classification of compact ancient solutions to the ricci … · theorem 1.2. let vbe an ancient...

CLASSIFICATION OF COMPACT ANCIENT SOLUTIONSTO THE RICCI FLOW ON SURFACES

PANAGIOTA DASKALOPOULOS∗, RICHARD HAMILTON,

AND NATASA SESUM∗∗

Abstract. We consider an ancient solution g(·, t) of the Ricci flow on

a compact surface that exists for t ∈ (−∞, T ) and becomes spherical at

time t = T . We prove that the metric g(·, t) is either a family of con-

tracting spheres, which is a type I ancient solution, or a King-Rosenau

solution, which is a type II ancient solution.

1. Introduction

We consider an ancient solution of the Ricci flow

∂gij∂t

= −2Rij (1.1)

on a compact two-dimensional surface that exists for time t ∈ (−∞, T ). In

two dimensions we have Rij = 12Rgij , where R is the scalar curvature of the

surface. Moreover, on an ancient non-flat solution we have R > 0. Such a

solution becomes singular at some finite time T , namely its scalar curvature

is blowing up at t = T . It is well known ([1], [5]) that the surface also

becomes extinct at T and it becomes spherical, which means that after a

normalization, the normalized flow converges to a spherical metric.

We can parametrize the Ricci flow by the limiting sphere at time T , that

is, we can write

g(·, t) = u(·, t) ds2p.

The spherical metric can be written as

ds2p = dψ2 + cos2 ψ dθ2. (1.2)

where ψ, θ denote the global coordinates on the sphere. An easy computation

shows that (1.1) is equivalent to the following evolution equation for the

∗ : Partially supported by NSF grant 0604657.

1

2 PANAGIOTA DASKALOPOULOS∗, RICHARD HAMILTON, AND NATASA SESUM∗∗

conformal factor u(·, t), namely

ut = ∆S2 log u− 2, on S2 × (−∞, T ) (1.3)

where ∆S2 denotes the Laplacian on the sphere. Let us recall, for future

references, that the only nonzero Christoffel symbols for the spherical metric

(1.2) are

Γ212 = Γ2

21 = − tanψ, Γ122 =

sin 2ψ2

, Γ222 = − tanψ.

It follows that for any function f on the sphere we have

∆S2f = fψψ − tanψ fψ + sec2 ψ (fθθ + tanψfθ)

which, in the case of a radially symmetric function f = f(ψ), becomes only

∆S2f = fψψ − tanψ fψ.

We next introduce Mercator’s projection, where

coshx = secψ and sinhx = tanψ (1.4)

to project the sphere S2 onto the cylinder, where (x, θ) denote the cylindrical

coordinates in R2. We can also express

g(·, t) = U(·, t) ds2, where ds2 = dx2 + dθ2

where U(x, θ, t) is a conformal factor on the cylinder. A simple computation

shows that U satisfies the logarithmic fast-diffusion equation

Ut = ∆c logU, on R× [0, 2π]× (−∞, T ) (1.5)

where ∆c is the cylindrical Laplacian, that is

∆cf = fxx + fθθ

for a function f defined on R× [0, 2π]. It follows that U is given in terms of

u by

U(x, θ, t) = u(ψ, θ, t) cos2 ψ , cosψ = (coshx)−1. (1.6)

We will assume, throughout this paper, that g = u ds2p is an ancient

solution to the Ricci flow (1.3) on the sphere which becomes extinct at time

T = 0.

It is natural to consider the pressure function v = u−1 which evolves by

vt = v2 (∆S2 log v + 2), on S2 × (−∞, 0) (1.7)

CLASSIFICATION OF COMPACT ANCIENT SOLUTIONSTO THE RICCI FLOW ON SURFACES3

or, after expanding the laplacian of log v,

vt = v∆S2v − |∇S2v|2 + 2v2, on S2 × (−∞, 0). (1.8)

Definition 1.1. We will say that an ancient solution to the Ricci flow (1.1)

on a compact surface M is of type I, if it satisfies

lim supt→−∞

(|t| maxM

R(·, t)) <∞.

A solution which is not of type I, will be called of type II.

Explicit examples of ancient solutions to the Ricci flow in two dimensions

are:

i. The contracting spheres

They are described on S2 by a pressure vS that is given by

vS(ψ, t) =1

2(−t)(1.9)

and they are examples of ancient type I shrinking Ricci solitons.

ii. The King-Rosenau solutions

They were discovered independently by J.R. King ([10], [11]) and P.

Rosenau ([14]) and are described on S2 by a pressure vR that has the

form

vR(ψ, t) = a(t)− b(t) sin2 ψ (1.10)

with a(t) = −µ coth(2µt), b(t) = −µ tanh(2µt), for some µ > 0. These

solutions are particularly interesting because they are not solitons. We

can visualize them as two cigars ”glued” together to form a compact

solution to the Ricci flow. They are type II ancient solutions.

Our goal in this paper is to prove the following classification result:

Theorem 1.2. Let v be an ancient compact solution to the Ricci flow (1.8).

Then, the solution v is either one of the contracting spheres or one of the

King-Rosenau solutions.

Remark 1.3. The classification of two-dimensional, complete, non-compact

ancient solutions of the Ricci flow was recently given in [4] (see also in [7],

[15]). The result in Theorem 1.2 together with the results in [4] and [15]

provide a complete classification of all ancient two-dimensional complete

solutions to the Ricci flow.


The outline of the paper is as follows:

i. In section 2 we will show a’ priori derivative estimates on any ancient

solution v of (1.8), which hold uniformly in time, up to t = −∞. These

estimates will play a crucial role throughout the rest of the paper.

ii. In section 3 we will introduce a suitable Lyapunov functional for our flow

and we will use it to show that the solution v(·, t) of (1.8) converges, as

t→ −∞, in the C1,α norm, to a steady state v.

iii. Section 4 will be devoted to the classification of all backward limits v.

We will show that there is a parametrization of the flow by a sphere in

which v = C cos2 ψ, for some C ≥ 0 (ψ, θ are the global coordinates on

S2).

iv. In section 5 we will show that if C = 0, then the solution v is one of the

contracting spheres.

v. In section 6 we will show that all ancient compact solutions of the Ricci

flow in two dimensions are radially symmetric.

vi. Finally, in section 7 we will show that any radially symmetric solution

v of (1.8) for which C > 0 must be one of the King-Rosenau solutions.

2. A’ priori estimates

We will assume, throughout this section, that v is an ancient solution of

the Ricci flow (1.8) on S2 × (−∞, 0) which becomes extinct at T = 0. We

fix t0 < 0. We will establish a’priori derivative estimates on v which hold

uniformly on S2 × (−∞, t0]. We will denote by C various constants which

may vary from line to line but they are always independent of time t.

Since our solution is ancient, the scalar curvature R = vt/v is strictly

positive. This in particular implies that vt > 0. Hence, we have the bound

v(·, t) ≤ C(t0), on (−∞, t0) (2.1)

for any t0 < 0. However, the backward limit v = limt→−∞ v(·, t), which

exists because of the inequality vt > 0, may vanish at some points on S2.

This actually happens on our model, the King-Rosenau solution. As a result,

the equation (1.8) fails to be uniformly parabolic near those points, as t→−∞, and the standard parabolic and elliptic derivative estimates fail as well.

Nevertheless, it is essential for our classification result, to establish a priori

derivative estimates which hold uniformly in time, as t→ −∞.


We recall that on our ancient solution, the Harnack estimate for the scalar

curvature shown in [5] takes the form

Rt ≥|∇R|2gR

which in particular implies that Rt ≥ 0. Hence, we also have

R(·, t) ≤ C, on (−∞, t0] (2.2)

because R(·, t) ≤ R(·, t0). We also know that the evolution equation for R

is

Rt = ∆gR+R2.

If we express g = v−1 ds2p, in which case ∆g = v∆S2 and |∇ · |2g = v|∇ · |2S2 ,

we can rewrite the Harnack estimate for R as

v∆S2R+R2 ≥ v |∇S2R|2

R

or, equivalently

∆S2R+R2

v≥ |∇S2R|2

R. (2.3)

The pressure v satisfies the elliptic equation

v∆S2v − |∇S2v|2 + 2v2 = Rv. (2.4)

We will next use this equation and the bounds (2.1) and (2.2) to establish

uniform first and second order derivative estimates on v.

Lemma 2.1. There exists a uniform constant C, independent of time, so

that

supS2

(|∆S2v|+

|∇S2v|2

v

)(·, t) ≤ C, for all t ≤ t0 < 0.

Proof. To simplify the notation we will set ∆ := ∆S2 and ∇ := ∇S2 . We

first differentiate (2.4) twice to compute the equation for ∆v. After some

direct calculations we find that

∆(∆v) +(∆v)2 − 2 |∇2v|2

v+ 4∆v + 2

|∇v|2

v=

∆(Rv)v

which implies that

∆(∆v + 4v) ≥ −2|∇v|2

v+ ∆R+ 2

∇R · ∇vv

+R∆vv

(2.5)

since by the trace formula we have

(∆v)2 ≤ 2|∇2v|2.


By (2.4) we also have

∆v =|∇v|2

v− 2v +R (2.6)

and if use it to replace ∆v from the last term on the right hand side of (2.5)

we obtain

∆(∆v + 4v) ≥ −2|∇v|2

v+ ∆R+ 2

∇R · ∇vv

+R

v

(|∇v|2

v− 2v +R

)= −2

|∇v|2

v+ (∆R+

R2

v) + 2

∇R · ∇vv

+R|∇v|2

v2− 2R.

By the Harnack estimate (2.3) we get

∆(∆v + 4v) ≥ −2|∇v|2

v+|∇R|2

R+ 2∇R · ∇v

v+R|∇v|2

v2− 2R

= −2|∇v|2

v+

1R

(|∇R|2 + 2∇R ·R∇v

v+R2 |∇v|2

v2

)− 2R

= −2|∇v|2

v+

1R

∣∣∣∣∇R+R∇vv

∣∣∣∣2 − 2R.

Since R > 0 we conclude the estimate

∆(∆v + 4v) ≥ −2|∇v|2

v− 2R. (2.7)

If we multiply (2.6) by M = 2 and add it to (2.7) we get

∆(∆v + 6 v) ≥ −4v ≥ −C

for a uniform constant C (independent of time). By (2.6) we also have

∆v ≥ −2v +R > −C (2.8)

and therefore

X := ∆v + C + 6v > 0

and

∆X ≥ −C. (2.9)

Standard Moser iteration applied to (2.9) yields to the bound

supX ≤ C1

∫S2

X da+ C2. (2.10)

Observe that∫S2

X da =∫S2

(∆v + C + 6v) da =∫S2

(C + 6v) da ≤ C.

The last estimate combined with (2.10) yields to the bound

∆v ≤ C.


This together with (2.8) imply

supS2

|∆v|(·, t) ≤ C, for all t ≤ t0 < 0 (2.11)

for C is a uniform constant. Since

|∇v|2

v= ∆v + 2v −R

the estimate (2.11) and the upper bound R ≤ C, readily imply the bound

supS2

|∇v|2

v≤ C, for all t ∈ (−∞, t0]. (2.12)

�

As a consequence of the previous lemma we have:

Corollary 2.2. For any p ≥ 1, we have

‖v(·, t)‖W 2,p(S2) ≤ C(p), for all t ∈ (−∞, t0]. (2.13)

It follows that for any α < 1, we have

‖v(·, t)‖C1,α(S2) ≤ C(α), for all t ∈ (−∞, t0]. (2.14)

Proof. Since ∆S2v = f in S2, with f ∈ L∞, standard W 2,p estimates for

Laplace’s equation imply that v ∈ W 2,p(S2) for all p ≥ 1. Hence, (2.14)

follows by the Sobolev embedding theorem. �

We will now use the estimates proven above to improve the regularity of

the function v.

Lemma 2.3. For every 0 < α < 1, there is a uniform constant C(α) so

that

‖|∇S2v(·, t)|2‖C1,α(S2) ≤ C(α) for all t ≤ t0 < 0. (2.15)

Proof. To simplify the notation we will set ∆ := ∆S2 and ∇ := ∇S2 . A

direct computation shows that |∇v|2 satisfies the evolution equation

∂

∂t|∇v|2 = v∆|∇v|2 − 2v |∇2v|2 + 2v|∇v|2 + 2|∇v|2 ∆v − 2∇(|∇v|2) · ∇v.

On the other hand, differentiating the equation vt = Rv gives

∂

∂t|∇v|2 = 2∇(Rv) · ∇v.

Combining the above gives that

∆(|∇v|2) = f (2.16)


with f given by

f = 2|∇2v|2−2|∇v|2−2|∇v|2

v∆v+

2∇(|∇v|2) · ∇vv

+2∇(Rv) · ∇v

v. (2.17)

We will show that for every p ≥ 1, we have

‖f(·, t)‖Lp(S2) ≤ C(p), for all t ≤ t0 < 0 (2.18)

with C(p) independent of t. We will denote in the sequel by C(p) various

constants that are independent of t. We begin by recalling that by (2.13),

we have

||∇2v(·, t)||Lp ≤ C(p), for all t ≤ t0 < 0.

Also, by Lemma 2.1, we have

‖2 |∇v|2

v∆v‖Lp(S2) + ‖∇v|2‖Lp(S2) ≤ C(p), for all t ≤ t0 < 0.

Since

|∇(|∇v|2) · ∇vv

| ≤ |∇2v| |∇v|2

v

by the previous estimates, we have

‖∇|(∇v|2) · ∇vv

‖Lp(S2) ≤ C(p), for all t ≤ t0 < 0.

We also have

|∇(Rv) · ∇vv

| ≤ R|∇v|2

v+ |∇R| |∇v|

≤ C + (√v|∇R|)

(|∇v|√v

)≤ C

for all t ≤ t0 < 0, since√v |∇R| = |∇R|g(t) ≤ C. We can now conclude

that (2.18) holds, for p ≥ 1. Standard elliptic regularity estimates applied

on (2.16) imply the bound

‖|∇v|2‖W 2,p ≤ C(p), for all t ≤ t0 < 0.

Since the previous estimate holds for any p ≥ 1, by the Sobolev embedding

theorem, we conclude (2.15). �

Lemma 2.4. For every 0 < α < 1, there is a uniform constant C(α), so

that

‖v∇3S2v‖C0,α(S2) + ‖

√v∇2

S2v‖C0,α(S2) ≤ C(α)

for all t ≤ t0 < 0.


Proof. To simplify the notation we will set ∆ := ∆S2 and ∇ := ∇S2 . To

prove the estimate on ‖√v∇2v‖C0,α(S2) we observe that we can rewrite (1.8)

in the form

∆v3/2 =9|∇v|2

4√v− 3v3/2 +

32R√v. (2.19)

We claim that the right hand side of the previous identity has uniformly in

time bounded C0,α-norm, for any 0 < α < 1. To see that, observe that for

every p ≥ 1, we have

‖∇(|∇v|2√

v)‖Lp(S2) ≤ ‖∇2v‖Lp(S2) ‖

|∇v|2√v‖L∞(S2) + ‖|∇v|

2

v‖3/2L∞(S2)

≤ C(p)

(2.20)

and also

‖∇(R√v)‖L∞(S2) ≤ C

since

|∇(R√v)| ≤ |∇R|

√v +R

|∇v|2√v≤ |∇R|g(t) + C ≤ C.

All of the above inequalities hold uniformly on t ≤ t0 < 0. By the Sobolev

embedding theorem we conclude that the right hand side of (2.19) has uni-

formly bounded C0,α norm, for any α < 1. Standard elliptic regularity

theory applied to (2.19) implies that

‖v3/2‖C2,α(S2) ≤ C(α)

which in particular yields to the estimate

‖√v∇2v‖C0,α(S2) ≤ C(α), for all t ≤ t0 < 0

since

∇∇v3/2 =34|∇v|2√

v+

32√v∇2v

and the first term on the right hand side is in C0,α by (2.20).

To prove the second estimate, we now rewrite (1.8) as

∆v2 = 4|∇v|2 − 4v2 + 2Rv. (2.21)

Lemma 2.3 implies that 4 |∇v|2 − 4v2 has uniformly in time bounded C1,α

norm. We claim the same is true for the term Rv. To see that, we differen-

tiate it twice and use the inequality

|∇2(Rv)| ≤ |∇2v|R+ |∇2R| v + 2|∇R| |∇v|.


By Lemmas 2.1 and 2.3 and the bounds

v |∇2R| = |∇2R|g ≤ C,√v |∇R| = |∇R|g ≤ C, R ≤ C

we conclude that for all p ≥ 1, we have

‖∇2(Rv)‖Lp(S2) ≤ C(p), for all t ≤ t0 < 0

The Sobolev embedding theorem now implies that ‖Rv‖C1,α(S2) is uniformly

bounded in time, for every α < 1. Standard elliptic theory applied to (2.21)

yields to the bound

‖v2‖C3,α ≤ C(α), for all t ≤ t0 < 0,

which in particular implies that

‖v∇3v‖C0,α(S2) ≤ C(α).

�

Recall that in the case of a spherical metric, given by ds2p = dψ2 +

cos2 ψ dθ2, we have

∇1∇1f =∂2f

∂ψ2− Γk11

∂f

∂ψ= fψψ

and

∇2∇2f =∂2f

∂θ2− Γk22

∂f

∂ψ= fθθ − fψ tanψ + fθ tanψ.

Using the above notation, we will next show the following bound.

Proposition 2.5. There is a uniform constant C, so that

|∇1∇1v|+ sec2 ψ |∇2∇2v| ≤ C


Proof. To simplify the notation we set ∆ := ∆S2 and ∇ := ∇S2 . We begin

by differentiating the equation

v∆v − |∇v|2 + 2v2 = Rv (2.22)

to get

v∇i ∆v +∇iv∆v − 2∇i∇jv∇jv + 4v∇iv = ∇i(Rv). (2.23)

Since

∇i∆v = ∆∇iv −12∇iv


we obtain that

∆∇iv =1v

(∇i(Rv)−∇iv∆v + 2∇i∇jv∇jv −

72v∇iv

). (2.24)

If we differentiate (2.23), using that

∇k∇i∆v = ∆(∇k∇iv)− 2(∇i∇kv −

12

∆v gki

)we obtain that

∇k∇i(Rv) = v∆(∇k∇iv)− 2v∇k∇iv + v∆v gik

+∇kv∇i∆v +∇k∇iv∆v +∇iv∇k∆v

− 2∇k∇i∇jv∇jv − 2∇i∇jv∇j∇kv + 4∇kv∇iv + 4v∇i∇kv.(2.25)

Take i = k = 1 and consider the maximum point of ∇1∇1v. We may assume

that at the maximum point the matrix ∇2v is diagonal. Hence, at that point

we have ∆(∇1∇1v) ≤ 0 and

∇1∇jv∇j∇1v = (∇1∇1v)2.

By (2.25) we then obtain

2(∇1∇1v)2 ≤ |∇1∇1(Rv)| − 2v∇1∇1v + v∆v g11 + 2∇1v∇1∆v

+∇1∇1v∆v − 2∇1∇1∇jv∇jv + 4|∇1v|2 + 4v∇1∇1v.(2.26)

We will now estimate the terms on the right hand side of (2.26) at the

maximum point of ∇1∇1v. We begin by estimating the term ∇1v∇1∆v. By

switching the order of differentiation, using also (2.24), we get

|∇1v∇1∆v| = |∇1v∆∇1v −12

(∇1v)2|

≤ C +|∇v|v|∇1(Rv)−∇1v∆v + 2∇1∇jv∇jv −

72v∇1v|

= C +|∇v|v|∇1(Rv)−∇1v∆v + 2∇1∇1v∇1v|

where we have used that |∇v| ≤ C for all t ≤ t0 < 0 (c.f. Lemma 2.1). To

estimate the terms above, we begin by observing that

|∇v|v|∇1∇1v∇1v| ≤ |∇1∇1v|

|∇v|2

v≤ C |∇1∇1v| ≤ Cε + ε |∇1∇1v|2

by the interpolation inequality and the estimate |∇v|2

v ≤ C shown in Lemma

2.1. In addition, we have

|∇v|v|∇1(Rv)| ≤ |∇v|√

v(√v|∇R|) +R

|∇v|2

v≤ C


and|∇v|v|∇1v∆v| ≤ |∆v| |∇v|

2

v≤ C

where we have used that R+|∇R|g+|∆v| ≤ C, for all t ≤ t0 < 0. Combining

the above yields to the bound

|∇1v∇1∆v| ≤ ε |∇1∇1|2 + Cε, for all t ≤ t0 < 0 (2.27)

where Cε is a uniform positive constant and ε > 0 is a small real number.

Moreover, after switching the order of differentiation, we get

|∇jv∇1∇1∇jv| = |∇jv||∇j∇1∇1v|+12

(∇jv)2

≤ 12|∇v|2 ≤ C (2.28)

since ∇j∇1∇1v = 0, for j ∈ {1, 2}, at the maximum point of ∇1∇1v. Also,

|∇1∇1(Rv)| ≤ v |∇1∇1R|+R |∇1∇1v|+ 2 |∇1R||∇1v|

≤ |∇2R|g + C |∇1∇1v|+ |∇R|g|∇v|√v

≤ C + ε |∇1∇1v|2. (2.29)

Combining the estimates (2.27), (2.28), (2.29) with (2.26) and choosing ε > 0

sufficiently small, we finally obtain the bound

2(∇1∇1v)2 ≤ (∇1∇1v)2 + C

which implies the estimate

|∇1∇1v| ≤ C, for all t ≤ t0 < 0. (2.30)

Since ∆v = gij∇i∇jv = ∇1∇1v+ sec2 ψ∇2∇2v and |∆v| ≤ C, by (2.30) we

also obtain the estimate

sec2 ψ |∇2∇2v| ≤ C, for all t ≤ t0 < 0.

This finishes the proof of the Proposition. �

Lemma 2.6. There is a uniform constant C, independent of time, such that

supS2

|vθ sec2 ψ| ≤ C, for all t ≤ t0 < 0.


Proof. To simplify the notation we will set ∆ := ∆S2 and ∇ := ∇S2 . Recall

that ∆v = vψψ − tanψ vψ + sec2 ψ (vθθ + tanψ vθ). Hence, we can rewrite

equation (2.22) as

| secψ (vθθ + tanψ vθ)| = | cosψ (|∇v|2

v− 2v +R)− cosψ (vψψ − tanψ vψ)|.

By Proposition 2.5 and our earlier estimates the right hand side of the pre-

vious identity is uniformly bounded in time. Fix ψ ∈ (−π2 ,

π2 ) and consider

the maximum (minimum) of vθ(ψ, ·, t) as a function of θ. At such a point

we have vθθ = 0 and therefore from the previous equation we get the bound

| tanψ secψ vθ| ≤ C, for all t ≤ t0 < 0.

We conclude that for δ > 0 small, we have

|vθ sec2 ψ| ≤ C, for (θ, ψ) ∈ [0, 2π]× (−π2,−δ) ∪ (δ,

π

2)

for all t ≤ t0 < 0. Notice that, due to our earlier estimates, we immediately

have that

|vθ sec2 ψ| ≤ C, for (θ, ψ) ∈ [0, 2π]× (−δ, δ)

for all t ≤ t0 < 0. This finishes the proof of our estimate. �

Lemma 2.7. There is a uniform constant C so that

|vψ tanψ| ≤ C, for all t ≤ t0 < 0.

Proof. To simplify the notation we will set ∆ := ∆S2 and ∇ := ∇S2 . We

start with the following two claims which justify the well posedness of certain

quantities at each time slice t < 0.

Claim 2.8. For every t < 0 there is a constant C(t) so that supS2 |vψ tanψ| ≤C(t).

Proof. If we integrate

∆v = vψψ − vψ tanψ + sec2 ψ (vθθ + vθ tanψ) (2.31)

in θ ∈ [0, 2π], using the uniform in time bound |∆v| ≤ C, we get∣∣∣∣∫ 2π

0vψψ dθ − tanψ

∫ 2π

0vψ dθ

∣∣∣∣ ≤ C. (2.32)

Consider

H(ψ, t) := secψ∫ 2π

0vψ dθ


a radial function on S2. For ψ ∈ (−δ, δ), where δ > 0 is a fixed small real

number, we have that

|H(ψ, t)| ≤ C,

where C is a uniform constant, independent of time. If ψ ∈ [−π2 ,−δ]∪ [δ, π2 ],

by Proposition 2.5 and (2.32), we have

|H(ψ, t)| ≤ C

where C > 0 is a time independent constant. Notice also that Lemma 2.6

implies that

vθ(π

2, θ, t) = vθ(−

π

2, θ, t) = 0 (2.33)

for all t < 0 and all θ ∈ [0, 2π]. We claim that

vθψ(π

2, θ, t) = vθψ(−π

2, θ, t) = 0.

Indeed, if there were some θ, t for which that were not true (say for ψ = π2 ),

then by using the Taylor expansion for vθ(·, θ, t) as a function of ψ and (2.33)

we would have

vθ(ψ, θ, t) = vθψ(π

2, θ, t) (ψ − π

2) +O((ψ − π

2)2), for ψ ∼ π

2

where vθψ(π2 , θ, t) 6= 0. Since ψ − π2 ∼ cosψ as ψ ∼ π

2 , Lemma 2.6 yields to

a contradiction, unless vθψ(π2 , θ, t) = 0.

Also, by the Taylor’s expansion around π2 (similarly around −π

2 ) we have

vθψ(ψ, θ, t) = vθψψ(π

2, θ, t) (ψ − π

2) + o(ψ − π

2), for ψ ∼ π

2.

Since at each time slice we are on a smooth compact surface, it follows that

|vθψ| ≤ C(t) |ψ − π

2| ≤ C(t) cosψ (2.34)

when |ψ− π2 | < δ, where δ > 0 is a small, time independent number and for

every θ ∈ [0, 2π].

Let us now use the above to prove our bound |vψ tanψ| ≤ C(t). For ψ in

the interval (−π2 + δ, π2 − δ) we have our bound, since |vψ secψ| ≤ C for t ≤

t0 < 0, for a uniform constant C. Assume that ψ ∈ [−π2 ,−

π2 +δ]∪ [π2 −δ,

π2 ].

We have shown above that for every ψ, we have

| secψ∫ 2π

0vψ dθ| ≤ C, for t ≤ t0 < 0.


By the integral mean value theorem, for every t, ψ there exists θ0 (that

depends on t, ψ) so that

|vψ(ψ, θ0, t) secψ| ≤ C

for a uniform constant C. By (2.34) we have

|vψ(ψ, θ, t)− vψ(ψ, θ0, t)| secψ = |vψθ(ψ, θ′, t)| |θ − θ0| secψ ≤ C(t)

for every θ ∈ [0, 2π] (θ′ is in between θ and θ0). We conclude the bound

|vψ(ψ, θ, t)| secψ ≤ C(t)

which finishes the proof of Claim 2.8. �

We would like to say that at the maximum (minimum) of vψ secψ, con-

sidered as a function on a sphere, its first derivative vanishes. Hence, we

need the following claim.

Claim 2.9. At each time slice t ≤ t0 < 0, we have

supS2

( |(vψ secψ)ψ|+ |(vψ secψ)θ| ) ≤ C(t).

Proof. By Proposition 2.5, Claim 2.8, (2.31) and the bound |∆v| ≤ C, we

have

sec2 ψ |vθθ + vθ tanψ| ≤ C(t).

Since at the maximum (minimum) in θ of sec3 ψ vθ we have vθθ = 0, similarly

as in Lemma 2.6, we obtain the bound

sec3 ψ |vθ| ≤ C(t), on S2.

This together with developing vθ in Taylor series around ψ = π2 (or ψ = −π

2 ),

similarly as in the proof of Claim 2.8, yield to

vθψψ(π

2, θ, t) = vθψψ(−π

2, θ, t) = 0 (2.35)

for every t < t0 and every θ. This implies the bounds

|(vψ sec2 ψ)θ| = |vψθ sec2 ψ| ≤ C(t) and |vψψθ| secψ ≤ C(t) (2.36)

which can be seen by expanding vψθ in Taylor series around ψ = π2 (or

ψ = −π2 ) and using (2.35). We have shown, in particular that |(vψ secψ)θ| ≤

C(t).


It remains to show that |(vψ secψ)ψ| ≤ C(t). To this end, we first inte-

grate (2.31) in θ ∈ [0, 2π] and then differentiate it in ψ to get∣∣∣∣∣(

tanψ∫ 2π

0vψ dθ

)ψ

∣∣∣∣∣ =∣∣∣∣∫ 2π

0vψψψ dθ −

∫ 2π

0(∆v)ψ dθ

∣∣∣∣ ≤ C(t).

The last bound holds because v is a smooth function on the sphere, at each

time slice t. Since we have the estimate we want away from π2 and −π

2 , we

only need to establish the estimate in the neighborhood of those two points.

Keeping that in mind, the previous estimate implies that∣∣∣∣∣(

secψ∫ 2π

0vψ dθ

)ψ

∣∣∣∣∣ ≤ C(t). (2.37)

By the integral mean value theorem, we have

12π

(secψ

∫ 2π

0vψ dθ

)ψ

=1

2πsecψ tanψ

∫ 2π

0vψ dθ +

12π

secψ∫ 2π

0vψψ dθ

= secψ tanψ vψ(ψ, θ1, t) + secψ vψψ(ψ, θ2, t)

where θ1, θ2 depend on ψ and t. Since

(secψ vψ)ψ = secψ tanψ vψ + secψ vψψ

the previous equality implies that (for some θ′i in between θ and θi and

i ∈ {1, 2})∣∣∣∣∣ 12π

(secψ

∫ 2π

0vψ dθ

)ψ

− (secψ vψ)ψ

∣∣∣∣∣≤ 2π (| secψ tanψ| |vψθ(ψ, θ′1, t)|+ secψ |vψψθ(ψ, θ′2, t)|)

≤ C(t)

where we have used (2.36). This together with (2.37) and (2.36) imply the

claim. �

To finish the proof of Lemma 2.7 consider the maximum (minimum) of

vψ secψ on S2 at each time-slice t < 0. By Claim 2.8 and Claim 2.9 this

function and its first derivatives are well defined on S2 at each time-slice.

At the maximum (minimum) point we have

vψψ secψ + vψ secψ tanψ = 0

which by Proposition 2.5 implies

|vψ tanψ| ≤ |vψψ| ≤ C, for t ≤ t0 < 0


at the maximum (minimum) point of vψ secψ, where C is a uniform constant.

Since we worry only when the maximum or the minimum occur close toπ2 and −π

2 (otherwise we have the Lemma 2.7 due to our earlier uniform

derivative estimates) we conclude from the previous bound that

supS2

|vψ tanψ| ≤ C, for t ≤ t0 < 0

where C is a uniform constant. This finishes the proof of our lemma. �

Once we have Lemma 2.7 we can improve the estimate on vθ shown in

Lemma 2.6.

Lemma 2.10. There exists a uniform in time constant C, so that

supS2

|vθ| ≤ C cos3 ψ, for t ≤ t0 < 0. (2.38)

Proof. Since

| sec2 ψ vθθ + sec2 ψ tanψ vθ| = |∆v − vψψ + vψ tanψ|

by Proposition 2.5, Lemma 2.7 and the estimate |∆v| ≤ C, it follows that

| sec2 ψ vθθ + sec2 ψ tanψ vθ| ≤ C, t ≤ t0 < 0.

Fix ψ ∈ (−π2 ,

π2 ) and look at the maximum (minimum) of vθ(ψ, θ, t) as a

function of θ. At that point vθθ = 0 and we have

| sec2 ψ tanψ vθ| ≤ C.

When ψ is away from zero (around which we have the estimate (2.38) any-

way), this also implies (2.38) with a possibly different uniform constant

C. �

Remark 2.11. In the case where v = v(ψ, t) is radially symmetric we can

improve estimates shown in Lemma 2.4. The reason is that in this case the

Hess(v) is given by the matrix(vψψ 0

0 − sin 2ψ2 vψ

)and therefore

|Hess(v)|g(t) = v2ψψ + sec4 ψ (

sin 2ψ2

)2 v2ψ = v2

ψψ + tan2 ψ v2ψ ≤ C

for a uniform constant C and all t ≤ t0 < 0.


From the above remark and the estimates shown previously we obtain the

following result which will be used in section 7.

Corollary 2.12. Let v be a radial solution of equation (1.8) on S2×(−∞, 0).

Then, there is a uniform constant C so that

‖√v∇2

S2v‖C0,1 ≤ C and ‖√v∇3

S2v‖C0 ≤ C


Proof. Fix t < t0 < 0. As in the proof of Lemma 2.4, we rewrite (1.8) in the

form

∆v3/2 =9|∇v|2

4√v− 3v3/2 +

32R√v.

By Remark 2.11 we have∣∣∣∣∇( |∇v|2√v

)∣∣∣∣ ≤ C |∇2v| |∇v|√v

+12

(|∇v|√v

)3

≤ C

thus

‖|∇v|2

√v‖C0,1(S2) ≤ C

for uniform in time constant C. Also, similarly as in the proof of Lemma

2.4, we have |∇(R√v)| ≤ C and |∇v3/2| ≤ C.

The above estimates imply the right hand side of (2.19) has uniformly in

time bounded C0,1 norm and standard elliptic regularity theory yields

‖v3/2‖C2,1(S2) ≤ C

which readily implies the bound

‖(√v∇2v)‖C0,1(S2) ≤ C.

We conclude that

|∇(√v∇2v)| ≤ C

that is,

|√v∇3v| ≤ C +

|∇v|2√v|∇2v| ≤ C

almost everywhere on S2. Since√v∇3v is a continuous function on S2 at

each time t, we obtain the estimate

||√v∇3v||L∞ ≤ C.

Since the constants C are independent on t, our proof is complete. �


3. Lyapunov functional and convergence

We introduce the Lyapunov functional

J(t) =∫S2

(|∇S2v|2

v− 4 v

)da. (3.1)

We will show next that J(t) is non-decreasing and bounded. In the sequel,

we will use these properties and the a priori estimates shown in the previous

section to prove that v(·, t) converges, as t → −∞, in the C1,α norm, to a

steady state solution v.

Lemma 3.1. The Lyapunov functional J(t) is monotone under (1.8) and

in particular, we have

d

dtJ(t) = −2

∫S2

v2t

v2da−

∫S2

|∇S2v|2

vvt da (3.2)

Proof. To simplify the notation we will set ∆ := ∆S2 and ∇ := ∇S2 . Equa-

tion (1.8) and a direct calculation show:

d

dt

∫S2

|∇v|2

vda = 2

∫S2

∇v∇vtv

da−∫S2

|∇v|2

v2vt da

= −2∫S2

∆vvvt da+

∫S2

|∇v|2

v2vt da

= −2∫S2

(vtv2

+|∇v|2

v2− 2) vt da+

∫S2

|∇v|2

v2vt da

= −2∫S2

v2t

v2da−

∫S2

|∇v|2

vvt da+ 4

∫S2

vt da.

We then conclude that

d

dt

∫S2

(|∇v|2

v− 4v

)da = −2

∫S2

v2t

v2da−

∫S2

|∇v|2

vvt da (3.3)

that is,d

dtJ(t) = −2

∫S2

v2t

v2da−

∫S2

|∇v|2

vvt da (3.4)

where both terms on the right hand side of (3.4) are nonnegative, since on

an ancient solution of (1.8) we have vt ≥ 0.

�

As an immediate consequence of the estimate in Lemma 2.1 and the

inequality

v∆v − |∇v|2 + 2 v2 ≥ 0

we have:


Lemma 3.2. There exists a uniform constant C so that −C ≤ J(t) ≤ 0 for

all t ∈ (−∞, t0] and t0 < 0.

We will next use Lemma 3.1 to show that on our ancient solution the

backward in time limit of the scalar curvature R is zero. We recall that the

scalar curvature R is given, in terms of the pressure function v, by R = vt/v.

We also recall that on an ancient solution we have R > 0 and by the Harnack

estimate on the curavture R, the inequality Rt ≥ 0. Hence, the point-wise

limit

R = limt→−∞

R(·, t)

exists.

Lemma 3.3. On an ancient solution v of equation (1.8), we have R = 0

a.e. on S2.

Proof. It is enough to show that∫S2

R2 da = 0.

Indeed, assume the opposite, namely that∫S2 R

2 da := c > 0. Then, since

Rt ≥ 0 we will have that∫S2 R

2(·, t) da ≥ c, i.e.∫S2

v2t

v2da ≥ c.

Integrating (3.2) in time while using the above inequality and the fact that

vt ≥ 0, gives

J(t2)− J(t1) ≤ −∫ t2

t1

∫S2

v2t

v2da ≤ −c (t2 − t1)

for every −∞ < t1 < t2 < t0 < 0. This obviously contradicts the uniform

bound −C ≤ J(t) ≤ 0, shown in Lemma 3.2. �

We will now combine the a priori estimates of the previous section with

Lemma 3.3 to prove the following convergence result.

Proposition 3.4. The solution v(·, t) of (1.8) converges, as t→ −∞, to a

limit v ∈ C1,α(S2), for any α < 1. Moreover, ‖v∇2S2 v‖Cα(S2) < ∞, for all

α < 1, and v satisfies the steady state equation

v∆S2 v − |∇S2 v|2 + 2 v2 = 0. (3.5)


Proof. Since vt ≥ 0 and v > 0, the pointwise limit

v := limt→−∞

v(·, t)

exists. Lemmas 2.3 and 2.4 ensure that for every α < 1 and every sequence

ti → −∞, along a subsequence still denoted by ti, we have v(·, ti)C1,α(S2)−→ v(·)

and v∇2S2v(·, ti)

Cα(S2)−→ v∇2S2 v. By the uniqueness of the limit, v = v, which

means that for every α < 1, we have

v(·, t) C1,α(S2)−→ v and (v∇2

S2v)(·, t) Cα(S2)−→ v∇2

S2 v, as t→ −∞.

We can now let t→ −∞ in equation

v∆S2v − |∇S2v|2 + 2 v2 = Rv

and use Lemma 3.3 to conclude that v satisfies equation (3.5).

�

4. The backward limit

Our goal in this section is to classify the backward limit v = limt→−∞ v(·, t),which has been shown to exist in the previous section.

Theorem 4.1. There exists a conformal change of S2 in which the limit

v(ψ, θ) := limt→−∞

v(ψ, θ, t) = C cos2 ψ

for some constant C ≥ 0, where ψ, θ are global coordinates on a conformally

changed sphere. Moreover, the convergence is in the C1,α-norm on S2 and

it is smooth, uniformly on every compact subset of S2\{S,N}, where S,N

are the south and the north pole on S2 (the points that correspond to ψ = π2

and ψ = −π2 ).

We have shown in the previous section that v(·, t) C1,α(S2)−→ v, for any

α ∈ (0, 1), where v satisfies the steady state equation

v∆v − |∇v|2 + 2 v2 = 0.

To classify the backward limit v, we will need the following lemma, which

constitutes the main step in the proof of Theorem 4.1.

Proposition 4.2. The limit v is either identically equal to zero, or it has

at most two zeros.


We will outline the main steps in the proof of Proposition 4.2. Their

detailed proofs will be given below. Recall that R(x) := limt→−∞R(x, t) = 0

almost everywhere, otherwise R ≥ 0 and Rt ≥ 0. Let

Z := {x ∈ S2 | R(x) > 0}.

Using a covering argument and the fact that the total curvature of our

evolving metric is equal to∫S2 Ruda = 8π < ∞, we will show that there

are at most finitely many points in Z, call them {p1, . . . , pN}, so that

(i) if Z = ∅, then v ≡ 0, and

(ii) if Z 6= ∅, on every compact set K ⊂ S2\{p1, . . . , pN}, u(·, t) is uni-

formly bounded by a constant that may depend on K, but is time

independent. This implies that v > 0 on S2\{p1, . . . , pN}.

In the case (ii) we will consider a dilated sequence of solutions around each

of the points pi and show that it converges to a cigar soliton. This will imply

that for a neighbourhood around each of them we have that∫URuda ≈ 4π. (4.1)

Since the total curvature∫S2 Ruda = 8π, we will conclude that there can

be at most two curvature concentration points. This readily implies v has

at most two zeros.

We will denote in the sequel by ∆δ(x) a ball of radius δ, centered at x,

computed in the standard spherical metric.

Lemma 4.3. Either v ≡ 0 or there are at most finitely many points p1, . . . , pN

so that for every δ > 0 there exists a C(δ) > 0 such that for all t < 0 we

have

sup∆δ/4(x)

u(·, t) ≤ C(δ), for all x ∈ S2\ ∪Ni=1 ∆2δ(pi). (4.2)

Moreover, R(x) = 0 for every x ∈ S2\{p1, . . . , pN}.

Proof. We will prove the Lemma in a few steps. For each t < 0 and r > 0,

we define the set

Ltr := {x ∈ S2 |∫

∆2r(x)(Ru)(·, t) da ≥ ε}.

Note that we can choose a cover of Ltr by finitely many balls {∆2r(ptjr)}Ntr

j=1

so that the balls ∆r(ptjr) are all disjoint and ptjr ∈ Ltr.


Step 1. For every ε > 0 there exists a uniform constant N so that N tr ≤ N ,

for all t < 0 and all r > 0.

Proof. Since ∆r(x) are balls in the spherical metric, there are uniform con-

stants C1, C2 so that

C1 r2 ≤ area(∆r(x)) ≤ C2 r

2

which implies that there is a uniform upper bound on the number of disjoint

balls of radius r contained in a ball of radius 4r. Then, since ptjr ∈ Ltr and∫S2 Ruda = 8π, we have

N tr ε ≤

Ntr∑

j=1

∫∆2r(ptjr)

(Ru)(·, t) da ≤ m∫S2

(Ru)(·, t) da = 8πm (4.3)

for a uniform constant m, which yields to the uniform upper bound on

N tr . �

Assume that v is not identically equal to zero and that it has at least

three different zeros (otherwise we are done). We can perform a conformal

change of coordinates (note that our evolution equations are invariant under

conformal changes) in which we bring two of the zeros of v to the poles of

S2. In other words, if ψ, θ are global coordinates on S2, two of the zeros of

v will correspond to ψ = π2 and ψ = −π

2 . Denote by w the pressure function

v in cylindrical coordinates, namely

w(x, θ, t) = v(ψ, θ, t) sec2 ψ, secψ = coshx (4.4)

and by U the conformal factor in cylindrical coordinates, namely

U(x, θ, t) =1

w(x, θ, t)=u(ψ, θ, t)cosh2 x

(4.5)

where u is the conformal factor of our evolving metric on S2. The function

U satisfies the equation

Ut = ∆c logU = −RU

where ∆c denotes the cylindrical Laplacian.

Step 2. Either v ≡ 0 or, for every compact set K ⊂ R× [0, 2π], there exists

a uniform constant C(K) so that∫ 2π

0(logU(x, θ, t))+ dθ ≤ C(K), for all t < 0. (4.6)


Proof. Let

F (x, t) =1

2π

∫ 2π

0logU(x, θ, t) dθ.

Claim 4.4. |Fx| ≤ 2, for all x ∈ R and all t < 0.

Proof. We begin by integrating in θ the equation

∆c logU = (logU)xx + (logU)θθ = −RU ≤ 0

to obtain (since U is periodic in θ) that Fxx ≤ 0, which implies that Fx is

decreasing in x and therefore

limx→∞

Fx ≤ F (x, t) ≤ limx→−∞

Fx.

At each time-slice t we have

logU(x, θ, t) = log u(ψ, θ, t)− 2 log coshx

which implies that

(logU)x = (log u)ψ cosψ − 2 tanhx

since dψ/dx = cosψ. We recall that |(log u)ψ| ≤ C(t), as each time t. Hence,

as x→ −∞ (or equivalently ψ → −π2 ), we have

limx→−∞

Fx = limx→−∞

12π

∫ 2π

0(logU)x(x, θ, t) dθ = 2.

Similarly,

limx→∞

Fx = limx→∞

12π

∫ 2π

0(logU)x(x, θ, t) dθ = −2

and therefore

|Fx| ≤ 2, for all x and all t ≤ t0 < 0.

�

Claim 4.5. There exist C > 0, x0 ∈ R so that

F (x0, t) :=1

2π

∫ 2π

0logU(x0, θ, t) dθ ≤ C, for all t ≤ t0 < 0.

Proof. We begin by choosing (x0, θ0) so that

U(x0, θ0, t) ≤ C0, for all t < 0. (4.7)

If there were no such a point, that would mean that for all points (x, θ) we

would have limt→−∞ U(x, θ, t) = ∞ (remember that Ut ≤ 0) which would


imply that the backward limit v = 0 and would end the proof of Theorem

4.1 for a constant C = 0.

We will prove that for this point x0 the statement of the claim holds.

We argue by contradiction. We assume that there is a sequence of times

tk → −∞ so that

F (x0, t) =1

2π

∫ 2π

0logU(x0, θ, tk) dθ ≥ 2C, for all k

for C a sufficiently large constant (to be chosen in the sequel). By Claim

4.4, for x ∈ (x0 − π, x0 + π) we have

12π

∫ 2π

0logU(x, θ, tk) dθ ≥

12π

∫ 2π

0logU(x0, θ, tk) dθ − 4π

≥ 2C − 4π ≥ C

if C ≥ 4π. Hence, by integrating in x, we have

14π2

∫ x0+π

x0−π

∫ 2π

0logU(x, θ, tk) dθ dx ≥ C. (4.8)

Notice that, since ut(ψ, θ, t) ≤ 0, we have

u(·, t) ≥ C1, for all t ≤ t0 < 0

for a uniform constant C1. Therefore

U(x, θ, t) =u(ψ, θ, t)cosh2 x

≥ C1

cosh2 x≥ C(K)

for all (x, θ) ∈ K, for any compact set K ⊂ R × [0, 2π]. Hence, there is a

uniform constant C2, independent of time, dependent on the compact set

K, so that

logU(x, θ, t) + C2 > 0, for all (x, θ) ∈ K and t ≤ t0 < 0. (4.9)

Let

Q = {(x, θ)| x ∈ [x0 − π, x0 + π], θ ∈ [θ0 − π, θ0 + π] }

and B = B√2π(x0, θ0) be the euclidean ball of radius√

2π, centered at

(x0, θ0). Then, Q ⊂ B. If we understand logU as a function which is

defined on R2 and is 2π periodic in the θ direction, then since

∆c(logU + C2) ≤ 0


by the mean value property and (4.8), we have

logU(x0, θ0, tk) + C2 ≥ 1|B|

∫B

(logU(·, tk) + C2) ≥ 1|B|

∫Q

(logU(·, tk) + C2)

=1|B|

∫ x0+π

x0−π

∫ 2π

0(logU(·, tk) + C2)

≥ 12π2

(4π2C + 4π2C2) = 2 (C + C2)

which implies that

U(x0, θ0, tk) ≥ e2C .

If C sufficiently large so that e2C ≥ 2C0 where C0 is taken from (4.7), we

contradict (4.7), finishing the proof of our claim. �

Claims 4.4 and 4.5 imply the bound

F (x, t) :=1

2π

∫ π

−πlogU(x, θ, t) dθ ≤ C(K)

for any compact set K ⊂ R× [−π, π]. By (4.9) we have∫ π

−π(logU(x, θ, t))+ dθ ≤ C(K) (4.10)

for all t ≤ t0 < 0 and all x so that {x} × [0, 2π] ⊂ K. �

Step 3. There is an ε0 > 0 so that for every compact set K ⊂ R × [0, 2π],

if ε < ε0 and∫K(Ru)(·, t) da < ε, then

supK

U(x, θ, t) ≤ C(K), for all K ⊂⊂ K. (4.11)

Proof. We have shown in Step 2 the bound∫ 2π

0 (logU(x, θ, t))+ dθ ≤ C(K)

for all (x, θ) ∈ K and t ≤ t0 < 0. Since U(x, θ, t) satisfies ∆c logU = −RU ,

we can apply the same arguments as in the proof of Lemma 2.3 in [4]. Notice

that in (4.11) we do not have a dependence on time (as in [4]) because our

smallness assumption (ii) is time independent. �

We will now conclude the proof of Lemma 4.3. Note that the estimate

(4.11) can be rewritten on a sphere as

supK

u(·, t) ≤ C(K) (4.12)

where K ⊂⊂ S2\{poles of S2} corresponds to a compact set K from (4.11)

under Mercator’s projection.


Choose ε0 from Step 3. By Step 1, for every r > 0 and t < 0 there are N

points pt1r, . . . , ptNr so that the balls {∆2r(ptjr)}Nj=1 cover the set Ltr and for

every point x ∈ S2\(∪Nj=1∆2r(ptjr)), we have∫∆2r(x)

(Ru)(·, t) da < ε0.

It follows form Step 3 that

sup∆r(x)

u(·, t) ≤ C(r) (4.13)

for every x ∈ S2\(∪Nj=1∆2r(ptjr)) and all t < 0.

Now fix a monotone sequence tk → −∞. For a sequence of radii rk → 0

we have (passing to a subsequence) that ptkjrk → pj as k →∞, for N points

pj ∈ S2 (these points may not be distinct).

Let x ∈ S2\(∪Nj=1∆2δ(pj). We may choose k sufficiently large, so that pkjrkis sufficiently close to pj , for all 1 ≤ j ≤ N and therefore x ∈ S2\(∪Nj=1∆δ/2(pkjrk)).

It follows by (4.13) with r = δ/4, that

sup∆δ/4(x)

u(x, tk) ≤ C(δ)

for all x ∈ S2\(∪Nj=1∆2δ(pj) and all k sufficiently large. Recalling that

ut ≤ 0, we conclude from the above that (4.2) holds for all t < 0.

This estimate implies v(·, t) is uniformly bounded from below on every

compact subset of S2 \ {p1, . . . , pN}, which means that the equation (1.8) is

uniformly parabolic there and the standard parabolic estimates imply that

R(·, t) converges to zero uniformly on those sets away from {p1, . . . , pN}. In

other words, R = 0 on S2\{p1, . . . , pN}. This finishes the proof of Lemma

4.3. �

Let p1, . . . , pN be the points as chosen in Lemma 4.3.

Corollary 4.6. For every 1 ≤ j ≤ N we have that limt→−∞ u(pj , t) = +∞.

Proof. If R(pj) = γ > 0, then since Rt ≥ 0 we have R(pj , t) ≥ γ for all t < 0.

Since ut = −Ru, at the particular point pj we have (lnu)t(pj , t) ≤ −γ and

therefore

u(pj , t) ≥ u(pj , t0) eγ (t0−t) = cj eγ|t|,

which implies the limt→−∞ u(pj , t) =∞.


If R(pj) = 0, assume the limt→−∞ u(pj , t) ≤ C < ∞, otherwise we are

done. Our construction of pj implies that for a sequence rk → 0 there are

points ptkjrk so that ∫∆rk

(ptkjrk

)(Ru)(·, tk) da ≥ ε0. (4.14)

On the other hand, since ut ≤ 0 we have c ≤ u(pj , t) ≤ C, which implies

that C−1 ≤ v(pj , t) ≤ c−1, for all t ≤ t0 < 0. Since v(·, t) converges as

t → −∞ to v in C1,α norm, there exists δ > 0 so that c/2 ≤ u(z, t) ≤ 2C,

for z ∈ ∆δ(pj) ⊂ S2 and all t ≤ t0 < 0. It follows that our equation is uni-

formly parabolic on ∆δ(pj) and therefore, by standard parabolic estimates,

R(·, t) → 0, as t → −∞ uniformly on ∆δ/2(pj) (since R(·, t) → 0 a.e. as

t→ −∞). This obviously contradicts (4.14), finishing the proof.

�

We will now conclude the proof of Proposition 4.2, based on Lemma 4.3

and Corollary 4.6.

Proof of Proposition 4.2. Assume that v is not identically equal to zero, and

let p1, · · · , pN be the points from Lemma 4.3. Denote by 2ρ the minimum

distance between any of those points. We will show that for any pj , j =

1, · · · , N , and any ε > 0, there exists a t0 = t0(ε) such that∫∆ρ(pj)

(Ru)(·, t) da ≥ 4π − ε, for all t ≤ t0

which will readily imply that there are only two of them, since we have∫S2

(Ru)(·, t) da = 8π, for all t ≤ t0.

Consider any of these points, say p1, and perform a stereographic projection

to map S2 onto R2 so that p1 is mapped to the origin in R2. Denote by u the

confomal factor of our evolving metric on the plane and by R its curvature.

It is well known that u satisfies the equation

ut = ∆ log u (4.15)

on R2 × (−∞, 0). Moreover, by Corollary 4.6, we have

limt→−∞

u(0, t) = +∞.


Each of the other points pj are mapped to points pj ∈ R2 ∪ {∞}. Denote

by 2ρ the minimum distance between zero and the points pj . It will be

sufficient to show that for any ε > 0, there exists a t0 = t0(ε) such that∫Bρ(0)

(R u)(x, t) dx ≥ 4π − ε, for all t ≤ t0. (4.16)

To this end, we will rescale u and show that the rescaled solution converges

to a cigar solution of the Ricci flow.

Fix a sequence tk → −∞. Since 0 may not need to be a local maximum

for u(·, t) and we will need to rescale u around local maximum points, we

let xk be such that

u(xk, tk) := maxBρ(0)

u(·, tk).

Since u(xk, tk) ≥ u(0, tk), we have limtk→−∞ u(xk, tk) =∞. It follows that

limk→∞

xk = 0,

otherwise a subsequence xkl would converge to a point x∞ ∈ Bρ(0) \ {0}and would contradict Lemma 4.3 .

Set αk = u(xk, tk)−1 and recall that αk → 0. In particular, we may

assume that αk < 1/4. Since xk → 0, we may also assume that |xk| < ρ2 .

Consider the rescaled solutions of equation (4.15) defined by

uk(x, t) = αk u(xk +√αk x, t+ tk), x ∈ Bρ(0), −∞ < t < −tk.

We observe that for |x| < ρ, we have |xk +√αk x| < ρ, hence uk(·, 0) ≤ 1,

from the definition of xk and αk. Moreover, since αk → 0, for every M > 0,

there is a k0 = k0(M) such that |xk+√αk x| < ρ for all |x| ≤M , which shows

that uk(·, 0) is uniformly bounded on any compact set of R2. By standard

estimates on equation (4.15), it follows that the sequence of solutions {uk}is uniformly bounded on compact subsets of R2 × R, hence equicontinuous.

We conclude, that passing to a subsequence, which we still denote by uk, we

have that uk → u uniformly on compact subsets of R2 ×R and that u is an

eternal solution of (4.15) (i.e. it is defined on R2 × (−∞,∞)). Moreover,

u(0, 0) = limk→∞ u(xk, 0) = 1. It follows, from the results in [4], that u is a

cigar solution which in particular implies that∫R2

R u dx = 4π, for all t ∈ (−∞,∞)

where R denotes the curvature of u.


From here the conclusion of our proposition is straightforward. For any

ε > 0, choose M > 0 sufficiently large so that∫|x|≤M

(R u)(x, 0) dx > 4π − 2ε.

From the uniform convergence of uk(·, 0) → u(·, 0) on |x| ≤ M , which also

implies uniform convergence of the corresponding scalar curvatures, it fol-

lows that for k sufficiently large∫|x|≤M

(Rk uk)(x, 0) dx > 4π − ε

or equivalently ∫|y|≤√αkM

(R u)(xk + y, tk) dy > 4π − ε

Recalling that αk → 0 and xk → 0, the above readily implies that (4.16)

holds for t = tk. Since tk is an arbitrary sequence, this finishes the proof of

the proposition.

�

Assume from now on that the backward limit v is not identically zero. We

have just shown in Proposition 4.2 that v has at most two zeros. Choose a

conformal change of S2 which brings those two zeros to two antipodal poles

on S2. Let ψ, θ be global coordinates on S2, where ψ = π2 and ψ = −π

2

correspond to the poles (denote them by S and N). Observe that equation

(1.8) is strictly parabolic away from the poles, uniformly as t → −∞. It

follows that the convergence v(·, t)→ v, as t→ −∞, is smooth on compact

subsets of S2\{S,N}. Perform the Mercator’s transformation (4.4) with

respect to the poles S,N at which v is potentially zero.

Since w(x, θ, t) = v(ψ, θ, t) cosh2 x, we conclude that

limt→−∞

w(x, θ, t) = w(x, θ) := v(ψ, θ) cosh2 x > 0

and the convergence is smooth on compact subsets of R× [0, 2π]. The limit

w satisfies

w∆w − |∇w|2 = 0

or (since w(x, θ) > 0 on R× [0, 2π]) equivalently,

∆c log w = 0 (4.17)


where ∆c is the cylindrical laplacian on R2. To finish the proof of Theorem

4.1 we need to classify the solutions of the steady state equation (4.17) that

come as limits of ancient solutions w(·, t).

Proof of Theorem 4.1. Denote by

W (x) :=∫ 2π

0log w(x, θ) dθ.

Keeping in mind that ∆c log w = (log w)xx+ (log w)θθ, if we integrate (4.17)

in θ we obtain

Wxx = 0

which implies that

W (x) =∫ 2π

0log w(x, θ) dθ = C1 + C2 x (4.18)

for some constants C1, C2. Denote by f := log w. We can view f as a

function on R2, after extending it in the θ direction so that it remains 2π-

periodic.

Since w(x, θ) = v(ψ, θ) cosh2 x and v = limt→−∞ v(·, t) ≤ C, it follows

that there are uniform constants C1, C2 > 0 so that

f ≤ C1 + C2 |x|. (4.19)

We define the function h(x, θ) = C1 + C2 |x| − log w > 0, where C1, C2 are

taken from (4.19). Since ∆ch = 0 on |x| > 0, if (x, θ) is an arbitrary point

with |x| ≥ 2π, by the mean value theorem applied to h(x, θ) we have

h(x, θ) =1

|B(x,θ)(π)|

∫B(x,θ)(π)

h(x′, θ) dx′dθ

where B(x,θ)(π) is a ball centered at (x, θ) and of radius π. If

Q(x,θ) := {(x′, θ′)| |x− x′| ≤ π, |θ − θ′| ≤ π}

then B(x,θ)(π) ⊂ Q(x,θ) and therefore

h(x, θ) ≤ C

∫Q(x,θ)

h(x′, θ′) dx′dθ′

= C

∫ x+π

x−π

∫ 2π

0(C1 + C2x

′) dθ dx′ − C∫ x+π

x−π

∫ 2π

0log w dθ dx′

≤ C1 + C2|x|


where we have used that∫ 2π

0 log w dθ = C1 + C2x, by (4.18). This implies

the bound

f ≥ −C3 − C4|x|, for |x| ≥ 2π (4.20)

for some uniform constants C3, C4. Combining the (4.19) and (4.20) yields

to the estimate

|f | ≤ A+B |x| ≤ A+B r, for r = |x| >> 1

where r =√x2 + θ2 and A,B are some uniform constants.

To finish our argument, we will need to classify all 2π-periodic functions

f on R2 so that

∆cf = 0 and f = O(r), for r >> 1.

By the result of Li and Tam in [13], the space H2 of harmonic functions with

linear growth in R2 is at most 3-dimensional. The harmonic polynomials

{1, x, θ} are linearly independent, hence they form a basis of H2. It follows

that

f(x, θ) = C1 + C2 x+ C3 θ.

Since f is a 2π-periodic function in θ, we must have C3 = 0 which implies

that

w(x, θ) = AeBx

for some constants A ≥ 0 and B. Since we have assumed that the function

w is not identically zero, we have A > 0.

By our previous estimates and the inequality wt ≤ 0, we have the bound

C1(K) ≤ w(x, θ, t) ≤ C2(K), for |x| ≤ K and all t ≤ t0 < 0 (4.21)

on any compact subset K of R × [0, 2π]. Hence, by the standard para-

bolic estimates w(x, θ, t) converges to w(x, θ), uniformly on compact sub-

sets of R× [0, 2π] in the C∞ norm. In particular, it follows that the metric

g(x, t) = w(x, t)−1ds2 converges to the metric g(x) = w(x)−1 ds2 in C∞

norm. Observe that the limiting metric

g(x, θ) = w−1 ds2 = A−1 e−Bx ds2

is not complete unless B = 0.

On the other hand, (4.21) implies the bound

injrad0(g(·, t)) ≥ δ > 0, for all t ≤ t0 < 0,


where the injrad0(g(·, t)) is the injectivity radius at x = 0 with respect to

the metric g(·, t). Since 0 < R(·, t) ≤ C for all t ≤ t0 < 0, by Hamilton’s

compactness theorem for the Ricci flow (c.f. in [9]), there exists a subse-

quence tk → −∞ so that (R2, g(tk), 0) converges as tk → −∞ to a complete

metric g∞ on R2. This contradicts the incompletness of g and implies that

B = 0 unless A = 0. We conclude that w(x, t) = A.

If we go back to a sphere S2 via Mercator’s transformation, this means

that

limt→−∞

v(ψ, θ, t) = C cos2 ψ

for some constant C ≥ 0. Moreover, the convergence is smooth on compact

subsets of S2\{S,N}. This finishes the proof of our proposition. �

5. The case when v ≡ 0

Throughout this section we will assume that the backward lmit

v := limt→−∞

v(·, t) ≡ 0. (5.1)

Our goal is to show that in this case the ancient solution v must be a family

of contracting spheres, as stated in the following proposition.

Proposition 5.1. If the backward limit v ≡ 0, then

v(·, t) =1

(−2t)

that is, our ancient solution is a family of contracting spheres.

To prove the proposition we will use an isoperimetric estimate for the Ricci

flow which was proven by R. Hamilton in [8]. Let M be any compact surface.

Any simple closed curve γ on M of length L(γ) divides the compact surface

M into two regions with areas A1(γ) and A2(γ). We define the isoperimetric

ratio as in [2], namely

I =1

4πinfγL2(γ)

(1

A1(γ)+

1A2(γ)

). (5.2)

It is well known that I ≤ 1 always, and that I ≡ 1 if and only if the surface

M is a sphere.

We will briefly outline the proof of Proposition 5.1 whose steps will be

proven in detail afterwards. We consider our evolving surfaces at each time

t < 0, and define the isoperimetric ratio I(t) as above. Our goal is to show


that our assumption (5.1) implies that I(t) ≡ 1, which forces (M, g(t)) to be

a family of contracting spheres. We will argue by contradiction and assume

that I(t0) < 1, for some t0 < 0. In that case we will show that there exists

a sequence tk → −∞ and closed curves βk on S2 so that simultaneously we

have

LS2(βk) ≥ δ > 0 and Lg(tk)(βk) ≤ C, ∀k (5.3)

where LS2 and Lg(tk) denote the length of a curve in the round metric on S2

and in the metric g(tk), respectively. This clearly contradicts the fact that

u(·, tk)→∞, uniformly in S2 ( implied by (5.1)) and finishes the proof.

We will now outline how we will find the curves βk. For each t < t0, let

γt be a curve for which the isoperimetric ratio I(t) is achieved.

i. If I(t0) < 1, for some t0 < 0, we will show that I(t) ≤ C|t| , for t < t0. We

will use that to show Lg(t)(γt) ≤ C, for all t < t0.

ii. For any sequence tk → −∞ and pk ∈ γtk , we will show that there exists a

subsequence such that (M, g(tk), pk) converges to (M∞, g∞, p∞), where

M∞ = S1×R and γ∞ := limk→∞ γk is a closed geodesic on M∞, one of

the cross circles of S1 × R.

iii. Let tk be as above. If A1(tk), A2(tk) are the areas of the two regions

into which γtk divides S2, we show that both of them are comparable

to τk = −tk.iv. We show that the maximal distances from γk to the points of the two re-

gions of areas A1(tk), A2(tk) respectively are both of length comparable

to τk.

v. The curves γk do not necessarily satisfy (5.3). However, we use them

and (ii) to define a foliation {βkw} of our surfaces (M, g(tk)) and we

choose the curve βk from this foliation that splits S2 into two parts of

equal areas with respect to the round metric. We prove that this is

the curve that satisfies (5.3) by using that IS2 = 1, the Bishop Gromov

volume comparison principle, (iii) and (iv).

Lemma 5.2. If I(t0) < 1, for some t0 < 0, then there exist positive con-

stants C1, C2 so that

I(t) ≤ C1

|t|+ C2, for all t < t0.


Moreover, if γt is the curve at which the infimum in (5.2) is attained then,

L(t) := L(γt) ≤ C, for all t < 0.

Proof. Let t < t0, with t0 as in the statement of the lemma. It has been

shown in [8] that

I ′(t) ≥ 4π (A21 +A2

2)A1A2(A1 +A2)

I (1− I2).

Since A1 + A2 = 8π|t| and A21 + A2

2 ≥ 2A1A2, we conclude the differential

inequality

I ′(t) ≥ 1|t|I (1− I2).

Since I(t0) < 1, the above inequality implies the bound

I(t) ≤ C1

|t|+ C2, for all t < t0

for uniform in time constants C1 and C2. Using that 1A1

+ 1A2≥ 1

4π|t| , we

will conclude that that the length L(t) of a curve γt at which the infimum

in (5.2) is attained satisfies L(t) ≤ C, for all t < t0.

�

We also have the following estimate from below on the length L(t) of the

curve at which the infimum in (5.2) is attained.

Lemma 5.3. There is a uniform constant c > 0, independent of time so

that

L(t) ≥ c, for all t ≤ t0 < 0.

Proof. Recall that for t0 < 0, the scalar curvature R satisfies 0 < R(·, t) ≤C, for all t ≤ t0. The Klingenberg injectivity radius estimate for even

dimensional manifolds implies the bound

injrad(g(t)) ≥ c√Rmax

≥ δ > 0, for all t ≤ t0 < 0 (5.4)

for a uniform in time constant c > 0. We will prove the Lemma by contra-

diction. Assume that there is a sequence ti → −∞, so that Li := L(ti)→ 0,

as i → ∞, and denote by γti a curve at which the isoperimetric ratio is

attained, i.e. L(ti) = L(γti).


Define a new sequence of re-scaled Ricci flows, gi(t) := L−2i g(ti + L2

i t)

and take a sequence of points pi ∈ γti . The bound (5.4) implies a lower

bound on the injectivity radius at pi with respect to metric gi, namely

injradgi(pi) =injradg(ti)(pi)

L2i

≥ δ

L2i

→∞, as i→∞. (5.5)

Also, since Ri(·, t) = L2i R(·, ti + L2

i t) ≤ C L2i and Li → 0, we get

maxRi(·, t)→ 0, as i→∞. (5.6)

Hamilton’s compactness theorem (c.f. in [9]) implies, passing to a sub-

sequence, the pointed smooth convergence of (M, gi(0), pi)) to a complete

manifold (M∞, g∞, p∞), which is, due to (5.5) and (5.6), a standard plane.

Moreover,

I(ti) = L2i

(1

A1(ti)+

1A2(ti)

)=

1A1(gi(0))

+1

A2(gi(0))

where A1(gi(0)) and A2(gi(0)) are the areas inside and outside the curve

γti , respectively, both computed with respect to metric gi(0). Since gi(0)

converges to the euclidean metric and γti converges to a curve of length 1,

it follows that limi→∞A1(gi(0)) = α > 0 and limi→∞A2(gi(0)) =∞, which

implies that

limi→∞

I(ti) ≥ δ > 0

and obviously contradicts Lemma 5.2. �

We recall that each time t, a curve γt at which the isoperimetric ratio is

achieved splits the surface into two regions of areas A1(t) and A2(t). Lemma

5.3 yields to the following conclusion.

Corollary 5.4. There are uniform constants c > 0 and C > 0 so that

c |t| ≤ A1(t) ≤ C |t| and c |t| ≤ A2(t) ≤ C |t|

for all t < t0 < 0.

Proof. It is well known that the total area of our evolving surface is A(t) =

8π |t|. Hence, A1(t) ≤ 8π |t| and A2(t) ≤ 8π |t|. On the other hand, by

lemmas 5.2 and 5.3, we have

c

Aj(t)≤ L2(t)Aj(t)

≤ I(t) ≤ C

|t|, j = 1, 2


for all t < t0, which shows that Aj(t) ≥ c |t|, j = 1, 2, for a uniform constant

c > 0, therefore proving the corollary. �

We will fix in the sequel a sequence tk → −∞. Let γk be, as before, a

curve at which the isoperimetric ratio is achieved. From now on we will refer

to γk as an isoperimetric curve at time tk. To simplify the notation, we will

set A1k := A1(tk), A2k := A2(tk) and Lk = L(tk). It follows from Corollary

5.4 that

limk→∞

A1k = +∞ and limk→∞

A2k = +∞. (5.7)

Pick a sequence of points pk ∈ γk and look at the pointed sequence of

solutions (M, g(tk + t), pk). Since the curvature is uniformly bounded and

since the injectivity radius at pk is uniformly bounded from below, by Hamil-

ton’s compactness theorem we can find a subsequence of pointed solutions

that converge, in the Cheeger-Gromov sense, to a complete smooth solution

(M∞, g∞, p∞). This means that for every compact set K ⊂ M∞ there are

compact sets Kk ⊂ M and diffeomorphisms φk : K → Kk so that φ∗kg(tk)

converges to g∞. From Lemma 5.2, L(tk) ≤ C, for all k, and therefore

our curves γk converge to a curve γ∞ (this convergence is induced by the

manifold convergence) which by (5.7) has the property that it splits M∞into two parts (call them M1∞ and M2∞), each of which has infinite area.

It follows that we can choose points xj ∈ M1∞ and yj ∈ M2∞ so that

distg∞(xj , p∞) = distg∞(p∞, yj) = ρj , where ρj is an arbitrary sequence so

that ρj →∞. Since (M∞, g∞) is complete, there exists a minimal geodesic

βj from xj to yj . This geodesic βj intersects γ∞ at some point qj . Since

qj ∈ γ∞ and γ∞ is a closed curve of finite length, the set {qj} is compact and

therefore there is a subsequence so that qj → q∞ ∈ γ∞. This implies that

there is a subsequence of geodesics {βj} so that, as j →∞, it converges to a

minimal geodesic β∞ : (−∞,∞)→M∞ (minimal geodesic means a globally

distance minimizing geodesic). It follows that our limiting manifold M∞

contains a straight line. Since the curvature of M∞ is zero, by the splitting

theorem our manifold splits off a line and therefore is diffeomorphic to the

cylinder S1 × R.

We next observe that the limiting curve γ∞ is a geodesic, as shown in the

following lemma.

Lemma 5.5. The geodesic curvature κ of the curve γ∞ is zero.


Proof. As in [8], at each time t < t0 < 0 we start with the isoperimetric

curve γt and we construct the one-parameter family of parallel curves γrt at

distance r from γt on either side. We take r > 0 when the curve moves from

the region of area A1(t) to the region of area A2(t), and r < 0 when it moves

the other way. We then regard L,A1, A2 and

I = I(γrt ) = L2(γrt )(

1A1(γrt )

+1

A2(γrt )

)as functions of r and t. By the computation in [8] we have

∂A1

∂r= L,

∂A2

∂r= −L, dL

dr=∫κ ds = κL

where κ is the geodesic curvature of the curve γrt . By a standard variational

argument κ is constant on γt. If A := A1 + A2 is the total surface area, we

have

log I = 2 logL+ logA− logA1 − logA2.

Since ∂I∂r |r=0 = 0, we conclude that

0 =2L

∂L

∂r+

1A

∂A

∂r− 1A1

∂A1

∂r− 1A2

∂A2

∂r=

2LκL− 1

A1L+

1A2

L

which leads to

κ =L

2(

1A1− 1A2

).

By lemmas 5.2 and 5.3 and (5.7) we conclude that

κ∞ := limt→−∞

κ = limt→−∞

L

2(

1A1− 1A2

) = 0

that is the geodesic curvature κ∞ of the limiting curve γ∞ is zero. �

We have just shown that our limiting manifold is a cylinder M∞ = S1×Rand γ∞ is a closed geodesic on M∞. Hence, γ∞ is one of the cross circles of

M∞.

We have the following picture assuming that the radius of γ∞ is 1.

Assume that we have a foliation of our limiting cylinder M∞ by circles

βw, where |w| is the distance from βw to γ∞, taking w > 0 if βw lies on the

upper side of the cylinder and w < 0 if βw lies on its lower side. Denote by

βkw the curve on M such that φ∗kβkw = βw.


One of the properties of the cylinder is that for every δ > 0 there is a

w0 > 0 so that for every |w| ≥ w0 we have

| supx∈βw,y∈γ∞

dist(x, y)− infx∈βw,y∈γ∞

dist(x, y)| ≤√w2 + π2−w ≤ C

|w|≤ C

w0<δ

2

where the distance is computed in the cylindrical metric on M∞.

Since for every sequence tk → −∞, there exists a subsequence for which

we have uniform convergence of our metrics {g(tk)} on bounded sets around

the points pk ∈ γtk , the previous observation implies the following claim

which will be used frequently from now on.

Claim 5.6. For every sequence tk → −∞ and every δ > 0 there exists k0

and w so that for k ≥ k0,

| supx∈βwk ,y∈γ(tk)

distg(tk)(x, y)− infx∈βwk ,y∈γ(tk)

distg(tk)(x, y)| < δ.

The variant of the Bishop-Gromov volume comparison principle (since

R ≥ 0) implies the following area comparison of the annuli, for each t < 0,

area(b1 ≤ s ≤ b2)area(a1 ≤ s ≤ a2)

≤ b22 − b21a2

2 − a21

(5.8)

where a1 ≤ a2 ≤ b1 ≤ b2 and s is the distance from a fixed point on (M, g(t)),

computed with respect to the metric g(t). We are going to use this fact in

the following lemma.

For each k, γtk splits our manifold in two parts, call them M1k and M2k

with areas A1k and A2k respectively. Choose points xk ∈M1k and yk ∈M2k

so that

distg(tk)(xk, γtk) = maxz∈M1k

distg(tk)(xk, z) =: ρk

and

distg(tk)(yk, γtk) = maxz∈M2k

distg(tk)(yk, z) =: σk.

By the definition of σk and ρk and from the convergence of (M, g(tk), pk) to

an infinite cylinder, we have

limk→∞

σk = +∞ and limk→∞

ρk = +∞.

Lemma 5.7. There are uniform constants k0 > 0 and c > 0 so that the

area (Bρk(xk)) ≥ c ρk and area (Bσk(yk)) ≥ c σk, for all k ≥ k0


where both the distance and the area are computed with respect to the metric

g(tk).

Proof. We take a1 = 0, a2 = b1 = σk ≥ 1 and b2 = σk + 1 in (5.8). Then, if

s is the distance from yk computed with respect to g(tk), we have

area (σk ≤ s ≤ σk + 1)area (0 ≤ s ≤ σk)

≤(σk + 1)2 − σ2

k

σ2k

≤ 3σk.

Hence,

area (Bσk(yk)) ≥σk3

area (σk ≤ s ≤ σk + 1). (5.9)

Claim 5.8. There are uniform constants c > 0 and k1 so that for k ≥ k1,

area (σk ≤ s ≤ σk + 1) ≥ c.

Proof. Denote by Uk := {z | σk ≤ s ≤ σk + 1}. We consider the set

Vk := {z | disttk(z, γk) ≤12, ykz ∩ γk 6= ∅}

where ykz denotes a geodesic connecting the points yk and z. It is enough to

show that Vk ⊂ Uk, for k sufficiently large, and that area (Vk) ≥ c > 0. To

prove that Vk ⊂ Uk, take z ∈ Vk and let wk ∈ γk be such that disttk(z, wk) =

disttk(z, γk) ≤ 12 . If qk := γk ∩ zyk, then

σk = disttk(yk, γk) ≤ disttk(yk, qk) ≤ disttk(z, yk)

which implies that σk ≤ disttk(z, yk). On the other hand, by Claim 5.6 we

have disttk(wk, yk) ≤ σk + 12 , for k sufficiently large. Hence

disttk(z, yk) ≤ disttk(yk, wk) + disttk(wk, z) ≤ σk +12

+12< σk + 1

for k sufficiently large. This proves that Vk ⊂ Uk and hence

area (Uk) ≥ area (Vk).

To estimate area (Vk) from below, we recall that for pk ∈ γk, we have pointed

convergence of (M, g(tk), pk) to a cylinder which is uniform on compact sets

around pk. To use this we need to show that there is a constant C > 0, for

which

Vk ⊂ Btk(pk, C), for all k ≥ k0.

Let z ∈ Vk and let qk ∈ ykz ∩ γk. Then by Claim 5.6 for k sufficiently large,

we have

σk − 1 ≤ distg(tk)(yk, qk) ≤ σk + 1. (5.10)


We also have

distg(tk)(pk, z) ≤ distg(tk)(pk, qk) + distg(tk)(qk, z).

Since, z ∈ Vk ⊂ Uk and (5.10) holds, we get

distg(tk)(qk, z) ≤ distg(tk)(yk, z)− distg(tk)(yk, qk) ≤ σk + 1− σk + 1 = 2

which combined with distg(tk)(pk, qk) ≤ L(γk) ≤ C gives us the bound

distg(tk)(pk, z) ≤ C.

This guarantees that, as k → ∞, Vk converges to a part of the cylinder

S1 × R, while (M, g(tk), pk) → (S1 × R, g∞, p∞) and g∞ is the cylindrical

metric. Recall that γtk → γ∞ and γ∞ is one of the cross circles on S1 × R.

It follows that Vk converges as k →∞ to the upper or lower part of the set

{z ∈ S1 × R | distg∞(z, γ∞) ≤ 12} with respect to γ∞. This implies that

c ≤ area ({σk ≤ s ≤ σk + 1}) ≤ C, for k ≥ k0, (5.11)

for some uniform constants c, C > 0 finishing the proof of the claim. �

The claim together with (5.9) yield to the estimate

area (Bσk(yk)) ≥ c σk

when k is sufficiently large. Similarly, area (Bρk(xk)) ≥ c ρk, for k sufficiently

large. This concludes the proof of the proposition. �

Let us denote briefly by Aσk := area (Bσk(yk)) and Aρk := area (Bρk(xk)).

Lemma 5.9. There exist a number k0 and constants c1 > 0, c2 > 0, so that

c1 τk ≤ Aρk ≤ c2 τk and c1 τk ≤ Aσk ≤ c2 τk, for all k ≥ k0

where τk = −tk.

Proof. Notice that

Aρk +Aσk ≤ 2A(tk) = 16π τk (5.12)

since A(tk) = 8πτk is the total surface area. Hence,

Aρk ≤ C τk and Aσk ≤ C τk. (5.13)


To establish the bounds from below, we will use Lemma 5.7 and show that

there is a uniform constant c so that

σk ≥ c τk and ρk ≥ c τk, for all k ≥ k0. (5.14)

Claim 5.10. There are uniform constants c > 0 and C <∞, so that

c ρk ≤ σk ≤ C ρk.

Proof. Recall that σk = distg(tk)(yk, γk). By our choice of points xk, ykand the figure we have that the diam(S2, g(tk)) ≤ σk + ρk + 1 for k ≥ k0,

sufficiently large. We also have that the subset of S2 that corresponds to

area A1(tk) contains a ball Bσk(yk). By Corollary 5.4 and the comparison

inequality (5.8), we have

c ≤ A2(tk)A1(tk)

≤ area(Bρk+σk+1(yk)\Bσk(yk))area(Bσk(yk))

=area(σk ≤ s ≤ ρk + σk + 1)

area(0 ≤ s ≤ σk)≤

(ρk + σk + 1)2 − σ2k

σ2k

.

Using the previous inequality we obtain the bound

c σ2k − 2 ρk − 2σk − 2ρk σk − 1 ≤ ρ2

k.

If there is a uniform constant c so that σk ≤ cρk we are done. If not, then

ρk << σk for k >> 1, and from the inequality above we get

c

2σ2k ≤ ρ2

k, for k >> 1.

In any case there are k1 and C1 > 0 so that

ρk ≤ C1σk, for k ≥ k1. (5.15)

By a similar analysis as above there are k2 ≥ k1 and C2 > 0 such that

σk ≤ C2ρk, for k ≥ k1. (5.16)

�

We will now conclude the proof of Lemma 5.9. By Lemma 5.7 and (5.12)

it follows that

ρk + σk ≤ Cτk, for k >> 1.

By (5.15) and (5.16) it follows that

ρk ≤ C τk and σk ≤ C τk, for k >> 1.


Moreover, by (5.8), we have

A2(tk)area (σk − 1 ≤ s ≤ σk)

≤ area (Bρk+σk+1(yk)\Bσk(yk))area (σk − 1 ≤ s ≤ σk)

≤(ρk + σk + 1)2 − σ2

k

σ2k − (σk − 1)2

≤(ρk + σk + 1)2 − σ2

k

2σk − 1

≤(ρk + σk + 1)2 − σ2

k

2σk − 1≤ C ρk (5.17)

where we have used (5.15) and (5.16). The same analysis that yielded to

(5.11) can be applied again to conclude that

area(σk − 1 ≤ s ≤ σk) ≤ C.

This together with Corollary 5.4 and (5.17) imply

ρk ≥ c τk, for k ≥ k0.

Claim 5.10 implies the same conclusion about σk. This is sufficient to con-

clude the proof of Lemma 5.9, as we have explained at the beginning of

it. �

We will now finish the proof of Proposition 5.1.

Proof of Proposition 5.1. If the isoperimetric constant I(t) ≡ 1, it follows

by a well known result that our solution is a family of contracting spheres.

Hence we will assume that I(t0) < 1, for some t0 < 0, which implies all the

results in this section are applicable. We will show that this contradicts the

fact that limt→−∞ v(·, t) = 0, uniformly on S2.

As explained at the beginning of this section, it suffices to find positive

constants δ, C and curves βk, so that

LS2(βk) ≥ δ > 0 and Lk(βk) ≤ C <∞ (5.18)

where LS2 denotes the length of a curve computed in the round spherical

metric and Lk denotes the length of a curve computed in the metric g(tk).

If we manage to find those curves βk that would imply

C ≥ Lk(βk) =∫βk

√u(tk) dS2 ≥M LS2(βk) ≥M δ, for k ≥ k0

where M > 0 is an arbitrary big constant and k0 is sufficiently large so that√u(tk) ≥ M , for k ≥ k0, uniformly on S2 (which is justified by the fact


v(·, t) converges uniformly to zero on S2, in C1,α norm). The last estimate

is impossible, when M is taken larger than C/δ, hence finishing the proof

of our proposition.

We will now prove (5.18). Our isoperimetric curves γtk have the property

that Lk(γtk) ≤ C for all k, but we do not know whether LS2(γtk) ≥ δ > 0,

uniformly in k. For each k, we will choose the curve βk which will satisfy

(5.18), from a constructed family of curves {βkα} that foliate our solution

(M, g(tk)). Define the foliation of (M, g(tk)) by the curves {βkα} so that for

every α and every x ∈ βkα, disttk(x, yk) = α. Choose a curve βk from that

foliation so that the corresponding curve βk on S2 splits S2 in two parts of

equal areas, where the area is computed with respect to the round metric.

Since the isperimetric constant for the sphere IS2 = 1, that is

1 ≤ LS2(βk) (1A1

+1A2

) = LS2(βk)4AS2

we have

LS2(βk) ≥ δ > 0, for all k.

To finish the proof of the proposition we will now show that there exists

a uniform constant C so that

Lk(βk) ≤ C, for all k.

To this end, we observe first that the area element of g(tk), when computed

in polar coordinates, is

dak = Jk(r, θ) r dr dθ

where Jk(r, θ) is the Jacobian and r is the radial distance from yk. The

length of βkr is given by

Lrk =∫ 2π

0Jk(r, θ) r dθ

which implies thatLrkr

=∫ 2π

0Jk(r, θ) dθ.

By the Jacobian comparison theorem, for each fixed θ, we have

J ′k(r, θ)Jk(r, θ)

≤ J ′a(r, θ)Ja(r, θ)

(5.19)

where the derivative is in the r direction and Ja(r, θ) denotes the Jacobian

for the model space and a refers to a lower bound on Ricci curvature (the


model space is a simply connected space of constant sectional curvature

equal to a). In our case a = 0 (since R ≥ 0) and the model space is the

euclidean plane, which implies that the right hand side of (5.19) is zero and

therefore Jk(r, θ) decreases in r. Hence Lrk/r decreases in r.

In the proof of Lemma 5.9 we showed that there are uniform constants

C1, C2 so that

C1τk ≤ ρk ≤ C2τk and c1τk ≤ σk ≤ C2τk.

We have shown that γtk → γ∞ and γ∞ is a circle in M∞. Let yk, pk

be the points which we have chosen previously. We may assume that

disttk(yk, pk) = σk. Choose a curve γk ∈ M so that pk ∈ γk and that

for every x ∈ γk we have the disttk(yk, x) = σk. Observe that for every

x ∈ γk, by the figure we have

σk ≤ disttk(x, yk) ≤ σk +C

σk,

for sufficiently big k. For x ∈ γk let z = γk ∩ ykx. Then

disttk(x, γk) ≤ disttk(x, z) ≤ disttk(yk, x)− dist(yk, z) ≤ σk +C

σk− σk =

C

σk.

This implies that the curves γk converge to γ∞ as k → ∞. Moreover,

this also implies the curve γk is at distance σk = O(τk) from yk and if

sk = disttk(βk, yk), then sk = O(τk) and we also know Lk(γk) ≤ C, for all

k. We may assume sk ≥ σk for infinitely many k, otherwise we can consider

point xk instead of yk and do the same analysis as above but with respect

to xk. Since Jk(r, θ) decreases in r we have

Lskksk≤Lσkkσk

that is

Lk(βk) = Lskk ≤skσk

Lσkk =skσk

Lk(γk) ≤ C, for all k

finishing the proof of (5.18) and the proposition.

�

6. Radial symmetry of an ancient compact solution

Let g(·, t) be a compact ancient solution to the Ricci flow in two dimen-

sions which becomes extinct at time t = 0. We denote, as in the previous

sections, by ψ, θ the global coordinates on S2 by which we parametrize


equation (1.8) (they are chosen as in Theorem 4.1, so that the zeros of the

backward limit v correspond to the poles of S2). We will show in this sec-

tion that our ancient solution needs to be radially symmetric with respect

to those coordinates.

Assuming that g = u ds2p, where ds2

p is the spherical metric on the limiting

sphere, we have shown in section 4 that the pressure v = u−1 satisfies

limt→−∞

v(ψ, θ, t) = v(ψ) := C cos2 ψ

for C ≥ 0, which in particular shows that the backward limit is radially

symmetric, that is,

vθ = 0.

Since we have discussed the case C = 0 in the previous section, we will

assume throughout this section that C > 0.

Theorem 6.1. The solution g(·, t) = u(·, t) ds2p, where ds2

p denotes the

spherical metric, is radially symmetric, that is, uθ(·, t) ≡ 0 for all t < 0.

Before we continue, let us recall our notation. We denoted by u(ψ, θ, t)

the conformal factor of our evolving metric on S2, by U(x, θ, t) the conformal

factor in cylindrical coordinates, by v(ψ, θ, t) = 1u(ψ,θ,t) the pressure func-

tion on S2 and by w(x, θ, t) = 1U(x,θ,t) the pressure function in cylindrical

coordinates. We recall that we have

v(ψ, θ, t) = w(x, θ, t) cos2 ψ , with cosψ = (coshx)−1.

Let us also denote by u(r, θ, t) the conformal factor of the metric in polar

coordinates on R2. The relation is as follows:

U(x, θ, t) = r2 u(r, θ, t), with x = log r.

It easily follows that u satisfies the equation

ut = ∆ log u, on R2 × (−∞, 0)

where ∆ is the Laplacian on R2 expressed in polar coordinates. If we differ-

entiate the previous equation in θ we get

(uθ)t = ∆(log u)θ = ∆( uθu

). (6.1)

Setting

I(t) :=∫

R2

|uθ| =∫

R×[0,2π]|uθ| dx dθ


we will show in the sequel that I(t) is well defined and decreasing in time.

Moreover, we will show that since the backward limit v is radially symmetric,

we have

limt→−∞

I(t) = 0.

Since I(t) ≥ 0 we will conclude that I(t) ≡ 0, which will prove that uθ ≡ 0,

therefore u is radially symmetric.

Proposition 6.2. For every t < 0, we have∫R2

|uθ| ≤ C(t).

Moreover,

d

dt

∫R2

|uθ| ≤ 0.

Proof. We recall Kato’s inequality, namely ∆f+ ≥ χf>0 ∆f , in the distri-

butional sense. Writing |f | = f+ + f−, it follows that ∆(|f |) ≥ signf ∆f .

Applying this inequality to the function uθ/u gives us the inequality

∆(|uθ|u

)≥ sign(uθ) ∆

( uθu

).

Multiplying (6.1) by sign(uθ) and using the previous inequality we obtain

|uθ|t ≤ ∆(|uθ|u

). (6.2)

Claim 6.3. For every t ≤ t0 there is a constant C(t), so that∫R2

|uθ| ≤ C(t).

Proof. Since w(x, θ, t) = v(ψ, θ, t) cosh2 x and since v is a positive function

on S2, for every time t there is a constant c(t) so that

w(x, θ, t) ≥ c(t) cosh2 x. (6.3)

Moreover,

|wθ| = |vθ| cosh2 x ≤ C cosh2 x (6.4)


since |∇S2v| ≤ C. Using (6.3), (6.4) and the change of coordinates, x = log r,

we obtain∫R2

|uθ| =∫ ∞

0

∫ 2π

0|uθ| r dθ dr =

∫ ∞−∞

∫ 2π

0

|Uθ|r2

r dθ r dx

=∫

R×[0,2π]|Uθ| dx dθ =

∫R×[0,2π]

|wθ|w2

dx dθ

≤ C(t)∫

R×[0,2π]

cosh2 x

cosh4 xdx dθ

= C(t)∫

R×[0,2π]

dx dθ

cosh2 x≤ C(t) <∞

which finishes the proof of the claim. �

Claim 6.4. For every t ≤ t0 < 0 there is a constant C(t) <∞ so that∫R2

|uθ|u≤ C(t).

Proof. We begin by noticing that by Lemma 2.10, we have

|wθ| =|vθ|

cos2 ψ≤ C cosψ =

C

coshx.

Hence, similarly as before, using also (6.3), we have∫R2

|uθ|u

=∫ ∞

0

∫ 2π

0

|uθ|u

dθ r dr =∫ ∞−∞

∫ 2π

0

|Uθ|U

r dθ r dx

=∫ ∞−∞

∫ 2π

0

|wθ|w

e2x dθ dx ≤ C∫

R×[0,2π]

e2x

w coshxdθ dx

≤ C(t)∫

R×[0,2π]

e2x

cosh3 xdθ dx ≤ C(t) <∞.

�

We will now conclude the proof of our proposition. We multiply (6.2) by

a cut off function η so that suppη ⊂ B2ρ := {(r, θ)| r < 2ρ}, η ≡ 1 for r ≤ ρ,

|∆η| ≤ C/ρ2 and then integrate the equation over R2 to get

d

dt

∫R2

|uθ| η ≤∫

R2

∆(|uθ|u

) η =∫

R2

|uθ|u

∆η

≤ C

ρ2‖ uθu‖L1(R2) ≤

C(t)ρ2

where we have used Claim 6.4. If we let ρ→∞ in the previous estimate we

getd

dt

∫R2

|uθ| ≤ 0.

�


Once we have Proposition 6.2 we can proceed with the proof of Theorem

6.1.

Proof of Theorem 6.1. Proposition 6.2 implies that the integral

I(t) :=∫

R2

|uθ| =∫

R×[0,2π]|Uθ| dx dθ

is decreasing in time. If we manage to show that the limt→−∞ I(t) = 0,

since I(t) ≥ 0, this will imply that I(t) ≡ 0, for all t < 0.

To this end, we recall that

limt→−∞

v(ψ, θ, t) = C cos2 ψ, C > 0 and limt→−∞

vθ(ψ, θ, t) = 0

which is equivalent to

limt→−∞

w(x, θ, t) = C > 0 and limt→−∞

wθ(x, θ, t) = 0.

Since wt ≥ 0, this means that w(x, θ, t) ≥ C for all t < 0 and therefore

using Lemma 2.10 we have

|Uθ| =|wθ|w2≤ |vθ| sec2 ψ

C2≤ C cosψ =

C

coshx(6.5)

for all t ≤ t0 < 0 and with C a uniform, time independent positive constant.

Since (coshx)−1 ∈ L1(R × [0, 2π]), by Lebesgue’s dominated convergence

theorem we conclude that

limt→−∞

∫R×[0,2π]

|Uθ| dx dθ =∫

R×[0,2π]lim

t→−∞|Uθ| dx dθ = 0

since limt→−∞ |Uθ| = limt→−∞|wθ|w2 = 0. Hence, I(t) ≡ 0 for all t < 0 and

therefore

uθ(x, θ, t) ≡ 0

which implies that u(x, θ, t) is independent of θ, that is, our solution u is

radially symmetric. �

7. Classification of radial ancient solutions in the case v 6= 0

In the previous section we have shown that our ancient solution is radially

symmetric with respect to the coordinates chosen in Theorem 4.1. In this

section we will prove Theorem 1.2, assuming that the solution v of (1.1) is

radial, hence v = v(ψ, t). Since the case v ≡ 0 has been discussed in section

5, we will assume v 6= 0, which by Proposition 4.1 means v = C cos2 ψ, or

equivalently w = limt→−∞w(·, t) = C, for C > 0.


Let us briefly outline the main steps of the proof of Theorem 1.2.

i. We define the positive quantity F (x, t) = (wxxx − 4wx)2 for which it

turns out that the supR F (·, t) decreases in time. Our goal is to show that

the limt→−∞ supR F (·, t) = 0, since this together with the monotonicity

yields to that F ≡ 0. We know that F (·, t) converges to zero pointwise,

however we will need to control F (·, t) near |x| → ∞ and uniformly in

time t to obtain the desired result.

ii. To obtain the results discussed above, we show the bound F (·, t) ≤ C√−t ,

for t ≤ t0 < 0 and a uniform constant C. We do so by comparing F

(which is a subsolution to a heat equation where the laplacian is com-

puted with respect to the changing metric g(t)) with the fundamental

solution to the heat equation on R2, centered at an arbitrary point

x0 ∈ R, namely with φ(x, t) = 1√−t e

|x−x0|2

b2(−t) , for b chosen appropriately.

We will do the comparison on the region

Sτ := {(x, θ) | |x− x0| ≤ b√τ log τ}, τ = −t.

iii. In order to compare our subsolution to the fundamental solution, we

need to show that our metric g = w−1 (dx2+dθ2) is uniformly equivalent

to the cylindrical metric ds2 = dx2 + dθ2 in Sτ , where the comparison

is being performed. We show that supSτ R(·, t) decays at a rate at most

τ−α for any α ∈ (0, 1). We conclude the bound C1 ≤ w(·, t) ≤ C2(x0),

which implies that all the derivatives of w decay nicely on Sτ , as τ →∞.

iv. The bound on F which was discussed above implies that limt→−∞ F (·, t) =

0, uniformly on R. Hence F (·, t) ≡ 0, for all t < 0, i.e. wxx − 4w ≡ 0.

It is now straightforward to conclude that our condition w must be a

King-Rosenau solution.

We will next proceed to the detailed proof of Theorem 1.2. To define our

quantity F , we will use cylindrical coordinates. To change to cylindrical co-

ordinates we use as before Mercator’s projection (1.4) and denote by w(x, t)

the function defined by

w(x, t) = v(ψ, t) cosh2 x, coshx = secψ.

The function w(x, t) solves the equation

wt = wwxx − w2x, for (x, t) ∈ R× (−∞, 0). (7.1)


Notice that the King-Rosenau solution given by (1.10), when expressed in

cylindrical coordinates, it takes the form

wR(x, t) = c(t) + d(t) cosh 2x

for some functions of time c(t) and b(t). Define

Q(·, t) := wxx(·, t)− 4w(·, t)

and observe that on both, the contracting spheres and the King-Rosenau

solution, Q is just a function of time and therefore Qx(·, t) ≡ 0. This

motivates our definition

F (·, t) := Q2x(·, t)

as our good monotone quantity. Indeed, we have:

Lemma 7.1. The function F := Q2x satisfies the differential inequality

Ft ≤ wFxx. (7.2)

Proof. We begin by differentiating (7.10) in x once, to get

(wx)t = wwxxx − wxwxx

and then once again, to also get

(wxx)t = wwxxxx − w2xx.

Hence

Qt = wQxx − w2xx + 4w2

x.

Furthermore,

(Qx)t = wQxxx + wxQxx − 2wxxQx

which implies that

(Q2x)t = w (Q2

x)xx − 2wQ2xx + 2wxQxQxx − 4wxxQ2

x.

By the interpolation inequality and the fact that

wxx =w2x +Rw

w≥ w2

x

w


since R > 0, we obtain

(Q2x)t = w (Q2

x)xx − 2wQ2xx + 2 (

√wQxx) (

wxQx√w

)− 4wxxQ2x (7.3)

≤ w (Q2x)xx − wQ2

xx +w2xQ

2x

w− 4

w2xQ

2x

w

≤ w (Q2x)xx

which proves (7.2). �

Our goal is to utilize the maximum principle on (7.2) to conclude that

F ≡ 0. To this end we will need some a priori bounds on F .

Lemma 7.2. The quantity F := Q2x is uniformly bounded in time, that is

there exists a constant C such that

F (·, t) ≤ C, for all t ≤ t0. (7.4)

Moreover, at each time-slice t, F is a smooth function with all the derivatives

uniformly bounded in space.

Proof. We will prove the lemma by considering Qx to be a function on the

sphere, that is we define G(ψ, t) to be equal to Qx(x, t) after the transfor-

mation

w(x, t) = v(ψ, t) sec2 ψ anddψ

dx=

1coshx

= cosψ.

Remembering that Qx = wxxx − 4wx, we compute

G = [[(v sec2 ψ)ψ cosψ]ψ cosψ]ψ cosψ − 4(v sec2 ψ)ψ cosψ

= (− cosψ + 2 secψ + sinψ tanψ) vψ + 3 sinψ vψψ + cosψ vψψψ.(7.5)

We see from the above expression that

|G(ψ, t)| ≤ C(δ), for ψ ∈ [−π2

+ δ,π

2− δ]

with C(δ) independent of time. To also see that G(ψ, t) is bounded near

the poles ψ = ±π/2, we recall the bounds in Proposition 2.5 which readily

imply that the first four terms on the right hand side of the last equation are

bounded there, independently of time. To see that the last term is bounded

as well, we observe that since vt ≥ 0, by Theorem 4.1, we have

v(ψ, t) ≥ limt→−∞

v(ψ, t) = C cos2 ψ


for some constant C > 0 (the case C = 0 has been discussed in section 5).

Hence, by Corollary 2.12 we obtain

cosψ vψψψ ≤√v

Cvψψψ ≤ C.

We conclude that G(ψ, t) is bounded near the poles as well, independently

of time. This proves the bound in (7.4).

Also, from (7.5) it follows that F (·, t) := Q2x(·, t) has bounded higher order

derivatives, at each time-slice t. �

By Lemma 7.2 the supF (·, t) is finite for every t. From the evolution

equation for F (·, t) we immediately obtain the following lemma.

Lemma 7.3. The supF (·, t) decreases in time.

Proof. Using that dψ/dx = cosψ, we can consider (7.3) as an equation on

S2. Hence

d

dtF ≤ w [Fψ cosψ]ψ cosψ (7.6)

≤ wFψψ cos2 ψ − wFψ sinψ cosψ

= w cos2 ψ (Fψψ − Fψ tanψ)

= v∆S2F.

We have seen in Lemma 7.2 that for each t, F (ψ, t) is a smooth function on

S2. Hence, by the maximum principle applied to (7.6) we easily obtain that

Fmax(t) is decreasing in time, or equivalently, supQ2x decreases in time. �

By Theorem 4.1, w(·, t) converges, as t → −∞, uniformly on compact

subsets of R, to a constant C ≥ 0 which we assume to be positive. Since

wt ≥ 0, we have w(·, t) ≥ C, for all t. It follows by standard parabolic

estimates applied to (7.10), that the derivatives of w converge, as t→ −∞,

uniformly on compact subsets of R to zero. Hence, F ≡ 0 and F (·, t) t→−∞−→ 0,

uniformly on compact subsets of R. We would like to show that actually the

convergence is uniform on R. This will follow from the following proposition.

Proposition 7.4. We have

F (·, t) ≤ 1√−t, for t ≤ t0 < 0. (7.7)


To prove the Proposition we will compare F (·, t) with the fundamen-

tal solution to the heat equation centered at x0, namely with φ(x, t) =

1√−t e

|x−x0|2

b2(−t) . To do so we will first need some extra estimates, which will be

shown in the next three lemmas. Fix x0 ∈ R to be an arbitrary point.

Lemma 7.5. There exists a uniform constant in time C(x0), so that the

scalar curvature R(x0, t) satisfies the estimate

R(x0, t) ≤C(x0)(−t)

, for t ≤ t0 < 0.

Proof. Notice that

R(x0, t) = R(ψ0, t) = (log v)t(ψ0, t)

where x0 ∈ R and ψ0 ∈ (−π2 ,

π2 ) are related via Mercator’s transformation.

Take a sequence of points ti = −2i ≤ t0. Then,∫ ti

ti+1

R(x0, s) ds = log v(ψ0, ti)− log v(ψ0, ti+1)

which, since 0 < c1(x0) ≤ v(x0, t) ≤ c2(x0) for all t < t0, implies the bound

R(x0, τi) ≤C(x0)

2i

for some τi ∈ (ti+1, ti). By the Harnack estimate for ancient solutions to the

Ricci flow, Rt ≥ 0. Using that property, for any t ∈ [τi+1, τi], we have

R(x0, t) ≤ R(x0, τi) ≤C(x0)

2i≤ C(x0)

(−t).

�

Let x0 ∈ R be arbitrary and let Sτ be as before. Then we have the

following lemma.

Lemma 7.6. For every α > 0 there exist b > 0, independent of x0 and

C(x0) > 0, so that

supSτ

R(·, t) ≤ C(x0)τα

, for t ≤ t0 < 0. (7.8)

Proof. To simplify the notation, set xτ = x0 ± b√τ log τ . Let x ∈ Sτ . We

will use the Harnack estimate for the curvature R shown in [6], which in our

case implies the bound

R(x, t) ≤ C R(x0,t

2) e

2dist2t (x0,x)

τ


for a uniform constant C. Using that wt ≥ 0 and limt→−∞w(x, t) = C > 0,

which implies w(x, t) ≥ C > 0, we can estimate the distance at time t, as

follows:

distt(x0, x) ≤∫ x

x0

dx√w(x, t)

≤ C |xτ − x0| ≤ C b√τ log τ

where the integral above is taken over a straight euclidean line connecting

the points x0 and xτ . Using Lemma 7.5, we obtain the bound

R(xτ , t) ≤C(x0)τ

τC2b2 , (7.9)

where C is a uniform constant, independent of time and x0. Choosing b so

that 1− C2b2 = α ∈ (0, 1) yields to the lemma. �

Remark 7.7. From the proof of Lemma 7.6, the estimate (7.8) holds also on

a ball Bg(t)(x0, b√τ log τ), where the ball is taken in metric g(t).

Let b be as in Lemma 7.6.

Lemma 7.8. For every x0 ∈ R there is a uniform constant C(x0) so that

w(x, t) ≤ C(x0), on Sτ (7.10)

for every t ≤ t0 < 0. Moreover, F → 0 as t→ −∞ uniformly on Sτ .

Proof. We will use Shi’s local derivative estimates to prove the Lemma. For

every t < t0 consider g(x, s) = g(x, 2t + s) for s ∈ [0,−t] and x ∈ Sτ . By

Lemma 7.6 we have

max[−2t,−t]×Sτ

R(x, s) ≤ C

τα,

where C depends on x0 but does not depend on τ and R is the scalar

curvature of g. Since Shi’s estimates are local in nature, to get derivative

estimates on all of Sτ we can initially take slightly bigger b in a definition

of Sτ if necessary. From Remark 7.7 and by Shi’s local derivative estimates,

there is a constant Cm so that for every (x, θ) ∈ Sτ , we have

supSτ

|∇mR|2g(s)(·, s) = supSτ

|∇mR|2g(s+2t)(·, s+ 2t) ≤ Cm

sα(2+m),

for all s ∈ (0,−t]. In particular, if s ∈ [−t/2,−t] this yields

supSτ

|∇mR|g(t′)(·, t′) ≤C

(−t′)α(1+m/2)


for all t′ ∈ [−3t/2,−t], where C is a universal constant. In particular it holds

for t′ = t. Note that the derivative and the norm in the previous estimate

are computed with respect to the metric g(·, t). Since c1 ≤ w(x, t) for all x

and since the previous analysis has been done for arbitrary t ≤ t0 < 0, we

get

supSτ

|∇mR|(·, t) ≤ C

(−t)α(1+m/2), for t ≤ t0 < 0. (7.11)

The number α can be taken to be any positive number less than one, C

is a universal constant depending only on x0 which is the center of the set

Sτ and the derivative and the norm in (7.11) are taken with respect to the

standard cylindrical metric on R2 (in our case of a rotationally symmetric

solution ∇ is just a derivative with respect to the x coordinate).

We will now conclude the proof of the lemma. The equation wt = Rw,

yields to∂

∂tlnw = R

and therefore for τ = −t implies

∂

∂τ∇m lnw = −∇mR.

If we integrate this over [τ, τ ′], using (7.11) we get

|∇m lnw(x, τ ′)−∇m lnw(x, τ)| ≤∫ τ ′

τ|∇mR| ds ≤

∫ τ ′

τ

C ds

sα(1+m/2)

=C

α(m/2 + 1)− 1

(1

τα(m/2+1)−1− 1τ ′α(m/2+1)−1

) (7.12)

for every x ∈ Sτ . We choose α ∈ (12 , 1) so that α(m/2+1) > 1 for every m ≥

1. Letting τ ′ →∞ in (7.12) and using that the limτ ′→∞∇m lnw(x, τ ′) = 0

we get

supSτ

|∇m lnw| ≤ Cm

τα(m/2+1)−1. (7.13)

This in particular implies that for every x ∈ Sτ , we have

| lnw(x, t)− lnw(x0, t)| ≤ supSτ

|(lnw)x| |x− x0| ≤C

τ3α2

√τ log τ ≤ C

and therefore

supSτ

w(x, t) ≤ C(x0), for all t ≤ t0 < 0.

Finally, by (7.13) and the previous estimate we obtain the bound

supSτ

|∇mw| ≤ Cm

τα(m/2+1)−1


which implies F → 0 uniformly as t→ −∞ on Sτ . �

We will now finish the proof of Proposition 7.4.

Proof of Proposition 7.4. We will compare F (·, t) with the fundamental so-

lution to the heat equation centered at x0. More precisely, we set

φε(x, t) =1√−t

e|x−x0|

2

b2(−t) + ε

where b is taken from Lemma 7.6. An easy computation shows that

φεt =b2

4φεxx.

Let C(x0) be the constant defined in 7.10. By (7.10), since φεxx ≥ 0, we have

∂φε

∂t=

b2

4C(x0)C(x0)φεxx ≥ C(x0)wφεxx (7.14)

for |x− x0| ≤ b√τ log τ , where C(x0) = b2

4C(x0) . Define

F (x, t) = F (x, C(x0) t).

It follows from (7.2) that F satisfies the differential inequality

∂F

∂t≤ C(x0)wFxx. (7.15)

By Lemma 7.2 we have that F ≤ C, for a uniform constant C. Hence

F (x, t) ≤ C = φε(x, t) (7.16)

for |x − x0| = b√2

√τ√

log τ + 2 log(C − ε). By Lemma 7.8, for every ε > 0

there is a tε so that for any t for which C(x0) t ≤ tε, we have

F (x, t) < ε ≤ φε(x, t), on Sτ

and in particular (for τ sufficiently large) on

|x− x0| ≤b√2

√τ√

log τ + 2 log(C − ε).

This together with (7.16), equations (7.14), (7.15) and the comparison prin-

ciple yield to the inequality

F (x, t) ≤ φε(x, t)

for all |x− x0| ≤ b√2

√τ√

log τ + 2 log(C − ε) and all t ≤ t0. In particular,

the estimate holds for x0, that is, after passing to the limit ε→ 0, we have

F (x0, t) ≤√C(x0)√−t

, for t ≤ C(x0) t0


where C(x0) = b2

C(x0) . Recall that b does not depend on x0 and that C(x0)

is the constant from Lemma 7.8 and therefore it can be taken bigger than 1.

Hence, C(x0) ≤ A, where A is a uniform constant, independent of x0. We

conclude that

F (x0, s) ≤√A√−t, for t ≤ A t0

since A t0 ≤ C(x0) t0 (recall that t0 < 0). The estimate (7.7) now readily

follows. �

As an easy consequence of Lemma 7.3 and Proposition 7.4 we have the

following Corollary.

Corollary 7.9. F (·, t) ≡ 0, for all t < 0.

Proof. Proposition 7.4 implies that F (·, t) converges to zero, as t → −∞,

uniformly on R. Hence

limt→−∞

supF (·, t) = sup limt→−∞

F (·, t) = 0.

Since F ≥ 0, by Lemma 7.3 we conclude that

F (·, t) ≡ 0, for t < 0.

�

Proof of Theorem 1.2 - Radial Case. Corollary 7.9 shows that

wxx − 4w = c(t), for every t < 0.

Solving the above ODE gives us the solution

w(x, t) = a(t) e2x + b(t) e−2x + d(t)

with d(t) = −c(t)/4. When we plug it in equation (7.10) we get the following

system of ODE’s:

a′(t) = 4 a(t) d(t)

b′(t) = 4 b(t) d(t)

d′(t) = 16 a(t) b(t).

Since w(x, t) > 0, we obviously have a(t) > 0 and b(t) > 0. Hence, the first

two equations imply that

(log a(t))′ = (log b(t))′


which shows that

b(t) = λ2 a(t)

for a constant λ > 0. Since b(t) = λ2 a(t), we may express w(x, t) as

w(x, t) = λ

(a(t)λ

e2x + λ a(t) e−2x

)+ d(t)

= λ a(t)(e2(x−x0) + e−2(x−x0)

)+ d(t)

= 2λ a(t) cosh(2 (x− x0)) + d(t)

with λ = e2x0 . Hence

w(x, t) = a(t) cosh(2 (x− x0)) + d(t)

with a(t) = 2λ a(t). The functions a(t) and d(t) satisfy the system

a′(t) = 4 a(t) d(t)

d′(t) = 4 a2(t).

Solving this system gives us

a(t) = −µ csch(4µt) and d(t) = −µ coth(4µt)

for a positive constant µ > 0. Combining the above, we conclude that

w(x, t) = −µ [ csch(4µt) cosh(2 (x− x0)) + coth(4µt)].

Assume for simplicity that x0 = 0. Then, we obtain the solution

w(x, t) = −µ [csch(4µt) cosh(2x) + coth(4µt)].

Let us now express w on a sphere. The corresponding pressure function v

on the sphere is defined by

v(ψ, t) = (coshx)−2w(x, t), secψ = coshx.

Using the formulas cosh(2x) = 2 cosh2 x − 1 and coth(4µt) − csch(4µt) =

tanh(2µt) we have

v(ψ, t) = −µ (coshx)−2 {csch(4µt) [2 cosh2 x− 1] + coth(4µ t)}

= −µ {2 csch(4µt) + [coth(4µt)− csch(4µt)] cos2 ψ}

= −µ {2 csch(4µt) + tanh(2µt) (1− sin2 ψ)

= −µ{[2 csch(4µt) + tanh(2µt)]− tanh(2µt) sin2 ψ}


Finally, since 2 csch(4µt) + tanh(2µt) = coth(2µt) we conclude that

v(ψ, t) = −µ coth(2µt) + µ tanh(2µt) sin2 ψ

and the proof of our theorem (in the radial case) is now complete.

�

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Department of Mathematics, Columbia University, New York, USA

E-mail address: [email protected]





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