+2a(1−β)(p−1)e(p−1)w|∇w|2+2b(1−β)(q−1)e(q−1)w|∇w|2+2(1−β)∇w∇wt−2k1β|∇w|2+e(p−1)w△a+2(p−1)e(p−1)w∇w∇a+a(p−1)2e(p−1)w|∇w|2+a(p−1)e(p−1)w[(β−1)|∇w|2−Ft]+e(q−1)w△b+2(q−1)e(q−1)w∇w∇b+b(q−1)2e(q−1)w|∇w|2+b(q−1)e(q−1)w[(β−1)|∇w|2−Ft]−[wtt−(|∇w|2)t−ate(p−1)w−a(p−1)e(p−1)wwt−bte(q−1)w−b(q−1)e(q−1)wwt+2s∇2w+2∇w(divs−12∇(trgs))]},$

and

$Ft=Ft+t{2β(|∇w|2)t+ate(p−1)w+a(p−1)e(p−1)wwt+bte(q−1)w +b(q−1)e(q−1)wwt−wtt} =Ft+t{2β∇w∇wt−2βs(∇w,∇w) +ate(p−1)w+a(p−1)e(p−1)wwt+bte(q−1)w +b(q−1)e(q−1)wwt−wtt}.$

This equation implies that

$(△−∂∂t)F≥−2∇w∇F+t{2β|∇2w|2+2(β−1)s(∇w,∇w) +a(p−β)(p−1)e(p−1)w|∇w|2+b(q−β)(q−1)e(q−1)w|∇w|2 +2(p−β)e(p−1)w∇w∇a+2(q−β)e(q−1)w∇w∇b +e(p−1)w△a+e(q−1)w△b −2K1β|∇w|2−2s∇2w−2∇w(divs−12∇(trgs))} −a(p−1)e(p−1)wF −b(q−1)e(q−1)wF− Ft.$

By our assumptions, we have

$−(K2+K3)g≤s≤(K2+K3)g$

which implies that

$|s|2≤(K2+K3)2|g|2=n(K2+K3)2.$

Using Young's inequality and applying those bounds yields

$|s∇2w|≤β2|∇2w|2+12β|s|2≤β2|∇2w|2+n2β(K2+K3)2.$

On the other hand,

$|divs−12∇(trgs)|=|gij∇isjl−12gij∇lsij|≤32|g||∇s|≤32nK4.$

Finally, with the help of the following inequality,

$|∇2w|2≥1n(tr∇2w)2=1n(△w)2=1n(−|∇w|2−ae(p−1)w−be(q−1)w+wt)2.$

We obtain

$(△−∂∂t)F≥−2∇w∇F+t{βn(wt−|∇w|2−ae(p−1)w−be(q−1)w)2 +a(p−β)(p−1)e(p−1)w|∇w|2+b(q−β)(q−1)e(q−1)w|∇w|2 +2(p−β)e(p−1)w∇w∇a+2(q−β)e(q−1)w∇w∇b+e(p−1)w△a +e(q−1)w△b+2(β−1)K3|∇w|2−2βK1|∇w|2−nβ(K2+K3)2 −3nK4|∇w|}−a(p−1)e(p−1)wF−b(q−1)e(q−1)wF−Ft.$

Applying AM-GM inequality, we can write

$3nK4|∇w|≤3K4(n2+|∇w|22),$

we get

$(△−∂∂t)F ≥−2∇w∇F+t{βn(wt−|∇w|2−ae(p−1)w−be(q−1)w)2 +[a(p−β)(p−1)e(p−1)w+b(q−β)(q−1)e(q−1)w+2(β−1)K3 −2βK1−32K4]|∇w|2+2(p−β)e(p−1)w∇w∇a+2(q−β)e(q−1)w∇w∇b +e(p−1)w△a+e(q−1)w△b−n(1β(K2+K3)2+32K4)} −a(p−1)e(p−1)wF−b(q−1)e(q−1)wF−Ft.$

This completes the proof.

Let's take a cut-off function $φ˜$ defined on $[0,∞)$ such that $0≤φ˜(r)≤1$, $φ˜(r)=1$ for $r∈[0,1]$ and, $φ˜(r)=0$ for $r∈[2,∞)$. Furthermore $φ˜$ satisfies the following inequalities for some positive constants c1 and c2.

$−φ˜′(r)φ˜12(r)≤c1, φ˜″(r)≥−c2.$

Define $r(x,t):=d(x,x0,t)$ and, set

$φ(x,t)=φ˜(r(x,t)R).$

Using Corollary in page 53 of [2], we can assume $φ(x,t)∈C2(M)$ with support in $Q2R,T$. A direct calculation indicates that on $Q2R,T$, we have

$|∇φ|2φ≤c12R2.$

According to the Laplace comparison theorem in [1], we can write

$△φ≥−(n−1)(1+K1R)c12+c2R2.$

For any $0, suppose that $φF$ attains it maximum value at the point (x0,t0) in the cube $Q2R,T1$. We can assume that this maximum value is positive (otherwise the proof of our main theorem will be trivial). At the maximum point (x0,t0), we have

$∇(φF)=0, △(φF)≤0, (φF)t≥0,$

which follows that

$0≥(△−∂∂t)(φF)=(△φ)F−φtF+φ(△−∂∂t)F+2∇φ∇F.$

So, we can write

$(△φ)F−φtF+φ(△−∂∂t)F−2Fφ−1|∇φ|2≤0.$

Also, we know (see [9], p. 494) there exists a positive constant c3 such that

$−φtF≥−c3K2F.$

The inequality (2.4) together with the inequalities (2.2) and (2.3) yield

$φ(△−∂∂t)F≤HF,$

where

$H=(n−1)(1+K1R)c12+c2+2c12R2+c3K2.$

Proof of Theorem 1.1. At the maximum point (x0,t0), by (2.5) and Lemma 2.2, we have

$0≥φ(△−∂∂t)F−HF≥−HF+φ{−2∇w∇F+βt0n(wt−|∇w|2−ae(p−1)w−be(q−1)w)2+t0[a(p−β)(p−1)e(p−1)w+b(q−β)(q−1)e(q−1)w+2(β−1)K3−2βK1−32K4]|∇w|2+2t0(p−β)e(p−1)w∇w∇a+2t0(q−β)e(q−1)w∇w∇b+t0e(p−1)w△a+t0e(q−1)w△b−nt0(1β(K2+K3)2+32K4)−a(p−1)e(p−1)wF−b(q−1)e(q−1)wF−Ft0}$
$≥−HF+2F∇w∇φ+βt0nφ(wt−|∇w|2−ae(p−1)w−be(q−1)w)2+t0φ[a(p−β)(p−1)e(p−1)w+b(q−β)(q−1)e(q−1)w+2(β−1)K3−2βK1−32K4]|∇w|2+2t0φ(p−β)e(p−1)w∇w∇a+2t0φ(q−β)e(q−1)w∇w∇b+t0φe(p−1)w△a+t0φe(q−1)w△b−nt0φ(1β(K2+K3)2+32K4)−a(p−1)e(p−1)wφF−b(q−1)e(q−1)wφF−φt0−1F$
$≥−HF+2F∇w∇φ+βt0nφ(wt−|∇w|2−ae(p−1)w−be(q−1)w)2−t0φ[|a|(p−β)(p−1)M1(p−1)+|b|(q−β)(q−1)M1(q−1)+2(1−β)K3+2βK1+32K4]|∇w|2+2t0φ(β−p)M1(p−1)γa|∇w|+2t0φ(β−q)M1(q−1)γb|∇w|−t0φM1(p−1)θa−t0φM1(q−1)θb−nt0φ(1β(K2+K3)2+32K4)+|a|(p−1)M1(p−1)φF+|b|(q−1)M1(q−1)φF−φt0−1F=−HF+2F∇w∇φ+βt0nφ(wt−|∇w|2−ae(p−1)w−be(q−1)w)2−t0φ[|a|(p−β)(p−1)M1(p−1)+|b|(q−β)(q−1)M1(q−1)]|∇w|2−t0φ[2(1−β)K3+2βK1+32K4]|∇w|2−t0φ[2(p−β)M1(p−1)γa+2(q−β)M1(q−1)γb]|∇w|−t0φ[M1(p−1)θa+M1(q−1)θb+n(1β(K2+K3)2+32K4)]+|a|(p−1)M1(p−1)φF+|b|(q−1)M1(q−1)φF−φt0−1F.$

For the sake of simplicity, set

$C˜1=2(1−β)K3+2βK1+32K4C˜2=M1(p−1)θa+M1(q−1)θb+n(1β(K2+K3)2+32K4)$

and

$C˜3=−[(p−β)M1(p−1)γa+(q−β)M1(q−1)γb]2|a|(p−β)(p−1)M1(p−1)+|b|(q−β)(q−1)M1(q−1).$

Using the inequality $ax2+bx≤−b24a$ which holds for $a<0$, we obtain

$0≥−HF+2F∇w∇φ+βt0nφ(wt−|∇w|2−ae(p−1)w−be(q−1)w)2−t0φ[C˜3+C˜2+C˜1|∇w|2]+|a|(p−1)M1(p−1)φF+|b|(q−1)M1(q−1)φF−φt0−1F.$

Noting the fact that $0<φ<1$ and multiplying both sides of the above inequality by $t0φ$, leads to

$0≥−Ht0φF+2t0φF∇w∇φ+βt02nφ2(wt−|∇w|2−ae(p−1)w−be(q−1)w)2−C˜1t02φ2|∇w|2−(C˜2+C˜3)t02φ2+|a|(p−1)M1(p−1)t0φF+|b|(q−1)M1(q−1)t0φF−φF≥−Ht0φF−2c1Rt0φF|∇w|φ32+|a|(p−1)M1(p−1)t0φF+|b|(q−1)M1(q−1)t0φF−φF+βt02nφ2[(wt−|∇w|2−ae(p−1)w−be(q−1)w)2−nβC˜1|∇w|2]−(C˜2+C˜3)t02φ2,$

where in the last inequality the following fact is applied

$−2φ∇w∇F=2F∇w∇φ≥−2F|∇w||∇φ|≥−2c1Rφ12F|∇w|.$

Assume that

$y=φ|∇w|2, z=φ(ae(p−1)w+be(q−1)w−wt).$

So, we can write

$0≥φF(−Ht0+|a|(p−1)M1(p−1)t0+|b|(q−1)M1(q−1)t0−1)−2c1Rt0F|∇w|φ32+βt02nφ2[(wt−|∇w|2−ae(p−1)w−be(q−1)w)2−nβC˜1|∇w|2]−(C˜2+C˜3)t02φ2≥φF(−Ht0+|a|(p−1)M1(p−1)t0+|b|(q−1)M1(q−1)t0−1)+βt02n{(y−z)2−nβC˜1y−2nc1R−1y12(y−1βz)}−(C˜2+C˜3)t02.$

For all $a,b>0$ the inequality $ax2−bx≥−b24a$ holds for every real number x. Using this inequality, we obtain

$βt02n{(y−z)2−nβC˜1y−2nc1R−1y12(y−1βz)}=βt02n{β2(y−zβ)2+(1−β2)y2−nβC˜1y+[2(β−β2)y−2nc1Ry12](y−zβ)}≥βt02n{β2(y−zβ)2−n2C˜124β2(1−β)2−n2c122R2(β−β2)(y−zβ)}=βn(φF)2−nC˜12t024β(1−β)2−nc12t02R2(β−β2)(φF).$

Hence,

$βn(φF)2+[−Ht0+|a|(p−1)M1(p−1)t0+|b|(q−1)M1(q−1)t0−1−nc12t02R2(β−β2)](ϕF)−nC˜12t024β(1−β)2−(C˜2+C˜3)t02≤0.$

As we know, the inequality $Ax2−2Bx≤C$, yields $x≤2BA+CA$. So, we get

$φF≤nβ(Ht0+|a|(1−p)M1(p−1)t0+|b|(1−q)M1(q−1)t0+1+nc12t02R2(β−β2)) +[nβ(nC˜124β(1−β)2+C˜2+C˜3)]12t0.$

If $d(x,x0,T1)≤2R$, we know that $φ(x,T1)=1$. Then

$F(x,T1)=T1(β|∇w|2+ae(p−1)w+be(q−1)w−wt) ≤φF(x0,t0) ≤nβ(Ht0+|a|(1−p)M1(p−1)t0+|b|(1−q)M1(q−1)t0+1+nc12t02R2(β−β2)) +[nβ(nC˜124β(1−β)2+C˜2+C˜3)]12t0.$

Since T1 was supposed to be arbitrary, we can get the assertion.

Proof of Corollary 1.3. For any points $(x1,t1)$ and (x2,t2) on $M×[0,T)$ with $0, we take a curve γ(t) parametrized with $γ(t1)=x1$ and $γ(t2)=x2$. In the ray of Corollary 1.2, one can get

$logu(x2,t2)−logu(x1,y1)=∫t1t2((logu)t+〈∇logu,γ′〉)dt≥∫t1t2(β|∇logu|2+aup−1+buq−1−H1¯−H2−nβt−|∇logu||γ′|)dt≥−∫t1t2(14β|γ′|2−aup−1−buq−1+H1¯+H2+nβt)dt≥−(log(t2t1)nβ+(H1¯+H2)(t2−t1)+∫t1t214β|γ′|2dt)$

which means

$logu(x1,t1)u(x2,t2)≤log(t2t1)nβ+(H1¯+H2)(t2−t1)+∫t1t2 1 4β |γ′|2dt.$

Hence,

$u(x1,t1)≤u(x2,t2)(t2 t1 )nβeΨ(x1,x2,t1,t2)+(H1 ¯+H2)(t2−t1),$

where $Ψ(x1,x2,t1,t2)=infγ∫t1 t2 1 4β|γ′|2dt$, and $H1¯=nβ(c3K2+a(1−p)M1(p−1)+b(1−q)M1(q−1))$, and

$H2=[n24β2(1−β)2(2(1−β)K3+2βK1+32K4)2+nβ{M1(p−1)θa+M1(q−1)θb+n(1β(K2+K3)2+32K4)}−nβ{[(p−β)M1(p−1)γa+(q−β)M1(q−1)γb]2a(p−β)(p−1)M1(p−1)+b(q−β)(q−1)M1(q−1)}]12.$

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