New York Journal of Mathematics
New York J. Math. 3 (1997) 103{118.
Harnack Inequalities For Curvature
Flows Depending On Mean Curvature
Knut Smoczyk
Abstract. We prove Harnack inequalities for parabolic ows of compact orientable
hypersurfaces in Rn+1 , where the normal velocity is given by a smooth function f
depending only on the mean curvature. We use these estimates to prove longtime
existence of solutions in some highly nonlinear cases. In addition we prove that
compact selfsimilar solutions with constant mean curvature must be spheres and
that compact selfsimilar solutions with nonconstant mean curvature can only occur
in the case, where f = Ax with two constants A and .
Contents
1. Introduction
2. Some relations and evolution equations
3. Homothetic solutions
4. The Harnack inequality
5. Longtime existence for some highly nonlinear ows
References
103
106
107
107
114
118
1. Introduction
Assume that M n is a compact orientable surface, smoothly immersed into Rn+1
by a smooth family of dieomorphisms Ft : M n ! Rn+1 that satisfy the PDE
@
@t F = ;f where denotes the outward pointing unit normal and f is a smooth function
depending only on the mean curvature H of the immersed surface, e.g. for f = H
we get the well-known mean curvature ow (MCF) and for f = ; H1 we obtain the
inverse mean curvature ow. Hamilton 4] proved a beautiful Harnack inequality
for the MCF. In 1] Harnack inequalities were derived for convex hypersurfaces in
cases where f may depend on the full second fundamental form. The case f = ; H1
Received September 22, 1997.
Mathematics Subject Classication. 53C21 (35K15).
Key words and phrases. Harnack, mean, curvature, ow, selfsimilar.
103
c 1997 State University of New York
ISSN 1076-9803/97
Knut Smoczyk
104
has not been studied yet nor are there results for arbitrary functions f depending
only on H . The aim of this paper is to address the following questions:
I. What conditions on f guarantee that the ow becomes parabolic so that
we have shorttime existence of a solution?
II. For which f do we get nice Harnack inequalities?
III. Since selfsimilar solutions play an important role in the Harnack inequality
for the MCF, we ask: For which f do selfsimilar solutions exist and can we
say something about the nature of these solutions?
An interesting case is the inverse mean curvature ow f = ; H1 since it is important in General Relativity 6]. There is some hope that one can generalize our
results to other target manifolds N n+1 .
Throughout this paper we will use the standard terminology, i.e., h i denotes the
euclidean inner product, gij dxi dxj is the induced metric on M n, xi coordinates for
M n and hij dxi dxj = hri F rj idxi dxj denotes the second fundamental form,
r the covariant derivative with respect to gij . Double latin indices are summed
from 1 to n and we set
jAj2 := hij hij C := hik hlk hil :
In this paper we will always assume that f : ! R is a smooth function dened
on an open subset in R and we dene
Denition. Let F : M n ! Rn+1 be an immersion. F is called admissible, if
H (x) 2 8x 2 M n .
The answer to question I is then given by
Proposition I. Let F0 : M n ! Rn+1 be an admissible smooth immersion of a
compact orientable surface M n and assume that f : ! R is strictly positive.
Then the PDE
0
@
@t F = ;f
F (x 0) = F0 (x) 8x 2 M n
(?)
has a smooth admissible solution on a maximal time interval 0 T ) T > 0.
Proof. This follows from the fact that the linearization of (?) diers from the linearization for the mean curvature ow only by a factor f which by assumption is
strictly positive. Therefore (?) is a (nonlinear) parabolic equation and the compactness of M n and the theory for parabolic equations imply shorttime existence.
0
In view of Proposition I we will always assume that f > 0. Then the main
theorem can be stated as follows
Theorem 1. Assume that F0 : M n ! Rn+1 is an admissible smooth and convex
immersion of an orientable compact M n and that f : ! R is a smooth function
such that for all x 2 we have
0
;
f > 0 ff x2 ax ff x
00
00
0
0
0
0
0 ff
00
x + ff
0
; (f ) 2 x 0
0
Harnack Inequalities
105
where a 2 R is a constant. Then we can nd a small positive constant d such that
@
i j
@t f + 2hrf V i + hij V V + cf H 0
holds for all tangent vectors V as long as t < T , d + (a + 2)t > 0 and Mt stays
convex, where we have set c(t) := d+(a1+2)t
Remark. We do have to make these assumptions on f to avoid negative terms in
the evolution equation for the basic Harnack expression Z (see below). f = x
satises the assumptions in Theorem 1 on = (0 1) with a = ; 1. The following
0
Propositions show that almost all functions satisfying the assumptions in Theorem 1
are of this form.
Proposition IIa. Assume that f : (0 a) ! R 0 < a 1 is a smooth function
that smoothly extends to x = 0 and satises all assumptions in Theorem 1. Let us
set if := minfl 0jf (l)(0) 6= 0g 1. If 0 < if < 1 then f = Aif xif with a
positive constant A.
Proof. Since f (0) = 0 and f (x) > 0 8x 2 = (0 a) we observe that f (x) >
0 8x 2 . By de l'Hospital's rule we obtain
0
lim f x = if
x 0 f
0
!
and
f x =i ;1
lim
f
x 0f
00
0
!
and then f x 0 and f 0 imply
(1.1)
f x (if ; 1)f 8x
Since f > 0 we have by assumption
; f 00 0
0
0
00
0
j
ff x + ff
00
and then also
(1.2)
But (1.1) and (1.2) imply
0 ff x + ff
and therefore
00
; (f )2x = f 2; ff x 0
0
0
0
0
f x if f 8x
0
0
; (f )2x (if ; 1)ff + ff ; if ff
0
0
;
which implies that
f x = 0
f
0
0
f x=i
f
f
0
and after integration
f = Aif xif
with a positive constant A (since f > 0).
0
0
0
=0
Knut Smoczyk
106
Proposition IIb. Assume that f : (a 1) ! R 0 < a < 1 is a smooth function that satises all assumptions in Theorem 1 and that g(x) := ;f ( x1 ) smoothly
extends to x = 0. If 0 < ig < 1 then f = ;Aig x ig with a positive constant A.
Proof. If we set y = x1 and g(y) = ;f (x) then the assumptions for f and x
translate into the same assumptions for g and y and as above we can derive the
desired result.
;
Remark. f (x) =
px does not satisfy the assumptions in Proposition IIa since
it does not extend smoothly to x = 0. f (x) = x + 1 satises all assumptions in
Theorem 1. f (x) = ln x satises almost all assumptions in Theorem 1. The only
condition which is violated is that ff x + ff ; (f )2 x = ; x1 < 0 on = (0 1).
00
0
0
2. Some relations and evolution equations
By assumption we have
@
@t F = ;f
(2.1)
In 7] we formally derived the evolution equations for various geometric objects on
M n , these are:
@
@t gij = ;2fhij
@
@t d
= ;Hfd
(2.2)
(2.3)
(d
denotes the volume form)
(2.4)
(2.5)
(2.6)
(2.7)
@
@t = rf
@
l
@t hij = ri rj f ; fhi hlj
@
2
@t H = f + f jAj
@ 2
@t jAj = 2hhij ri rj f i + 2fC
In addition we have the well-known Gau-Weingarten-Codazzi-Mainardi equations
(2.8)
(2.9)
(2.10)
(2.11)
(2.12)
(2.13)
ri rj F = ;hij F = gij ri rj F = ;H
rihjk = rj hik
ri = hil rl F
ri rj = rlhij rlF ; hil hlj = rH ; jAj2 Harnack Inequalities
107
Note that in these equations we assume that F are sets of n + 1 functions on
M n.
We also have the Simons identity
(2.14)
ri rj H = hij ; Hhilhlj + jAj2hij
3. Homothetic solutions
A homothetic solution Ft is a family of dieomorphisms such that the surfaces
given by the rescaled dieomorphisms Fet := F are stationary in Rn+1 , where denotes a function depending only on time t. The assumption that Fet represents a
stationary surface means that the normal velocity must be zero. So we have
@ F
@ hF i ; f e 0 = h @t
ei =
@t
Let us dene c := ; @t@ ln . Then we have shown that for a homothetic solution
of (?) we have
(H.1)
f = ;chF i
Taking covariant derivatives of f and using (2.11), (2.10), (2.8) we obtain with
e
Vi := chF ri F i
(H.2)
(H.3)
(H.4)
(H.5)
(H.6)
ri f = ;hil Vel
ri Vej = cgij + fhij
ri rj f = ;rl hij Vel ; chij ; fhilhlj
f + f jAj2 = ;hrH Ve i ; cH
Vej = f rj H + hji ri f
4. The Harnack inequality
In the sequel we have to calculate many evolution equations. To avoid too
complicated formulas it is most convenient to work with coordinates associated to
a moving frame. We use similar moving frame coordinates as in 4]. Here the
moving frame fEa ga=1:::n evolves according to
@ i
i j
@t Ea = fh j Ea
and we denote the coordinates of a vector V with respect to the moving frame by
yai . The following calculations closely follow the procedure in 4]. As in this paper
(4.1)
we have
(4.2)
(4.3)
(4.4)
rba := yai @y@ b
i
@
k
i
Da := ya ( @xi ; ;ij ybj @y@ k )
b
ab := gac rcb ; gbc rca
Knut Smoczyk
108
In addition we dene
@ + fh b ra
Dt := @t
a b
(4.5)
Then straightforward computations give the commutator relations
r
(4.12)
;
r
;
@ D ] = D (fh d ) + D (fh d ) ; Dd (fh )
@t
a
a
b
ab
b
a
;
(4.9)
(4.10)
(4.11)
r
Da Db ] = Rdeba ed
a Dc ] = I c Da a D ] = I a D
b
b
b c
c b
bc Da ] = Iba Dc Ica Db
(4.6)
(4.7)
(4.8)
rd
b
Dt Da ] = Db (fhac )bc + fhab Db
Da ] = Db (Rdlab rld) + Rdlab rldDb
Dt ; f Da ] = f Ra bcd Db cd + f Dahbc hbd cd + ff Da f 00
0
0
0
+(f ; f H )ha
0
(4.13)
Dt ] = ( ff
0
0
lD
l+f
0
han hnl Dl
; H )DnfDn + 2fhabDbDa + 2hanDafDn + Da(Db (fhac)bc)
In the moving frame the evolution equations reduce to
(4.14)
Dt gab = 0 gab = Iab
(4.15)
Dt hab = Da Db f + fhan hnb
(4.16)
Dt f = f (f + f jAj2 )
Let us now dene the following tensors, where we assume that Va is an arbitrary
tangent vector on M n and c a smooth function to be determined later and only
depending on time t.
0
Xa := Da f + hal Vl
Yab := Da Vb ; fhab ; cgab
Z := Dt f + 2hDf V i + hab V a V b + cf H
Wab := Dt hab + Dl hab Vl + chab
W := Dt f + hDf V i + cf H
0
0
By (H.1){(H.6) these tensors vanish on homothetic solutions if we take Va = Vea
and the induced c. On the other hand we have
@ (chF D F i) + fh d rc Ve
Dt Vea = @t
a
c d a
@ (lnc)Ve ; chF D (f )i + fh d Ve
= @t
a
a
a d
@ (lnc)Ve ; chF iD f
= @t
a
a
Harnack Inequalities
109
and therefore
@ (ln c)Ve + f h b h cVe
(Dt ; f )Vea = @t
a
a b c
(4.17)
0
0
and on a homothetic solution
@ (ln c)Ve ; f h Db f
(Dt ; f )Vea = @t
a
ab
(H.7)
0
0
In view of (H.7) we dene
@ (ln c)V + f h Db f
Ua := (Dt ; f )Va ; @t
a
ab
0
0
which also vanishes on a homothetic solution.
We want to calculate (Dt ; f )Z . We do this in several steps. Using (4.12)
and (4.16) we obtain
0
(Dt ; f )Da f = ff Da f f + (f ; f H )hab Db f + f han hnl Dl f + Da(ff jAj2 )
00
0
0
0
0
0
and therefore
(4.18)
(Dt ; f )Da f =f jAj2 Da f + 2ff hbcDa hbc + f han hnl Dl f
0
0
0
0
+ (f ; f H )hab Db f + (ff )2 Da fDtf
00
0
0
Further, we use Simons identity to rewrite (4.15)
Dt hab = Da Db f + fhan hnb
= f (hab ; Hhan hnb + jAj2 hab ) + (ff )2 Da fDb f + fhan hnb
00
0
0
This gives
(Dt ; f )hab = f jAj2 hab + (f ; f H )han hnb + (ff )2 Da fDbf
00
(4.19)
0
0
0
0
(4.18), (4.19) and the denition of U give
(Dt ; f )Xa = f jAj2 Da f + 2ff hbc Da hbc + f han hnl Dl f + (f ; f H )hab Db f
0
0
0
0
0
+ (ff )2 Da fDtf + V b (f jAj2 hab + (f ; f H )han hnb + (ff )2 Da fDbf )
00
00
0
0
0
0
@ (ln c)V ; f h Dc f ) ; 2f Dc h b D V
+ hab (Ub + @t
b
bc
a c b
= f jAj2 Xa + (f ; f H )hab Xb + hab Ub ; 2f Da hbc(Y bc + cgbc)
0
0
0
0
0
@ (ln c)(X ; D f ) + f D f (W ; cf H )
+ @t
a
a
(f )2 a
00
0
0
Knut Smoczyk
110
and nally
(4.20)
(Dt ; f )Xa =f jAj2 Xa + (f ; f H )hab Xb + hab Ub ; 2f Da hbc Y bc
0
0
0
0
@ (ln c)X + f D fW ; (2c + @ (ln c) + c f H )D f
+ @t
a
a
(f )2 a
@t
f
00
00
0
0
Next we compute
(Dt ; f )Dt f =Dt f ]f + Dt (ff jAj2 )
=f Dt ]f + Dt f f + Dt f f jAj2 + f jAj2 Dt f + ff Dt jAj2
=(f ; f H )jDf j2 + 2ff hab Da Db f + 2f hab Da fDb f
0
0
0
0
0
0
0
0
0
0
0
+ (ff )2 (Dt f )2 + f jAj2 Dt f + 2ff hab Dt hab
00
0
0
0
giving
(4.21)
(Dt ; f )Dt f =f jAj2 Dt f + 4ff hab Dt hab ; 2f 2f C + (ff )2 (Dt f )2
00
0
0
0
0
+ (f ; f H )jDf j
0
0
2 + 2f 0 h
ab
Da fDb f
Furthermore
@ (ln c)cf H + c f HD f + cD f
(Dt ; f )(cf H ) = @t
t
t
f
00
0
0
0
; cf (H (f
0
H + f jDH j2 ) + f H + 2f jDH j2 )
@ (ln c)cf H + c(f H + f )(f + f jAj2 )
= @t
; c(f H + f )(f H ) ; cf (Hf + 2f )jDH j2
@ (ln c)cf H + cf (f + f H )jAj2 + c((f )2 H ; f f
= @t
0
00
000
0
00
00
0
0
0
00
0
0
0
0
000
00
00
00
0
00
;f f
0
000
H )jDH j2
which gives
@ (ln c)cf H + cf (f + f H )jAj2 ; c; f H jDf j2
(4.22) (Dt ; f )(cf H ) = @t
f
00
0
0
0
0
00
0
0
Harnack Inequalities
111
(4.21), (4.22) and (4.18) give
(Dt ; f )W =f jAj2 Dt f + 4ff hab Dt hab ; 2f 2f C + (ff )2 (Dt f )2 + (f ; f H )jDf j2
00
0
0
0
0
0
0
@ (ln c)cf H + cf (f + f H )jAj2 ; c; f H jDf j2
+ 2f hab Da fDb f + @t
f
00
0
0
0
00
+ V a (f jAj2 Da f + 2ff hbc Da hbc + f han hnl Dl f
0
0
0
0
0
+ (f ; f H )hab Db f + (ff )2 Da fDt f )
00
0
0
@ (ln c)V ; f h Db f ) ; 2f Da V b D D f
+ Da f (Ua + @t
a
ab
a b
0
0
; ff H )jAj2 + @t@ (ln c)W ; @t@ (ln c)Dtf
+ (ff )2 Dt fW ; c ff HDt f + (f ; f H )hX Df i
+ f (4fhab ; 2Da V b )Dt hab + 2ff Da V b han hnb ; 2f 2f C
; c; ff H jDf j2 + 2ff V a Dahbchbc + hDf U i + f habX aDbf
@ (ln c)W
=f jAj2 W ; c((f )2 H ; ff ; ff H )jAj2 + @t
; (c ff H + @t@ (ln c))Dt f + (ff )2 DtfW + (f ; f H )hX Df i
+ f (4fhab ; 2Da V b )Wab ; f (4fhab ; 2Da V b )(Dl hab Vl + chab )
;
+ 2ff Da V b han hnb ; 2f 2 f C ; c ff H jDf j2
+ 2ff V a Da hbc hbc + f hhab Da fXb i + hDf U i
=f jAj2 W ; c((f )2 H ; ff
0
0
00
00
0
0
0
00
0
0
0
00
0
0
0
0
0
0
0
0
00
00
00
0
0
0
0
0
00
0
0
0
0
0
0
and after rearranging terms we conclude
(4.23)
(Dt ; f )W =f jAj2 W + f (4fhab ; 2Da V b )Wab + f hab Xa Db f + hDf U i
0
0
0
0
@ (ln c))W
+ 2f (fhan hnb + Dl hab Vl )Y ab + (2c + @t
0
; c((f )2 H ; ff ; ff H )jAj2 ; (c ff H + @t@ (ln c) + 2c)Dtf
;
+ (ff )2 Dt fW + (f ; f H )hX Df i ; c ff H jDf j2 + 2cf hab Y ab
00
0
0
00
0
00
00
0
0
Eventually the evolution equation for Z is given by
0
0
0
Knut Smoczyk
112
(Dt ; f )Z =f jAj2 W + f (4fhab ; 2Da V b )Wab + f hab Xa Db f + hDf U i
@ (ln c))W
+ 2f (fhan hnb + Dl hab Vl )Y ab + (2c + @t
0
0
0
0
0
; c((f )2 H ; ff ; ff H )jAj2 ; (c ff H + @t@ (ln c) + 2c)Dtf
;
+ (ff )2 Dt fW + (f ; f H )hX Df i ; c ff H jDf j2 + 2cf hab Y ab
+ V a (f jAj2 Xa + (f ; f H )hab Xb + hab Ub ; 2f Da hbc Y bc
@ (ln c)X + f D fW ; (2c + @ (ln c) + c f H )D f )
+ @t
a
a
(f )2 a
@t
f
@ (ln c)V ; f h Db f ) ; 2f D V (W ab + han Y b )
+ X a(Ua + @t
a
ab
a b
n
2
ab
ab
an
b
=f jAj Z + 2hX U i ; 4f (Y + cg )Wab ; 2f h Yn (Yab + cgab )
@ (ln c) + 2c)(D f + hDf V i)
+ (f ; f H )jX j2 + (ff )2 W (Dt f + hV Df i) ; (c ff H + @t
t
@ (ln c)hX V i + (2c + @ (ln c))W ; c((f )2 H ; ff ; ff H )jAj2
+ 2 @t
@t
;f
; c f H jDf j2 + 2cf habY ab
=f jAj2 Z + 2hX U i ; 2f han Ynb Y ab ; 4f Wab Y ab ; 4cW
@ (ln c) + 2c)(W ; cf H )
+ (f ; f H )jX j2 + (ff )2 W (W ; cf H ) ; (c ff H + @t
@ (ln c)hX V i + (2c + @ (ln c))W ; c((f )2 H ; ff ; ff H )jAj2 ; c; f H jDf j2
+ 2 @t
@t
f
00
0
0
00
0
00
00
0
0
0
0
0
0
0
0
00
00
0
0
0
0
0
0
0
00
00
0
0
0
0
00
0
0
00
0
0
0
0
0
00
00
0
0
0
0
0
00
0
0
00
0
0
So we nally arrive at
(4.24)
;
(Dt ;f )Z = f jAj2 + (ff )2 (Z ; 2hX V i) ; 2(c ff H + 2c) Z
00
0
00
0
0
0
@ (ln c) + 2c)hX V i
+ 2hX U i + (f ; f H )jX j2 + (ff )2 hX V i2 + 2(c ff H + @t
00
00
0
0
0
; 2f han YnbY ab ; 4f Wab Y ab + (c ff H + @t@ (ln c) + 2c)cf H
;
+ c(ff H + ff ; (f )2 H )jAj2 ; c ff H jDf j2
00
0
0
0
0
00
00
0
0
0
0
Now we are able to prove Theorem 1.
Proof of Theorem 1. From (4.24) and the assumptions in Theorem 1 we conclude, with c := d+(a1+2)t
(Dt ; f )Z f jAj2 + (ff )2 (Z ; 2hX V i) ; 2(c ff H + 2c) Z ; 2f han Ynb Y ab ; 4f Wab Y ab
0
;
00
00
0
0
0
0
0
@ (ln c) + 2c)hX V i
+ 2hX U i + (f ; f H )jX j2 + (ff )2 hX V i2 + 2(c ff H + @t
00
00
0
0
0
Harnack Inequalities
113
Now choose d so small such that Z > > 0 for t = 0 and for all tangent vectors V .
This is possible since the initial surface is convex. On any compact time interval
0 t0 ] with t0 < T we can therefore estimate
(Dt ;f )Z ;bZ ; 2f han Ynb Y ab ; 4f Wab Y ab
@ (ln c) + 2c)hX V i
+ 2hX U i + (f ; f H )jX j2 + (ff )2 hX V i2 + 2(c ff H + @t
0
0
0
00
00
0
0
0
with a large constant b depending on t0 . Additionally for t > 0 and an arbitrary
positive constant (Dt ; f )(ebt Z + t) >ebt ;2f han Ynb Y ab ; 4f Wab Y ab + 2hX U i + (f ; f H )jX j2
0
;
0
0
0
@ (ln c) + 2c)hX V i
+ (ff )2 hX V i2 + 2(c ff H + @t
00
00
0
0
If t1 t0 would be the rst time where ebt Z + t would become zero at some point
x 2 M n and for some tangent vector V then we must have Xa = 0 since it is the
rst variation with respect to V and we can also extend V in spacetime such that
Yab = 0. This implies a contradiction and therefore ebt Z + t > 0 8t t0 . Since and t0 are arbitrary we conclude Z 0 whenever t < T and d + (a + 2)t > 0.
Now we want to answer the question for which f one can expect selfsimilar
solutions of (?). First we remark that if x 2 and x > 0, then the sphere of
constant radius nx and with constant mean curvature H = x gives always rise to a
selfsimilar solution for a short time.
For any subset A M n let us dene
H (A) := fx 2 j9p 2 A : H (p) = xg
and
Pt := fp 2 MtnjrH 6= 0g
The answer to question III is then given by
Proposition IIIa. If P0 6= and Ft : M n ! Rn+1 is a selfsimilar solution of
(?) for a compact connected M n , then we have f = Ax 8x 2 H (Mt ) with
nonvanishing constants A and .
Proof. Since Ft is selfsimilar we have Pt = P0 8t 2 0 T ). Since Xa W vanish on
selfsimilar solutions so must their time derivatives. From (4.20) we conclude that
on H (Pt ) and by continuity also on H (Pt ) we must have c ff x + 2c + @t@ (ln c) = 0.
Since P0 6= and M n is connected we have H (M n) H (Pt ) and we derive
00
0
@ (ln c) = 0
c ff x + 2c + @t
00
0
for all x 2 H (Mtn). Since f > 0 there can be at most one point x 2 such that
f (x) = 0. At all other points we have
0
ff x + ff
00
; (f )2 x = f 2; ff x
0
0
0
0
Knut Smoczyk
114
Then (4.23) implies that also
;
f x = 0
f
0
0
in all points x 2 H (Mt) where f (x) 6= 0. But after integration we obtain that at
these points
f = Ax
with constants A (nonvanishing since we must have f > 0) and again by continuity this also holds on all of H (Mt ).
0
We observe that by (2.9)
jF j2 = 2(n ; H hF i)
and with (H.1) we obtain on a homothetic solution
jF j2 = 2(n ; fH
c )
(H.8)
This implies
Proposition IIIb. If P0 = and M nn is a compact1 orientable selfsimilar solution
of (?), then Mt is a sphere of radius H . If f = ; H , then any compact orientable
homothetic solution is a sphere of radius Hn .
Proof. P0 = and (H.8) imply jF j2 = const. By the assumptions on M n this
constant must be zero and then consequently jF j2 = const. This implies that Mt
is a sphere.
5. Longtime existence for some highly nonlinear ows
In this paragraph we are going to show that one can use the Harnack inequality
to prove longtime existence of solutions for some highly nonlinear ows. To be
precise
Theorem 2. Assume that f : (0 1) ! R is a smooth negative function that
satises the assumptions in Theorem 1 with a > ;2 and that limx 0 f (x) = ;1
and F0 : M 2 ! R3 is a smooth convex immersion of an orientable compact surface
M 2 . Then (?) has a smooth immortal solution and we have
lim H = 0 tlim jFt j2 = 1
t
!
!1
!1
;
Remark. The functions f = x with 1 < < 0 satisfy the assumptions in
Theorem 2 on = (0 1). These speed functions are not homogenous of degree
one and are not included in the class of functions considered in 2]
To prove Theorem 2 we need some lemmas that are interesting on their own.
Lemma 1. Assume f > 0 and that F0 : M n ! Rn+1 is a smooth immersion of an
orientable compact surface and that on M0 = F0 (M n ) we have f 2 minM0 f 2 > 0.
Then this is also true on the maximal time interval 0 T ) where a smooth solution
of (?) exists.
0
Harnack Inequalities
115
Proof. The evolution equation for f 2 is given by
(Dt ; f )f 2 = f
0
0
;2jrf j2 + 2f 2jAj2
;
and the result follows from the parabolic maximum principle.
Corollary 1. With the assumptions in Lemma 1 we have that for negative f
H max
H
M
0
and for positive f
H min
H
M0
We can do even better
Lemma 2. Assume = (0 1) f < 0 f > 0 and that F0 : M n ! Rn+1 is an
admissible immersion of an orientable compact manifold M n. Then we can nd a
positive such that on 0 T )
0
M0 H
max
H 1 +max
Mt
t maxM0 H
Proof. Since Ft is admissible on 0 T ) we must always have H > 0. At a point
where maxMt H is attained we have H 0 and rH = 0. The evolution equation
for H (2.6), Lemma 1 and the fact that jAj2 Hn2 give us
@
H ;(max
H )2
Mt
Mt
@t max
with =
;
maxM0 f
n
and after integration we obtain the result.
(2.7) and Simons identity give
(5.1) (Dt ; f )jAj2 = ;2f jrAj2 + 2f jAj4 + 2(f ; f H )C + 2f hij ri H rj H
0
0
0
0
00
Let R = H 2 ; jAj2 be the Scalar curvature of Mt. Now we turn our attention
to the case where n = 2. In this case we can decompose C into
C = H2 (3jAj2 ; H 2 )
Then (5.1) and (2.6) imply that for n = 2
(5.2)
(Dt ; f )R =2f (jrAj2 ; jrH j2 ) + 2f (H jrH j2 ; hij ri H rj H )
; (2f jAj2 + (f ; f H )H )R
0
0
00
0
0
Knut Smoczyk
116
Lemma 3. Under the assumptions in Theorem 2 we have
R > 0 8t 2 0 T )
Proof. Let t0 T be the maximal time such that R > 0 on all of Mt for t 2 0 t0),
i.e., the maximal time for which Mt stays convex. If t0 = T we are done. So assume
that t0 < T . Since Mt is admissible for t 2 0 t0 ] we must have H > 0 on 0 t0 ].
By denition of t0 we also have R > 0 on 0 t0 ) and R 0 on 0 t0 ]. Since M and
0 t0 ] are both compact and R 0 we can nd a constant c such that
(5.3) (Dt ; f )R 2f (jrAj2 ; jrH j2 ) + 2f (H jrH j2 ; hij ri H rj H ) ; cR
0
0
00
Now assume that x is any point on Mt where the minimum of R is attained.
At this point we must have f R 0 and rR = 0. Let and denote the two
principal curvatures in a neighborhood of x. We have
0
0 = rR = 2r + 2 r
Since H > 0 and 0 we must either have > 0 or > 0. Let us assume that
> 0. First we compute that
H jrH j2 ; hij ri H rj H = jr2 H j2 + jr1 H j2 0
where we choose normal coordinates such that hij = diag( ). Since f x af
and H > 0 a > ;2 we can estimate
00
0
f (jr H j2 + jr H j2 )
2f (H jrH j2 ; hij ri H rj H ) ;4 H
2
1
0
00
On the other hand Codazzi's equation gives us that
jrAj2 = jr1j2 + jr2j2 + 3jr2j2 + 3jr1 j2
Combining the last two statements we see that at a point where rR = 0 we must
always have
2f (jrAj2 ; jrH j2 ) + 2f (H jrH j2 ; hij ri H rj H ) 0
0
00
and consequently at any point where the minimum of R is attained
Dt R ;cR
which implies that
min
R (min
R)e
M
M
t
;
0
This proves that t0 = T .
We come to the proof of Theorem 2
ct
> 0 8t 2 0 t0 ]
Harnack Inequalities
117
Proof of Theorem 2. On 0 T ) we must have H > 0: Otherwise (?) is not well-
dened, i.e., the surfaces would not be admissible. If T is nite, then this can only
happen for two reasons. Either the solutions converge to a surface that is no longer
admissible, i.e., the mean curvature would vanish somewhere, or the surfaces must
develop a singularity. Since Mt is convex on 0 T ) we can apply Theorem 1 with
V = 0 to obtain
@
1
@t H ; d + (a + 2)t H
and since by assumption a > ;2 we can estimate
@ H ;1H
which implies that
@t
d
H e d1 t min
H
M
;
0
So H cannot become zero in nite time. This estimate and Corollary 1 imply
that the surfaces stay admissible in nite time. Therefore we conclude that if T
is nite the surfaces must develop a singularity. It is well-known that the second fundamental form jAj2 must then blow up for t ! T (compare 2] and 5]).
By Lemma 3 we conclude that in the case where T < 1, H must also blow up
which, in view of Corollary 1, proves that T = 1. By Lemma 2 we conclude that
limt (maxMt H ) = 0. It remains to prove that jFt j ! 1. An easy calculation
gives the following evolution equation
!1
(Dt ; f )( hFf i + (2 + a)t) = 2 ff hr( hFf i + (2 + a)t) rf i + a + 1 ; f fH
0
0
0
The assumptions on f imply that limx 0 ff x exists and that limx 0 ff x a.
Since limx 0 f = ;1 we can apply de l'Hospital's rule and get limx 0 ffx =
;
1 + limx 0 ff x a + 1. On the other hand the assumption that ffx 0 implies
for any x 2 a + 1 ; ffx 0
00
!
00
!
0
0
0
!
!
0
00
!
0
0
0
This gives
(Dt ; f )( hFf i + (2 + a)t) 2 ff hr( hFf i + (2 + a)t) rf i
0
0
and with the maximum principle we conclude
hF i c ; (2 + a)t
f
for a constant c. Using Schwarz' inequality we have proven that
jF j (2 + a)t ; c
jf j
and since limt jf j ! 1 and 2+ a > 0, this can only hold when limt
proving the rest of Theorem 2.
!1
!1
jF j = 1,
Knut Smoczyk
118
References
1] B. Andrews, Harnack inequalities for evolving hypersurfaces, Math. Zeitschrift 217 (1994),
179{197.
2] C. Gerhardt, Flow of nonconvex hypersurfaces into spheres, J.D.G 32 (1990), 299{314.
3] R.S. Hamilton, The Harnack estimate for the Ricci Flow, J.D.G 37 (1993), 225{243.
4] R.S. Hamilton, Harnack estimate for the mean curvature ow, J.D.G 41 (1995), 215{226.
5] G. Huisken, Flow by mean curvature of convex surfaces into spheres, J.D.G 20 (1984), 237{
268.
6] G. Huisken, T. Ilmanen, Proof of the Penrose inequality, to appear.
7] K. Smoczyk, Symmetric hypersurfaces in Riemannian manifolds contracting to Lie-groups by
their mean curvature, Calc. Var. 4 (1996), 155{170.
8] S.T. Yau, On the Harnack inequalities of partial dierential equations, Comm. Anal. Geom.
2 (1994), 431{430.
ETH Zurich, Math. Department, CH-8092 Zurich, Switzerland
ksmoczyk@math.ethz.ch
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