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PacificJournal ofMathematics
GRADIENT ESTIMATES FOR POSITIVE SOLUTIONS OF THEHEAT EQUATION UNDER GEOMETRIC FLOW
JUN SUN
Volume 253 No. 2 October 2011
PACIFIC JOURNAL OF MATHEMATICSVol. 253, No. 2, 2011
GRADIENT ESTIMATES FOR POSITIVE SOLUTIONS OF THEHEAT EQUATION UNDER GEOMETRIC FLOW
JUN SUN
We establish first- and second-order gradient estimates for positive solu-tions of the heat equations under general geometric flows. Our results gen-eralize the recent work of S. Liu, who established similar results for theRicci flow. Both results can also be considered as the generalization of P. Li,S. T. Yau, and J. Li’s gradient estimates under geometric flow setting. Wealso give an application to the mean curvature flow.
1. Introduction
Starting with the pioneering work of P. Li and S. T. Yau [1986], gradient esti-mates are also called differential Harnack inequalities, because we can obtain theclassical Harnack inequality after integrating along the space-time curve. Theyare very powerful tools in geometric analysis. For example, R. Hamilton [1993;1995b] established differential Harnack inequalities for the scalar curvature alongthe Ricci flow and for the mean curvature along the mean curvature flow. Bothhave important applications in the singularity analysis.
In Perelman’s breakthrough work [2002] on the Poincaré conjecture and thegeometrization conjecture, an important role was played by a differential Harnackinequality. Since then, there have been many works on gradient estimates along theRicci flow or the conjugate Ricci flow for the solution of the heat equation or theconjugate heat equation; examples include [Cao 2008; Cao and Hamilton 2009;Kuang and Zhang 2008; Zhang 2006].
Under some curvature constraints, Guenther [2002] has established gradient es-timates for positive solutions of the heat equation under general geometric flowon a closed manifold. Using this result, she derived a Harnack-type inequalityand found a lower bound for the heat kernel under Ricci flow. As mentioned in[Liu 2009] (see also Section 4 of this paper), we can weaken the assumption ofGuenther’s results by removing the bound on the gradient of scalar curvature whenrestricting to the Ricci flow case. We can also obtain a local gradient estimate forcomplete case.
MSC2000: primary 53C44, 58J35; secondary 53C21, 35K55.Keywords: gradient estimate, geometric flow, heat equation, Harnack inequality.
489
490 JUN SUN
While most of the works deal with the first-order case, higher-order gradient es-timates have their own interest. Indeed, they are closely related to the boundednessof the Riesz transform and the Sobolev inequality. J. Li [1991] obtained second-order gradient estimates for heat kernels on complete noncompact Riemannianmanifolds. In [Li 1994], he used the boundedness of Riesz transform to prove theSobolev inequality on Riemannian manifolds with some constraints.
S. Liu [2009] obtained the first and the second order gradient estimates for posi-tive solutions of the heat equations under Ricci flow. His work generalized [Li andYau 1986] and [Li 1991].
In this paper, we generalize Liu’s work to general geometric flow. Of course,we need impose stronger conditions on the flow and the curvature. Compared togeneral geometric flow, there are two advantages to the Ricci flow: the contractedsecond Bianchi identity, which gives a nice expression for the commuting formula(Section 4), and the fact that the Ricci curvature arises when we use the Bochnerformula to compute the Laplacian of |∇u|2. Sometimes, the Ricci curvature willbe canceled with the time derivative of the metric under the Ricci flow.
K. Ecker, D. Knopf, L. Ni and P. Topping [Ecker et al. 2008] recently obtained alocal gradient estimate for bounded positive solution of the conjugate heat equationfor general geometric flow.
Our paper is organized as follows: We prove first-order gradient estimates inSection 2 and second-order gradient estimates in Section 3. We give two applica-tions to the Ricci flow and the mean curvature flow in Section 4.
2. First-order gradient estimates
For a function f on M ×[0, T ], where T is a positive constant, we write
ft = ∂t f =∂ f (x, t)∂t
.
Theorem 1 (gradient estimate: local version). Let (M, g(t)) be a smooth one-parameter family of complete Riemannian manifolds evolving by
(2-1)∂
∂tg = 2h,
for t in some time interval [0, T ]. Let M be complete under the initial metric g(0).Given x0 ∈ M and R > 0, let u be a positive solution to the equation
(1− ∂t)u(x, t)= 0
in the cube Q2R,T := {(x, t) | d(x, x0, t) ≤ 2R, 0 ≤ t ≤ T }. Suppose that thereexist constants K1, K2, K3, K4 ≥ 0 such that
Ric≥−K1g, −K2g ≤ h ≤ K3g, |∇h| ≤ K4,
GRADIENT ESTIMATES UNDER GEOMETRIC FLOW 491
on Q2R,T . Then for (x, t) ∈ Q R,T , we have
(2-2)|∇u(x, t)|2
u2(x, t)−α
ut(x, t)u(x, t)
≤ C(
K1+ K2+ K3+ K4+√
K4+1t+
1R2
)for any α > 1, where C depends on n, α only.
More explicitly, we have
(2-3)|∇u(x, t)|2
u2(x, t)−α
ut(x, t)u(x, t)
≤nα2
t+
Cα2
R2
(R√
K1+α2
α−1
)+Cα2K2
+nα2
α−1(K1+ (α−1)K3+ K4)+ nα2(K2+ K3+
√2K4),
for any α > 1, where C depends only on n.
Remark 2. When h = −Ric, (2-1) is the Ricci flow equation. In this case ourresults reduce to [Liu 2009]. Note that for Ricci flow the assumption |∇ Ric | ≤ K4
is not needed because of the contracted second Bianchi identity (see Section 4).
As in [Li and Yau 1986], let f = log u; then
(2-4) (1− ∂t) f =−|∇ f |2.
Set
F = t(|∇u(x, t)|2
u2(x, t)−α
ut(x, t)u(x, t)
)= t (|∇ f |2−α ft).
To prepare the ground for the proof of the theorem we need some lemmas.
Lemma 3. Suppose the metric evolves by (2-1). Then, for any smooth function f ,we have
∂
∂t|∇ f |2 =−2h(∇ f,∇ f )+ 2〈∇ f,∇ ft 〉
and
(2-5)∂
∂t1 f =1
∂
∂tf − 2〈h,∇2 f 〉− 2〈div h− 1
2∇(trg h),∇ f 〉.
Here, div h is the divergence of h.
Proof. To prove the first equation, write
∂
∂t|∇ f |2 =
∂
∂t
(gi j ∂ f∂xi
∂ f∂x j
)=−2h(∇ f,∇ f )+ 2〈∇ f,∇ ft 〉.
For the second, recall that
∂
∂t0k
i j = gkl{∇i h jl +∇ j hil −∇lhi j }.
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Thus,
∂
∂t1 f =
∂
∂t
{gi j(∂2 f∂xi∂x j
−0ki j∂ f∂xk
)}=−2hi j
(∂2 f∂xi∂x j
−0ki j∂ f∂xk
)+1
∂ f∂t− gi j
(∂
∂t0k
i j
)∂ f∂xk
=−2〈h,∇2 f 〉+1∂ f∂t− gi j gkl
{∇i h jl +∇ j hil −∇lhi j }∇k f
=1∂ f∂t− 2〈h,∇2 f 〉− 2gkl
{gi j∇i h jl −
12∇l(trg h)}∇k f
=1∂
∂tf − 2〈h,∇2 f 〉− 2〈div h−
12∇(trg h),∇ f 〉. �
Lemma 4. Suppose (M, g(t)) satisfies the hypotheses of Theorem 1. We have
(1− ∂t)F ≥−2〈∇ f,∇F〉+ tn(|∇ f |2− ft)
2− (|∇ f |2−α ft)
− 2(K1+ (α−1)K3)t |∇ f |2− 3√
nαK4t |∇ f | −α2nt (K2+ K3)2.
Proof. For a given time t , choose {x1, x2, . . . , xn} to be a normal coordinate sys-tem at a fixed point. Subscripts i, j will denote covariant derivatives in the xi , x j
directions. We will compute at the fixed point.Using the Bochner formula, (2-4) and Lemma 3, we calculate
1F = t(2|∇2 f |2+ 2 Ric(∇ f,∇ f )+ 2〈∇ f,∇1 f 〉−α1( ft)
)= t
(2|∇2 f |2+ 2 Ric(∇ f,∇ f )+ 2〈∇ f,∇( ft − |∇ f |2)〉
)−αt
((1 f )t + 2〈h,∇2 f 〉+ 2〈div h− 1
2∇(trg h),∇ f 〉)
=−2〈∇ f,∇F〉+ 2t (|∇2 f |2+Ric(∇ f,∇ f )+ (1−α)h(∇ f,∇ f )
−α〈h,∇2 f 〉−α〈div h− 12∇(trg h),∇ f 〉)+ t (|∇ f |2)t −αt ft t .
On the other hand, we have
Ft = (|∇ f |2−α ft)+ t((|∇ f |2)t −α ft t
).
Therefore, we arrive at
(1− ∂t)F =−2〈∇ f,∇F〉+ 2t (|∇2 f |2+Ric(∇ f,∇ f )+ (1−α)h(∇ f,∇ f )
−α〈h,∇2 f 〉−α〈div h− 12∇(trg h),∇ f 〉)− (|∇ f |2−α ft).
By our assumption, we have
−(K2+ K3)g ≤ h ≤ (K2+ K3),
GRADIENT ESTIMATES UNDER GEOMETRIC FLOW 493
which implies that
|h|2 ≤ (K2+ K3)2|g|2 = n(K2+ K3)
2.
Applying those bounds and Young’s inequality yields
|α〈h,∇2 f 〉| ≤ 12 |∇
2 f |2+ 12α
2|h|2 ≤ 1
2 |∇2 f |2+ 1
2α2n(K2+ K3)
2.
On the other hand,
|div h− 12∇(trg h)| = |gi j
∇i h jl −12 gi j∇lhi j | ≤
32 |g||∇h| ≤ 3
2
√nK4.
We conclude by our assumptions that
(1− ∂t)F ≥−2〈∇ f,∇F〉+ t |∇2 f |2− (|∇ f |2−α ft)
− 2(K1+ (α−1)K3)t |∇ f |2− 3√
nαK4t |∇ f | −α2nt (K2+ K3)2.
Finally, with the help of the inequality
|∇2 f |2 ≥ 1
n(tr∇2 f )2 = 1
n(1 f )2 = 1
n(|∇ f |2− ft)
2,
we complete the proof of the lemma. �
Proof of Theorem 1. By our assumption of the bounds of h and the evolution of themetric, we know that g(t) is uniformly equivalent to the initial metric g(0), that is,
e−2K2T g(0)≤ g(t)≤ e2K3T g(0).
Thus we know that (M, g(t)) is also complete for t ∈ [0, T ].Now let ψ(r) be a C2 function on [0,+∞) such that
ψ(r)={
1 if r ∈ [0, 1],0 if r ∈ [2,+∞),
(2-6)
0≤ ψ(r)≤ 1, ψ ′(r)≤ 0, ψ ′′(r)≥−C,|ψ ′(r)|2
ψ(r)≤ C,(2-7)
where C is an absolute constant. Define
ϕ(x, t)= ϕ(d(x, x0, t))= ψ(
d(x, x0, t)R
)= ψ
(ρ(x, t)
R
),
where ρ(x, t) = d(x, x0, t). For the purpose of applying the maximum principle,the argument of [Calabi 1958] allows us to assume that the function ϕ(x, t), withsupport in Q2R,T , is C2 at the maximum point.
For any 0 < T1 ≤ T , let (x1, t1) be the point in Q2R,T1 , at which ϕF achievesits maximum value. We can assume that this value is positive, because in the other
494 JUN SUN
case the proof is trivial. As F(x, 0) = 0, we know that t1 > 0. Then at the point(x1, t1), we have
(2-8) ∇(ϕF)= F∇ϕ+ϕ∇F = 0, 1(ϕF)≤ 0,∂
∂t(ϕF)≥ 0.
Therefore,
(2-9) 0≥ (1− ∂t)(ϕF)= (1ϕ)F −ϕt F +ϕ(1− ∂t)F + 2∇ϕ · ∇F.
Using the Laplacian comparison theorem, we have
1ϕ = ψ ′1ρ
R+ψ ′′
|∇ρ|2
R2 ≥−CR2 −
CR
√K1.
Furthermore, we have
|∇ϕ|2
ϕ=(ψ ′)2
ψ
|∇ρ|2
R2 ≤CR2 .
By our assumption, F(x1, t1) > 0. By the evolution formula of the geodesic lengthunder geometric flow [Hamilton 1995a], we calculate at the point (x1, t1)
−ϕt F =−ψ ′(ρ
R
) 1R
dρdt
F =−ψ ′(ρ
R
) 1R
∫γt1
h(S, S) ds F
≥ ψ ′(ρ
R
) 1R
K2ρ F ≥−√
C K2 F,
where γt1 is the geodesic connecting x and x0 under the metric g(t1), S is the unitetangent vector to γt1 and ds is the element of arc length. Substituting the threeinequalities above into (2-9) and using (2-8), we obtain
0≥(−
CR2 −
CR
√K1
)F −√
C K2 F +ϕ(1− ∂t)F.
Applying Lemma 4 and Young’s inequality to this inequality yields
(2-10) 0≥(−
CR2 −
CR
√K1
)F−√
C K2 F−2
√C
R√ϕ|∇ f |F+
t1nϕ(|∇ f |2− ft)
2
−ϕ(|∇ f |2−α ft)− 2(K1+ (α−1)K3+ K4)ϕt1|∇ f |2
−α2nt1ϕ[(K2+ K3)2+ 2K4].
Multiplying through by ϕt1 and setting y=ϕ|∇ f |2 and z=ϕ ft , (2-10) becomes
(2-11) 0≥ t1
(−
CR2 −
CR
√K1
)(ϕF)−
√C K2t1(ϕF)− 2
√C
Rt21 y1/2(y−αz)
+t21
n(y−z)2−ϕ2 F−2(K1+(α−1)K3+K4)t2
1 y−α2nt21ϕ
2[(K2+K3)
2+2K4].
GRADIENT ESTIMATES UNDER GEOMETRIC FLOW 495
Using the inequality ax2− bx ≥−b2/(4a), valid for a, b > 0, one obtains
t21
n(y− z)2− 2
√C
Rt21 y1/2(y−αz)− 2(K1+ (α−1)K3+ K4)t2
1 y
=t21
n
[1α2 (y−αz)2+
(α−1α
)2y2− 2n(K1+ (α−1)K3+ K4)y
+
(2α−1α2 y−
2n√
CR
y1/2)(y−αz)
]≥
t21
n
[1α2 (y−αz)2−
α2n2(K1+ (α−1)K3+ K4)2
(α−1)2−
α2n2C2(α−1)R2 (y−αz)
].
Hence (2-11) becomes
1nα2 (ϕF)2− (ϕF)
(1+
CR2 t1+
CR
√K1t1+
Cnα2t12(α−1)R2 +
√C K2t1
)−
(n(K1+ (α−1)K3+ K4)
2α2t21
(α−1)2+ t2
1α2nϕ2[(K2+ K3)
2+ 2K4]
)≤ 0.
We apply the quadratic formula and then arrive at
ϕF(x1, t1)≤ nα2+
Cnα2
R2
(R√
K1+α2
α−1
)t1+√
Cnα2K2t1
+nα2t1α−1
(K1+ (α−1)K3+ K4)+ n(K2+ K3+√
2K4)α2t1.
If d(x, x0, T1) < R, we have ϕ(x, T1)= 1. Then
F(x, T1)= T1(|∇ f |2−α ft)≤ ϕF(x1, t1)
≤ nα2+
Cnα2
R2
(R√
K1+α2
α−1
)T1+√
Cnα2K2T1
+nα2T1
α−1(K1+ (α−1)K3+ K4)+ n(K2+ K3+
√2K4)α
2T1.
As T1 is arbitrary, we obtain the result. �
From the local result above, we get a global one:
Corollary 5. Let (M, g(0)) be a complete noncompact Riemannian manifold with-out boundary, and let g(t) evolves by (2-1) for t ∈ [0, T ] and satisfy
Ric≥−K1g, −K2g ≤ h ≤ K3g, |∇h| ≤ K4.
If u is a positive solution to the equation (1 − ∂t)u(x, t) = 0, then for (x, t) ∈M × (0, T ], we have
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(2-12)|∇u(x, t)|2
u2(x, t)−α
ut(x, t)u(x, t)
≤nα2
t+C
(K1+ K2+ K3+ K4+
√K4),
for any α > 1, where C depends only on n, α.
Proof. By the uniform equivalence of g(t), we know that (M, g(t)) is completenoncompact for t ∈ [0, T ]. Letting R→+∞ in (2-3) completes the proof. �
Using Lemma 3, we can also derive a similar gradient estimate on a closedRiemannian manifold.
Theorem 6. Let (M, g(t)) be a closed Riemannian manifold, where g(t) evolvesby (2-1) for t ∈ [0, T ] and satisfies
Ric≥−K1g, −K2g ≤ h ≤ K3g, |∇h| ≤ K4.
If u is a positive solution to the equation (1 − ∂t)u(x, t) = 0, then for (x, t) ∈M × (0, T ], we have
(2-13)|∇u(x, t)|2
u2(x, t)−α
ut(x, t)u(x, t)
≤nα2
t+
nα2
α−1(K1+ (α−1)K3+ K4)+ nα2(K2+ K3+
√2K4),
for any α > 1, where C depends only on n, α.
Proof. We use the same symbols F, f as above. Set
F̄(x, t)= F(x, t)−nα2
α−1(K1+ (α−1)K3+ K4) t − nα2(K2+ K3+
√2K4)t.
If F̄(x, t)≤ nα2 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 F̄(x, t), we have
F̄(x0, t0) > nα2.
Since F̄(x, 0)=0, we know that t0>0 here. Then applying the maximum principle,we have at the point (x0, t0),
∇ F̄(x0, t0)= 0, 1F̄(x0, t0)≤ 0,∂
∂tF̄(x0, t0)≥ 0.
Therefore we obtain0≥ (1− ∂t)F̄ ≥ (1− ∂t)F.
Using Lemma 4 and the fact that
(|∇ f |2− ft)2=
( 1α(|∇ f |2−α ft)+
α−1α|∇ f |2
)2
=1α2
(Ft0
)2+ 2α−1
α2 |∇ f |2(F
t0
)+(α−1)2
α2 |∇ f |4,
GRADIENT ESTIMATES UNDER GEOMETRIC FLOW 497
we get that
0≥t0
nα2
(Ft0
)2−
(Ft0
)−
nα2
(α−1)2(K1+ (α−1)K3+ K4)
2t0
− nα2((K2+ K3)2+ 2K4
)t0+
2t0nα−1α2 |∇ f |2
Ft0.
Since
Ft0=
F̄t0+
nα2
α−1(K1+ (α−1)K3+ K4)+ nα2(K2+ K3+
√2K4) > 0,
we get
t0nα2
(Ft0
)−
Ft0−
nα2
(α−1)2(K1+(α−1)K3+K4)
2t0−nα2((K2+K3)2+2K4
)t0≤ 0.
Solving this quadratic inequality yields
Ft0≤
nα2
t0+
nα2
α−1(K1+ (α−1)K3+ K4)+ nα2(K2+ K3+
√2K4).
This implies that F̄(x0, t0)≤ nα2, in contradiction with our assumption. So (2-13)holds. �
Remark 7. In Corollary 5 and Theorem 9, if K1 = K4 = 0, we can let α→ 1.
Integrating the gradient estimate in space-time as in [Li and Yau 1986] or [Guen-ther 2002], we can derive the following Harnack-type inequality.
Corollary 8. Let (M, g(0)) be a complete noncompact Riemannian manifold with-out boundary or a closed Riemannian manifold. Assume g(t) evolves by (2-1) fort ∈ [0, T ] and satisfies
Ric≥−K1g, −K2g ≤ h ≤ K3g, |∇h| ≤ K4.
If u is a positive solution to the equation (1− ∂t)u(x, t) = 0, then for any pair ofpoints (x, t1), (y, t2) in M × (0, T ] such that t1 < t2 we have
u(x, t1)≤ u(y, t2)(
t2t1
)2nε
exp(
ε3
2(t2− t1)+C
t2− t12ε
K), for any ε > 1
2 ,
where K = K1+ K2+ K3+ K4+√
K4, the constant C depends only on n and ε,and
3= infγ
∫ 1
0|γ ′(s)|2σ(s) ds
is the infimum over smooth curves γ joining y to x (γ (0) = y, γ (1) = x) of theaveraged square velocity of γ measured at time σ(s)= (1− s)t2+ st1.
498 JUN SUN
Proof. The gradient estimates in Corollary 8 and Theorem 9 can both be written as
|∇u(x, t)|2
u2(x, t)−α
ut(x, t)u(x, t)
≤nα2
t+Cn,αK .
Fix ε > 12 , take any curve γ satisfying the assumption and set
l(s)= ln u(γ (s),σ (s)).
Then l(0)= ln u(y, t2) and l(1)= ln u(x, t1). Direct calculation shows that
∂l(s)∂s= (t2− t1)
(∇uuγ ′(s)t2− t1
−ut
u
)≤ε|γ ′(s)|2σ2(t2− t1)
+t2− t1
2ε
(C K +
4ε2nσ(s)
).
Integrating this inequality over γ (s), we have
(2-14) lnu(x, t1)u(y, t2)
=
∫ 1
0
∂l(s)∂s
ds ≤∫ 1
0
ε|γ ′(s)|2σ2(t2− t1)
ds+Ct2− t1
2εK + 2εn ln
t2t1,
which implies the corollary. �
3. Second-order gradient estimates
In this section we derive the second order gradient estimate for the positive solutionof the heat equation along a general geometric flow, which generalizes the resultsin [Li 1991; Liu 2009].
Theorem 9. Let g(t) be a solution to (2-1) on a Riemannian manifold Mn for tin some time interval [0, T ]. Assume that (M, g(0)) is a complete noncompactmanifold without boundary. Suppose that (M, g(t)) satisfies
|Rm | ≤ k1, |∇ Rm | ≤ k2, −k3g ≤ h ≤ k3g, |∇h| ≤ k4,
for some nonnegative constants k1, k2, k3, k4. Let u be a positive solution to theequation (1−∂t)u(x, t)= 0. Then, for any (x, t)∈M×(0, T ] and α > 1, we have
|∇2u(x, t)|u(x, t)
+α|∇u(x, t)|2
u2(x, t)− 5α
ut(x, t)u(x, t)
≤ C(
k1+ k2/32 + k3+ k4+
√k4+
1t
),
where C depends only on n and α.
Before proving the theorem, we need a lemma. Set
F(x, y, t)= t F1 = t(|∇
2u(x, t)|u(x, t)
+α|∇u(x, t)|2
u2(x, t)−β
ut(x, t)u(x, t)
),
where β is a constant to be fixed.
GRADIENT ESTIMATES UNDER GEOMETRIC FLOW 499
Lemma 10. Suppose (M, g(t)) satisfies the hypotheses of Theorem 9. Then forsufficiently small δ > 0, γ − 1> 0, and ε > 0, we have, with β = 5α,
(1− ∂t)F ≥−2〈∇F,∇ log u〉+δα
tF2+ 2δαβF
ut
u− 2δα2 F
|∇u|2
u2
−C(k1+ k3)F −C(k1+ k3)
2
4(γ − 1)2t − εnk2
3βt −Ft
−2(n− 1)2
δα(k1+ k3)
2t −Cβ4/3
δ1/3αk4/3
4 t −C
δ1/3α(k2+ k4)
4/3t
− 2Ct(
4δα3+
δ
2α(1−δ)2
)(1t+ k1+ k3+ k4+
√k4
)2
,
where C depends on n and α.
Proof. As in the proof of Lemma 4, choose {x1, x2, . . . , xn} to be a normal coor-dinate system at a fixed point. Subscript i, j, k will denote covariant derivatives inthe xi , x j , xk directions.
We will first calculate the evolution equation for F1 and divide it into three parts.In the calculation, we will use the following formula a few times:
(3-1) (1− ∂t)fg=
1g(1− ∂t) f − f
g2 (1− ∂t)g−2g
⟨∇
fg,∇g
⟩.
Part 1. We first calculate a parabolic inequality for |∇2u|/u.Using (3-1) and the fact that (1− ∂t)u = 0, we obtain that
(1− ∂t)
(|∇
2u|u
)=−2
⟨∇|∇
2u|u
,∇u⟩+
1u(1− ∂t)|∇
2u|.
Note that1|∇2u|2 = 2|∇2u|1|∇2u| + 2
∣∣∇|∇2u|∣∣2
and1|∇2u|2 =
∑i jk
(u2i j )kk = 2|∇3u|2+ 2
∑i jk
ui j ui jkk .
We get that
1|∇2u| =|∇
3u|2+∑
i jk ui j ui jkk −∣∣∇|∇2u|
∣∣2|∇2u|
≥
∑i jk ui j ui jkk
|∇2u|.
Here we have used the fact that |∇3u| ≥∣∣∇|∇2u|
∣∣.The Ricci identity gives
ui jkk =
ukki j +∑
l
Rk jkl,i ul +∑
l
Rki jl,kul +∑
l
Rkiklul j +∑
l
Rk jkluli + 2∑
l
Rki jlukl .
500 JUN SUN
By our assumption on the curvature, we have
1|∇2u| ≥〈∇
2u,∇2(1u)〉|∇2u|
−Ck1|∇2u| −Ck2|∇u|.
Noting that the metric evolves by (2-1), we have
∂
∂t∇i∇ j u =
∂
∂t
(∂2u∂xi∂x j
−0pi j∂u∂x p
)=∇i∇ j ut −
(∂
∂t0
pi j
)∇pu
=∇i∇ j ut − (∇i h j p +∇ j hi p −∇phi j )∇pu.
This leads to
∂t(|∇2u|2)=
∂
∂t(gik g jl
∇i∇ j u∇k∇lu)
=−4hik∇i∇ j u∇k∇ j u+ 2〈∇i∇ j ut ,∇i∇ j u〉
− 2(∇i h j p +∇ j hi p −∇phi j )∇pu∇i∇ j u,
∂t |∇2u| ≥
〈∇2u,∇2(∂t u)〉|∇2u|
−Ck3|∇2u| −Ck4|∇u|.
Combining together all of the above, we conclude that
(3-2) (1− ∂t)
(|∇
2u|u
)≥−2
⟨∇
(|∇
2u|u
),∇ log u
⟩−C(k1+ k3)
|∇2u|u−C(k2+ k4)
|∇u|u.
Part 2. We next calculate a parabolic inequality for |∇u|2/u2. Using (3-1) and thefact that (1− ∂t)u = 0, we obtain
(1− ∂t)
(|∇u|2
u2
)=−2
⟨∇
(|∇u|2
u2
),∇ log u
⟩+(1− ∂t)|∇u|2
u2 + 2|∇u|4
u4 −2u3 〈∇|∇u|2,∇u〉.
Using Bochner’s formula and Lemma 3, we obtain
(1− ∂t)|∇u|2 = 2|∇2u|2+ 2 Ric(∇u,∇u)+ 2h(∇u,∇u).
GRADIENT ESTIMATES UNDER GEOMETRIC FLOW 501
Therefore,
(1− ∂t)
(|∇u|2
u2
)≥−2〈∇
(|∇u|2
u2
),∇ log u〉+ 2
|∇2u|2
u2 + 2|∇u|4
u4
−4u3 |∇
2u| |∇u|2+2u2 (Ric+h)(∇u,∇u)
≥−2〈∇(|∇u|2
u2
),∇ log u〉+2δ
|∇2u|2
u2 −2δ
1−δ|∇u|4
u4 −2(n−1)(k1+k3)|∇u|2
u2 .
Here we have used Young’s inequality to obtain
4u3 |∇
2u| |∇u|2 ≤ 2(1−δ)|∇
2u|2
u2 +2
1−δ|∇u|4
u4 .
Thus we have
(3-3) (1− ∂t)
(α|∇u|2
u2
)≥−2
⟨∇
(α|∇u|2
u2
),∇ log u
⟩+2δα
|∇2u|2
u2 −2δα1−δ|∇u|4
u4 − 2(n− 1)(k1+ k3)α|∇u|2
u2 .
Part 3. Finally, using (3-1) and Lemma 3, we get, for any ε > 0,
(1− ∂t)(ut
u
)=(1−∂t)ut
u− 2
⟨∇
(utu
),∇ log u
⟩=
1u[(1u)t + 2〈h,∇2u〉+ 2〈div h− 1
2∇(trg h),∇u〉− ut t ] − 2⟨∇
(utu
),∇ log u
⟩=−2
⟨∇
(utu
),∇ log u
⟩+
2u〈h,∇2u〉+ 2
u〈div h− 1
2∇(trg h),∇u〉
≤ −2⟨∇
(utu
),∇ log u
⟩+ 2√
nk3|∇
2u|u+ 3√
nk4|∇u|
u
≤−2⟨∇
(utu
),∇ log u
⟩+
1ε
|∇2u|2
u2 + εnk23 + 3√
nk4|∇u|
u.
Therefore, we have
(3-4) (1− ∂t)(−β
utu
)≥−2
⟨∇
(−β
utu
),∇ log u
⟩−β
ε
|∇2u|2
u2 − εnk23β − 3
√nk4β|∇u|
u.
502 JUN SUN
Combining the results from parts 1, 2, and 3, we obtain that for any 0 < δ < 1and ε > 0,
(3-5) (1− ∂t)F1 ≥−2〈∇F1,∇ log u〉−C(k1+ k3)|∇
2u|u+
(δα−
β
ε
)|∇
2u|2
u2
+ δα|∇
2u|2
u2 −2δα1−δ|∇u|4
u4 − 2(n− 1)(k1+ k3)α|∇u|2
u2
− 3√
nk4β|∇u|
u− εnk2
3β −C(k2+ k4)|∇u|
u.
By the definition of F1, we have
(3-6)|∇
2u|u≤ F1+β
ut
u,
and
|∇2u|2
u2 =
(F1−α
|∇u|2
u2 +βut
u
)2
(3-7)
= F21 +α
2 |∇u|4
u4 +β2 u2
t
u2 + 2βF1ut
u− 2αF1
|∇u|2
u2 − 2αβ|∇u|2
u2
ut
u.
Inserting (3-6) and (3-7) into (3-5) and applying Young’s inequality, we arrive at
(3-8) (1− ∂t)F1 ≥−2〈∇F1,∇ log u〉−C(k1+ k3)F1−C(k1+ k3)
2
4(γ − 1)2
+[ 1
2δαβ2−Cβ2(γ − 1)2
] u2t
u2 +
(δα−
β
ε
)|∇
2u|2
u2
−
(4δα3+
δ
2α(1−δ)2
)|∇u|4
u4 + δαF21 + 2δαβF1
ut
u
− 2δα2 F1|∇u|2
u2 − εnk23β −
2(n− 1)2
δα(k1+ k3)
2
−Cβ4/3
δ1/3αk4/3
4 −C
δ1/3α(k2+ k4)
4/3,
for any γ − 1> 0.Using the inequality
(|∇u|2
u2
)2
≤ 2(|∇u|2
u2 − γut
u
)2
+ 2γ 2 u2t
u2 ,
GRADIENT ESTIMATES UNDER GEOMETRIC FLOW 503
we calculate[ 12δαβ
2−Cβ2(γ − 1)2
] u2t
u2 −
(4δα3+
δ
2α(1−δ)2
)(|∇u|2
u2
)2
≥
[12δαβ
2− 2γ 2
(4δα3+
δ
2α(1−δ)2
)−Cβ2(γ − 1)2
]u2
t
u2
− 2(
4δα3+
δ
2α(1−δ)2
)(|∇u|2
u2 − γut
u
)2
.
Setting β = 5α, we check that
12δαβ
2− 2γ 2
(4δα3+
δ
2α(1−δ)2
)−Cβ2(γ − 1)2
= 8δα3( 2516 − γ
2)−
δ
α(1−δ)2γ 2−Cβ2(γ − 1)2,
which is nonnegative when δ > 0, γ − 1> 0 are sufficiently small.Now we take ε ≥ 5/δ such that δα−β/ε ≥ 0. Then (3-8) becomes
(1− ∂t)F1 ≥−2〈∇F1,∇ log u〉−C(k1+ k3)F1−C(k1+ k3)
2
4(γ − 1)2
− 2(
4δα3+
δ
2α(1−δ)2
)(|∇u|2
u2 − γut
u
)2
+ δαF21
+ 2δαβF1ut
u− 2δα2 F1
|∇u|2
u2 − εnk23β −
2(n− 1)2
δα(k1+ k3)
2
−Cβ4/3
δ1/3αk4/3
4 −C
δ1/3α(k2+ k4)
4/3.
Applying Corollary 5 and noting that
(3-9) (1− ∂t)F = t (1− ∂t)F1− F1,
we complete the proof of the lemma. �
Proof of Theorem 9. As in the proof of Theorem 1, we see that (M, g(t)) is com-plete for t ∈ [0, T ]. Let ρ(x, t)= d(x, x0, t) and
ϕ(x, t)= ψ(ρ(x, t)
R
).
SetϕF(x, t) := ψ
(ρ(x, t)
R
)F(x, t),
where (x, t)∈Q2R,T . Suppose (x1, t1) is the point where ϕF achieves its maximumin Q2R,T1 , where 0< T1 ≤ T .
504 JUN SUN
If |∇2u(x1, t1)| = 0, Corollary 5 yields
(3-10) (ϕF)(x1, t1)=ϕt1
(α|∇u|2
u2 −βut
u
)≤Cn,α,β
((k1+k3+k4+
√k4)t1+1
),
which implies the result.Using arguments from [Calabi 1958; Li 1991], we can assume ϕF to be smooth
at (x1, t1) and ϕF(x1, t1) > 0.As in the proof of Theorem 1, using Lemma 10, we obtain at the point (x1, t1)
(3-11) 0≥ (1− ∂t)(ϕF)
≥
(−
CR2 −
CR
√k1
)F −Ck1 F +ϕ(1− ∂t)F
≥
(−
CR2 −
CR
√k1
)F −Ck1 F −
F |∇ϕ|2
2sϕ− 2Fsϕ
|∇u|2
u2
+δα
t1ϕF2+ 2δαβϕF
ut
u− 2δα2ϕF
|∇u|2
u2 −C(k1+ k3)ϕF
−C(k1+ k3)
2
4(γ − 1)2ϕt1− εnk2
3βϕt1−ϕFt1−
2(n− 1)2
δα(k1+ k3)
2ϕt1
−Cβ4/3
δ1/3αk4/3
4 ϕt1−C
δ1/3α(k2+ k4)
4/3ϕt1
− 2Cϕt1
(4δα3+
δ
2α(1−δ)2
)( 1t1+ k1+ k3+ k4+
√k4
)2.
Using Corollary 5, we have
2δαβϕFut
u− 2δα2ϕF
|∇u|2
u2 − 2Fsϕ|∇u|2
u2
≥ (2δαβ − 2sγ − 2δα2γ )ϕFut
u−C(2δα2
+ 2s)ϕF( 1
t1+ k1+ k3+ k4+
√k4
).
Observe that 2δαβ−2sγ −2δα2γ = 0 when we set s = δα2( 5γ− 1
). Then (3-11)
becomes
(3-12) 0≥δα
t1ϕF2−ϕFC(2s+ 2δα2)
( 1t1+ k1+ k3+ k4+
√k4
)+
(−
CR2 −
CR
√k1
)F −
C F2s R2 −Ck1 F −C(k1+ k3)ϕF −
ϕFt1
−C(k1+ k3)
2
4(γ − 1)2ϕt1− εnk2
3βϕt1−2(n− 1)2
δα(k1+ k3)
2ϕt1−Cβ4/3
δ1/3α
− 2Cϕt1
(4δα3+
δ
2α(1−δ)2
)( 1t1+ k1+ k3+ k4+
√k4
)2.
GRADIENT ESTIMATES UNDER GEOMETRIC FLOW 505
Multiplying through by ϕt1 and using 0≤ ϕ ≤ 1, we have
0≥ δα(ϕF)2− (ϕF)(( C
R2 +CR
√k1
)t1+
10Cδα2
γ
( 1t1+ k1+ k3+ k4+
√k4
)t1
)− (ϕF)
(Ct1
2δα2( 5γ− 1
)R2+ (1+C(k1+ k3)t1)
)−
C(k1+ k3)2
4(γ − 1)2t21 − εnk2
3βt21
−2(n− 1)2
δα(k1+ k3)
2t21 −
Cβ4/3
δ1/3αk4/3
4 t21 −
Cδ1/3α
(k2+ k4)4/3t2
1
− 2C(
4δα3+
δ
2α(1−δ)2
)( 1t1+ k1+ k3+ k4+
√k4
)2t21 .
Solving this quadratic inequality, one obtains
ϕF(x1, t1)≤ C(
1+ (k1+ k2/32 + k3+ k4+
√k4)t1+
1R2 t1
).
By a similar argument as that in the proof of Theorem 1, we conclude in Q R,T ,
F1(x, t)≤ C(
k1+ k2/32 + k3+ k4+
√k4+
1t+
1R2
).
where C depends on n and α. Because M is noncompact, we can let R→+∞.This completes the proof of Theorem 9. �
4. Applications to Ricci flow and mean curvature flow
In this section, we apply our results to the special cases of the Ricci flow and themean curvature flow.
4.1. The Ricci flow. When h =−Ric, (2-1) is the Ricci flow equation introducedin [Hamilton 1982]. In this situation our results reduced to those in [Liu 2009].In this case our results in Section 2 do not need the assumption |∇ Ric | ≤ K4,because of the second contracted Bianchi identity. Indeed, checking the proof ofTheorem 1 carefully, we find that we need the bound on |∇h| because we want tocontrol the term
div h− 12∇(trg h)
in (2-5). But the contracted second Bianchi identity says that when h =−Ric,
div Ric− 12∇R = 0.
Now (2-5) becomes∂
∂t1 f =1
∂
∂tf + 2〈Ric,∇2 f 〉.
506 JUN SUN
4.2. The mean curvature flow. Let M be an n-dimensional closed smooth sub-manifold in N n+p. Given an embedding F0 :M→ N , we consider a one-parameterfamily of smooth maps Ft = F( · , t) : M → N n+p with corresponding imagesMt = Ft(M), where F satisfies the mean curvature flow equation
(4-1) ddt
F(x, t)= H(x, t), F(x, 0)= F0(x).
Here H(x, t) is the mean curvature vector of Mt = Ft(M) at F(x, t) in N .It is easy to check (or see [Huisken 1984; Chen and Li 2001]) that the induced
metric on Mt evolves by∂
∂tgi j =−2HαAαi j ,
where {Aαi j } is the second fundamental form of Mt in N . In this case, the tensor hin (2-1) becomes
(4-2) hi j =−HαAαi j .
In this section, we will always assume that
|KN | + |∇KN | + |∇2KN | ≤ L ,
for some constant L . Here, KN is the curvature tensor of N .Using the evolution of the second fundamental form and the standard maximum
principle, we can obtain a derivative estimate:
Proposition 11. Let {Mt }0≤t≤T be a closed smooth solution of mean curvatureflow in a Riemannian manifold N. Suppose that there exist constants 30 and 31
such that|A| ≤30 on M ×[0, T ],
|∇A| ≤31 on M0.
Then there is a constant K depending only on n, 30, 31 and L such that
|∇A| ≤ K on M ×[0, T ].
Remark 12. Another version of derivative estimate for the second fundamentalform along the mean curvature flow, similar to the derivative estimate along theRicci flow given in [Shi 1989], is proved in [Han and Sun 2012].
Proof of Proposition 11. Our proof follows [Huisken 1990]. Ci will denote variousconstants depending only on n, 30 and L . By our assumption and the evolutionequations of the second fundamental form and its derivative [Han and Sun 2012,Corollary 3.5], we have
(4-3)(∂
∂t−1
)|A|2 ≤−2|∇A|2+C1
GRADIENT ESTIMATES UNDER GEOMETRIC FLOW 507
and
(4-4)(∂
∂t−1
)|∇A|2 ≤−2|∇2 A|2+C2(|∇A|2+ 1).
Therefore,
(4-5)(∂
∂t−1
)(|∇A|2+C2|A|2
)≤−C2
(|∇A|2+C2|A|2
)+C3.
As Mt is closed for each t , we obtain by the maximum principle that(|∇A|2+C2|A|2
)(x, t)≤ e−C2t sup
M0
(|∇A|2+C2|A|2
)+
C3
C2(1− e−C2t)≤ K ,
where K depends on n, 30, 31 and L . This proves the proposition. �
Theorem 13. Let {Mt }0≤t≤T be a closed smooth solution of mean curvature flowin a Riemannian manifold N. Suppose that there exist constants 30 and 31 suchthat
|A| ≤30 on M ×[0, T ],
|∇A| ≤31 on M0.
If u is a positive solution to the equation (1 − ∂t)u(x, t) = 0, then for (x, t) ∈M × (0, T ], we have
|∇u(x, t)|2
u2(x, t)−α
ut(x, t)u(x, t)
≤nα2
t+C K ,
for any α > 1. Here C depends on n, α only and K depends 30, 31 and L.
Proof. During the proof of this theorem, the constant K will denote a constantdepending only on 30, 31 and L which may vary from one line to the next.
By Proposition 11 and (4-2), we see that |h| ≤ K and |∇h| ≤ K . On the otherhand, using the Gauss equation, we have
Ri jkl − Ki jkl = Aαik Aαjl − Aαil Aαjk,
where Ki jkl is the curvature tensor on N . Hence our assumption and Proposition 11imply that |Rm| ≤ K and |∇Rm| ≤ K . This shows that all the assumptions ofTheorem 6 are satisfied, and the conclusion follows. �
Remark 14. K. Smoczyk [1999] proved a similar gradient estimate for the positivesolution of the heat equation along the Lagrangian mean curvature flow, and ob-tained a Harnack inequality for the Lagrangian angle. Indeed, under the Lagrangianmean curvature flow, the Lagrangian angle evolves by (1− ∂t)θ = 0.
Next we deal with the complete noncompact case.
508 JUN SUN
Proposition 15. Let {Mt }0≤t≤T be a smooth solution of mean curvature flow in aRiemannian manifold N. Assume that M0 is complete noncompact without bound-ary. Suppose that there exist two constants 30 and 31 such that
|A| ≤30 on M ×[0, T ],
|∇A| ≤31 on M0.
Then there is a constant K depending only on n, 30, 31 and L such that
|∇A| ≤ K on M ×[0, T ].
Proof. Recall that in the proof of Proposition 11, we obtained(∂
∂t−1
)(|∇A|2+C2|A|2
)≤−C2
(|∇A|2+C2|A|2
)+C3.
Set F = |∇A|2+C2|A|2. Then
(4-6) (1−∂
∂t)F ≥ C2 F −C3.
Let Q R,T and ϕ be defined as in the proof of Theorem 1. We consider ϕF . Suppose
(ϕF)(x1, t1)= supM×[0,T ]
(ϕF).
Then (x1, t1) ∈ Q2R,T . We consider two cases.• If t1=0, then (ϕF)(x, t)≤ (ϕF)(x1, 0)≤ F(x1, 0)≤ supM0
F ≤31+C230≤ K .In particular, for any (x, t) ∈ Q R,T , we have
F(x, t)≤ K .
• If, on the contrary, t1 > 0, the maximum principle gives(1−
∂
∂t
)(ϕF)(x1, t1)≤ 0.
By the Gauss equation and our assumption, the Ricci curvature is bounded. Simi-larly to the proof of Theorem 1 and using (4-6), we get that, at (x1, t1),
0≥(1−
∂
∂t
)(ϕF)(x1, t1)≥−
CR2 −
C√
KR+C2ϕF −C3.
Thus we obtain that
(ϕF)(x1, t1)≤C3
C2+
CC2 R2 +
C√
KC2 R
.
In particular, for any (x, t) ∈ Q R,T , we have
F(x, t)≤C3
C2+
CC2 R2 +
C√
KC2 R
.
GRADIENT ESTIMATES UNDER GEOMETRIC FLOW 509
Combining the two cases above and letting R →∞ we get F(x, t) ≤ K , forsome constant K depends on n, 30, 31 and L . This proves the proposition. �
Arguing as in the proof of Theorem 13 and using Corollary 5 and Theorem 9,we obtain
Theorem 16. Let {Mt }0≤t≤T be a smooth solution of mean curvature flow in a Rie-mannian manifold N. Assume that M0 is complete noncompact without boundary.Suppose that there exist two constants 30 and 31 such that
|A| ≤30 on M ×[0, T ],
|∇A| ≤31 on M0.
If u is a positive solution to the equation (1 − ∂t)u(x, t) = 0, then for (x, t) ∈M × (0, T ] we have
|∇u(x, t)|2
u2(x, t)−α
ut(x, t)u(x, t)
≤nα2
t+C K
and|∇
2u(x, t)|u(x, t)
+α|∇u(x, t)|2
u2(x, t)− 5α
ut(x, t)u(x, t)
≤ C(
K + 1t
),
for any α > 1. Here C depends only on n, α and K depends on n, 30, 31 and L.
Acknowledgement
We thank Professor Jiayu Li for his guidance and encouragement. We also wouldlike to thank Doctor Shiping Liu for helpful discussions.
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PACIFIC JOURNAL OF MATHEMATICS
Volume 253 No. 2 October 2011
257Fusion rules on a parametrized series of graphsMARTA ASAEDA and UFFE HAAGERUP
289Group gradings on restricted Cartan-type Lie algebrasYURI BAHTURIN and MIKHAIL KOCHETOV
321B2-convexity implies strong and weak lower semicontinuity of partitionsof Rn
DAVID G. CARABALLO
349Testing the functional equation of a high-degree Euler productDAVID W. FARMER, NATHAN C. RYAN and RALF SCHMIDT
367Asymptotic structure of a Leray solution to the Navier–Stokes flow around arotating body
REINHARD FARWIG, GIOVANNI P. GALDI and MADS KYED
383Type II almost-homogeneous manifolds of cohomogeneity oneDANIEL GUAN
423Cell decompositions of Teichmüller spaces of surfaces with boundaryREN GUO and FENG LUO
439A system of third-order differential operators conformally invariant undersl(3, C) and so(8, C)
TOSHIHISA KUBO
455Axial symmetry and regularity of solutions to an integral equation in ahalf-space
GUOZHEN LU and JIUYI ZHU
475Braiding knots in contact 3-manifoldsELENA PAVELESCU
489Gradient estimates for positive solutions of the heat equation undergeometric flow
JUN SUN
0030-8730(201110)253:2;1-D
PacificJournalofM
athematics
2011Vol.253,N
o.2
