Gradient estimates and Harnack inequalities of a parabolic equation under geometric flow

J. Math. Anal. Appl. 483 (2020) 123631

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

Gradient estimates and Harnack inequalities of a parabolic equation under geometric flow Guangwen Zhao 1 School of Mathematical Sciences, Fudan University, Shanghai 200433, China

a r t i c l e

i n f o

Article history: Received 14 May 2019 Available online 31 October 2019 Submitted by J. Lenells Keywords: Parabolic equation Gradient estimate Harnack inequality Geometric flow

a b s t r a c t In this paper, we consider a manifold evolving by a general geometric flow and study parabolic equation (Δ − q(x, t) − ∂t )u(x, t) = A(u(x, t)),

(x, t) ∈ M × [0, T ].

We establish space-time gradient estimates for positive solutions and elliptic type gradient estimates for bounded positive solutions of this equation. By integrating the gradient estimates, we derive the corresponding Harnack inequalities. Finally, as applications, we give gradient estimates of some specific parabolic equations. © 2019 Elsevier Inc. All rights reserved.

1. Introduction The paper study parabolic equation (Δ − q(x, t) − ∂t )u(x, t) = A(u(x, t))

(1.1)

on Riemannian manifold M evolving by the geometric flow ∂ g(t) = 2h(t), ∂t

(1.2)

where (x, t) ∈ M × [0, T ], q(x, t) is a function on M × [0, T ] of C 2 in x-variables and C 1 in t-variable, A(u) is a function of C 2 in u, and h(t) is a symmetric (0, 2)-tensor field on (M, g(t)). A important example would be the case where h(t) = − Ric(t) and g(t) is a solution of the Ricci flow introduced by R.S. Hamilton [12]. We will give some gradient estimates and Harnack inequalities for positive solutions of equation (1.1).

1

E-mail address: [email protected]. Current address: Department of Mathematics, Wuhan University of Technology, Wuhan 430070, P.R. China.

https://doi.org/10.1016/j.jmaa.2019.123631 0022-247X/© 2019 Elsevier Inc. All rights reserved.

2

G. Zhao / J. Math. Anal. Appl. 483 (2020) 123631

The study of gradient estimates for parabolic equations originated with the work of P. Li and S.-T. Yau [19]. They prove a space-time gradient estimate for positive solutions of the heat equation on a complete manifold. By integrating the gradient estimate along a space-time path, a Harnack inequality was derived. Therefore, Li–Yau inequality is often called differential Harnack inequality. It is easy to see that the above space-time estimate will become an elliptic type gradient estimate for a time-independent solution (see [7]). But the elliptic type estimate cannot hold for a time-dependent solution in general, this can be seen from the form of the fundamental solution of the heat equation in Rn . However, in 1993, R.S. Hamilton [14] established an elliptic type gradient estimate for positive solutions of the heat equation on compact manifolds. It is worth noting that the noncompact version of Hamilton’s estimate is not true even for Rn (see [29, Remark 1.1]). Nevertheless, for complete noncompact manifolds, P. Souplet and Q.S. Zhang [29] obtained an elliptic type gradient estimate for a bounded positive solution of the heat equation after inserting a necessary logarithmic correction term. Li–Yau type and Hamilton–Souplet–Zhang type gradient estimates have been obtained for other nonlinear equations on manifolds, see for example [4,5,9,17,18,23,24,27,31–33] and the references therein. On the other hand, gradient estimates are very powerful tools in geometric analysis. For instance, R.S. Hamilton [13,15] established differential Harnack inequalities for the Ricci flow and the mean curvature flow. These results have important applications in the singularity analysis. Over the past two decades, many authors used similar techniques to prove gradient estimates and Harnack inequalities for geometric flows. The list of relevant references includes but is not limited to [1,3,11,16,21,20,22,25,30,34,35]. In this paper, we follow the work of J. Sun [30] and M. Bailesteanu et al. [1], and focus on the system (1.1)–(1.2). Now we give some remarks on equation (1.1). When A(u) = au log u, the nonlinear elliptic equation corresponding to (1.1) is related to the gradient Ricci soliton. When A(u) = auβ , the nonlinear elliptic equation corresponding to (1.1) is related to the Yamabe-type equation. In general, the parabolic equation (1.1) is the so-called reaction-diffusion equation, which can be found in many mathematical models in physics, chemistry and biology (see [26,28]), where qu + A(u) and Δu are the reaction term and the diffusion term, respectively. The reaction-diffusion equations are very important objects in pure and applied mathematics. In [6], Q. Chen and the author studied the equation (1.1) with a convection term on a complete manifold with a fixed metric. Here, we establish some gradient estimates for positive solutions of (1.1) under geometric flow (1.2), which are richer and sharper than [6]. The rest of this paper is organized as follows. In Section 2, we establish space-time gradient estimates for positive solution of (1.1). We firstly consider that M is a complete noncompact manifold without boundary. A local and a global estimate were established, see Theorem 2.1 and Corollary 2.6. Next, the case that M is closed is also deal with. In this case, inspired by [1], we obtain a sharper estimate than [30, Theorem 6], see Theorem 2.7. We also give the corresponding Harnack inequalities in the above two cases, see Corollary 2.10. In Section 3, we consider the case that the solution is bounded, and establish elliptic type gradient estimates of local and global versions, see Theorem 3.1 and Corollary 3.5. The elliptic type Harnack inequality is also obtained, see Corollary 3.6. Finally, in Section 4, we give some applications and explanations of these gradient estimates in some specific cases. For the case of A(u) = au log u with a ∈ R, we can derive the gradient estimate for positive solutions. In particular, we deal with the case that the manifold evolving by the Ricci flow, see Corollary 4.2, 4.3, 4.4. For the case of A(u) = auβ with a ∈ R and β ∈ (−∞, 0] ∪ [1, +∞), we give the gradient estimate for bounded positive solutions, see Corollary 4.5, 4.6. Throughout the paper, we denote by n the dimension of the manifold M , and by d(x, y, t) the geodesic distance between x, y ∈ M under g(t). When we say that u(x, t) is a solution to the equation (1.1), we mean

G. Zhao / J. Math. Anal. Appl. 483 (2020) 123631

3

u is a solution which is smooth in x-variables and t-variable. In addition, we have to give some notations ˆ ) = A(u) . Then for the convenience of writing. Let f = log u and A(f u Aˆf = A (u) − A(u)/u,

Aˆf f = uA (u) − A (u) + A(u)/u.

For u > 0 we define several nonnegative real numbers (some of λ, Λ, Σ, κ are allowed to be infinite) as follows: λ2R

  −  ˆ := − min Af = − min 0, min (A (u) − A(u)/u) , Q2R,T

Q2R,T



  Λ2R := max Aˆ+ = max 0, max (A (u) − A(u)/u) , f Q2R,T



Q2R,T

Σ2R

 +   ˆ := max Af f = max 0, max (uA (u) − A (u) + A(u)/u)

κ2R

    := − min 0, min (A (u) − A(u)/u) , min A (u)

Q2R,T

Q2R,T

Q2R,T

Q2R,T

and λ := − Λ :=

inf

M ×[0,T ]

sup M ×[0,T ]

Σ :=

sup M ×[0,T ]

 − ˆ Af = − min 0, 

Aˆ+ f

inf

M ×[0,T ]





= max 0, sup (A (u) − A(u)/u) , 

Aˆ+ ff

 (A (u) − A(u)/u) , 

M ×[0,T ]

 



= max 0, sup (uA (u) − A (u) + A(u)/u) M ×[0,T ]









κ := − min 0, inf (A (u) − A(u)/u) , inf A (u) . Q2R,T

Q2R,T

Here, we denote by v + = max{0, v} and v − = min{0, v} the positive part and the negative part of a function v. Notice that if M is compact, then λ, Λ, Σ and κ must be finite. 2. Space-time gradient estimates for positive solutions Firstly, we have the following local space-time gradient estimate for (1.1)–(1.2). Theorem 2.1. Let (M, g(0)) be a complete Riemannian manifold, and let g(t) evolves by (1.2) for t ∈ [0, T ]. Given x0 and R > 0, let u be a positive solution to (1.1) in the cube Q2R,T := {(x, t) : d(x, x0 , t) ≤ 2R, 0 ≤ t ≤ T }. Suppose that there exist constants K1 , K2 , K3 , K4 , γ, θ ≥ 0 such that Ric ≥ −K1 g,

−K2 g ≤ h ≤ K3 g,

and |∇q| ≤ γ2R , on Q2R,T . Then for any α > 1 and 0 < ε < 1, we have

Δq ≤ θ2R

|∇h| ≤ K4

G. Zhao / J. Math. Anal. Appl. 483 (2020) 123631

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ut (x, t) A(u(x, t)) |∇u(x, t)|2 −α − αq(x, t) − α 2 u (x, t) u(x, t) u(x, t)  2  2 2  nα Cα α ≤ + 2 + K1 R + Cα2 K2 + nα2 λ2R t R α−1  +

nα2 αθ2R + nα2 max{K22 , K32 }

(2.1)

2

nα2 (α − 1)Λ2R + αΣ2R + 2(K1 + (α − 1)K3 + K4 ) 2 4(1 − ε)(α − 1)  1  12 4 2 9 2 3 4nα2 3 3 + nα K4 + (α − 1) 3 γ2R 8 4 ε

+

on QR,T , where C is a constant that depends only on n. Remark 2.2. We see that Theorem 2.1 covers [30, Theorem 1]. In fact, when q(x, t) = A(u) = 0, from Theorem 2.1 we can get  2   α ut (x, t) nα2 Cα2 |∇u(x, t)|2 −α ≤ + 2 + K1 R + Cα2 K2 u2 (x, t) u(x, t) t R α−1  9 2 + nα nα2 max{K22 , K32 } + nα2 K4 8  12 nα2 2 + (K1 + (α − 1)K3 + K4 ) . (1 − ε)(α − 1)2 Let ε → 0+, we thus get ut (x, t) nα2 Cα2 |∇u(x, t)|2 − α ≤ + u2 (x, t) u(x, t) t R2

   α2 R K1 + + Cα2 K2 α−1

nα2 (K1 + (α − 1)K3 + K4 ) α−1 2

 + nα2 max{K2 , K3 } + 9K4 /8 . +

To prove Theorem 2.1, we need the following two lemmas. Let f = log u, by (1.1) we know that f satisfies Δf = ft + q +

A(u) ˆ ) − |∇f |2 − |∇f |2 = ft + q + A(f u

(2.2)

ˆ We have Set F = t(|∇f |2 − αft − αq − αA). Lemma 2.3 (Lemma 3 in [30]). Suppose the metric evolves by (1.2). Then for any smooth function f , we have ∂ |∇f |2 = −2h(∇f, ∇f ) + 2 ∇f, ∇(ft )

∂t and (Δf )t = Δ(ft ) − 2 h, Hess f − 2 div h − 12 ∇(trg h), ∇f , where div h is the divergence of h.

G. Zhao / J. Math. Anal. Appl. 483 (2020) 123631

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Lemma 2.4. Let (M, g(t)) satisfies the hypotheses of Theorem 2.1. Then for any δ ∈ (0, α1 ), we have (Δ − ∂t )F ≥

2(1 − δα)t ˆ 2 − F − 2 ∇f, ∇F

(|∇f |2 − ft − q − A) n t ˆ − 2(α − 1)t ∇f, ∇q

+ αtAˆf (|∇f |2 − ft − q − A) − t(2(α − 1)Aˆf + αAˆf f + 2K1 + 2(α − 1)K3 )|∇f |2

(2.3)

√ αtn max{K22 , K32 } − αtΔq. − 3αt nK4 |∇f | − 2δ Proof. By the Bochner formula, (2.2) and Lemma 2.3, we calculate ΔF =2t| Hess f |2 + 2t Ric(∇f, ∇f ) + 2t ∇f, ∇Δf

− αtΔ(ft ) − αtΔq − αtAˆf Δf − αtAˆf f |∇f |2 =2t| Hess f |2 + 2t Ric(∇f, ∇f ) + 2t ∇f, ∇Δf

− αt(Δf )t − 2αt h, Hess f − 2αt div h − 12 ∇(trg h), ∇f

− αtΔq − αtaAˆf Δf − αtaAˆf f |∇f |2 . By (2.2) and the definition of F we have ∇Δf = −

∇F − (α − 1)(∇(ft ) + ∇q + Aˆf ∇f ) t

(Δf )t =

F Ft − (α − 1)(ftt + qt + Aˆf ft ). − 2 t t

and

By Lemma 2.3 we also have Ft =|∇f |2 − αft − αq − αAˆ + 2t ∇f, (∇f )t − αt(ftt + qt + Aˆf ft ) =|∇f |2 − αft − αq − αAˆ + 2t ∇f, ∇(ft ) − 2th(∇f, ∇f ) − αt(ftt + qt + Aˆf ft ). It follows the above equalities that (Δ − ∂t )F =2t| Hess f |2 + 2t Ric(∇f, ∇f ) −

F − 2 ∇f, ∇F

t

− 2αt h, Hess f − 2αt div h − 12 ∇(trg h), ∇f

ˆ − αtΔq − 2(α − 1)t ∇f, ∇q

+ αtAˆf (|∇f |2 − ft − q − A) − 2(α − 1)th(∇f, ∇f ) − 2(α − 1)tAˆf |∇f |2 − αtAˆf f |∇f |2 The assumption −K2 g ≤ h ≤ K3 g implies |h|2 ≤ n max{K22 , K32 }. By Young’s inequality,

(2.4)

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G. Zhao / J. Math. Anal. Appl. 483 (2020) 123631

1 2 |h| 4δ n max{K22 , K32 } ≤δ| Hess f |2 + 4δ

h, Hess f ≤δ| Hess f |2 +

(2.5)

for any δ ∈ (0, α1 ). We also have | div h − 12 ∇(trg h)| =|g ij ∇i hjl − 12 g ij ∇l hij | 3√ 3 nK4 . ≤ |g||∇h| ≤ 2 2

(2.6)

On the other hand, | Hess f |2 ≥

1 1 ˆ 2. (Δf )2 = (|∇f |2 − ft − q − A) n n

(2.7)

Substituting (2.5), (2.6) and (2.7) into (2.4) and using the assumptions on bounds of Ric and h, we obtain the final inequality (2.3). 2 The proof of Theorem 2.1. By the assumption of bounds of Ricci tensor and the evolution of the metric, we know that g(t) is uniformly equivalent to the initial metric g(0) (see [8, Corollary 6.11]), that is, e−2K2 T g(0) ≤ g(t) ≤ e2K3 T g(0). Then we know that (M, g(t)) is also complete for t ∈ [0, T ]. Let φ ∈ C ∞ ((0, +∞]), φ(s) =

 1,

satisfies φ(s) ∈ [0, 1], φ (s) ≤ 0, φ (s) ≥ −C1 and

0,

s ∈ [0, 1], s ∈ [2, +∞)

|φ (s)|2 φ(s)

 η(x, t) = φ

≤ C1 , where C1 is an absolute constant. Define

r(x, t) R

 ,

where r(x, t) = d(x, x0 , t). Using the argument of [2], we can assume that the function η(x, t) is C 2 with support in Q2R,T . Define G = ηF . For any T1 ∈ (0, T ], let (x1 , t1 ) ∈ Q2R,T1 at which G attains its maximum, and without loss of generality, we can assume G(x1 , t1 ) > 0, and then η(x1 , t1 ) > 0 and F (x1 , t1 ) > 0. Hence, at (x1 , t1 ), we have ∇G = 0,

ΔG ≤ 0,

∂t G ≥ 0.

Hence, we obtain ∇F = −

F ∇η η

(2.8)

and 0 ≥(Δ − ∂t )G =F (Δ − ∂t )η + η(Δ − ∂t )F + 2 ∇η, ∇F .

(2.9)

G. Zhao / J. Math. Anal. Appl. 483 (2020) 123631

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By the properties of φ and the Laplacian comparison theorem, we have C1 |∇η|2 ≤ 2 η R and Δr |∇r|2 + φ 2 R R √    C1 C1 K1 ≥− (n − 1)K1 coth n−1 R − R2 R  √ (n − 1)C1 K1 C1 (n − 1) C1 − 2. ≥− − R2 R R

Δη =φ

By [30, p. 494], there exist a constant C2 such that −F ηt ≥ −C2 K2 F. Substituting the above three inequalities into (2.9) and using (2.8), we obtain 

  √ (n − 1)C1 K1 3C1 (n − 1) C1 + 2 + C2 K2 F + η(Δ − ∂t )F 0≥− + R2 R R   C3 (n) C4 (n)  + K1 + C2 K2 F. =:η(Δ − ∂t )F − R2 R Let μ =

|∇f (x1 ,t1 )|2 F (x1 ,t1 ) ,

(2.10)

then, at (x1 , t1 ) we have √ C1 1 η 2 F |∇f | η ∇f, ∇F = −F ∇f, ∇η ≤ R

and  |∇f |2 − ft − q − aAˆ =

μ−

t1 μ − 1 t1 α

 F.

Therefore, at (x1 , t1 ), by Lemma 2.4 and (2.10), and using the inequality √ 9 3α nK4 |∇f | ≤ 2K4 |∇f |2 + nα2 K4 , 8 we obtain 0≥

√  2 t1 μ − 1 2(1 − δα)t1 ηF 2 C1 1 1 3 η μ− η 2 μ2 F 2 F2 − − n t1 α t1 R   t1 μ − 1 ˆ F − αt1 ηΔq − 2(α − 1)t1 η ∇f, ∇q

+ αt1 η Af μ − t1 α − t1 η(2(α − 1)Aˆf + αAˆf f + 2(K1 + (α − 1)K3 + K4 ))μF t1 ηnα 9 max{K22 , K32 } − t1 ηnα2 K4 − 8 2δ 

 3 (n) + C4R(n) K1 + C2 K2 F. − CR 2

(2.11)

G. Zhao / J. Math. Anal. Appl. 483 (2020) 123631

8

Multiplying through by t1 η, we conclude that 0≥

√ 1 3 2(1 − δα) 2 C1 2 2 (1 + (α − 1)t μ) G − ηG − t1 μ 2 G 2 1 2 nα R 2 2 ˆ + t1 η Af (1 + (α − 1)t1 μ)G − αt η Δq − 2(α − 1)t2 η 2 ∇f, ∇q

1

−

t21 η[2(α

1

− 1)Aˆf + αAˆf f + 2(K1 + (α − 1)K3 + K4 )]μG

9 t2 η 2 nα max{K22 , K32 } − t21 η 2 nα2 K4 − 1 8 2δ

  3 (n) + C4R(n) K1 + C2 K2 G − t1 CR 2

=

√ 1 3 2 C1 2(1 − δα) 2 2 t1 μ 2 G 2 (1 + (α − 1)t μ) G − ηG − 1 nα2 R + t1 η Aˆf G − αt2 η 2 Δq − 2(α − 1)t2 η 2 ∇f, ∇q

1

1

− t21 η[(α − 1)Aˆf + αAˆf f + 2(K1 + (α − 1)K3 + K4 )]μG t2 η 2 nα 9 − t21 η 2 nα2 K4 − 1 max{K22 , K32 } 8 2δ

  3 (n) − t1 CR + C4R(n) K1 + C2 K2 G 2 Noticing that 0 < η(x1 , t1 ) ≤ 1, from the above inequalities we obtain 0≥

2(1 − δα) 2 4(1 − δα) G + (α − 1)t1 μG2 nα2 nα2 √ 1 3 2(1 − δα) 2 C1 2 2 2 2 + (α − 1) t1 μ G − G − t1 λ2R G − t1 μ 2 G 2 nα2 R   − t21 (α − 1)Λ2R + αΣ2R + 2(K1 + (α − 1)K3 + K4 ) μG 1 1 9 t2 nα − 2t21 (α − 1)γ2R μ 2 G 2 − t21 nα2 K4 − 1 max{K22 , K32 } 8 2δ

  3 (n) + C4R(n) K1 + C2 K2 G. − αt21 θ2R − t1 CR 2

By Young’s inequality, we have √ 2 C1 1 3 4(1 − δα) nα2 C1 G μ2 G2 ≤ (α − 1)μG2 + , 2 R nα 8(1 − δα)(α − 1)R2   13 4 1 1 2 2(1 − δα)ε 3 2nα2 2 2 2 3 2(α − 1)γ2R μ 2 G 2 ≤ (α − 1) μ G + (α − 1) 3 γ2R 2 nα 4 (1 − δα)ε and   (α − 1)Λ2R + αΣ2R + 2(K1 + (α − 1)K3 + K4 ) μG ≤

2(1 − δα)(1 − ε) (α − 1)2 μ2 G2 nα2 2  nα2 + (α − 1)Λ + αΣ + 2(K + (α − 1)K + K ) , 2R 2R 1 3 4 8(1 − δα)(1 − ε)(α − 1)2

where ε ∈ (0, 1) is an arbitrary constant. Combining the above four inequalities, there exists a constant C5 (n) that depends only on n, such that

G. Zhao / J. Math. Anal. Appl. 483 (2020) 123631

0≥

9

2(1 − δα) 2 G nα2      C5 (n) α2 − 1 + t1 λ2R + + K1 R + C2 K2 G R2 (1 − δα)(α − 1) − t21 αθ2R − t21 nα2 max{K22 , K32 } − t21

nα2 8(1 − δα)(1 − ε)(α − 1)2

 2 · (α − 1)Λ2R + αΣ2R + 2(K1 + (α − 1)K3 + K4 ) 9 3 − t21 nα2 K4 − t21 8 4



2nα2 (1 − δα)ε

 13

2

4

3 (α − 1) 3 γ2R .

For a positive number a and two nonnegative numbers b, c, from the inequality ax2 − bx − c ≤ 0 we have  x ≤ ab + ac . Hence, we obtain G≤

     nα2 C5 (n) α2 1 + t1 λ2R + + K R + C K 1 2 2 2(1 − δα) R2 (1 − δα)(α − 1)  + t1 nα2 αθ2R + nα2 max{K22 , K32 } +

nα2 8(1 − δα)(1 − ε)(α − 1)2

 2 · (α − 1)Λ2R + αΣ2R + 2(K1 + (α − 1)K3 + K4 ) 9 3 + nα2 K4 + 8 4 Now, by taking δ =

1 2α ,



2nα2 (1 − δα)ε

 13

2 3

4 3

(α − 1) γ2R

 12 .

and noticing that d(x, x0 , T1 ) ≤ R implies η(x, T1 ) = 1, we can get

F (x, T1 ) G(x1 , t1 ) ˆ (|∇f |2 − αf1 − αqt − αA)(x, T1 ) = ≤ T1 T1  2  2 2  nα C(n)α α ≤ + + K1 R + C(n)α2 K2 + nα2 λ2R 2 T1 R α−1  + nα2 αθ2R + nα2 max{K22 , K32 }

2 nα2 (α − 1)Λ + αΣ + 2(K + (α − 1)K + K ) 2R 2R 1 3 4 4(1 − ε)(α − 1)2  1  12 4 2 9 2 3 4nα2 3 3 + nα K4 + (α − 1) 3 γ2R , 8 4 ε

+

where C(n) is an appropriate constant that depends only n. Since T1 is arbitrary, we complete the proof. 2 √   1 Remark 2.5. In the above proof, if we use x ≤ 2a b + b2 + 4ac instead of x ≤ with ax2 − bx − c ≤ 0, then a more appropriate δ may give a sharper estimate. From the above local estimate, we get a global one:

b a

+

c a

when we deal

G. Zhao / J. Math. Anal. Appl. 483 (2020) 123631

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Corollary 2.6. Let (M, g(0)) be a complete noncompact Riemannian manifold without boundary, and let g(t) evolves by (1.2) for t ∈ [0, T ]. Suppose that there exist constant K1 , K2 , K3 , K4 , γ, θ ≥ 0 such that Ric ≥ −K1 g,

−K2 g ≤ h ≤ K3 g,

|∇h| ≤ K4

and |∇q| ≤ γ,

Δq ≤ θ.

If u is a positive solution to (1.1), then for any α > 1, we have ut (x, t) A(u(x, t)) |∇u(x, t)|2 −α − αq(x, t) − α 2 u (x, t) u(x, t) u(x, t) 

2  √ 2 nα ≤ + C  K1 + K 2 + K 3 + K 4 + K 4 + θ + γ 3 + λ + Λ + Σ t

(2.12)

on M × [0, T ], where C  is a constant that depends only on n, α. Proof. By the uniform equivalence of g(t), we know that (M, g(t)) is complete noncompact for t ∈ [0, T ]. Now we choose ε = ε0 in (2.1), where ε0 is an arbitrary fixed number in (0, 1). Let R → +∞ in (2.1), and √ √ √ using the inequality x + y ≤ x + y holds for any x, y ≥ 0, we complete the proof. 2 We now consider the case that the manifold M is closed. By Lemma 2.4, we have a global gradient estimate on a closed Riemannian manifold. Theorem 2.7. Let (M, g(t)) be a closed Riemannian manifold, where g(t) evolves by (1.2) for t ∈ [0, T ] and satisfies Ric ≥ −K1 g,

−K2 g ≤ h ≤ K3 g,

|∇h| ≤ K4 .

If u is a positive solution to (1.1), and q(x, t) satisfies |∇q| ≤ γ,

Δq ≤ θ.

Then for any α > 1, we have ut (x, t) A(u(x, t)) |∇u(x, t)|2 −α − αq(x, t) − α u2 (x, t) u(x, t) u(x, t)   2 2 nα nα 3 ≤ + (K1 + (α − 1)K3 + K4 ) + nα2 max{K2 , K3 } + 2K4 2t α−1 4   2 2 √ √ 3 1 1 2 nα3 nα nα (α − 1) 2 + nα(α − 1) 4 γ 3 + (λ + Λ) + Σ + α 2 nθ + 2 2 2(α − 1) on M × (0, T ].

(2.13)

G. Zhao / J. Math. Anal. Appl. 483 (2020) 123631

11

Proof. We use the same symbols f, F as above. Set 2

nα (K1 + (α − 1)K3 + K4 )t F¯ (x, t) =F (x, t) − α−1   3√ 3 2K4 t − α 2 nθt − nα2 max{K2 , K3 } + 4  2  √ 1 1 2 nα3 nα2 nα − (α − 1) 2 + nα(α − 1) 4 γ 3 t − (λ + Λ)t − Σt. 2 2 2(α − 1) 2 If F¯ (x, t) ≤ nα 2 for any (x, t) ∈ M × (0, T ], the proof is complete. If (2.13) doesn’t hold, then at the maximal point (x0 , t0 ) of F¯ (x, t), we have

nα2 . F¯ (x0 , t0 ) > 2 As F¯ (x, 0) = 0, we know that t0 > 0 here. Then applying the maximum principle, we have ∇F¯ (x0 , t0 ) = 0,

ΔF¯ (x0 , t0 ) = 0,

∂t F¯ (x0 , t0 ) = 0.

Therefore, we obtain 0 ≥ (Δ − ∂t )F¯ ≥ (Δ − ∂t )F. Using Lemma 2.4, inequality (2.11) and the fact that 1 ˆ + α − 1 |∇f |2 |∇f |2 − ft − q − Aˆ = (|∇f |2 − αft − αq − αA) α α 1F α−1 |∇f |2 , = + α t0 α we obtain 2(1 − δα)t0 0≥ nα2



F t0

2 +

4(1 − δα)(α − 1)t0 F |∇f |2 nα2 t0

2(1 − δα)(α − 1)2 t0 F F |∇f |4 − − t0 λ nα2 t0 t0 9 t0 nα max{K22 , K32 } − 2t0 (α − 1)γ|∇f | − t0 nα2 K4 − t0 αθ − 8 2δ +

− t0 ((α − 1)Λ + αΣ + 2(K1 + (α − 1)K3 + K4 )) |∇f |2 . By  nα2 3 F F¯ (K1 + (α − 1)K3 + K4 ) + nα2 K4 = + t0 t0 α−1 2   2 √ 1 1 2 nα (α − 1) 2 + 2nα(α − 1) 4 γ 3 + 2 + nα2 A1 +

nα3 A2 > 0, 2(α − 1) 3

2

1

4

2t0 (α − 1)γ|∇f |2 ≤ t0 (α − 1) 2 γ 3 |∇f |2 + t0 (α − 1) 2 γ 3 ,

G. Zhao / J. Math. Anal. Appl. 483 (2020) 123631

12

2

b and using the inequality ax2 − bx ≥ − 4a holds for a > 0, b ≥ 0, we obtain

0≥

2(1 − δα)t0 nα2 −



F t0

2 −

F F 9 − t0 λ − t0 nα2 K4 − t0 αθ t0 t0 8

1 4 t0 nα t0 nα2 max{K22 , K32 } − t0 (α − 1) 2 γ 3 − E2, 2δ 8(1 − δα)(α − 1)2

where 3

2

E = (α − 1) 2 γ 3 + (α − 1)Λ + αΣ + 2(K1 + (α − 1)K3 + K4 ). For a positive number a and two nonnegative numbers b, c, from the inequality ax2 − bx − c ≤ 0 we have x≤

  1

b + b2 + 4ac . 2a

Hence, we obtain  nα2 F 1 + t0 λ + (1 + t0 λ)2 ≤ t0 4(1 − δα)t0  nα2 t0 8(1 − δα)t0 + E2 nα2 8(1 − δα)(α − 1)2

 12 

9 t0 nα max{K22 , K32 } + t0 (α − 1) γ + t0 nα2 K4 + t0 αθ + 8 2δ 1 2

Using the inequality

4 3

.

√ √ √ x + y ≤ x + y holds for any x, y ≥ 0, we obtain

 12  nα2 4(1 − δα)t20 F 2 2 2 max{K2 , K3 } ≤ 1 + t0 λ + (1 + t0 λ) + t0 4(1 − δα)t0 δα  nα2 nα2 8(1 − δα)t20 + E2 2 4(1 − δα)t0 nα 8(1 − δα)(α − 1)2 9 + (α − 1) γ + nα2 K4 + αθ 8 1 2

 12

4 3

.

0 max{K2 ,K3 } Now, by taking δ = 1+t0tλ+2t · α1 ∈ (0, α1 ) for max{K2 , K3 } = 0. However, from Lemma 2.4 we 0 max{K2 ,K3 } know that we can choose δ = 0 if K2 = K3 = 0. Therefore, in any case, we can get

 12  4(1 − δα)t20 nα2 2 2 2 max{K2 , K3 } 1 + t0 λ + (1 + t0 λ) + 4(1 − δα)t0 δα =

nα2 nα2 λ + nα2 max{K2 , K3 } + 2t0 2

and nα2 4(1 − δα)t0



8(1 − δα)t20 nα2

 12

 =

1 + t0 λ + 2t0 max{K2 , K3 } 2(1 + t0 λ + t0 max{K2 , K3 })

 12

√

nα ≤

√

nα.

G. Zhao / J. Math. Anal. Appl. 483 (2020) 123631

Therefore, again according to

13

√ √ √ x + y ≤ x + y, we obtain

nα2 F nα2 λ + nα2 max{K2 , K3 } ≤ + t0 2t0 2   √ √ nα 1 + t0 λ + 2t0 max{K2 , K3 } E + nα √ 2 2(α − 1) 1 + t0 λ + t0 max{K2 , K3 }  √ 1 2 3 √  4 3 + (α − 1) γ + √ nα K4 + αθ 2 2 ≤

nα2 nα2 λ + nα2 max{K2 , K3 } + 2t0 2  √ 1 2 3√ nα2 3 E + nα(α − 1) 4 γ 3 + nα2 2K4 + α 2 nθ. + 2(α − 1) 4

Substituting E into the above inequality yields nα2 F nα2 (K1 + (α − 1)K3 + K4 ) ≤ + t0 2t0 α−1   3√ 3 2 2K4 + α 2 nθ + nα max{K2 , K3 } + 4   2 √ 1 1 2 nα3 nα2 nα + (α − 1) 2 + nα(α − 1) 4 γ 3 + (λ + Λ) + Σ. 2 2 2(α − 1) This implies that F¯ (x0 , t0 ) ≤

nα2 2 ,

in contradiction with our assumption. So (2.13) holds. 2

Remark 2.8. In [30, Theorem 6], the coefficient of 1t in the right hand side of the gradient inequality is nα2 . We see Theorem 2.7 extends and improves Sun’s estimate. Remark 2.9. In Theorem 2.1 if K1 = K4 = Σ2R = 0, we can let α → 1. Similarly, in Corollary 2.6 and Theorem 2.7, if K1 = K4 = Σ = 0, we can also let α → 1. Similar to [30, Corollary 8], integrating the gradient estimate in space-time as in [19] or [10], we can derive the following parabolic Harnack type inequality. Corollary 2.10. Let (M, g(0)) be a complete noncompact Riemannian manifold without boundary or a closed Riemannian manifold. Assume that g(t) evolves by (1.2) for t ∈ [0, T ] and satisfies Ric ≥ −K1 g,

−K2 g ≤ h ≤ K3 g,

|∇h| ≤ K4 .

If u is a positive solution to (1.1), and q(x, t) satisfies |∇q| ≤ γ,

Δq ≤ θ.

Then for any (x1 , t1 ), (x2 , t2 ) in M × (0, T ] such that t1 < t2 , we have  u(x1 , t1 ) ≤ u(x2 , t2 ) for any α > 1, where

t2 t1

τ nα

 exp

 t 2 − t1 αZ +C K , 4(t2 − t1 ) α

(2.14)

G. Zhao / J. Math. Anal. Appl. 483 (2020) 123631

14

τ=

 1,

if (M, g(0)) is complete noncompact without boundary,

1 2,

if (M, g(0)) is closed,  √ 2 K = K1 + K2 + K3 + K4 + K4 + γ + θ + γ 3 + λ + Λ + Σ,

C is a constant that depends only on n, α, and 1 Z = inf ζ

|ζ  (s)|2σ(s) ds

0

is the infimum over smooth curves ζ jointing x2 and x1 (ζ(0) = x2 , ζ(1) = x1 ) of the averaged square velocity of ζ measured at time σ(s) = (1 − s)t2 + st1 . Proof. The gradient estimate in Corollary 2.6 and Theorem 2.7 can both be written as ut (x, t) τ nα2 |∇u(x, t)|2 −α ≤ + C(n, α)K, 2 u (x, t) u(x, t) t for any α > 1. Take any curve ζ satisfying the assumption and define l(s) = log u(ζ(s), σ(s)). Then l(0) = log u(x2 , t2 ) and l(1) = log u(x1 , t1 ). A direct computation yields  ∇u ζ  (s) ut − u t2 − t1 u    2 t2 − t1 τ nα2 α|ζ (s)|σ + + CK . ≤ 4(t2 − t1 ) α σ(s)

dl(s) =(t2 − t1 ) ds



Integrating this inequality over ζ(s), we have u(x1 , t1 ) = log u(x2 , t2 )

1

dl(s) ds ds

0

1 ≤ 0

t2 − t1 t2 α|ζ  (s)|2σ +C K + τ nα log , 4(t2 − t1 ) α t1

which implies the corollary. 2 3. Elliptic type gradient estimates for bounded positive solutions Now we establish elliptic type gradient estimates for (1.1)–(1.2). Firstly we give the local version. Theorem 3.1. Let (M, g(t)) be a complete solution to (1.2) for t ∈ [0, T ] and let u be a positive solution to (1.1). Suppose that there exist constants L > 0 and K ≥ 0, such that u ≤ L and Ric ≥ −K1 g, on Q2R,T . Then we have

h ≥ −K2 g

G. Zhao / J. Math. Anal. Appl. 483 (2020) 123631

|∇u(x, t)| ≤ C˜ u(x, t)



1 1 √ √ + + H t R

 1 + log

L u(x, t)

15

 (3.1)

on QR,T , where C˜ is a constant that depends only on n and |∇q| H = K1 + K2 + max (|q| − q) + max  + κ2R . Q2R,T Q2R,T |q| Remark 3.2. In [6, Theorem 1.9], the authors gave a elliptic type gradient estimate for bounded positive solutions of (1.1) with a convection term on a complete manifold, where the metric does not depend on time. In the estimate of [6, Theorem 1.9], the upper bound induced by the term A(u) is     − min 0, min (A (u) − A(u)/u) − min 0, min (A(u)/u) Q2R,T

Q2R,T

instead of κ2R here. Compare with [6, Theorem 1.9], we see that our estimate (3.1) is sharper. In fact, in general, for real numbers x, y, a direct calculation yields min{0, x} + min{0, y} ≤ min{0, x + y}. Similarly, for two functions f, g on the same domain D, the following obvious fact holds: min f + min g ≤ min(f + g). D

D

D

By the above two inequalities we obtain      min 0, min (A (u) − A(u)/u) + min 0, min (A(u)/u) Q2R,T

Q2R,T

   ≤ min 0, min (A (u) − A(u)/u) + min (A(u)/u) Q2R,T

Q2R,T

   ≤ min 0, min A (u) . Q2R,T

On the other hand, it is obvious that      min 0, min (A (u) − A(u)/u) + min 0, min (A(u)/u) Q2R,T

  ≤ min 0, min (A (u) − A(u)/u) .

Q2R,T

Q2R,T

In conclusion,     min 0, min (A (u) − A(u)/u) + min 0, min (A(u)/u) Q2R,T

Q2R,T

    ≤ min 0, min (A (u) − A(u)/u), min A (u) = −κ2R . Q2R,T

Q2R,T

That is, 

κ2R

   ≤ − min 0, min (A (u) − A(u)/u) − min 0, min (A(u)/u) . 

Q2R,T

Q2R,T

G. Zhao / J. Math. Anal. Appl. 483 (2020) 123631

16

And we will see in the proof of Theorem 3.1 that this sharper estimate comes from a more careful treatment ˆ ) of the term Aˆf + A(f 1−f . However, the treatment we give here is not necessarily optimal. It is possible that a sharper estimate will be applied to more equations. Now we are ready to prove Theorem 3.1. Noticing that if 0 < u ≤ L is a solution to (1.1), then u ˜= a solution to the equation (Δ − q(x, t) − ∂t )u(x, t) =

u L

is

1 A(L˜ u(x, t)) L

and 0 < u ˜ ≤ 1. Hence, we can assume that 0 < u ≤ 1 in the proof of Theorem 3.1. Similar to the proof ˆ ) = A(u) . In this case, we of Theorem 2.1, we need a auxiliary lemma. We still set f = log u ≤ 0 and A(f u 2 define w = |∇ log(1 − f )| and F (x, t) = tw(x, t). Lemma 3.3. Let (M, g(t)) be a complete solution to (1.2) for t ∈ [0, T ] and let u ∈ (0, 1] be a solution to (1.1). Suppose that there exists a constant K ≥ 0, such that Ric +h ≥ −Kg on Q2R,T . Then we have   ˆ ) A(f 2(1 − f ) 2 q ˆ F − 2 K − Af − − F (Δ − ∂t )F ≥ t 1−f 1−f 2t F ∇ log(1 − f ), ∇q

− − 2f ∇ log(1 − f ), ∇F − t 1−f on Q2R,T . Proof. By the Bochner formula we have ΔF =2t| Hess log(1 − f )|2 + 2t Ric(∇ log(1 − f ), ∇ log(1 − f )) + 2t ∇ log(1 − f ), ∇Δ log(1 − f ) . However, by (2.2), Δ log(1 − f ) = −

|∇f |2 Δf ft + q + Aˆ − . = −f w − 2 1−f (1 − f ) 1−f

Therefore, we obtain ΔF =2t| Hess log(1 − f )|2 + 2t Ric(∇ log(1 − f ), ∇ log(1 − f )) 2(1 − f ) 2 F − 2f ∇ log(1 − f ), ∇F

t 2 ˆ − 2t ∇ log(1 − f ), ∇(ft )

+ (ft + q + A)F 1−f 1−f 2tAˆf 2t ∇ log(1 − f ), ∇q − ∇ log(1 − f ), ∇f . − 1−f 1−f +

On the other hand, by the first equality of Lemma 2.3,

(3.2)

G. Zhao / J. Math. Anal. Appl. 483 (2020) 123631

17

Ft = − 2th(∇ log(1 − f ), ∇ log(1 − f )) + 2t ∇ log(1 − f ), ∇((log(1 − f ))t ) +

F t

= − 2th(∇ log(1 − f ), ∇ log(1 − f )) −

2t F 2ft F ∇ log(1 − f ), ∇(ft ) + + . 1−f 1−f t

Combining the above two equalities, we get (Δ − ∂t )F ≥

2(1 − f ) 2 F + 2t(Ric +h)(∇ log(1 − f ), ∇ log(1 − f )) t 2 ˆ − 2f ∇ log(1 − f ), ∇F

(q + A)F + 2Aˆf F + 1−f F 2t ∇ log(1 − f ), ∇q − . − 1−f t

The lemma follows from the assumption on bound of Ric +h.

2

Remark 3.4. It is easy to see that we don’t need any assumption on the Ricci tensor if geometric flow (1.2) is the Ricci flow, i.e., h = − Ric. The proof of Theorem 3.1. Choosing φ and η as in the proof of Theorem 2.1. For any T1 ∈ (0, T ], let (x1 , t1 ) ∈ Q2R,T1 , at which G(x, t) = η(x, t)F (x, t) attains its maximum, and without loss of generality, we can assume G(x1 , t1 ) > 0, and then η(x1 , t1 ) > 0 and F (x1 , t1 ) > 0. By Lemma 3.3 and a similar argument as in the proof of Theorem 2.1, we have at (x1 , t1 ),   ˆ ) A(f 2η(1 − f ) 2 q − F 0≥ F − 2η K1 + K2 − Aˆf − t1 1−f 1−f 2t1 η ∇ log(1 − f ), ∇q

1−f   C3 (n) C4 (n)  ηF − + K1 + C2 K2 F − t1 R2 R   ˆ ) A(f q 2η(1 − f ) 2 − F F − 2η K1 + K2 − Aˆf − = t1 1−f 1−f − 2ηf ∇ log(1 − f ), ∇F −

2t1 η ∇ log(1 − f ), ∇q

1−f   C3 (n) C4 (n)  ηF − + K + C K − 1 2 2 F. t1 R2 R

+ 2f ∇ log(1 − f ), ∇η F −

Multiplying both sides of the above inequality by t1 η, we have 

ˆ ) A(f q − 0 ≥2(1 − f )G − 2t1 η K1 + K2 − Aˆf − 1−f 1−f



2

2t21 η 2 ∇ log(1 − f ), ∇q

1−f   C3 (n) C4 (n)  − ηG − t1 + K1 + C2 K2 G. R2 R + 2t1 f ∇ log(1 − f ), ∇η G −

G

G. Zhao / J. Math. Anal. Appl. 483 (2020) 123631

18

Noticing that f ≤ 0, by Young’s inequality, 1

−2t1 f ∇ log(1 − f ), ∇η G ≤ − 2f t12

√ |∇η| 3 C1 3 2 ≤ −2f t G G2 1 1 R η2

≤(1 − f )G2 +

27t21 C12 f4 4 16 R (1 − f )3

and 1

∇ log(1 − f ), ∇q ≤ |∇q|w 2 ≤ |q|w +

|∇q|2 . 4|q|

Combining the above three inequalities we have  0 ≥(1 − f )G − 2t1 η K1 + K2 − Aˆf − 2

ˆ ) q A(f − 1−f 1−f

 G

f4 27t21 C12 2t1 η t21 η2 |∇q|2 − |q|G − 4 3 16 R (1 − f ) 1−f 2(1 − f ) |q|    C3 (n) C4 (n) − ηG − t1 + K1 + C2 K2 G. R2 R

−

From 0 ≤

1 1−f

< 1 and 0 <

−f 1−f

≤ 1, we see that

 ˆ ) A(f 1 ˆ −f ˆ ˆ ) Aˆf + = Af + Af + A(f 1−f 1−f 1−f     −f 1  ˆ ≥ min 0, min Af + min 0, min A (u) Q2R,T Q2R,T 1−f 1−f   ≥ min 0, min Aˆf , min A (u) Q2R,T

Q2R,T

= − κ2R . By 0 < η(x1 , t1 ) ≤ 1, we get   0 ≥(1 − f )G2 − G − 2t1 K1 + K2 + max (|q| − q) + κ2R G  − t1

Q2R,T

 f4 |∇q|2 C3 (n) C4 (n)  27t21 C12 t21 G − + K + C K − max . 1 2 2 R2 R 16 R4 (1 − f )3 2 Q2R,T |q|

Applying the quadratic formula and the inequality of arithmetic and geometric means √

K1 1 K1 ≤ , + R 2R2 2

and noticing the fact 0 ≤

−f 1−f

< 1 again, we obtain    1 G ≤C6 (n) 1 + t1 + H , R2

where C6 (n) is a constant that depends only on n and

G. Zhao / J. Math. Anal. Appl. 483 (2020) 123631

19

|∇q| H = K1 + K2 + max (|q| − q) + max  + κ2R . Q2R,T Q2R,T |q| Noticing that d(x, x0 , T1 ) ≤ R implies η(x, T1 ) = 1, we can get F (x, T1 ) G(x1 , t1 ) w(x, T1 ) = ≤ ≤ C6 (n) T1 T1 Since T1 is arbitrary, and using



 1 1 + 2 +H . T1 R

√ √ √ x + y ≤ x + y, we complete the proof. 2

Similar to Corollary 2.6, when (M, g(0)) is a complete noncompact Riemannian manifold without boundary and g(t) evolves by (1.2), we can obtain a global estimate from Theorem 3.1 by taking R → 0. Corollary 3.5. Let (M, g(t)) be a complete solution to (1.2) for t ∈ [0, T ] and (M, g(0)) be a complete noncompact Riemannian manifold without boundary. Let u be a positive solution to (1.1). Suppose that there exist constants L > 0 and K ≥ 0, such that u ≤ L and Ric ≥ −K1 g,

h ≥ −K2 g.

Then we have |∇u(x, t)| ≤ C˜ u(x, t)



√ 1 √ + H t

 1 + log

L u(x, t)

 (3.3)

on M × [0, T ], where C˜ as in Theorem 3.1 and H = K1 + K 2 +

sup (|q| − q) +

M ×[0,T ]

|∇q|  + κ. |q| M ×[0,T ] sup

The following corollary gives a elliptic Harnack inequality by integrating the elliptic type gradient estimate (3.3) in space only. Unlike Corollary 2.10, this inequality can compare the function values at two spatial points at the same time, but inequality (2.14) cannot. Corollary 3.6. Let (M, g(t)) be a complete solution to (1.2) for t ∈ [0, T ] and (M, g(0)) be a complete noncompact Riemannian manifold without boundary. Let u be a positive solution to (1.1). Suppose that there exist constants L > 0 and K ≥ 0, such that u ≤ L and Ric ≥ −K1 g,

h ≥ −K2 g.

Then for any x1 , x2 ∈ M , we have u(x1 , t) ≥ u(x2 , t)H exp((1 − H)(1 + log L))

(3.4)



 √  in each t ∈ [0, T ]. Here, H = exp C˜ √1t + H d(x1 , x2 , t) , where C˜ as in Theorem 3.1 and H = K1 + K 2 +

sup (|q| − q) +

M ×[0,T ]

sup M ×[0,T ]

|∇q|  + κ. |q|

Proof. For any fixed t and any x1 , x2 ∈ M , let ζ : [0, 1] → M be the geodesic of minimal length, which connecting x2 and x1 , ζ(0) = x2 and ζ(1) = x1 . Let f = log u and

G. Zhao / J. Math. Anal. Appl. 483 (2020) 123631

20

l(s) = log(1 + log L − f (ζ(s), t)). By Corollary 3.5 we have dl(s) − (∇f )(ζ(s), t), ζ  (s)

= ds 1 + log L − f (ζ(s), t) |∇f |(ζ(s), t) 1 + log L − f (ζ(s), t)   √ 1  ˜ ≤C|ζ (s)| √ + H . t

≤|ζ  (s)| ·

Integrating this inequality over ζ(s), we have log

1 + log L − f (x1 , t) ≤ C˜ 1 + log L − f (x2 , t)



 √ 1 √ + H d(x1 , x2 , t). t

From this inequality, inequality (3.4) can be obtained through a simple calculation. 2 4. Applications We will give some applications of gradient estimates in section 2 and section 3 to some special equations. In some cases, we also take the geometric flow as the Ricci flow, i.e., h = −Ric in (1.2). 4.1. Applications of space-time gradient estimates In this subsection, we focus on applications of space-time gradient estimate for positive solutions. In this case, we see that λ, Λ, Σ are not necessarily finite in (2.12). For example, if we choose A(u) = up for p > 0, then λ, Λ, Σ are all multiples of supM ×[0,T ] up−1 and they are not all zero if p = 1. But supM ×[0,T ] up−1 is not necessarily finite, unless u is bounded. For the case that q = 0 and A(u) = up , the reader can also refer to [20,21,35]. Naturally, for general unbounded positive function u, we want to know when λ, Λ and Σ are finite. Since 2 u is a positive solution to (1.1), A(u) can be written as u · A(u) u , which is A(u) = uH(u) for some C function H(u). As pointed by Q. Chen and the author in [6, Remark 1.8], λ, Λ, Σ < +∞ implies |H(u)| ≤ C0 log u,

as u → +∞.

When A(u) = au log u. A direct computation yields Σ = 0 and 

λ2R = λ ≡ 0, λ2R = λ ≡ −a,

Λ2R = Λ ≡ a, Λ2R = Λ ≡ 0,

if a ≥ 0, if a ≤ 0.

Therefore, we can obtain that local and global gradient estimates for positive solutions of the equation (Δ − q(x, t) − ∂t )u(x, t) = au(x, t) log(u(x, t)),

a∈R

(4.1)

from Theorem 2.1 and Corollary 2.6. Remark 4.1. By the asymptotic behavior of H, we can find many examples that satisfy λ, Λ, Σ < +∞. Such (u) as H(u) = PPkl (u) , where Pk , Pl are polynomials of degree k, l, respectively and k < l. Hence we can obtain gradient estimates for positive solution of the following series of equations

G. Zhao / J. Math. Anal. Appl. 483 (2020) 123631

(Δ − q(x, t) − ∂t )u(x, t) = au(x, t) For instance, if we choose k = 0, l = 1, so A(u) =

u u+1 ,

Pk (u(x, t)) , Pl (u(x, t))

21

a ∈ R, k < l.

then λ = 14 , Λ = 0 and Σ =

8 27 .

On the other hand, as mentioned in Remark 2.9, if A(u) = au log u, then Σ = 0. In addition, we take h = − Ric. In this case, we don’t need the assumption on the bound |∇h| since the contracted second Bianchi identity. At this time, when α → 1, we can also let ε → 0 in local estimate (2.1), and then we are arriving at Corollary 4.2. Let (M, g(0)) be a complete Riemannian manifold, and let g(t) evolves by the Ricci flow for t ∈ [0, T ]. Suppose that there exist constants K2 , θ ≥ 0 such that 0 ≤ Ric ≤ K2 g and Δq ≤ θ2R on Q2R,T . If u is a positive solution to (4.1). Then on QR,T , we have 1. for a ≥ 0, √ ut (x, t) A(u(x, t)) n n |∇u(x, t)|2 − − q(x, t) − ≤ + (C + n)K2 + nθ + a; 2 u (x, t) u(x, t) u(x, t) t 2 2. for a < 0, √ ut (x, t) A(u(x, t)) n |∇u(x, t)|2 − − q(x, t) − ≤ + (C + n)K + nθ − na, 2 u2 (x, t) u(x, t) u(x, t) t where C as in Theorem 2.1. From the above local estimate, we have immediately Corollary 4.3. Let (M, g(0)) be a complete noncompact Riemannian manifold without boundary, and let g(t) evolves by the Ricci flow for t ∈ [0, T ]. Suppose that there exist constants K2 , θ ≥ 0 such that 0 ≤ Ric ≤ K2 g and Δq ≤ θ. If u is a positive solution to (4.1). Then we have 

√ ut (x, t) A(u(x, t)) n |∇u(x, t)|2  − − q(x, t) − ≤ + C + θ + |a| K 2 u2 (x, t) u(x, t) u(x, t) t on M × [0, T ], where C  is a constant that depends only on n. When the manifold is closed, we also have

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22

Corollary 4.4. Let (M, g(t)) be a closed Riemannian manifold, where g(t) evolves by (1.2) for t ∈ [0, T ] and satisfies 0 ≤ Ric ≤ K2 g. If u is a positive solution to the equation (Δ − q(x, t) − ∂t )u(x, t) = au(x, t) log(u(x, t)), and q(x, t) satisfies Δq ≤ θ. Then we have √ ut (x, t) A(u(x, t)) n n |∇u(x, t)|2 − − q(x, t) − ≤ + nK + nθ + |a| 2 u2 (x, t) u(x, t) u(x, t) 2t 2 on M × (0, T ]. 4.2. Applications of elliptic type gradient estimates Now we give some applications of elliptic type gradient estimates for bounded positive solutions. Since we are dealing with bounded positive solutions, A(u) that satisfies the conditions κ < +∞ is easy to find. We will consider that elliptic type gradient estimates for bounded positive solutions of the equation (Δ − q(x, t) − ∂t )u(x, t) = au(x, t)β ,

a ∈ R, β ∈ (−∞, 0] ∪ [1, +∞).

(4.2)

In order not to be redundant, we only give the global estimate here, and the local one is omitted. Corollary 4.5. Let (M, g(t)) be a complete solution to (1.2) for t ∈ [0, T ] and (M, g(0)) be a complete noncompact Riemannian manifold without boundary. Let u be a positive solution to (4.2). Suppose that there exist constants L > 0 and K ≥ 0, such that u ≤ L and Ric ≥ −K1 g,

h ≥ −K2 g.

Then on M × [0, T ], we have |∇u(x, t)| ≤ C˜ u(x, t)



√ 1 √ + H t



L 1 + log u(x, t)

 ,

where C˜ as in Theorem 3.1 and H  = K1 + K 2 +

sup (|q| − q) +

M ×[0,T ]

|∇q|  + κ0 |q| M ×[0,T ] sup

with ⎧ sign a−1 β−1 ⎪ , ⎪ ⎨ 2 aβL κ0 = 0, ⎪ ⎪ ⎩a(1 − β) inf

M ×[0,T ]

if a ∈ R, β ≥ 1, u(x, t)

β−1

if a ≤ 0, β ≤ 0, ,

if a ≥ 0, β ≤ 0.

G. Zhao / J. Math. Anal. Appl. 483 (2020) 123631

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Here, sign a is the sign function, which is 1, 0, −1 if a > 0, = 0, < 0, respectively. Proof. From Corollary 3.5, we just have to compute κ. By the definition, we have  κ = − min 0,

 inf

M ×[0,T ]

β−1

(aβu

),

inf

M ×[0,T ]

⎧ ⎪ 0, ⎪ ⎪ ⎪ ⎨−aβLβ−1 , = ⎪ 0, ⎪ ⎪ ⎪ β−1  ⎩ , a(1 − β) inf M ×[0,T ] u(x, t)

(a(β − 1)u

β−1

)

if a ≥ 0, β ≥ 1, if a ≤ 0, β ≥ 1, if a ≤ 0, β ≤ 0, if a ≥ 0, β ≤ 0.

Therefore, we obtain the corollary. 2 In particular, when q(x, t) = const., the term qu(x, t) can be combined by au(x, t)β , so we get Corollary 4.6. Let (M, g(t)) be a complete solution to (1.2) for t ∈ [0, T ] and (M, g(0)) be a complete noncompact Riemannian manifold without boundary. Let u be a positive solution to (Δ − ∂t )u(x, t) = au(x, t)β ,

a ∈ R, β ∈ (−∞, 0] ∪ [1, +∞).

Suppose that there exist constants L > 0 and K ≥ 0, such that u ≤ L and Ric ≥ −K1 g,

h ≥ −K2 g.

Then on M × [0, T ], we have |∇u(x, t)| ≤ C˜ u(x, t)



 1 √ + K1 + K2 + κ0 t

where C˜ as in Theorem 3.1 and ⎧ sign a−1 β−1 ⎪ , ⎪ ⎨ 2 aβL κ0 = 0, ⎪ ⎪ ⎩a(1 − β) inf

M ×[0,T ]



L 1 + log u(x, t)

 ,

if a ∈ R, β ≥ 1, u(x, t)

β−1

if a ≤ 0, β ≤ 0, ,

if a ≥ 0, β ≤ 0.

Here, sign a is the sign function, which is 1, 0, −1 if a > 0, = 0, < 0, respectively. Remark 4.7. For each of these specific equations that appear in this section, we also have the corresponding Harnack inequality, which we will not write them all down here. References [1] M. Bailesteanu, X. Cao, A. Pulemotov, Gradient estimates for the heat equation under the Ricci flow, J. Funct. Anal. 258 (10) (2010) 3517–3542. [2] E. Calabi, An extension of E. Hopf’s maximum principle with an application to Riemannian geometry, Duke Math. J. 25 (1) (1958) 45–56. [3] X. Cao, R.S. Hamilton, Differential Harnack estimates for time-dependent heat equations with potentials, Geom. Funct. Anal. 19 (4) (2009) 989–1000. [4] L. Chen, W. Chen, Gradient estimates for a nonlinear parabolic equation on complete non-compact Riemannian manifolds, Ann. Global Anal. Geom. 35 (4) (2009) 397–404.

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