Local Aronson-Benolan type gradient estimates for the porous medium type equation under Ricci

0.16 MB·

by  Wen Wang

January 10, 2017
Local Aronson-B´enolan type gradient estimates for the porous
medium type equation under Ricci flow
WenWang HuiZhou DapengXie
7
1
0 Abstract. Inthispaper,weinvestigatesomenewlocalAronson-B´enilantype
2 gradientestimatesforpositivesolutionsoftheporousmediumequation

n ut=∆um,
a
underRicciflow. Asapplication,therelatedHarnackinequalitiesarederived.
J
Ourresultsgeneralizeknownresults. Theseresultsinthepapercanberegard
9 as generalizing the gradient estimates of Lu-Ni-Va´zquez-Villani and Huang-
Huang-LitotheRicciflow.
]
G
D
. 1. Introduction
h
t
a In the paper, we mainly derive the parabolic version gradient estimates and
m Harnackinequality forpositive solutionsto the porousmedium equation(PME for
[ short)
ut =∆um, m>1 (1.1)
1
under Ricci flow.
v
3 Let (Mn,g) be a complete Riemannian manifold. Li and Yau [9] established a
0 famous gradientestimate for positive solutionsto the heat equation. In 1991,Liin
2 [10] deduced gradient estimates and Harnack inequalities for positive solutions to
2
thenonlinearparabolicequationonM [0, ).In1993,Hamiltonin[5]generalized
0 × ∞
the constant α of Li and Yau’s result to the function α(t) = e2Kt. In 2006, Sun
.
1
[17] also proved gradient estimates of different coefficient. In 2011, Li and Xu
0
in [11] further promoted Li and Yau’s result, and found two new functions α(t).
7
1 Recently, first author and Zhang in [18] further generalized Li and Xu’s results to
: the nonlinear parabolic equation. Related results can be found in [4, 14, 19].
v
In 2009,Lu, Ni, Va´zquez andVillaniin [13]studied the PMEonmanifolds, and
i
X obtained the results below.
r TheoremA(Lu,Ni,V´azquezandVillani). Let(Mn,g)beann-dimensional
a
complete Riemannnian manifold with Ric(B (2R)) K, K > 0. Assume that
p
u is a positive solution of (1.1). Let v = m um−1≥an−d M = max v.
m−1 Bp(2R)×[0,T]
Then for any α>1, we have
v 2 v CMaα2 α2
|∇ | α t am2+(m 1)(1+√KR)
v − v ≤ R2 α 1 −
(cid:18) − (cid:19)
α2 aα2
+ a(m 1)MK+
α 1 − t
−
on the ball B (2R), where a= n(m−1) and the constant Cdepends only on n.
p n(m−1)+2
2010 Mathematics Subject Classification. 58J35, 35K05,53C21.
Key words and phrases. porousmediumequation; Gradientestimate;Harnackinequality.
ThispaperwastypesetusingAMS-LATEX.
Correspondingauthor: WenWang,E-mail: [email protected].
1
2 W.WANG,H.ZHOU,D.XIE
Moreover, when R , the following gradient estimate on complete noncaom-
→∞
pact Riemannian manifold (Mn,g) can be deduced:
v 2 v α2 aα2
|∇ | α t a(m 1)MK+
v − v ≤ α 1 − t
−
Huang,HuangandLiin[7]generalizedtheresultsofLu,Ni,Va´zquezandVillani,
obtained Li-Yau type, Hamilton type and Li-Xu type gradient estimates.
Recently, above these results had been generalized to the Ricci flow.
The Ricci flow
∂ g(x,t)= 2Ric(x,t) (1.2)
t
−
wasfirstintroducedbyHamilton[6]toinvestigatethe Poincar´econjectureoncom-
pact three dimensional manifolds of positive Ricci curvature. In 2008, Kuang and
Zhang [8] proved a gradient estimate for positive solutions to the conjugate heat
equationunderRicciflowonaclosedmanifold. Soonafterwards,gradientestimate
for positive solutions to the heat equation under Ricci flow were further studied,
one can see [1, 12, 14, 16]. Recently, Cao and Zhu [3] derived some Aronson and
B´enilan estimates for PME (1.1) under Ricci flow.
We first introduce three C1 functions α(t), ϕ(t) and γ(t) :(0,+ ) (0,+ ).
∞ → ∞
SupposethatthreeC1 functionsα(t),ϕ(t)andγ(t)satisfythefollowingconditions:
(C1) α(t)>1, ϕ(t) and γ(t).
(C2) α(t) and ϕ(t) satisfy the following system
2ϕ 2ϕ 1
2(m 1)MK ( α′) ,
n(m 1) − − ≥ n(m 1) − α
 − −
 n(m2ϕ−1) −α′ >0,
ϕ2
+αϕ′ 0.
(C3) γ(t) satisfies n(m−1) ≥
γ′ 2ϕ 1
( α′) 0.
γ − n(m 1) − α ≤
−
(C4) α(t) and γ(t) are non-decreasing.
2. Main Results
Our results state as follows.
Theorem2.1. Let(Mn,g(x,t)) beacompletesolutiontotheRicciflow (1.2).
t∈[0,T]
Let Mn be complete under the initial metric g(x,0). Assume that Ric(x,t) K
| | ≤
for some K > 0 and all t [0,T]. Suppose that there exist three functions α(t),
∈
ϕ(t) and γ(t) satisfy conditions (C1), (C2), (C3) and (C4).
Given x M and R>0, let u be a positive solution of the nonlinear parabolic
0
∈
equation (1.1) in the cube B := (x,t)d(x,x ,t) 2R,0 t T . Let
2R,T 0
v = m um−1 and v M. { | ≤ ≤ ≤ }
m−1 ≤
If γα4 C for some constant C , Then for any (x,t) B ,
α−1 ≤ 2 2 ∈ 2R,T
v 2 v 1 CaMm2
|∇ | α t Caα2 1+√KR +K +
v − v ≤ R2 R2γ
(cid:20) (cid:16) (cid:17) (cid:21)
Local Aronson-B´enolan type gradient estimates 3
Kα2
+α2K a(m 1)+ √an. (2.1)
− m 1
−
If γ C for some constant Cp, Then for any (x,t) B ,
α−1 ≤ 2 2 ∈ 2R,T
v 2 v 1 CaMm2α4
|∇ | α t Caα2 1+√KR +K +
v − v ≤ R2 R2γ
(cid:20) (cid:16) (cid:17) (cid:21)
Kα2
+α2K a(m 1)+ √an. (2.2)
− m 1
−
holds on B , where a= n(m−1)p, and the constant C depends only on n.
2R,T n(m−1)+1
Letusshowsomespecialfunctionstoillustratethetheorem2.1holdsfordifferent
circumstances and see appendix in section 5 for detailed calculation process.
Remarks: 1. Li-Yau type:
αn(m 1) n(m 1)2MK
α(t)=constant, ϕ(t)= − + − ,
t α 1
−
γ(t)=tθ with 0<θ 2.
≤
Then
v 2 v 1 Cam2α4M
|∇ | α t Caα2 1+√KR +K +
v − v ≤ R2 R2γ
(cid:20) (cid:16) (cid:17) (cid:21)
Kα2
+α2K a(m 1)+ √an
− m 1
−
2. Hamilton type: p
n(m 1)
α(t)=e2(m−1)MKt,ϕ(t)= − e4(m−1)MKt,γ(t)=te2(m−1)MKt.
t
Then
v 2 v 1 Cam2α4M
|∇ | α t Caα2 1+√KR +K +
v − v ≤ R2 R2γ
(cid:20) (cid:16) (cid:17) (cid:21)
Kα2
+α2K a(m 1)+ √an.
− m 1
−
3. Li-Xu type: p
sinh((m 1)MKt)cosh((m 1)MKt) (m 1)MKt
α(t)=1+ − − − − ,
sinh2((m 1)MKt)
−
ϕ(t)=2n(m 1)2MK[1+coth((m 1)MKt)], γ(t)=tanh((m 1)MKt).
− − −
Then
v 2 v 1 Cam2M
|∇ | α t Ca 1+√KR +K +
v − v ≤ R2 R2γ
(cid:20) (cid:16) (cid:17) (cid:21)
Kα2
+α2K a(m 1)+ √an,
− m 1
−
where α(t) is bounded uniformly. p
4. Linear Li-Xu type:
n(m 1)
α(t)=1+(m 1)MKt, ϕ(t)= − +n(m 1)2MK,
− t −
γ(t)=(m 1)MKt.
−
4 W.WANG,H.ZHOU,D.XIE
Then
v 2 v 1 Cam2α4M
|∇ | α t Caα2 1+√KR +K +
v − v ≤ R2 R2γ
(cid:20) (cid:16) (cid:17) (cid:21)
Kα2
+α2K a(m 1)+ √an.
− m 1
−
The local estimates above impply global estimates.
Corollary 2.1. Let (Mn,g(0)) be a complete noncompact Riemannian manifold
without boundary, and asuume g(t) evolves by Ricci flow in such a way that Ric
| |≤
K for t [0,T]. Let u(x,t) be a positive solution to the equation (1.1), and let
v = m ∈um−1 and v M.
m−1 ≤
If l 1 and for (x,t) Mn (0,T], then
≤ ∈ ×
v 2 v Kα2
|∇ | α t Caα2K+α2K a(m 1)+ √an.
v − v ≤ − m 1
−
p
3. Preliminary
Let v = m um−1 and put into equation (1.1), we get
m−1
v =(m 1)v∆v+ v 2, (3.1)
t
− |∇ |
which is equivalent to the following form:
v v 2
t =(m 1)∆v+ |∇ | . (3.2)
v − v
Lemma 3.1. Assume that (Mn,g(x,t)) satisfies the hypotheses of Theorem 2.1.
We introduce the differential operator
=∂ (m 1)v∆.
t
L − −
Let F = |∇v|2 αvt αϕ, where α=α(t)>1. Then we have
v − v −
n
(F) (m 1) v2 +(m 1)α2K2+2(m 1)K v 2+2m v F
L ≤ − − ij − − |∇ | ∇ ∇
i,j
X
v 2 v
[(m 1)∆v]2+2(α 1)K|∇ | α′ t α′ϕ αϕ′. (3.3)
− − − v − v − −
Proof. Simple calculation shows
v v 2
∂ t = ∂ (m 1)∆v+ |∇ |
t t
v − v
(cid:16) (cid:17) (cid:20) (cid:21)
2 v v 2 v 2v
= (m 1)(∆v) + ∇ ∇ t + Ric( v, v) |∇ | t
− t v v ∇ ∇ − v2
2 v
= (m 1)(∆v) + ∇ (m 1)v∆v+ v 2
t
− v ∇ − |∇ |
v 2 h 2 i
|∇ | (m 1)v∆v+ v 2 + Ric( v, v)
− v − |∇ | v ∇ ∇
h v 2i∆v
= (m 1)(∆v) +(m 1)|∇ | +2(m 1) v (∆v)
t
− − v − ∇ ∇
2 v v 2 (m 1)λ v 2 v 4 2
+ ∇ ∇|∇ | + − |∇ | |∇ | + Ric( v, v),(3.4)
v v − v2 v ∇ ∇
Local Aronson-B´enolan type gradient estimates 5
and
∆ vt = ∆(vt)+2 i,jRijfij 2∇v∇vt vt∆v + 2|∇v|2vt, (3.5)
v v − v2 − v2 v3
P
(cid:16) (cid:17)
where we use the fact that
n
(∆v) =∆(v )+2 R v , (3.6)
t t ij ij
i,j
X
( v 2) =2 v (v )+2Ric( v,v). (3.7)
t t
|∇ | ∇ ∇ ∇
Combining (3.4) and (3.5), we have
v v 2∆v 2 v v 2
t = (m 1)|∇ | +2(m 1) v (∆v)+ ∇ ∇|∇ |
L v − v − ∇ ∇ v
(cid:16) (cid:17) v 4 v v v ∆v v 2v
|∇ | +2(m 1)∇ ∇ t +(m 1) t 2(m 1)|∇ | t
− v2 − v − v − − v2
2
+2(m 1) R v + Ric( v, v). (3.8)
ij ij
− v ∇ ∇
i,j
X
Since
v =(m 1) v∆v+(m 1)v (∆v)+ v 2,
t
∇ − ∇ − ∇ ∇|∇ |
then
2 v v 2 v 2∆v v 4
2(m 1) v (∆v)+ ∇ ∇|∇ | +(m 1)|∇ | |∇ |
− ∇ ∇ v − v − v2
2 v 2v
t
= v v |∇ | .
v∇ ∇ t− v v
Substituting above identity and the following identity
v v v 2v v
2(m 1)∇ ∇ t 2(m 1)|∇ | t = 2(m 1)v t logv
− v − − v2 − ∇ v ∇
(cid:16) (cid:17)
into (3.8), we have
v 2 v 2v v
t = v v |∇ | t +2(m 1)v t logv
t
L v v∇ ∇ − v v − ∇ v ∇
(cid:16) (cid:17) v ∆v (cid:16) (cid:17) 2
+(m 1) t ++2(m 1) R v + Ric( v, v). (3.9)
ij ij
− v − v ∇ ∇
i,j
X
On the other hand, similar calculations show
v 2 2 v 2Ric( v, v)
∂ |∇ | = ∇ (m 1)v∆v+ v 2 + ∇ ∇
t
v v ∇ − |∇ | v
(cid:18) (cid:19)
h i
v 2 v 2
|∇ | (m 1)∆v+ |∇ | +
− v − v
(cid:20) (cid:21)
v 2 2
= 2(m 1)∆v|∇ | +2(m 1) v (∆v)+ v v 2
− v − ∇ ∇ v∇ ∇|∇ |
v 2 v 4 2Ric( v, v)
(m 1)∆v|∇ | |∇ | + ∇ ∇ , (3.10)
− − v − v2 v
6 W.WANG,H.ZHOU,D.XIE
where we utilize the formula (3.7) in (3.10).
By utilize Bochner’s formula, we have
v 2 2v2 2 v∆( v) v v 2 v 2 2 v 4
∆ |∇ | = ij + ∇ ∇ 2∇ ∇|∇ | ∆v|∇ | + |∇ |
v v v − v2 − v2 v3
(cid:18) (cid:19)
2v2 v 2 2 v 2
= ij +2R |∇ | + v (∆v) ∆v|∇ |
v ij v v∇ ∇ − v2
v 2
2 |∇ | (logv). (3.11)
− ∇ v ∇
(cid:18) (cid:19)
From (3.10) and (3.11), we obtain
v 2 v 2 2
|∇ | = 2(m 1)∆v|∇ | + v v 2 2(m 1)R v 2
ij
L v − v v∇ ∇|∇ | − − |∇ |
(cid:18) (cid:19)
v 4 v 2
2(m 1)v2 |∇ | +2(m 1)v |∇ | (logv)
− − ij − v2 − ∇ v ∇
(cid:18) (cid:19)
2Ric( v, v)
+ ∇ ∇ . (3.12)
v
By utilize (3.9) and (3.12), we have
v 2 v v
(F)= |∇ | α t α′ t α′ϕ αϕ′
L L v − L v − v − −
(cid:18) (cid:19) (cid:16) (cid:17)
v 2 2
= 2(m 1)∆v|∇ | + v v 2 2(m 1)(v2 +αR v )
− v v∇ ∇|∇ | − − ij ij ij
v 4 v 2
2(m 1)R v 2 |∇ | +2(m 1)v |∇ | (logv)
− − ij|∇ | − v2 − ∇ v ∇
(cid:18) (cid:19)
2 v 2v v
α v v +α|∇ | t 2α(m 1)v t (logv)
t
− v∇ ∇ v v − − ∇ v ∇
v 2(α 1) (cid:16) (cid:17)
α(m 1)∆v t − Ric( v, v) α′ϕ αϕ′. (3.13)
− − v − v ∇ ∇ − −
It is not difficult to calculate that
v 2 v
2(m 1)v |∇ | (logv) 2α(m 1) t (logv)
− ∇ v ∇ − − ∇ v ∇
(cid:18) (cid:19) (cid:16) (cid:17)
v 2 v
= 2(m 1) v |∇ | α t
− ∇ ∇ v − v
(cid:20) (cid:21)
= 2(m 1) v F, (3.14)
− ∇ ∇
and
2 2 2
v v 2 α v v = v ( v 2 αv )
t t
v∇ ∇|∇ | − v∇ ∇ v∇ ∇ |∇ | −
2
= v (Fv)
v∇ ∇
v 2
= 2 v F +2F|∇ | . (3.15)
∇ ∇ v
We deduce from (3.14) and (3.15) that
v 2 v
2(m 1)v |∇ | (logv) 2α(m 1) t (logv)
− ∇ v ∇ − − ∇ v ∇
(cid:18) (cid:19) (cid:16) (cid:17)
Local Aronson-B´enolan type gradient estimates 7
2 2
+ v v 2 α v v
t
v∇ ∇|∇ | − v∇ ∇
v 2
= 2m v F +2F|∇ |
∇ ∇ v
v 2 v v 2
t
= 2m v F +2 |∇ | α |∇ | , (3.16)
∇ ∇ v − v v
(cid:18) (cid:19)
and
v 2 v 4 v v v 2
2(m 1)∆v|∇ | |∇ | α(m 1)∆v t +α t|∇ |
− v − v2 − − v v v
v 2 v v 2 v 4 v v v 2 v v 2
t t t t
= 2|∇ | |∇ | |∇ | α |∇ | +α |∇ |
v v − v − v2 − v v − v v v
(cid:18) (cid:19) (cid:18) (cid:19)
v v 2 v 4 v 2
= (2α+2) t|∇ | 3|∇ | α t . (3.17)
v v − v2 − v
(cid:16) (cid:17)
From (3.16) and (3.17), we have
v 2 v 2
2(m 1)v |∇ | (logv) 2α(m 1) t (logv)+ v v 2
− ∇ v ∇ − − ∇ v ∇ v∇ ∇|∇ |
(cid:18) (cid:19) (cid:16) (cid:17)
2 v 2 v 4 v v v 2
α v v +2(m 1)∆v|∇ | |∇ | α(m 1)∆v t +α t|∇ |
− v∇ ∇ t − v − v2 − − v v v
v v 2 2 v 2
= 2m v F t |∇ | +(1 α) t
∇ ∇ − v − v − v
(cid:18) (cid:19) (cid:16) (cid:17)
2m v F [(m 1)∆v]2 for α>1. (3.18)
≤ ∇ ∇ − −
Substituting (3.18) into (3.13), we arrive at
(F) 2(m 1)(v2 +αR v ) 2(m 1)R v 2
L ≤− − ij ij ij − − ij|∇ |
v 2
+2m v F [(m 1)∆v]2 2(α 1)R |∇ | α′ϕ αϕ′. (3.19)
ij
∇ ∇ − − − − v − −
Further, applying Young’s inequality
α 1
R v R2 + v2
| ij|| ij|≤ 2 ij 2α ij
to (3.19), we conclude We complete the proof of Lemma 3.1. (cid:3)
Lemma 3.2. Suppose that (Mn,g(t)) satisfies the hypotheses of Theorem
t∈[0,T]
2.1. We also assume that α(t)>1 and ϕ(t)>0 satisfy the following system
2ϕ 2ϕ 1
2(m 1)MK ( α′) ,
n(m 1) − − ≥ n(m 1) − α
 − −
 n(m2ϕ−1) −α′ >0, (3.20)
ϕ2
+αϕ′ 0,
and α(t) is non-decreasing. Then n(m−1) ≥
n ϕ 2 2ϕ 1
F (m 1) v + δ α′ F
ij ij
L ≤− − n(m 1) − n(m 1) − α
i,j (cid:20) − (cid:21) (cid:20) − (cid:21)
X
8 W.WANG,H.ZHOU,D.XIE
v 2
+(m 1)α2K2+2(α 1)K|∇ | +2m v F [(m 1)∆v]2. (3.21)
− − v ∇ ∇ − −
Proof. By utilizing the unit matrix (δ ) and (3.3), we obtain
ij n×n
n ϕ ϕ2 2ϕ
(F) (m 1) v2 + δ2 + + ∆v
L ≤ − − ij n(m 1) ij n(m 1) n
Xi,j h − i −
v 2
+(m 1)α2K2+2(m 1)KM|∇ | +2m v F
− − v ∇ ∇
v 2 v
[(m 1)∆v]2+2(α 1)K|∇ | α′ t α′ϕ αϕ′.
− − − v − v − −
Applying (3.2) to above inequality, we have
n ϕ 2ϕ v 2
(F) (m 1) v2 + δ2 2(m 1)MK |∇ |
L ≤ − − ij n(m 1) ij − n(m 1) − − v
Xi,j h − i h − i
2ϕ v 2ϕ αϕ ϕ2
+ α′ t + α′ +
n(m 1) − v n(m 1) − α n(m 1)
h − i h − i −
2ϕ αϕ v 2
α′ +(m 1)α2K2+2(m 1)KM|∇ |
− n(m 1) − α − − v
h − i
v 2
+2m v F [(m 1)∆v]2+2(α 1)K|∇ | α′ϕ αϕ′.
∇ ∇ − − − v − −
Again using (3.20), we follows (3.21). (cid:3)
Lemma 3.3. Let G=γ(t)F. Then
1 γ′ 2ϕ 1
G G2+ α′ G
L ≤ −aα2γ γ − n(m 1) − α
(cid:20) (cid:18) − (cid:19) (cid:21)
2(α 1) v 2 γ(m 1)(α 1)2 v 4
− |∇ | G − − |∇ |
− nα2 v − nα2 v2
v 2
+(m 1)α2γK2+2γ(α 1)K|∇ | +2m v G, (3.22)
− − v ∇ ∇
where a= n(m−1) .
n(m−1)+1
Proof. Sample calculation gives
G = γ F +γ′F
L L
ϕ 2ϕ 1 γ′
(m 1)γ v2 + δ2 + α′ + G
≤ − − ij n(m 1) ij − n(m 1) − α γ
(cid:20) − (cid:21) (cid:20) (cid:18) − (cid:19) (cid:21)
v 2
+(m 1)α2γK2+2γ(α 1)K|∇ | +2m v G
− − v ∇ ∇
γ[(m 1)∆v]2. (3.23)
− −
Since
ϕ 1
v2 + δ2 (∆v+ϕ)
ij n(m 1) ij ≥ n
(cid:20) − (cid:21)
1 v 2 2
= F +(m 1)(α 1)|∇ | , (3.24)
nα2(m 1)2 − − v
− (cid:20) (cid:21)
Local Aronson-B´enolan type gradient estimates 9
and
F α 1 v 2
(m 1)∆v = − |∇ | ϕ
− −α − α v −
F
. (3.25)
≤ −α
Therefore, we follow that from (3.23), (3.24) and (3.25)
γ v 2 2
G F +(m 1)(α 1)|∇ |
L ≤ −nα2(m 1) − − v
− (cid:20) (cid:21)
2ϕ 1 γ′
+ α′ + G+(m 1)α2γK2
− n(m 1) − α γ −
(cid:20) (cid:18) − (cid:19) (cid:21)
v 2 G2
+2γ(α 1)K|∇ | +2m v G . (3.26)
− v ∇ ∇ − α2γ
From (3.26), we infer (3.22). The proof is complete. (cid:3)
4. Proof of Main Results
In this section, we will prove our main results.
Proof of Theorem 2.1. Now let ϕ(r) be a C2 function on [0, ) such that
∞
1 if r [0,1],
ϕ(r)= ∈
(0 if r [2, ),
∈ ∞
and
ϕ′(r)
0 ϕ(r) 1, ϕ′(r) 0, ϕ′′(r) 0, | | C,
≤ ≤ ≤ ≤ ϕ(r) ≤
where C is an absolute constant. Let define by
d(x,x ,t) ρ(x,t)
φ(x,t)=ϕ(d(x,x ,t))=ϕ 0 =ϕ ,
0 R R
(cid:18) (cid:19) (cid:18) (cid:19)
whereρ(x,t)=d(x,x ,t). Byusingthemaximumprinciple,theargumentofCalabi
0
[2] allows us to suppose that the function φ(x,t) with support in B , is C2 at
2R,T
the maximum point. By utilize the Laplacian theorem, we deduce that
φ2 C
|∇ | , (4.1)
φ ≤ R2
C
∆φ (1+√KR), (4.2)
− ≤ R2
For any 0 T T, let H = φG and (x ,t ) be the point in B at which
≤ 1 ≤ 1 1 2R,T1
G attain its maximum value. We can suppose that the value is positive, because
otherwise the proof is trivial. Then at the point (x ,t ), we infer
1 1
G
(H) 0, G= φ. (4.3)
L ≥ ∇ −φ∇
BytheevolutionformulaofthegeodesiclengthundertheRicciflow[4],wecalculate
ρ 1 dρ ρ
φ G= Gφ′ =Gφ′ Ric(S,S)ds
t
− R R dt R
(cid:16) (cid:17) (cid:16) (cid:17)Zγt1
ρ 1 ρ
Gφ′ K ρ Gφ′ K G√CK ,
≤ R R 2 ≤ R 2 ≤ 2
(cid:16) (cid:17) (cid:16) (cid:17)
10 W.WANG,H.ZHOU,D.XIE
whereγ isthegeodesicconnectingxandx underthemetricg(t ), S isthe unite
t1 0 1
tangent vector to γ , and ds is the element of the arc length. Hence, by applying
t1
(4.2), we have
φ2
0 (H) φ G (m 1)G ∆φ 2|∇ | +φ G
t
≤ L ≤ L − − − φ
(cid:18) (cid:19)
1 γ′ 2ϕ 1
φG2+ α′ φG
≤ −aα2γ γ − n(m 1) − α
(cid:20) (cid:18) − (cid:19) (cid:21)
2(α 1) v 2 γ(m 1)(α 1)2 v 4
− |∇ | φG − − |∇ | φ
− nα2 v − nα2 v2
v 2
+(m 1)α2γφK2++2γφ(α 1)K|∇ | +2mφ v G
− − v ∇ ∇
φ2
(m 1)G ∆φ 2|∇ | +√CKG (4.4)
− − − φ
(cid:18) (cid:19)
Multiply φ, we have
1 γ′ 2ϕ φ
0 φ2G2+ φ α′ φG
≤ −aα2γ γ − n(m 1) − α
(cid:20) (cid:18) − (cid:19) (cid:21)
2(α 1) v 2 γ(m 1)(α 1)2 v 4
− |∇ | φ2G − − |∇ | φ2
− nα2 v − nα2 v2
v 2
+(m 1)α2γφ2K2+2γφ2(α 1)K|∇ |
− − v
φ φ2
2mφ2∇ G v (m 1)φG ∆φ 2|∇ | +√CKφG
− φ ∇ − − − φ
(cid:18) (cid:19)
Further using the inequality Ax2+Bx B2 with A>0, we have
≥−4A
2(α 1) v 2 φ nm2α2 φ2
− |∇ | φ2G 2mφ2∇ G v |∇ | φG,
− nα2 v − φ ∇ ≤ 2(α 1) φ
−
γ(m 1)(α 1)2 v 4 v 2 nα2K2
− − |∇ | φ2++2γφ2(α 1)K|∇ | φ2γ.
− nα2 v2 − v ≤ m 1
−
Hence, we deduce that
1 γ′ 2ϕ φ nm2α2 φ2
0 φ2G2+ φ α′ + |∇ |
≤ −aα2γ γ − n(m 1) − α 2(α 1) φ
(cid:20) (cid:18) − (cid:19) −
φ2
+(m 1) ∆φ 2|∇ | +√CK φG
− − φ
(cid:18) (cid:19) (cid:21)
nα2K2
+(m 1)α2K2φ2γ+ φ2γ. (4.5)
− m 1
−
Combine (4.1), (4.2) and (4.5), we have
1 γ′ 2ϕ φ Cm2α2
0 φ2G2+ φ α′ +
≤ −aα2γ γ − n(m 1) − α R2(α 1)
C(m 1) h (cid:0) − (cid:1) −
+ − (1+√KR)+√CK φG
R2
nα2K2 i
+(m 1)α2K2φ2γ+ φ2γ.
− m 1
−