Harnack Inequalities For Curvature Flows Depending On Mean Curvature

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 flows of compact orientable hypersurfaces inRⁿ⁺¹, where the normal velocity is given by a smooth functionf 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, wheref=Aαx^αwith two constantsAandα.

Contents

1. Introduction 103

2. Some relations and evolution equations 106

3. Homothetic solutions 107

4. The Harnack inequality 107

5. Longtime existence for some highly nonlinear flows 114

References 118

1. Introduction

Assume thatMⁿis a compact orientable surface, smoothly immersed intoRⁿ⁺¹ by a smooth family of diffeomorphismsFt:Mⁿ→Rⁿ⁺¹ that satisfy the PDE

∂

∂tF =−fν ,

where ν denotes the outward pointing unit normal and f is a smooth function depending only on the mean curvatureH of the immersed surface, e.g. forf =H we get the well-known mean curvature flow (MCF) and forf =−_H¹ we obtain the inverse mean curvature flow. Hamilton [4] proved a beautiful Harnack inequality for the MCF. In [1] Harnack inequalities were derived for convex hypersurfaces in cases wheref may depend on the full second fundamental form. The casef =−_H¹

Received September 22, 1997.

Mathematics Subject Classification. 53C21 (35K15).

Key words and phrases. Harnack, mean, curvature, flow, selfsimilar.

c

1997 State University of New York ISSN 1076-9803/97

103

has not been studied yet nor are there results for arbitrary functionsf depending only onH. The aim of this paper is to address the following questions:

I. What conditions on f guarantee that the flow 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 whichf do selfsimilar solutions exist and can we say something about the nature of these solutions?

An interesting case is the inverse mean curvature flowf =−_H¹ since it is im- portant in General Relativity [6]. There is some hope that one can generalize our results to other target manifoldsNⁿ⁺¹.

Throughout this paper we will use the standard terminology, i.e.,h·,·idenotes the euclidean inner product,g_ijdxⁱ⊗dx^jis the induced metric onMⁿ,xⁱcoordinates for Mⁿandhijdxⁱ⊗dx^j=h∇iF,∇jνidxⁱ⊗dx^j denotes the second fundamental form,

∇ the covariant derivative with respect to g_ij. Double latin indices are summed from 1 tonand we set

|A|²:=h_ijh^ij , C :=h_ikh_l^kh^il.

In this paper we will always assume thatf : Ω→Ris a smooth function defined on an open subset inRand we define

Definition. Let F : Mⁿ → Rⁿ⁺¹ be an immersion. F is called admissible, if H(x)∈Ω, ∀x∈Mⁿ.

The answer to questionIis then given by

Proposition I. Let F0 : Mⁿ → Rⁿ⁺¹ be an admissible smooth immersion of a compact orientable surface Mⁿ and assume that f⁰ : Ω → R is strictly positive.

Then the PDE

∂

∂tF =−fν (?)

F(x,0) =F0(x), ∀x∈Mⁿ

has a smooth admissible solution on a maximal time interval[0, T), T >0.

Proof. This follows from the fact that the linearization of (?) differs from the lin- earization for the mean curvature flow only by a factorf⁰ which by assumption is strictly positive. Therefore (?) is a (nonlinear) parabolic equation and the compact- ness ofMⁿ and the theory for parabolic equations imply shorttime existence.

In view of Proposition I we will always assume that f⁰ > 0. Then the main theorem can be stated as follows

Theorem 1. Assume that F0:Mⁿ→Rⁿ⁺¹ is an admissible smooth and convex immersion of an orientable compactMⁿ and that f : Ω→R is a smooth function such that for allx∈Ωwe have

f⁰>0, f⁰⁰

f⁰x²≥ax, f⁰⁰ f⁰x₀

≤0, ff⁰⁰x+ff⁰−(f⁰)²x≥0

wherea∈R is a constant. Then we can find a small positive constantdsuch that

∂

∂tf+ 2h∇f, Vi+hijVⁱV^j+cf⁰H ≥0

holds for all tangent vectors V as long as t < T, d+ (a+ 2)t > 0 and M_t stays convex, where we have setc(t) := _d+(a+2)t¹

Remark. We do have to make these assumptions onf to avoid negative terms in the evolution equation for the basic Harnack expressionZ (see below). f =αx^α satisfies the assumptions in Theorem1on Ω = (0,∞) witha=α−1. The following Propositions show that almost all functions satisfying the assumptions in Theorem1 are of this form.

Proposition IIa. Assume thatf : (0, a)→R , 0< a≤ ∞is a smooth function that smoothly extends tox= 0and satisfies all assumptions in Theorem1. Let us set if := min{l ≥ 0|f^(l)(0) 6= 0} ≤ ∞. If 0 < if < ∞ then f =Aifx^if with a positive constantA.

Proof. Since f(0) = 0 and f⁰(x)> 0 , ∀x∈Ω = (0, a) we observe thatf(x)>

0, ∀x∈Ω. By de l’Hospital’s rule we obtain

x→0lim f⁰

f x=if

and

x→0lim f⁰⁰

f⁰x=if−1 and then ^f_f⁰⁰0x₀

≤0 andf⁰≥0 imply

(1.1) f⁰⁰x≤(if−1)f⁰ , ∀x Sincef_|Ω>0 we have by assumption

ff⁰⁰x+ff⁰−(f⁰)²x=f² f⁰ f x₀

≥0 and then also

(1.2) f⁰x≥iff , ∀x

But (1.1) and (1.2) imply

0≤ff⁰⁰x+ff⁰−(f⁰)²x≤(if−1)ff⁰+ff⁰−ifff⁰= 0 and therefore

f⁰ f x₀

= 0 which implies that

f⁰ fx=i_f and after integration

f =Aifx^if

with a positive constantA(sincef⁰ >0).

Proposition IIb. Assume that f : (a,∞)→R , 0< a < ∞is a smooth func- tion that satisfies all assumptions in Theorem1 and that g(x) :=−f(¹_x)smoothly extends tox= 0. If 0< i_g<∞thenf =−Ai_gx^−ig with a positive constantA.

Proof. If we set y = ¹_x 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) = √

x does not satisfy the assumptions in Proposition IIa since it does not extend smoothly to x = 0. f(x) = x+ 1 satisfies all assumptions in Theorem 1. f(x) = lnxsatisfies almost all assumptions in Theorem 1. The only condition which is violated is thatff⁰⁰x+ff⁰−(f⁰)²x=−_x¹ <0 on Ω = (0,∞).

2. Some relations and evolution equations

By assumption we have

(2.1) ∂

∂tF =−fν

In [7] we formally derived the evolution equations for various geometric objects on Mⁿ, these are:

∂

∂tg_ij =−2fh_ij (2.2)

∂

∂tdµ=−Hfdµ (2.3)

(dµdenotes the volume form)

∂

∂tν=∇f (2.4)

∂

∂thij =∇i∇jf−fh_i^lhlj

(2.5)

∂

∂tH = ∆f +f|A|² (2.6)

∂

∂t|A|²= 2hh_ij,∇_i∇_jfi+ 2fC (2.7)

In addition we have the well-known Gauß-Weingarten-Codazzi-Mainardi equa- tions

∇_i∇_jF =−h_ijν (2.8)

∆F =g^ij∇i∇jF =−Hν (2.9)

∇ihjk=∇jhik

(2.10)

∇_iν =h_i^l∇_lF (2.11)

∇_i∇jν=∇^lhij∇lF−h_i^lhljν (2.12)

∆ν =∇H− |A|²ν (2.13)

Note that in these equations we assume thatF, ν are sets ofn+ 1 functions on Mⁿ.

We also have the Simons identity

(2.14) ∇_i∇_jH = ∆h_ij−Hh_i^lh_lj+|A|²h_ij

3. Homothetic solutions

A homothetic solutionF_t is a family of diffeomorphisms such that the surfaces given by the rescaled diffeomorphismsFe_t:= ΨF are stationary in Rⁿ⁺¹, where Ψ denotes a function depending only on timet. The assumption thatFe_trepresents a stationary surface means that the normal velocity must be zero. So we have

0 =h∂

∂tF ,e eνi= ∂

∂tΨhF, νi −fΨ

Let us definec:=−_∂t^∂ ln Ψ. Then we have shown that for a homothetic solution of (?) we have

(H.1) f =−chF, νi

Taking covariant derivatives off and using (2.11), (2.10), (2.8) we obtain with Vei :=chF,∇iFi

∇_if =−h_i^lVe_l (H.2)

∇iVej =cgij+fhij

(H.3)

∇_i∇_jf =−∇^lh_ijVe_l−ch_ij−fh_i^lh_lj (H.4)

∆f+f|A|²=−h∇H,Vei −cH (H.5)

∆Ve_j=f∇_jH+h_ji∇ⁱf (H.6)

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{Ea}a=1,...,n evolves according to

(4.1) ∂

∂tE_aⁱ =fhⁱ_jE^j_a

and we denote the coordinates of a vectorV with respect to the moving frame by y_aⁱ. The following calculations closely follow the procedure in [4]. As in this paper we have

∇^b_a:=yⁱ_a ∂

∂y^b_i (4.2)

Da :=yⁱ_a( ∂

∂xⁱ −Γ^k_ijy_b^j ∂

∂y_b^k) (4.3)

δ_ab:=g_ac∇^c_b−g_bc∇^c_a (4.4)

In addition we define

D_t:= ∂

∂t +fh_a^b∇^a_b (4.5)

Then straightforward computations give the commutator relations [D_a, D_b] =R^d_eba∇^e_d

(4.6)

[∇^a_b, D^c] =−I_b^cD^a ; [∇^a_b, D_c] =I_c^aD_b (4.7)

[δbc, D^a] =I_b^aDc−I_c^aDb

(4.8)

[∂

∂t, D_a] = D_a(fh_b^d) +D_b(fh_a^d)−D^d(fh_ab)

∇^b_d (4.9)

[Dt, Da] =D^b(fh_a^c)δbc+fh_a^bDb

(4.10)

[∆, D_a] =D^b(R^d_lab∇^l_d) +R^d_lab∇^l_dD^b (4.11)

[Dt−f⁰∆, Da] =f⁰R_a^bcdDbδcd+f⁰Dah^bch_b^dδcd+f⁰⁰ f⁰Daf∆ (4.12)

+(f−f⁰H)h_a^lD_l+f⁰h_aⁿh_n^lD_l

[D_t,∆] = (f

f⁰ −H)D_nfDⁿ+ 2fh_a^bD_bD^a+ 2h_anD^afDⁿ+D^a(D^b(fh_a^c)δ_bc) (4.13)

In the moving frame the evolution equations reduce to Dtgab= 0 ; gab=Iab

(4.14)

D_th_ab=D_aD_bf +fh_aⁿh_nb (4.15)

D_tf =f⁰(∆f+f|A|²) (4.16)

Let us now define the following tensors, where we assume thatV_ais an arbitrary tangent vector on Mⁿ and c a smooth function to be determined later and only depending on timet.

X_a :=D_af+h_a^lV_l Yab:=DaVb−fhab−cgab

Z :=D_tf+ 2hDf, Vi+h_abV^aV^b+cf⁰H Wab:=Dthab+D^lhabVl+chab

W :=Dtf+hDf, Vi+cf⁰H

By (H.1)–(H.6) these tensors vanish on homothetic solutions if we takeVa=Vea

and the inducedc. On the other hand we have DtVea = ∂

∂t(chF, DaFi) +fh_c^d∇^c_dVea

= ∂

∂t(lnc)Ve_a−chF, D_a(fν)i+fh_a^dVe_d

= ∂

∂t(lnc)Ve_a−chF, νiD_af

and therefore

(Dt−f⁰∆)Vea= ∂

∂t(lnc)Vea+f⁰h_a^bh_b^cVec

(4.17)

and on a homothetic solution

(D_t−f⁰∆)Ve_a= ∂

∂t(lnc)Ve_a−f⁰h_abD^bf (H.7)

In view of (H.7) we define

Ua:= (Dt−f⁰∆)Va− ∂

∂t(lnc)Va+f⁰habD^bf which also vanishes on a homothetic solution.

We want to calculate (D_t−f⁰∆)Z. We do this in several steps. Using (4.12) and (4.16) we obtain

(Dt−f⁰∆)Daf = f⁰⁰

f⁰Daf∆f+ (f−f⁰H)h_a^bDbf+f⁰h_aⁿh_n^lDlf +Da(ff⁰|A|²) and therefore

(Dt−f⁰∆)Daf =f⁰|A|²Daf+ 2ff⁰h^bcDahbc+f⁰h_aⁿh_n^lDlf (4.18)

+ (f −f⁰H)h_a^bD_bf+ f⁰⁰

(f⁰)²D_afD_tf Further, we use Simons identity to rewrite (4.15)

D_th_ab=D_aD_bf+fh_aⁿh_nb

=f⁰(∆hab−Hh_aⁿhnb+|A|²hab) + f⁰⁰

(f⁰)²DafDbf+fh_aⁿhnb

This gives

(4.19) (Dt−f⁰∆)hab=f⁰|A|²hab+ (f−f⁰H)h_aⁿhnb+ f⁰⁰

(f⁰)²DafDbf (4.18), (4.19) and the definition ofU give

(Dt−f⁰∆)Xa =f⁰|A|²Daf+ 2ff⁰h^bcDahbc+f⁰h_aⁿh_n^lDlf+ (f−f⁰H)h_a^bDbf + f⁰⁰

(f⁰)²DafDtf+V^b(f⁰|A|²hab+ (f −f⁰H)h_aⁿhnb+ f⁰⁰

(f⁰)²DafDbf) +h_a^b(U_b+ ∂

∂t(lnc)V_b−f⁰h_bcD^cf)−2f⁰D^ch_a^bD_cV_b

=f⁰|A|²X_a+ (f −f⁰H)h_a^bX_b+h_a^bU_b−2f⁰D_ah_bc(Y^bc+cg^bc) + ∂

∂t(lnc)(X_a−D_af) + f⁰⁰

(f⁰)²D_af(W−cf⁰H)

and finally

(Dt−f⁰∆)Xa=f⁰|A|²Xa+ (f−f⁰H)h_a^bXb+h_a^bUb−2f⁰DahbcY^bc (4.20)

+ ∂

∂t(lnc)Xa+ f⁰⁰

(f⁰)²DafW −(2c+ ∂

∂t(lnc) +cf⁰⁰ f⁰H)Daf

Next we compute

(Dt−f⁰∆)Dtf =[Dt, f⁰∆]f+Dt(ff⁰|A|²)

=f⁰[D_t,∆]f+D_tf⁰∆f+D_tf⁰f|A|²+f⁰|A|²D_tf +ff⁰D_t|A|²

=(f−f⁰H)|Df|²+ 2ff⁰habD^aD^bf + 2f⁰habD^afD^bf + f⁰⁰

(f⁰)²(Dtf)²+f⁰|A|²Dtf+ 2ff⁰h^abDthab

giving

(D_t−f⁰∆)D_tf =f⁰|A|²D_tf + 4ff⁰h^abD_th_ab−2f²f⁰C+ f⁰⁰

(f⁰)²(D_tf)² (4.21)

+ (f−f⁰H)|Df|²+ 2f⁰habD^afD^bf

Furthermore

(Dt−f⁰∆)(cf⁰H) =∂

∂t(lnc)cf⁰H+cf⁰⁰

f⁰HDtf+cDtf

−cf⁰(H(f⁰⁰∆H+f⁰⁰⁰|DH|²) +f⁰∆H+ 2f⁰⁰|DH|²)

=∂

∂t(lnc)cf⁰H+c(f⁰⁰H+f⁰)(∆f+f|A|²)

−c(f⁰⁰H+f⁰)(f⁰∆H)−cf⁰(Hf⁰⁰⁰+ 2f⁰⁰)|DH|²

=∂

∂t(lnc)cf⁰H+cf(f⁰+f⁰⁰H)|A|²+c((f⁰⁰)²H−f⁰f⁰⁰−f⁰f⁰⁰⁰H)|DH|² which gives

(4.22) (D_t−f⁰∆)(cf⁰H) = ∂

∂t(lnc)cf⁰H+cf(f⁰+f⁰⁰H)|A|²−c f⁰⁰ f⁰H₀

|Df|²

(4.21), (4.22) and (4.18) give

(D_t−f⁰∆)W =f⁰|A|²D_tf+ 4ff⁰h^abD_th_ab−2f²f⁰C+ f⁰⁰

(f⁰)²(D_tf)²+ (f−f⁰H)|Df|² + 2f⁰h_abD^afD^bf + ∂

∂t(lnc)cf⁰H+cf(f⁰+f⁰⁰H)|A|²−c f⁰⁰ f⁰H₀

|Df|² +V^a(f⁰|A|²D_af+ 2ff⁰h^bcD_ah_bc+f⁰h_aⁿh_n^lD_lf

+ (f−f⁰H)h_a^bD_bf+ f⁰⁰

(f⁰)²D_afD_tf) +D^af(Ua+ ∂

∂t(lnc)Va−f⁰habD^bf)−2f⁰D^aV^bDaDbf

=f⁰|A|²W −c((f⁰)²H−ff⁰−ff⁰⁰H)|A|²+ ∂

∂t(lnc)W− ∂

∂t(lnc)Dtf + f⁰⁰

(f⁰)²D_tfW−cf⁰⁰

f⁰HD_tf+ (f−f⁰H)hX, Dfi

+f⁰(4fh^ab−2D^aV^b)Dthab+ 2ff⁰D^aV^bh_aⁿhnb−2f²f⁰C

−c f⁰⁰ f⁰H₀

|Df|²+ 2ff⁰V^aDahbch^bc+hDf, Ui+f⁰habX^aD^bf

=f⁰|A|²W −c((f⁰)²H−ff⁰−ff⁰⁰H)|A|²+ ∂

∂t(lnc)W

−(cf⁰⁰ f⁰H+ ∂

∂t(lnc))D_tf+ f⁰⁰

(f⁰)²D_tfW + (f−f⁰H)hX, Dfi +f⁰(4fh^ab−2D^aV^b)W_ab−f⁰(4fh^ab−2D^aV^b)(D^lh_abV_l+ch_ab) + 2ff⁰D^aV^bh_aⁿhnb−2f²f⁰C−c f⁰⁰

f⁰H₀

|Df|² + 2ff⁰V^aD_ah_bch^bc+f⁰hh_ab, D_afX_bi+hDf, Ui

and after rearranging terms we conclude

(Dt−f⁰∆)W =f⁰|A|²W +f⁰(4fh^ab−2D^aV^b)Wab+f⁰h^abXaDbf+hDf, Ui (4.23)

+ 2f⁰(fh_aⁿh_nb+D^lh_abV_l)Y^ab+ (2c+ ∂

∂t(lnc))W

−c((f⁰)²H−ff⁰−ff⁰⁰H)|A|²−(cf⁰⁰ f⁰H+ ∂

∂t(lnc) + 2c)Dtf + f⁰⁰

(f⁰)²D_tfW+ (f−f⁰H)hX, Dfi −c f⁰⁰ f⁰H₀

|Df|²+ 2cf⁰h_abY^ab

Eventually the evolution equation forZ is given by

(Dt−f⁰∆)Z =f⁰|A|²W+f⁰(4fh^ab−2D^aV^b)Wab+f⁰h^abXaDbf+hDf, Ui + 2f⁰(fh_aⁿhnb+D^lhabVl)Y^ab+ (2c+ ∂

∂t(lnc))W

−c((f⁰)²H−ff⁰−ff⁰⁰H)|A|²−(cf⁰⁰ f⁰H+ ∂

∂t(lnc) + 2c)Dtf + f⁰⁰

(f⁰)²DtfW + (f −f⁰H)hX, Dfi −c f⁰⁰ f⁰H₀

|Df|²+ 2cf⁰habY^ab +V^a(f⁰|A|²X_a+ (f−f⁰H)h_a^bX_b+h_a^bU_b−2f⁰D_ah_bcY^bc

+ ∂

∂t(lnc)X_a+ f⁰⁰

(f⁰)²D_afW−(2c+ ∂

∂t(lnc) +cf⁰⁰

f⁰H)D_af) +X^a(Ua+ ∂

∂t(lnc)Va−f⁰habD^bf)−2f⁰DaVb(W^ab+h^anY_n^b)

=f⁰|A|²Z+ 2hX, Ui −4f⁰(Y^ab+cg^ab)Wab−2f⁰h^anY_n^b(Yab+cgab) + (f −f⁰H)|X|²+ f⁰⁰

(f⁰)²W(Dtf+hV, Dfi)−(cf⁰⁰ f⁰H+ ∂

∂t(lnc) + 2c)(Dtf+hDf, Vi) + 2∂

∂t(lnc)hX, Vi+ (2c+ ∂

∂t(lnc))W−c((f⁰)²H−ff⁰−ff⁰⁰H)|A|²

−c f⁰⁰ f⁰H₀

|Df|²+ 2cf⁰habY^ab

=f⁰|A|²Z+ 2hX, Ui −2f⁰h_aⁿY_nbY^ab−4f⁰W_abY^ab−4cW + (f −f⁰H)|X|²+ f⁰⁰

(f⁰)²W(W−cf⁰H)−(cf⁰⁰ f⁰H+ ∂

∂t(lnc) + 2c)(W−cf⁰H) + 2∂

∂t(lnc)hX, Vi+ (2c+ ∂

∂t(lnc))W−c((f⁰)²H−ff⁰−ff⁰⁰H)|A|²−c f⁰⁰ f⁰H₀

|Df|² So we finally arrive at

(Dt−f⁰∆)Z = f⁰|A|²+ f⁰⁰

(f⁰)²(Z−2hX, Vi)−2(cf⁰⁰

f⁰H+ 2c) Z (4.24)

+ 2hX, Ui+ (f −f⁰H)|X|²+ f⁰⁰

(f⁰)²hX, Vi²+ 2(cf⁰⁰ f⁰H+ ∂

∂t(lnc) + 2c)hX, Vi

−2f⁰h_aⁿY_nbY^ab−4f⁰W_abY^ab+ (cf⁰⁰ f⁰H+ ∂

∂t(lnc) + 2c)cf⁰H +c(ff⁰⁰H+ff⁰−(f⁰)²H)|A|²−c f⁰⁰

f⁰H₀

|Df|² Now we are able to prove Theorem1.

Proof of Theorem 1. From (4.24) and the assumptions in Theorem1 we con- clude, withc:=_d+(a+2)t¹

(D_t−f⁰∆)Z≥ f⁰|A|²+ f⁰⁰

(f⁰)²(Z−2hX, Vi)−2(cf⁰⁰

f⁰H+ 2c)

Z−2f⁰h_aⁿY_nbY^ab−4f⁰W_abY^ab + 2hX, Ui+ (f−f⁰H)|X|²+ f⁰⁰

(f⁰)²hX, Vi²+ 2(cf⁰⁰ f⁰H+ ∂

∂t(lnc) + 2c)hX, Vi

Now choosedso small such thatZ > >0 fort= 0 and for all tangent vectorsV. This is possible since the initial surface is convex. On any compact time interval [0, t₀] witht₀< T we can therefore estimate

(D_t−f⁰∆)Z≥ −bZ−2f⁰h_aⁿY_nbY^ab−4f⁰W_abY^ab + 2hX, Ui+ (f−f⁰H)|X|²+ f⁰⁰

(f⁰)²hX, Vi²+ 2(cf⁰⁰ f⁰H+ ∂

∂t(lnc) + 2c)hX, Vi with a large constant b depending on t₀. Additionally fort > 0 and an arbitrary positive constantδ

(Dt−f⁰∆)(e^btZ+δt)>e^bt −2f⁰h_aⁿYnbY^ab−4f⁰WabY^ab+ 2hX, Ui+ (f−f⁰H)|X|² + f⁰⁰

(f⁰)²hX, Vi²+ 2(cf⁰⁰ f⁰H+ ∂

∂t(lnc) + 2c)hX, Vi

Ift1≤t0would be the first time wheree^btZ+δtwould become zero at some point x∈Mⁿ and for some tangent vectorV then we must haveXa = 0 since it is the first variation with respect toV and we can also extendV in spacetime such that Yab= 0. This implies a contradiction and thereforee^btZ+δt >0, ∀t≤t0. Sinceδ andt₀ are arbitrary we concludeZ≥0 whenevert < T andd+ (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 ∈ Ω and x > 0, then the sphere of constant radius ⁿ_x and with constant mean curvatureH =xgives always rise to a selfsimilar solution for a short time.

For any subsetA⊂Mⁿ let us define

H(A) :={x∈Ω|∃p∈A:H(p) =x}

and P_t:={p∈M_tⁿ|∇H 6= 0}

The answer to questionIIIis then given by

Proposition IIIa. If P0 6= ∅ and Ft: Mⁿ → Rⁿ⁺¹ is a selfsimilar solution of (?) for a compact connected Mⁿ, then we have f = Aαx^α , ∀x ∈ H(Mt) with nonvanishing constantsA andα.

Proof. SinceFtis selfsimilar we havePt=P0, ∀t∈[0, T). SinceXa, W vanish on selfsimilar solutions so must their time derivatives. From (4.20) we conclude that onH(P_t) and by continuity also onH(P_t) we must havec^f_f⁰⁰0x+ 2c+_∂t^∂(lnc) = 0.

SinceP₀6=∅andMⁿ is connected we haveH(Mⁿ)⊂H(P_t) and we derive cf⁰⁰

f⁰x+ 2c+ ∂

∂t(lnc) = 0

for allx∈H(M_tⁿ). Since f⁰ >0 there can be at most one point x∈Ω such that f(x) = 0. At all other points we have

ff⁰⁰x+ff⁰−(f⁰)²x=f² f⁰ f x₀

Then (4.23) implies that also

f⁰ f x₀

= 0

in all points x∈H(Mt) wheref(x)6= 0. But after integration we obtain that at these points

f =Aαx^α

with constantsA, α (nonvanishing since we must havef⁰>0) and again by conti-

nuity this also holds on all ofH(M_t).

We observe that by (2.9)

∆|F|²= 2(n−HhF, νi) and with (H.1) we obtain on a homothetic solution

(H.8) ∆|F|²= 2(n−fH

c ) This implies

Proposition IIIb. IfP₀=∅andMⁿ is a compact orientable selfsimilar solution of (?), thenM_t is a sphere of radius _Hⁿ. Iff =−_H¹, then any compact orientable homothetic solution is a sphere of radius _Hⁿ.

Proof. P0 =∅ and (H.8) imply ∆|F|² =const. By the assumptions on Mⁿ this constant must be zero and then consequently|F|²=const. This implies that Mt

is a sphere.

5. Longtime existence for some highly nonlinear flows

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 flows. To be precise

Theorem 2. Assume that f : (0,∞) → R is a smooth negative function that satisfies the assumptions in Theorem 1 with a >−2 and that limx→0f(x) =−∞

andF0:M²→R³ is a smooth convex immersion of an orientable compact surface M². Then(?)has a smooth immortal solution and we have

t→∞lim H = 0, lim

t→∞|Ft|²=∞

Remark. The functions f = αx^α with −1 < α < 0 satisfy the assumptions in Theorem 2 on Ω = (0,∞). These speed functions are not homogenous of degree one and are not included in the class of functions considered in [2]

To prove Theorem2 we need some lemmas that are interesting on their own.

Lemma 1. Assumef⁰>0and thatF₀:Mⁿ→Rⁿ⁺¹is a smooth immersion of an orientable compact surface and that onM₀=F₀(Mⁿ)we havef²≥min_M0f²>0.

Then this is also true on the maximal time interval[0, T)where a smooth solution of(?)exists.

Proof. The evolution equation forf² is given by

(Dt−f⁰∆)f²=f⁰ −2|∇f|²+ 2f²|A|²

and the result follows from the parabolic maximum principle.

Corollary 1. With the assumptions in Lemma1 we have that for negative f H ≤max

M0 H and for positivef

H ≥min

M0 H We can do even better

Lemma 2. Assume Ω = (0,∞), f <0, f⁰>0and thatF₀:Mⁿ →Rⁿ⁺¹ is an admissible immersion of an orientable compact manifoldMⁿ. Then we can find a positivesuch that on[0, T)

maxMt H ≤ maxM0H 1 +tmax_M0H

Proof. Since Ft is admissible on [0, T) we must always have H >0. At a point where maxMtH is attained we have ∆H ≤0 and∇H = 0. The evolution equation forH (2.6), Lemma1 and the fact that|A|²≥^H_n² give us

∂

∂tmax

Mt H ≤ −(max

Mt H)²

with= ^−max_n^M0f and after integration we obtain the result.

(2.7) and Simons identity give

(5.1) (Dt−f⁰∆)|A|²=−2f⁰|∇A|²+ 2f⁰|A|⁴+ 2(f−f⁰H)C+ 2f⁰⁰hij∇ⁱH∇^jH LetR =H²− |A|² be the Scalar curvature of Mt. Now we turn our attention to the case wheren= 2. In this case we can decomposeC into

C= H

2(3|A|²−H²) Then (5.1) and (2.6) imply that forn= 2

(D_t−f⁰∆)R=2f⁰(|∇A|²− |∇H|²) + 2f⁰⁰(H|∇H|²−h_ij∇ⁱH∇^jH) (5.2)

−(2f⁰|A|²+ (f−f⁰H)H)R

Lemma 3. Under the assumptions in Theorem 2 we have R >0, ∀t∈[0, T)

Proof. Lett₀≤T be the maximal time such thatR >0 on all ofM_tfort∈[0, t₀), i.e., the maximal time for whichM_tstays convex. Ift₀=T we are done. So assume that t₀ < T. SinceM_t is admissible for t∈[0, t₀] we must haveH > 0 on [0, t₀].

By definition oft₀we also haveR >0 on [0, t₀) andR≥0 on [0, t₀]. SinceM and [0, t₀] are both compact andR≥0 we can find a constantc such that

(5.3) (Dt−f⁰∆)R≥2f⁰(|∇A|²− |∇H|²) + 2f⁰⁰(H|∇H|²−hij∇ⁱH∇^jH)−cR Now assume thatx is any point on M_t where the minimum of R is attained.

At this point we must havef⁰∆R≥0 and∇R= 0. Let αand β denote the two principal curvatures in a neighborhood ofx. We have

0 =∇R= 2α∇β+ 2β∇α

SinceH >0 andα, β≥0 we must either haveα >0 orβ >0. Let us assume that α >0. First we compute that

H|∇H|²−hij∇ⁱH∇^jH =α|∇2H|²+β|∇1H|²≥0,

where we choose normal coordinates such thath_ij = diag(α, β). Since f⁰⁰x≥af⁰ andH >0, a >−2 we can estimate

2f⁰⁰(H|∇H|²−hij∇ⁱH∇^jH)≥ −4f⁰

H(α|∇2H|²+β|∇1H|²) On the other hand Codazzi’s equation gives us that

|∇A|²=|∇1α|²+|∇2β|²+ 3|∇2α|²+ 3|∇1β|²

Combining the last two statements we see that at a point where∇R= 0 we must always have

2f⁰(|∇A|²− |∇H|²) + 2f⁰⁰(H|∇H|²−hij∇ⁱH∇^jH)≥0 and consequently at any point where the minimum ofRis attained

DtR≥ −cR which implies that

minMt R≥(min

M0 R)e^−ct>0, ∀t∈[0, t₀]

This proves thatt₀=T.

We come to the proof of Theorem2