Harnack inequality, heat kernel bounds and eigenvalue estimates under integral Ricci curvature bounds

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Harnack inequality, heat kernel bounds and eigenvalue estimates under integral Ricci curvature bounds Wen Wang School of Mathematics and Statistics, Hefei Normal University, Hefei 230601, PR China Received 7 December 2019; accepted 1 January 2020

Abstract Let (Mn , gij ) be a complete Riemannian manifold. We prove that for any p > n2 , when k(p, 1) is small enough, some parabolic type gradient bounds hold for the positive solutions of a nonlinear parabolic equation

ut = u + au log u, on geodesic balls B(O, r) in M with 0 < r ≤ 1. We can also derive the gradient estimates for any solutions to the above nonlinear parabolic equation along the Ricci flow on a closed manifold without any curvature conditions. As its application, we derive some parabolic type gradient estimates for a positive solution u(x, t) of the heat equation ut = u. Moreover, our estimate is stronger than Zhang and Zhang’s estimate (see Remark 2 in Section 4). Those results are generalizations of Li-Yau, Hamilton, Li-Xu type gradient estimates under the integral Ricci curvature bounds. By utilizing the gradient estimates of the heat equation, we obtain Harnack inequalities, the upper bound and the lower bound for the heat kernel, eigenvalue estimate and the lower bound of Green’s function on Riemannian manifolds under the integral Ricci curvature bounds. © 2020 Elsevier Inc. All rights reserved. MSC: 58J35; 35K08; 53C21 Keywords: Heat equation; Gradient estimate; Harnack inequality; Heat kernel; Eigenvalue estimate; Green’s function

E-mail addresses: [email protected], [email protected]. https://doi.org/10.1016/j.jde.2020.01.003 0022-0396/© 2020 Elsevier Inc. All rights reserved.

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Contents 1. Introduction and main results . . . . . . . . . . 2. Some lemmas . . . . . . . . . . . . . . . . . . . . 3. The proof of main results . . . . . . . . . . . . . 4. Gradient estimates for the heat equation . . . 5. Harnack inequality . . . . . . . . . . . . . . . . . 6. Upper and lower bounds for the heat kernel 7. Local eigenvalue estimate . . . . . . . . . . . . 8. Lower bound of Green’s function . . . . . . . 9. Appendix . . . . . . . . . . . . . . . . . . . . . . . Acknowledgments . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . .

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2 10 13 18 20 26 29 30 31 33 33

1. Introduction and main results Let (Mn , gij ) be a complete Riemannian manifold. P. Petersen and G. F. Wei [25] introduced an integral Ricci curvature condition. For x ∈ Mn , let r(x) denote the smallest eigenvalue for the Ricci tensor Ric : Tx M → Tx M, Ric− (x) = (−r(x))+ = max{0, −r(x)}. For any constants p, r > 0, define ⎛ ⎜ k(p, r) = sup r 2 ⎝ x∈M



⎞1/p ⎟ |Ric− |p dV ⎠

.

B(x,r)

The gradient estimate is an important tool in study of elliptic and parabolic type equations. It was originated by Yau [45], Li-Yau [13], and was generalized by many researchers, such as Li [11,12], Hamilton [7], Li-Xu [14], and [8,10,15,28,31,33,35,42,44,48,49,49]. First, let’s recall some results related to this article. Let (Mn , g) be a complete Riemannian manifold with Ricci(Mn ) ≥ −K. Assume that u(x, t) is any positive solution to the heat equation ut = u

(1.1)

on Mn . Li and Yau [13] proved for any α > 1, |∇u|2 ut nα 2 K nα 2 − α ≤ + . u 2(α − 1) 2t u2

(1.2)

In 1993, Hamilton [7] proved the following parabolic type gradient estimate to the equation (1.1),

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|∇u|2 ut n − e2Kt ≤ e4Kt . u 2t u2

3

(1.3)

In 2006, H. J. Sun [30] obtained a different parabolic type gradient estimate to the heat equation than Hamilton’s estimate (1.3), |∇u|2 2 ut n nK 2 nK − (1 + Kt) ≤ ( + )(1 + Kt) − . 2 u 3 u 2t 4 3 12

(1.4)

In 2011, J. F. Li and X. J. Xu [14] further considered Li-Yau’s estimate, and obtained |∇u|2 − 1+ u2 |∇u|2 − 1+ u2

sinh(Kt) cosh(Kt) − Kt 2

2 Kt 3



sinh (Kt)



nK ut ≤ (coth(Kt) + 1), u 2

ut n nK 1 ≤ + (1 + Kt). u 2t 2 3

(1.5) (1.6)

In 2015, D. G. Chen and C. W. Xiong [4] generalized those estimates (1.2)–(1.6) to the positive solutions of the doubly nonlinear diffusion equations ut = p uγ = div(|∇uγ |p−2 ∇uγ ), on a Riemannian manifold with Ric ≥ −K, where γ > 0, p > 1, and p is p-Laplace. Recently, the first author and Zhang in [36] further generalized Li and Xu’s results to the nonlinear parabolic equation. Related results can be referred to [6,11,26,34,39]. Huang, Huang and Li in [9] generalized the results of Lu, Ni, Vázquez and Villani [20] on the porous medium equation ut = um ,

m > 1,

and obtained Li-Yau type, Hamilton type and Li-Xu type gradient estimates. The author and coauthors [37] had uniformly promoted these results to the Ricci flow. In [43], Yang generalized Ma’s result [21] and derived a local gradient estimates for positive solutions to the equation ut = u + au log u + bu in M n × (0, T ], where a, b ∈ R are constants for complete noncompact manifolds with a fixed metric and curvature locally bounded below. In recent years, some parabolic type and elliptic type gradient estimates of positive solutions to some parabolic equations along the Ricci flow are also derived, see [1–3,16–18,29] In 2017–2018, Q. S. Zhang and M. Zhu [46,47] derived a Li-Yau type gradient estimate under the integral Ricci curvature bounds introduced by P. Petersen and Wei [24,25]. In [46,47], Q. S. Zhang and M. Zhu proved the Li-Yau type gradient bounds for any positive solution u(x, t) to the heat equation (1.1) under the integral Ricci curvature

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⎛ ⎜ k(p, r) = sup r 2 ⎝ x∈M



⎞1/p ⎟ |Ric− |p dV ⎠

≤ κ,

(1.7)

B(x,r)

that is  C 1 C2 |∇u|2 ∂t u C1 αJ 2 − ≤ 1+ + 3 u u J J J t

(1.8)

where C1 = C1 (n, p, α), C2 = C2 (n, p, α), C3 = C3 (n, p, α) and J = J (t) is a decreasing exponential function (see [46,47] for more details) Afterwards, Christian Rose [23] generalized (1.8) to compact manifolds with negative part of Ricci curvature in the Kato class. Recently, Xavier Ramos Olivé [22] generalized the estimate of Q. S. Zhang and M. Zhu [46, 47] to the positive solutions to the heat equation, with Neumann boundary conditions, on a compact Riemannian submanifold with boundary Mn ⊆ Nn , satisfying the integral Ricci curvature assumption: ⎛ ⎜ D sup ⎝



2

x∈N

⎞1

p

⎟ |Ric | dy ⎠ < K − p

B(x,D)

for K(n, p) small enough, p > n/2, where diam(M) ≤ D. Recently, the author [38] derived two certain elliptic type gradient estimates for any positive solutions of the heat equation (1.1) and the nonlinear parabolic equation (1.9). Motivated by [5,14,22,23,26,43,46,47], in this paper, we continue to consider the nonlinear parabolic equation ut = u + au log u,

(1.9)

on a Riemannian manifold under the integral Ricci curvature (1.7). We first define two C 1 functions α(t) and ϕ(t) : (0, +∞) → (0, +∞) that satisfy the following conditions: (C1) α(t) > 1. (C2) α(t) and ϕ(t) satisfy the following system ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

1 ϕ ϕ  ≥ 2(2 − δ) − α  , n n α ϕ 2(2 − δ) J − α  > 0, n ϕ2 αϕ  + (2 − δ) J ≥ 0. n 2(2 − δ)

(C3) There exist two positive real numbers λ and δ satisfying

(1.10)

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⎧ α ⎪ ⎨0 < α − 1 ≤ λ, 2 ⎪ ⎩0 < δ ≤ . 1 + nλ2

5

(1.11)

dϕ  Where α  = dα dt and ϕ = dt . Our paper is organized as follows: we show our main results in section 1. Some lemmas need in proof of the main theorems are provided in section 2. The proof of the main results in section 3. As its application, we derive some parabolic type gradient estimates for a positive solution u(x, t) of the heat equation ut = u. Those results are generalizations of Li-Yau, Hamilton, Li-Xu type gradient estimates under the integral Ricci curvature bounds in section 4. By utilizing the gradient estimates of the heat equation, we obtain Harnack inequalities in section 5. We can derive the upper bounds and the lower bounds for the heat kernel on Riemannian manifolds under the integral Ricci curvature bounds in section 6. Eigenvalue estimate on Riemannian manifolds under the integral Ricci curvature bounds is given in section 7. And Li-Yau estimate of Green’s function on Riemannian manifolds under the integral Ricci curvature bounds in section 8. Detailed calculation of some specific functions α(t) and ϕ(t) are given in section 9.

Theorem 1.1. Let (Mn , g) be an n-dimensional complete Riemannian manifold. Suppose that there exist two positive functions α(t) and ϕ(t) satisfying conditions (C1) and (C2), and two positive constants δ and λ satisfying condition (C3). Let u(x, t) be any positive solution to the equation (1.9) on Mn . For p > n2 , there exists a constant κ = κ(n, p) such that κ(p, 1) ≤ κ, then for any point O ∈ Mn , we have the following estimates. (1) If a ≥ 0, then |∇u|2 ut − α(t) + α(t)a log u 2 u u

C1 α 4 1 nα 2 1 +a+1 + ≤ (2 − δ)J 2t (2 − δ)J (2 − δ)(α − 1)J  anϕα 3 + + αϕ, (2 − δ)J

J

(1.12)

in B(O, 12 ) × (0, ∞). (2) If a < 0, then |∇u|2 ut − α(t) + α(t)a log u u u2

2 1 C2 α 4 1 nα +1 + ≤ (2 − δ)J 2t (2 − δ)J (2 − δ)(α − 1)J √ nα|a|(α − J ) + + αϕ, (2−δ)2 J 2 − δ(2 − δ) (1 − α) nα 2

J

in B(O, 12 ) × (0, ∞). Where

(1.13)

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n   1 J = J (t) = 2− b−1 exp{−2Cκ 1 + [2C(b − 1)κ] 2p−n t},

and b = 5δ −1 ,

C = C(p, n),

C1 = C1 (n, p),

and

C2 = C2 (n, p).

As some special cases, we can derive the gradient estimates of Li-Yau type, Hamilton type, Li-Xu type, and other types, as follows, Corollary 1.2. 1. Li-Yau type: α(t) = constant,

ϕ(t) =

n 1 , (2 − δ)α J

and

0<δ≤

2(α − 1)2 . (α − 1)2 + nα 2

Then (1) If a ≥ 0 and C3 = C3 (n, p, α), then J

|∇u|2 ut − α + αa log u 2 u u

≤

nα 2 1 C3 α 4 + (2 − δ)J 2t (2 − δ)J



√

anα 1 +a+1 + , (2 − δ)(α − 1)J (2 − δ)J

(1.14)

in B(O, 12 ) × (0, ∞). (2) If a < 0, then

J

|∇u|2 ut − α + αa log u u2 u

nα 2 1 Cα 4 1 +1 + (2 − δ)J 2t (2 − δ)J (2 − δ)(α − 1)J √ nα|a|(α − J ) + , 2 (2−δ) J 2 − δ(2 − δ) (1 − α) nα 2 ≤

in B(O, 12 ) × (0, ∞). 2. Hamilton type: α(t) = eκt , ϕ(t) =

and

0<δ<

κneκt 2(2 − δ)J

for

Then (1) If a ≥ 0 and C4 = C4 (n, p, ), then

2( − 1)2 , ( − 1)2 + n 2 any > 1.

(1.15)

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|∇u|2 ut − α(t) + α(t)a log u u u2

 nα 2 1 anϕα 3 C3 α 4 1 ≤ +a+1 + , + (2 − δ)J 2t (2 − δ)J (2 − δ)(α − 1)J (2 − δ)J

7

J

(1.16)

in B(O, 12 ) × (0, ∞). (2) If a < 0, then |∇u|2 ut − α(t) + α(t)a log u u2 u

2 1 C2 α 4 1 nα +1 + ≤ (2 − δ)J 2t (2 − δ)J (2 − δ)(α − 1)J √ nα|a|(α − J ) + , (2−δ)2 J 2 − δ(2 − δ) (1 − α) 2 nα

J

(1.17)

in B(O, 12 ) × (0, ∞). 3. Linear Li-Xu type: α(t) = κt + , ϕ(t) =

and 0 < δ <

nκ 2(2 − δ)J

f or

2( − 1)2 , ( − 1)2 + n 2

any

> 1.

Then (1) If a ≥ 0 and C4 = C4 (n, p, ), then |∇u|2 ut − α(t) + α(t)a log u 2 u u

 anϕα 3 C3 α 4 1 nα 2 1 +a+1 + , + ≤ (2 − δ)J 2t (2 − δ)J (2 − δ)(α − 1)J (2 − δ)J

J

(1.18)

in B(O, 12 ) × (0, ∞). (2) If a < 0, then |∇u|2 ut − α(t) + α(t)a log u 2 u u

2 1 C2 α 4 1 nα +1 + ≤ (2 − δ)J 2t (2 − δ)J (2 − δ)(α − 1)J √ nα|a|(α − J ) + , 2 (2−δ) J 2 − δ(2 − δ) (1 − α) nα 2

J

in B(O, 12 ) × (0, ∞).

(1.19)

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4. Li-Xu type: α(t) = +

sinh κt cosh κt − κt , sinh κt

and 0 < δ <

ϕ(t) =

nκ ( + coth κt) (2 − δ)J

for any > 1.

,

−1

Then (1) If a ≥ 0, then |∇u|2 ut − α(t) + α(t)a log u u2 u

 anϕα 3 C1 α 4 1 nα 2 1 +a+1 + , + ≤ (2 − δ)J 2t (2 − δ)J (2 − δ)(α − 1)J (2 − δ)J

J

(1.20)

in B(O, 12 ) × (0, ∞). (2) If a < 0, then |∇u|2 ut − α(t) + α(t)a log u 2 u u

2 1 C2 α 4 1 nα +1 + ≤ (2 − δ)J 2t (2 − δ)J (2 − δ)(α − 1)J √ nα|a|(α − J ) + , (2−δ)2 J 2 − δ(2 − δ) (1 − α) nα 2

J

(1.21)

in B(O, 12 ) × (0, ∞). 5. α(t) is a decreasing function. α(t) = e−κt + , ϕ(t) =

and

nκ 2(2 − δ)

0<δ<

2( − 1)2 , ( − 1)2 + n 2

for any > 1.

Then (1) If a ≥ 0 and C4 = C4 (n, p, α), then |∇u|2 ut − α(t) + α(t)a log u 2 u u

2 1 1 C3 α 4 1 nα + +a+1 , + ≤ (2 − δ)J 2t (2 − δ)J (2 − δ)(α − 1)J t

J

in B(O, 12 ) × (0, ∞).

(1.22)

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(2) If a < 0, then |∇u|2 ut − α(t) + α(t)a log u 2 u u

2 1 C2 α 4 1 nα +1 + ≤ (2 − δ)J 2t (2 − δ)J (2 − δ)(α − 1)J √ nα|a|(α − J ) + , (2−δ)2 J 2 − δ(2 − δ) (1 − α) 2 nα

J

(1.23)

in B(O, 12 ) × (0, ∞). Remark 1. 1. These above results can be regarded as generalizations the gradient estimates of Yang [43] under the integral Ricci bounds. 2. These above results can be regarded as generalizations of Zhang and Zhu’s results on the nonlinear parabolic equation. 3. α(t) is increasing in some known results, however, α(t) is not only decreasing but also bounded uniformly in our result (1.22) and (1.23). We may also derive the gradient estimates for any solutions to the equation ut = u + au log u, along the Ricci flow on a closed Riemannian manifold without arbitrary curvature conditions. The proof mainly adopts the argument of Hamilton [7], Shi [27] and Liu [19]. Theorem 1.3. Let (M n , g(t)) be a closed Riemannian manifold, where g(t) solves the Ricci ∂g(t) = −2Rij , ∂t

(x, t) ∈ M n × [0, T ],

for t ∈ [0, T ]. Let u(x, t) be any solution of the equation (1.9). (1) If a > 0 and u ≤ e−1 , then |∇u(x, t)|2 ≤

1 2t



maxn u2 (x, 0) − u2 (x, t) .

(1.24)

maxn u (x, 0) − u (x, t) .

(1.25)

x∈M

(2) If a < 0 and u ≥ 1, then 1 |∇u(x, t)| ≤ 2t 2



2

x∈M

2

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2. Some lemmas To prove the Theorem 1.1, we need to some lemmas. Let f = log u, then ( − ∂t )f = −|∇f |2 − af.

(2.1)

Lemma 2.1. Let J = J (x, t) be a positive smooth function, and F = J |∇f |2 − αft + αaf − αϕ, where α = α(t) > 1 and ϕ = ϕ(t) > 0. Then

|∇J |2 ( − ∂t )F = (2 − δ)J |fij |2 + J − 5δ −1 − ∂t J − 2V J |∇f |2 J −2∇f · ∇F + 2a(α − J )|∇f |2 + αaf − δJ |∇f |4 +α  ft − α  af + α  ϕ + αϕ 

(2.2)

Proof. By the Bochner-Weitzenböck formula and (2.1), we have F = 2J |fij |2 + 2J Ric(∇f, ∇f ) + 2J ∇f · ∇f − α(ft ) +αaf + J |∇f |2 + 2∇J · ∇|∇f |2 = 2J |fij |2 + 2J Ric(∇f, ∇f ) − α(ft )   +2J ∇f · ∇ ft − |∇f |2 − af +αaf + J · |∇f |2 + 2∇J · ∇|∇f |2 ,

(2.3)

and ∂t F = 2J ∇f · ∇ft + (∂t J )|∇f |2 − αftt − α  ft +αaft + α  af − α  ϕ − αϕ  = 2J ∇f · ∇ft + (∂t J )|∇f |2 − α(ft ) − 2α∇f · ∇ft −aαft − α  ft + αaft + α  af − α  ϕ − αϕ  , dϕ(t)  where α  = dα(t) dt and ϕ = dt . Combine (2.3) with (2.4), we obtain

( − ∂t )F = 2J |fij |2 + 2J Ric(∇f, ∇f )   −2∇f J ∇|∇f |2 + ∇J · |∇f |2 − α∇ft + αa∇f − ∇(αϕ) +2∇J · ∇f · |∇f |2 + 2a(α − J )|∇f |2 + αaf + J · |∇f |2 +2∇J · ∇|∇f |2 − (∂t J )|∇f |2 + α  ft − α  af + α  ϕ + αϕ  = 2J |fij |2 + 2J Ric(∇f, ∇f ) − 2∇f ∇F + 2∇J · ∇f |∇f |2 +2a(α − J )|∇f |2 + αaf + J |∇f |2 + 2∇J · ∇|∇f |2

(2.4)

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−(∂t J )|∇f |2 + α  ft − α  af + α  ϕ + αϕ 

11

(2.5)

By Cauchy inequality, we have for any real number δ > 0, 2∇J · ∇|∇f |2 ≥ −δJ |fij |2 − 4δ −1

|∇J |2 |∇f |2 , J

(2.6)

|∇J |2 |∇f |2 . J

(2.7)

and 2∇J · ∇f |∇f |2 ≥ −δJ |∇f |4 − δ −1 Plugging (2.6) and (2.7) into (2.5), we have

|∇J |2 ( − ∂t )F = (2 − δ)J |fij |2 + J − 5δ −1 − ∂t J − 2V J |∇f |2 J −2∇f · ∇F + 2a(α − J )|∇f |2 + αaf − δJ |∇f |4 +α  ft − α  af + α  ϕ + αϕ  So, the proof is complete. 2 To simplify the equation (2.2), we need a Lemma (see [47]). Lemma 2.2 (Zhang-Zhu [47], Claim). There exists a κ = κ(p, n) such that when k(p, 1) ≤ κ, for any 0 < r ≤ 1, the system ⎧ |∇J |2 ⎪ ⎪ − ∂t J = 0, ⎨J − 2V J − 5δ −1 J J (·, 0) = 1, on B(O, r), ⎪ ⎪ ⎩ J (·, t) = 1., on ∂B(O, r)

on B(O, r) × (0, ∞);

has a unique solution for t ∈ [0, ∞), which satisfies J r (t) ≤ J (x, t) ≤ 1, where n   1 J r (t) = 2− b−1 exp{−2Cκr −2 1 + [2C(b − 1)κ] 2p−n t}

for some constant C = C(n, p) and b = 5δ −1 . Together Lemmas 2.2 and 2.1, we can deduce the following inequality.

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Lemma 2.3. If F ≥ 0, then ( − ∂t )F ≥

(2 − δ)J 2 2(2 − δ)(α − J )J F + F |∇f |2 nα 2 nα 2  (2 − δ)J 2 + (J − α) − δ |∇f |4 nα 2 −2∇f · ∇F + 2a(α − J )|∇f |2 + aαf.

(2.8)

Proof. Applying Lemma 2.2, we have ( − ∂t )F = (2 − δ)J |fij |2 − 2∇f · ∇F + 2a(α − J )|∇f |2 +αaf − δJ · |∇f |4 + α  ft − α  af + α  ϕ + αϕ   ϕ 2 ϕ2 ϕ  = (2 − δ)J fij + gij  − (2 − δ)J − 2(2 − δ)J f n n n −2∇f · ∇F + 2a(α − J )|∇f |2 + αaf − δJ · |∇f |4 +α  ft − α  af + α  ϕ + αϕ  .

(2.9)

By (2.1), we have  ϕ 2 ϕ  ( − ∂t )F = (2 − δ)J fij + gij  + 2(2 − δ) J |∇f |2 n n     ϕ ϕ − 2(2 − δ) J − α  ft + 2(2 − δ) J − α  af n n   ϕ − 2(2 − δ) J − α  ϕ − 2∇f · ∇F + 2a(α − J )|∇f |2 n +αaf − δJ · |∇f |4 + αϕ  + (2 − δ)J By conditions ⎧   ϕ ϕ  1 ⎪ ⎪2(2 − δ) n ≥ 2(2 − δ) n − α α , ⎨ 2(2 − δ) ϕn J − α  > 0, ⎪ ⎪ 2 ⎩  αϕ + (2 − δ) ϕn J ≥ 0, we have ⎧   ϕ ϕ  1 ⎪ ⎪2(2 − δ) n ≥ 2(2 − δ) n J − α α , ⎨ 2(2 − δ) ϕn J − α  > 0, ⎪ ⎪ 2 ⎩  αϕ + (2 − δ) ϕn J ≥ 0. Then, (2.10) can be written as

ϕ2 . n

(2.10)

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 1 ϕ 2  ϕ  ( − ∂t )F = (2 − δ)J fij + gij  + 2(2 − δ) J − α  F n n α −2∇f · ∇F + 2(α − J )a|∇f |2 + αaf − δJ |∇f |4 .

(2.11)

Applying Cauchy inequality, we have  ϕ 2 1  fij + gij  ≥ (f + ϕ)2 n n 1 = (ft − |∇f |2 − af + ϕ)2 n 2 1  = 2 (J − α)|∇f |2 − F nα 2 1  = 2 F + (α − J )|∇f |2 . nα

(2.12)

Therefore, by (2.11) and (2.12), we obtain ( − ∂t )F ≥

(2 − δ)J 2 2(2 − δ)(α − J )J F + F |∇f |2 nα 2 nα 2  (2 − δ)J 2(2 − δ)J 2 4  1 + (J − α) |∇f | + ϕ − α F nα 2 n α −2∇f · ∇F + 2a(α − J )|∇f |2 − δJ |∇f |4 + aαf.

We complete the proof.

2

To prove our results, we need a cut-off function below. Lemma 2.4 (Dai-Wei-Zhang [5], Lemma 5.4). Let (Mn , g) be a complete Riemannian manifold. Then for any p > n2 , there exist constants κ = κ(p, n) and C = C(n, p) such that if k(p, 1) ≤ κ, the for any geodesic ball B(x, r) and 0 < r ≤ 1 there exists φ ∈ C0∞ (B(x, r)) satisfying 0 ≤ φ ≤ 1, φ ≡ 1 in B(x, 2r ), and |∇φ|2 + |φ| ≤

C . r2

3. The proof of main results In this section, we will prove Theorem 1.1 by the method of Li-Yau [13] and Zhang-Zhu [46, 47]. The proof of Theorem 1.1. According to Lemma 2.4, we may choose a cut-off function φ satisfying

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⎧ 0 ≤ φ ≤ 1; ⎪ ⎪ ⎪ ⎨ suppφ ⊂⊂ B(O, 1), ⎪ φ = 1 in B(O, 12 ), ⎪ ⎪ ⎩ |∇φ|2 + |φ| ≤ C(n, p) = C.

(3.1)

√ For any T > 0, let (x1 , t1 ) be a maximum point of tφ 2 F in B(O, 1) × (0, T ]. We can suppose that the value is positive, because otherwise the proof is trivial. Then at the point (x1, t1 ), we infer f = ft − |∇f |2 − af α−J F |∇f |2 − ϕ. =− − α α

(3.2)

By ∇(φ 2 F ) = 0, we have φ∇F = −2F ∇φ,

(3.3)

therefore, by (2.8), one has √ 2 √ tφ ( − ∂t )( tφ 2 F ) √ √ √  1 = tφ 4 ( − ∂t )F + tφ 2 tF φ 2 + 2 t∇F · ∇φ 2 − √ φ 2 F 2 t (2 − δ)J √ 2 2 2(2 − δ)(α − J )J 4 = ( tφ F ) + tφ F |∇f |2 nα 2 nα 2  (2 − δ)J 2 + (J − α) − δ tφ 4 |∇f |4 nα 2 − 2tφ 4 ∇f · ∇F + 2a(α − J )|∇f |2 tφ 4 + aαf tφ 4   1 + φ 2 tF φ 2 + 2t∇F · ∇φ 2 − φ 2 F . 2

(3.4)

By using (3.3) and (3.4), we have √ 2 √ tφ ( − ∂t )( tφ 2 F ) =

(2 − δ)J √ 2 2 2(2 − δ)(α − J )J 4 ( tφ F ) + tφ F |∇f |2 nα 2 nα 2  (2 − δ)J 2 + (J − α) − δ tφ 4 |∇f |4 nα 2 + 4tφ 3 F ∇f · ∇φ + a(α − J )|∇f |2 tφ 4 − atφ 4 F − atφ 4 αϕ 1 + 2tφ 3 F φ − 6tφ 2 F |∇φ|2 − φ 4 F. 2

Choosing 0 < δ ≤

2 α and by using ≤ λ, we have 1 + nλ2 α−1

(3.5)

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δ≤

1+n

2 

 α 2 α−1

15

,

that is (2 − δ) (1 − α)2 − δ ≥ 0. nα 2 Since J = J (x, t) ≤ 1, we deduce from above inequality that (2 − δ) (J − α)2 − δ ≥ 0. nα 2 So, (3.5) can be written as √ 2 √ tφ ( − ∂t )( tφ 2 F ) =

(2 − δ)J √ 2 2 2(2 − δ)(α − J )J 4 ( tφ F ) + tφ F |∇f |2 nα 2 nα 2 + 4tφ 3 F ∇f · ∇φ + a(α − J )|∇f |2 tφ 4 − atφ 4 F − atφ 4 αϕ 1 + 2tφ 3 F φ − 6tφ 2 F |∇φ|2 − φ 4 F. 2

(3.6)

Next, the discussion is divided into two situations. Case 1. (a ≥ 0) Applying (3.1) to (3.6), we can infer √

√ tφ 2 ( − ∂t )( tφ 2 F ) ≥

(2 − δ)J √ 2 2 2(2 − δ)(α − J )J 4 ( tφ F ) + tφ F |∇f |2 nα 2 nα 2 1 √ − C1 tφ 4 F |∇f | − atφ 4 F − atφ 4 αϕ − C1 tφ 2 F − √ tφ 4 F, 2 t

where we drop two terms a(α − J )|∇f |2 tφ 4 and Applying Cauchy inequality, we have



(2−δ)J (J nα 2

(3.7)

 − α)2 − δ tφ 4 |∇f |4 .

2(2 − δ)(α − J )J 4 tφ F |∇f |2 − C1 tφ 4 F ∇f nα 2 ≥−

C1 nα 2 tφ 4 F, (2 − δ)(α − 1)J

where we use α − J ≥ α − 1, and C = C(n).

(3.8)

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By (3.7) and (3.8), we have  √ 2 √ (2 − δ)J √ 2 2 C1 nα 2 tφ ( − ∂t )( tφ 2 F ) ≥ ( tφ F ) − + a tφ 4 F (2 − δ)(α − 1)J nα 2 1 √ −atφ 4 αϕ − C1 tφ 2 F − √ tφ 4 F. 2 t By max principle, we have  √ (2 − δ)J √ 2 2 C1 nα 2 ( tφ F ) ≤ + a ( t)2 φ 4 F 2 (2 − δ)(α − 1)J nα √ √ 1 √ +a( t)2 φ 4 αϕ + C1 ( t)2 φ 2 F + √ tφ 4 F. 2 t So, we have nα 2 1 C1 α 4 F≤ + (2 − δ)J 2t (2 − δ)J



 anϕα 3 1 +a+1 + . (2 − δ)(α − 1)J (2 − δ)J

Case 2. (a < 0) By (3.2), we infer that f ≤ 0. Then from (3.3), (3.4) and (3.8), and apply the Cauchy inequality, we get 

(2 − δ)J 2 (J − α) − δ tφ 4 |∇f |4 + +2a(α − J )|∇f |2 tφ 4 nα 2

a 2 (α − J )2 ≥ − (2−δ)J tφ 4 , 2−δ (J − α) nα 2 we have √ 2 √ tφ ( − ∂t )( tφ 2 F ) ≥

(2 − δ)J √ 2 2 C2 nα 2 ( tφ F ) − tφ 4 F 2 nα (2 − δ)(α − 1)J 1 √ − C2 tφ 2 F − √ tφ 4 F − 2 t

a 2 (α − J )2 (2−δ)J nα 2

(J − α)2 − δ

tφ 4 .

By max principle, we have (2 − δ)J √ 2 2 C2 nα 2 ( tφ F ) ≤ tφ 4 F + C2 t 2 φ 2 F 2 (2 − δ)(α − 1)J nα 1 √ + √ tφ 4 F + 2 t

a 2 (α − J )2 (2−δ)J (J nα 2

− α)2 − δ

t.

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So, we have

nα 2 1 C2 α 4 1 F≤ +1 + (2 − δ)J 2t (2 − δ)J (2 − δ)(α − 1)J √ nα|a|(α − J ) + . (2−δ)2 J 2 − δ(2 − δ) (1 − α) 2 nα So, the proof is complete. 2 The proof of Theorem 1.3. Since ut = u + au log u, we have ∂t (|∇u|2 ) = 2Ric(∇u, ∇u) + 2 < ∇u, ∇(ut ) > = 2Ric(∇u, ∇u) + 2 < ∇u, ∇(u) > +2a(1 + log u)|∇u|2 . Applying the Bochner’s formula, the above equation becomes ∂t (|∇u|2 ) = (|∇u|2 ) − 2|∇ 2 u|2 + 2a(1 + log u)|∇u|2 .

(3.9)

∂t (u2 ) = (u2 ) − 2|∇u|2 + 2au2 log u.

(3.10)

Besides,

Let F = t|∇u|2 + Xu2 , where X decided later. Then combining (3.9) with (3.10), we obtain   ∂t F = |∇u|2 + t (|∇u|2 ) − 2|∇ 2 u|2 + 2a(1 + log u)|∇u|2   +X (u2 ) − 2|∇u|2 + 2au2 log u = F + (1 − 2X)|∇u|2 − 2t|∇ 2 u|2 + 2at (1 + log u)|∇u|2 + 2aXu2 log u. So, if a > 0 and u ≤ e−1 , then

∂t F ≤

Selecting X =

1 2

⎧ ⎨F ,

if a > 0

and

u ≤ e−1 ,

⎩ F ,

if a < 0

and

u ≥ 1.

and using the maximum principle, we infer F (x, t) ≤ maxn F (x, 0) = x∈M

which implies the theorem is valid. 2

1 max u2 (x, 0), 2 x∈M n

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18

4. Gradient estimates for the heat equation When a = 0, we derive the following estimate for the heat equation ut = u. Theorem 4.1. Let (Mn , g) be an n-dimensional complete Riemannian manifold. Suppose that there exist two functions α(t) and ϕ(t) satisfying conditions (C1), (C2) and (C3), and there exists a positive constant δ satisfying condition (C3). Assume that u(x, t) is any positive solution to the heat equation ut = u

(4.1)

on Mn × (0, ∞). For p > n2 , there exists a constant κ = κ(n, p) such that κ(p, 1) ≤ κ, then for any point O ∈ Mn , we have the following estimates, J

|∇u|2 ut 1 nα 2 − α(t) · ≤ 2 u u (2 − δ)J 2t

C1 α 2 nα 2 + + 1 + αϕ, (2 − δ)J (2 − δ)(α − 1)J

(4.2)

in B(O, 12 ) × (0, ∞). Where n   1 J = J (t) = 2− b−1 exp{−2Cκ 1 + [2C(b − 1)κ] 2p−n t},

and b = 5δ −1 ,

C = C(p, n),

and

C1 = C1 (n, p).

Let us list some special functions α(t) and ϕ(t) satisfying conditions (C1, C2, C3) to illustrate the Theorem 4.1 holds and see appendix in section 9 for detailed calculation process. Corollary 4.2. 1. Li-Yau type: α(t) = constant,

ϕ(t) =

2−δ 1 , nα J

and

0<δ≤

2(α − 1)2 . (α − 1)2 + nα 2

Then |∇u|2 ut 1 nα 2 C3 α 2 J 2 −α ≤ · + u (2 − δ)J 2t (2 − δ)J u



nα 2 +1 . (2 − δ)(α − 1)J

2. Hamilton type: α(t) = eκt , ϕ(t) =

and

0<δ<

κneκt 2(2 − δ)J

for

2( − 1)2 , ( − 1)2 + n 2 any > 1.

(4.3)

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19

Then J

|∇u|2 n 2 e2κt 1 C4 2 e2κt κt ut −

e · ≤ + u2 u (2 − δ)J 2t (2 − δ)J



nλ eκt +1 . (2 − δ)J

(4.4)

3. Linear Li-Xu type: α(t) = κt + , ϕ(t) =

and

n κ 2(2 − δ)J

0<δ<

1 ,

−1

for any > 1.

Then J

|∇u|2 (κt + )2 ut n − (κt + ) ≤ 2 u u (2 − δ)J 2t

C4 (κt + )2 nλ(κt + ) + +1 . (2 − δ)J (2 − δ)J

(4.5)

4. Li-Xu type: sinh κt cosh κt − κt

, and 0 < δ < , sinh κt

−1 nκ ϕ(t) = ( + coth κt) for any > 1. (2 − δ)J

α(t) = +

Then (1) If a ≥ 0, then |∇u|2 ut − α(t) + α(t)a log u u2 u

 anϕα 3 C4 α 4 1 nα 2 1 +a+1 + , + ≤ (2 − δ)J 2t (2 − δ)J (2 − δ)(α − 1)J (2 − δ)J

J

in B(O, 12 ) × (0, ∞). 5. α(t) is a decreasing function. α(t) = e−κt + , ϕ(t) = Then

and

nκ 2(2 − δ)

0<δ<

2( − 1)2 , ( − 1)2 + n 2

for any > 1.

(4.6)

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20

J

|∇u|2 ut n(e−κt + )2 1 − (e−κt + ) ≤ · 2 u (2 − δ)J 2t u

C4 (e−κt + )2 nλ(e−κt + ) +1 , + (2 − δ)J (2 − δ)J

(4.7)

where α(t) is bounded uniformly. Remark 2. Zhang and Zhang [46,47] obtain the first term on the right of Li-Yau gradient esti1 1 nα 2 nα 2 · . The first term on the right of (4.3) is · . So, our estimate is mates is (2 − δ)J t (2 − δ)J 2 · t stronger. 5. Harnack inequality Following the standard argument in [4] we obtain Harnack inequalities. The geodesic distance on M n will be denoted by d(·, ·). Theorem 5.1. Let (Mn , g) be an n-dimensional complete Riemannian manifold. Assume that u(x, t) is any positive solution to the heat equation ut = u on Mn × (0, ∞). For p > n2 , there exist a constant κ = κ(n, p) such that κ(p, 1) ≤ κ, then for any point O ∈ Mn , and for any positive solution satisfies for all x1 , x2 ∈ B(O, 12 ) and 0 < t1 < t2 < ∞, we have   u(x1 , t1 ) ≤ u(x2 , t2 ) × exp 1 (t1 , t2 , δ, J (T ), d(x1 , x2 )) ,

(5.1)

where γ (s) is a smooth curve connecting x1 and x2 with γ (1) = x1 and γ (0) = x2 , and 1 (t1 , t2 , δ, J (T ), d(x1 , x2 )) 1 d 2 (x1 , x2 ) = 4J (T ) t2 − t1 C(t2 − t1 ) + (2 − δ)J (T )

t2

t2

t2 α(t)dt + (t2 − t1 )

t1

ϕ(t)dt t1



1 1 α (t) + + 1 dt. (2 − δ)(α − 1)J (T ) t 3

t1

Proof. We follow the proof of Theorem 2.1 of [13]. Let γ : [0, 1] → Mn be any curve with γ (0) = x2 and γ (1) = x1 and define l(s) = log u(γ (s), (1 − s)t2 + st1 ). Obviously, we infer that l(0) = log u(x2 , t2 ) and l(1) = log u(x1 , t1 ). Direct calculation shows

∂l(s) ∇u γ  (s) ut − = (t2 − t1 ) ∂s u t 2 − t1 u

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 ≤ (t2 − t1 ) Regarding

|∇u| u

∇u γ  (s) J (t) |∇u|2 Cα 3 − + 2 u t2 − t1 α(t) u (2 − δ)J



21

1 1 + +1 +ϕ . (2 − δ)(α − 1)J t

as a variable and the integrand as a quadratic in it, we observe that



∂l(s) Cα 3 1 α(t) |γ  (s)|2 1 + (t2 − t1 ) + +1 +ϕ . ≤ ∂s 4J (t) t2 − t1 (2 − δ)J (2 − δ)(α − 1)J t By the definition of t = (1 − s)t2 + st1 and the monotono decreasing of J (t) max

s∈[0,1]

1 1 ≤ . J (t) J (T )

(5.2)

Integrating the above inequality over γ (s), we obtain

log

u(x1 , t1 ) = u(x2 , t2 )

1 ≤ 0

1

∂l(s) ds ∂s

0





Cα 3 1 α(t) |γ  (s)|2 1 + (t2 − t1 ) + + 1 + ϕ ds 4J (t) t2 − t1 (2 − δ)J (2 − δ)(α − 1)J t

1 d 2 (x1 , x2 ) ≤ 4J (T ) t2 − t1 C(t2 − t1 ) + (2 − δ)J (T )

t2

t2

t2 α(t)dt + (t2 − t1 )

t1

ϕ(t)dt t1



1 1 α (t) + + 1 dt. (2 − δ)(α − 1)J (T ) t 3

t1

Taking exponentials yields the claim. The proof is complete. 2 Theorem 5.2. Let (Mn , g) be an n-dimensional complete Riemannian manifold. Assume that u(x, t) is any positive solution to the heat equation ut = u on Mn × (0, ∞). For p > n2 , there exist a constant κ = κ(n, p) such that κ(p, 1) ≤ κ, then any point O ∈ Mn , and for any positive solution satisfies for all x1 , x2 ∈ B(O, 12 ) and 0 < t1 < t2 < ∞, we have 2

b−1 n 2   t2 2(2 − δ) × exp 2 (t1 , t2 , δ, κ, n, p, , λ, d(x1 , x2 )) , u(x1 , t1 ) ≤ u(x2 , t2 ) t1

(5.3)

where γ (s) is a smooth curve connecting x1 and x2 with γ (1) = x1 and γ (0) = x2 , and = n  2Cκ 1 + [2C(b − 1)κ] 2p−n ,

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22

 2 b−1  (κ+ )t2 2 (t1 , t2 , δ, , λ, d(x1 , x2 )) = − e(κ+ )t1 e (κ + )   2 C

C 2 λn  (κ+ )t2 (κ + ) d (x1 , x2 ) (κ+ )t1 + +e + e . × 2 − δ (2 − δ)2 4(t2 − t1 )2 1

In particular, for 1 < <

1 3

12

· 4 13 , we have

n   t2 2 × exp (t1 , t2 , δ, , λ, d(x1 , x2 )) , u(x1 , t1 ) ≤ u(x2 , t2 ) t1

(5.4)

Proof. We follow the proof of Theorem 2.1 of [4]. Let γ : [0, 1] → Mn be any curve with γ (0) = x2 and γ (1) = x1 and define l(s) = log u(γ (s), (1 − s)t2 + st1 ). Obviously, we infer that l(0) = log u(x2 , t2 ) and l(1) = log u(x1 , t1 ). By (4.4), we have

∂l(s) ∇u γ  (s) ut − = (t2 − t1 ) ∂s u t 2 − t1 u 

 ∇u γ (s) J (t) |∇u|2 n eκt 1 C eκt nλ eκt − + · + +1 . ≤ (t2 − t1 ) u t2 − t1 eκt u2 (2 − δ)J 2t (2 − δ)J (2 − δ)J Regarding

|∇u| u

as a variable and the integrand as a quadratic in it, we observe that



∂l(s) n eκt 1

eκt |γ  (s)|2 C eκt nλ eκt + (t2 − t1 ) · + +1 . ≤ ∂s 4J (t) t2 − t1 (2 − δ)J 2t (2 − δ)J (2 − δ)J By the definition of t = (1 − s)t2 + st1 and integrating the above inequality over γ (s), we obtain u(x1 , t1 ) log = u(x2 , t2 ) 1 ≤ 0

1

∂l(s) ds ∂s

0





n eκt 1 C eκt

eκt |γ  (s)|2 nλ eκt + (t2 − t1 ) · + +1 ds. (5.5) 4J t2 − t1 (2 − δ)J 2t (2 − δ)J (2 − δ)J

We next estimate the right side of (5.5), 1 0

eκt |γ  (s)|2 ds = 4J t2 − t1 ≤ 1 0

1

eκt 1 − b−1

0

4·2

e− t

|γ  (s)|2 ds t2 − t1

  1

d 2 (x1 , x2 ) b−1 e (κ+ )t2 − e (κ+ )t1 . · 2 4(t2 − t1 )2

n eκt 1 (t2 − t1 ) · ds = (2 − δ)J 2t

t2 t1

1 n

1 2 b−1 e(κ+ )t · dt. 2−δ 2t

(5.6)

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23

Since eu 1 ≤ + eu u u we can estimate 1 (t2 − t1 ) 0

n eκt 1 · ds (2 − δ)J 2 · t

n (κ + ) 1 ≤ 2 b−1 2(2 − δ) =

t2 e(κ+ )t + t1

1 (κ + )t

dt



t2 (κ+ )t2 (κ+ )t1 −e log + e t1

1 n

2 b−1 2(2 − δ)

(5.7)

and 

1 (t2 − t1 ) 0

C eκt (2 − δ)J





nλ eκt +1 (2 − δ)J

ds

 Cn 2 λ 2 b−1  2(κ+ )t2 2(κ+ )t1 = − e e (2 − δ)2 2(κ + ) 2

 C 2 b−1  (κ+ )t2 − e(κ+ )t1 . e 2−δ κ + 1

+

(5.8)

Substituting (5.6)–(5.8) into (5.5), we have

log

1 1  u(x1 , t1 ) 2 b−1  (κ+ )t2 n 2 b−1 t2 − e(κ+ )t1 ≤ log + e u(x2 , t2 ) 2(2 − δ) t1 κ +   2 (κ + ) d (x1 , x2 ) C

C 2 λ  (κ+ )t2 (κ+ )t1 × + + e + e . 4(t2 − t1 )2 2 − δ (2 − δ)2

Since λ ≥

α α−1

> 1, then 0 < δ < 0<

2 1+nλ2

1 , b−1

<

2 1+n

≤

2 3

and b = 5δ −1 >

4 < 2 − δ < 2. 3

Further, we have 1

2

· 2 b−1 3

· 2 13 < 4 = 12 . 2−δ 4 13 3

15 2 .

Therefore,

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Since

3 12

4 13

1

1 12

· 2 b−1 < 1, so we choose 1 < < · 4 13 , then < 1, and 3 2−δ log

  1 u(x1 , t1 ) n t2 ≤ log + 2 b−1 e(1+ )t2 − e(1+ )t1 u(x2 , t2 ) 2 t1  2  C

Cn 2 λ  (1+ )t2

d (x1 , x2 ) (1+ )t1 + +e + e . × 4(t2 − t1 )2 2 − δ (2 − δ)2

The proof is complete.

2

Theorem 5.3. Let (Mn , g) be an n-dimensional complete Riemannian manifold. Assume that u(x, t) is any positive solution to the heat equation ut = u on Mn × (0, ∞). For p > n2 , there exist a constant κ = κ(n, p) such that κ(p, 1) ≤ κ, then any point O ∈ Mn , and for any positive solution satisfies for all x1 , x2 ∈ B(O, 12 ) and 0 < t1 < t2 < ∞, we have 2

b−1 n 2   t2 2(2 − δ) × exp 3 (t1 , t2 , δ, κ, n, p, , λ, d(x1 , x2 )) , u(x1 , t1 ) ≤ u(x2 , t2 ) t1

(5.9)

where γ (s) is a smooth curve connecting x1 and x2 with γ (1) = x1 and γ (0) = x2 , and = n  2Cκ 1 + [2C(b − 1)κ] 2p−n , 2

 2 b−1  t2 e − e t1

 2

d (x1 , x2 ) · (κt2 + ) n(1 + ) + × 2(2 − δ) 4(t2 − t1 )2   2 2 C λ b−1 e 2(1+ )t2 − e 2(1+ )t1 . + 2 (2 − δ)2

3 (t1 , t2 , δ, , λ, d(x1 , x2 )) =

In particular, for 1 < <

1 3

12

· 4 13 , we have

n   t2 2 × exp 2 (t1 , t2 , δ, , λ, d(x1 , x2 )) , u(x1 , t1 ) ≤ u(x2 , t2 ) t1

(5.10)

Proof. We follow the proof of Theorem 2.1 of [13]. Let γ : [0, 1] → Mn be any curve with γ (0) = x2 and γ (1) = x1 and define l(s) = log u(γ (s), (1 − s)t2 + st1 ). Obviously, we infer that l(0) = log u(x2 , t2 ) and l(1) = log u(x1 , t1 ). By (4.4), we have

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∂l(s) ∇u γ  (s) ut − = (t2 − t1 ) ∂s u t 2 − t1 u 

 ∇u γ (s) J (t) |∇u|2 n(κt + ) 1 C(κt + ) nλ(κt + ) ≤ (t2 − t1 ) − + · + 1 . + u t2 − t1 κt + u2 (2 − δ)J 2 · t (2 − δ)J (2 − δ)J Regarding

|∇u| u

as a variable and the integrand as a quadratic in it, we observe that 

n(κt + ) 1 κt + |γ  (s)|2 C(κt + ) nλ(κt + ) + (t2 − t1 ) · + +1 . 4J (t) t2 − t1 (2 − δ)J 2t (2 − δ)J (2 − δ)J

∂l(s) ≤ ∂s

By the definition of t = (1 − s)t2 + st1 and integrating the above inequality over γ (s), we obtain u(x1 , t1 ) log = u(x2 , t2 ) 1 ≤ 0

1

∂l(s) ds ∂s

0





n(κt + ) 1 C(κt + ) nλ(κt + ) κt + |γ  (s)|2 + (t2 − t1 ) · + +1 ds. 4J t2 − t1 (2 − δ)J 2t (2 − δ)J (2 − δ)J (5.11)

We next estimate the right side of (5.11), 1 0

κt + |γ  (s)|2 ds = 4J t2 − t1

1 0

1 κt + |γ  (s)|2 · 2 b−1 e t ds 4 t 2 − t1 1

≤

 2 b−1 · d 2 (x1 , x2 ) κt2 +  t2 · · e − e t1 , 2 4(t2 − t1 )

and 1 0

n(κt + ) 1 (t2 − t1 ) · ds = (2 − δ)J 2 · t

t2 t1

n(κt + ) 1 t 1 2 b−1 e · dt 2−δ 2·t

Since eu 1 ≤ + eu u u we can estimate 1 (t2 − t1 ) 0

n(κt + ) 1 · ds (2 − δ)J 2t

(5.12)

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1 n ≤ 2 b−1 2(2 − δ)

=

t2

t

t dt κe + e +

t t1

1

t2 κ + t2 n2 b−1

log + (e − e t1 ) 2(2 − δ) t1

(5.13)

and 

1 (t2 − t1 ) 0

t2  = t1

C(κt + ) (2 − δ)J



nλ(κt + ) +1 (2 − δ)J

ds

 2 1 C2 b−1 Cλ2 b−1 2 2 t

t (κt + ) e + (κt + )e dt 2−δ (2 − δ)2

 2 b−1  2(1+ )t2 C 2 λ − e2(1+ )t1 e 2 (2 − δ) 2(1 + ) 2

=

 C 2 b−1  (1+ )t2 − e(1+ )t1 e + 2−δ 1+ 2  C 2 λ2 b−1  2(1+ )t2 2(1+ )t1 ≤ − e e . (2 − δ)2 2

(5.14)

Substituting (5.12)–(5.14) into (5.5), we have 2

log

The proof is complete.

2

 u(x1 , t1 ) n 2 b−1 t2 2 b−1  t2 ≤ log + e − e t1 u(x2 , t2 ) 2(2 − δ) t1

 2

d (x1 , x2 ) · t2 + n(1 + ) + × 4(t2 − t1 )2 2(2 − δ)   2 2 C λ b−1 e 2(1+ )t2 − e 2(1+ )t1 . 2 + (2 − δ)2 2

6. Upper and lower bounds for the heat kernel Motivated by [13,32,40,41], and applied the Harnack estimate in Corollary 5.2, we will prove upper and lower for the heat kernel on Riemannian manifold under the integral Ricci curvature bounds. Theorem 6.1. Let (Mn , g) be an n-dimensional complete Riemannian manifold. Assume that u(x, t) is any positive solution to the heat equation ut = u

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on Mn × (0, ∞). For p > n2 , there exist a constant κ = κ(n, p) such that κ(p, 1) ≤ κ, then for any point O ∈ Mn , there exists an explicit constant C5 = C5 (n, p, T , δ, , κ) > 0 such that for any t ∈ (0, T ] and x ∈ B(O, 12 ), the heat kernel H (·, ·, t) of M n has an upper bound by H (x, x, t) ≤

n C1 √ t− 2 . Vol(Bx ( t))

(6.1)

In particular, such that for all x, y ∈ B(O, 12 ) and t ∈ (0, T ] H (x, y, t) ≤

n C1 √ t− 2 . Vol(Bx ( t))

(6.2)

Proof. Now let x ∈ B(O, 12 ), t < T and s > 0 such that t + s ≤ T . Choose u(·, t) = H (x, ·, t). Inserting the two estimates above in inequality (5.3) or (5.10) implies for all y ∈ B(O, 12 )

n t +s 2 × exp { 2 } . H (x, x, t) ≤H (x, y, t + s) (6.3) t  Integrating (6.3) with respect to the y-variable, using M H (x, y, t)dvol(y) = 1 for all x ∈ B(O, 12 ) and all t > 0 and t + s ≤ T yields



n √ t +s 2 Vol(Bx ( t))H (x, x, t) ≤ × exp { 2 } t n T 2 ≤ × exp { 2 } t

(6.4)

So, we get H (x, x, t) ≤

n C1 √ t− 2 , Vol(Bx ( t))

(6.5)

where C5 = C5 (T , 2 ). By Harnack inequality and (6.5), we can derive that H (x, y, t) ≤ H (x, x, t + s) ≤ ≤

n C1 √ (t + s)− 2 Vol(Bx ( t))

n C1 √ t− 2 . Vol(Bx ( t))

So, we complete the proof. 2 Theorem 6.2. Let (Mn , g) be an n-dimensional complete Riemannian manifold. Assume that u(x, t) is any positive solution to the heat equation ut = u

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on Mn × (0, ∞). For p > n2 , there exist a constant κ = κ(n, p) such that κ(p, 1) ≤ κ, then for any point O ∈ Mn , there exists an explicit constant C = C(n, α, δ, β, b, diamMn ) > 0 such that for any t > 0 and x ∈ B(O, 12 ) the heat kernel H (·, ·, t) of Mn has a lower bound by n

H (x, y, t) ≥ (4πt)− 2 × exp {− 4 (t, δ, , κ, λ, d(x, y))} , n

H (x, y, t) ≥ (4πt)− 2 × exp {− 5 (t, δ, , κ, λ, d(x, y))} , where 1  2 b−1  (κ+ )t 4 (t, δ, , κ, λ, d(x, y)) = −1 e (κ + )   C

C 2 λn  (κ+ )t (κ + ) d 2 (x, y) + +1 , + e × 4t 2 2 − δ (2 − δ)2 2   2 b−1  t

(κt + ) d 2 (x, y) n(1 + ) + e −1 × 5 (t, δ, , κ, λ, d(x, y)) = ( ) 4t 2 2(2 − δ)   2 C λ e2(1+κ)t − 1 + (2 − δ)2

In particular, such that for all x, y ∈ B(O, 12 ) and t > 0  2 b−1  (1+ )t H (x, x, t) ≥ (4πt) × exp − −1 e κ +   ! C

C 2 λ  (κ+ )t × +1 . + e 2 − δ (2 − δ)2 1

− n2

 2 b−1  (κ+ )t H (x, x, t) ≥ (4πt) × exp − −1 e κ +   ! C

C 2 λ  (κ+ )t × + 1 . + e 2 − δ (2 − δ)2 1

− n2

Proof. By (5.3), we have

t +s H (x, x, s) ≤ H (x, y, t + s) s

n 2

× exp { 2 } ,

for all s > 0 and t > 0, where  2 b−1  (κ+ )(t+s) 3 (s, t + s, δ, , λ, d(x, y)) = − e(κ+ )s e (κ + )   2 C

C 2 λn  (κ+ )(t+s) (κ + ) d (x, y) (κ+ )s + +e + e . × 2 − δ (2 − δ)2 4t 2 1

(6.6) (6.7)

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By calculations we get n

lim (4πs) 2 H (x, x, s) = 1,

s→0

 2 b−1  (κ+ )t −1 e lim 3 (s, t + s, δ, , κ, λ, d(x, y)) = s→0 (κ + )   C

C 2 λn  (κ+ )t (κ + ) d 2 (x, y) + + 1 + e × 4t 2 2 − δ (2 − δ)2 1

= 4 (t, δ, , κ, λ, d(x, y)). n

Thus, multiplying by lims→0 (4πs) 2 , we have n

n 2

n 2

(4πs) H (x, x, s) ≤ (4πt) H (x, y, t + s)

s2 n



n t +s 2 s

t2 × exp { 4 (t, δ, , κ, λ, d(x, y))} .

Letting s → 0, we get n

1 ≤ (4πt) 2 H (x, y, t) × exp { 4 (t, δ, , κ, λ, d(x, y))} . So, we complete the proof of (6.6). Besides, by using (5.10), we can prove (6.7) in a similar way.

2

7. Local eigenvalue estimate Theorem 7.1. Let (Mn , g) be an n-dimensional complete Riemannian manifold. For p > n2 , there exists a constant κ = κ(n, p) such that κ(p, 1) ≤ κ, then any point O ∈ Mn . Let λ0 ≤ λ1 ≤ · · · be eigenvalues of the Laplacian. Then there exists a constant depending only on n, such that for k≥1 1

λk ≥ C(k + 1) n ,

(7.1)

in B(O, 12 ). Proof. Since H (x, x, t) ≤

n C √ t− 2 , V (Bx ( t))

where C a constant depending only on n. Notice that the heat kernel can be written as H (x, y, t) =

∞ "

e−λi t ϕi (x)ϕi (y),

i=0

where ϕi is the eigenfunction of  corresponding to λi and ||ϕi ||L2 = 1.

(7.2)

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By the volume comparison theorem, we get, for any t ≤ d 2 , n V (Bx (d)) d . √ ≤2 √ V (Bx ( t)) t

(7.3)

Integrating both sides and (7.3), we have ∞ " i=0

e−λi t ≤ C1



√ V −1 (Bx ( t))dt ≤ C1

B(O, 12 )

 f (t)dt,

(7.4)

B(O, 12 )

where n ⎧ d ⎪ ⎪ V −1 (Bx (d)), ⎨ 2 t n f (t) = ⎪ 1 ⎪ ⎩ 2 V −1 (Bx (d)), d

if

t ≤ d 2,

if

t > d 2,

which implies that (k + 1)e−λi t ≤ Ch(t) for any t > 0, that is Ceλk t h(t) ≥ (k + 1),

for any t > 0,

(7.5)

where n ⎧ d ⎪ ⎪ , 2 ⎨ t n h(t) = ⎪ 1 ⎪ ⎩ 2 , d

if t ≤ d 2 , if t > d 2 .

It is easy to find that eλk t h(t) achieves its minimum at t0 = obtain the lower bound for λk ,

n λk .

Substituting into (7.5), we

1

λk ≥ where d is the diameter of B(O, 12 ).

C(k + 1) n , d

2

8. Lower bound of Green’s function In a complete manifold, let H (x, y, t) be a heat kernel, recal the Green’s function ∞ G(x, y) =

H (x, y, t)dt 0

if the integral on the right hand side converges [13]. Li-Yau proved a estimate of Green’s function for a complete manifold with nonnegative Ricci curvature. Jiayong Wu generalized Li-Yau

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estimate of Green’s function to the f -Green’s function for a complete metric measure space with nonnegative Ricci curvature and |f | ≤ C. In this section, we will continue to study this problem, and derive a lower estimate of Green’s function for a complete manifold with the integral Ricci curvature bounds. Theorem 8.1. Let (Mn , g) be an n-dimensional complete Riemannian manifold. For p > n2 , there exists a constant κ = κ(n, p) such that the following inequalities hold. If κ(p, 1) ≤ κ, then any point O ∈ M n , there exists an explicit C1 = C1 (n, α, δ, diamMn ) > 0 such that for any t ∈ (0, 14 ) and x ∈ Mn , the Green’s function has a lower bound by G(x, y) ≥ C5 r 2−n .

(8.1)

Proof. Since the function exp {− 4 (t, δ, , κ, λ)} for r 2 < t < ∞ is bound, then we have from (6.6) ∞ G(x, y) ≥

∞ H (x, y, t)dt ≥ C5

r2

n

t − 2 dt

r2

=C5 r 2−n . Hence, (8.1) holds.

(8.2)

2

9. Appendix We will check some special functions α(t) > 1 and ϕ(t) > 0 satisfy the following two systems We first define two C 1 functions α(t) and ϕ(t) : (0, +∞) → (0, +∞) that satisfy the following conditions: (C1) α(t) > 1. (C2) α(t) and ϕ(t) satisfy the following system ⎧  ϕ ϕ  ⎪  1, ⎪ ≥ 2(2 − δ) − α 2(2 − δ) ⎪ ⎪ n n α ⎪ ⎨ ϕ 2(2 − δ) J − α  > 0, ⎪ n ⎪ ⎪ ⎪ ϕ2 ⎪  ⎩ αϕ + (2 − δ) J ≥ 0. n

(9.1)

(C3) There exist two real numbers λ and δ satisfying ⎧ α ⎪ ⎨0 < α − 1 ≤ λ, 2 ⎪ ⎩ 0<δ≤ . 1 + nλ2 Where α  =

dα dt

and ϕ  =

dϕ dt .

(9.2)

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(1) Let α(t) = constant,

ϕ(t) =

2−δ 1 , nα J · t

and

0<δ≤

2(α − 1)2 . (α − 1)2 + nα 2

Obviously, conditions valid. (2) Let α(t) = eκt ,

and

ϕ(t) =

κneκt 2(2 − δ)J

for

any > 1.

Direct calculation satisfies system (9.1). Besides, we have d α(t) d eκt − κeκt < 0, = = dt α(t) − 1 dt eκt − 1 ( eκt − 1)2 α(t)

eκt

α(t)

eκt = κt → , as t → +∞, = κt → 1. This α(t) − 1

e − 1

−1 α(t) − 1

e − 1 α(t)

2( − 1)2 implies 1 < . So, (9.2) is also satisfied. ≤ and 0 < δ < α(t) − 1 − 1 ( − 1)2 + n 2 (3) Let

and as t → 0+ ,

α(t) = κt + ,

and ϕ(t) =

nκ +

2(2 − δ)J

for

any > 1.

Direct calculation gives satisfy system (9.1). Besides, we have d α(t) d κt +

−κ < 0, = = dt α(t) − 1 dt κt + − 1 (κt + − 1)2 α(t) κt +

α(t) κt +

= → , as t → +∞, = → 1. α(t) − 1 κt + − 1

−1 α(t) − 1 κt + − 1 α(t)

2( − 1)2 This implies 1 < . So, (9.2) is also satisfied. ≤ and 0 < δ < α(t) − 1 − 1 ( − 1)2 + n 2 (4) Let

and as t → 0+ ,

α(t) = +

sinh κt cosh κt − κt , sinh κt

and ϕ(t) =

nκ ( + coth κt) (2 − δ)J

Since α  > 0, direct calculation gives satisfy system (9.1). Besides, we have d α(t) α  (t) < 0, = dt α(t) − 1 (α(t) − 1)2

for any > 1.

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and as t → 0+ ,

33

α(t)

α(t)

→ . This implies 0 < ≤ and 0 < δ < α(t) − 1

−1 α(t) − 1

−1

2( − 1)2 . So, (9.2) is also satisfied. ( − 1)2 + n 2 (5) Let α(t) = e−κt + ,

and

ϕ(t) =

nκ 2(2 − δ)

for

any > 1.

Direct calculation gives satisfy system (9.1). Besides, we have d α(t) d e−t +

e−t > 0, = = dt α(t) − 1 dt e−t + − 1 (et + − 1)2 α(t) e−t +

1+

α(t) e−t +

= −t → , as t → +∞, = −t → α(t) − 1 e + −1

α(t) − 1 e + −1 2

1+

α(t)

2( − 1) . So, (9.2) is also . This implies < ≤ and 0 < δ <

−1

α(t) − 1

−1 ( − 1)2 + n 2 satisfied. and as t → 0+ ,

Acknowledgments We are grateful to Professor Jiayu Li for their support and encouragement. The author also thanks Professor Qi S Zhang for introduction of this problem in the summer course. This work is partially supported by NSFC (No. 11721101) and the NSF of Anhui Province (1908085QA04). References [1] M. Bailesteanu, X.D. Cao, A. Pulemotov, Gradient estimates for the heat equation under the Ricci flow, J. Funct. Anal. 258 (2010) 3517–3542. [2] Xiaodong Cao, Qi S. Zhang, The conjugate heat equation and ancient solutions of the Ricci flow, Adv. Math. 228 (5) (2011) 2891–2919. [3] H. Cao, M. Zhu, Aronson-Bénilan estimates for the fast diffusion equation under the Ricci flow, Nonlinear Anal. 170 (2018) 258–281. [4] D.G. Chen, C.W. Xiong, Gradient estimates for doubly nonlinear diffusion equations, Nonlinear Anal. 112 (2015) 156–164. [5] X.Z. Dai, G.F. Wei, Z.L. Zhang, Local Sobolev constant estimate for integral Ricci curvature bounds, Adv. Math. 325 (2018) 1–33. [6] Ha Tuan Dung, Nguyen Thac Dung, Sharp gradient estimates for a heat equation in Riemannian manifolds, Proc. Am. Math. Soc. (2019), https://doi.org/10.1090/proc/14645. [7] R.S. Hamilton, A matrix Harnack estimates for the heat equation, Commun. Anal. Geom. 1 (1993) 113–126. [8] S.B. Hou, L. Zou, Harnack estimate for a semilinear parabolic equation, Sci. China Math. 60 (2017) 833–840, https://doi.org/10.1007/s11425-016-0270-6. [9] G. Huang, Z. Huang, H. Li, Gradient estimates for the porous medium equations on Riemannian manifolds, J. Geom. Anal. 23 (2013) 1851–1875. [10] X.R. Jiang, Gradient estimates for a nonlinear heat equation on Riemannian manifolds, Proc. Am. Math. Soc. 144 (2016) 3635–3642. [11] J.Y. Li, Gradient estimates and Harnack inequalities for nonlinear parabolic and nonlinear elliptic equations on Riemannian manifolds, J. Funct. Anal. 100 (1991) 233–256. [12] J.Y. Li, Gradient estimate for the heat kernel of a complete Riemannian manifold and its applications, J. Funct. Anal. 97 (2) (1991) 293–310.

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