Harnack inequality for the negative power Gaussian curvature flow

AMERICAN MATHEMATICAL SOCIETY

Volume 139, Number 10, October 2011, Pages 3707–3717 S 0002-9939(2011)11039-6

Article electronically published on April 1, 2011

HARNACK INEQUALITY

FOR THE NEGATIVE POWER GAUSSIAN CURVATURE FLOW

YI LI

(Communicated by Jianguo Cao)

Abstract. In this paper, we study the power of Gaussian curvature flow of a compact convex hypersurface and establish its Harnack inequality when the power is negative. In the Harnack inequality, we require that the absolute value of the power is strictly positive and strictly less than the inverse of the dimension of the hypersurface.

1. Introduction

The Harnack estimate or Harnack inequality plays an important role in geometric flows. For the heat equation, P. Li and S.-T. Yau [6] obtained the corresponding Harnack inequality by using the parabolic maximum principle. Hamilton [4, 5]

proved a Harnack inequality for the Ricci flow and mean curvature flow for all dimensions. For a Harnack inequality form-power mean curvature flow, we refer to [1], [9] and [10], wheremis positive. B. Chow [3] considered a Harnack inequality form-power Gaussian curvature flow form >0.

In this paper, we consider the negative power Gaussian curvature flow of a com- pact convex hypersurfaceF0:Mⁿ →Rⁿ⁺¹,

(1.1) ∂

∂tF(x, t) = 1

K(x, t)^b ·ν(x, t), 0< b < 1

n; F(x,0) =F0(x), x∈Mⁿ. Here K is the Gaussian curvature andν denotes the outward unit normal vector field. Using the similar argument in [3], we obtain a Harnack inequality for the flow (1.1).

Theorem 1.1. Suppose thatF0 :Mⁿ →Rⁿ⁺¹ is a compact convex hypersurface.

If 0< b < _n¹, then

(1.2) ∂

∂t 1

K(x, t)^b

+ ∇

1 K(x, t)^b

²

h

− nb (1−nb)t

1 K(x, t)^b

≤0, where the notation | · |h is defined in the next section.

When _n¹ ≤b ≤1, some interesting results have been derived in [7], where the author consideredn= 2.

Received by the editors August 29, 2010.

2010Mathematics Subject Classification. Primary 53C44, 53C40.

Key words and phrases. Harnack inequality, negative power Gaussian curvature flow.

c2011 American Mathematical Society Reverts to public domain 28 years from publication 3707

2. Notation and evolution equations

2.1. Notation. Suppose that F : Mⁿ → Rⁿ⁺¹ is a hypersurface. The second fundamental form is given by

(2.1) hij =−

∂²F

∂xⁱ∂x^j, ν

,

where·,·denotes the standard metric onRⁿ⁺¹. The induced metric

(2.2) gij =

∂F

∂xⁱ, ∂F

∂x^j

onM gives us the mean curvature

(2.3) H=g^ijhij

and the Gaussian curvature

(2.4) K= det(hij)

det(gij).

If α = {αi} and β ={βi} are 1-forms and s = {sij} is a symmetric positive definite covariant 2-tensor, we use the short notation

α, βs≡ αi, βis:=s⁻_ij¹αiβj,

where (s⁻_ij¹) is the inverse matrix of (sij). Similarly, ifA={αijk}andB={βpqr} are covariant 3-tensors, we define

A, Bs≡ Aijk, Bijks:=s⁻_ip¹s⁻_jq¹s⁻_kr¹AijkBpqr. Finally, we define the Laplacian-type operator by

(2.5) :=h⁻_ij¹∇i∇j.

Here∇ denotes the Levi-Civita connection of the induced metricg onM.

LetMⁿ be a convex hypersurface inRⁿ⁺¹,α={αi}a 1-form onMⁿ, andφ a smooth function onM. We have the following identities (see [2] or [3]):

Rijk = hihjk−hikhj, (2.6)

∇ihjk = ∇jhik, (2.7)

(∇i∇j− ∇j∇i)αk = −Rijkg^pαp, (2.8)

∇i∇jF = −hijν, (2.9)

∇iν = hijg^jk∇kF, (2.10)

∇i∇jν = g^k∇khij· ∇F−g^khihkjν, (2.11)

∇i

Kh⁻_ij¹

= 0, (2.12)

(∇i−∇i)φ = −∇ihjk,∇j∇kφh−(n−1)hijg^jk∇kφ.

(2.13)

2.2. Evolution equations. Now we consider a generalized Gaussian curvature flow

(2.14) ∂

∂tF(x, t) =−f(K(x, t))·ν(x, t), F(x,0) =F0(x), x∈Mⁿ, where f : (0,+∞) → R is a smooth function depending only on the Gaussian curvatureK, which satisfies f>0 everywhere in order to guarantee a short time existence. Such a type of Gaussian curvature flow is called thef-Gaussian curvature flow.

Remark 2.1. For convenience, in what follows, we writeft=f(Kt) and∂2= _∂^∂

t. Under thef-Gaussian curvature flow, it is easy to verify the following evolution equations (compared with Lemma 3.1 in [3]), whereht={(ht)ij}:

∂t(gt)ij = −2ft(ht)ij, (2.15)

∂tνt = ∇ft = f_t· ∇Kt, (2.16)

∂t(ht)ij = ∇i∇jft−ft(gt)^k(ht)ik(ht)j, (2.17)

∂tKt = f_tKt·

tKt+f_t ft

|∇Kt|²h_t+ ft

ftKt

HtKt

, (2.18)

∂tft = f_tKt·[tft+Htft], (2.19)

∂tHt = Δtft+ft|ht|²g_t, (2.20)

∂tt = −∇∇ft,∇∇h_t+ftΔt+

2−n+ ft

f_tKt

Δtft,Δtg. (2.21)

Remark 2.2. IfMⁿ is compact and convex, then H =H0>0; using the evolution equation (2.20), we see that H(x, t) = Ht(x) > 0 under the f-Gaussian curva- ture flow. According to (2.4), we conclude thatK(x, t)>0 along the f-Gaussian curvature flow. Therefore _K¹

t is well-defined.

3. Harnack inequality

Motivated by the self-similar solutions in [3], we define a time-dependent tensor fieldPt={(Pt)ij}by

(3.1) (Pt)ij =∇i∇jft−(ht)⁻_k¹∇k(ht)ij· ∇ft+ft(gt)^k(ht)ik(ht)j. Taking the trace of (Pt)ij with respect to (ht)ij, we set

(3.2) P_t= (ht)⁻_ij¹(Pt)ij.

Since∇Kt=Kt(ht)_pq⁻¹∇(ht)pq by (2.12), we can rewritePtas

(3.3) Pt=tft+ftHt−(ht)⁻_ij¹(ht)⁻_k¹∇k(ht)ij· ∇ft=tft+ftHt−|∇ft|²ht

f_tKt

.

3.1. Evolution equation for Pt. In this subsection our task is to find the evo- lution equation for Pt. Before doing this, we first write down some elementary formulas which will be used in our complicated and tedious computation. Sincef is smooth depending only onKt, we have∇ft=f_t∇Ktand

tft = (ht)_ij⁻¹∇i∇jft = (ht)⁻_ij¹∇i(f_t∇jKt)

= (ht)⁻_ij¹[f_t∇iKt· ∇jKt+f_t∇i∇jKt] = f_ttft+f_t|∇Kt|²h_t. (3.4)

Using∇iKt=∇ift/f_t, we obtain (3.5) ∇i∇jKt=∇i

ft−1∇ift

=−ft−3

ft∇ift· ∇jft+ft−1∇i∇jft.

The next useful formula is

t(f_t)Kt) = (ht)⁻_ij¹∇i(f_t∇jKt·Kt+f_t· ∇jKt)

= (ht)⁻_ij¹[f_tKt∇i∇jKt + (f_tKt+ 2f_t)∇iKt· ∇jKt+f_t∇i∇jKt]

= (ht)⁻_ij¹[f_t+f_tKt] 1

f_t∇i∇jft− ft

f_t³∇ift· ∇jft

+ (ht)⁻_ij¹[f_tKt+ 2f_t]∇ift· ∇jft

ft2

=

1 +f_tKt

f_t

tft+

f_tKt

f_t² + f_t

f_t²−f_t²Kt

f_t³

|∇ft|²h_t. (3.6)

Lemma 3.1. Under the f-Gaussian curvature flow, we have

∂t(tft) = f_tKt·t(tft) + 2

1 + f_tKt

f_t

∇ft,∇(tfth_t

+

1 +f_tKt

f_t

(tft)²+f_tKt·t(Htf(Kt)) + 2

1 +ftKt

f_t

∇ft,∇(Htft)h_t+

1 +ftKt

f_t

Htft·tft

(3.7)

+

f_tKt

f_t² + f_t

f_t²−f_t²Kt

f_t³

|∇ft|²h_t[tft+Htft]

− |∇∇ft|²ht+ftΔtft+

2−n+ ft

f_tKt

|∇ft|²gt. Proof. From∂t(tft) = (∂tt)ft+t(∂tft), we get

∂t(tft) =−|∇∇ft|²ht+ftΔtft+

2−n+ ft

f_tKt

|∇ft|²gt+t[f_tKt(tft+Htft)]. Now we evaluate the last term:

t[f_tKt(tft+Htft)] = t(f_tKt)·(tft+Htft)

+f_tKt·t(tft+Htft) + 2∇(ftKt),∇(tft+Htft)h_t

=

1 +f_tK ft

tft +

f_tKt

ft2 + f_t

ft2 −f_t²Ky

ft3

|∇ft|²h_t

(tft+Htft) +f_tKt·t(tft) +f_tKt·t(Htft)

+ 2

1 + f_tKt

f_t

∇ft,∇(tft) +∇(Htft)

h_t

.

Simplifying the above and plugging into the expression of∂t(tft), we obtain the

required result.

Lemma 3.2. Under the f-Gaussian curvature flow, we have

∂t

|∇ft|²h

f_tKt

= (ftKt)t

|∇ft|²h_t

f_tKt

+ 2

1 +f_tKt

f_t ∇ft,∇ |∇ft|²h_t

f_tKt

h_t

− 2|∇∇ft|²h_t+ 2∇i(ht)jk,∇ift· ∇j∇kftht

+

1 + f_tKt

ft

2tft|∇ft|²g

ftKt

+

1 +

1 +f_tKt

ft

ft

ftKt

Ht|∇ft|²h_t

− ∇if · ∇jhkl,∇jf· ∇ihklh+ f

fK + 2−n

|∇f|²g

+ 2ft∇Ht,∇fth_t− ft2

f_t⁴ −ft

f_t³ + 1 (f_tKt)²

∇ft|²ht. Proof. The proof is similar to that in [3]. We observe first that

∂t

(f_tKt)⁻¹|∇ft|²h_t

= ∂t

f_t⁻¹K_t⁻¹(ht)⁻_ij¹∇ift∇jft

=− 1 f_t²

f_t f_tKt

+ 1 K_t²

f_tKt(tft+Htft)|∇ft|²h_t

+ 2(f_tKt)⁻¹∇(ftKt(tft+Htft)),∇ftht

−(f_tKt)⁻¹(ht)⁻_ik¹(ht)⁻_j¹(∇k∇ft−ftg^pqhkphq)∇ift∇jft

=− 1 f_t²

ft

f_tKt

+ 1 K_t²

ftKt(tft+Htft)|∇ft|²h_t

−(f_tKt)⁻¹∇i∇jft,∇ift∇jfth_t+ (f_tKt)⁻¹ft|∇ft|²gt

+ 2∇(tft+Htft),∇fth_t+ 2(f_tKt)⁻¹∇(ftKt),∇fth_t(tft+Htft)

=

1 + f_tKt

f_t 1

f_tKt

(tft+Htft)|∇ft|²h_t− 1

f_tKt∇i∇jft,∇ift∇jfth

+ ft

f_tKt

|∇ft|²g_t+ 2∇(tft+Htft),∇ftht.

On the other hand, we compute the Laplacian of (f_tKt)⁻¹|∇ft|²h_t with respect to (ht)ij:

t

(f_tKt)⁻¹|∇ft|²h_t

= t

f_t⁻¹K_t⁻¹(ht)⁻_ij¹∇ift∇jft

=t

(f_tKt)⁻¹

|∇ft|²ht+ (f_tKt)⁻¹·t(ht)⁻_ij¹· ∇ift∇jft

+ (f_tKt)⁻¹(ht)⁻_ij¹

2t(∇ift)· ∇jft+ 2∇k∇ift,∇k∇jft_ht + 4

∇kh⁻_ij¹,∇k∇if· ∇jf

h(fK)⁻¹ + 2

∇k(f_tKt)⁻¹,∇k(ht)⁻_ij¹· ∇ift∇jft

h_t

+ 4

∇k(f_tKt)⁻¹,(ht)⁻_ij¹∇k∇ift· ∇jft

ht

=t

(f_tKt)⁻¹

|∇ft|²h_t+ (f_tKt)⁻¹·(ht)⁻_ij¹· ∇ift∇jft

+ 2(f_tKt)⁻¹t(∇ft),∇ftht

+ 2(f_tKt)⁻¹|∇i∇jft|²h_t+ 2

∇k(f_tKt)⁻¹,∇k(ht)⁻_ij¹

h_t· ∇ift∇jft

+ 4

∇k(f_tKt)⁻¹∇ift,∇k∇ift

ht−4(f_tKt)⁻¹∇k(ht)ij,∇k∇ift· ∇jfth_t. We compute some elementary formulas which will be used later. Note that

∇(Kt⁻¹) = −Kt⁻²∇Kt = − ∇ft

f_tK_t²,

∇(ft−1

) = −ft−2∇ft = −f_t f_t³∇ft.

Therefore

t(K_t⁻¹) = (ht)⁻_ij¹

2K_t⁻³∇iKt· ∇jKt−K_t⁻²∇i∇jKt

= (ht)⁻_ij¹ 2

Kt³

∇ift· ∇jft

f₂² − 1 Kt²

∇i∇jff

ft

− f_t

ft3∇ift∇jft

= 2

f_t²K_t³|∇ft|²ht− 1

f_tK_t²tft+ f_t

f_t³K_t²|∇ft|²ht

= 2

Kt

+f_t f_t

1

(f_tKt)²|∇ft|²h_t− 1 f_tK_t²tft, t(f_t⁻¹) = (ht)⁻_ij¹

2f_t²

f_t³∇iKt∇jKt−f_t

f_t²∇iKt∇jKt− f_t

f_t²∇i∇jKt

= 3f_t²

ft5 −f_t ft4

|∇ft|²h_t− f_t ft3tft, ∇(ft−1

),∇(Kt⁻¹)

h_t = f_t

f_t⁴K_t²|∇ft|²ht. Using these equations, we arrive at

t

(f_tKt)⁻¹

= t(f_t⁻¹)·K_t⁻¹+f_t⁻¹·t(K_t⁻¹) + 2

∇f_t⁻¹,∇K_t⁻¹

h_t

=

3f_t²

Ktf_t⁵ − f_t Ktf_t⁴

|∇ft|²h_t− f_t Ktf_t³tft

+

2

f_t³K_t³ + f_t f_t⁴K_t²

|∇ft|²ht− 1

f_t²K_t²tft+ 2f_t

f_t⁴K_t²|∇ft|²ht

= −

1 + ftKt

f_t

1 f_t²K_t²tft

+

3ft2

f_t⁴ −ft

f_t³ + 2

f_t²K_t² + 3ft

f_t³Kt

|∇ft|²h_t

f_tKt

. The Laplacian of (ht)⁻_ij¹ with respect to (ht)ij is given by

(ht)⁻_ij¹

= (ht)⁻_k¹∇k

−(ht)⁻_ip¹(ht)⁻_jq¹∇(ht)pq

= −(ht)⁻_ip¹(ht)⁻_jq¹t(ht)pq

+ (ht)⁻_k¹(ht)⁻_ip¹(ht)_jr⁻¹h⁻_qs¹∇k(ht)rs∇(ht)pq

+ (ht)⁻_k¹(ht)_jq⁻¹(ht)⁻_ir¹(ht)⁻_ps¹∇k(ht)rs∇(ht)pq. So the second term in the expression oft((f_tKt)⁻¹|∇ft|²ht) is t((ht)⁻_ij¹)· ∇ift∇jft=−(ht)_ip⁻¹(ht)⁻_jq¹t(ht)pq· ∇ift∇jft

+ 2(ht)⁻_k¹(ht)_ip⁻¹(ht)⁻_jr¹(ht)_qs⁻¹∇k(ht)rs∇(ht)pq· ∇ift∇jft

=− t(ht)ij,∇ift∇jft_ht

+ 2∇ift∇j(ht)k,∇jft· ∇i(ht)k_ht. Combining those identities, we have

t

(f_tKt)⁻¹|∇ft|²h_t

=−

1 +f_tKt

ft

1

(ftKt)²tft· |∇ft|²h_t

+ 3f_t²

ft4 −f_t ft3 + 2

ft2Kt²

+ 3f_t ft3Kt

|∇ft|⁴h_t

ftKt

−(f_tKt)⁻¹t(ht)ij,∇ift∇jft_ht