# Strict type-II blowup in harmonic map flow

Alex Waldron  University of Wisconsin, Madison<EMAIL_ADDRESS>
###### Abstract.

A finite-time singularity of 2D harmonic map flow will be called “strictly
type-II” if the outer energy scale satisfies

$\lambda(t)=O(T-t)^{\frac{1+\alpha}{2}}.$

We prove that the body map at a strict type-II blowup is Hölder continuous.

##  1\. Introduction

Let $M$ and $N$ be compact Riemannian manifolds. For differentiable maps
$u:M\to N,$ we may define the Dirichlet energy:

$\int_{M}|du|^{2}\,dV.$

Its downward gradient flow is given by

(1.1) $\frac{{\partial}u}{{\partial}t}=\mathcal{T}(u).$

Here, ${\mathcal{T}}(u)$ is the tension field of $u,$ a generalization of the
Laplace-Beltrami operator to maps between manifolds. The evolution equation
(1.1), known as _harmonic map flow_ , was introduced in 1964 by Eells and
Sampson [5] and has been studied since then almost without interruption.

When $M$ has dimension two, the Dirichlet functional is conformally invariant;
we shall be concerned exclusively with this case. Struwe [11] constructed a
global weak solution of (1.1) starting from any initial map of Sobolev class
$W^{1,2}.$ The solution is smooth away from finitely many singular times, when
isolated point singularities (“bubbles”) may form. The trace at a singular
time, $u(T),$ is referred to as the _body map_.

Topping [12] demonstrated that for certain target manifolds and initial data,
the body map at the first singular time may be discontinuous—more precisely,
$u(T)$ can have an essential singularity at one of the bubble points. At the
same time, he conjectured that for a well-behaved (specifically, real-
analytic) metric on $N,$ the body map must be continuous. Topping’s conjecture
is the sine qua non for future geometric applications of harmonic map flow.

In previous joint work with C. Song [10], we proved continuity of the body map
when $N$ is compact Kähler with nonnegative holomorphic bisectional curvature
and the energy of the initial map is near the holomorphic energy. The argument
relied partly on establishing a bound on the outer energy scale of the form

(1.2) $\lambda(t)=O(T-t)^{\frac{1+\alpha}{2}},$

where $0<\alpha\leq 1.$ We shall refer to a singularity satisfying (1.2) as a
_strict type-II blowup_. Note that (1.2) is a refinement of the ordinary type-
II estimate

(1.3) $\lambda(t)=o(T-t)^{\frac{1}{2}},$

which automatically holds for 2D harmonic map flow—see e.g. [12], Theorem
1.6v.

With $\alpha=1,$ the bound (1.2) is best known from the rotationally symmetric
setting. Angenent, Hulshof, and Matano [1] proved that the finite-time blowup
first constructed by Chang, Ding, and Ye [3] occurs with rate
$\lambda(t)=o(T-t).$ Raphael and Schweyer [9] determined a large set of
rotationally symmetric maps that blow up under (1.1) with the precise rate
$\lambda(t)\sim\frac{T-t}{|\ln T-t|^{2}};$ they also proved in this context
that the body map is $W^{2,2},$ hence $C^{\beta}$ for each $\beta<1.$111This
also follows from the main theorem of [10]. Davila, Del Pino, and Wei [4]
constructed a larger set of examples which blow up at the same rate, with
continuous body map.

Accordingly, all known examples of strict type-II blowup have continuous body
map. On the other hand, the only known example with discontinuous body map,
due to Topping, fails to be strictly type-II—see [12], Theorem 1.14e. Our main
theorem confirms the implication, as follows.

######  Theorem 1.1.

For any strict type-II blowup of (1.1) in dimension two, the body map is
$C^{\frac{\alpha}{3}}.$

The proof depends on obtaining decay estimates for the differential of $u$ in
the “neck region,” i.e., the area close to the singularity but outside the
energy scale. The required estimate on the angular component of $du$ is known
from [10], so it remains only to estimate the radial component. We obtain a
preliminary bound on the difference between the radial and angular components
from a well-known integral identity (2.3). Using a parabolic estimate on the
radial component, we are able to promote this integral bound to pointwise
decay throughout the neck region.

### Acknowledgement

The author is partially supported by DMS-2004661.

##  2\. Technical results

For an introduction to harmonic map flow, we refer the reader to §2 of our
previous paper [10] or to the textbook of Lin and Wang [7].

Given any smooth map $u:M\to N,$ the _stress-energy tensor_

$S_{ij}=\left\langle\partial_{i}u,\partial_{j}u\right\rangle-\frac{1}{2}g_{ij}|du|^{2}$

satisfies

(2.1)
$\nabla^{i}S_{ij}=\left\langle{\mathcal{T}}(u),\partial_{j}u\right\rangle,$

where ${\mathcal{T}}(u)=\operatorname{tr}_{g}\nabla du.$ See [10], §2.2, for a
derivation.

Suppose that $\dim(M)=2.$ Choose conformal coordinates $X^{i}$ centered at
$x_{0}\in M,$ and let $r=\sqrt{\delta_{ij}X^{i}X^{j}}.$ Since the radial
vector field is conformally Killing, we have

$\nabla^{i}X^{j}+\nabla^{j}X^{i}=\lambda g^{ij}$

for a scalar function $\lambda.$ Contracting with $X^{i}$ in (2.1), we obtain

(2.2)
$\begin{split}\nabla^{i}(X^{j}S_{ij})&=\left\langle{\mathcal{T}}(u),X^{j}\partial_{j}u\right\rangle+\frac{1}{2}\left(\nabla^{i}X^{j}+\nabla^{j}X^{i}\right)S_{ij},\\\
&=\left\langle{\mathcal{T}}(u),X^{j}\partial_{j}u\right\rangle,\end{split}$

by tracelessness of the stress-energy tensor in dimension two. Integrating
over $B_{r}(x_{0})$ and applying the divergence theorem, we have

(2.3)
$\int_{S^{1}_{r}(x_{0})}X^{i}X^{j}S_{ij}\,d\theta=\int_{B_{r}(x_{0})}\left\langle{\mathcal{T}}(u),X^{j}\partial_{j}u\right\rangle\,dV.$

This identity is well known from the theory of approximate harmonic maps, and
will be used in a crucial way below.

Next, we need the following parabolic estimates. Given $0<\rho<1,$ let

$U_{\rho}^{1}=\\{(r,\theta)\mid\rho\leq r\leq 1\\}$

be an annulus in the plane. Assume that $g$ is a metric on $U^{1}_{\rho}$ of
the form

(2.4) $g=\xi^{2}(dr^{2}+r^{2}d\theta^{2}),$

where $\xi$ be a positive smooth function with

(2.5)
$\left|\xi^{2}-1\right|+r|\nabla\xi|+r^{2}|\nabla^{2}\xi|\leq\xi_{0}r^{2}$

for a constant $0\leq\xi_{0}\leq\frac{1}{2}.$ Note that

$|r-{\mbox{dist}}_{g}(x_{0},\cdot)|\leq C\xi_{0}r^{2}.$

In view of this bound, the difference between conformal and geodesic
coordinates will be of no significance for our results.

For $0\leq\nu\leq 1,$ let

$\square_{\nu}=\partial_{t}-\left(\partial_{r}^{2}+\frac{1}{r}\partial_{r}-\frac{\nu^{2}}{r^{2}}\right).$

Notice that

$\square_{\nu}r^{\beta}=\left(\nu^{2}-\beta^{2}\right)r^{\beta-2},$

and

(2.6)
$\square_{\nu}\left(r^{\nu}y\right)=r^{\nu}\left(\partial_{t}-\Delta_{2(\nu+1)}\right)y,$

where $\Delta_{\mu}=\partial_{r}^{2}+\frac{(\mu-1)}{r}\partial_{r}.$

######  Lemma 2.1.

Let $u$ be a solution of (1.1) on $U^{1}_{\rho}\times[0,T)$ with respect to a
metric $g$ of the above form. Suppose that for some
$0<\eta^{2}<\varepsilon_{0},$ we have

(2.7) $r|du|+r^{2}|\nabla du|+r^{3}|\nabla^{2}du|\leq\eta.$

Then the angular energy

$f=f(u;r,t):=\sqrt{\int_{\\{r\\}\times
S^{1}}|u_{\theta}(r,\theta,t)|^{2}d\theta+\int_{\\{r\\}\times
S^{1}}|\nabla_{\theta}u_{\theta}(r,\theta,t)|^{2}d\theta}$

satisfies a differential inequality

(2.8) $\square_{\nu}\\!f\leq C\xi_{0}\eta,$

where $\nu=\sqrt{1-C\eta}.$ The radial energy

(2.9) $g=g(u;r,t):=\sqrt{\int_{\\{r\\}\times
S^{1}}r^{2}|u_{r}(r,\theta,t)|^{2}d\theta}$

satisfies

(2.10)
$\square_{\nu}\\!\left(\frac{g}{r}\right)\leq\frac{6f}{r^{3}}+\frac{C\xi_{0}\eta}{r}.$

Here, $\varepsilon_{0}$ depends on the geometry of $N,$ and $\xi_{0}$ is the
constant of (2.5).

###### Proof.

The proof is an elementary extension of prior calculations; see Appendix A. ∎

######  Proposition 2.2.

Let $-\nu<\beta_{i}<\nu\leq 1,$ for $i=1,2.$ Suppose that $f(r,t)$ is
continuous on $\left[\rho,1\right]\times\left[\tau,T\right)$ and satisfies

(2.11) $\Box_{\nu}\\!f\leq A$

with

$\begin{split}f(\rho,t)\leq A,\qquad f(1,t)\leq A,\qquad|f(r,\tau)|\leq
A\left(\left(\frac{\rho}{r}\right)^{\beta_{0}}+r^{\beta_{1}}\right).\end{split}$

Then, given $0<\kappa\leq 1/2,$ for

$\frac{\rho}{\kappa}\leq
r\leq\kappa\quad\text{and}\quad\tau+\frac{r^{2}}{\kappa^{2}}\leq t<T,$

we have

(2.12) $|f(r,t)|\leq
C_{\ref{festimate}}A\left(\kappa^{\nu-\beta_{0}}\left(\frac{\rho}{r}\right)^{\beta_{0}}+\kappa^{\nu-\beta_{1}}r^{\beta_{1}}\right).$

Here $C_{\ref{festimate}}$ depends on $\beta_{0},\beta_{1},$ and $\nu.$

###### Proof.

Using the results of Appendix B.1, we can construct a supersolution for (2.11)
of the form

$\bar{v}(r,t)=r^{\nu}v_{0}(r,t)+2A\left(\left(\frac{\rho}{r}\right)^{\nu}+r^{\nu}\right)-\frac{A}{4-\nu^{2}}r^{2},$

where

$\left(\partial_{t}-\Delta_{2(\nu+1)}\right)v_{0}=0,$
$v_{0}(r,0)=r^{-\nu}f(r,0),\qquad v_{0}(\rho,t)=0=v_{0}(1,t).$

Applying the comparison principle to (2.11), we have

(2.13) $g(r,t)\leq\bar{v}(r,t).$

By Proposition B.2 and (2.2), we have

$v_{0}(r,t)\leq
CA\left(\frac{\rho^{\beta_{0}}}{r^{\beta_{0}+\nu}}w^{\beta_{0}+\nu}\left(r,t-\tau\right)+r^{\beta_{1}-\nu}w^{\nu-\beta_{1}}(r,t-\tau)\right),$

where $w^{a}(r,t)$ is defined by

$w^{a}(r,t)=\left(\frac{r^{2}}{r^{2}+t}\right)^{a/2}.$

Overall, from (2.13), we obtain

(2.14) $|f(r,t)|\leq
CA\left(\left(\frac{\rho}{r}\right)^{\beta_{0}}w^{\beta_{0}+\nu}(r,t-\tau)+r^{\beta_{1}}w^{\nu-\beta_{1}}(r,t-\tau)+\left(\frac{\rho}{r}\right)^{\nu}+r^{\nu}+r^{2}\right).$

For $r\geq\rho/\kappa,$ we have

$\left(\frac{\rho}{r}\right)^{\nu}\leq\kappa^{\nu-\beta_{0}}\left(\frac{\rho}{r}\right)^{\beta_{0}}.$

For $r\leq\kappa$ and $t-\tau\geq r^{2}/\kappa^{2},$ we have

$w^{\beta_{0}+\nu}(r,t-\tau)\leq\kappa^{\beta_{0}+\nu},\qquad
w^{\nu-\beta_{1}}(r,t-\tau)\leq\kappa^{\nu-\beta_{1}},\qquad
r^{\nu}\leq\kappa^{\nu-\beta_{1}}r^{\beta_{1}}.$

Hence (2.12) clearly follows from (2.14). ∎

######  Proposition 2.3.

Let $-\nu<\gamma_{i}<\beta_{i}<\nu\leq 1,$ for $i=0,1.$ Given $0<\kappa\leq
1/2,$ let $\rho\leq\rho_{1}\leq\kappa.$

Suppose that $g(r,t)$ is continuous on
$\left[\rho_{1},1\right]\times\left[\tau,T\right)$ and satisfies

(2.15)
$\square_{\nu}\\!\left(\frac{g}{r}\right)\leq\frac{A}{r^{3}}\left(\left(\frac{\rho}{r}\right)^{\beta_{0}}+r^{\beta_{1}}\right),$

with

(2.16) $\begin{split}\int_{\tau}^{T}\\!\\!g(\rho_{1},t)^{2}\,dt&\leq
B^{2},\qquad|g(1,t)|\leq A,\\\ |g(r,\tau)|&\leq
A\left(\left(\frac{\rho}{r}\right)^{\gamma_{0}}+r^{\gamma_{1}}\right).\end{split}$

Then, for

(2.17) $2\rho_{1}\leq
r\leq\kappa\quad\text{and}\quad\tau+\frac{r^{2}}{\kappa^{2}}\leq t<T,$

we have

(2.18) $|g(r,t)|\leq
C_{\ref{gestimate}}\left(B\frac{\rho_{1}^{\nu-1}}{r^{\nu}}+A\left(\kappa^{\nu+\gamma_{0}+1}\left(\frac{\rho}{r}\right)^{\gamma_{0}}+\kappa^{\nu-\gamma_{1}+1}r^{\gamma_{1}}+\left(\frac{\rho}{r}\right)^{\beta_{0}}+r^{\beta_{1}}\right)\right).$

Here, $C_{\ref{gestimate}}$ depends on
$\gamma_{0},\gamma_{1},\beta_{0},\beta_{1},$ and $\nu.$

###### Proof.

As in the previous proof, we construct a supersolution for (2.15) of the form

$\bar{v}(r,t)=r^{\nu}\left(v_{1}(r,t)+v_{2}(r,t)\right)+CA\left(\frac{\rho^{\beta_{0}}}{r^{\beta_{0}+1}}+r^{\beta_{1}-1}\right),$

where

$\left(\partial_{t}-\Delta_{\mu}\right)v_{i}=0,\quad i=1,2,$

with

$v_{1}(r,0)=r^{-\nu-1}g(r,0),\qquad v_{1}(\rho,t)=0=v_{2}(1,t),$

and

$v_{2}(\rho_{1},t)=\rho_{1}^{-\nu-1}g(\rho_{1},t),\qquad
v_{2}(r,0)=0=v_{3}(1,t).$

Applying the comparison principle to (2.15), we have

(2.19) $g(r,t)\leq r\bar{v}(r,t).$

By Proposition B.4 and (2.16), we have

$v_{1}(r,t)\leq
CA\left(\frac{\rho_{1}^{\gamma_{0}}}{r^{\gamma_{0}+\nu+1}}w^{\gamma_{0}+\nu+1}\left(r,t\right)+r^{\gamma_{1}-\nu-1}w^{\nu-\gamma_{1}+1}(r,t)\right).$

By Proposition B.4 and (2.16), since $\mu-2=2\nu,$ we have

$v_{2}(r,t)\leq
CB\rho_{1}^{-\nu-1}\left(\frac{\rho_{1}^{2\nu}}{r^{2\nu+1}}\right)\leq
CB\frac{\rho_{1}^{\nu-1}}{r^{2\nu+1}}.$

Overall, from (2.19), we obtain

$|g(r,t)|\leq
C\left(\frac{B\rho_{1}^{\nu-1}}{r^{\nu}}+A\left(\left(\frac{\rho_{1}}{r}\right)^{\gamma_{0}}w^{\gamma_{0}+\nu+1}(r,t)+r^{\gamma_{1}}w^{\nu-\gamma_{1}+1}(r,t)+\left(\frac{\rho}{r}\right)^{\beta_{0}}+r^{\beta_{1}}\right)\right).$

As in the previous proof, under the assumptions (2.17) on $r$ and $t,$ this
implies (2.18). ∎

##  3\. Main Theorem

Given a $W^{1,2}$ map $u:B_{R}(x_{0})\to N,$ the _outer energy scale_
$\lambda_{\varepsilon,R,x_{0}}(u)$ is, by definition, the smallest number
$0\leq\lambda\leq R$ such that

$\sup_{\lambda<r<R}E\left(u,U^{r}_{r/2}(x_{0})\right)<\varepsilon.$

Cf. Definition 4.2 of [10].

We first establish the following “baby case” of the main theorem.

######  Lemma 3.1.

Given $E,\lambda>0$ and $0<\varepsilon<\varepsilon_{0},$ there exists
$\delta>0$ as follows. Suppose that $u$ is a solution of (1.1) on
$B_{R}\times(-R^{2},T),$ with $R<R_{0}$ and $T>0,$ which satisfies

(3.1) $\sup_{-R^{2}<t<T}\int_{B_{R}}|du(t)|^{2}\,dV\leq E,$ (3.2)
$\int_{-R^{2}}^{T}\\!\\!\int_{B_{R}}|{\mathcal{T}}|^{2}\,dVdt<\delta^{2},$

and

(3.3) $\lambda_{\varepsilon,R,x_{0}}(u(t))\leq R\lambda.$

Then, writing $r=\text{dist}(x,x_{0}),$ for $2R\lambda\leq r\leq R/2$ and each
$k\geq 0,$ we have

(3.4) $r^{1+k}|\nabla^{(k)}du(x,0)|\leq
C_{k}\sqrt{\varepsilon}\left(\frac{R\lambda}{r}+\frac{r}{R}\right).$

Here, $R_{0}>0$ depends on the geometry of $M,$ $\varepsilon_{0}$ depends on
the geometry of $N,$ and $\delta$ depends on $E,\varepsilon,$ and $\lambda.$

###### Proof.

Assuming that $R_{0}$ is sufficiently small, we may rescale so that $R=1$ and
our metric $g$ takes the form (2.4-2.5), with $\xi_{0}\leq\frac{1}{2}.$

In view of (3.3), the standard $\varepsilon$-regularity lemma (see e.g.
Theorem 3.4 of [10]) implies

(3.5) $\sup_{\lambda\leq r\leq R}\left(r|du|+r^{2}|\nabla
du|+r^{3}|\nabla^{2}du|\right)\leq C\sqrt{\varepsilon}.$

Hence, by Lemma 2.1, $f$ satisfies an evolution equation

$\square_{\nu}f\leq C\sqrt{\varepsilon},$

where $\nu=\sqrt{1-C\sqrt{\varepsilon}}.$ We can apply the proof of
Proposition 2.2 on $U_{\lambda}^{1},$ with $A=C\sqrt{\varepsilon},$
$\beta_{0}=\beta_{1}=0,$ and $\tau=-\frac{1}{2}.$ Since $w^{a}(r,s)\leq
Cr^{a}$ for $s\geq\frac{1}{4},$ (2.14) gives

(3.6) $\begin{split}\sup_{-\frac{1}{4}\leq t\leq 0}|f(r,t)|&\leq
C\sqrt{\varepsilon}\left(r^{\nu}+r^{\beta_{1}}r^{\nu-\beta_{1}}+\left(\frac{\lambda}{r}\right)^{\nu}+r^{\nu}+r^{2}\right)\\\
&\leq
C\sqrt{\varepsilon}\left(\left(\frac{\lambda}{r}\right)^{\nu}+r^{\nu}\right).\end{split}$

Applying Hölder’s inequality identity to (2.3), we have

(3.7)
$\begin{split}\left|\int_{-\frac{1}{2}}^{0}\left(g^{2}(r,t)-f^{2}(r,t)\right)\,dt\right|&=\left|\int_{t_{0}}^{t_{1}}\\!\\!\\!\\!\\!\int_{S^{1}_{\rho_{1}}}X^{i}X^{j}S_{ij}(x,t)\,d\theta
dt\right|\\\
&\leq\left(\int_{t_{0}}^{t_{1}}\\!\\!\\!\\!\\!\int_{B_{\rho_{1}}}|{\mathcal{T}}(u)|^{2}\,dVdt\right)^{\frac{1}{2}}\left(\int_{t_{0}}^{t_{1}}\\!\\!\\!\\!\\!\int_{B_{\rho_{1}}}r^{\prime
2}\,|du|^{2}\,dV_{r^{\prime}}dt\right)^{\frac{1}{2}}\\\ &\leq
C\delta\sqrt{E}.\end{split}$

Since $|du|^{2}\leq f^{2}+g^{2},$ for $\delta$ sufficiently small, (3.6-3.7)
imply

$\begin{split}\int_{-\frac{1}{2}}^{0}\int_{r/2}^{r}|du|^{2}\,dVdt&\leq
C\varepsilon\left(\left(\frac{\lambda}{r}\right)^{\nu}+r^{\nu}\right)^{2}.\end{split}$

Applying $\varepsilon$-regularity (Theorem 3.4 of [10]) again, we obtain

(3.8) $\begin{split}r^{1+k}|\nabla^{(k)}du(x,0)|&\leq
C_{k}\sqrt{\varepsilon}\left(\left(\frac{\lambda}{r}\right)^{\nu}+r^{\nu}\right).\end{split}$

To obtain the same estimate with $\nu=1,$ one can apply a supersolution
argument as in Propositions 2.2-2.3, with (3.8) in place of (3.5). Since the
sharp result will not be used below, we leave the proof to the reader. ∎

######  Theorem 3.2.

Let $0<\varepsilon<\varepsilon_{0}$ and $0<R<R_{0}.$ Suppose that
$u:B_{R}\times\left(-R^{2},T\right)\to N,$ with $T>0,$ is a smooth solution of
(1.1). Assume that

(3.9) $\sup_{-R^{2}<t<T}\int_{B_{R}}|du(t)|^{2}\,dV\leq E,$ (3.10)
$\int_{-R^{2}}^{T}\\!\\!\int_{B_{R}}|{\mathcal{T}}|^{2}\,dVdt<\delta_{0}^{2},$

and, for some $0<\lambda_{1}\leq\frac{1}{2}$ and $0\leq t_{1}<T,$ we have

(3.11) $\lambda_{\varepsilon,R,x_{0}}(u(t))\leq
R\left(\lambda_{1}+\left(\frac{t_{1}-t}{R^{2}}\right)^{\frac{1+\alpha}{2}}\right)$

for all $-R^{2}<t\leq t_{1}.$ Then, writing $r=\text{dist}(x,x_{0}),$ we have

(3.12) $r|du(x,t_{1})|\leq
C_{\alpha}\sqrt{\varepsilon}\left(\frac{R\lambda_{1}}{r}+\left(\frac{r}{R}\right)^{\alpha}\right)^{\frac{1}{3}}$

for $2R\lambda_{1}\leq r\leq R/2.$

Here, $R_{0}>0$ depends on the geometry of $M,$ $\varepsilon_{0}$ depends on
the geometry of $N,$ $\delta_{0}$ depends on $E,\varepsilon,$ and $\alpha,$
and $C_{\alpha}$ depends only on $\alpha.$

###### Proof.

For convenience, in place of (3.11), we shall assume

(3.13) $\lambda_{\varepsilon,R,x_{0}}(u(t))\leq
R\left(\lambda_{1}+\kappa^{a}\left(\frac{t_{1}-t}{R^{2}}\right)^{\frac{1+\alpha}{2}}\right),$

where $0<\kappa\leq\frac{1}{2}$ is a constant to be determined by the argument
below (depending only on $\alpha$ and $\nu$), and $a$ is sufficiently large,
for instance

(3.14) $a=\frac{72(1+\alpha)}{\alpha}.$

The assumptions (3.13) and (3.11) are equivalent after rescaling.

We will also prove the more detailed estimates

(3.15) $f(u;r,t_{1})\leq
C_{0}\sqrt{\varepsilon}\left(\frac{\lambda_{1}}{r}+r^{\alpha}\right)^{2\nu-1},$
(3.16) $g(u;r,t_{1})\leq
C_{0}\sqrt{\varepsilon}\left(\frac{\lambda_{1}}{r}+r^{\alpha}\right)^{\frac{1}{3}},$

which clearly imply (3.12). Here, we assume

(3.17) $\frac{26}{27}\leq\nu\leq\sqrt{1-C\sqrt{\varepsilon}}.$

It suffices to prove the theorem for $\lambda_{1}$ of the form
$\lambda_{1}=2^{-n}\kappa^{a},$ so we may proceed by induction on $n.$ For the
base case, we have $\lambda_{1}=\kappa^{a}.$ According to Lemma 3.1, there
exists $\delta_{0}=\delta>0$ such that (3.15-3.16) hold, with a universal
constant $C_{0}>1.$

For induction, suppose that (3.15-3.16) hold for all $\lambda_{1}\geq
2\bar{\lambda}_{1},$ where $0<\bar{\lambda}_{1}\leq\kappa^{a};$ i.e., the
conclusion of the theorem holds for such $\lambda_{1}$ and for all possible
solutions $u$ satisfying the hypotheses. We must establish the Theorem for
$\lambda_{1}=\bar{\lambda}_{1}.$

Let $u(t)$ be a solution satisfying the hypotheses, with
$\lambda_{1}=\bar{\lambda}_{1}.$ By rescaling, it suffices to assume $R=1.$

In view of (3.11), the standard $\varepsilon$-regularity lemma (see e.g.
Theorem 3.4 of [10]) implies

$\sup_{\lambda(t)\leq r\leq R}\left(r|du|+r^{2}|\nabla
du|+r^{3}|\nabla^{2}du|\right)\leq C\sqrt{\varepsilon}.$

Hence, by Lemma 2.1, $f$ and $g$ satisfy the evolution equations

$\begin{split}\square_{\nu}f&\leq C\sqrt{\varepsilon}\\\
\square_{\nu}\left(\frac{g}{r}\right)&\leq\frac{6f}{r^{3}}+C\frac{\sqrt{\varepsilon}}{r},\end{split}$

for $\lambda_{1}+\kappa^{a}\left(t_{1}-t\right)^{\frac{1+\alpha}{2}}\leq r\leq
1.$ Let

$\begin{split}\rho=2\bar{\lambda}_{1},\qquad\zeta=\rho^{\frac{1}{1+\alpha}}.\end{split}$

Since $\bar{\lambda}_{1}\leq\kappa^{a}$ by the base case, we have

(3.18) $\frac{\rho}{\kappa}\leq\zeta\leq 2\kappa^{\frac{a}{1+\alpha}}.$

Also notice that $\rho=\zeta^{1+\alpha}$ and

(3.19) $\frac{\rho}{\zeta}=\zeta^{\alpha}.$

We first apply the induction hypotheses to $u$ with $R=1/2$ and
$\lambda_{1}=\rho,$ hence $R\rho=\bar{\lambda}_{1}.$ In view of (3.19), we
clearly have (3.15-3.16) for $r\leq\zeta/2.$ Applying the induction hypothesis
with $R=1$ and $\lambda_{1}=\rho,$ we also have (3.15-3.16) for all $r\geq
2\zeta.$ Hence, it remains only to establish (3.15-3.16) for

(3.20) $\frac{1}{2}\zeta\leq r\leq 2\zeta.$

In other words, for $r$ as in (3.20), we must show

(3.21) $f(r,t_{1})\leq\frac{C_{0}}{2}\sqrt{\varepsilon}\zeta^{\alpha(2\nu-1)}$

and

(3.22)
$g(r,t_{1})\leq\frac{C_{0}}{2}\sqrt{\varepsilon}\zeta^{\frac{\alpha}{3}}.$

Let

$t_{0}=t_{1}-\frac{\zeta^{2}}{\kappa^{2}},$

which, by (3.13), satisfies $\lambda(t_{0})\leq\rho.$ By the induction
hypothesis, we have

(3.23) $f(r,t)\leq
C\sqrt{\varepsilon}\left(\frac{\rho}{r}+r^{\alpha}\right)^{2\nu-1}$ (3.24)
$g(r,t)\leq
C\sqrt{\varepsilon}\left(\frac{\rho}{r}+r^{\alpha}\right)^{\frac{1}{3}}.$

for all $t_{0}\leq t\leq t_{1}.$ Combining these, we also have

(3.25) $r|du(x,t)|\leq
C\sqrt{\varepsilon}\left(\frac{\rho}{r}+r^{\alpha}\right)^{\frac{1}{3}}$

for all $t_{0}\leq t\leq t_{1}.$

To obtain the estimate (3.21) on $f,$ we apply Proposition 2.2. From
(3.23-3.24), we obtain

$\begin{split}\sup_{\zeta/2\leq r\leq 2\zeta}f(r,t_{1})&\leq
CC_{\ref{festimate}}\sqrt{\varepsilon}\left(\kappa^{1-\nu}\left(\frac{\rho}{\zeta}\right)^{2\nu-1}+\kappa^{\nu-\alpha(2\nu-1)}\zeta^{\alpha(2\nu-1)}\right)\\\
&\leq
C\sqrt{\varepsilon}\left(\kappa^{1-\nu}+\kappa^{\nu-\alpha(2\nu-1)}\right)\zeta^{\alpha(2\nu-1)},\end{split}$

where we have used (3.18-3.19). Hence, assuming that $\kappa$ is sufficiently
small, (3.21) is satisfied at $t=t_{1},$ as desired.222The decay estimate on
$f$ can also be obtained directly from [10], Lemma 5.4. We have re-proven it
here (by a different method) for the sake of exposition.

Next, to obtain the estimate on $g,$ let

$\rho_{1}=\zeta^{1+\frac{9}{16}\alpha}.$

By the induction hypothesis, we have

$\sup_{t_{0}\leq t\leq t_{1}}f(\rho_{1},t)\leq
C\sqrt{\varepsilon}\left(\frac{\rho}{\rho_{1}}\right)^{2\nu-1}\leq
C\sqrt{\varepsilon}\zeta^{\frac{7\cdot 25}{16\cdot 27}\alpha}\leq
C\sqrt{\varepsilon}\zeta^{\frac{3}{8}\alpha}.$

Using (3.17), we obtain

(3.26)
$\int_{t_{0}}^{t_{1}}f(\rho_{1},t)^{2}\,dt\leq\frac{C\varepsilon}{\kappa^{2}}\zeta^{2+\frac{3}{4}\alpha}.$

We now integrate (2.3) in time and apply Hölder’s inequality:

(3.27)
$\begin{split}\left|\int_{t_{0}}^{t_{1}}\left(g^{2}(\rho_{1},t)-f^{2}(\rho_{1},t)\right)\,dt\right|&=\left|\int_{t_{0}}^{t_{1}}\\!\\!\\!\\!\\!\int_{S^{1}_{\rho_{1}}}X^{i}X^{j}S_{ij}(x,t)\,d\theta
dt\right|\\\
&\leq\left(\int_{t_{0}}^{t_{1}}\\!\\!\\!\\!\\!\int_{B_{\rho_{1}}}|{\mathcal{T}}(u)|^{2}\,dVdt\right)^{\frac{1}{2}}\left(\int_{t_{0}}^{t_{1}}\\!\\!\\!\\!\\!\int_{B_{\rho_{1}}}r^{\prime
2}\,|du|^{2}\,dV_{r^{\prime}}dt\right)^{\frac{1}{2}}\\\ &\leq
C\delta_{0}\left((t_{1}-t_{0})\left(\rho^{2}E+C\varepsilon\rho_{1}^{2}\left(\frac{\rho}{\rho_{1}}\right)^{\frac{2}{3}}\right)\right)^{\frac{1}{2}}.\end{split}$

Here we have used the assumptions (3.9-3.10) and (3.25). We have

$(t_{1}-t_{0})\rho^{2}E=\frac{\zeta^{4+2\alpha}}{\kappa^{2}}E$

and

$(t_{1}-t_{0})\rho_{1}^{2}\left(\frac{\rho}{\rho_{1}}\right)^{\frac{2}{3}\alpha}=\frac{\zeta^{2}}{\kappa^{2}}\zeta^{2+\frac{9}{8}\alpha}\zeta^{\frac{7}{16}\frac{2}{3}\alpha}=\frac{\zeta^{4+\frac{17}{12}\alpha}}{\kappa^{2}}.$

Hence, for $\delta_{0}$ sufficiently small (independently of
$\bar{\lambda}_{1}$), (3.27) reduces to

$\begin{split}\left|\int_{t_{0}}^{t_{1}}\left(g^{2}(\rho_{1},t)-f^{2}(\rho_{1},t)\right)\,dt\right|&\leq\kappa\varepsilon\zeta^{2+\frac{17}{24}\alpha}.\end{split}$

Combining this with (3.26), we obtain

$\begin{split}\int_{t_{0}}^{t_{1}}g(\rho_{1},t)^{2}\,dt&\leq\kappa\varepsilon\zeta^{2+\frac{17}{24}\alpha}+\frac{C\varepsilon}{\kappa^{2}}\zeta^{2+\frac{3}{4}\alpha}\\\
&\leq
C\varepsilon\left(\kappa+C\kappa^{\frac{1}{24}\frac{a\alpha}{1+\alpha}-2}\right)\zeta^{2+\frac{17}{24}\alpha}\\\
&\leq C\kappa\varepsilon\zeta^{2+\frac{17}{24}\alpha},\end{split}$

where we have used (3.14) and (3.18).

We may now apply Proposition 2.3, to obtain

$\begin{split}\sup_{\zeta/2\leq r\leq 2\zeta}g(r,t_{1})&\leq
CC_{\ref{gestimate}}\left(\begin{split}&\frac{\sqrt{\kappa\varepsilon}\zeta^{1+\frac{17}{48}\alpha}}{\zeta^{1+\frac{9}{16}\alpha}}\left(\frac{\zeta^{1+\frac{9}{16}\alpha}}{\zeta}\right)^{\nu}+\sqrt{\varepsilon}\left(\kappa^{\nu+\frac{1}{3}+1}\left(\frac{\zeta^{1+\alpha}}{\zeta}\right)^{\frac{1}{3}}+\kappa^{\nu-\frac{\alpha}{3}+1}\zeta^{\frac{\alpha}{3}}\right)\\\
&\quad+\sqrt{\varepsilon}\left(\frac{\zeta^{1+\alpha}}{\zeta}+\zeta^{\alpha}\right)^{2\nu-1}\end{split}\right)\\\
&\leq
C\left(\sqrt{\kappa\varepsilon}\zeta^{\frac{\alpha}{48}\left(17+27(\nu-1)\right)}+\sqrt{\varepsilon}\kappa\zeta^{\frac{\alpha}{3}}+\sqrt{\varepsilon}\zeta^{\frac{25}{27}\alpha}\right)\\\
&\leq
C\sqrt{\varepsilon}\left(\sqrt{\kappa}+\kappa+\kappa^{\alpha\left(\frac{25a}{27(1+\alpha)}-\frac{1}{3}\right)}\right)\zeta^{\frac{\alpha}{3}},\end{split}$

where we have used (3.17) and (3.18). For $\kappa$ sufficiently small, this
implies the desired bound (3.22), completing the induction. ∎

######  Corollary 3.3 (Cf. Theorem 1.1).

Let $M$ be any Riemannian surface and suppose that $u$ is a classical solution
of (1.1) on $M\times\left[0,T\right)$ with bounded energy. Given $x_{0}\in M$
and $0<\varepsilon<\varepsilon_{0},$ choose
$0<R<\min\left[R_{0},{\mathrm{inj}}_{x_{0}}(M)\right]$ small enough that

(3.28) $\qquad\lambda_{\varepsilon,R,x_{0}}(u(t))\to 0\quad(t\nearrow T).$

If also

(3.29)
$\lambda_{\varepsilon,R,x_{0}}(u(t))=O(T-t)^{\frac{1+\alpha}{2}},\qquad(t\nearrow
T),$

where $0<\alpha\leq 1,$ then $u(T)$ is $C^{\frac{\alpha}{3}}$ on
$B_{R/2}(x_{0}).$

###### Proof.

First note that since the energy is bounded, the stress-energy tensor is also
bounded in $L^{1}.$ By Corollary 4.5 of [10], with $q=1,$ (3.28) is true for
$R>0$ sufficiently small.

Now, given any $\lambda_{1}>0,$ in view of (3.29), (3.11) will be satisfied
for all $t_{1}$ sufficiently close to $T.$ For $x\in
U_{2\lambda_{1}}^{R}(x_{0}),$ by Theorem 3.2, we have

$r|du(x,T)|=\lim_{t_{1}\nearrow T}r|du(x,t_{1})|\leq
C\sqrt{\varepsilon}\left(\frac{R\lambda_{1}}{r}+\left(\frac{r}{R}\right)^{\alpha}\right)^{\frac{1}{3}}.$

Letting $\lambda_{1}\searrow 0,$ we obtain

$r|du(x,T)|\leq
C\sqrt{\varepsilon}\left(\frac{r}{R}\right)^{\frac{\alpha}{3}}$

for all $x\in B_{R/2}\setminus\\{x_{0}\\}.$ Integrating radially, this gives
$u(T)\in C^{\frac{\alpha}{3}}.$ ∎

##  Appendix A Proof of Lemma 2.1

Let

$f_{0}=\sqrt{\int_{\\{r\\}\times
S^{1}}|u_{\theta}(r,\theta,t)|^{2}d\theta},\qquad
f_{1}=\sqrt{\int_{\\{r\\}\times
S^{1}}|\nabla_{\theta}u_{\theta}(r,\theta,t)|^{2}d\theta}.$

In our previous paper [10], Lemma 5.1, we calculated the desired evolution of
$f_{0};$ we now apply a similar analysis to $f_{1}.$ These calculations go
back to Lin-Wang [6] and Parker [8].

For convenience, we shall work below in cylindrical coordinates $(s=\ln
r,\theta).$ Letting

$g_{0}=ds^{2}+d\theta^{2}$

denote the flat cylindrical metric, we have $g=\xi^{2}e^{2s}g_{0}.$ The
differential of $u$ is given by

$du=u_{s}ds+u_{\theta}d\theta.$

The tension field with respect to $g_{0}$ is given by

$\mathcal{T}_{0}(u)=\nabla_{s}u_{s}+\nabla_{\theta}u_{\theta},$

where $\nabla$ denotes the pullback connection on $u^{*}TN,$ as above. The
heat-flow equation (1.1) with respect to the metric $g$ becomes

(A.1) $u_{t}=\mathcal{T}(u)=\xi^{-2}e^{-2s}\mathcal{T}_{0}(u).$

We start from the identity

(A.2)
$\frac{1}{2}\partial_{s}^{2}f_{1}^{2}=\int_{S^{1}}\left(|\nabla_{s}\nabla_{\theta}u_{\theta}|^{2}+\left\langle\nabla_{s}^{2}\nabla_{\theta}u_{\theta},\nabla_{\theta}u_{\theta}\right\rangle\right).$

We have

(A.3)
$\nabla_{s}u_{s}+\nabla_{\theta}u_{\theta}={\mathcal{T}}_{0}(u)=\xi^{2}e^{2s}u_{t}.$

Applying $\nabla_{\theta},$ we obtain

$\begin{split}\nabla_{\theta}\nabla_{s}u_{s}+\nabla^{2}_{\theta}u_{\theta}&=2\xi\nabla_{\theta}\xi
e^{2s}u_{t}+\xi^{2}e^{2s}\nabla_{\theta}u_{t}\\\ &=2\xi\nabla_{\theta}\xi
e^{2s}u_{t}+\xi^{2}e^{2s}\nabla_{t}u_{\theta},\end{split}$

and

(A.4)
$\begin{split}\nabla_{\theta}^{2}\nabla_{s}u_{s}&=-\nabla^{3}_{\theta}u_{\theta}+\left(2(\nabla_{\theta}\xi)^{2}+2\xi\nabla^{2}_{\theta}\xi\right)e^{2s}u_{t}\\\
&\quad+2\xi\nabla_{\theta}\xi
e^{2s}\nabla_{t}u_{\theta}+\xi^{2}e^{2s}\nabla_{\theta}\nabla_{t}u_{\theta}\\\
&=-\nabla^{3}_{\theta}u_{\theta}+e^{2s}\nabla_{t}\nabla_{\theta}u_{\theta}+I,\end{split}$

where

(A.5)
$I=(\xi^{2}-1)e^{2s}\nabla_{\theta}^{2}u_{t}+e^{2s}R(u_{\theta},u_{t})u_{\theta}+\left(2(\nabla_{\theta}\xi)^{2}+2\xi\nabla^{2}_{\theta}\xi\right)e^{2s}u_{t}+2\xi\nabla_{\theta}\xi
e^{2s}\nabla_{\theta}u_{t}.$

We may also commute derivatives to obtain

(A.6)
$\begin{split}\nabla_{s}^{2}\nabla_{\theta}u_{\theta}&=\nabla_{s}\left(\nabla_{\theta}\nabla_{s}u_{\theta}+R(u_{s},u_{\theta})u_{\theta}\right)\\\
&=\nabla_{\theta}\nabla_{s}^{2}u_{\theta}+\nabla_{s}\left(R(u_{s},u_{\theta})u_{\theta}\right)\\\
&=\nabla_{\theta}\nabla_{s}(\nabla_{\theta}u_{s})+\nabla_{s}\left(R(u_{s},u_{\theta})u_{\theta}\right)\\\
&=\nabla_{\theta}^{2}\nabla_{s}u_{s}+I\\!I,\end{split}$

where

(A.7)
$\begin{split}I\\!I&=\nabla_{\theta}\left(R(u_{s},u_{\theta})u_{s}\right)+\nabla_{s}\left(R(u_{s},u_{\theta})u_{\theta}\right)\\\
&=\nabla
R\left(u_{\theta},u_{s},u_{\theta}\right)u_{s}+R(\nabla_{\theta}u_{s},u_{\theta})u_{s}+R(u_{s},\nabla_{\theta}u_{\theta})u_{s}+R(u_{s},u_{\theta})\nabla_{\theta}u_{s}\\\
&\quad+\nabla
R\left(u_{s},u_{s},u_{\theta}\right)u_{\theta}+R(\nabla_{s}u_{s},u_{\theta})u_{\theta}+R(u_{s},\nabla_{s}u_{\theta})u_{\theta}+R(u_{s},u_{\theta})\nabla_{s}u_{\theta}.\end{split}$

Inserting (A.4) and (A.6) into (A.2), integrating by parts, and rearranging,
we obtain

(A.8)
$\frac{1}{2}\left(e^{2s}\partial_{t}f_{1}^{2}-\partial_{s}^{2}f_{1}^{2}\right)=-\int_{S^{1}}\left(|\nabla_{s}\nabla_{\theta}u_{\theta}|^{2}+|\nabla_{\theta}^{2}u_{\theta}|^{2}\right)-\int_{S^{1}}\left\langle
I+I\\!I,\nabla_{\theta}u_{\theta}\right\rangle.$

We need the following simple estimates. Since $|u_{\theta}|\leq\eta$ is small,
we may assume that the image of the curve $u(s,\cdot,t):S^{1}\to N$ lies in a
coordinate chart of $N$ where the Christoffel symbol $\Gamma$ is bounded by
$C_{N}$. Then, in local coordinates, we have

$|\partial_{\theta}u_{\theta}|=|\nabla_{\theta}u_{\theta}-\Gamma(u_{\theta},u_{\theta})|\leq|\nabla_{\theta}u_{\theta}|+|\Gamma(u_{\theta},\nabla_{\theta}u_{\theta})|\leq\eta+C_{N}\eta^{2}\leq
2\eta,$

assuming that $\eta$ is sufficiently small (depending on $N$). Thus

$\displaystyle|\nabla_{\theta}u_{\theta}|^{2}$
$\displaystyle=|\partial_{\theta}u_{\theta}+\Gamma(u_{\theta},u_{\theta})|^{2}$
$\displaystyle\geq|\partial_{\theta}u_{\theta}|^{2}-2|\partial_{\theta}u_{\theta}||\Gamma(u_{\theta},u_{\theta})|+|\Gamma(u_{\theta},u_{\theta})|^{2}$
$\displaystyle\geq|\partial_{\theta}u_{\theta}|^{2}-C\eta|u_{\theta}|^{2}.$

Then the ordinary Poincaré inequality on $S^{1}$ yields

(A.9)
$f_{1}^{2}=\int_{S^{1}}|\nabla_{\theta}u_{\theta}|^{2}\geq\int_{S^{2}}|u_{\theta}|^{2}-C\eta\int_{S^{1}}|u_{\theta}|^{2}=(1-C\eta)f_{0}^{2}.$

A similar argument, applied to $\nabla_{\theta}^{2}u,$ gives

(A.10)
$\int_{S^{1}}|\nabla_{\theta}^{2}u_{\theta}|^{2}\geq(1-C\eta)f_{1}^{2}.$

We first apply (A.10) and Hölder’s inequality to (A.8), to obtain

(A.11)
$\begin{split}\frac{1}{2}\left(e^{2s}\partial_{t}f_{1}^{2}-\partial_{s}^{2}f_{1}^{2}+(1-C\eta)f_{1}^{2}\right)&\leq-\int_{S^{1}}|\nabla_{s}\nabla_{\theta}u_{\theta}|^{2}-\int_{S^{1}}\left\langle
I+II,\nabla_{\theta}u_{\theta}\right\rangle\\\
&\leq-\int_{S^{1}}|\nabla_{s}\nabla_{\theta}u_{\theta}|^{2}+\left(\|I\|_{L^{2}(S^{1})}+\|II\|_{L^{2}(S^{1})}\right)f_{1}.\end{split}$

Note that

$\frac{1}{2}\partial_{s}f_{1}^{2}=f_{1}\partial_{s}f_{1}=\int_{S^{1}}\left\langle\nabla_{s}\nabla_{\theta}u_{\theta},\nabla_{\theta}u_{\theta}\right\rangle\leq\left(\int_{S^{1}}|\nabla_{s}\nabla_{\theta}u_{\theta}|^{2}\right)^{1/2}f_{1},$

so we have

$|\partial_{s}f_{1}|^{2}\leq\int|\nabla_{s}\nabla_{\theta}u_{\theta}|^{2}.$

On the other hand,

$\frac{1}{2}\partial_{s}^{2}(f_{1}^{2})=f_{1}\cdot\partial_{s}^{2}f_{1}+|\partial_{s}f_{1}|^{2}.$

Hence, “dividing out” by $f_{1}$ in (A.11) (which is justified in the
distribution sense), we obtain

(A.12)
$\begin{split}e^{2s}\partial_{t}f_{1}-\partial_{s}^{2}f_{1}+(1-C\eta)f_{1}&\leq
2\left(\|I\|_{L^{2}(S^{1})}+\|II\|_{L^{2}(S^{1})}\right).\end{split}$

It remains to estimate the terms on the RHS. By (2.7), we have

$e^{2s}\left(|u_{t}|+|\nabla_{\theta}u_{t}|+|\nabla^{2}_{\theta}u_{t}|\right)\leq
C\eta.$

Combining this with (2.5), from (A.5), we obtain

$\left\|I\right\|_{L^{2}(S^{1})}\leq C\xi_{0}\eta e^{2s}+\eta^{2}f_{0}\leq
C\xi_{0}\eta e^{2s}+\eta f_{1},$

where we have used (A.9). From (A.7), since each term has at least one factor
of $u_{\theta}$ or $\nabla_{\theta}u_{\theta},$ we also obtain

$\left\|I\\!I\right\|_{L^{2}(S^{1})}\leq C_{N}\eta^{2}f_{1}\leq\eta f_{1},$

for $\eta$ sufficiently small. Inserting these estimates into (A.12), and
absorbing the $C\eta f_{1}$ terms in the LHS, we obtain

(A.13)
$\begin{split}e^{2s}\partial_{t}f_{1}-\partial_{s}^{2}f_{1}+(1-C\eta)f_{1}&\leq
C\xi_{0}\eta e^{2s}.\end{split}$

Translating the above equation back to polar coordinates, we get (2.8).

To estimate the radial energy

$g=\sqrt{\int_{\\{e^{s}\\}\times S^{1}}|u_{s}|^{2}d\theta},$

we start from the identity

(A.14)
$\frac{1}{2}\partial_{s}^{2}g^{2}=\int_{S^{1}}\left(|\nabla_{s}u_{s}|^{2}+\left\langle\nabla_{s}^{2}u_{s},u_{s}\right\rangle\right).$

Applying $\nabla_{s}$ to (A.3), we obtain

$\begin{split}\nabla^{2}_{s}u_{s}+\nabla_{s}\nabla_{\theta}u_{\theta}&=\left(2\xi\partial_{s}\xi+2\xi^{2}\right)e^{2s}u_{t}+\xi^{2}e^{2s}\nabla_{s}u_{t}\\\
&=2\left(\xi^{-1}\partial_{s}\xi+1\right){\mathcal{T}}_{0}(u)+\xi^{2}e^{2s}\nabla_{t}u_{s}\\\
&=2\left(\nabla_{s}u_{s}+\nabla_{\theta}u_{\theta}\right)+e^{2s}\nabla_{t}u_{s}+2\xi^{-1}\partial_{s}\xi(\nabla_{s}u_{s}+\nabla_{\theta}u_{\theta})+\left(\xi^{2}-1\right)e^{2s}\nabla_{s}u_{t}.\end{split}$

We also have

$\nabla_{s}\nabla_{\theta}u_{\theta}=\nabla_{\theta}^{2}u_{s}+R(u_{s},u_{\theta})u_{\theta}.$

Returning to (A.14), we have

$\begin{split}\frac{1}{2}\partial_{s}^{2}g^{2}&=\int|\nabla_{s}u_{s}|^{2}+\int\left\langle
e^{2s}\nabla_{t}u_{s}+2\nabla_{s}u_{s}+2\nabla_{\theta}u_{\theta},u_{s}\right\rangle\\\
&\qquad+\int|\nabla_{\theta}u_{s}|^{2}+\int\left\langle
2\xi^{-1}\partial_{s}\xi\left(\nabla_{s}u_{s}+\nabla_{\theta}u_{\theta}\right)+(\xi^{2}-1)e^{2s}\nabla_{s}u_{t},u_{s}\right\rangle-\int
R(u_{s},u_{\theta})u_{\theta},\end{split}$

where we have integrated by parts once. Applying Hölder’s inequality,
rearranging, and using (2.5), we obtain

$\begin{split}\frac{1}{2}\left(e^{2s}\partial_{t}-\partial_{s}^{2}+2\partial_{s}\right)g^{2}&\leq-\int|\nabla_{s}u_{s}|^{2}+2f_{1}g+C\xi_{0}\eta
e^{2s}g\\\ &\qquad-\int|\nabla_{\theta}u_{s}|^{2}+C_{N}\eta
f_{0}g.\end{split}$

We choose $\eta$ small enough that $C_{N}\eta\leq 1,$ and discard the
$-\int|\nabla_{\theta}u_{s}|^{2}$ term. After “dividing out” by $g$ as above,
we obtain

(A.15)
$\begin{split}\left(e^{2s}\partial_{t}-\partial_{s}^{2}+2\partial_{s}\right)g\leq
6f_{1}+C\xi_{0}\eta e^{2s}.\end{split}$

Changing back to polar coordinates and dividing by $r,$ we get the desired
evolution equation.

##  Appendix B Radial heat kernel

In this appendix, we extract several results from the appendix of [13],
replacing the integer dimension by a real number $\mu>1.$ The proofs of
Propositions 2.2-2.3 are based on these results.

Let

$\Delta_{\mu}=\partial_{r}^{2}+\frac{\mu-1}{r}\partial_{r}.$

In the case that $\mu$ is an integer, the spherical average of the Euclidean
heat kernel is given by

(B.1)
$H(r,s,t)=\frac{c_{\mu}e^{\frac{-(r^{2}+s^{2})}{4t}}}{t^{\mu/2}}I\left(\frac{rs}{2t}\right),$

where $c_{\mu}$ is an appropriate constant, and333Since $I$ satisfies the ODE
(B.4), we in fact have  $I(x)=x^{1-\frac{\mu}{2}}I_{\frac{\mu}{2}-1}(x),$
where $I_{\frac{\mu}{2}-1}$ is the modified Bessel function. This recovers
formula (3.3) of Bragg [2].

(B.2) $I(x)=\int_{0}^{\pi}e^{x\cos\theta}\sin^{\mu-2}\theta\,d\theta.$

######  Lemma B.1.

For any real $\mu>1,$ the above function $H$ satisfies

$\begin{split}\left(\partial_{t}-\Delta_{\mu}\right)H(\cdot,s,t)=0,\qquad
t>0,\\\ H(r,s,t)>0\mbox{ for }0<r,s,t<\infty,\\\
H(r,s,t)\to\frac{1}{s^{\mu-1}}\delta(r-s)\qquad t\searrow 0,\end{split}$

and

(B.3)
$\frac{C_{\mu}^{-1}e^{\frac{-(r-s)^{2}}{4t}}}{t^{1/2}(rs+t)^{\frac{n-1}{2}}}\leq
H(r,s,t)\leq\frac{C_{\mu}e^{\frac{-(r-s)^{2}}{4t}}}{t^{1/2}(rs+t)^{\frac{n-1}{2}}}.$

###### Proof.

We calculate

$\left(\partial_{t}-\Delta_{\mu}\right)H(r,s,t)=\frac{c_{\mu}e^{\frac{-(r^{2}+s^{2})}{4t}}}{t^{\mu/2}}\left(\frac{-s^{2}}{4t^{2}}\right)\left(I^{\prime\prime}(x)+\frac{(\mu-1)}{x}I^{\prime}(x)-I(x)\right).$

For $I(x)$ given by (B.2), we have

(B.4)
$\begin{split}I^{\prime\prime}(x)+\frac{(\mu-1)}{x}I^{\prime}(x)-I(x)&=\int_{0}^{\pi}e^{x\cos\theta}\left(\cos^{2}\theta+\frac{(\mu-1)}{x}\cos\theta-1\right)\sin^{\mu-2}\theta\,d\theta\\\
&=\int_{0}^{\pi}e^{x\cos\theta}\left(-\sin^{2}\theta+\frac{(\mu-1)}{x}\cos\theta\right)\sin^{\mu-2}\theta\,d\theta\\\
&=0,\end{split}$

after integrating by parts. Hence, $H$ solves the PDE as required for any real
$\mu.$

Borrowing a factor of $e^{\frac{rs}{2t}}$ in (B.2), we have

(B.5)
$H(r,s,t)=\frac{c_{\mu}e^{\frac{-(r-s)^{2}}{4t}}}{t^{\mu/2}}I_{1}\left(\frac{rs}{2t}\right),$

where

$I_{1}(x)=\int_{0}^{\pi}e^{x(\cos\theta-1)}\sin^{\mu-2}\theta d\theta.$

Then $I_{1}(x)$ clearly tends to a positive constant as $x\to 0.$ It is clear
form the substitution $u=\sqrt{x(1-\cos\theta)}$ that the integral is bounded
by a constant times $x^{-\frac{\mu-1}{2}}.$ Hence

$I_{1}(x)\leq\frac{C}{(1+x)^{\frac{\mu-1}{2}}}\leq\left(\frac{t}{rs+t}\right)^{\frac{\mu-1}{2}}.$

Substituting into (B.5), we obtain the desired bound. ∎

###  B.1. Initial data

Let $H_{\left[\rho,R\right]}(r,s,t)$ be the Dirichlet kernel for the operator
$\partial_{t}-\Delta_{\mu}$ on the interval $\left[\rho,R\right],$ satisfying

$\begin{split}\left(\partial_{t}-\Delta_{\mu}\right)H_{\left[\rho,R\right]}(\cdot,s,t)=0,\qquad
t>0,\\\
H_{\left[\rho,R\right]}(\rho,s,t)=0=H_{\left[\rho,R\right]}(R,s,t)\,\,\forall\,\,\rho\leq
s\leq R,t>0,\\\
H_{\left[\rho,R\right]}(r,s,t)\to\frac{1}{s^{\mu-1}}\delta(r-s)\qquad
t\searrow 0.\end{split}$

By the maximum principle, we have $0\leq H_{\left[\rho,R\right]}(r,s,t)\leq
H(r,s,t),$ and so

(B.6) $0\leq
H_{\left[\rho,R\right]}(r,s,t)\leq\frac{C_{\mu}e^{\frac{-(r-s)^{2}}{4t}}}{t^{1/2}(rs+t)^{\frac{\mu-1}{2}}}.$

Given an initial function $\varphi(r)$ on $\left[\rho,R\right],$ the solution
of the initial-value problem is given by

(B.7)
$v_{0}(r,t)=\int_{\rho}^{R}H_{\left[\rho,R\right]}(r,s,t)\varphi(s)s^{\mu-1}\,ds.$

Let

$w^{a}(r,t)=\left(\frac{r^{2}}{r^{2}+t}\right)^{a/2}.$

######  Proposition B.2.

For $0\leq k\leq\mu-1,$ assuming that $|\varphi(r)|\leq Ar^{-k},$ we have

$|v_{0}(r,t)|\leq C_{\mu}Ar^{-k}w^{k}(r,t)w^{\mu-k}(R,t).$

###### Proof.

From (B.6) and (B.7), we have

$\begin{split}|v_{0}(r,t)|&\leq
CA\int_{\rho}^{R}e^{-(r-s)^{2}/4t}\frac{s^{-k+\mu-1}}{(rs+t)^{\frac{\mu-1}{2}}}\,\frac{ds}{t^{1/2}}\\\
&\leq
CAr^{-k}\int_{\rho}^{R}e^{-(r-s)^{2}/4t}\frac{r^{k}s^{-k+\mu-1}}{(rs+t)^{\frac{\mu-1}{2}}}\,\frac{ds}{t^{1/2}}.\end{split}$

By Lemma A.1a of [13], applied with $a=k,b=\mu-k-1,$ and $c=d=0,$ we have

$\begin{split}\int_{\rho}^{R}e^{-(r-s)^{2}/4t}\frac{r^{k}s^{\mu-k-1}}{(rs+t)^{\frac{\mu-1}{2}}}\,\frac{ds}{t^{1/2}}&\leq
C\frac{R-\rho}{R-\rho+\sqrt{t}}w^{k}(r,t)w^{\mu-k-1}(R,t)\\\ &\leq
Cw^{k}(r,t)w^{\mu-k}(R,t).\end{split}$

The result follows. ∎

###  B.2. Boundary data

To construct a kernel for the boundary data at the inner radius $\rho=1,$ we
follow the argument of [13], Appendix A.3. Suppose $R>1,$ and let

$h(r)=\frac{r^{2-\mu}-R^{2-\mu}}{1-R^{2-\mu}}.$

Let

$y_{1}(r,t)=h(r)-\int_{1}^{R}H_{\left[1,R\right]}(r,s,t)h(s)\,s^{\mu-1}ds.$

This satisfies

$\begin{split}(\partial_{t}-\Delta_{\mu})y_{1}&=0\\\ y_{1}(r,0)&=0,\quad
1<r<R\\\ y_{1}(1,t)&=1,\,y_{1}(R,t)=0,\quad t>0.\end{split}$

The function

$G_{\left[1,R\right]}(r,t)=\partial_{t}y_{1}(r,t)$

satisfies

$\lim_{r\searrow 1}G_{\left[1,R\right]}(r,t)=\delta(t).$

######  Lemma B.3.

We have

$0\leq
G_{\left[1,R\right]}(r,t)\leq\frac{C_{\mu}e^{\frac{-(r-1)^{2}}{5t}}}{t(t+1)^{\frac{\mu}{2}-1}}\cdot\begin{cases}\min\left[(r-1)/\sqrt{t},1\right]&(t\leq
1)\\\ \min\left[r-1,1\right]&(t\geq 1).\end{cases}$

###### Proof.

Replacing $n$ by $\mu,$ the bound is identical to that of Lemma A.4b of [13],
and the proof there carries over. ∎

To obtain an inner boundary kernel for $\left[\rho,R\right],$ we let

$G_{\left[\rho,R\right]}(r,t)=\frac{1}{\rho^{2}}G_{\left[1,R/\rho\right]}(r/\rho,t/\rho^{2}).$

By Lemma B.3, this satisfies

(B.8)
$G_{\left[\rho,R\right]}(r,t)\leq\frac{C_{\mu}e^{\frac{-(r-\rho)^{2}}{5t}}\rho^{\mu-2}}{t(t+\rho^{2})^{\frac{\mu}{2}-1}}\cdot\begin{cases}\min\left[(r-\rho)/\sqrt{t},1\right]&(t\leq
1)\\\ \min\left[r-\rho,1\right]&(t\geq 1).\end{cases}$

The solution of the boundary problem with data $\psi(t)$ at $r=\rho$ is given
by

(B.9)
$v_{1}(r,t)=\int_{0}^{t}\psi(\tau)G_{\left[\rho,R\right]}(r,t-\tau)\,d\tau.$

######  Proposition B.4.

For $2\rho\leq r\leq R$ and $t\geq 0,$ we have

$|v_{1}(r,t)|\leq
C_{\mu}e^{-(r-\rho)^{2}/6t}\left(\frac{\rho^{\mu-2}}{r^{\mu-1}}\right)\sqrt{\int_{0}^{t}\psi^{2}(\tau)d\tau}.$

###### Proof.

We apply Hölder’s inequality as in the proof of Proposition A.5b of [13]. From
(B.8-B.9), we have:

$\begin{split}r^{\mu-1}|v_{1}(r,t)|&\leq
C_{\mu}\rho^{\mu-2}\int_{0}^{t}|\psi(\tau)|\frac{r^{\mu-1}e^{\frac{-(r-\rho)^{2}}{5(t-\tau)}}}{(t-\tau)(t-\tau+\rho^{2})^{\frac{\mu}{2}-1}}\,d\tau\\\
&\leq C_{\mu}\rho^{\mu-2}\sqrt{\int_{0}^{t}\psi^{2}(\tau)\,d\tau}\\\
&\qquad\cdot\sqrt{\int_{0}^{t}\frac{r^{2\mu-2}(r-\rho)^{2}}{(t-\tau)^{2}(t-\tau+\rho^{2})^{\mu-2}}e^{\frac{-2(r-\rho)^{2}}{5(t-\tau)}}\,\frac{d\tau}{(r-\rho)^{2}}}.\end{split}$

The result follows by changing variables $u=\frac{\tau}{(r-\rho)^{2}}.$ ∎

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